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exists","display":"\\exists","kind":"use","occurrences":[90,642,749]},{"id":"symbol-4bab02d584adcb84","tex":"\\varnothing","aliases":"∅ varnothing","display":"\\varnothing","kind":"use","occurrences":[3,339,356]},{"id":"symbol-4471c87ad5a9031e","tex":"\\notin","aliases":"∉ notin","display":"\\notin","kind":"use","occurrences":[1023]},{"id":"symbol-56dadbdf217ad92b","tex":"\\sum","aliases":"∑ sum","display":"\\sum","kind":"use","occurrences":[141,142,146,342,371,439]},{"id":"symbol-f7b37e8032ebe855","tex":"\\sum_{j=0}^{n-1}","aliases":"∑_j=0^n-1 sum","display":"\\sum_{j=0}^{n-1}","kind":"use","occurrences":[141,142,146,371]},{"id":"symbol-db50c77870a75686","tex":"\\sum_{j=1}^n","aliases":"∑_j=1^n sum","display":"\\sum_{j=1}^n","kind":"use","occurrences":[342]},{"id":"symbol-802368a4fc45105f","tex":"\\sum_n","aliases":"∑_n sum","display":"\\sum_n","kind":"use","occurrences":[439]},{"id":"symbol-0e301b22b90725c4","tex":"\\setminus","aliases":"∖ setminus","display":"\\setminus","kind":"use","occurrences":[842,1008]},{"id":"symbol-db8e84427c2e1e02","tex":"\\infty","aliases":"∞ infty","display":"\\infty","kind":"use","occurrences":[185,243]},{"id":"symbol-795a720997b72208","tex":"\\cap","aliases":"∩ cap","display":"\\cap","kind":"use","occurrences":[246,575,597,605,690,705,782]},{"id":"symbol-e0395c0dd9f66f1e","tex":"\\int","aliases":"∫ int","display":"\\int","kind":"use","occurrences":[417,446,448,493,521,533]},{"id":"symbol-a2972272b24e9c45","tex":"\\int_0^t","aliases":"∫_0^t int","display":"\\int_0^t","kind":"use","occurrences":[417,446,448,493,521,533]},{"id":"symbol-eed4f7039b9ba49f","tex":"\\cong","aliases":"≅ cong","display":"\\cong","kind":"use","occurrences":[272,292,297,309,782]},{"id":"symbol-4d3a285fd7941bf1","tex":"\\ne","aliases":"≠ ne","display":"\\ne","kind":"use","occurrences":[372,521,1030,1108]},{"id":"symbol-39c3d5a5ec4b06af","tex":"\\subseteq","aliases":"⊆ subseteq","display":"\\subseteq","kind":"use","occurrences":[76,91,111,115,246,258,259,283,301,314,566,643,688,777]},{"id":"symbol-c04523e0e91c10a1","tex":"\\perp","aliases":"⊥ perp","display":"\\perp","kind":"use","occurrences":[575,576,597,605,621,623,631,647,649,673,678,685,782,873,891,906,948,952,954,960,971,979,989,998,1014,1051,1092,1094,1100,1130,1450]},{"id":"symbol-0ac7e4b05dea168a","tex":"\\bigcap","aliases":"⋂ bigcap","display":"\\bigcap","kind":"use","occurrences":[103,244,246]},{"id":"symbol-892cecbf386dea45","tex":"\\bigcap_i","aliases":"⋂_i bigcap","display":"\\bigcap_i","kind":"use","occurrences":[246]},{"id":"symbol-27b1d2fc9bc15338","tex":"\\bigcap_w","aliases":"⋂_w bigcap","display":"\\bigcap_w","kind":"use","occurrences":[103]},{"id":"symbol-930bb050c29b1bb0","tex":"\\bigcap_{|w|\\le r}","aliases":"⋂_|w|≤ r bigcap le","display":"\\bigcap_{|w|\\le r}","kind":"use","occurrences":[244]},{"id":"symbol-4f3867cf1346eed7","tex":"\\cdots","aliases":"⋯ cdots","display":"\\cdots","kind":"use","occurrences":[2,403]},{"id":"symbol-942faac63e3ce23f","tex":"\\longrightarrow","aliases":"⟶ longrightarrow","display":"\\longrightarrow","kind":"use","occurrences":[952,979,1006,1181]},{"id":"symbol-22449108dc31ebd8","tex":"\\longmapsto","aliases":"⟼ longmapsto","display":"\\longmapsto","kind":"use","occurrences":[209,598]}],"contexts":["\\subsection{Conventions and domains}\nOperator multiplication applies the right factor first: $STx=S(Tx)$.\nA word $w=[i_1,\\ldots,i_n]$ is chronological, so\n$T_w=T_{i_n}\\cdots T_{i_1}$; $T_\\varnothing=I$. Concatenating $u$ then $v$\ngives $T_{u\\mathbin{+\\!+}v}=T_vT_u$. Empty sums are zero and empty products are one.\nAll equalities of operators are equalities on the entire stated domain.","For a subspace $L$ admitting orthogonal projection, $P_L$ denotes that actual\nprojection. On a Hilbert space, closed subspaces admit it. The notation\n$\\dist(v,L)=\\inf_{\\ell\\in L}\\nn{v-\\ell}$ always refers to the actual subspace,\nnot a replacement chosen to make a theorem true. Closure is written $\\cl L$.","The first contract supplies two maps rather than identifying a coordinate\nspace with the object by convention. Let $a_X:X\\to V_X$ and $s_X:V_X\\to X$\nbe linear with $s_Xa_X=I_X$. The ambient space $V_X$ may have redundant\ncoordinates. Define $E_X=a_Xs_X$. For another such resolution of $Y$, put\n\\[\n\\mathcal K(T)=a_YTs_X,\\qquad \\mathcal D(K)=s_YKa_X.\n\\]\nWe call a proposed coordinate operator compatible when $E_YKE_X=K$.","contract supplies two maps rather than identifying a coordinate\nspace with the object by convention. Let $a_X:X\\to V_X$ and $s_X:V_X\\to X$\nbe linear with $s_Xa_X=I_X$. The ambient space $V_X$ may have redundant\ncoordinates. Define $E_X=a_Xs_X$. For another such resolution of $Y$, put\n\\[\n\\mathcal K(T)=a_YTs_X,\\qquad \\mathcal D(K)=s_YKa_X.\n\\]\nWe call a proposed coordinate operator compatible when $E_YKE_X=K$.","$E_X^2=E_X$, and $E_Xv=v$ exactly when $v\\in\\Ran a_X$. Moreover\n\\[\n\\mathcal D\\mathcal K(T)=T,\\qquad\n\\mathcal K\\mathcal D(K)=E_YKE_X.\n\\]\nConsequently a compatible $K$ is the encoding of a unique map $T:X\\to Y$.\nEncoding preserves composition. Its identity on resolved coordinates is $E_X$.","The splitting gives $E_X^2=a_X(s_Xa_X)s_X=a_Xs_X$. If $E_Xv=v$, then\n$v=a_X(s_Xv)$; conversely $E_Xa_Xx=a_Xx$. Substituting the definitions yields\n\\[\ns_Y(a_YTs_X)a_X=T,\\qquad\na_Y(s_YKa_X)s_X=E_YKE_X.\n\\]\nFor compatible $K$, take $T=\\mathcal D(K)$. If another $T'$ encodes to $K$,\ndecoding proves $T'=\\mathcal D(K)$. Finally, an intermediate resolution of $Y$\nhas $s_Ya_Y=I_Y$, so\n$\\mathcal K(S)\\mathcal K(T)=a_ZS(s_Ya_Y)Ts_X=a_ZSTs_X$.\nEncoding $I_X$ gives $a_Xs_X=E_X$, which acts as identity on compatible kernels.\nNo inverse on all of $V_X$ has been assumed.","s_X$. If $E_Xv=v$, then\n$v=a_X(s_Xv)$; conversely $E_Xa_Xx=a_Xx$. Substituting the definitions yields\n\\[\ns_Y(a_YTs_X)a_X=T,\\qquad\na_Y(s_YKa_X)s_X=E_YKE_X.\n\\]\nFor compatible $K$, take $T=\\mathcal D(K)$. If another $T'$ encodes to $K$,\ndecoding proves $T'=\\mathcal D(K)$. Finally, an intermediate resolution of $Y$\nhas $s_Ya_Y=I_Y$, so\n$\\mathcal K(S)\\mathcal K(T)=a_ZS(s_Ya_Y)Ts_X=a_ZSTs_X$.\nEncoding $I_X$ gives $a_Xs_X=E_X$, which acts as identity on compatible kernels.\nNo inverse on all of $V_X$ has been assumed.","Xv=v$, then\n$v=a_X(s_Xv)$; conversely $E_Xa_Xx=a_Xx$. Substituting the definitions yields\n\\[\ns_Y(a_YTs_X)a_X=T,\\qquad\na_Y(s_YKa_X)s_X=E_YKE_X.\n\\]\nFor compatible $K$, take $T=\\mathcal D(K)$. If another $T'$ encodes to $K$,\ndecoding proves $T'=\\mathcal D(K)$. Finally, an intermediate resolution of $Y$\nhas $s_Ya_Y=I_Y$, so\n$\\mathcal K(S)\\mathcal K(T)=a_ZS(s_Ya_Y)Ts_X=a_ZSTs_X$.\nEncoding $I_X$ gives $a_Xs_X=E_X$, which acts as identity on compatible kernels.\nNo inverse on all of $V_X$ has been assumed.","s yields\n\\[\ns_Y(a_YTs_X)a_X=T,\\qquad\na_Y(s_YKa_X)s_X=E_YKE_X.\n\\]\nFor compatible $K$, take $T=\\mathcal D(K)$. If another $T'$ encodes to $K$,\ndecoding proves $T'=\\mathcal D(K)$. Finally, an intermediate resolution of $Y$\nhas $s_Ya_Y=I_Y$, so\n$\\mathcal K(S)\\mathcal K(T)=a_ZS(s_Ya_Y)Ts_X=a_ZSTs_X$.\nEncoding $I_X$ gives $a_Xs_X=E_X$, which acts as identity on compatible kernels.\nNo inverse on all of $V_X$ has been assumed.","\\begin{example}[Redundant coordinates]\nFor $X=\\RR$ and $V_X=\\RR^2$, take $a(x)=(x,x)$ and\n$s(u,v)=(u+v)/2$. Then $E(u,v)=((u+v)/2,(u+v)/2)$.\nThe encoded scalar transformation $x\\mapsto cx$ has matrix\n$\\frac c2\\bigl(\\begin{smallmatrix}1&1","\\begin{example}[Redundant coordinates]\nFor $X=\\RR$ and $V_X=\\RR^2$, take $a(x)=(x,x)$ and\n$s(u,v)=(u+v)/2$. Then $E(u,v)=((u+v)/2,(u+v)/2)$.\nThe encoded scalar transformation $x\\mapsto cx$ has matrix\n$\\frac c2\\bigl(\\begin{smallmatrix}1&1\\\\1&1\\end{smallmatrix}\\bigr)$.\nIts redundant coordinate identity is $E$, while the ambient identity matrix\nalso acts on an unrealizable anti-diagonal sector. The distinction is exact.\n\\end{example}","1&1\\end{smallmatrix}\\bigr)$.\nIts redundant coordinate identity is $E$, while the ambient identity matrix\nalso acts on an unrealizable anti-diagonal sector. The distinction is exact.\n\\end{example}","\\par{\\small\\textbf{Hypotheses.} Commutative-ring modules; linear $C:X\\to Z$, $R:Z\\to X$ with $CR=I_Z$; arbitrary endomorphisms $S,T,U$ of $X$.}\\par","Set $P=RC$, $Q=I-P$, $\\Phi(T)=CTR$, and\n$\\Delta(S,T)=\\Phi(ST)-\\Phi(S)\\Phi(T)$. Then\n\\[\n\\Delta(S,T)=CSQTR,\n\\]\nand associativity gives\n\\[\n\\Delta(S,T)\\Phi(U)+\\Delta(ST,U)\n=\\Phi(S)\\Delta(T,U)+\\Delta(S,TU).\n\\]\nFor two successive split compressions,\n\\[\n\\Delta_{21}(S,T)=\\Phi_2(\\Delta_1(S,T))\n+\\Delta_2(\\Phi_1(S),\\Phi_1(T)).\n\\]","Insert $I=P+Q$ between $S$ and $T$:\n$CSTR=CS(P+Q)TR=(CSR)(CTR)+CSQTR$. This proves the factorization.\nExpanding either side of the coherence identity cancels the terms\n$\\Phi(S)\\Phi(T)\\Phi(U)$ and leaves\n$\\Phi(STU)-\\Phi(S)\\Phi(TU)$ together with the same compensating term.\nFor the two-stage formula, add and subtract\n$\\Phi_2(\\Phi_1(S)\\Phi_1(T))$ in the composite defect. Linearity of $\\Phi_2$\nidentifies the first difference with $\\Phi_2(\\Delta_1(S,T))$; the second is\nthe second-stage defect. Thus a later representation transports the earlier\nerror and can introduce another one. It does not remove the first term by definition.","$T$:\n$CSTR=CS(P+Q)TR=(CSR)(CTR)+CSQTR$. This proves the factorization.\nExpanding either side of the coherence identity cancels the terms\n$\\Phi(S)\\Phi(T)\\Phi(U)$ and leaves\n$\\Phi(STU)-\\Phi(S)\\Phi(TU)$ together with the same compensating term.\nFor the two-stage formula, add and subtract\n$\\Phi_2(\\Phi_1(S)\\Phi_1(T))$ in the composite defect. Linearity of $\\Phi_2$\nidentifies the first difference with $\\Phi_2(\\Delta_1(S,T))$; the second is\nthe second-stage defect. Thus a later representation transports the earlier\nerror and can introduce another one. It does not remove the first term by definition.","\\par{\\small\\textbf{Hypotheses.} Commutative-ring modules and a linear presentation $G\\xrightarrow{A}V\\xrightarrow{\\delta}W$ with $\\delta A=0$; a compatible vector $v\\in\\ker\\delta$.}\\par","For $\\delta A=0$, define $K=\\ker\\delta$, $R=\\Ran A\\subseteq K$, and\n$\\mathcal H=K/R$. A compatible $v$ has class zero exactly when $v=Ag$\nfor some $g$. All classes vanish exactly when $R=K$. If $A$ is injective,\na zero class has a unique source.","Commuting presentation maps $LA=A'F$ and $\\delta'L=M\\delta$ induce a map\n$\\overline L:\\mathcal H\\to\\mathcal H'$ by $[v]\\mapsto[Lv]$.\nThese maps compose and preserve zero classes. Reflection of zero follows\nwhen $\\overline L$ is injective.","The chain condition puts the actual range of $A$ inside $K$. The zero class\ncriterion for a quotient is membership in the submodule divided out, giving\n$[v]=0\\iff v\\in R\\iff\\exists g,Ag=v$. Every compatible class is zero exactly\nwhen $K\\subseteq R$; the reverse inclusion is already known. Injectivity of\n$A$ gives uniqueness of any preimage.","For transport, if $v\\in K$, then $\\delta'Lv=M\\delta v=0$.\nIf $v=Ag$, then $Lv=A'Fg$ lies in $R'$. Hence $L$ descends to the actual\nquotients. Evaluating on representatives proves the composition law.\nA linear map sends zero to zero. If its induced quotient map is injective,\n$\\overline L[v]=0=\\overline L[0]$ implies $[v]=0$.\nThe commuting squares alone provide no such injectivity and do not assert\nthat either obstruction quotient vanishes.","\\par{\\small\\textbf{Hypotheses.} Commutative-ring modules; a linear observation $O:X\\to Y$ and an arbitrary family of allowed endomorphisms $T_i:X\\to X$; every finite chronological word is included.}\\par","Let $N=\\bigcap_w\\ker(OT_w)$. It is the largest submodule contained in\n$\\ker O$ and invariant under every $T_i$. The operations and observation\ndescend to $X/N$, preserving every future observation. In particular,\n\\[\n[x]=[y]\\iff OT_wx=OT_wy\\quad\\text{for every finite word }w.\n\\]\nIf $q:X\\to Z$ satisfies $qT_i=U_iq$ and $Vq=O$, then $\\ker q\\subseteq N$.","An intersection of kernels is a submodule. If $x\\in N$, then\n$OT_wT_ix=OT_{i::w}x=0$, so $T_ix\\in N$. If an invariant submodule\n$L\\subseteq\\ker O$ contains $x$, induction on word length gives $T_wx\\in L$;\ntherefore $x\\in N$. This establishes maximality.","Invariance defines quotient operations $[x]\\mapsto[T_ix]$; containment in\n$\\ker O$ defines $[x]\\mapsto Ox$. Induction gives their all-word fidelity.\nEquality $[x]=[y]$ means $x-y\\in N$, which is precisely equality of the\ndisplayed observations by linearity. In an exact competing model, induction\ngives $qT_w=U_wq$ and hence $OT_w=VU_wq$. If $qx=0$, every future observation\nof $x$ is zero. Thus a competing model can retain extra distinctions, but\ncannot erase any distinction visible to the specified future task.","Let $V$ be visible state and $H$ hidden state. Over a commutative ring, specify\nfour maps and form the actual full dynamics\n\\[\nT(y,z)=(Ay+Bz,Gy+Dz).\n\\]\nPut $C(y,z)=y$, $Ry=(y,0)$, $E_n=CT^nR$, and $M_k=BD^kG$.\nThe return $M_k$ contains entry into $H$, $k$ hidden transitions, and return\nto $V$. Its first effect occurs after $k+2$ full transitions.","For the trajectory $(y_n,z_n)=T^n(y_0,z_0)$,\n\\[\nz_n=D^nz_0+\\sum_{j=0}^{n-1}D^{n-1-j}Gy_j,\n\\]\nand\n\\[\ny_{n+1}=Ay_n+BD^nz_0+\\sum_{j=0}^{n-1}M_{n-1-j}y_j.\n\\]\nPrepared evolution ($z_0=0$) satisfies\n$E_0=I$, $E_1=A$, and\n\\[\nE_{n+1}=AE_n+\\sum_{j=0}^{n-1}M_{n-1-j}E_j.\n\\]","At $n=0$, the first formula is $z_0=z_0$ with an empty sum. If it holds at\n$n$, the actual recurrence $z_{n+1}=Gy_n+Dz_n$ gives the same formula at\n$n+1$ after distributing $D$ and appending the newest forcing term.\nSubstituting this expression into $y_{n+1}=Ay_n+Bz_n$ proves the visible\nmemory equation. For a prepared input $y_0$, each $y_j$ equals $E_jy_0$.\nThe pointwise memory equation is therefore an operator identity because it\nholds for every $y_0$. The equations are derived from powers of $T$; the\ntrajectory recurrence is not an extra certificate assumption.","sum. If it holds at\n$n$, the actual recurrence $z_{n+1}=Gy_n+Dz_n$ gives the same formula at\n$n+1$ after distributing $D$ and appending the newest forcing term.\nSubstituting this expression into $y_{n+1}=Ay_n+Bz_n$ proves the visible\nmemory equation. For a prepared input $y_0$, each $y_j$ equals $E_jy_0$.\nThe pointwise memory equation is therefore an operator identity because it\nholds for every $y_0$. The equations are derived from powers of $T$; the\ntrajectory recurrence is not an extra certificate assumption.","$z_{n+1}=Gy_n+Dz_n$ gives the same formula at\n$n+1$ after distributing $D$ and appending the newest forcing term.\nSubstituting this expression into $y_{n+1}=Ay_n+Bz_n$ proves the visible\nmemory equation. For a prepared input $y_0$, each $y_j$ equals $E_jy_0$.\nThe pointwise memory equation is therefore an operator identity because it\nholds for every $y_0$. The equations are derived from powers of $T$; the\ntrajectory recurrence is not an extra certificate assumption.","The following conditions are equivalent:\n\\[\nE_n=A^n\\ (\\forall n),\\qquad M_k=0\\ (\\forall k),\\qquad\nE_{m+n}=E_mE_n\\ (\\forall m,n).\n\\]\nMore precisely,\n\\[\nE_n=A^n\\ (n\\le N+1)\\iff M_k=0\\ (k<N).\n\\]\nIf $M_j=0$ for $j<k$, then $E_{k+2}-A^{k+2}=M_k$.\nClosure for every initial hidden state is equivalent to $B=0$.","If all memory maps vanish, the recurrence in SR-05 reduces to\n$E_{n+1}=AE_n$, proving $E_n=A^n$ by induction. Conversely, suppose exact\npowers hold. Induct on $k$. In the recurrence for $E_{k+2}$, every term\nexcept $M_kE_0$ involves a previously vanishing memory map. Thus\n$E_{k+2}=AE_{k+1}+M_k$, forcing $M_k=0$.\nThe same induction restricted to $k<N$ proves the finite-prefix equivalence\nand, before setting the last term to zero, the first-return identity.","Exact powers imply the semigroup law. The law conversely gives\n$E_{n+1}=E_nE_1=E_nA$, so it implies exact powers. For arbitrary hidden\ninitialization, evaluating closure at $n=1$, $y_0=0$ gives $Bz_0=0$ for\nevery $z_0$, hence $B=0$. That condition also suffices directly from the\nvisible recurrence. Vanishing $BD^kG$ imposes a weaker prepared-state requirement.","If a monic polynomial $p$ annihilates $D$, vanishing of $M_k$ for\n$k<\\deg p$ implies vanishing for all $k$. If $\\dim H=d<\\infty$, then\n\\[\nE_n=A^n\\ (\\forall n)\n\\iff M_k=0\\ (0\\le k<d)\n\\iff E_n=A^n\\ (0\\le n\\le d+1).\n\\]\nThe visible space need not be finite-dimensional. The case $d=0$ is included.","Divide $X^n$ by the monic $p$ to obtain a remainder $q_n$ of degree below\n$\\deg p$. Evaluation at $D$ gives $D^n=q_n(D)$ because $p(D)=0$.\nWriting $q_n(D)$ as a finite linear combination of the powers $D^k$ below\n$\\deg p$ shows $BD^nG=0$. For finite-dimensional $H$, Cayley--Hamilton\nprovides the monic characteristic polynomial of $D$ with degree $d$.\nSR-06 then yields both displayed equivalences. In dimension zero,\nall maps through $H$ vanish and the empty list of memory tests is sufficient.\nIf exact closure fails, the contrapositive gives a nonzero memory map with\nindex below $d$. These tests concern complete linear maps, not sampled state\nvectors or approximate floating-point agreement.","in a remainder $q_n$ of degree below\n$\\deg p$. Evaluation at $D$ gives $D^n=q_n(D)$ because $p(D)=0$.\nWriting $q_n(D)$ as a finite linear combination of the powers $D^k$ below\n$\\deg p$ shows $BD^nG=0$. For finite-dimensional $H$, Cayley--Hamilton\nprovides the monic characteristic polynomial of $D$ with degree $d$.\nSR-06 then yields both displayed equivalences. In dimension zero,\nall maps through $H$ vanish and the empty list of memory tests is sufficient.\nIf exact closure fails, the contrapositive gives a nonzero memory map with\nindex below $d$. These tests concern complete linear maps, not sampled state\nvectors or approximate floating-point agreement.","n of the powers $D^k$ below\n$\\deg p$ shows $BD^nG=0$. For finite-dimensional $H$, Cayley--Hamilton\nprovides the monic characteristic polynomial of $D$ with degree $d$.\nSR-06 then yields both displayed equivalences. In dimension zero,\nall maps through $H$ vanish and the empty list of memory tests is sufficient.\nIf exact closure fails, the contrapositive gives a nonzero memory map with\nindex below $d$. These tests concern complete linear maps, not sampled state\nvectors or approximate floating-point agreement.","For $C:X\\to Z$, $R:Z\\to X$, $CR=I$, let $Q=I-RC$ and $H=\\ker C$.\nThe maps\n\\[\nx\\longmapsto(Cx,Qx),\\qquad (y,z)\\longmapsto Ry+z\n\\]\nare inverse linear maps between $X$ and $Z\\times H$.\nAn endomorphism $T$ is conjugate to block dynamics with\n\\[\nA=CTR,\\quad B=CT\\iota_H,\\quad G=QTR,\\quad D=QT\\iota_H.\n\\]\nTheir prepared maps equal $CT^nR$. Hence exact compressed powers are\nequivalent to zero canonical memory. If $\\dim\\ker C=d$, complete compressed\nmap tests through $d+1$ decide all depths.","to Z$, $R:Z\\to X$, $CR=I$, let $Q=I-RC$ and $H=\\ker C$.\nThe maps\n\\[\nx\\longmapsto(Cx,Qx),\\qquad (y,z)\\longmapsto Ry+z\n\\]\nare inverse linear maps between $X$ and $Z\\times H$.\nAn endomorphism $T$ is conjugate to block dynamics with\n\\[\nA=CTR,\\quad B=CT\\iota_H,\\quad G=QTR,\\quad D=QT\\iota_H.\n\\]\nTheir prepared maps equal $CT^nR$. Hence exact compressed powers are\nequivalent to zero canonical memory. If $\\dim\\ker C=d$, complete compressed\nmap tests through $d+1$ decide all depths.",".\nThe maps\n\\[\nx\\longmapsto(Cx,Qx),\\qquad (y,z)\\longmapsto Ry+z\n\\]\nare inverse linear maps between $X$ and $Z\\times H$.\nAn endomorphism $T$ is conjugate to block dynamics with\n\\[\nA=CTR,\\quad B=CT\\iota_H,\\quad G=QTR,\\quad D=QT\\iota_H.\n\\]\nTheir prepared maps equal $CT^nR$. Hence exact compressed powers are\nequivalent to zero canonical memory. If $\\dim\\ker C=d$, complete compressed\nmap tests through $d+1$ decide all depths.","$CQ=C-CRC=0$, so $Qx$ belongs to $H$. Synthesis after analysis gives\n$RCx+Qx=x$. Conversely, for $z\\in\\ker C$, one has $C(Ry+z)=y$ and\n$Q(Ry+z)=z$, using $QR=0$ and $Qz=z$. Conjugating $T$ by these inverse maps\ngives the four displayed blocks; in the hidden blocks, $Q$ supplies the\nrequired codomain. Induction transports powers under conjugacy.\nThe lift of $y$ is $(y,0)$ in these coordinates, so the prepared visible\nmap is precisely $CT^nR$. Apply SR-06 and SR-07. The construction uses the\nactual discarded subspace and imposes no orthogonality on the chosen splitting.","Synthesis after analysis gives\n$RCx+Qx=x$. Conversely, for $z\\in\\ker C$, one has $C(Ry+z)=y$ and\n$Q(Ry+z)=z$, using $QR=0$ and $Qz=z$. Conjugating $T$ by these inverse maps\ngives the four displayed blocks; in the hidden blocks, $Q$ supplies the\nrequired codomain. Induction transports powers under conjugacy.\nThe lift of $y$ is $(y,0)$ in these coordinates, so the prepared visible\nmap is precisely $CT^nR$. Apply SR-06 and SR-07. The construction uses the\nactual discarded subspace and imposes no orthogonality on the chosen splitting.","\\begin{example}[Delayed return and disconnected return]\nOn $\\QQ^3$, set $T(y,z_1,z_2)=(z_1,z_2,y)$ and retain $y$.\nThen $A=0$, $M_0=0$, $M_1=I$, and $E_1=E_2=0$ while $E_3=I$.\nTwo shallow exact checks cannot certify all depths without the dimension budget.\nFor another system, let $A=D=I$, $B(z_1,z_2)=z_1$, and $G(y)=(0,y)$.\nBoth couplings are nonzero but every $BD^kG$ vanishes. Prepared closure is exact,\nwhile an arbitrary hidden initial first coordinate affects the next visible state.\n\\end{example}","Now work over a field and, for dimension statements, assume $\\dim X<\\infty$.\nThe allowed operations need not commute. Define\n\\[\nN_r=\\bigcap_{|w|\\le r}\\ker(OT_w),\\qquad k=\\dim\\ker O.\n\\]\nA finite horizon means bounded word length. It gives a finite enumeration of\nconstraints only when the operation alphabet is finite.","$N_0=\\ker O$ and\n\\[\nN_{r+1}=\\ker O\\cap\\bigcap_i T_i^{-1}N_r\\subseteq N_r.\n\\]\nOne plateau $N_{r+1}=N_r$ implies $N_r=N$. In all cases $N_k=N$.\nThus equality of future observations through word length $k$ is equivalent\nto equality for every finite word.","The empty word enforces $Ox=0$. A nonempty word $i::w$ of length at most\n$r+1$ enforces $OT_wT_ix=0$, yielding the recursion. At a plateau, that\nrecursion shows $N_r$ is invariant under every $T_i$ and lies in $\\ker O$.\nBy SR-04, $N_r\\subseteq N$, while $N\\subseteq N_r$ always holds.","Before reaching $N$, each successive inclusion must be strict, since an\nequality would already certify the final subspace. Every strict inclusion\nreduces dimension by at least one. Starting from dimension $k$ allows at\nmost $k$ such reductions; if $N_k$ has dimension zero it is already $N$,\nand otherwise an earlier plateau has occurred. Applying the kernel identity\nto $x-y$ proves the all-future comparison. No word-order commutativity is used.","The model $X/N$ is observable. Every exact competitor induces a unique\nsurjection $\\Ran q\\to X/N$, taking $qx$ to $[x]$. Therefore\n\\[\n\\dim(X/N)\\le\\dim\\Ran q,\n\\]\nwith equality exactly when $\\ker q=N$. In that case\n$\\Ran q\\cong X/N$. If $Z$ is finite-dimensional, also\n$\\dim\\Ran q\\le\\dim Z$. Surjectivity of $q$ onto $Z$ is not required.","If $qx=qy$, SR-04 gives $x-y\\in N$, so $qx\\mapsto[x]$ is well-defined.\nIt is linear and surjective because every class has a representative $x$.\nIts values on the actual range force uniqueness.\nRank-nullity gives $\\dim\\Ran q=\\dim X-\\dim\\ker q$ and\n$\\dim(X/N)=\\dim X-\\dim N$. Since $\\ker q\\subseteq N$, this proves the\ninequality. Equality of dimensions forces equality of the nested finite-dimensional\nsubspaces; conversely their equality makes the comparison injective and hence\nan equivalence. In the quotient model, an invisible class $[x]$ has every\n$OT_wx=0$, so $x\\in N$ and $[x]=0$. This establishes observability.\nUnused coordinates in $Z$ do not change the reachable-image comparison.","$x$.