Realization Calculus: twin edition

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1 \partnote{Resolution, repair, and retained detail}{Resolution gives a concrete family of views. Its compatibility differential connects exact maps, protected information, and residual coordinates. Recovering every finite coordinate is then separated from convergence in a chosen norm.}
2 \section{Resolution towers and exact transport}\label{sec:towers}
3 \subsection{The sequential model}
4 A tower in a category \(\mathcal C\) is a sequence
5 \[
6  \cdots\xrightarrow{\rho_2}X_2\xrightarrow{\