\nIts values on the actual range force uniqueness.\nRank-nullity gives $\\dim\\Ran q=\\dim X-\\dim\\ker q$ and\n$\\dim(X/N)=\\dim X-\\dim N$. Since $\\ker q\\subseteq N$, this proves the\ninequality. Equality of dimensions forces equality of the nested finite-dimensional\nsubspaces; conversely their equality makes the comparison injective and hence\nan equivalence. In the quotient model, an invisible class $[x]$ has every\n$OT_wx=0$, so $x\\in N$ and $[x]=0$. This establishes observability.\nUnused coordinates in $Z$ do not change the reachable-image comparison.","on the actual range force uniqueness.\nRank-nullity gives $\\dim\\Ran q=\\dim X-\\dim\\ker q$ and\n$\\dim(X/N)=\\dim X-\\dim N$. Since $\\ker q\\subseteq N$, this proves the\ninequality. Equality of dimensions forces equality of the nested finite-dimensional\nsubspaces; conversely their equality makes the comparison injective and hence\nan equivalence. In the quotient model, an invisible class $[x]$ has every\n$OT_wx=0$, so $x\\in N$ and $[x]=0$. This establishes observability.\nUnused coordinates in $Z$ do not change the reachable-image comparison.","range force uniqueness.\nRank-nullity gives $\\dim\\Ran q=\\dim X-\\dim\\ker q$ and\n$\\dim(X/N)=\\dim X-\\dim N$. Since $\\ker q\\subseteq N$, this proves the\ninequality. Equality of dimensions forces equality of the nested finite-dimensional\nsubspaces; conversely their equality makes the comparison injective and hence\nan equivalence. In the quotient model, an invisible class $[x]$ has every\n$OT_wx=0$, so $x\\in N$ and $[x]=0$. This establishes observability.\nUnused coordinates in $Z$ do not change the reachable-image comparison.","uniqueness.\nRank-nullity gives $\\dim\\Ran q=\\dim X-\\dim\\ker q$ and\n$\\dim(X/N)=\\dim X-\\dim N$. Since $\\ker q\\subseteq N$, this proves the\ninequality. Equality of dimensions forces equality of the nested finite-dimensional\nsubspaces; conversely their equality makes the comparison injective and hence\nan equivalence. In the quotient model, an invisible class $[x]$ has every\n$OT_wx=0$, so $x\\in N$ and $[x]=0$. This establishes observability.\nUnused coordinates in $Z$ do not change the reachable-image comparison.","nd\n$\\dim(X/N)=\\dim X-\\dim N$. Since $\\ker q\\subseteq N$, this proves the\ninequality. Equality of dimensions forces equality of the nested finite-dimensional\nsubspaces; conversely their equality makes the comparison injective and hence\nan equivalence. In the quotient model, an invisible class $[x]$ has every\n$OT_wx=0$, so $x\\in N$ and $[x]=0$. This establishes observability.\nUnused coordinates in $Z$ do not change the reachable-image comparison.","Let $H_kx=(OT_wx)_{|w|\\le k}$ and $Z_k=\\Ran H_k$. Then\n$\\ker H_k=N$ and $X/N\\cong Z_k$. The formulas\n\\[\nU_i(H_kx)=H_kT_ix,\\qquad V(H_kx)=Ox\n\\]\ndefine actual linear maps on $Z_k$ and preserve every finite future word.\nThis model has minimum exact dimension.","The recursively enumerated words are exactly those of length at most $k$.\nSR-09 identifies the kernel of their combined observation map with $N$.\nThe first isomorphism theorem gives $X/N\\cong\\Ran H_k$. If $H_kx=H_ky$,\nthen $x-y\\in N$; its invariance gives $H_kT_ix=H_kT_iy$, and\n$N\\subseteq\\ker O$ gives $Ox=Oy$. Hence the displayed operations and readout\nare well-defined on realizable tables. Induction proves\n$U_wH_kx=H_kT_wx$ and $VU_wH_kx=OT_wx$ for words of any length.\nApply SR-10 to get minimality. An arbitrary table in the ambient coordinate\nproduct need not be realizable and is not part of this constructed state space.","ly enumerated words are exactly those of length at most $k$.\nSR-09 identifies the kernel of their combined observation map with $N$.\nThe first isomorphism theorem gives $X/N\\cong\\Ran H_k$. If $H_kx=H_ky$,\nthen $x-y\\in N$; its invariance gives $H_kT_ix=H_kT_iy$, and\n$N\\subseteq\\ker O$ gives $Ox=Oy$. Hence the displayed operations and readout\nare well-defined on realizable tables. Induction proves\n$U_wH_kx=H_kT_wx$ and $VU_wH_kx=OT_wx$ for words of any length.\nApply SR-10 to get minimality. An arbitrary table in the ambient coordinate\nproduct need not be realizable and is not part of this constructed state space.","e exactly those of length at most $k$.\nSR-09 identifies the kernel of their combined observation map with $N$.\nThe first isomorphism theorem gives $X/N\\cong\\Ran H_k$. If $H_kx=H_ky$,\nthen $x-y\\in N$; its invariance gives $H_kT_ix=H_kT_iy$, and\n$N\\subseteq\\ker O$ gives $Ox=Oy$. Hence the displayed operations and readout\nare well-defined on realizable tables. Induction proves\n$U_wH_kx=H_kT_wx$ and $VU_wH_kx=OT_wx$ for words of any length.\nApply SR-10 to get minimality. An arbitrary table in the ambient coordinate\nproduct need not be realizable and is not part of this constructed state space.","For a finite operation alphabet, suppose a proposed representation satisfies\n\\[\nq=F H_r,\\qquad qT_i=U_iq,\\qquad Vq=O.\n\\]\nThen $\\ker q=N$, regardless of whether $r$ reaches the a priori horizon.\nIf $q$ is surjective, it gives an equivalence $Z\\cong X/N$ preserving\nthe observation and intertwining every operation. Without surjectivity,\nthe equivalence applies to $\\Ran q$.","If $x\\in N$, then every coordinate of $H_rx$ is zero, so $qx=FH_rx=0$.\nThis gives $N\\subseteq\\ker q$. Exact intertwining and output give the reverse\ninclusion by SR-04. The equality makes $qx\\mapsto[x]$ injective on the range\nand surjective onto the quotient. Its inverse is the corresponding quotient\nmap induced by $q$. Evaluating both sides on $qx$ proves that the equivalence\nintertwines $U_i$ with the quotient operations and preserves the readout.\nThe certificate has two roles: generation by observations prevents retention\nof invisible distinctions, while exact dynamics and output prevent loss of\nfuture-visible distinctions. Finite dimension is needed to guarantee that\nsuch a certificate can be foun","If $x\\in N$, then every coordinate of $H_rx$ is zero, so $qx=FH_rx=0$.\nThis gives $N\\subseteq\\ker q$. Exact intertwining and output give the reverse\ninclusion by SR-04. The equality makes $qx\\mapsto[x]$ injective on the range\nand surjective onto the quotient. Its inverse is the corresponding quotient\nmap induced by $q$. Evaluating both sides on $qx$ proves that the equivalence\nintertwines $U_i$ with the quotient operations and preserves the readout.\nThe certificate has two roles: generation by observations prevents retention\nof invisible distinctions, while exact dynamics and output prevent loss of\nfuture-visible distinctions. Finite dimension is needed to guarantee that\nsuch a certificate can be found at a bounded depth, not for its","If $x\\in N$, then every coordinate of $H_rx$ is zero, so $qx=FH_rx=0$.\nThis gives $N\\subseteq\\ker q$. Exact intertwining and output give the reverse\ninclusion by SR-04. The equality makes $qx\\mapsto[x]$ injective on the range\nand surjective onto the quotient. Its inverse is the corresponding quotient\nmap induced by $q$. Evaluating both sides on $qx$ proves that the equivalence\nintertwines $U_i$ with the quotient operations and preserves the readout.\nThe certificate has two roles: generation by observations prevents retention\nof invisible distinctions, while exact dynamics and output prevent loss of\nfuture-visible distinctions. Finite dimension is needed to guarantee that\nsuch a certificate can be found at a bounded depth, not for its soundness.","\\begin{example}[A noncommuting minimal model]\nOn $\\QQ^3$, let $O(x,y,h)=x$,\n$T_A(x,y,h)=(y,0,h)$, and $T_B(x,y,h)=(0,x,h)$.\nThe two operations do not commute: on $(1,0,0)$ their two compositions\ngive $(1,0,0)$ and $(0,0,0)$. The map $q(x,y,h)=(x,y)$ is generated by\n$O$ and $OT_A$, intertwines both operations, and factors $O$.\nIt therefore has a depth-one certificate. The hidden coordinate $h$ is\ninvisible to every future word, and the minimum exact state dimension is two.\nThe current observation alone has dimension one and misses the distinction\nbetween $(0,1,0)$ and zero, which one application of $T_A$ reveals.\n\\end{example}","$(1,0,0)$ their two compositions\ngive $(1,0,0)$ and $(0,0,0)$. The map $q(x,y,h)=(x,y)$ is generated by\n$O$ and $OT_A$, intertwines both operations, and factors $O$.\nIt therefore has a depth-one certificate. The hidden coordinate $h$ is\ninvisible to every future word, and the minimum exact state dimension is two.\nThe current observation alone has dimension one and misses the distinction\nbetween $(0,1,0)$ and zero, which one application of $T_A$ reveals.\n\\end{example}","0)$ and $(0,0,0)$. The map $q(x,y,h)=(x,y)$ is generated by\n$O$ and $OT_A$, intertwines both operations, and factors $O$.\nIt therefore has a depth-one certificate. The hidden coordinate $h$ is\ninvisible to every future word, and the minimum exact state dimension is two.\nThe current observation alone has dimension one and misses the distinction\nbetween $(0,1,0)$ and zero, which one application of $T_A$ reveals.\n\\end{example}","Let $q:X\\to Z$ be a proposed representation, with true operations $T_i$,\nreduced operations $U_i$, and readouts $O:X\\to Y$, $V:Z\\to Y$.\nDefine the step and word defects\n\\[\n\\delta_i=qT_i-U_iq,\\qquad E_w=qT_w-U_wq.\n\\]\nAlgebraic identities require only the stated commutative-ring structure.\nQuantitative estimates below use continuous linear maps over a nontrivially\nnormed field and their whole-map operator norms.","One has $E_\\varnothing=0$,\n\\[\nE_{i::w}=E_wT_i+U_w\\delta_i,\\qquad\nE_{u\\mathbin{+\\!+}v}=E_vT_u+U_vE_u.\n\\]\nFor $w=[i_1,\\ldots,i_n]$ this yields\n\\[\nE_w=\\sum_{j=1}^n\nU_{[i_{j+1},\\ldots,i_n]}\\delta_{i_j}T_{[i_1,\\ldots,i_{j-1}]}.\n\\]\nThe output defect is $(O-Vq)T_w+VE_w$.","The recursive identity follows by adding and subtracting $U_wqT_i$ in\n$qT_wT_i-U_wU_iq$. For concatenation, add and subtract $U_vqT_u$;\nchronological order gives $T_{u\\mathbin{+\\!+}v}=T_vT_u$ and the analogous\nidentity for $U$. Inducting on the word expands one term per letter:\neach step defect is transported by its true prefix and reduced suffix.\nThe empty word contributes no terms. Finally\n$OT_w-VU_wq=(O-Vq)T_w+V(qT_w-U_wq)$.\nNeither output mismatch nor accumulated dynamics mismatch is discarded.","\\par{\\small\\textbf{Hypotheses.} Normed spaces over nontrivially normed field; bounded maps; nonnegative letter bounds a\\_i,b\\_i,epsilon\\_i and eta\\(\\ge\\Vert\\)O-Vq\\(\\Vert\\).}\\par","Suppose $\\nn{T_i}\\le a_i$, $\\nn{U_i}\\le b_i$, $\\nn{\\delta_i}\\le\\epsilon_i$.\nLet $A_w,B_w$ be the products of the respective letter bounds and define\n\\[\nD_\\varnothing=0,\\qquad D_{i::w}=D_wa_i+B_w\\epsilon_i.\n\\]\nThen $\\nn{T_w}\\le A_w$, $\\nn{U_w}\\le B_w$, $\\nn{E_w}\\le D_w$, and\n\\[\nD_{u\\mathbin{+\\!+}v}=D_vA_u+B_vD_u.\n\\]\nIf $\\nn{O-Vq}\\le\\eta$, then\n\\[\n\\nn{OT_w-VU_wq}\\le\\eta A_w+\\nn V D_w.\n\\]\nFor arbitrary initial $x\\in X$, $z\\in Z$,\n$\\nn{qT_wx-U_wz}\\le D_w\\nn x+B_w\\nn{qx-z}$.","{T_i}\\le a_i$, $\\nn{U_i}\\le b_i$, $\\nn{\\delta_i}\\le\\epsilon_i$.\nLet $A_w,B_w$ be the products of the respective letter bounds and define\n\\[\nD_\\varnothing=0,\\qquad D_{i::w}=D_wa_i+B_w\\epsilon_i.\n\\]\nThen $\\nn{T_w}\\le A_w$, $\\nn{U_w}\\le B_w$, $\\nn{E_w}\\le D_w$, and\n\\[\nD_{u\\mathbin{+\\!+}v}=D_vA_u+B_vD_u.\n\\]\nIf $\\nn{O-Vq}\\le\\eta$, then\n\\[\n\\nn{OT_w-VU_wq}\\le\\eta A_w+\\nn V D_w.\n\\]\nFor arbitrary initial $x\\in X$, $z\\in Z$,\n$\\nn{qT_wx-U_wz}\\le D_w\\nn x+B_w\\nn{qx-z}$.","_i$, $\\nn{U_i}\\le b_i$, $\\nn{\\delta_i}\\le\\epsilon_i$.\nLet $A_w,B_w$ be the products of the respective letter bounds and define\n\\[\nD_\\varnothing=0,\\qquad D_{i::w}=D_wa_i+B_w\\epsilon_i.\n\\]\nThen $\\nn{T_w}\\le A_w$, $\\nn{U_w}\\le B_w$, $\\nn{E_w}\\le D_w$, and\n\\[\nD_{u\\mathbin{+\\!+}v}=D_vA_u+B_vD_u.\n\\]\nIf $\\nn{O-Vq}\\le\\eta$, then\n\\[\n\\nn{OT_w-VU_wq}\\le\\eta A_w+\\nn V D_w.\n\\]\nFor arbitrary initial $x\\in X$, $z\\in Z$,\n$\\nn{qT_wx-U_wz}\\le D_w\\nn x+B_w\\nn{qx-z}$.","U_i}\\le b_i$, $\\nn{\\delta_i}\\le\\epsilon_i$.\nLet $A_w,B_w$ be the products of the respective letter bounds and define\n\\[\nD_\\varnothing=0,\\qquad D_{i::w}=D_wa_i+B_w\\epsilon_i.\n\\]\nThen $\\nn{T_w}\\le A_w$, $\\nn{U_w}\\le B_w$, $\\nn{E_w}\\le D_w$, and\n\\[\nD_{u\\mathbin{+\\!+}v}=D_vA_u+B_vD_u.\n\\]\nIf $\\nn{O-Vq}\\le\\eta$, then\n\\[\n\\nn{OT_w-VU_wq}\\le\\eta A_w+\\nn V D_w.\n\\]\nFor arbitrary initial $x\\in X$, $z\\in Z$,\n$\\nn{qT_wx-U_wz}\\le D_w\\nn x+B_w\\nn{qx-z}$.","Induction and submultiplicativity bound the two word operators by the\nproducts. Apply the triangle inequality and submultiplicativity to the\nfirst identity in SR-13 to obtain exactly the recursion for $D_w$.\nInduction on $u$ proves the displayed scalar concatenation identity; it\ntracks the same algebra as the exact defect identity, with nonnegative\nfactors preventing an inequality reversal. The output bound follows from\nits exact decomposition in SR-13. For initialization, write\n$qT_wx-U_wz=E_wx+U_w(qx-z)$ and bound both terms. These are operator budgets;\nagreement on a finite set of sample vectors does not supply them.","the triangle inequality and submultiplicativity to the\nfirst identity in SR-13 to obtain exactly the recursion for $D_w$.\nInduction on $u$ proves the displayed scalar concatenation identity; it\ntracks the same algebra as the exact defect identity, with nonnegative\nfactors preventing an inequality reversal. The output bound follows from\nits exact decomposition in SR-13. For initialization, write\n$qT_wx-U_wz=E_wx+U_w(qx-z)$ and bound both terms. These are operator budgets;\nagreement on a finite set of sample vectors does not supply them.","For $H_0(a,b)=0$ and\n$H_{n+1}(a,b)=aH_n(a,b)+b^n$,\n\\[\nH_n(a,b)=\\sum_{j=0}^{n-1}a^jb^{n-1-j},\\qquad\n(a-b)H_n=a^n-b^n.\n\\]\nIf $a\\ne b$, $H_n=(a^n-b^n)/(a-b)$; for equal rates,\n$H_{n+1}(a,a)=(n+1)a^n$. Uniform step bounds give\n\\[\n\\nn{E_w}\\le\\epsilon H_n(a,b),\\qquad\n\\nn{OT_w-VU_wq}\\le\\eta a^n+\\nn V\\epsilon H_n(a,b).\n\\]","Induction proves the finite-sum formula: multiplying the old sum by $a$\nand adjoining $b^n$ generates every term of the new sum. Multiplication\nby $a-b$ telescopes, leaving $a^n-b^n$. Division is used only when the\nrates are distinct. If they coincide, all $n+1$ terms of $H_{n+1}$ equal\n$a^n$, including the zero-rate boundary with the usual $a^0=1$.\nFor uniform bounds, the heterogeneous budget becomes\n$D_{i::w}=aD_w+b^{|w|}\\epsilon$, so induction gives\n$D_w=\\epsilon H_{|w|}(a,b)$. Apply SR-14. The empty word has zero dynamics\ndefect and is handled by $H_0=0$, with no negative exponent required.","ion proves the finite-sum formula: multiplying the old sum by $a$\nand adjoining $b^n$ generates every term of the new sum. Multiplication\nby $a-b$ telescopes, leaving $a^n-b^n$. Division is used only when the\nrates are distinct. If they coincide, all $n+1$ terms of $H_{n+1}$ equal\n$a^n$, including the zero-rate boundary with the usual $a^0=1$.\nFor uniform bounds, the heterogeneous budget becomes\n$D_{i::w}=aD_w+b^{|w|}\\epsilon$, so induction gives\n$D_w=\\epsilon H_{|w|}(a,b)$. Apply SR-14. The empty word has zero dynamics\ndefect and is handled by $H_0=0$, with no negative exponent required.","ld sum by $a$\nand adjoining $b^n$ generates every term of the new sum. Multiplication\nby $a-b$ telescopes, leaving $a^n-b^n$. Division is used only when the\nrates are distinct. If they coincide, all $n+1$ terms of $H_{n+1}$ equal\n$a^n$, including the zero-rate boundary with the usual $a^0=1$.\nFor uniform bounds, the heterogeneous budget becomes\n$D_{i::w}=aD_w+b^{|w|}\\epsilon$, so induction gives\n$D_w=\\epsilon H_{|w|}(a,b)$. Apply SR-14. The empty word has zero dynamics\ndefect and is handled by $H_0=0$, with no negative exponent required.","copes, leaving $a^n-b^n$. Division is used only when the\nrates are distinct. If they coincide, all $n+1$ terms of $H_{n+1}$ equal\n$a^n$, including the zero-rate boundary with the usual $a^0=1$.\nFor uniform bounds, the heterogeneous budget becomes\n$D_{i::w}=aD_w+b^{|w|}\\epsilon$, so induction gives\n$D_w=\\epsilon H_{|w|}(a,b)$. Apply SR-14. The empty word has zero dynamics\ndefect and is handled by $H_0=0$, with no negative exponent required.","\\par{\\small\\textbf{Hypotheses.} Bounded maps on the normed spaces of SR-14; $\\nn{T_i}\\le1$ for every true operation; $\\nn{U_i}\\le1$ in the nonexpansive case or a common $0\\le b<1$ with $\\nn{U_i}\\le b$ in the contractive case; $\\nn{qT_i-U_iq}\\le\\epsilon$ and $\\nn{O-Vq}\\le\\eta$ with $\\epsilon,\\eta\\ge0$.}\\par","If both families are nonexpansive, $\\nn{E_w}\\le n\\epsilon$ and the output\ndefect is at most $\\eta+\\nn V n\\epsilon$.\nIf the reduced family has a common bound $0\\le b<1$, then\n\\[\n\\nn{E_w}\\le\\frac\\epsilon{1-b},\\qquad\n\\nn{OT_w-VU_wq}\\le\\eta+\\nn V\\frac\\epsilon{1-b}.\n\\]\nFor arbitrary reduced initialization,\n\\[\n\\nn{OT_wx-VU_wz}\n\\le\\left(\\eta+\\nn V\\frac\\epsilon{1-b}\\right)\\nn x\n+\\nn V b^n\\nn{qx-z}.\n\\]","For $a=b=1$, the response sum has $n$ unit terms. For $a=1$ and $b<1$,\nit is $1+b+\\cdots+b^{n-1}\\le1/(1-b)$. SR-15 gives the operator estimates.\nFor initialization, add $VU_w(qx-z)$ to the operator output error\napplied to $x$, and use $\\nn{U_w}\\le b^n$.\nThe common bound concerns every allowed operation in the chosen norm.\nIndividual spectral radii do not imply a common product bound for a\nnoncommuting family. The nonexpansive estimate is an upper bound and does\nnot state that every such system accumulates error linearly.","Let $L$ be a bounded real operator on a Banach space. Its flow is the actual\noperator exponential $e^{tL}$. Boundedness allows differentiation of the\nnorm-convergent exponential series and integration of continuous operator\ncurves. Equalities in this section hold for all real $t$, using oriented\ninterval integrals when $t<0$.","\\par{\\small\\textbf{Hypotheses.} Real Banach space; bounded autonomous D; continuous forcing f; differentiable actual z with z\\({}^{\\prime}\\)=Dz+f at every real time.}\\par","$\\frac d{dt}e^{tD}=De^{tD}=e^{tD}D$,\n$e^{(s+t)D}=e^{sD}e^{tD}$, and $e^{-tD}$ is the inverse flow.\nEvery actual curve satisfying $z'=Dz+f$ with continuous $f$ obeys\n\\[\nz(t)=e^{tD}z(0)+\\int_0^t e^{(t-s)D}f(s)\\,ds.\n\\]\nThe initial value uniquely determines the homogeneous solution; two forced\nsolutions with the same forcing and initial value also agree.","Termwise differentiation of the exponential series gives both derivative\norders, since all powers of $D$ commute with $D$. The exponential addition\nidentity for commuting multiples of one bounded operator gives the flow law\nand inverse. For fixed $t$, differentiate $e^{(t-s)D}z(s)$. The product rule\nproduces $-De^{(t-s)D}z(s)+e^{(t-s)D}(Dz(s)+f(s))$; commutation cancels\nthe generator terms. The remaining derivative is the continuous integrand\n$e^{(t-s)D}f(s)$. The fundamental theorem of calculus on the oriented interval\ngives the displayed identity. Setting $f=0$ yields the homogeneous formula.\nSubtracting two forced solutions reduces uniqueness to that case.\nThe theorem starts from the differential equation, not from an assumed\nintegral representation.","es gives both derivative\norders, since all powers of $D$ commute with $D$. The exponential addition\nidentity for commuting multiples of one bounded operator gives the flow law\nand inverse. For fixed $t$, differentiate $e^{(t-s)D}z(s)$. The product rule\nproduces $-De^{(t-s)D}z(s)+e^{(t-s)D}(Dz(s)+f(s))$; commutation cancels\nthe generator terms. The remaining derivative is the continuous integrand\n$e^{(t-s)D}f(s)$. The fundamental theorem of calculus on the oriented interval\ngives the displayed identity. Setting $f=0$ yields the homogeneous formula.\nSubtracting two forced solutions reduces uniqueness to that case.\nThe theorem starts from the differential equation, not from an assumed\nintegral representation.","uting multiples of one bounded operator gives the flow law\nand inverse. For fixed $t$, differentiate $e^{(t-s)D}z(s)$. The product rule\nproduces $-De^{(t-s)D}z(s)+e^{(t-s)D}(Dz(s)+f(s))$; commutation cancels\nthe generator terms. The remaining derivative is the continuous integrand\n$e^{(t-s)D}f(s)$. The fundamental theorem of calculus on the oriented interval\ngives the displayed identity. Setting $f=0$ yields the homogeneous formula.\nSubtracting two forced solutions reduces uniqueness to that case.\nThe theorem starts from the differential equation, not from an assumed\nintegral representation.","For bounded linear observation maps $K,J$ on bounded generator algebras,\n\\[\nK(e^{tL})=J(e^{tD})\\ (\\forall t)\n\\iff K(L^n)=J(D^n)\\ (\\forall n).\n\\]\nIn particular, $Be^{tD}G=0$ for every $t$ exactly when $BD^nG=0$ for every $n$.\nFor $CR=I$, $A=CLR$, one has\n\\[\nCe^{tL}R=e^{tA}\\ (\\forall t)\n\\iff CL^nR=A^n\\ (\\forall n),\n\\]\nand, with $Q=I-RC$,\n\\[\nCe^{(s+t)L}R-(Ce^{sL}R)(Ce^{tL}R)=Ce^{sL}Qe^{tL}R.\n\\]","The $n$th derivative at zero of $e^{tL}$ is $L^n$. Bounded linear maps\ncommute with these derivatives, so equality for all times implies equality\nof every observed power. Conversely bounded linear maps commute with the\nsum of the norm-convergent exponential series:\n$K(e^{tL})=\\sum_n(t^n/n!)K(L^n)$. Equality term by term gives equality of\nthe sums. Sandwiching and compression are bounded linear maps on the\ncorresponding operator spaces, so they give the specializations. For the\nlast identity, use the flow law and insert $I=RC+Q$ between the two flows.\nThe defect factorization itself does not need $CR=I$; that condition makes\n$Q$ the discarded-state projector and ensures $E(0)=I$.","es implies equality\nof every observed power. Conversely bounded linear maps commute with the\nsum of the norm-convergent exponential series:\n$K(e^{tL})=\\sum_n(t^n/n!)K(L^n)$. Equality term by term gives equality of\nthe sums. Sandwiching and compression are bounded linear maps on the\ncorresponding operator spaces, so they give the specializations. For the\nlast identity, use the flow law and insert $I=RC+Q$ between the two flows.\nThe defect factorization itself does not need $CR=I$; that condition makes\n$Q$ the discarded-state projector and ensures $E(0)=I$.","mmute with the\nsum of the norm-convergent exponential series:\n$K(e^{tL})=\\sum_n(t^n/n!)K(L^n)$. Equality term by term gives equality of\nthe sums. Sandwiching and compression are bounded linear maps on the\ncorresponding operator spaces, so they give the specializations. For the\nlast identity, use the flow law and insert $I=RC+Q$ between the two flows.\nThe defect factorization itself does not need $CR=I$; that condition makes\n$Q$ the discarded-state projector and ensures $E(0)=I$.","m-convergent exponential series:\n$K(e^{tL})=\\sum_n(t^n/n!)K(L^n)$. Equality term by term gives equality of\nthe sums. Sandwiching and compression are bounded linear maps on the\ncorresponding operator spaces, so they give the specializations. For the\nlast identity, use the flow law and insert $I=RC+Q$ between the two flows.\nThe defect factorization itself does not need $CR=I$; that condition makes\n$Q$ the discarded-state projector and ensures $E(0)=I$.","um_n(t^n/n!)K(L^n)$. Equality term by term gives equality of\nthe sums. Sandwiching and compression are bounded linear maps on the\ncorresponding operator spaces, so they give the specializations. For the\nlast identity, use the flow law and insert $I=RC+Q$ between the two flows.\nThe defect factorization itself does not need $CR=I$; that condition makes\n$Q$ the discarded-state projector and ensures $E(0)=I$.","For $L(y,z)=(Ay+Bz,Gy+Dz)$ and\n$(y(t),z(t))=e^{tL}(y_0,z_0)$,\n\\[\nz(t)=e^{tD}z_0+\\int_0^t e^{(t-s)D}Gy(s)\\,ds.\n\\]\nWriting $M(t)=Be^{tD}G$, the visible dynamics satisfy\n\\[\ny'(t)=Ay(t)+Be^{tD}z_0+\\int_0^t M(t-s)y(s)\\,ds.\n\\]\nPrepared initialization removes the term involving $z_0$.","Projecting the derivative of $e^{tL}$ onto the visible and hidden components\ngives $y'=Ay+Bz$ and $z'=Gy+Dz$. The visible curve is continuous, so\n$Gy$ is continuous forcing for the hidden equation. SR-17 yields the hidden\nelimination formula. Bounded $B$ commutes with the interval integral,\ngiving the memory kernel in the visible equation. Taking $z_0=0$ removes\nonly the initialization term. No converse construction of a full trajectory\nfrom an arbitrary Volterra solution is asserted here, and no unbounded\ngenerator domain is present.","Let $E(t)=Ce^{tL}R$. The following are equivalent:\n\\[\nE(t)=e^{tA}\\ (\\forall t),\\quad M(t)=0\\ (\\forall t),\\quad\nBD^kG=0\\ (\\forall k),\\quad E(s+t)=E(s)E(t)\\ (\\forall s,t).\n\\]\nIf $\\dim H=d$, each is equivalent to $BD^kG=0$ for $k<d$, or to\n$E^{(n)}(0)=A^n$ for $n\\le d+1$. If the first $k$ return moments vanish,\n\\[\nE^{(k+2)}(0)-A^{k+2}=BD^kG.\n\\]","SR-18 converts prepared flow equality into prepared power equality and\nmemory flow vanishing into memory moment vanishing. SR-06 equates the\ncorresponding algebraic conditions, and SR-07 supplies the finite moment\ncriterion. Exact prepared flow gives its semigroup law. Conversely, if\n$E$ is a semigroup, differentiating $E(s+t)=E(s)E(t)$ in $s$ at zero\ngives $E'(t)=AE(t)$ because $E'(0)=CLR=A$ and $E(0)=I$.\nThe bounded-generator uniqueness theorem yields $E(t)=e^{tA}$.","R-18 converts prepared flow equality into prepared power equality and\nmemory flow vanishing into memory moment vanishing. SR-06 equates the\ncorresponding algebraic conditions, and SR-07 supplies the finite moment\ncriterion. Exact prepared flow gives its semigroup law. Conversely, if\n$E$ is a semigroup, differentiating $E(s+t)=E(s)E(t)$ in $s$ at zero\ngives $E'(t)=AE(t)$ because $E'(0)=CLR=A$ and $E(0)=I$.\nThe bounded-generator uniqueness theorem yields $E(t)=e^{tA}$.","quality and\nmemory flow vanishing into memory moment vanishing. SR-06 equates the\ncorresponding algebraic conditions, and SR-07 supplies the finite moment\ncriterion. Exact prepared flow gives its semigroup law. Conversely, if\n$E$ is a semigroup, differentiating $E(s+t)=E(s)E(t)$ in $s$ at zero\ngives $E'(t)=AE(t)$ because $E'(0)=CLR=A$ and $E(0)=I$.\nThe bounded-generator uniqueness theorem yields $E(t)=e^{tA}$.","Actual iterated derivatives of the exponential satisfy\n$E^{(n)}(0)=CL^nR$. Applying the finite algebraic horizon and first-return\nidentity proves the jet statements. The leading Taylor coefficient has\nthe additional denominator $(k+2)!$; the derivative itself does not.\nThe visible Banach space can be infinite-dimensional. The finite tests\nare whole-operator jet identities, not a finite list of sampled times.","\\begin{example}[A linear continuous memory kernel]\nTake $V=\\RR$, $H=\\RR^2$, $A=0$, $B(z_1,z_2)=z_1$,\n$G(y)=(0,y)$, and $D(z_1,z_2)=(z_2,0)$. Since $D^2=0$,\n$e^{tD}(z_1,z_2)=(z_1+tz_2,z_2)$ and $M(t)=tI$.\nThe first two prepared jets agree with the reduced generator, while the\nthird derivative differs by $I$. Changing $D$ to zero leaves both boundary\ncouplings nonzero but makes the memory kernel identically zero.\nArbitrary hidden initialization still affects the visible initial derivative.\n\\end{example}","For bounded real generators $L$ on $X$ and $A$ on $Z$, and bounded $q:X\\to Z$,\nwrite $\\delta=qL-Aq$ and $E(t)=qe^{tL}-e^{tA}q$.\nAlgebraic evolution identities hold for all real times. Norm bounds in this\nsection restrict to $t\\ge0$, with growth assumptions on the nonnegative-time flows.","One has the operator identity\n\\[\nE(t)=\\int_0^t e^{(t-s)A}\\delta e^{sL}\\,ds,\n\\]\nand $\\delta=0$ exactly when $E(t)=0$ for every $t$. Also\n\\[\nE(s+t)=E(s)e^{tL}+e^{sA}E(t).\n\\]\nFor a second representation $p:Z\\to W$ with generator $B$,\n\\[\n\\delta_{pq}=p\\delta_q+\\delta_pq,\\qquad\nE_{pq}(t)=pE_q(t)+E_p(t)q.\n\\]","Differentiate $F(s)=e^{(t-s)A}qe^{sL}$ using the product rule. Its derivative\nis $e^{(t-s)A}(qL-Aq)e^{sL}$. Integrating from zero to $t$ gives\n$F(t)-F(0)=E(t)$ and proves the formula. Zero generator defect therefore\nforces zero evolution defect; differentiating $E$ at zero gives the\nconverse. The addition identity follows by adding and subtracting\n$e^{sA}qe^{tL}$ and using the two flow laws.\nFor a composite representation, add and subtract $pAq$ in its generator\ndefect and $pe^{tA}q$ in its evolution defect. Both stages' contributions\nremain explicit. Boundedness guarantees the continuous operator integrand.","the product rule. Its derivative\nis $e^{(t-s)A}(qL-Aq)e^{sL}$. Integrating from zero to $t$ gives\n$F(t)-F(0)=E(t)$ and proves the formula. Zero generator defect therefore\nforces zero evolution defect; differentiating $E$ at zero gives the\nconverse. The addition identity follows by adding and subtracting\n$e^{sA}qe^{tL}$ and using the two flow laws.\nFor a composite representation, add and subtract $pAq$ in its generator\ndefect and $pe^{tA}q$ in its evolution defect. Both stages' contributions\nremain explicit. Boundedness guarantees the continuous operator integrand.","s $e^{(t-s)A}(qL-Aq)e^{sL}$. Integrating from zero to $t$ gives\n$F(t)-F(0)=E(t)$ and proves the formula. Zero generator defect therefore\nforces zero evolution defect; differentiating $E$ at zero gives the\nconverse. The addition identity follows by adding and subtracting\n$e^{sA}qe^{tL}$ and using the two flow laws.\nFor a composite representation, add and subtract $pAq$ in its generator\ndefect and $pe^{tA}q$ in its evolution defect. Both stages' contributions\nremain explicit. Boundedness guarantees the continuous operator integrand.","\\par{\\small\\textbf{Hypotheses.} Real Banach spaces $X,Z$; bounded autonomous real generators $L,A$ and bounded representation/readouts $q,O,V$; real rates $a,b$ of either sign; $t\\ge0$ and $M_L,M_A,\\epsilon,\\eta\\ge0$. For every $s\\ge0$, $\\nn{e^{sL}}\\le M_Le^{as}$ and $\\nn{e^{sA}}\\le M_Ae^{bs}$; also $\\nn{qL-Aq}\\le\\epsilon$ and $\\nn{O-Vq}\\le\\eta$.}\\par","For $\\nn\\delta\\le\\epsilon$, put\n\\[\nR(a,b,t)=\\int_0^t e^{b(t-s)}e^{as}\\,ds\n=\\begin{cases}(e^{at}-e^{bt})/(a-b),&a\\ne b,","For $\\nn\\delta\\le\\epsilon$, put\n\\[\nR(a,b,t)=\\int_0^t e^{b(t-s)}e^{as}\\,ds\n=\\begin{cases}(e^{at}-e^{bt})/(a-b),&a\\ne b,\\\\ te^{at},&a=b.\\end{cases}\n\\]\nThen\n\\[\n\\nn{E(t)}\\le M_LM_A\\epsilon R(a,b,t).\n\\]\nIf $\\nn{O-Vq}\\le\\eta$, the output defect $F(t)=Oe^{tL}-Ve^{tA}q$ obeys\n\\[\n\\nn{F(t)}\\le\\eta M_Le^{at}+\\nn V M_LM_A\\epsilon R(a,b,t).\n\\]","te^{at},&a=b.\\end{cases}\n\\]\nThen\n\\[\n\\nn{E(t)}\\le M_LM_A\\epsilon R(a,b,t).\n\\]\nIf $\\nn{O-Vq}\\le\\eta$, the output defect $F(t)=Oe^{tL}-Ve^{tA}q$ obeys\n\\[\n\\nn{F(t)}\\le\\eta M_Le^{at}+\\nn V M_LM_A\\epsilon R(a,b,t).\n\\]","Apply the integral norm inequality and submultiplicativity to SR-21.\nFor $0\\le s\\le t$, the integrand norm is at most\n$M_Ae^{b(t-s)}\\epsilon M_Le^{as}$. Integration gives the first estimate.\nFactoring $e^{bt}$ and integrating $e^{(a-b)s}$ gives the distinct-rate\nformula; for equal rates the integrand is constant $e^{at}$.\nThe output decomposition is\n$F(t)=(O-Vq)e^{tL}+VE(t)$, giving the stated bound.\nThe same argument works with continuous nonnegative envelope functions\n$u,v$ on the interval, replacing the scalar response by\n$\\int_0^tv(t-s)u(s)\\,ds$. In particular these assumptions concern actual\nflow norms, not just eigenvalues of the generators.","le s\\le t$, the integrand norm is at most\n$M_Ae^{b(t-s)}\\epsilon M_Le^{as}$. Integration gives the first estimate.\nFactoring $e^{bt}$ and integrating $e^{(a-b)s}$ gives the distinct-rate\nformula; for equal rates the integrand is constant $e^{at}$.\nThe output decomposition is\n$F(t)=(O-Vq)e^{tL}+VE(t)$, giving the stated bound.\nThe same argument works with continuous nonnegative envelope functions\n$u,v$ on the interval, replacing the scalar response by\n$\\int_0^tv(t-s)u(s)\\,ds$. In particular these assumptions concern actual\nflow norms, not just eigenvalues of the generators.","\\epsilon M_Le^{as}$. Integration gives the first estimate.\nFactoring $e^{bt}$ and integrating $e^{(a-b)s}$ gives the distinct-rate\nformula; for equal rates the integrand is constant $e^{at}$.\nThe output decomposition is\n$F(t)=(O-Vq)e^{tL}+VE(t)$, giving the stated bound.\nThe same argument works with continuous nonnegative envelope functions\n$u,v$ on the interval, replacing the scalar response by\n$\\int_0^tv(t-s)u(s)\\,ds$. In particular these assumptions concern actual\nflow norms, not just eigenvalues of the generators.","\\par{\\small\\textbf{Hypotheses.} True flow bounded by M\\_L; reduced flow bounded by M\\_A exp(-rt), r>0, for t\\(\\ge\\)0; generator/\\allowbreak readout allowances; arbitrary initial reduced z.}\\par","If $\\nn{e^{tL}}\\le M_L$ and $\\nn{e^{tA}}\\le M_Ae^{-rt}$ with $r>0$, then\n\\[\n\\nn{E(t)}\\le\\frac{M_LM_A\\epsilon}{r},\\qquad\n\\nn{F(t)}\\le\\eta M_L+\\nn V\\frac{M_LM_A\\epsilon}{r}.\n\\]\nFor arbitrary $x,z$,\n\\begin{align*}\n\\nn{Oe^{tL}x-Ve^{tA}z}\n&\\le\\left(\\eta M_L+\\nn V\\frac{M_LM_A\\epsilon}{r}\\right)\\nn x","If $\\nn{e^{tL}}\\le M_L$ and $\\nn{e^{tA}}\\le M_Ae^{-rt}$ with $r>0$, then\n\\[\n\\nn{E(t)}\\le\\frac{M_LM_A\\epsilon}{r},\\qquad\n\\nn{F(t)}\\le\\eta M_L+\\nn V\\frac{M_LM_A\\epsilon}{r}.\n\\]\nFor arbitrary $x,z$,\n\\begin{align*}\n\\nn{Oe^{tL}x-Ve^{tA}z}\n&\\le\\left(\\eta M_L+\\nn V\\frac{M_LM_A\\epsilon}{r}\\right)\\nn x\\\\\n&\\quad+\\nn V M_Ae^{-rt}\\nn{qx-z}.\n\\end{align*}\nComposition through a second representation propagates allowances with\nits explicit map norm, as in SR-21.","Use SR-22 with $a=0$, $b=-r$. Its response equals\n$(1-e^{-rt})/r\\le1/r$ for $t\\ge0$. This gives the evolution and output\noperator bounds. The initialization error decomposes into $F(t)x$ plus\n$Ve^{tA}(qx-z)$, which proves the displayed estimate. The first allowance\nmust be multiplied by $\\nn x$ because it is an operator estimate.\nFor another representation $p$ with its own evolution defect, SR-21 gives\n$\\nn{E_{pq}(t)}\\le\\nn p\\nn{E_q(t)}+\\nn{E_p(t)}\\nn q$.\nThis retains error introduced at both stages, rather than identifying\ntheir budgets with a single unqualified small quantity.","22 with $a=0$, $b=-r$. Its response equals\n$(1-e^{-rt})/r\\le1/r$ for $t\\ge0$. This gives the evolution and output\noperator bounds. The initialization error decomposes into $F(t)x$ plus\n$Ve^{tA}(qx-z)$, which proves the displayed estimate. The first allowance\nmust be multiplied by $\\nn x$ because it is an operator estimate.\nFor another representation $p$ with its own evolution defect, SR-21 gives\n$\\nn{E_{pq}(t)}\\le\\nn p\\nn{E_q(t)}+\\nn{E_p(t)}\\nn q$.\nThis retains error introduced at both stages, rather than identifying\ntheir budgets with a single unqualified small quantity.","\\par{\\small\\textbf{Hypotheses.} Actual real scalar bounded generators, q=I; t\\(\\ge\\)0; b\\(\\le\\)a for equality and epsilon>0 for counterexample.}\\par","For scalar generators $a\\ge b$,\n\\[\n|e^{at}-e^{bt}|=(a-b)R(a,b,t),\n\\]\nso the response estimate is attained. For every $\\epsilon>0$, choosing\n$L=\\epsilon I$, $A=0$, $q=I$ gives\n\\[\n\\nn\\delta=\\epsilon,\\qquad \\nn{E(t)}=e^{\\epsilon t}-1\\ge\\epsilon t.\n\\]\nIts evolution defect exceeds every fixed bound at some nonnegative time.","The scalar exponential is the actual flow by the homogeneous ODE uniqueness\ntheorem. For $a>b$, the response formula immediately gives the equality;\nfor $a=b$, both sides vanish. In the second example, the generator defect\nis multiplication by $\\epsilon$, while the evolution defect is multiplication\nby $e^{\\epsilon t}-1$. The elementary exponential inequality\n$e^u\\ge1+u$ for $u\\ge0$ gives the lower bound. Taking $t$ sufficiently\nlarge makes it exceed any prescribed real bound. Thus a small generator\ndefect alone does not supply uniform accuracy. The hypothesis of a bounded\ntrue flow in SR-23 fails here.","ntial is the actual flow by the homogeneous ODE uniqueness\ntheorem. For $a>b$, the response formula immediately gives the equality;\nfor $a=b$, both sides vanish. In the second example, the generator defect\nis multiplication by $\\epsilon$, while the evolution defect is multiplication\nby $e^{\\epsilon t}-1$. The elementary exponential inequality\n$e^u\\ge1+u$ for $u\\ge0$ gives the lower bound. Taking $t$ sufficiently\nlarge makes it exceed any prescribed real bound. Thus a small generator\ndefect alone does not supply uniform accuracy. The hypothesis of a bounded\ntrue flow in SR-23 fails here.","Let $R\\subseteq K$ be subspaces of a real inner-product space $V$, both admitting\northogonal projections. Define the actual maps\n\\[\nr(v)=P_Rv,\\quad o(v)=(P_K-P_R)v,\\quad i(v)=(I-P_K)v.\n\\]\nThey distinguish realized data, compatible obstruction, and incompatibility.\nIn a globalization presentation use $R=\\Ran A$, $K=\\ker\\delta$ with $\\delta A=0$.","\\par{\\small\\textbf{Hypotheses.} Real inner-product V; nested subspaces R\\(\\le\\)K both with orthogonal projections; arbitrary v and represented candidate u\\(\\in\\)R.}\\par","The components lie in $R$, $K\\cap R^\\perp$, and $K^\\perp$, respectively,\nare pairwise orthogonal, and give the unique decomposition\n$v=r(v)+o(v)+i(v)$. Furthermore,\n\\[\n\\nn v^2=\\nn r^2+\\nn o^2+\\nn i^2,\\qquad\n\\dist(v,R)^2=\\nn o^2+\\nn i^2.\n\\]\nFor any $u\\in R$,\n\\[\n\\nn{v-u}^2=\\dist(v,R)^2+\\nn{r(v)-u}^2.\n\\]\nThus $r(v)$ is the unique nearest represented vector.","Nested projections satisfy $P_RP_K=P_R$. Hence\n$P_Kv-P_Rv$ lies in $K$ and is orthogonal to $R$; the last component is\northogonal to $K$. Their definitions sum to $v$ and yield pairwise\northogonality. In any competing decomposition, projection onto $R$\nselects its first term and projection onto $K$ selects the first two,\nso all components are forced. Pythagoras gives the norm identities.\nFor $u\\in R$, the residual $v-r(v)$ is orthogonal to $r(v)-u$,\nso the candidate identity follows. It proves optimality and uniqueness,\nand taking the infimum over actual $u\\in R$ proves the distance formula.\nEach component map is an orthogonal projection onto its displayed sector\nand consequently is nonexpansive.","satisfy $P_RP_K=P_R$. Hence\n$P_Kv-P_Rv$ lies in $K$ and is orthogonal to $R$; the last component is\northogonal to $K$. Their definitions sum to $v$ and yield pairwise\northogonality. In any competing decomposition, projection onto $R$\nselects its first term and projection onto $K$ selects the first two,\nso all components are forced. Pythagoras gives the norm identities.\nFor $u\\in R$, the residual $v-r(v)$ is orthogonal to $r(v)-u$,\nso the candidate identity follows. It proves optimality and uniqueness,\nand taking the infimum over actual $u\\in R$ proves the distance formula.\nEach component map is an orthogonal projection onto its displayed sector\nand consequently is nonexpansive.","e\n$P_Kv-P_Rv$ lies in $K$ and is orthogonal to $R$; the last component is\northogonal to $K$. Their definitions sum to $v$ and yield pairwise\northogonality. In any competing decomposition, projection onto $R$\nselects its first term and projection onto $K$ selects the first two,\nso all components are forced. Pythagoras gives the norm identities.\nFor $u\\in R$, the residual $v-r(v)$ is orthogonal to $r(v)-u$,\nso the candidate identity follows. It proves optimality and uniqueness,\nand taking the infimum over actual $u\\in R$ proves the distance formula.\nEach component map is an orthogonal projection onto its displayed sector\nand consequently is nonexpansive.","$v$ and yield pairwise\northogonality. In any competing decomposition, projection onto $R$\nselects its first term and projection onto $K$ selects the first two,\nso all components are forced. Pythagoras gives the norm identities.\nFor $u\\in R$, the residual $v-r(v)$ is orthogonal to $r(v)-u$,\nso the candidate identity follows. It proves optimality and uniqueness,\nand taking the infimum over actual $u\\in R$ proves the distance formula.\nEach component map is an orthogonal projection onto its displayed sector\nand consequently is nonexpansive.","\\par{\\small\\textbf{Hypotheses.} Real linear presentation \\(\\delta\\)A=0; projections on actual R=range A and K=ker\\(\\delta\\); source G only needs real module structure.}\\par","The same quotient $\\mathcal H=K/R$ from SR-03 is linearly equivalent to\n$K\\cap R^\\perp$ by\n\\[\n[v]\\longmapsto v-P_Rv\\quad(v\\in K).\n\\]\nIt is inverse to the quotient map restricted to the harmonic subspace.\nFor compatible data, the class is zero exactly when $o(v)=0$; equality\nof classes is equality of harmonic representatives. The best image $P_Rv$\nhas an actual source under $A$.","Adding a vector of $R$ leaves $v-P_Rv$ unchanged, so the map descends to\n$K/R$. Its values lie in $K\\cap R^\\perp$ by SR-25. For harmonic $h$,\n$P_Rh=0$, so its class maps back to $h$. For compatible $v$, the difference\n$v-(v-P_Rv)=P_Rv$ lies in $R$, so the reverse roundtrip returns $[v]$.\nThe inverse laws prove the equivalence and both class criteria.\nSince $P_Rv$ belongs to the actual range, some $g$ satisfies $Ag=P_Rv$.\nThis is image existence; uniqueness of $g$ requires the separate injectivity\ncondition in SR-03. No norm isometry is inferred from linear equivalence\nalone; the metric statement is proved in SR-33.","dding a vector of $R$ leaves $v-P_Rv$ unchanged, so the map descends to\n$K/R$. Its values lie in $K\\cap R^\\perp$ by SR-25. For harmonic $h$,\n$P_Rh=0$, so its class maps back to $h$. For compatible $v$, the difference\n$v-(v-P_Rv)=P_Rv$ lies in $R$, so the reverse roundtrip returns $[v]$.\nThe inverse laws prove the equivalence and both class criteria.\nSince $P_Rv$ belongs to the actual range, some $g$ satisfies $Ag=P_Rv$.\nThis is image existence; uniqueness of $g$ requires the separate injectivity\ncondition in SR-03. No norm isometry is inferred from linear equivalence\nalone; the metric statement is proved in SR-33.","of $R$ leaves $v-P_Rv$ unchanged, so the map descends to\n$K/R$. Its values lie in $K\\cap R^\\perp$ by SR-25. For harmonic $h$,\n$P_Rh=0$, so its class maps back to $h$. For compatible $v$, the difference\n$v-(v-P_Rv)=P_Rv$ lies in $R$, so the reverse roundtrip returns $[v]$.\nThe inverse laws prove the equivalence and both class criteria.\nSince $P_Rv$ belongs to the actual range, some $g$ satisfies $Ag=P_Rv$.\nThis is image existence; uniqueness of $g$ requires the separate injectivity\ncondition in SR-03. No norm isometry is inferred from linear equivalence\nalone; the metric statement is proved in SR-33.","ends to\n$K/R$. Its values lie in $K\\cap R^\\perp$ by SR-25. For harmonic $h$,\n$P_Rh=0$, so its class maps back to $h$. For compatible $v$, the difference\n$v-(v-P_Rv)=P_Rv$ lies in $R$, so the reverse roundtrip returns $[v]$.\nThe inverse laws prove the equivalence and both class criteria.\nSince $P_Rv$ belongs to the actual range, some $g$ satisfies $Ag=P_Rv$.\nThis is image existence; uniqueness of $g$ requires the separate injectivity\ncondition in SR-03. No norm isometry is inferred from linear equivalence\nalone; the metric statement is proved in SR-33.","\\par{\\small\\textbf{Hypotheses.} Same projection presentation; bounded \\(\\delta\\); sigma>0 with sigma\\(\\Vert\\)x\\(\\Vert\\le\\Vert\\delta\\)x\\(\\Vert\\) on K\\(\\perp\\); arbitrary data.}\\par","If $\\sigma\\nn x\\le\\nn{\\delta x}$ for $x\\in K^\\perp$ and $\\sigma>0$, then\n\\[\n\\delta i(v)=\\delta v,\\qquad\n\\nn{i(v)}\\le\\frac{\\nn{\\delta v}}\\sigma,\n\\]\nand\n\\[\n\\nn{o(v)}\\le\\dist(v,R),\\qquad\n\\dist(v,R)^2\\le\\nn{o(v)}^2+\\left(\\frac{\\nn{\\delta v}}\\sigma\\right)^2.\n\\]\nFor compatible data, $\\dist(v,R)=\\nn{o(v)}$.","The component $P_Kv$ is in $\\ker\\delta$, so\n$\\delta(v-P_Kv)=\\delta v$. Apply coercivity to $i(v)\\in K^\\perp$ and divide\nby the positive $\\sigma$. SR-25 gives the exact sum of two nonnegative\nsquares, proving both the lower bound and conditioned upper bound.\nIf $\\delta v=0$, then $v\\in K$ and $i(v)=0$, giving equality.\nIn particular, a compatibility residual bounds only inconsistency. A\nresidual-only global bound needs a vanished obstruction or exactness\n$R=K$. The condition on $\\sigma$ is independent of the size of a particular\nmeasured residual and cannot be inferred from that measurement.","=\\delta v$. Apply coercivity to $i(v)\\in K^\\perp$ and divide\nby the positive $\\sigma$. SR-25 gives the exact sum of two nonnegative\nsquares, proving both the lower bound and conditioned upper bound.\nIf $\\delta v=0$, then $v\\in K$ and $i(v)=0$, giving equality.\nIn particular, a compatibility residual bounds only inconsistency. A\nresidual-only global bound needs a vanished obstruction or exactness\n$R=K$. The condition on $\\sigma$ is independent of the size of a particular\nmeasured residual and cannot be inferred from that measurement.","vity to $i(v)\\in K^\\perp$ and divide\nby the positive $\\sigma$. SR-25 gives the exact sum of two nonnegative\nsquares, proving both the lower bound and conditioned upper bound.\nIf $\\delta v=0$, then $v\\in K$ and $i(v)=0$, giving equality.\nIn particular, a compatibility residual bounds only inconsistency. A\nresidual-only global bound needs a vanished obstruction or exactness\n$R=K$. The condition on $\\sigma$ is independent of the size of a particular\nmeasured residual and cannot be inferred from that measurement.","\\par{\\small\\textbf{Hypotheses.} Finite-dimensional real V for existence/\\allowbreak characterization; fixed metric and K with projected R0\\(\\le\\)R1\\(\\le\\)K for refinement.}\\par","In finite dimension some $\\sigma>0$ satisfies SR-27. Under $\\delta A=0$,\n\\[\n\\exists c>0\\ \\forall v:\\dist(v,\\Ran A)\\le c\\nn{\\delta v}\n\\iff\\Ran A=\\ker\\delta.\n\\]\nFor projected $R_0\\subseteq R_1\\subseteq K$ in a fixed metric, put\n$g(v)=(P_{R_1}-P_{R_0})v$. Then\n\\[\n\\nn{o_0(v)}^2=\\nn{o_1(v)}^2+\\nn{g(v)}^2,\n\\quad\n\\dist(v,R_0)^2=\\dist(v,R_1)^2+\\nn{g(v)}^2.\n\\]\nNo uniform conditioning constant exists across all nonzero scalar compatibility maps.","Restrict $\\delta$ to $K^\\perp$. It is injective: a zero image lies in both\n$K$ and $K^\\perp$. A finite-dimensional injective linear map has a positive\nlower norm bound, giving $\\sigma$; the zero domain is harmless and permits\nany positive choice. If the residual-only inequality holds, every $v\\in K$\nhas zero distance to $\\Ran A$. Finite-dimensional ranges are closed, so\n$v\\in\\Ran A$. Conversely exactness removes $o(v)$ in SR-27.","to $K^\\perp$. It is injective: a zero image lies in both\n$K$ and $K^\\perp$. A finite-dimensional injective linear map has a positive\nlower norm bound, giving $\\sigma$; the zero domain is harmless and permits\nany positive choice. If the residual-only inequality holds, every $v\\in K$\nhas zero distance to $\\Ran A$. Finite-dimensional ranges are closed, so\n$v\\in\\Ran A$. Conversely exactness removes $o(v)$ in SR-27.","Nested orthogonal projections split $o_0(v)=o_1(v)+g(v)$ into orthogonal\nsectors, and the inconsistency component stays fixed. Pythagoras gives\nboth refinement identities. For the scalar family $\\delta_s(x)=sx$, $s>0$,\ntake $R=0$ and $v=1$. The reconstruction distance is one and the residual\nis $s$. For any proposed common $c$, choosing $s$ sufficiently small\nviolates $1\\le cs$. The theorem guarantees a constant per presentation,\nnot a common constant across degenerating presentations.","\\begin{example}[A complete error split]\nFor $A(g)=(g,0,0)$, $\\delta(x,y,z)=2z$, and $v=(3,4,5)$,\nthe components are $(3,0,0)$, $(0,4,0)$, and $(0,0,5)$.\nThe squared reconstruction distance is $41=16+25$. The compatibility\nresidual has squared norm $100$, and $\\sigma=2$ is sharp on $K^\\perp$.\nThus the conditioned upper bound is attained: $16+100/4=41$.\n\\end{example}","Let $A:G\\to V$ be bounded between real Hilbert spaces, with actual closed\nrange $R=\\Ran A$. Write $N_A=\\ker A$ and $S=N_A^\\perp$.\nClosedness is a substantive hypothesis for a bounded exact inverse. It will\nbe removed for regularized recovery in the following sections.","Restriction $A_S:S\\to R$ is a bounded linear bijection with bounded inverse.\nDefine\n\\[\nB=\\iota_S A_S^{-1}P_R.\n\\]\nThen $AB=P_R$, $BA=P_S$, $ABA=A$, and $BAB=B$.\nIts two projections are symmetric, $\\ker B=R^\\perp$, and $\\Ran B=S$.\nThe best-image identity $AB=P_R$ and the condition $\\Ran B\\subseteq S$\nuniquely determine $B$.","The restriction is injective because $S\\cap\\ker A=0$. Every $Ag\\in R$\nequals $AP_Sg$, so it is surjective. Closed $R$ and closed $S$ are Banach\nspaces; the bounded inverse theorem supplies $A_S^{-1}$ as an actual\ncontinuous linear inverse. The definition then gives $AB=P_R$.\nFor $g$, $BAg$ is the unique element of $S$ with image $Ag$, namely\n$P_Sg$. The two inverse identities imply the generalized inverse laws.\nOrthogonal projections are symmetric, proving the symmetry conditions.\nThe definition and injectivity of $A_S^{-1}$ identify the kernel and range.\nFinally, if $B'$ has the same best image and values in $S$, then\n$B'v-Bv\\in S\\cap\\ker A$ for each $v$, hence it is zero.\nThis construction is the classical Moore--Penrose inverse with the\na","The restriction is injective because $S\\cap\\ker A=0$. Every $Ag\\in R$\nequals $AP_Sg$, so it is surjective. Closed $R$ and closed $S$ are Banach\nspaces; the bounded inverse theorem supplies $A_S^{-1}$ as an actual\ncontinuous linear inverse. The definition then gives $AB=P_R$.\nFor $g$, $BAg$ is the unique element of $S$ with image $Ag$, namely\n$P_Sg$. The two inverse identities imply the generalized inverse laws.\nOrthogonal projections are symmetric, proving the symmetry conditions.\nThe definition and injectivity of $A_S^{-1}$ identify the kernel and range.\nFinally, if $B'$ has the same best image and values in $S$, then\n$B'v-Bv\\in S\\cap\\ker A$ for each $v$, hence it is zero.\nThis construction is the classical Moore--Penrose inverse with the\nactual range and s","The restriction is injective because $S\\cap\\ker A=0$. Every $Ag\\in R$\nequals $AP_Sg$, so it is surjective. Closed $R$ and closed $S$ are Banach\nspaces; the bounded inverse theorem supplies $A_S^{-1}$ as an actual\ncontinuous linear inverse. The definition then gives $AB=P_R$.\nFor $g$, $BAg$ is the unique element of $S$ with image $Ag$, namely\n$P_Sg$. The two inverse identities imply the generalized inverse laws.\nOrthogonal projections are symmetric, proving the symmetry conditions.\nThe definition and injectivity of $A_S^{-1}$ identify the kernel and range.\nFinally, if $B'$ has the same best image and values in $S$, then\n$B'v-Bv\\in S\\cap\\ker A$ for each $v$, hence it is zero.\nThis construction is the classical Moore--Penrose inverse with the\nactual range and source ambiguity","The restriction is injective because $S\\cap\\ker A=0$. Every $Ag\\in R$\nequals $AP_Sg$, so it is surjective. Closed $R$ and closed $S$ are Banach\nspaces; the bounded inverse theorem supplies $A_S^{-1}$ as an actual\ncontinuous linear inverse. The definition then gives $AB=P_R$.\nFor $g$, $BAg$ is the unique element of $S$ with image $Ag$, namely\n$P_Sg$. The two inverse identities imply the generalized inverse laws.\nOrthogonal projections are symmetric, proving the symmetry conditions.\nThe definition and injectivity of $A_S^{-1}$ identify the kernel and range.\nFinally, if $B'$ has the same best image and values in $S$, then\n$B'v-Bv\\in S\\cap\\ker A$ for each $v$, hence it is zero.\nThis construction is the classical Moore--Penrose inverse with the\nactual range and source ambiguity retained.","closed $S$ are Banach\nspaces; the bounded inverse theorem supplies $A_S^{-1}$ as an actual\ncontinuous linear inverse. The definition then gives $AB=P_R$.\nFor $g$, $BAg$ is the unique element of $S$ with image $Ag$, namely\n$P_Sg$. The two inverse identities imply the generalized inverse laws.\nOrthogonal projections are symmetric, proving the symmetry conditions.\nThe definition and injectivity of $A_S^{-1}$ identify the kernel and range.\nFinally, if $B'$ has the same best image and values in $S$, then\n$B'v-Bv\\in S\\cap\\ker A$ for each $v$, hence it is zero.\nThis construction is the classical Moore--Penrose inverse with the\nactual range and source ambiguity retained.","orem supplies $A_S^{-1}$ as an actual\ncontinuous linear inverse. The definition then gives $AB=P_R$.\nFor $g$, $BAg$ is the unique element of $S$ with image $Ag$, namely\n$P_Sg$. The two inverse identities imply the generalized inverse laws.\nOrthogonal projections are symmetric, proving the symmetry conditions.\nThe definition and injectivity of $A_S^{-1}$ identify the kernel and range.\nFinally, if $B'$ has the same best image and values in $S$, then\n$B'v-Bv\\in S\\cap\\ker A$ for each $v$, hence it is zero.\nThis construction is the classical Moore--Penrose inverse with the\nactual range and source ambiguity retained.","nuous linear inverse. The definition then gives $AB=P_R$.\nFor $g$, $BAg$ is the unique element of $S$ with image $Ag$, namely\n$P_Sg$. The two inverse identities imply the generalized inverse laws.\nOrthogonal projections are symmetric, proving the symmetry conditions.\nThe definition and injectivity of $A_S^{-1}$ identify the kernel and range.\nFinally, if $B'$ has the same best image and values in $S$, then\n$B'v-Bv\\in S\\cap\\ker A$ for each $v$, hence it is zero.\nThis construction is the classical Moore--Penrose inverse with the\nactual range and source ambiguity retained.","ar inverse. The definition then gives $AB=P_R$.\nFor $g$, $BAg$ is the unique element of $S$ with image $Ag$, namely\n$P_Sg$. The two inverse identities imply the generalized inverse laws.\nOrthogonal projections are symmetric, proving the symmetry conditions.\nThe definition and injectivity of $A_S^{-1}$ identify the kernel and range.\nFinally, if $B'$ has the same best image and values in $S$, then\n$B'v-Bv\\in S\\cap\\ker A$ for each $v$, hence it is zero.\nThis construction is the classical Moore--Penrose inverse with the\nactual range and source ambiguity retained.","ives $AB=P_R$.\nFor $g$, $BAg$ is the unique element of $S$ with image $Ag$, namely\n$P_Sg$. The two inverse identities imply the generalized inverse laws.\nOrthogonal projections are symmetric, proving the symmetry conditions.\nThe definition and injectivity of $A_S^{-1}$ identify the kernel and range.\nFinally, if $B'$ has the same best image and values in $S$, then\n$B'v-Bv\\in S\\cap\\ker A$ for each $v$, hence it is zero.\nThis construction is the classical Moore--Penrose inverse with the\nactual range and source ambiguity retained.","For every source $g$,\n\\[\n\\nn{v-Ag}^2=\\dist(v,R)^2+\\nn{A(g-Bv)}^2.\n\\]\nAll least-squares minimizers are $Bv+k$ with $k\\in\\ker A$.\nAmong them $Bv$ is the unique minimum-norm source. Exact source recovery\n$BAg=g$ holds exactly for $g\\in S$. For data $Ag+e$,\n\\[\n\\nn{B(Ag+e)-g}^2=\\nn{Be}^2+\\nn{P_{N_A}g}^2.\n\\]","$ABv=P_Rv$, so SR-25's candidate identity with $u=Ag$ proves the first\nformula. Equality to the minimum image error is equivalent to\n$A(g-Bv)=0$, yielding exactly the affine family of minimizers.\nSince $Bv\\in S$ and $k\\in N_A$, Pythagoras gives\n$\\nn{Bv+k}^2=\\nn{Bv}^2+\\nn k^2$. It proves minimum norm and uniqueness.\nThe identity $BA=P_S$ proves the exact recovery criterion.\nFor noisy data, linearity gives $B(Ag+e)-g=Be-P_{N_A}g$.\nThe first term belongs to $S$ and the second to $N_A$, so they are\northogonal. This last identity does not assume a statistical noise model.\nIt shows explicitly which source content the operator cannot identify.","$, so SR-25's candidate identity with $u=Ag$ proves the first\nformula. Equality to the minimum image error is equivalent to\n$A(g-Bv)=0$, yielding exactly the affine family of minimizers.\nSince $Bv\\in S$ and $k\\in N_A$, Pythagoras gives\n$\\nn{Bv+k}^2=\\nn{Bv}^2+\\nn k^2$. It proves minimum norm and uniqueness.\nThe identity $BA=P_S$ proves the exact recovery criterion.\nFor noisy data, linearity gives $B(Ag+e)-g=Be-P_{N_A}g$.\nThe first term belongs to $S$ and the second to $N_A$, so they are\northogonal. This last identity does not assume a statistical noise model.\nIt shows explicitly which source content the operator cannot identify.","the first\nformula. Equality to the minimum image error is equivalent to\n$A(g-Bv)=0$, yielding exactly the affine family of minimizers.\nSince $Bv\\in S$ and $k\\in N_A$, Pythagoras gives\n$\\nn{Bv+k}^2=\\nn{Bv}^2+\\nn k^2$. It proves minimum norm and uniqueness.\nThe identity $BA=P_S$ proves the exact recovery criterion.\nFor noisy data, linearity gives $B(Ag+e)-g=Be-P_{N_A}g$.\nThe first term belongs to $S$ and the second to $N_A$, so they are\northogonal. This last identity does not assume a statistical noise model.\nIt shows explicitly which source content the operator cannot identify.","uality to the minimum image error is equivalent to\n$A(g-Bv)=0$, yielding exactly the affine family of minimizers.\nSince $Bv\\in S$ and $k\\in N_A$, Pythagoras gives\n$\\nn{Bv+k}^2=\\nn{Bv}^2+\\nn k^2$. It proves minimum norm and uniqueness.\nThe identity $BA=P_S$ proves the exact recovery criterion.\nFor noisy data, linearity gives $B(Ag+e)-g=Be-P_{N_A}g$.\nThe first term belongs to $S$ and the second to $N_A$, so they are\northogonal. This last identity does not assume a statistical noise model.\nIt shows explicitly which source content the operator cannot identify.","\\par{\\small\\textbf{Hypotheses.} Closed-range A; alpha>0 with alpha\\(\\Vert\\)g\\(\\Vert\\le\\Vert\\)Ag\\(\\Vert\\) on S; computed source h may contain a kernel component.}\\par","Some $\\alpha>0$ satisfies $\\alpha\\nn g\\le\\nn{Ag}$ on $S$, and any supplied\nsuch constant gives $\\nn B\\le1/\\alpha$. Therefore\n\\[\n\\nn{B(Ag+e)-g}^2\\le(\\nn e/\\alpha)^2+\\nn{P_{N_A}g}^2.\n\\]\nFor an arbitrary computed source $h$,\n\\[\n\\nn{h-Bv}^2\n\\le\\left(\\frac{\\nn{Ah-P_Rv}}\\alpha\\right)^2+\\nn{P_{N_A}h}^2.\n\\]\nIf also $\\delta A=0$, a compatibility constant $\\sigma$ bounds the local\nerror as in SR-27. The constants $\\alpha$ and $\\sigma$ have separate roles.","lpha>0$ satisfies $\\alpha\\nn g\\le\\nn{Ag}$ on $S$, and any supplied\nsuch constant gives $\\nn B\\le1/\\alpha$. Therefore\n\\[\n\\nn{B(Ag+e)-g}^2\\le(\\nn e/\\alpha)^2+\\nn{P_{N_A}g}^2.\n\\]\nFor an arbitrary computed source $h$,\n\\[\n\\nn{h-Bv}^2\n\\le\\left(\\frac{\\nn{Ah-P_Rv}}\\alpha\\right)^2+\\nn{P_{N_A}h}^2.\n\\]\nIf also $\\delta A=0$, a compatibility constant $\\sigma$ bounds the local\nerror as in SR-27. The constants $\\alpha$ and $\\sigma$ have separate roles.","fies $\\alpha\\nn g\\le\\nn{Ag}$ on $S$, and any supplied\nsuch constant gives $\\nn B\\le1/\\alpha$. Therefore\n\\[\n\\nn{B(Ag+e)-g}^2\\le(\\nn e/\\alpha)^2+\\nn{P_{N_A}g}^2.\n\\]\nFor an arbitrary computed source $h$,\n\\[\n\\nn{h-Bv}^2\n\\le\\left(\\frac{\\nn{Ah-P_Rv}}\\alpha\\right)^2+\\nn{P_{N_A}h}^2.\n\\]\nIf also $\\delta A=0$, a compatibility constant $\\sigma$ bounds the local\nerror as in SR-27. The constants $\\alpha$ and $\\sigma$ have separate roles.","The bounded inverse on $S$ gives\n$\\nn g\\le\\nn B\\nn{Ag}$. Taking, for example,\n$\\alpha=1/(\\nn B+1)>0$ handles also the zero complement and proves\nexistence. For a supplied constant, apply it to $Bv\\in S$:\n$\\alpha\\nn{Bv}\\le\\nn{ABv}=\\nn{P_Rv}\\le\\nn v$.\nTake the operator norm and combine with SR-30 for the noisy bound.\nFor a computed $h$, split\n$h-Bv=(P_Sh-Bv)+P_{N_A}h$ orthogonally. Coercivity bounds the first term\nby $\\nn{Ah-P_Rv}/\\alpha$ because $AP_Sh=Ah$.\nThe compatibility operator bounds distance of local data to realizability;\nthe forward operator bounds source amplification. Neither condition\nestablishes the other. A small fit without control of the computed kernel\ncomponent does not certify the minimum-norm sou","The bounded inverse on $S$ gives\n$\\nn g\\le\\nn B\\nn{Ag}$. Taking, for example,\n$\\alpha=1/(\\nn B+1)>0$ handles also the zero complement and proves\nexistence. For a supplied constant, apply it to $Bv\\in S$:\n$\\alpha\\nn{Bv}\\le\\nn{ABv}=\\nn{P_Rv}\\le\\nn v$.\nTake the operator norm and combine with SR-30 for the noisy bound.\nFor a computed $h$, split\n$h-Bv=(P_Sh-Bv)+P_{N_A}h$ orthogonally. Coercivity bounds the first term\nby $\\nn{Ah-P_Rv}/\\alpha$ because $AP_Sh=Ah$.\nThe compatibility operator bounds distance of local data to realizability;\nthe forward operator bounds source amplification. Neither condition\nestablishes the other. A small fit without control of the computed kernel\ncomponent does not certify the minimum-norm source.","on $S$ gives\n$\\nn g\\le\\nn B\\nn{Ag}$. Taking, for example,\n$\\alpha=1/(\\nn B+1)>0$ handles also the zero complement and proves\nexistence. For a supplied constant, apply it to $Bv\\in S$:\n$\\alpha\\nn{Bv}\\le\\nn{ABv}=\\nn{P_Rv}\\le\\nn v$.\nTake the operator norm and combine with SR-30 for the noisy bound.\nFor a computed $h$, split\n$h-Bv=(P_Sh-Bv)+P_{N_A}h$ orthogonally. Coercivity bounds the first term\nby $\\nn{Ah-P_Rv}/\\alpha$ because $AP_Sh=Ah$.\nThe compatibility operator bounds distance of local data to realizability;\nthe forward operator bounds source amplification. Neither condition\nestablishes the other. A small fit without control of the computed kernel\ncomponent does not certify the minimum-norm source.","{Ag}$. Taking, for example,\n$\\alpha=1/(\\nn B+1)>0$ handles also the zero complement and proves\nexistence. For a supplied constant, apply it to $Bv\\in S$:\n$\\alpha\\nn{Bv}\\le\\nn{ABv}=\\nn{P_Rv}\\le\\nn v$.\nTake the operator norm and combine with SR-30 for the noisy bound.\nFor a computed $h$, split\n$h-Bv=(P_Sh-Bv)+P_{N_A}h$ orthogonally. Coercivity bounds the first term\nby $\\nn{Ah-P_Rv}/\\alpha$ because $AP_Sh=Ah$.\nThe compatibility operator bounds distance of local data to realizability;\nthe forward operator bounds source amplification. Neither condition\nestablishes the other. A small fit without control of the computed kernel\ncomponent does not certify the minimum-norm source.","The following conditions are equivalent:\n\\[\n\\Ran A\\text{ is closed};\\qquad\n\\exists\\alpha>0:\\alpha\\nn g\\le\\nn{Ag}\\ (g\\in S);\n\\qquad \\exists B\\text{ bounded}:BA=P_S.\n\\]\nThus a nonclosed range admits no bounded exact identifiable-source recovery.\nFinite-dimensional source or target supplies closed range. There is no common\ninverse-sensitivity constant across all nonzero real scalar forward maps.","Closed range gives the bounded constructor and coercivity by SR-29--SR-31.\nConversely suppose coercivity holds. If $Ag_n$ converges in $V$, replace\n$g_n$ by $P_Sg_n$ without changing its image. Coercivity makes this source\nsequence Cauchy. Completeness gives a limit $g\\in S$, and continuity gives\nthe image limit $Ag\\in\\Ran A$. Hence the range is closed.\nIf bounded $B$ satisfies $BA=P_S$, then for $g\\in S$,\n$\\nn g=\\nn{BAg}\\le\\nn B\\nn{Ag}$. The same positive choice\n$1/(\\nn B+1)$ gives coercivity. Finite-dimensional image subspaces are\nclosed in a normed space. Finally $A_cg=cg$, $c>0$, has canonical inverse\n$B_cv=v/c$. At $v=c$, data tend to zero with $c$ while the recovered\nsource remains one, ruling out a common constant as $c\\downarrow0$.","losed range gives the bounded constructor and coercivity by SR-29--SR-31.\nConversely suppose coercivity holds. If $Ag_n$ converges in $V$, replace\n$g_n$ by $P_Sg_n$ without changing its image. Coercivity makes this source\nsequence Cauchy. Completeness gives a limit $g\\in S$, and continuity gives\nthe image limit $Ag\\in\\Ran A$. Hence the range is closed.\nIf bounded $B$ satisfies $BA=P_S$, then for $g\\in S$,\n$\\nn g=\\nn{BAg}\\le\\nn B\\nn{Ag}$. The same positive choice\n$1/(\\nn B+1)$ gives coercivity. Finite-dimensional image subspaces are\nclosed in a normed space. Finally $A_cg=cg$, $c>0$, has canonical inverse\n$B_cv=v/c$. At $v=c$, data tend to zero with $c$ while the recovered\nsource remains one, ruling out a common constant as $c\\downarrow0$.","e gives the bounded constructor and coercivity by SR-29--SR-31.\nConversely suppose coercivity holds. If $Ag_n$ converges in $V$, replace\n$g_n$ by $P_Sg_n$ without changing its image. Coercivity makes this source\nsequence Cauchy. Completeness gives a limit $g\\in S$, and continuity gives\nthe image limit $Ag\\in\\Ran A$. Hence the range is closed.\nIf bounded $B$ satisfies $BA=P_S$, then for $g\\in S$,\n$\\nn g=\\nn{BAg}\\le\\nn B\\nn{Ag}$. The same positive choice\n$1/(\\nn B+1)$ gives coercivity. Finite-dimensional image subspaces are\nclosed in a normed space. Finally $A_cg=cg$, $c>0$, has canonical inverse\n$B_cv=v/c$. At $v=c$, data tend to zero with $c$ while the recovered\nsource remains one, ruling out a common constant as $c\\downarrow0$.","R-31.\nConversely suppose coercivity holds. If $Ag_n$ converges in $V$, replace\n$g_n$ by $P_Sg_n$ without changing its image. Coercivity makes this source\nsequence Cauchy. Completeness gives a limit $g\\in S$, and continuity gives\nthe image limit $Ag\\in\\Ran A$. Hence the range is closed.\nIf bounded $B$ satisfies $BA=P_S$, then for $g\\in S$,\n$\\nn g=\\nn{BAg}\\le\\nn B\\nn{Ag}$. The same positive choice\n$1/(\\nn B+1)$ gives coercivity. Finite-dimensional image subspaces are\nclosed in a normed space. Finally $A_cg=cg$, $c>0$, has canonical inverse\n$B_cv=v/c$. At $v=c$, data tend to zero with $c$ while the recovered\nsource remains one, ruling out a common constant as $c\\downarrow0$.","changing its image. Coercivity makes this source\nsequence Cauchy. Completeness gives a limit $g\\in S$, and continuity gives\nthe image limit $Ag\\in\\Ran A$. Hence the range is closed.\nIf bounded $B$ satisfies $BA=P_S$, then for $g\\in S$,\n$\\nn g=\\nn{BAg}\\le\\nn B\\nn{Ag}$. The same positive choice\n$1/(\\nn B+1)$ gives coercivity. Finite-dimensional image subspaces are\nclosed in a normed space. Finally $A_cg=cg$, $c>0$, has canonical inverse\n$B_cv=v/c$. At $v=c$, data tend to zero with $c$ while the recovered\nsource remains one, ruling out a common constant as $c\\downarrow0$.","s image. Coercivity makes this source\nsequence Cauchy. Completeness gives a limit $g\\in S$, and continuity gives\nthe image limit $Ag\\in\\Ran A$. Hence the range is closed.\nIf bounded $B$ satisfies $BA=P_S$, then for $g\\in S$,\n$\\nn g=\\nn{BAg}\\le\\nn B\\nn{Ag}$. The same positive choice\n$1/(\\nn B+1)$ gives coercivity. Finite-dimensional image subspaces are\nclosed in a normed space. Finally $A_cg=cg$, $c>0$, has canonical inverse\n$B_cv=v/c$. At $v=c$, data tend to zero with $c$ while the recovered\nsource remains one, ruling out a common constant as $c\\downarrow0$.","s source\nsequence Cauchy. Completeness gives a limit $g\\in S$, and continuity gives\nthe image limit $Ag\\in\\Ran A$. Hence the range is closed.\nIf bounded $B$ satisfies $BA=P_S$, then for $g\\in S$,\n$\\nn g=\\nn{BAg}\\le\\nn B\\nn{Ag}$. The same positive choice\n$1/(\\nn B+1)$ gives coercivity. Finite-dimensional image subspaces are\nclosed in a normed space. Finally $A_cg=cg$, $c>0$, has canonical inverse\n$B_cv=v/c$. At $v=c$, data tend to zero with $c$ while the recovered\nsource remains one, ruling out a common constant as $c\\downarrow0$.","ce Cauchy. Completeness gives a limit $g\\in S$, and continuity gives\nthe image limit $Ag\\in\\Ran A$. Hence the range is closed.\nIf bounded $B$ satisfies $BA=P_S$, then for $g\\in S$,\n$\\nn g=\\nn{BAg}\\le\\nn B\\nn{Ag}$. The same positive choice\n$1/(\\nn B+1)$ gives coercivity. Finite-dimensional image subspaces are\nclosed in a normed space. Finally $A_cg=cg$, $c>0$, has canonical inverse\n$B_cv=v/c$. At $v=c$, data tend to zero with $c$ while the recovered\nsource remains one, ruling out a common constant as $c\\downarrow0$.","a limit $g\\in S$, and continuity gives\nthe image limit $Ag\\in\\Ran A$. Hence the range is closed.\nIf bounded $B$ satisfies $BA=P_S$, then for $g\\in S$,\n$\\nn g=\\nn{BAg}\\le\\nn B\\nn{Ag}$. The same positive choice\n$1/(\\nn B+1)$ gives coercivity. Finite-dimensional image subspaces are\nclosed in a normed space. Finally $A_cg=cg$, $c>0$, has canonical inverse\n$B_cv=v/c$. At $v=c$, data tend to zero with $c$ while the recovered\nsource remains one, ruling out a common constant as $c\\downarrow0$.","the range is closed.\nIf bounded $B$ satisfies $BA=P_S$, then for $g\\in S$,\n$\\nn g=\\nn{BAg}\\le\\nn B\\nn{Ag}$. The same positive choice\n$1/(\\nn B+1)$ gives coercivity. Finite-dimensional image subspaces are\nclosed in a normed space. Finally $A_cg=cg$, $c>0$, has canonical inverse\n$B_cv=v/c$. At $v=c$, data tend to zero with $c$ while the recovered\nsource remains one, ruling out a common constant as $c\\downarrow0$.","\\begin{example}[Source and image are different optima]\nLet $A(g_1,g_2)=(g_1+g_2,0,0)$ and $v=(3,4,5)$.\nThe best image is $(3,0,0)$, with the same squared local error $41$.\nThe minimum-norm source is $(3/2,3/2)$ of squared norm $9/2$.\nThe source $(3,0)$ gives the same best image but has squared norm $9$.\nTheir difference $(3/2,-3/2)$ is kernel ambiguity, not local obstruction.\n\\end{example}","Retain the globalization presentation $\\delta A=0$, the actual projected\nsubspaces $R=\\Ran A\\subseteq K=\\ker\\delta$, and the same algebraic quotient\n$\\mathcal H=K/R$. Its seminorm is the quotient seminorm induced by the\nambient metric. We prove the harmonic equivalence respects that preexisting\nseminorm, rather than defining a new norm to match its answer.","\\par{\\small\\textbf{Hypotheses.} Real inner-product V; R=range A,K=ker\\(\\delta\\) both projected; \\(\\delta\\) bounded, \\(\\delta\\)A=0; native quotient seminorm.}\\par","The equivalence $K/R\\cong K\\cap R^\\perp$ in SR-26 is a linear isometry.\nFor compatible $v,w$,\n\\[\n\\nn{[v]}=\\nn{o(v)}=\\dist(v,R),\\qquad\n\\dist([v],[w])=\\dist(v-w,R).\n\\]\nThe quotient norm detects zero classes. Its standard quotient inner\nproduct is preserved when the induced projection structure in $K$ is used.","Let $h$ be the harmonic representative of a class $z$. For any compatible\nrepresentative $v$ of $z$, its harmonic component is $h$, and the\nnonexpansive harmonic projection gives $\\nn h\\le\\nn v$.\nTaking the infimum over representatives proves $\\nn h\\le\\nn z$.\nConversely $h$ itself represents $z$, so the quotient map's norm bound\ngives $\\nn z\\le\\nn h$. This proves isometry against the existing quotient\nseminorm. The compatible distance formula follows from SR-25, and\ndistances of classes follow by applying it to $v-w$.\nZero norm implies zero harmonic vector and hence zero class by the inverse\nlaws. Orthogonal projection onto $R$ restricts to $K$, constructing the\nquotient inner-product structure th","Let $h$ be the harmonic representative of a class $z$. For any compatible\nrepresentative $v$ of $z$, its harmonic component is $h$, and the\nnonexpansive harmonic projection gives $\\nn h\\le\\nn v$.\nTaking the infimum over representatives proves $\\nn h\\le\\nn z$.\nConversely $h$ itself represents $z$, so the quotient map's norm bound\ngives $\\nn z\\le\\nn h$. This proves isometry against the existing quotient\nseminorm. The compatible distance formula follows from SR-25, and\ndistances of classes follow by applying it to $v-w$.\nZero norm implies zero harmonic vector and hence zero class by the inverse\nlaws. Orthogonal projection onto $R$ restricts to $K$, constructing the\nquotient inner-product structure there. A real linear isometry preserves\nits inne","Let $h$ be the harmonic representative of a class $z$. For any compatible\nrepresentative $v$ of $z$, its harmonic component is $h$, and the\nnonexpansive harmonic projection gives $\\nn h\\le\\nn v$.\nTaking the infimum over representatives proves $\\nn h\\le\\nn z$.\nConversely $h$ itself represents $z$, so the quotient map's norm bound\ngives $\\nn z\\le\\nn h$. This proves isometry against the existing quotient\nseminorm. The compatible distance formula follows from SR-25, and\ndistances of classes follow by applying it to $v-w$.\nZero norm implies zero harmonic vector and hence zero class by the inverse\nlaws. Orthogonal projection onto $R$ restricts to $K$, constructing the\nquotient inner-product structure there. A real linear isometry preserves\nits inner product by the polarization identity.","onent is $h$, and the\nnonexpansive harmonic projection gives $\\nn h\\le\\nn v$.\nTaking the infimum over representatives proves $\\nn h\\le\\nn z$.\nConversely $h$ itself represents $z$, so the quotient map's norm bound\ngives $\\nn z\\le\\nn h$. This proves isometry against the existing quotient\nseminorm. The compatible distance formula follows from SR-25, and\ndistances of classes follow by applying it to $v-w$.\nZero norm implies zero harmonic vector and hence zero class by the inverse\nlaws. Orthogonal projection onto $R$ restricts to $K$, constructing the\nquotient inner-product structure there. A real linear isometry preserves\nits inner product by the polarization identity.","es proves $\\nn h\\le\\nn z$.\nConversely $h$ itself represents $z$, so the quotient map's norm bound\ngives $\\nn z\\le\\nn h$. This proves isometry against the existing quotient\nseminorm. The compatible distance formula follows from SR-25, and\ndistances of classes follow by applying it to $v-w$.\nZero norm implies zero harmonic vector and hence zero class by the inverse\nlaws. Orthogonal projection onto $R$ restricts to $K$, constructing the\nquotient inner-product structure there. A real linear isometry preserves\nits inner product by the polarization identity.","le\\nn z$.\nConversely $h$ itself represents $z$, so the quotient map's norm bound\ngives $\\nn z\\le\\nn h$. This proves isometry against the existing quotient\nseminorm. The compatible distance formula follows from SR-25, and\ndistances of classes follow by applying it to $v-w$.\nZero norm implies zero harmonic vector and hence zero class by the inverse\nlaws. Orthogonal projection onto $R$ restricts to $K$, constructing the\nquotient inner-product structure there. A real linear isometry preserves\nits inner product by the polarization identity.","Every class $z$ has a unique compatible minimum-norm representative $s(z)$,\nand every compatible representative $v$ satisfies\n\\[\n\\nn v^2=\\nn z^2+\\nn{v-s(z)}^2.\n\\]\nFor arbitrary data define $c(v)=[P_Kv]$. This is nonexpansive and\n\\[\n\\dist(v,R)^2=\\nn{c(v)}^2+\\nn{i(v)}^2.\n\\]\nThe datum is realizable exactly when both $c(v)=0$ and $\\delta v=0$.\nWhen the canonical inverse exists,\n\\[\n\\nn{v-Ag}^2=\\nn{c(v)}^2+\\nn{i(v)}^2+\\nn{A(g-Bv)}^2.\n\\]","Take $s(z)$ to be its harmonic representative from SR-33. The difference\nbetween any other compatible representative and $s(z)$ lies in $R$,\nwhich is orthogonal to the harmonic sector. Pythagoras gives the first\nformula, including unique attainment of the minimum.\nThe composition of $P_K$ and the quotient map is bounded by one; its\nharmonic representative is $o(v)$. SR-25 and SR-33 give the intrinsic\ndistance identity. If both displayed tests vanish, then $v\\in K$ and\nits obstruction class is zero, so SR-03 supplies a source. Realizability\nimplies both tests directly. Finally combine the candidate identity\nof SR-30 with the intrinsic distance split. The minimum representative\nof an obstruction class b","Take $s(z)$ to be its harmonic representative from SR-33. The difference\nbetween any other compatible representative and $s(z)$ lies in $R$,\nwhich is orthogonal to the harmonic sector. Pythagoras gives the first\nformula, including unique attainment of the minimum.\nThe composition of $P_K$ and the quotient map is bounded by one; its\nharmonic representative is $o(v)$. SR-25 and SR-33 give the intrinsic\ndistance identity. If both displayed tests vanish, then $v\\in K$ and\nits obstruction class is zero, so SR-03 supplies a source. Realizability\nimplies both tests directly. Finally combine the candidate identity\nof SR-30 with the intrinsic distance split. The minimum representative\nof an obstruction class belongs to local data space; it is a different\nconstruction from the minimum-norm global source $Bv$.","difference\nbetween any other compatible representative and $s(z)$ lies in $R$,\nwhich is orthogonal to the harmonic sector. Pythagoras gives the first\nformula, including unique attainment of the minimum.\nThe composition of $P_K$ and the quotient map is bounded by one; its\nharmonic representative is $o(v)$. SR-25 and SR-33 give the intrinsic\ndistance identity. If both displayed tests vanish, then $v\\in K$ and\nits obstruction class is zero, so SR-03 supplies a source. Realizability\nimplies both tests directly. Finally combine the candidate identity\nof SR-30 with the intrinsic distance split. The minimum representative\nof an obstruction class belongs to local data space; it is a different\nconstruction from the minimum-norm global source $Bv$.","stance identity. If both displayed tests vanish, then $v\\in K$ and\nits obstruction class is zero, so SR-03 supplies a source. Realizability\nimplies both tests directly. Finally combine the candidate identity\nof SR-30 with the intrinsic distance split. The minimum representative\nof an obstruction class belongs to local data space; it is a different\nconstruction from the minimum-norm global source $Bv$.","\\par{\\small\\textbf{Hypotheses.} Commuting linear presentation maps; whole local-map norm bound C\\(\\ge\\)0; reverse presentation with bound M>0 and induced left inverse for two-sided estimate.}\\par","A presentation map with $\\nn{Lv}\\le C\\nn v$ induces the same quotient map\nas in SR-03, now satisfying\n\\[\n\\nn{\\overline Lz}\\le C\\nn z.\n\\]\nIf a reverse presentation has bound $M>0$ and its quotient map is a left\ninverse of $\\overline L$, then\n\\[\n\\frac{\\nn z}{M}\\le\\nn{\\overline Lz}\\le C\\nn z.\n\\]\nThe corresponding class distances satisfy the same bounds.","For any representative $v$ of $z$, the quotient norm of $[Lv]$ is at most\n$\\nn{Lv}\\le C\\nn v$. Infimizing over representatives gives the upper bound\nand supplies a bounded linear map with exactly the algebraic action of\nSR-03. Alternatively, the minimum representative from SR-34 realizes the\ninfimum and yields it immediately. Apply the reverse bound to\n$\\overline Lz$: the left-inverse identity gives\n$\\nn z\\le M\\nn{\\overline Lz}$. Division by positive $M$ proves the lower\nbound. Apply the bounds to $z-w$ for distances. Forward boundedness alone\ndoes not prove reflection; the reverse identity is genuine additional data.","any representative $v$ of $z$, the quotient norm of $[Lv]$ is at most\n$\\nn{Lv}\\le C\\nn v$. Infimizing over representatives gives the upper bound\nand supplies a bounded linear map with exactly the algebraic action of\nSR-03. Alternatively, the minimum representative from SR-34 realizes the\ninfimum and yields it immediately. Apply the reverse bound to\n$\\overline Lz$: the left-inverse identity gives\n$\\nn z\\le M\\nn{\\overline Lz}$. Division by positive $M$ proves the lower\nbound. Apply the bounds to $z-w$ for distances. Forward boundedness alone\ndoes not prove reflection; the reverse identity is genuine additional data.","$[Lv]$ is at most\n$\\nn{Lv}\\le C\\nn v$. Infimizing over representatives gives the upper bound\nand supplies a bounded linear map with exactly the algebraic action of\nSR-03. Alternatively, the minimum representative from SR-34 realizes the\ninfimum and yields it immediately. Apply the reverse bound to\n$\\overline Lz$: the left-inverse identity gives\n$\\nn z\\le M\\nn{\\overline Lz}$. Division by positive $M$ proves the lower\nbound. Apply the bounds to $z-w$ for distances. Forward boundedness alone\ndoes not prove reflection; the reverse identity is genuine additional data.","g over representatives gives the upper bound\nand supplies a bounded linear map with exactly the algebraic action of\nSR-03. Alternatively, the minimum representative from SR-34 realizes the\ninfimum and yields it immediately. Apply the reverse bound to\n$\\overline Lz$: the left-inverse identity gives\n$\\nn z\\le M\\nn{\\overline Lz}$. Division by positive $M$ proves the lower\nbound. Apply the bounds to $z-w$ for distances. Forward boundedness alone\ndoes not prove reflection; the reverse identity is genuine additional data.","\\par{\\small\\textbf{Hypotheses.} Projected nested R0\\(\\le\\)R1\\(\\le\\)K in fixed real metric for energy; arbitrary submodule R in real normed space for nonclosed seminorm boundary.}\\par","Range enlargement induces a surjective contraction $F:K/R_0\\to K/R_1$.\nFor the old minimum representative $s_0(z)$,\n\\[\n\\nn z^2=\\nn{Fz}^2+\\nn{(P_{R_1}-P_{R_0})s_0(z)}^2.\n\\]\nWithout a closedness or projection hypothesis, a general quotient instead has\n\\[\n\\nn{[v]}=0\\iff v\\in\\cl R,\\qquad [v]=0\\iff v\\in R.\n\\]\nThus its seminorm detects zero exactly when $R$ is closed. Nonclosed range\ncan leave a nonzero class of zero seminorm.","The larger realization submodule contains the smaller, so the identity\non $K$ descends to a surjection of quotients. Every old representative is\na new representative, proving contraction by the quotient infimum.\nFor the energy identity apply the orthogonal refinement identity in\nSR-28 to $s_0(z)$, whose old norm is $\\nn z$. The new harmonic component\nrepresents $Fz$ and has norm $\\nn{Fz}$ by SR-33.","For any submodule, the native quotient seminorm equals distance to that\nsubmodule. Zero distance is equivalent to membership in its closure by\nthe metric closure criterion; algebraic quotient zero remains actual\nmembership. If $R$ is nonclosed, choose $v\\in\\cl R\\setminus R$.\nIts class is nonzero with zero seminorm. Conversely closedness makes\nzero distance imply membership. This is why the projection hypotheses\nin the preceding metric theory cannot be discarded silently.","\\begin{example}[A metric must survive its coordinate change]\nFor $K=\\{(x,y,0)\\}$ and $R=\\spanof\\{(1,0,0)\\}$, take the compatible basis\n$b_1=(1,0,0)$, $b_2=(1,2,0)$. Its Gram matrix is\n$\\bigl(\\begin{smallmatrix}1&1","\\begin{example}[A metric must survive its coordinate change]\nFor $K=\\{(x,y,0)\\}$ and $R=\\spanof\\{(1,0,0)\\}$, take the compatible basis\n$b_1=(1,0,0)$, $b_2=(1,2,0)$. Its Gram matrix is\n$\\bigl(\\begin{smallmatrix}1&1\\\\1&5\\end{smallmatrix}\\bigr)$.\nThe vector $(3,4,0)$ has coordinates $(1,2)$; its harmonic representative\n$(0,4,0)$ has coordinates $(-2,2)$ and squared norm $16$ in that Gram metric.\nComputing with an untransported coordinate Euclidean metric would change\nthe problem. Redundant or nonorthogonal coordinates do not license such a change.\n\\end{example}","1&5\\end{smallmatrix}\\bigr)$.\nThe vector $(3,4,0)$ has coordinates $(1,2)$; its harmonic representative\n$(0,4,0)$ has coordinates $(-2,2)$ and squared norm $16$ in that Gram metric.\nComputing with an untransported coordinate Euclidean metric would change\nthe problem. Redundant or nonorthogonal coordinates do not license such a change.\n\\end{example}","Fix a bounded operator $A:G\\to V$ between real Hilbert spaces, without\nassuming closed range. For $\\lambda>0$, put\n\\[\nN_\\lambda=A^*A+\\lambda I,\\qquad R_\\lambda=N_\\lambda^{-1},\\qquad\nB_\\lambda=R_\\lambda A^*.\n\\]\nThe inverse is constructed from coercivity. It is not supplied as a\ncertificate premise. Here $A^*$ is the actual Hilbert adjoint.","$N_\\lambda$ is a bounded bijection with bounded inverse $R_\\lambda$.\nFor $x_\\lambda=B_\\lambda v$,\n\\[\nN_\\lambda x_\\lambda=A^*v,\\qquad\nA^*(Ax_\\lambda-v)+\\lambda x_\\lambda=0,\n\\]\nand\n\\[\n\\nn{Ax_\\lambda}^2+\\lambda\\nn{x_\\lambda}^2\n=\\ip v{Ax_\\lambda}.\n\\]","The bilinear form\n$b(x,y)=\\ip{Ax}{Ay}+\\lambda\\ip xy$ is bounded and satisfies\n$b(x,x)\\ge\\lambda\\nn x^2$. Lax--Milgram gives a continuous linear\nequivalence from the source to itself under the Riesz identification.\nThe adjoint identity identifies its map with $A^*A+\\lambda I$.\nThe two inverse laws give an actual resolvent. Applying one inverse law\nto $A^*v$ yields the normal equation. Taking its inner product with\n$x_\\lambda$ and using $\\ip{A^*v}{x_\\lambda}=\\ip v{Ax_\\lambda}$ gives\nthe energy identity. Strictly positive $\\lambda$ supplies coercivity\neven if $A$ has a nontrivial kernel or a nonclosed range.","$b(x,y)=\\ip{Ax}{Ay}+\\lambda\\ip xy$ is bounded and satisfies\n$b(x,x)\\ge\\lambda\\nn x^2$. Lax--Milgram gives a continuous linear\nequivalence from the source to itself under the Riesz identification.\nThe adjoint identity identifies its map with $A^*A+\\lambda I$.\nThe two inverse laws give an actual resolvent. Applying one inverse law\nto $A^*v$ yields the normal equation. Taking its inner product with\n$x_\\lambda$ and using $\\ip{A^*v}{x_\\lambda}=\\ip v{Ax_\\lambda}$ gives\nthe energy identity. Strictly positive $\\lambda$ supplies coercivity\neven if $A$ has a nontrivial kernel or a nonclosed range.","ambda\\ip xy$ is bounded and satisfies\n$b(x,x)\\ge\\lambda\\nn x^2$. Lax--Milgram gives a continuous linear\nequivalence from the source to itself under the Riesz identification.\nThe adjoint identity identifies its map with $A^*A+\\lambda I$.\nThe two inverse laws give an actual resolvent. Applying one inverse law\nto $A^*v$ yields the normal equation. Taking its inner product with\n$x_\\lambda$ and using $\\ip{A^*v}{x_\\lambda}=\\ip v{Ax_\\lambda}$ gives\nthe energy identity. Strictly positive $\\lambda$ supplies coercivity\neven if $A$ has a nontrivial kernel or a nonclosed range.","continuous linear\nequivalence from the source to itself under the Riesz identification.\nThe adjoint identity identifies its map with $A^*A+\\lambda I$.\nThe two inverse laws give an actual resolvent. Applying one inverse law\nto $A^*v$ yields the normal equation. Taking its inner product with\n$x_\\lambda$ and using $\\ip{A^*v}{x_\\lambda}=\\ip v{Ax_\\lambda}$ gives\nthe energy identity. Strictly positive $\\lambda$ supplies coercivity\neven if $A$ has a nontrivial kernel or a nonclosed range.","source to itself under the Riesz identification.\nThe adjoint identity identifies its map with $A^*A+\\lambda I$.\nThe two inverse laws give an actual resolvent. Applying one inverse law\nto $A^*v$ yields the normal equation. Taking its inner product with\n$x_\\lambda$ and using $\\ip{A^*v}{x_\\lambda}=\\ip v{Ax_\\lambda}$ gives\nthe energy identity. Strictly positive $\\lambda$ supplies coercivity\neven if $A$ has a nontrivial kernel or a nonclosed range.","For $F_\\lambda(v,x)=\\nn{Ax-v}^2+\\lambda\\nn x^2$,\n\\[\nF_\\lambda(v,x)=F_\\lambda(v,x_\\lambda)\n+\\nn{A(x-x_\\lambda)}^2+\\lambda\\nn{x-x_\\lambda}^2.\n\\]\nThus $x_\\lambda$ is the unique minimizer and lies in $(\\ker A)^\\perp$.\nThe whole-map bounds are\n\\[\n\\nn{R_\\lambda}\\le1/\\lambda,\\qquad\n\\nn{B_\\lambda}\\le\\frac1{2\\sqrt\\lambda},\\qquad\n\\nn{AB_\\lambda v}\\le\\nn v,\\quad\\nn{AB_\\lambda v-v}\\le\\nn v.\n\\]\nFor $r=N_\\lambda x-A^*v$, one has\n$x-x_\\lambda=R_\\lambda r$ and $\\nn{x-x_\\lambda}\\le\\nn r/\\lambda$.","\\nn{Ax-v}^2+\\lambda\\nn x^2$,\n\\[\nF_\\lambda(v,x)=F_\\lambda(v,x_\\lambda)\n+\\nn{A(x-x_\\lambda)}^2+\\lambda\\nn{x-x_\\lambda}^2.\n\\]\nThus $x_\\lambda$ is the unique minimizer and lies in $(\\ker A)^\\perp$.\nThe whole-map bounds are\n\\[\n\\nn{R_\\lambda}\\le1/\\lambda,\\qquad\n\\nn{B_\\lambda}\\le\\frac1{2\\sqrt\\lambda},\\qquad\n\\nn{AB_\\lambda v}\\le\\nn v,\\quad\\nn{AB_\\lambda v-v}\\le\\nn v.\n\\]\nFor $r=N_\\lambda x-A^*v$, one has\n$x-x_\\lambda=R_\\lambda r$ and $\\nn{x-x_\\lambda}\\le\\nn r/\\lambda$.","[\nF_\\lambda(v,x)=F_\\lambda(v,x_\\lambda)\n+\\nn{A(x-x_\\lambda)}^2+\\lambda\\nn{x-x_\\lambda}^2.\n\\]\nThus $x_\\lambda$ is the unique minimizer and lies in $(\\ker A)^\\perp$.\nThe whole-map bounds are\n\\[\n\\nn{R_\\lambda}\\le1/\\lambda,\\qquad\n\\nn{B_\\lambda}\\le\\frac1{2\\sqrt\\lambda},\\qquad\n\\nn{AB_\\lambda v}\\le\\nn v,\\quad\\nn{AB_\\lambda v-v}\\le\\nn v.\n\\]\nFor $r=N_\\lambda x-A^*v$, one has\n$x-x_\\lambda=R_\\lambda r$ and $\\nn{x-x_\\lambda}\\le\\nn r/\\lambda$.","Expand both squared norms around $x_\\lambda$. Their cross terms combine to\n\\[\n2\\ip{A^*(Ax_\\lambda-v)+\\lambda x_\\lambda}{x-x_\\lambda},\n\\]\nwhich vanishes by SR-37. The remaining positive terms prove the increment\nidentity and uniqueness. Taking the normal equation against a kernel\nvector $k$ yields $\\lambda\\ip{x_\\lambda}k=0$, proving source orthogonality.","For $y=R_\\lambda r$, its energy gives\n$\\lambda\\nn y^2\\le\\ip ry\\le\\nn r\\nn y$, hence the resolvent bound.\nFor $x_\\lambda$, set $u=\\nn{Ax_\\lambda}$. The energy gives\n$u^2+\\lambda\\nn{x_\\lambda}^2\\le\\nn v\\,u$.\nCompleting the square in $u$ yields\n$4\\lambda\\nn{x_\\lambda}^2\\le\\nn v^2$ and also $u\\le\\nn v$.\nComparing the minimum objective with $F_\\lambda(v,0)=\\nn v^2$ bounds the\nfitting residual. Subtract the exact normal equation from the computed\none and apply the resolvent for the correction identity.","Write $p=P_{(\\ker A)^\\perp}g$, $k=P_{\\ker A}g$. Then\n\\[\nB_\\lambda Ag+\\lambda R_\\lambda g=g,\n\\quad B_\\lambda(Ag+e)-g=-\\lambda R_\\lambda g+B_\\lambda e,\n\\]\nand\n\\[\n\\nn{B_\\lambda(Ag+e)-g}^2\n=\\nn{-\\lambda R_\\lambda p+B_\\lambda e}^2+\\nn k^2.\n\\]\nIf $A$ has source coercivity $\\alpha>0$, this is at most\n\\[\n\\left(\\frac\\lambda{\\alpha^2+\\lambda}\\nn p\n+\\frac{\\nn e}{2\\sqrt\\lambda}\\right)^2+\\nn k^2.\n\\]\nFor a supplied source condition $g=A^*Aw$,\n$\\nn{B_\\lambda Ag-g}\\le\\lambda\\nn w$ without requiring closed range.","erp}g$, $k=P_{\\ker A}g$. Then\n\\[\nB_\\lambda Ag+\\lambda R_\\lambda g=g,\n\\quad B_\\lambda(Ag+e)-g=-\\lambda R_\\lambda g+B_\\lambda e,\n\\]\nand\n\\[\n\\nn{B_\\lambda(Ag+e)-g}^2\n=\\nn{-\\lambda R_\\lambda p+B_\\lambda e}^2+\\nn k^2.\n\\]\nIf $A$ has source coercivity $\\alpha>0$, this is at most\n\\[\n\\left(\\frac\\lambda{\\alpha^2+\\lambda}\\nn p\n+\\frac{\\nn e}{2\\sqrt\\lambda}\\right)^2+\\nn k^2.\n\\]\nFor a supplied source condition $g=A^*Aw$,\n$\\nn{B_\\lambda Ag-g}\\le\\lambda\\nn w$ without requiring closed range.","P_{\\ker A}g$. Then\n\\[\nB_\\lambda Ag+\\lambda R_\\lambda g=g,\n\\quad B_\\lambda(Ag+e)-g=-\\lambda R_\\lambda g+B_\\lambda e,\n\\]\nand\n\\[\n\\nn{B_\\lambda(Ag+e)-g}^2\n=\\nn{-\\lambda R_\\lambda p+B_\\lambda e}^2+\\nn k^2.\n\\]\nIf $A$ has source coercivity $\\alpha>0$, this is at most\n\\[\n\\left(\\frac\\lambda{\\alpha^2+\\lambda}\\nn p\n+\\frac{\\nn e}{2\\sqrt\\lambda}\\right)^2+\\nn k^2.\n\\]\nFor a supplied source condition $g=A^*Aw$,\n$\\nn{B_\\lambda Ag-g}\\le\\lambda\\nn w$ without requiring closed range.","The inverse identity $R_\\lambda(A^*A+\\lambda I)=I$ gives the filter\ndecomposition. Linearity adds the propagated noise. On $\\ker A$,\n$N_\\lambda$ acts as $\\lambda I$, so $\\lambda R_\\lambda k=k$.\nThe resolvent preserves the orthogonal complement: pairing its normal\nequation with every kernel vector proves this directly. Thus the combined\nidentifiable error is orthogonal to $k$, giving the exact squared formula.\nBias and noise inside that first term need not be orthogonal to each other.","For $y=R_\\lambda p\\in(\\ker A)^\\perp$, source coercivity and its energy\ngive $(\\alpha^2+\\lambda)\\nn y^2\\le\\nn p\\nn y$.\nMultiply by $\\lambda$ and combine with the noise bound in SR-38.\nFor $g=A^*Aw$, the normal-range source filter is a contraction, so\n$\\nn{R_\\lambda A^*Aw}\\le\\nn w$. Multiplication by $\\lambda$ proves\nthe source-condition bias rate. The representation by $w$ is supplied;\nno controlled $w$ is asserted for every identifiable source.","$y=R_\\lambda p\\in(\\ker A)^\\perp$, source coercivity and its energy\ngive $(\\alpha^2+\\lambda)\\nn y^2\\le\\nn p\\nn y$.\nMultiply by $\\lambda$ and combine with the noise bound in SR-38.\nFor $g=A^*Aw$, the normal-range source filter is a contraction, so\n$\\nn{R_\\lambda A^*Aw}\\le\\nn w$. Multiplication by $\\lambda$ proves\nthe source-condition bias rate. The representation by $w$ is supplied;\nno controlled $w$ is asserted for every identifiable source.","For two operators and parameters,\n\\[\nR_{A,\\lambda}-R_{C,\\mu}\n=R_{A,\\lambda}(N_{C,\\mu}-N_{A,\\lambda})R_{C,\\mu}.\n\\]\nAt fixed positive $\\lambda$, with $\\epsilon=\\nn{A-C}$,\n\\[\n\\nn{B_{A,\\lambda}-B_{C,\\lambda}}\n\\le\\frac\\epsilon\\lambda+\n\\frac{(\\nn A+\\nn C)\\epsilon\\nn C}{\\lambda^2}.\n\\]\nIf $A$ has closed range and canonical inverse $B$, then\n$B_\\lambda v-Bv=-\\lambda R_\\lambda Bv$. For $\\delta A=0$,\n\\[\n\\nn{v-AB_\\lambda v}^2\n=\\nn{c(v)}^2+\\nn{i(v)}^2+\\nn{A(B_\\lambda v-Bv)}^2.\n\\]","Multiply the difference of inverses on the left and right by their\nnormal operators, or insert their inverse laws, to obtain the resolvent\nidentity. SR-38 bounds its norm by\n$\\nn{N_{C,\\mu}-N_{A,\\lambda}}/(\\lambda\\mu)$.\nAt equal parameters,\n$C^*C-A^*A=(C^*-A^*)C+A^*(C-A)$, whose norm is at most\n$(\\nn A+\\nn C)\\epsilon$. Splitting\n$R_AA^*-R_CC^*=R_A(A^*-C^*)+(R_A-R_C)C^*$ proves the stated bound.\nIt permits rank changes and is not uniform as $\\lambda$ vanishes.","eft and right by their\nnormal operators, or insert their inverse laws, to obtain the resolvent\nidentity. SR-38 bounds its norm by\n$\\nn{N_{C,\\mu}-N_{A,\\lambda}}/(\\lambda\\mu)$.\nAt equal parameters,\n$C^*C-A^*A=(C^*-A^*)C+A^*(C-A)$, whose norm is at most\n$(\\nn A+\\nn C)\\epsilon$. Splitting\n$R_AA^*-R_CC^*=R_A(A^*-C^*)+(R_A-R_C)C^*$ proves the stated bound.\nIt permits rank changes and is not uniform as $\\lambda$ vanishes.","Under closed range, $A^*$ ignores the part of data orthogonal to\n$\\Ran A$, so $B_\\lambda v=B_\\lambda ABv$. Apply the filter identity\nto $Bv$ for its bias. The intrinsic candidate identity SR-34 with\n$g=B_\\lambda v$ yields the local error split. Regularization introduces\nimage bias while retaining the same obstruction and inconsistency.","\\begin{example}[Exact scalar response and a two-coordinate example]\nFor $A_cx=cx$, $B_\\lambda v=cv/(c^2+\\lambda)$.\nAt $c>0$, $\\lambda=c^2$, the value is $v/(2c)$, attaining the noise bound.\nFor $A(g_1,g_2)=(g_1+g_2,0,0)$, $v=(3,4,5)$, and $\\lambda=1$,\nthe source is $(1,1)$. Its squared fit is $42=16+25+1$, separating\nobstruction, inconsistency, and regularization image bias. Its objective\nis $44$, whereas the unregularized source $(3/2,3/2)$ has penalized\nobjective $91/2$. Positive regularization solves its own stated objective.\n\\end{example}","se and a two-coordinate example]\nFor $A_cx=cx$, $B_\\lambda v=cv/(c^2+\\lambda)$.\nAt $c>0$, $\\lambda=c^2$, the value is $v/(2c)$, attaining the noise bound.\nFor $A(g_1,g_2)=(g_1+g_2,0,0)$, $v=(3,4,5)$, and $\\lambda=1$,\nthe source is $(1,1)$. Its squared fit is $42=16+25+1$, separating\nobstruction, inconsistency, and regularization image bias. Its objective\nis $44$, whereas the unregularized source $(3/2,3/2)$ has penalized\nobjective $91/2$. Positive regularization solves its own stated objective.\n\\end{example}","$A_cx=cx$, $B_\\lambda v=cv/(c^2+\\lambda)$.\nAt $c>0$, $\\lambda=c^2$, the value is $v/(2c)$, attaining the noise bound.\nFor $A(g_1,g_2)=(g_1+g_2,0,0)$, $v=(3,4,5)$, and $\\lambda=1$,\nthe source is $(1,1)$. Its squared fit is $42=16+25+1$, separating\nobstruction, inconsistency, and regularization image bias. Its objective\nis $44$, whereas the unregularized source $(3/2,3/2)$ has penalized\nobjective $91/2$. Positive regularization solves its own stated objective.\n\\end{example}","The following limits fix one bounded $A$ between real Hilbert spaces.\nStatements can be indexed by an arbitrary filter. When uniqueness of\nlimits or a converse is used, the filter is required to be nontrivial.\nClosed range is unnecessary for strong convergence on each fixed source.","\\par{\\small\\textbf{Hypotheses.} Fixed bounded real Hilbert A; positive lambda\\_i\\(\\to\\)0 along a filter; arbitrary fixed g; nontrivial filter for exact-limit iff identifiability.}\\par","One has $\\cl\\Ran(A^*A)=(\\ker A)^\\perp$, and for every $g,w$,\n\\[\n\\nn{B_\\lambda Ag-g}\\le\\nn{g-A^*Aw}+\\lambda\\nn w.\n\\]\nAs $\\lambda\\downarrow0$,\n\\[\nB_\\lambda Ag\\longrightarrow P_{(\\ker A)^\\perp}g,\n\\qquad \\nn{B_\\lambda Ag-g}\\longrightarrow\\nn{P_{\\ker A}g}.\n\\]\nFor a nontrivial parameter filter, convergence to $g$ itself holds exactly\nwhen $g\\in(\\ker A)^\\perp$.","If $x$ is orthogonal to $\\Ran(A^*A)$, pairing with $A^*Ax$ gives\n$\\nn{Ax}^2=0$. Conversely a kernel vector is orthogonal to every\n$A^*Aw$ by the adjoint identity. Thus\n$(\\Ran(A^*A))^\\perp=\\ker A$, and taking orthogonal complements gives\nthe closure identity. Decompose the bias into the bias of\n$g-A^*Aw$ and of $A^*Aw$. The first is bounded by its input norm because\n$\\lambda R_\\lambda$ is contractive; the second is bounded by\n$\\lambda\\nn w$ from SR-39.","A^*A)$, pairing with $A^*Ax$ gives\n$\\nn{Ax}^2=0$. Conversely a kernel vector is orthogonal to every\n$A^*Aw$ by the adjoint identity. Thus\n$(\\Ran(A^*A))^\\perp=\\ker A$, and taking orthogonal complements gives\nthe closure identity. Decompose the bias into the bias of\n$g-A^*Aw$ and of $A^*Aw$. The first is bounded by its input norm because\n$\\lambda R_\\lambda$ is contractive; the second is bounded by\n$\\lambda\\nn w$ from SR-39.","For identifiable $p$, density lets us choose $w$ making the first term\narbitrarily small, independently of $\\lambda$. Then let $\\lambda$ tend\nto zero to remove the second. For arbitrary $g$, $Ag=Ap$ with\n$p=P_{(\\ker A)^\\perp}g$, yielding the claimed source limit.\nSubtracting $g$ and taking norms gives the kernel-error limit.\nUniqueness of limits on a nontrivial filter gives the final equivalence.","\\par{\\small\\textbf{Hypotheses.} Fixed real Hilbert bounded A; lambda\\_i>0\\(\\to\\)0; deterministic noise with \\(\\Vert\\)e\\_i\\(\\Vert\\)/\\allowbreak sqrt(lambda\\_i)\\(\\to\\)0; actual normal residual scale vanishes for computed sources.}\\par","If $\\lambda_i>0$ tends to zero and\n$\\nn{e_i}/\\sqrt{\\lambda_i}\\to0$, then\n\\[\nB_{\\lambda_i}(Ag+e_i)\\longrightarrow P_{(\\ker A)^\\perp}g.\n\\]\nThe same limit holds for computed $x_i$ when additionally\n\\[\n\\frac{\\nn{(A^*A+\\lambda_iI)x_i-A^*(Ag+e_i)}}{\\lambda_i}\\to0.\n\\]\nFor example, $\\lambda_i=s_i^2$, $s_i>0\\to0$, and $\\nn{e_i}\\le s_i^2$\ngive the exact-solve consistency condition.","The source difference between noisy and exact-data reconstruction is\n$B_{\\lambda_i}e_i$, whose norm is at most\n$\\nn{e_i}/(2\\sqrt{\\lambda_i})$ by SR-38. Combine its vanishing with\nthe exact-source limit in SR-41. For computed sources, the normal-residual\ncorrection in SR-38 bounds their difference from the actual regularized\nsource by precisely the second displayed scale. The triangle inequality\nthen proves the same limit. In the square schedule, the noise ratio is\nat most $s_i$, hence vanishes. These schedules are sufficient conditions;\nthey are not asserted necessary for every favorable operator or noise direction.","da_i}e_i$, whose norm is at most\n$\\nn{e_i}/(2\\sqrt{\\lambda_i})$ by SR-38. Combine its vanishing with\nthe exact-source limit in SR-41. For computed sources, the normal-residual\ncorrection in SR-38 bounds their difference from the actual regularized\nsource by precisely the second displayed scale. The triangle inequality\nthen proves the same limit. In the square schedule, the noise ratio is\nat most $s_i$, hence vanishes. These schedules are sufficient conditions;\nthey are not asserted necessary for every favorable operator or noise direction.","\\par{\\small\\textbf{Hypotheses.} Fixed real Hilbert bounded A; positive parameter family\\(\\to\\)0; nontrivial filter for iff; source conditioning alpha only for positive estimate.}\\par","Let $E_\\lambda=B_\\lambda A-P_{(\\ker A)^\\perp}$. Always $\\nn{E_\\lambda}\\le1$.\nIf $\\Ran A$ is nonclosed, then $\\nn{E_\\lambda}=1$ for every positive parameter.\nFor a positive family tending to zero along a nontrivial filter,\n\\[\nE_\\lambda\\to0\\text{ in operator norm}\\iff\\Ran A\\text{ is closed}.\n\\]\nUnder source coercivity $\\alpha>0$,\n$\\nn{E_\\lambda}\\le\\lambda/(\\alpha^2+\\lambda)$.","The projected bias bound from SR-39 gives the universal upper bound one.\nSuppose any bounded $B'$ satisfies\n$c=\\nn{B'A-P_S}<1$, where $S=(\\ker A)^\\perp$.\nFor $x\\in S$,\n\\[\n\\nn x\\le\\nn{x-B'Ax}+\\nn{B'Ax}\n\\le c\\nn x+\\nn{B'}\\nn{Ax}.\n\\]\nTherefore $\\alpha=(1-c)/(\\nn{B'}+1)>0$ supplies off-kernel coercivity,\nforcing closed range by SR-32. A nonclosed range thus cannot have\n$\\nn{E_\\lambda}<1$, proving equality. For closed range, source coercivity\nexists and the conditioned bias bound yields the displayed quantity tending\nto zero. Conversely convergence to zero contradicts the constant norm one\non any nontrivial filter. Strong convergence in SR-41 remains valid in\nthe nonclosed case because it fixes a source before taking the limit.","\\par{\\small\\textbf{Hypotheses.} Fixed real Hilbert bounded A; positive lambda\\_i\\(\\to\\)0; data noise\\(\\to\\)0 for fitted limit; nontrivial filter for impossible source limit.}\\par","For arbitrary data $v$,\n\\[\nAB_\\lambda^Av=B_\\lambda^{A^*}A^*v\n\\longrightarrow P_{\\cl\\Ran A}v.\n\\]\nAdding data noise $e_i\\to0$ preserves this fitted-data limit.\nIf $v\\in\\cl\\Ran A\\setminus\\Ran A$, fitted data converge to $v$ but the\ncorresponding reconstructed source family has no norm limit in $G$.","The identity\n$(AA^*+\\lambda I)A=A(A^*A+\\lambda I)$ and uniqueness of the positive\nnormal inverse prove the dual formula. Apply SR-41 to the bounded\noperator $A^*$ with source $v$. Its identifiable space is\n$(\\ker A^*)^\\perp=\\cl\\Ran A$, yielding the fitted limit.\nSR-38 shows $AB_\\lambda$ is contractive, so changing data by $e_i$\nchanges the fitted value by at most $\\nn{e_i}$.\nIf reconstructed sources converged in norm to some $g$, continuity of\n$A$ would make their images converge to $Ag$. On a nontrivial filter,\nthe already established image limit $v$ would force $Ag=v$, contrary\nto $v\\notin\\Ran A$. This proves absence of a source norm limit without\nadding an unproved assertion that source norms diverge.","A=A(A^*A+\\lambda I)$ and uniqueness of the positive\nnormal inverse prove the dual formula. Apply SR-41 to the bounded\noperator $A^*$ with source $v$. Its identifiable space is\n$(\\ker A^*)^\\perp=\\cl\\Ran A$, yielding the fitted limit.\nSR-38 shows $AB_\\lambda$ is contractive, so changing data by $e_i$\nchanges the fitted value by at most $\\nn{e_i}$.\nIf reconstructed sources converged in norm to some $g$, continuity of\n$A$ would make their images converge to $Ag$. On a nontrivial filter,\nthe already established image limit $v$ would force $Ag=v$, contrary\nto $v\\notin\\Ran A$. This proves absence of a source norm limit without\nadding an unproved assertion that source norms diverge.","$ and uniqueness of the positive\nnormal inverse prove the dual formula. Apply SR-41 to the bounded\noperator $A^*$ with source $v$. Its identifiable space is\n$(\\ker A^*)^\\perp=\\cl\\Ran A$, yielding the fitted limit.\nSR-38 shows $AB_\\lambda$ is contractive, so changing data by $e_i$\nchanges the fitted value by at most $\\nn{e_i}$.\nIf reconstructed sources converged in norm to some $g$, continuity of\n$A$ would make their images converge to $Ag$. On a nontrivial filter,\nthe already established image limit $v$ would force $Ag=v$, contrary\nto $v\\notin\\Ran A$. This proves absence of a source norm limit without\nadding an unproved assertion that source norms diverge.","inverse prove the dual formula. Apply SR-41 to the bounded\noperator $A^*$ with source $v$. Its identifiable space is\n$(\\ker A^*)^\\perp=\\cl\\Ran A$, yielding the fitted limit.\nSR-38 shows $AB_\\lambda$ is contractive, so changing data by $e_i$\nchanges the fitted value by at most $\\nn{e_i}$.\nIf reconstructed sources converged in norm to some $g$, continuity of\n$A$ would make their images converge to $Ag$. On a nontrivial filter,\nthe already established image limit $v$ would force $Ag=v$, contrary\nto $v\\notin\\Ran A$. This proves absence of a source norm limit without\nadding an unproved assertion that source norms diverge.","r $A^*$ with source $v$. Its identifiable space is\n$(\\ker A^*)^\\perp=\\cl\\Ran A$, yielding the fitted limit.\nSR-38 shows $AB_\\lambda$ is contractive, so changing data by $e_i$\nchanges the fitted value by at most $\\nn{e_i}$.\nIf reconstructed sources converged in norm to some $g$, continuity of\n$A$ would make their images converge to $Ag$. On a nontrivial filter,\nthe already established image limit $v$ would force $Ag=v$, contrary\nto $v\\notin\\Ran A$. This proves absence of a source norm limit without\nadding an unproved assertion that source norms diverge.","rce $v$. Its identifiable space is\n$(\\ker A^*)^\\perp=\\cl\\Ran A$, yielding the fitted limit.\nSR-38 shows $AB_\\lambda$ is contractive, so changing data by $e_i$\nchanges the fitted value by at most $\\nn{e_i}$.\nIf reconstructed sources converged in norm to some $g$, continuity of\n$A$ would make their images converge to $Ag$. On a nontrivial filter,\nthe already established image limit $v$ would force $Ag=v$, contrary\nto $v\\notin\\Ran A$. This proves absence of a source norm limit without\nadding an unproved assertion that source norms diverge.","iable space is\n$(\\ker A^*)^\\perp=\\cl\\Ran A$, yielding the fitted limit.\nSR-38 shows $AB_\\lambda$ is contractive, so changing data by $e_i$\nchanges the fitted value by at most $\\nn{e_i}$.\nIf reconstructed sources converged in norm to some $g$, continuity of\n$A$ would make their images converge to $Ag$. On a nontrivial filter,\nthe already established image limit $v$ would force $Ag=v$, contrary\nto $v\\notin\\Ran A$. This proves absence of a source norm limit without\nadding an unproved assertion that source norms diverge.","Still fix a known bounded $A$ between real Hilbert spaces. Define the actual\nresidual curve on positive parameters\n$r_v(\\lambda)=\\nn{AB_\\lambda v-v}$. Its parameter is mathematical data,\nnot an assumed noise estimate. Let $\\eta$ be a justified true-data allowance\nand $\\tau>1$ a band multiplier. A total rule must include a small-data branch.","\\par{\\small\\textbf{Hypotheses.} Real Hilbert bounded A; lambda>0; strict floor dist(v,range A)<rho<\\(\\Vert\\)v\\(\\Vert\\) for existence; actual positive parameter only.}\\par","The residual is continuous and nondecreasing on positive parameters;\nthe source norm is nonincreasing. It is strictly increasing when $A^*v\\ne0$,\nand constant at $\\nn v$ when $A^*v=0$.\nFor\n\\[\n\\dist(v,\\Ran A)<\\rho<\\nn v,\n\\]\nthere is exactly one $\\lambda>0$ with $r_v(\\lambda)=\\rho$.\nThe resulting exact real parameter is a noncomputable choice with a proved\nspecification.","Compare the minimum objectives at $0<\\lambda<\\mu$ using each minimizer\nas a competitor for the other. Adding the two inequalities gives\n$(\\mu-\\lambda)(\\nn{x_\\mu}^2-\\nn{x_\\lambda}^2)\\le0$, so source norms\nare antitone. Substitution back into the smaller-parameter comparison\nshows residual monotonicity. If the two residuals were equal, both\nnorms and objective values would be equal, and unique minimization would\ngive $x_\\lambda=x_\\mu$. Subtracting their normal equations gives\n$(\\mu-\\lambda)x_\\lambda=0$, hence $x_\\lambda=0$ and $A^*v=0$.\nThis proves strictness and the constant case.","objectives at $0<\\lambda<\\mu$ using each minimizer\nas a competitor for the other. Adding the two inequalities gives\n$(\\mu-\\lambda)(\\nn{x_\\mu}^2-\\nn{x_\\lambda}^2)\\le0$, so source norms\nare antitone. Substitution back into the smaller-parameter comparison\nshows residual monotonicity. If the two residuals were equal, both\nnorms and objective values would be equal, and unique minimization would\ngive $x_\\lambda=x_\\mu$. Subtracting their normal equations gives\n$(\\mu-\\lambda)x_\\lambda=0$, hence $x_\\lambda=0$ and $A^*v=0$.\nThis proves strictness and the constant case.","etitor for the other. Adding the two inequalities gives\n$(\\mu-\\lambda)(\\nn{x_\\mu}^2-\\nn{x_\\lambda}^2)\\le0$, so source norms\nare antitone. Substitution back into the smaller-parameter comparison\nshows residual monotonicity. If the two residuals were equal, both\nnorms and objective values would be equal, and unique minimization would\ngive $x_\\lambda=x_\\mu$. Subtracting their normal equations gives\n$(\\mu-\\lambda)x_\\lambda=0$, hence $x_\\lambda=0$ and $A^*v=0$.\nThis proves strictness and the constant case.","wo inequalities gives\n$(\\mu-\\lambda)(\\nn{x_\\mu}^2-\\nn{x_\\lambda}^2)\\le0$, so source norms\nare antitone. Substitution back into the smaller-parameter comparison\nshows residual monotonicity. If the two residuals were equal, both\nnorms and objective values would be equal, and unique minimization would\ngive $x_\\lambda=x_\\mu$. Subtracting their normal equations gives\n$(\\mu-\\lambda)x_\\lambda=0$, hence $x_\\lambda=0$ and $A^*v=0$.\nThis proves strictness and the constant case.","ves\n$(\\mu-\\lambda)(\\nn{x_\\mu}^2-\\nn{x_\\lambda}^2)\\le0$, so source norms\nare antitone. Substitution back into the smaller-parameter comparison\nshows residual monotonicity. If the two residuals were equal, both\nnorms and objective values would be equal, and unique minimization would\ngive $x_\\lambda=x_\\mu$. Subtracting their normal equations gives\n$(\\mu-\\lambda)x_\\lambda=0$, hence $x_\\lambda=0$ and $A^*v=0$.\nThis proves strictness and the constant case.","The resolvent parameter identity gives, for $s,t\\ge a>0$,\n\\[\n|r_v(s)-r_v(t)|\\le\n\\frac{\\nn A\\nn{A^*v}}{a^2}|s-t|,\n\\]\nproving continuity on positive compact intervals. The strict floor permits\na comparator $g$ with $\\nn{Ag-v}<\\rho$; objective minimization at sufficiently\nsmall positive $\\lambda$ gives a lower residual bracket. For large\n$\\lambda$, $\\nn{B_\\lambda v}\\le\\nn{A^*v}/\\lambda$, so the residual\napproaches $\\nn v>\\rho$ and gives an upper bracket. The intermediate\nvalue theorem provides a root; strictness gives uniqueness. The floor\ncondition itself excludes the constant orthogonal-data case.","rameter identity gives, for $s,t\\ge a>0$,\n\\[\n|r_v(s)-r_v(t)|\\le\n\\frac{\\nn A\\nn{A^*v}}{a^2}|s-t|,\n\\]\nproving continuity on positive compact intervals. The strict floor permits\na comparator $g$ with $\\nn{Ag-v}<\\rho$; objective minimization at sufficiently\nsmall positive $\\lambda$ gives a lower residual bracket. For large\n$\\lambda$, $\\nn{B_\\lambda v}\\le\\nn{A^*v}/\\lambda$, so the residual\napproaches $\\nn v>\\rho$ and gives an upper bracket. The intermediate\nvalue theorem provides a root; strictness gives uniqueness. The floor\ncondition itself excludes the constant orthogonal-data case.","\\par{\\small\\textbf{Hypotheses.} Fixed bounded real Hilbert A; p\\(\\in\\)(ker A)\\(\\perp\\); family norm bounds \\(\\Vert\\)x\\_i\\(\\Vert\\le\\Vert\\)p\\(\\Vert\\) and true residual \\(\\Vert\\)Ax\\_i-Ap\\(\\Vert\\to\\)0.}\\par","If $\\nn{Ag-v}\\le r_v(\\lambda)$, then $\\nn{B_\\lambda v}\\le\\nn g$.\nFor any $x,p,w$ with $\\nn x\\le\\nn p$,\n\\[\n\\nn{x-p}^2\\le4\\nn p\\nn{p-A^*w}+2\\nn w\\nn{Ax-Ap}.\n\\]\nConsequently, for identifiable fixed $p$, the conditions\n$\\nn{x_i}\\le\\nn p$ and $\\nn{Ax_i-Ap}\\to0$ imply $x_i\\to p$ strongly,\nwithout assuming closed range.","Objective minimality gives\n$r_v(\\lambda)^2+\\lambda\\nn{B_\\lambda v}^2\n\\le\\nn{Ag-v}^2+\\lambda\\nn g^2$. The residual ordering and positive\n$\\lambda$ imply norm domination.\nFor the source estimate, expand $\\nn{x-p}^2$ and use $\\nn x\\le\\nn p$\nto obtain $\\nn{x-p}^2\\le2\\ip p{p-x}$.\nWrite $p=(p-A^*w)+A^*w$. The first pairing is at most\n$\\nn{p-A^*w}\\nn{p-x}\\le2\\nn p\\nn{p-A^*w}$; the adjoint identity bounds\nthe second by $\\nn w\\nn{Ap-Ax}$. This yields the displayed estimate.\nFor identifiable $p$, the density in SR-41 permits $w$ with\n$p-A^*w$ arbitrarily small (take $w=A u$ from the normal range).\nFixing this $w$, the residual term tends to zero. An epsilon argument\nthen proves norm convergence. No common inverse bound is invoked.","lity gives\n$r_v(\\lambda)^2+\\lambda\\nn{B_\\lambda v}^2\n\\le\\nn{Ag-v}^2+\\lambda\\nn g^2$. The residual ordering and positive\n$\\lambda$ imply norm domination.\nFor the source estimate, expand $\\nn{x-p}^2$ and use $\\nn x\\le\\nn p$\nto obtain $\\nn{x-p}^2\\le2\\ip p{p-x}$.\nWrite $p=(p-A^*w)+A^*w$. The first pairing is at most\n$\\nn{p-A^*w}\\nn{p-x}\\le2\\nn p\\nn{p-A^*w}$; the adjoint identity bounds\nthe second by $\\nn w\\nn{Ap-Ax}$. This yields the displayed estimate.\nFor identifiable $p$, the density in SR-41 permits $w$ with\n$p-A^*w$ arbitrarily small (take $w=A u$ from the normal range).\nFixing this $w$, the residual term tends to zero. An epsilon argument\nthen proves norm convergence. No common inverse bound is invoked.","bda\\nn g^2$. The residual ordering and positive\n$\\lambda$ imply norm domination.\nFor the source estimate, expand $\\nn{x-p}^2$ and use $\\nn x\\le\\nn p$\nto obtain $\\nn{x-p}^2\\le2\\ip p{p-x}$.\nWrite $p=(p-A^*w)+A^*w$. The first pairing is at most\n$\\nn{p-A^*w}\\nn{p-x}\\le2\\nn p\\nn{p-A^*w}$; the adjoint identity bounds\nthe second by $\\nn w\\nn{Ap-Ax}$. This yields the displayed estimate.\nFor identifiable $p$, the density in SR-41 permits $w$ with\n$p-A^*w$ arbitrarily small (take $w=A u$ from the normal range).\nFixing this $w$, the residual term tends to zero. An epsilon argument\nthen proves norm convergence. No common inverse bound is invoked.","and positive\n$\\lambda$ imply norm domination.\nFor the source estimate, expand $\\nn{x-p}^2$ and use $\\nn x\\le\\nn p$\nto obtain $\\nn{x-p}^2\\le2\\ip p{p-x}$.\nWrite $p=(p-A^*w)+A^*w$. The first pairing is at most\n$\\nn{p-A^*w}\\nn{p-x}\\le2\\nn p\\nn{p-A^*w}$; the adjoint identity bounds\nthe second by $\\nn w\\nn{Ap-Ax}$. This yields the displayed estimate.\nFor identifiable $p$, the density in SR-41 permits $w$ with\n$p-A^*w$ arbitrarily small (take $w=A u$ from the normal range).\nFixing this $w$, the residual term tends to zero. An epsilon argument\nthen proves norm convergence. No common inverse bound is invoked.","tive\n$\\lambda$ imply norm domination.\nFor the source estimate, expand $\\nn{x-p}^2$ and use $\\nn x\\le\\nn p$\nto obtain $\\nn{x-p}^2\\le2\\ip p{p-x}$.\nWrite $p=(p-A^*w)+A^*w$. The first pairing is at most\n$\\nn{p-A^*w}\\nn{p-x}\\le2\\nn p\\nn{p-A^*w}$; the adjoint identity bounds\nthe second by $\\nn w\\nn{Ap-Ax}$. This yields the displayed estimate.\nFor identifiable $p$, the density in SR-41 permits $w$ with\n$p-A^*w$ arbitrarily small (take $w=A u$ from the normal range).\nFixing this $w$, the residual term tends to zero. An epsilon argument\nthen proves norm convergence. No common inverse bound is invoked.","ion.\nFor the source estimate, expand $\\nn{x-p}^2$ and use $\\nn x\\le\\nn p$\nto obtain $\\nn{x-p}^2\\le2\\ip p{p-x}$.\nWrite $p=(p-A^*w)+A^*w$. The first pairing is at most\n$\\nn{p-A^*w}\\nn{p-x}\\le2\\nn p\\nn{p-A^*w}$; the adjoint identity bounds\nthe second by $\\nn w\\nn{Ap-Ax}$. This yields the displayed estimate.\nFor identifiable $p$, the density in SR-41 permits $w$ with\n$p-A^*w$ arbitrarily small (take $w=A u$ from the normal range).\nFixing this $w$, the residual term tends to zero. An epsilon argument\nthen proves norm convergence. No common inverse bound is invoked.","p}^2$ and use $\\nn x\\le\\nn p$\nto obtain $\\nn{x-p}^2\\le2\\ip p{p-x}$.\nWrite $p=(p-A^*w)+A^*w$. The first pairing is at most\n$\\nn{p-A^*w}\\nn{p-x}\\le2\\nn p\\nn{p-A^*w}$; the adjoint identity bounds\nthe second by $\\nn w\\nn{Ap-Ax}$. This yields the displayed estimate.\nFor identifiable $p$, the density in SR-41 permits $w$ with\n$p-A^*w$ arbitrarily small (take $w=A u$ from the normal range).\nFixing this $w$, the residual term tends to zero. An epsilon argument\nthen proves norm convergence. No common inverse bound is invoked.","\\par{\\small\\textbf{Hypotheses.} Known bounded $A:G\\to V$ between real Hilbert spaces and a fixed true source $g$; a valid pointwise allowance $\\nn{Ag-v}\\le\\eta$. Accepted-source consistency uses a fixed $\\tau\\ge1$ and genuine bounds $\\nn{Ag-v_i}\\le\\eta_i$ with $\\eta_i\\to0$. Total selector existence additionally requires $\\eta>0$ and $\\tau>1$.}\\par","Call $x$ accepted when either\n\\[\n\\nn v\\le\\tau\\eta,\\quad x=0,\n\\]\nor $x=B_\\lambda v$ for some $\\lambda>0$ with\n$\\eta\\le r_v(\\lambda)\\le\\tau\\eta$.\nWith a valid noise allowance, every accepted source satisfies\n\\[\n\\nn x\\le\\nn{P_{(\\ker A)^\\perp}g},\\qquad\n\\nn{Ax-Ag}\\le(\\tau+1)\\eta.\n\\]\nFor accepted families with $\\eta_i\\to0$,\n\\[\nx_i\\to P_{(\\ker A)^\\perp}g,\n\\qquad \\nn{x_i-g}\\to\\nn{P_{\\ker A}g}.\n\\]\nWith $\\eta>0$, $\\tau>1$, the total selector chooses zero on small data\nand otherwise the root $r_v(\\lambda)=\\tau\\eta$. If $g=A^*w$, accepted\nsources obey $\\nn{x-g}^2\\le2\\nn w(\\tau+1)\\eta$.","accepted when either\n\\[\n\\nn v\\le\\tau\\eta,\\quad x=0,\n\\]\nor $x=B_\\lambda v$ for some $\\lambda>0$ with\n$\\eta\\le r_v(\\lambda)\\le\\tau\\eta$.\nWith a valid noise allowance, every accepted source satisfies\n\\[\n\\nn x\\le\\nn{P_{(\\ker A)^\\perp}g},\\qquad\n\\nn{Ax-Ag}\\le(\\tau+1)\\eta.\n\\]\nFor accepted families with $\\eta_i\\to0$,\n\\[\nx_i\\to P_{(\\ker A)^\\perp}g,\n\\qquad \\nn{x_i-g}\\to\\nn{P_{\\ker A}g}.\n\\]\nWith $\\eta>0$, $\\tau>1$, the total selector chooses zero on small data\nand otherwise the root $r_v(\\lambda)=\\tau\\eta$. If $g=A^*w$, accepted\nsources obey $\\nn{x-g}^2\\le2\\nn w(\\tau+1)\\eta$.","some $\\lambda>0$ with\n$\\eta\\le r_v(\\lambda)\\le\\tau\\eta$.\nWith a valid noise allowance, every accepted source satisfies\n\\[\n\\nn x\\le\\nn{P_{(\\ker A)^\\perp}g},\\qquad\n\\nn{Ax-Ag}\\le(\\tau+1)\\eta.\n\\]\nFor accepted families with $\\eta_i\\to0$,\n\\[\nx_i\\to P_{(\\ker A)^\\perp}g,\n\\qquad \\nn{x_i-g}\\to\\nn{P_{\\ker A}g}.\n\\]\nWith $\\eta>0$, $\\tau>1$, the total selector chooses zero on small data\nand otherwise the root $r_v(\\lambda)=\\tau\\eta$. If $g=A^*w$, accepted\nsources obey $\\nn{x-g}^2\\le2\\nn w(\\tau+1)\\eta$.","le r_v(\\lambda)\\le\\tau\\eta$.\nWith a valid noise allowance, every accepted source satisfies\n\\[\n\\nn x\\le\\nn{P_{(\\ker A)^\\perp}g},\\qquad\n\\nn{Ax-Ag}\\le(\\tau+1)\\eta.\n\\]\nFor accepted families with $\\eta_i\\to0$,\n\\[\nx_i\\to P_{(\\ker A)^\\perp}g,\n\\qquad \\nn{x_i-g}\\to\\nn{P_{\\ker A}g}.\n\\]\nWith $\\eta>0$, $\\tau>1$, the total selector chooses zero on small data\nand otherwise the root $r_v(\\lambda)=\\tau\\eta$. If $g=A^*w$, accepted\nsources obey $\\nn{x-g}^2\\le2\\nn w(\\tau+1)\\eta$.","h a valid noise allowance, every accepted source satisfies\n\\[\n\\nn x\\le\\nn{P_{(\\ker A)^\\perp}g},\\qquad\n\\nn{Ax-Ag}\\le(\\tau+1)\\eta.\n\\]\nFor accepted families with $\\eta_i\\to0$,\n\\[\nx_i\\to P_{(\\ker A)^\\perp}g,\n\\qquad \\nn{x_i-g}\\to\\nn{P_{\\ker A}g}.\n\\]\nWith $\\eta>0$, $\\tau>1$, the total selector chooses zero on small data\nand otherwise the root $r_v(\\lambda)=\\tau\\eta$. If $g=A^*w$, accepted\nsources obey $\\nn{x-g}^2\\le2\\nn w(\\tau+1)\\eta$.","Let $p=P_{(\\ker A)^\\perp}g$, so $Ap=Ag$.\nIn the positive branch, $\\nn{Ap-v}\\le\\eta\\le r_v(\\lambda)$ gives norm\ndomination by SR-46; the triangle inequality gives the true residual.\nIn the zero branch, norm domination is immediate and\n$\\nn{Ag}\\le\\nn{Ag-v}+\\nn v\\le(\\tau+1)\\eta$.\nSR-46 now gives strong convergence to $p$, and continuity of the error\nnorm gives the kernel limit. For large data, feasibility supplies\n$\\dist(v,\\Ran A)\\le\\eta<\\tau\\eta<\\nn v$, so SR-45 provides the exact root.\nFor $g=A^*w$, the approximation term in SR-46 is zero, and the residual\nbound proves the squared source-condition rate. General identifiable\nsources have no asserted common rate. The rule controls actual Tikhonov\nsources; numerical solv","Let $p=P_{(\\ker A)^\\perp}g$, so $Ap=Ag$.\nIn the positive branch, $\\nn{Ap-v}\\le\\eta\\le r_v(\\lambda)$ gives norm\ndomination by SR-46; the triangle inequality gives the true residual.\nIn the zero branch, norm domination is immediate and\n$\\nn{Ag}\\le\\nn{Ag-v}+\\nn v\\le(\\tau+1)\\eta$.\nSR-46 now gives strong convergence to $p$, and continuity of the error\nnorm gives the kernel limit. For large data, feasibility supplies\n$\\dist(v,\\Ran A)\\le\\eta<\\tau\\eta<\\nn v$, so SR-45 provides the exact root.\nFor $g=A^*w$, the approximation term in SR-46 is zero, and the residual\nbound proves the squared source-condition rate. General identifiable\nsources have no asserted common rate. The rule controls actual Tikhonov\nsources; numerical solve errors req","Let $p=P_{(\\ker A)^\\perp}g$, so $Ap=Ag$.\nIn the positive branch, $\\nn{Ap-v}\\le\\eta\\le r_v(\\lambda)$ gives norm\ndomination by SR-46; the triangle inequality gives the true residual.\nIn the zero branch, norm domination is immediate and\n$\\nn{Ag}\\le\\nn{Ag-v}+\\nn v\\le(\\tau+1)\\eta$.\nSR-46 now gives strong convergence to $p$, and continuity of the error\nnorm gives the kernel limit. For large data, feasibility supplies\n$\\dist(v,\\Ran A)\\le\\eta<\\tau\\eta<\\nn v$, so SR-45 provides the exact root.\nFor $g=A^*w$, the approximation term in SR-46 is zero, and the residual\nbound proves the squared source-condition rate. General identifiable\nsources have no asserted common rate. The rule controls actual Tikhonov\nsources; numerical solve errors require the additional certificates below.","A)^\\perp}g$, so $Ap=Ag$.\nIn the positive branch, $\\nn{Ap-v}\\le\\eta\\le r_v(\\lambda)$ gives norm\ndomination by SR-46; the triangle inequality gives the true residual.\nIn the zero branch, norm domination is immediate and\n$\\nn{Ag}\\le\\nn{Ag-v}+\\nn v\\le(\\tau+1)\\eta$.\nSR-46 now gives strong convergence to $p$, and continuity of the error\nnorm gives the kernel limit. For large data, feasibility supplies\n$\\dist(v,\\Ran A)\\le\\eta<\\tau\\eta<\\nn v$, so SR-45 provides the exact root.\nFor $g=A^*w$, the approximation term in SR-46 is zero, and the residual\nbound proves the squared source-condition rate. General identifiable\nsources have no asserted common rate. The rule controls actual Tikhonov\nsources; numerical solve errors require the additional certificates below.","da)$ gives norm\ndomination by SR-46; the triangle inequality gives the true residual.\nIn the zero branch, norm domination is immediate and\n$\\nn{Ag}\\le\\nn{Ag-v}+\\nn v\\le(\\tau+1)\\eta$.\nSR-46 now gives strong convergence to $p$, and continuity of the error\nnorm gives the kernel limit. For large data, feasibility supplies\n$\\dist(v,\\Ran A)\\le\\eta<\\tau\\eta<\\nn v$, so SR-45 provides the exact root.\nFor $g=A^*w$, the approximation term in SR-46 is zero, and the residual\nbound proves the squared source-condition rate. General identifiable\nsources have no asserted common rate. The rule controls actual Tikhonov\nsources; numerical solve errors require the additional certificates below.","No positive parameter has residual larger than $\\nn v$.\nWhen $A^*v\\ne0$, none has residual equal to $\\nn v$.\nFor every positive parameter, $r_v(\\lambda)=0$ exactly when $v=0$.\nFor the real scalar identity model and $0<\\rho<v$,\n\\[\n\\lambda=\\frac\\rho{v-\\rho},\\qquad B_\\lambda v=v-\\rho\n\\]\nis the unique root at residual $\\rho$.","The first assertion is the residual contraction in SR-38.\nEquality at $\\nn v$ would give\n$F_\\lambda(v,x_\\lambda)\\le F_\\lambda(v,0)=\\nn v^2$ while its fit term\nalready equals $\\nn v^2$. Hence $x_\\lambda=0$, forcing $A^*v=0$.\nIf the residual is zero, then $Ax_\\lambda=v$, and the normal equation\nreduces to $\\lambda x_\\lambda=0$. Thus $x_\\lambda=0$ and $v=0$.\nThe converse is direct. For $A=I$, the normal equation gives\n$B_\\lambda v=v/(1+\\lambda)$ and residual\n$\\lambda v/(1+\\lambda)$ for $v>0$. Substitution proves the formula and\nSR-45 supplies uniqueness. These endpoints explain the strict root\ninequalities and the separate small-data branch.","is the residual contraction in SR-38.\nEquality at $\\nn v$ would give\n$F_\\lambda(v,x_\\lambda)\\le F_\\lambda(v,0)=\\nn v^2$ while its fit term\nalready equals $\\nn v^2$. Hence $x_\\lambda=0$, forcing $A^*v=0$.\nIf the residual is zero, then $Ax_\\lambda=v$, and the normal equation\nreduces to $\\lambda x_\\lambda=0$. Thus $x_\\lambda=0$ and $v=0$.\nThe converse is direct. For $A=I$, the normal equation gives\n$B_\\lambda v=v/(1+\\lambda)$ and residual\n$\\lambda v/(1+\\lambda)$ for $v>0$. Substitution proves the formula and\nSR-45 supplies uniqueness. These endpoints explain the strict root\ninequalities and the separate small-data branch.","ality at $\\nn v$ would give\n$F_\\lambda(v,x_\\lambda)\\le F_\\lambda(v,0)=\\nn v^2$ while its fit term\nalready equals $\\nn v^2$. Hence $x_\\lambda=0$, forcing $A^*v=0$.\nIf the residual is zero, then $Ax_\\lambda=v$, and the normal equation\nreduces to $\\lambda x_\\lambda=0$. Thus $x_\\lambda=0$ and $v=0$.\nThe converse is direct. For $A=I$, the normal equation gives\n$B_\\lambda v=v/(1+\\lambda)$ and residual\n$\\lambda v/(1+\\lambda)$ for $v>0$. Substitution proves the formula and\nSR-45 supplies uniqueness. These endpoints explain the strict root\ninequalities and the separate small-data branch.","$F_\\lambda(v,x_\\lambda)\\le F_\\lambda(v,0)=\\nn v^2$ while its fit term\nalready equals $\\nn v^2$. Hence $x_\\lambda=0$, forcing $A^*v=0$.\nIf the residual is zero, then $Ax_\\lambda=v$, and the normal equation\nreduces to $\\lambda x_\\lambda=0$. Thus $x_\\lambda=0$ and $v=0$.\nThe converse is direct. For $A=I$, the normal equation gives\n$B_\\lambda v=v/(1+\\lambda)$ and residual\n$\\lambda v/(1+\\lambda)$ for $v>0$. Substitution proves the formula and\nSR-45 supplies uniqueness. These endpoints explain the strict root\ninequalities and the separate small-data branch.","Throughout this section $A:G\\to V$ is bounded between real Hilbert spaces,\n$N_A=\\ker A$, $S_A=N_A^\\perp$, and $\\lambda>0$. Write\n\\[\nN_\\lambda=A^*A+\\lambda I,\\qquad\nx_\\lambda=N_\\lambda^{-1}A^*v,\\qquad\nr_\\lambda(x)=N_\\lambda x-A^*v .\n\\]\nThe inverse is the actual positive-parameter constructor already established.\nThe vector $r_\\lambda(x)$ is the normal-equation residual of the computed\nsource $x$. Its size measures source correction and image correction on\ndifferent scales. A fitting residual $\\nn{Ax-v}$ alone does not supply\neither correction.","^\\perp$, and $\\lambda>0$. Write\n\\[\nN_\\lambda=A^*A+\\lambda I,\\qquad\nx_\\lambda=N_\\lambda^{-1}A^*v,\\qquad\nr_\\lambda(x)=N_\\lambda x-A^*v .\n\\]\nThe inverse is the actual positive-parameter constructor already established.\nThe vector $r_\\lambda(x)$ is the normal-equation residual of the computed\nsource $x$. Its size measures source correction and image correction on\ndifferent scales. A fitting residual $\\nn{Ax-v}$ alone does not supply\neither correction.","For every computed source $x$,\n\\[\nx-x_\\lambda=N_\\lambda^{-1}r_\\lambda(x),\\qquad\n\\nn{x-x_\\lambda}\\le\\frac{\\nn{r_\\lambda(x)}}{\\lambda},\n\\qquad\n\\nn{Ax-Ax_\\lambda}\\le\n\\frac{\\nn{r_\\lambda(x)}}{2\\sqrt\\lambda}.\n\\]\nConsequently\n\\[\n\\left|\\nn{Ax-v}-\\nn{Ax_\\lambda-v}\\right|\n\\le\\frac{\\nn{r_\\lambda(x)}}{2\\sqrt\\lambda}.\n\\]\nThe bounds use the same actual residual and the existing Hilbert norms.","Subtract $N_\\lambda x_\\lambda=A^*v$ from the definition of the normal\nresidual and apply the inverse. For $u=N_\\lambda^{-1}r$, the energy identity\ngives\n\\[\n\\nn{Au}^2+\\lambda\\nn u^2=\\ip ru\n\\le\\nn r\\,\\nn u .\n\\]\nDropping the first term shows $\\nn u\\le\\nn r/\\lambda$, including $u=0$.\nTo control the image, complete the scalar square:\n\\[\n\\nn{Au}^2\\le\\nn r\\,\\nn u-\\lambda\\nn u^2\n\\le\\frac{\\nn r^2}{4\\lambda}.\n\\]\nBoth sides of the desired norm inequality are nonnegative, so taking square\nroots gives the image bound. Set $r=r_\\lambda(x)$ and use linearity of\n$A$. Finally the reverse triangle inequality bounds the difference of the\ntwo fitting norms by $\\nn{A(x-x_\\lambda)}$.","da^{-1}r$, the energy identity\ngives\n\\[\n\\nn{Au}^2+\\lambda\\nn u^2=\\ip ru\n\\le\\nn r\\,\\nn u .\n\\]\nDropping the first term shows $\\nn u\\le\\nn r/\\lambda$, including $u=0$.\nTo control the image, complete the scalar square:\n\\[\n\\nn{Au}^2\\le\\nn r\\,\\nn u-\\lambda\\nn u^2\n\\le\\frac{\\nn r^2}{4\\lambda}.\n\\]\nBoth sides of the desired norm inequality are nonnegative, so taking square\nroots gives the image bound. Set $r=r_\\lambda(x)$ and use linearity of\n$A$. Finally the reverse triangle inequality bounds the difference of the\ntwo fitting norms by $\\nn{A(x-x_\\lambda)}$.","[\n\\nn{Au}^2+\\lambda\\nn u^2=\\ip ru\n\\le\\nn r\\,\\nn u .\n\\]\nDropping the first term shows $\\nn u\\le\\nn r/\\lambda$, including $u=0$.\nTo control the image, complete the scalar square:\n\\[\n\\nn{Au}^2\\le\\nn r\\,\\nn u-\\lambda\\nn u^2\n\\le\\frac{\\nn r^2}{4\\lambda}.\n\\]\nBoth sides of the desired norm inequality are nonnegative, so taking square\nroots gives the image bound. Set $r=r_\\lambda(x)$ and use linearity of\n$A$. Finally the reverse triangle inequality bounds the difference of the\ntwo fitting norms by $\\nn{A(x-x_\\lambda)}$.","nn r/\\lambda$, including $u=0$.\nTo control the image, complete the scalar square:\n\\[\n\\nn{Au}^2\\le\\nn r\\,\\nn u-\\lambda\\nn u^2\n\\le\\frac{\\nn r^2}{4\\lambda}.\n\\]\nBoth sides of the desired norm inequality are nonnegative, so taking square\nroots gives the image bound. Set $r=r_\\lambda(x)$ and use linearity of\n$A$. Finally the reverse triangle inequality bounds the difference of the\ntwo fitting norms by $\\nn{A(x-x_\\lambda)}$.","The distinction matters when $\\lambda$ is small. A normal residual sufficient\nfor a prescribed image tolerance may be insufficient for a prescribed\nsource tolerance. If source correction must be at most $\\kappa$ and image\ncorrection at most $\\xi$, a sound pair of conditions is\n$\\nn{r_\\lambda(x)}\\le\\lambda\\kappa$ and\n$\\nn{r_\\lambda(x)}\\le2\\sqrt\\lambda\\,\\xi$.\nThese are parameter-dependent requirements, not a uniform inverse bound\nas $\\lambda$ tends to zero.","a$ is small. A normal residual sufficient\nfor a prescribed image tolerance may be insufficient for a prescribed\nsource tolerance. If source correction must be at most $\\kappa$ and image\ncorrection at most $\\xi$, a sound pair of conditions is\n$\\nn{r_\\lambda(x)}\\le\\lambda\\kappa$ and\n$\\nn{r_\\lambda(x)}\\le2\\sqrt\\lambda\\,\\xi$.\nThese are parameter-dependent requirements, not a uniform inverse bound\nas $\\lambda$ tends to zero.","Suppose $m$ is a measured fitting norm with\n\\[\n|m-\\nn{Ax-v}|\\le\\omega,\\quad\n\\nn{r_\\lambda(x)}\\le\\lambda\\kappa,\\quad\n\\nn{r_\\lambda(x)}\\le2\\sqrt\\lambda\\,\\xi .\n\\]\nIf\n\\[\n\\eta+\\omega+\\xi\\le m\\le\\tau\\eta-\\omega-\\xi,\n\\]\nthen the actual exact source $x_\\lambda$ has discrepancy in\n$[\\eta,\\tau\\eta]$, and $\\nn{x-x_\\lambda}\\le\\kappa$.\nA protected scalar window exists exactly when\n\\[\n2(\\omega+\\xi)\\le(\\tau-1)\\eta .\n\\]\nIts midpoint $(\\tau+1)\\eta/2$ is then admissible.","red fitting norm with\n\\[\n|m-\\nn{Ax-v}|\\le\\omega,\\quad\n\\nn{r_\\lambda(x)}\\le\\lambda\\kappa,\\quad\n\\nn{r_\\lambda(x)}\\le2\\sqrt\\lambda\\,\\xi .\n\\]\nIf\n\\[\n\\eta+\\omega+\\xi\\le m\\le\\tau\\eta-\\omega-\\xi,\n\\]\nthen the actual exact source $x_\\lambda$ has discrepancy in\n$[\\eta,\\tau\\eta]$, and $\\nn{x-x_\\lambda}\\le\\kappa$.\nA protected scalar window exists exactly when\n\\[\n2(\\omega+\\xi)\\le(\\tau-1)\\eta .\n\\]\nIts midpoint $(\\tau+1)\\eta/2$ is then admissible.","SR-49 and the measurement enclosure imply\n\\[\n|m-\\nn{Ax_\\lambda-v}|\n\\le |m-\\nn{Ax-v}|+\n|\\nn{Ax-v}-\\nn{Ax_\\lambda-v}|\n\\le\\omega+\\xi .\n\\]\nSubtracting this total allowance from the lower protected inequality gives\n$\\eta\\le\\nn{Ax_\\lambda-v}$. Adding it to the upper inequality gives\n$\\nn{Ax_\\lambda-v}\\le\\tau\\eta$. The source-correction bound follows\nfrom the other residual condition in SR-49. Hence the computed source is\nwithin $\\kappa$ of an actual accepted discrepancy source.","nclosure imply\n\\[\n|m-\\nn{Ax_\\lambda-v}|\n\\le |m-\\nn{Ax-v}|+\n|\\nn{Ax-v}-\\nn{Ax_\\lambda-v}|\n\\le\\omega+\\xi .\n\\]\nSubtracting this total allowance from the lower protected inequality gives\n$\\eta\\le\\nn{Ax_\\lambda-v}$. Adding it to the upper inequality gives\n$\\nn{Ax_\\lambda-v}\\le\\tau\\eta$. The source-correction bound follows\nfrom the other residual condition in SR-49. Hence the computed source is\nwithin $\\kappa$ of an actual accepted discrepancy source.","For example, let $\\eta=1/10$, $\\tau=2$, and\n$\\omega=\\xi=1/100$. The total protected margin is $1/50$,\nand the interval is $[3/25,9/50]$. Its midpoint is $3/20$.\nThis allocation proves room for the measurement and image-correction\nerrors; it does not prove that a particular computed source attains the\ninterval or that $1/10$ is a true data-noise allowance.","Call $x$ near accepted with budget $\\kappa$ when some exact accepted source\n$y$ satisfies $\\nn{x-y}\\le\\kappa$. If\n$\\nn{Ag-v_i}\\le\\eta_i$, $\\eta_i\\to0$, and near-accepted $x_i$ have\n$\\kappa_i\\to0$, then\n\\[\nx_i\\longrightarrow P_{S_A}g,\\qquad\n\\nn{x_i-g}\\longrightarrow\\nn{P_{N_A}g}.\n\\]\nAt every index $\\nn{P_{N_A}x_i}\\le\\kappa_i$; therefore the computed true-kernel\ncomponent tends to zero. No separate parameter schedule is assumed.","Choose accepted comparison sources $y_i$ from the definition. Exact\naccepted consistency gives $y_i\\to p=P_{S_A}g$, while\n$\\nn{x_i-y_i}\\le\\kappa_i\\to0$ gives $x_i-y_i\\to0$. Adding the two limits\nproves $x_i\\to p$. Continuity of the norm gives the source-error limit,\nbecause $p-g=-P_{N_A}g$.","The exact accepted source belongs to $S_A$, including the small-data zero\nbranch. Orthogonal projection is nonexpansive, so\n\\[\n\\nn{P_{N_A}x_i}\n=\\nn{P_{N_A}(x_i-y_i)}\n\\le\\nn{x_i-y_i}\\le\\kappa_i .\n\\]\nThis pointwise estimate only needs near acceptance, not a truth-noise\npremise. Source consistency additionally needs the stated true-noise\nallowance tending to zero.","The underlying exact consistency theorem has substantive content even\nfor a nonclosed range. Accepted sources satisfy norm domination\n$\\nn{y_i}\\le\\nn p$ and $\\nn{Ay_i-Ap}\\to0$. Since\n$p\\in S_A=\\cl\\Ran A^*$, approximate $p$ by $A^*w$ and use\n\\[\n\\nn{y_i-p}^2\n\\le2\\ip{p}{p-y_i}\n\\le2\\nn{p-A^*w}\\nn{p-y_i}\n+2\\nn w\\,\\nn{A(p-y_i)} .\n\\]\nThe first factor $\\nn{p-y_i}$ is bounded by $2\\nn p$.\nFirst choose the approximation accurately, then let $i$ tend along the\nfilter. This proves strong convergence without a bounded inverse.","orem has substantive content even\nfor a nonclosed range. Accepted sources satisfy norm domination\n$\\nn{y_i}\\le\\nn p$ and $\\nn{Ay_i-Ap}\\to0$. Since\n$p\\in S_A=\\cl\\Ran A^*$, approximate $p$ by $A^*w$ and use\n\\[\n\\nn{y_i-p}^2\n\\le2\\ip{p}{p-y_i}\n\\le2\\nn{p-A^*w}\\nn{p-y_i}\n+2\\nn w\\,\\nn{A(p-y_i)} .\n\\]\nThe first factor $\\nn{p-y_i}$ is bounded by $2\\nn p$.\nFirst choose the approximation accurately, then let $i$ tend along the\nfilter. This proves strong convergence without a bounded inverse.","Suppose a record supplies $\\lambda>0$, nonnegative\n$\\rho,\\xi,\\kappa,m,\\omega$, and\n\\[\n\\nn{r_\\lambda(x)}^2\\le\\rho^2,\\quad\n\\rho^2\\le(\\lambda\\kappa)^2,\\quad\n\\rho^2\\le4\\lambda\\xi^2,\n\\]\n\\[\n\\max(0,m-\\omega)^2\\le\\nn{Ax-v}^2\\le(m+\\omega)^2,\n\\qquad\n\\eta+\\omega+\\xi\\le m\\le\\tau\\eta-\\omega-\\xi .\n\\]\nThen $x$ is near accepted with source budget $\\kappa$. Families of such\nrecords give SR-51 when the genuine data allowances and source budgets\ntend to zero.","All quantities whose squares are compared are nonnegative. The first\ntwo inequalities therefore give\n$\\nn{r_\\lambda(x)}\\le\\rho\\le\\lambda\\kappa$.\nThe third gives $\\rho\\le2\\sqrt\\lambda\\,\\xi$ because\n$(2\\sqrt\\lambda\\,\\xi)^2=4\\lambda\\xi^2$.","Put $d=\\nn{Ax-v}\\ge0$. The squared fitting enclosure is equivalent to\nthe unsquared enclosure\n$\\max(0,m-\\omega)\\le d\\le m+\\omega$.\nSince $d\\ge0$, its lower part is equivalent to $m-\\omega\\le d$.\nTogether these say $|m-d|\\le\\omega$. Thus every premise of SR-50 has\nbeen obtained from the supplied record, and its conclusion proves soundness.\nApply SR-51 pointwise for the family statement.","If the normal residual itself is evaluated approximately as $\\widehat r$,\nthe distinct assumptions\n$\\nn{r_\\lambda(x)-\\widehat r}\\le\\nu$ and\n$\\nn{\\widehat r}\\le\\widehat\\rho$ give the allowance\n$\\widehat\\rho+\\nu$. Substituting that sum into both correction tests retains\nthe evaluation error rather than silently identifying it with zero.","Data noise, solver error and model error are different inputs. Let $A$ be\nthe true forward operator and $C$ the available operator, acting between the\nsame real Hilbert spaces. The premise $\\nn{C-A}\\le\\delta$ is a whole-operator\nallowance. A small residual for one source does not establish it.\nThis section first uses a fixed cap $M$ on a true comparison source.\nThe next section removes that cap from the acceptance rule by making the\nallowance depend on the candidate itself.","If $\\nn{C-A}\\le\\delta$, $\\nn p\\le M$ and $\\nn{Ap-v}\\le\\eta$, then\n\\[\n\\nn{Cp-v}\\le\\epsilon:=\\eta+\\delta M.\n\\]\nFor every candidate $x$,\n\\[\n|\\nn{Ax-v}-\\nn{Cx-v}|\\le\\delta\\nn x.\n\\]\nIf $|m-\\nn{Cx-v}|\\le\\omega$, then\n$|m-\\nn{Ax-v}|\\le\\omega+\\delta\\nn x$.\nThe effective-noise allowance is sharp in scalar examples.","The definition of operator norm gives\n$\\nn{Cp-Ap}\\le\\delta\\nn p$. Because a norm bound implies\n$\\delta\\ge0$, the source cap gives\n\\[\n\\nn{Cp-v}\\le\\nn{Cp-Ap}+\\nn{Ap-v}\n\\le\\delta M+\\eta .\n\\]\nFor the fitting-norm comparison, apply the reverse triangle inequality\nto $Ax-v$ and $Cx-v$, whose difference is $(A-C)x$. Combining this\nwith the measurement enclosure gives the last claim.","To see the possible equality, take real scalar maps\n$A=1$, $C=1+\\delta$, $p=M$ and $v=M-\\eta$, with\n$\\eta,\\delta,M\\ge0$. Then\n$Cp-v=\\delta M+\\eta\\ge0$, so its absolute value is exactly $\\epsilon$.\nThe source-size factor cannot generally be omitted: for fixed\n$\\delta>0$, the action difference $\\delta|p|$ grows without bound as\n$|p|$ grows. Thus replacing $\\delta M$ by an operator-only constant\nwould strengthen the contract incorrectly.","For the normal residuals formed with the same parameter,\n\\[\nr^A_\\lambda(x)-r^C_\\lambda(x)\n=(A^*-C^*)(Cx-v)+A^*(A-C)x .\n\\]\nIf $\\nn{C-A}\\le\\delta$, then\n\\[\n\\nn{r^A_\\lambda(x)-r^C_\\lambda(x)}\n\\le\\delta\\bigl(\\nn{Cx-v}+(\\nn C+\\delta)\\nn x\\bigr).\n\\]\nHence a computed-model normal allowance $\\nn{r^C_\\lambda(x)}\\le\\rho$\ngives the corresponding true normal allowance by adding this expression.","Expand the two residuals. The $\\lambda x$ terms cancel. Insert and\nsubtract $A^*Cx$ to obtain the displayed identity. Adjoint isometry\ngives $\\nn{A^*-C^*}=\\nn{A-C}\\le\\delta$ and $\\nn{A^*}=\\nn A$.\nSubmultiplicativity therefore bounds its two terms by\n$\\delta\\nn{Cx-v}$ and $\\nn A\\,\\delta\\nn x$.\nThe triangle inequality gives $\\nn A\\le\\nn C+\\delta$, yielding a\nbound written in terms of available-model quantities and the allowance.\nFinally apply $\\nn{r^A}\\le\\nn{r^C}+\\nn{r^A-r^C}$.","s cancel. Insert and\nsubtract $A^*Cx$ to obtain the displayed identity. Adjoint isometry\ngives $\\nn{A^*-C^*}=\\nn{A-C}\\le\\delta$ and $\\nn{A^*}=\\nn A$.\nSubmultiplicativity therefore bounds its two terms by\n$\\delta\\nn{Cx-v}$ and $\\nn A\\,\\delta\\nn x$.\nThe triangle inequality gives $\\nn A\\le\\nn C+\\delta$, yielding a\nbound written in terms of available-model quantities and the allowance.\nFinally apply $\\nn{r^A}\\le\\nn{r^C}+\\nn{r^A-r^C}$.","The identity itself does not require a positive parameter; positivity\nis needed when this transported residual is used in SR-49 to control\nan inverse correction. The model error remains distinct from the solver's\nnormal residual. Both contributions are amplified by the corresponding\npositive-parameter correction scale. A small fixed model allowance is\ntherefore not a uniform guarantee when $\\lambda$ tends to zero.","Suppose $p\\in S_A$, $\\nn p\\le M$, and\n\\[\n\\nn{C_i-A}\\le\\delta_i,\\qquad\n\\nn{Ap-v_i}\\le\\eta_i .\n\\]\nIf $x_i$ lies within $\\kappa_i$ of an exact accepted source for\n$C_i,v_i$ with allowance $\\epsilon_i=\\eta_i+\\delta_iM$, and\n$\\eta_i,\\delta_i,\\kappa_i\\to0$, then $x_i\\to p$.\nThe model kernels and ranks may change. For arbitrary $g$, using\n$p=P_{S_A}g$ yields\n$\\nn{x_i-g}\\to\\nn{P_{N_A}g}$ and $P_{N_A}x_i\\to0$.","Let $y_i$ be the exact accepted comparison. SR-53 makes $p$ feasible\nfor the available model at allowance $\\epsilon_i$. In the positive\nbranch, regularized optimality and the lower discrepancy bound give\n$\\nn{y_i}\\le\\nn p$: compare the objective at $y_i$ and $p$, then\ncancel the fitting terms using\n$\\nn{C_ip-v_i}\\le\\epsilon_i\\le\\nn{C_iy_i-v_i}$.\nThe small-data branch is zero and satisfies the same domination.","The upper discrepancy bound gives\n$\\nn{C_iy_i-v_i}\\le\\tau\\epsilon_i$. Returning to the true operator,\n\\[\n\\nn{Ay_i-Ap}\n\\le\\delta_i\\nn{y_i}+\\tau\\epsilon_i+\\eta_i\n\\le(\\tau+1)\\epsilon_i .\n\\]\nThus $y_i$ is norm dominated by the fixed identifiable $p$ and has true\nimage residual tending to zero. The density argument printed in SR-51\nproves $y_i\\to p$ without projecting onto the changing model kernels.\nThe near-source difference then tends to zero, proving $x_i\\to p$.","au\\epsilon_i$. Returning to the true operator,\n\\[\n\\nn{Ay_i-Ap}\n\\le\\delta_i\\nn{y_i}+\\tau\\epsilon_i+\\eta_i\n\\le(\\tau+1)\\epsilon_i .\n\\]\nThus $y_i$ is norm dominated by the fixed identifiable $p$ and has true\nimage residual tending to zero. The density argument printed in SR-51\nproves $y_i\\to p$ without projecting onto the changing model kernels.\nThe near-source difference then tends to zero, proving $x_i\\to p$.","For arbitrary $g$, $AP_{S_A}g=Ag$, so the same true-data bound applies\nto $p=P_{S_A}g$. Continuity of norm and true-kernel projection gives\nthe remaining limits. The fixed cap concerns this limiting identifiable\nsource. It is not a claim that the minimum-norm sources for unrelated\nmodels have already converged.","A source is certified using $C$ when either\n$\\nn v\\le\\tau\\epsilon$ and $\\nn x\\le\\kappa$, or it has an SR-52\npositive certificate for $C$ with allowance $\\epsilon=\\eta+\\delta M$.\nSuch a source is near accepted for the available model.\nWith the external truth premises of SR-55 and vanishing allowances,\ncertified sources converge to the true identifiable source.","If $\\epsilon>0$ and $\\tau>1$, those truth premises also give feasibility\nof the total exact discrepancy selector for $C$; the selected exact\nsources are consistent under the same vanishing data/operator bounds.","In the small-data branch compare $x$ with the accepted zero source:\n$\\nn{x-0}\\le\\kappa$. In the positive branch, SR-52 supplies an actual\naccepted comparison source at distance at most $\\kappa$. This proves\noperational soundness without using $A$.","For the true conclusion, separately supply $\\nn{C-A}\\le\\delta$,\nthe true-data bound, and $\\nn p\\le M$. These hypotheses are not fields\nestablished by an observed fitting certificate. They allow SR-55 to\nbe applied to every certified source.","SR-53 gives the feasible witness $p$ for the available model. A positive\neffective allowance and $\\tau>1$ ensure a strict residual target above\nits attainable floor in the non-small-data case; the total discrepancy\nconstruction chooses an actual positive-parameter source there and zero\nin the other branch. Each selected source is exact accepted, so SR-55\napplies with $\\kappa=0$. This selected object is noncomputable in the\ngeneric formalization. Its existence theorem and soundness are distinct\nfrom a finite numerical search implementation.","The acceptance rule can avoid an externally supplied number $M$ by using\nthe candidate itself in the uncertainty allowance. Put\n\\[\nc_{\\eta,\\delta}(x)=\\eta+\\delta\\nn x .\n\\]\nA generalized accepted source is either zero in the small-data branch\n$\\nn v\\le\\tau\\eta$, or an actual Tikhonov source whose residual lies between\n$c_{\\eta,\\delta}(x)$ and $\\tau c_{\\eta,\\delta}(x)$.\nThis definition contains $C,v,\\eta,\\delta,\\tau,x$ and the positive parameter;\nit does not contain a true-source cap. Feasibility remains a separate\nquestion and retains a quantitative relation between model accuracy and\nthe true noise scale.","\\par{\\small\\textbf{Hypotheses.} Real Hilbert spaces; bounded C; lambda>0; delta\\(\\ge\\)0; feasible comparison p for candidate noise; actual regularized candidate.}\\par","Let $x$ be the actual regularized source for $C,v,\\lambda$. If\n\\[\n\\nn{Cp-v}\\le c_{\\eta,\\delta}(p),\\qquad\nc_{\\eta,\\delta}(x)\\le\\nn{Cx-v},\n\\]\nthen $\\nn x\\le\\nn p$.\nIn particular, if $\\nn{C-A}\\le\\delta$ and $\\nn{Ap-v}\\le\\eta$,\nevery generalized accepted source satisfies $\\nn x\\le\\nn p$.","Suppose instead $\\nn x>\\nn p$. Since $\\delta\\ge0$, candidate noise\nis monotone in source norm:\n$c_{\\eta,\\delta}(p)\\le c_{\\eta,\\delta}(x)$.\nThe two displayed hypotheses imply\n$\\nn{Cp-v}\\le\\nn{Cx-v}$.\nComparing Tikhonov objectives, the fitting-square term at $p$ is no larger,\nwhereas its positive weighted source-square term is strictly smaller.\nThe total objective at $p$ is strictly smaller than the objective at its\nactual unique minimizer $x$, a contradiction.","a\\ge0$, candidate noise\nis monotone in source norm:\n$c_{\\eta,\\delta}(p)\\le c_{\\eta,\\delta}(x)$.\nThe two displayed hypotheses imply\n$\\nn{Cp-v}\\le\\nn{Cx-v}$.\nComparing Tikhonov objectives, the fitting-square term at $p$ is no larger,\nwhereas its positive weighted source-square term is strictly smaller.\nThe total objective at $p$ is strictly smaller than the objective at its\nactual unique minimizer $x$, a contradiction.","With true operator bounds, SR-53 applied using $M=\\nn p$ gives the\nrequired candidate-noise feasibility. The operator allowance implies\n$\\delta\\ge0$. The positive accepted branch therefore satisfies the lemma;\nthe small-data branch $x=0$ satisfies the norm conclusion directly.","\\par{\\small\\textbf{Hypotheses.} Real Hilbert spaces; true whole-operator/\\allowbreak data bounds; tau\\(\\ge\\)0; generalized accepted or near-accepted sources; vanishing allowances.}\\par","A generalized accepted $x$ obeys\n\\[\n\\nn{Ax-Ap}\\le(\\tau+1)c_{\\eta,\\delta}(p).\n\\]\nA source within $\\kappa$ of such an $x$ obeys the same bound with\n$\\nn A\\,\\kappa$ added. For $p\\in S_A$, vanishing data/operator\nallowances and near-source budgets imply strong convergence to $p$.\nIf $p=A^*w$, an inexact accepted candidate also satisfies\n\\[\n\\nn{x-p}^2\\le\n2\\kappa^2+4\\nn w(\\tau+1)c_{\\eta,\\delta}(p).\n\\]","SR-57 gives $\\nn x\\le\\nn p$, so\n$c_{\\eta,\\delta}(x)\\le c_{\\eta,\\delta}(p)$.\nThe upper generalized discrepancy bound and $\\tau\\ge0$ yield\n$\\nn{Cx-v}\\le\\tau c_{\\eta,\\delta}(p)$. Insert $Cx$ and $v$ into\nthe true residual:\n\\[\n\\nn{Ax-Ap}\\le\n\\delta\\nn x+\\nn{Cx-v}+\\eta\n\\le(\\tau+1)c_{\\eta,\\delta}(p).\n\\]\nFor a near candidate, split its image difference through the exact accepted\ncomparison and apply the operator norm bound to the correction.","As $\\eta_i,\\delta_i\\to0$, the comparison-source noise\n$\\eta_i+\\delta_i\\nn p$ tends to zero. Exact candidates are norm dominated\nby $p$ and have true image error tending to zero; the same density argument\nas SR-51 proves their strong convergence. Add a vanishing near-source\ndifference for inexact candidates. No relative feasibility condition is\nneeded in this implication once accepted candidates have been supplied.","For the source-condition rate, norm domination and $p=A^*w$ give\n\\[\n\\nn{y-p}^2\\le2\\ip{p}{p-y}\\le2\\nn w\\,\\nn{A(p-y)}\n\\]\nfor an exact comparison $y$. Bound its image by the first conclusion.\nThen $\\nn{x-p}\\le\\nn{x-y}+\\nn{y-p}$ and\n$(a+b)^2\\le2a^2+2b^2$ prove the stated inexact rate.","\\par{\\small\\textbf{Hypotheses.} Real Hilbert spaces; eta>0,tau>1; true operator/\\allowbreak data bounds; delta\\(\\Vert\\)p\\(\\Vert\\)<(tau-1)eta.}\\par","If $\\nn{C-A}\\le\\delta$, $\\nn{Ap-v}\\le\\eta$ and\n\\[\n\\delta\\nn p<(\\tau-1)\\eta,\n\\]\nthen an accepted source exists with norm at most $\\nn p$. In the large-data\ncase one can choose a positive parameter whose residual is exactly\n$\\tau\\eta$. A sufficient ratio condition is\n$\\delta\\le\\gamma\\eta$ and $\\gamma\\nn p<\\tau-1$.\nNoncomputably selected families converge to $p\\in S_A$ when the data/operator\nallowances vanish and the pointwise feasibility conditions hold.","The true bounds imply\n$\\nn{Cp-v}\\le\\eta+\\delta\\nn p<\\tau\\eta$.\nIf $\\nn v\\le\\tau\\eta$, the zero source is accepted. Otherwise the strict\nattainable-floor bound and $\\tau\\eta<\\nn v$ give a positive exact\ndiscrepancy root for $C$ with residual $\\tau\\eta$.\nLet $x$ be its actual Tikhonov source. Objective comparison, or the\nfixed-noise domination lemma, gives $\\nn x\\le\\nn p$.\nConsequently\n\\[\nc_{\\eta,\\delta}(x)\\le\\eta+\\delta\\nn p<\\tau\\eta\n=\\nn{Cx-v}.\n\\]\nBecause $\\delta\\ge0$ and $\\tau>1$,\n$\\tau\\eta\\le\\tau c_{\\eta,\\delta}(x)$.\nThus both generalized band inequalities hold.",".\nIf $\\nn v\\le\\tau\\eta$, the zero source is accepted. Otherwise the strict\nattainable-floor bound and $\\tau\\eta<\\nn v$ give a positive exact\ndiscrepancy root for $C$ with residual $\\tau\\eta$.\nLet $x$ be its actual Tikhonov source. Objective comparison, or the\nfixed-noise domination lemma, gives $\\nn x\\le\\nn p$.\nConsequently\n\\[\nc_{\\eta,\\delta}(x)\\le\\eta+\\delta\\nn p<\\tau\\eta\n=\\nn{Cx-v}.\n\\]\nBecause $\\delta\\ge0$ and $\\tau>1$,\n$\\tau\\eta\\le\\tau c_{\\eta,\\delta}(x)$.\nThus both generalized band inequalities hold.","u\\eta$, the zero source is accepted. Otherwise the strict\nattainable-floor bound and $\\tau\\eta<\\nn v$ give a positive exact\ndiscrepancy root for $C$ with residual $\\tau\\eta$.\nLet $x$ be its actual Tikhonov source. Objective comparison, or the\nfixed-noise domination lemma, gives $\\nn x\\le\\nn p$.\nConsequently\n\\[\nc_{\\eta,\\delta}(x)\\le\\eta+\\delta\\nn p<\\tau\\eta\n=\\nn{Cx-v}.\n\\]\nBecause $\\delta\\ge0$ and $\\tau>1$,\n$\\tau\\eta\\le\\tau c_{\\eta,\\delta}(x)$.\nThus both generalized band inequalities hold.","e zero source is accepted. Otherwise the strict\nattainable-floor bound and $\\tau\\eta<\\nn v$ give a positive exact\ndiscrepancy root for $C$ with residual $\\tau\\eta$.\nLet $x$ be its actual Tikhonov source. Objective comparison, or the\nfixed-noise domination lemma, gives $\\nn x\\le\\nn p$.\nConsequently\n\\[\nc_{\\eta,\\delta}(x)\\le\\eta+\\delta\\nn p<\\tau\\eta\n=\\nn{Cx-v}.\n\\]\nBecause $\\delta\\ge0$ and $\\tau>1$,\n$\\tau\\eta\\le\\tau c_{\\eta,\\delta}(x)$.\nThus both generalized band inequalities hold.","Multiplying $\\delta\\le\\gamma\\eta$ by $\\nn p\\ge0$ and the strict margin\n$\\gamma\\nn p<\\tau-1$ by $\\eta>0$ proves the ratio specialization.\nThe chosen source is defined by classical choice from the existence\ntheorem. Its raw definition takes $A,p$ and the truth/feasibility proofs\nas arguments; it is a theorem-level witness rather than an extracted\nalgorithm using only available measurements. Each chosen source is\naccepted, so SR-58 proves the family convergence. The selection theorem\nretains the feasibility premises even though the acceptance predicate has\nno supplied scalar source cap.","Near-generalized acceptance is preserved by enlarging a source-correction\nbudget and gives the consistency conclusion of SR-58 when that budget\ntends to zero.\nNevertheless valid true-noise and operator allowances alone do not ensure\ngeneralized feasibility. With\n$A=1$, $C=0$, $p=v=1$, $\\eta=1/10$, $\\delta=1$ and $\\tau=2$,\nno source is generalized accepted.","Near acceptance means there exists an exact generalized accepted comparison\n$y$ with $\\nn{x-y}\\le\\kappa$. Increasing $\\kappa$ preserves that\ninequality. Along a family, choosing the comparisons and adding the\nvanishing difference proves consistency by SR-58.","For the scalar boundary, the true data error is zero and\n$\\nn{C-A}=1\\le\\delta$, so both stated truth allowances are valid.\nThe small-data branch would require $1\\le2/10$, which is false.\nFor every positive parameter, the actual regularized source of $C=0$\nis zero and its fitting residual is one. Its candidate noise is\n$1/10+1\\cdot0=1/10$, so the upper generalized band would again require\n$1\\le2/10$. Every branch fails. The relative feasibility condition also\nfails, since $1<(2-1)/10$ is false. Thus the boundary refutes an\nunconditional existence interpretation, not the conditional theorem.","$, so both stated truth allowances are valid.\nThe small-data branch would require $1\\le2/10$, which is false.\nFor every positive parameter, the actual regularized source of $C=0$\nis zero and its fitting residual is one. Its candidate noise is\n$1/10+1\\cdot0=1/10$, so the upper generalized band would again require\n$1\\le2/10$. Every branch fails. The relative feasibility condition also\nfails, since $1<(2-1)/10$ is false. Thus the boundary refutes an\nunconditional existence interpretation, not the conditional theorem.","For comparison, take $A=C=1$, $v=1$, $\\eta=1/10$, $\\delta=0$,\n$\\tau=2$ and $\\lambda=1/4$. Then\n$x_\\lambda=4/5$ and the residual is $1/5$.\nIt lies in $[1/10,2/10]$, so the candidate is accepted, and\n$0<(2-1)/10$ verifies the strict feasibility margin.\nThese examples use the actual scalar normal inverse and exact norms,\nnot a substituted approximate residual.","The final link connects justified calibration measurements with the\nwhole-operator premise used above. It does not infer the calibration truth\nallowance from the model's own fit. Let $K$ be a real normed calibration\ncoordinate space, $G$ a real normed source space and $V$ a real normed\nresponse space. A split frame has bounded synthesis $U:K\\to G$ and\nanalysis $R:G\\to K$ with $UR=I_G$. Its coordinates may be redundant:\n$U$ need not be injective or square. The reconstruction identity requires\ncoverage of every source direction.","justified calibration measurements with the\nwhole-operator premise used above. It does not infer the calibration truth\nallowance from the model's own fit. Let $K$ be a real normed calibration\ncoordinate space, $G$ a real normed source space and $V$ a real normed\nresponse space. A split frame has bounded synthesis $U:K\\to G$ and\nanalysis $R:G\\to K$ with $UR=I_G$. Its coordinates may be redundant:\n$U$ need not be injective or square. The reconstruction identity requires\ncoverage of every source direction.","The observed response $Y:K\\to V$ is an operator, and the assumptions\n$\\nn{Y-AU}\\le\\eta_{\\rm cal}$ and $\\nn{Y-CU}\\le f$ are operator-norm\nenclosures. Individual calibration samples do not automatically give these\nenclosures. A mathematical or experimental justification of\n$\\eta_{\\rm cal}$ remains a separate input.","Every bounded source operator $T$ factors as $T=(TU)R$. Therefore\n\\[\nC-A=((CU)-(AU))R,\\qquad\n\\nn{C-A}\\le(\\eta_{\\rm cal}+f)\\nn R .\n\\]\nIf a supplied analysis bound is $\\nn R\\le k$, the allowance is at most\n$(\\eta_{\\rm cal}+f)k$. Exact calibration $Y=AU=CU$ forces $A=C$.","For every source $g$, $URg=g$, so $Tg=TU(Rg)$. This proves the actual\nfactorization, including its application to $C-A$.\nThe triangle inequality through the common measured operator gives\n\\[\n\\nn{CU-AU}\\le\\nn{CU-Y}+\\nn{Y-AU}\n\\le f+\\eta_{\\rm cal}.\n\\]\nSubmultiplicativity then yields the displayed allowance.\nBoth measured bounds imply their scalar allowances are nonnegative;\nhence replacing $\\nn R$ by a larger $k$ preserves the inequality.\nWith exact calibration, $CU-AU=0$ and the factorization gives $C-A=0$.","ource $g$, $URg=g$, so $Tg=TU(Rg)$. This proves the actual\nfactorization, including its application to $C-A$.\nThe triangle inequality through the common measured operator gives\n\\[\n\\nn{CU-AU}\\le\\nn{CU-Y}+\\nn{Y-AU}\n\\le f+\\eta_{\\rm cal}.\n\\]\nSubmultiplicativity then yields the displayed allowance.\nBoth measured bounds imply their scalar allowances are nonnegative;\nhence replacing $\\nn R$ by a larger $k$ preserves the inequality.\nWith exact calibration, $CU-AU=0$ and the factorization gives $C-A=0$.","tual\nfactorization, including its application to $C-A$.\nThe triangle inequality through the common measured operator gives\n\\[\n\\nn{CU-AU}\\le\\nn{CU-Y}+\\nn{Y-AU}\n\\le f+\\eta_{\\rm cal}.\n\\]\nSubmultiplicativity then yields the displayed allowance.\nBoth measured bounds imply their scalar allowances are nonnegative;\nhence replacing $\\nn R$ by a larger $k$ preserves the inequality.\nWith exact calibration, $CU-AU=0$ and the factorization gives $C-A=0$.","ication to $C-A$.\nThe triangle inequality through the common measured operator gives\n\\[\n\\nn{CU-AU}\\le\\nn{CU-Y}+\\nn{Y-AU}\n\\le f+\\eta_{\\rm cal}.\n\\]\nSubmultiplicativity then yields the displayed allowance.\nBoth measured bounds imply their scalar allowances are nonnegative;\nhence replacing $\\nn R$ by a larger $k$ preserves the inequality.\nWith exact calibration, $CU-AU=0$ and the factorization gives $C-A=0$.","Redundancy causes no logical problem. For instance, let\n$K=\\RR^2$, $G=\\RR$, $U(k_1,k_2)=k_1$ and $Rg=(g,0)$, with the\nstandard product norm. Then $UR=I$, $\\nn U=\\nn R=1$, but\n$U$ has a nontrivial kernel. The frame spans the source even though\nit has an unused coordinate. A true scalar map $A=a$ and model $C=c$\nwith measured row $Y(k_1,k_2)=yk_1$ give\n$|y-a|\\le\\eta_{\\rm cal}$ and $|y-c|\\le f$,\nhence $|c-a|\\le\\eta_{\\rm cal}+f$.\nThe same estimate comes from the general factorization rather than\nassuming the coordinates are square and invertible.","undancy causes no logical problem. For instance, let\n$K=\\RR^2$, $G=\\RR$, $U(k_1,k_2)=k_1$ and $Rg=(g,0)$, with the\nstandard product norm. Then $UR=I$, $\\nn U=\\nn R=1$, but\n$U$ has a nontrivial kernel. The frame spans the source even though\nit has an unused coordinate. A true scalar map $A=a$ and model $C=c$\nwith measured row $Y(k_1,k_2)=yk_1$ give\n$|y-a|\\le\\eta_{\\rm cal}$ and $|y-c|\\le f$,\nhence $|c-a|\\le\\eta_{\\rm cal}+f$.\nThe same estimate comes from the general factorization rather than\nassuming the coordinates are square and invertible.","\\par{\\small\\textbf{Hypotheses.} Fixed split calibration frame; real Hilbert G,V; fixed bounded true A and tau\\(\\ge\\)0; fixed identifiable p; whole calibration truth/\\allowbreak model fit and true data allowances; generalized acceptance; relative feasibility for choice.}\\par","For recovery assume $G,V$ are real Hilbert spaces, $A$ is a fixed bounded\ntrue operator, $\\tau\\ge0$ is fixed, and $p\\in(\\ker A)^\\perp$ is fixed.\nRetain the whole calibration bounds and the true data bounds\n$\\nn{Ap-v_i}\\le\\eta_i$. Put $\\delta_i=(\\eta_{{\\rm cal},i}+f_i)\\nn R$.\nIf the true data,\ncalibration-truth and model-fit allowances tend to zero, then\n$\\delta_i\\to0$. Generalized accepted sources using this generated\nallowance converge to a fixed identifiable true source $p$.\nFor an arbitrary true source $g$, apply this with $p=P_{S_A}g$ and genuine\ndata bounds $\\nn{Ag-v_i}\\le\\eta_i$; then the source-error norm tends to\n$\\nn{P_{N_A}g}$.","operator, $\\tau\\ge0$ is fixed, and $p\\in(\\ker A)^\\perp$ is fixed.\nRetain the whole calibration bounds and the true data bounds\n$\\nn{Ap-v_i}\\le\\eta_i$. Put $\\delta_i=(\\eta_{{\\rm cal},i}+f_i)\\nn R$.\nIf the true data,\ncalibration-truth and model-fit allowances tend to zero, then\n$\\delta_i\\to0$. Generalized accepted sources using this generated\nallowance converge to a fixed identifiable true source $p$.\nFor an arbitrary true source $g$, apply this with $p=P_{S_A}g$ and genuine\ndata bounds $\\nn{Ag-v_i}\\le\\eta_i$; then the source-error norm tends to\n$\\nn{P_{N_A}g}$.","d $p\\in(\\ker A)^\\perp$ is fixed.\nRetain the whole calibration bounds and the true data bounds\n$\\nn{Ap-v_i}\\le\\eta_i$. Put $\\delta_i=(\\eta_{{\\rm cal},i}+f_i)\\nn R$.\nIf the true data,\ncalibration-truth and model-fit allowances tend to zero, then\n$\\delta_i\\to0$. Generalized accepted sources using this generated\nallowance converge to a fixed identifiable true source $p$.\nFor an arbitrary true source $g$, apply this with $p=P_{S_A}g$ and genuine\ndata bounds $\\nn{Ag-v_i}\\le\\eta_i$; then the source-error norm tends to\n$\\nn{P_{N_A}g}$.","$ is fixed.\nRetain the whole calibration bounds and the true data bounds\n$\\nn{Ap-v_i}\\le\\eta_i$. Put $\\delta_i=(\\eta_{{\\rm cal},i}+f_i)\\nn R$.\nIf the true data,\ncalibration-truth and model-fit allowances tend to zero, then\n$\\delta_i\\to0$. Generalized accepted sources using this generated\nallowance converge to a fixed identifiable true source $p$.\nFor an arbitrary true source $g$, apply this with $p=P_{S_A}g$ and genuine\ndata bounds $\\nn{Ag-v_i}\\le\\eta_i$; then the source-error norm tends to\n$\\nn{P_{N_A}g}$.","ion bounds and the true data bounds\n$\\nn{Ap-v_i}\\le\\eta_i$. Put $\\delta_i=(\\eta_{{\\rm cal},i}+f_i)\\nn R$.\nIf the true data,\ncalibration-truth and model-fit allowances tend to zero, then\n$\\delta_i\\to0$. Generalized accepted sources using this generated\nallowance converge to a fixed identifiable true source $p$.\nFor an arbitrary true source $g$, apply this with $p=P_{S_A}g$ and genuine\ndata bounds $\\nn{Ag-v_i}\\le\\eta_i$; then the source-error norm tends to\n$\\nn{P_{N_A}g}$.","Put $\\delta_i=(\\eta_{{\\rm cal},i}+f_i)\\nn R$.\nIf the true data,\ncalibration-truth and model-fit allowances tend to zero, then\n$\\delta_i\\to0$. Generalized accepted sources using this generated\nallowance converge to a fixed identifiable true source $p$.\nFor an arbitrary true source $g$, apply this with $p=P_{S_A}g$ and genuine\ndata bounds $\\nn{Ag-v_i}\\le\\eta_i$; then the source-error norm tends to\n$\\nn{P_{N_A}g}$.","If additionally $\\eta_i>0$, $\\tau>1$ and\n$\\delta_i\\nn p<(\\tau-1)\\eta_i$ pointwise, accepted sources exist and the\nnoncomputable selected sources have the same limit.","Apply SR-61 at every index to obtain the actual operator bound\n$\\nn{C_i-A}\\le\\delta_i$. Because the frame is fixed, its finite norm\n$\\nn R$ is a constant. The sum of the two vanishing calibration\nallowances tends to zero, so multiplying by that constant proves\n$\\delta_i\\to0$. With the genuine data-noise bounds, every premise of\nthe accepted-source consistency theorem SR-58 is now available.","For arbitrary $g$, use $p=P_{S_A}g$. Its forward image is $Ag$, so the\nsame data bounds hold. The strong source limit is $p$, and continuity\ngives $\\nn{x_i-g}\\to\\nn{p-g}=\\nn{P_{N_A}g}$.\nNo calibration can remove source ambiguity in the true forward kernel\nwithout an additional identifying condition.","For split frames $(U_E,R_E):J\\rightleftarrows K$ and\n$(U_F,R_F):K\\rightleftarrows G$, the composite is split, with analysis\n$R_ER_F$ and\n$\\nn{R_ER_F}\\le\\nn{R_E}\\nn{R_F}$.\nThe corresponding calibration allowance is at most\n$(\\eta_{\\rm cal}+f)\\nn{R_E}\\nn{R_F}$.","More generally, if $\\nn{I-UR}\\le\\rho<1$, then\n\\[\n\\nn T\\le\\frac{\\nn{TU}\\nn R}{1-\\rho},\\qquad\n\\nn{C-A}\\le\\frac{(\\eta_{\\rm cal}+f)\\nn R}{1-\\rho}.\n\\]","The composite synthesis is $U_FU_E$ and analysis is $R_ER_F$.\nTheir product is\n$U_F(U_ER_E)R_F=U_FR_F=I_G$.\nSubmultiplicativity bounds the composite analysis norm; applying\nSR-61 and that bound proves the hierarchical allowance. There is no\nrequirement that either synthesis be square or injective.","For the approximate frame, use the exact identity\n$T=(TU)R+T(I-UR)$.\nIt gives\n\\[\n\\nn T\\le\\nn{TU}\\nn R+\\nn T\\,\\nn{I-UR}\n\\le\\nn{TU}\\nn R+\\rho\\nn T .\n\\]\nSince $1-\\rho>0$, absorb the final term and divide.\nFor $T=C-A$, the measured calibration bounds give\n$\\nn{TU}\\le\\eta_{\\rm cal}+f$, producing the operator allowance.\nThis argument only uses norm inequalities; it does not substitute an\nassumed exact inverse for the approximate reconstruction.","The denominator records the loss of control. At $\\rho=0$ the exact\nframe estimate is recovered. At $\\rho=1$ the absorption step gives no\nbound, and it is not legitimate to reuse the formula by division.\nLikewise the product of hierarchical analysis norms is a valid\nconservative bound, not generally an equality or the optimal sensitivity\nof the composite. These conditioning factors belong in the certificate.","\\par{\\small\\textbf{Hypotheses.} Actual partial calibration from R to R\\({}^2\\) with product norm; bounded unseen scalar operators; arbitrary real norm threshold.}\\par","Let $U(t)=(t,0)$ and $C_s(x,y)=sy$. Then $C_sU=0$ and\n$\\nn{C_s}=|s|$. For every real $M$ there is an operator agreeing\nperfectly with zero on the entire calibration subspace yet satisfying\n$\\nn{C_s}>M$.\nThe calibration synthesis is not surjective. Thus perfect agreement on a\nproper source subspace does not control whole-source operator accuracy.","Direct substitution gives $C_sU(t)=s\\cdot0=0$.\nThe second-coordinate projection has norm one, so scalar multiplication\ngives $\\nn{C_s}=|s|$; the vector $(0,1)$ also verifies the lower bound.\nChoose $s=|M|+1>0$. Then $\\nn{C_s}=|M|+1>M$ while its calibrated\nresponse remains zero. Taking the true operator and measured response\nboth zero shows that both calibration residuals vanish, even though\nthe whole operator discrepancy exceeds $M$.","=s\\cdot0=0$.\nThe second-coordinate projection has norm one, so scalar multiplication\ngives $\\nn{C_s}=|s|$; the vector $(0,1)$ also verifies the lower bound.\nChoose $s=|M|+1>0$. Then $\\nn{C_s}=|M|+1>M$ while its calibrated\nresponse remains zero. Taking the true operator and measured response\nboth zero shows that both calibration residuals vanish, even though\nthe whole operator discrepancy exceeds $M$.","The vector $(0,1)$ is outside the range of $U$, proving nonsurjectivity.\nA split reconstruction $UR=I_{\\RR^2}$ would make every source a range\nvector, so no such reconstruction exists here. Nor can any approximate\nreconstruction for this synthesis have whole defect below one:\nthe range of $UR$ has second coordinate zero, and therefore\n$(I-UR)(0,1)$ has second coordinate one and norm at least one.\nThis last observation explains the direction-coverage role of the\nstrict absorption threshold in SR-63.","\\begin{enumerate}\n\\item \\textbf{Stable target-only recovery from behavioral minimality.}\nThe exact quotient $X/N$ preserves all specified future observations.\nThe Hilbert theory stably recovers an identifiable source under its\nreconstruction or regularization assumptions. A general theorem connecting\nthese two constructions must identify a task target, its quotient metric,\nthe relevant forward map, and sufficient conditioning. It should recover\nthat target without requiring reconstruction of the full original source.\nNo such unrestricted bridge is asserted in this edition.\n\\item \\textbf{Calibration truth allowances from individual samples.}\nThe calibration theorems start from the whole-operator inequality\n$\\nn{Y-AU}\\le\\eta_{\\rm cal}$. Deriving it from individual noisy probes\nrequires a specified sampl","tructions must identify a task target, its quotient metric,\nthe relevant forward map, and sufficient conditioning. It should recover\nthat target without requiring reconstruction of the full original source.\nNo such unrestricted bridge is asserted in this edition.\n\\item \\textbf{Calibration truth allowances from individual samples.}\nThe calibration theorems start from the whole-operator inequality\n$\\nn{Y-AU}\\le\\eta_{\\rm cal}$. Deriving it from individual noisy probes\nrequires a specified sample model, a justified aggregation bound, and\ncontrol of untested directions. A measured model fit supplies neither\nthe true operator nor that truth allowance by itself.\n\\item \\textbf{Verified numerical procedures.}\nThe formalization proves the certificate contracts. The rational Python\nimplementations compute and check records under explicit input limits.\nTheir parsers, arithmetic helpers, row-space saturation, parameter\nbracketing, and stopping procedures are not formally verified.\nVerified extraction or an independent formal certificate importer would\nadd a new layer of assurance.\n\\item \\textbf{A…"]}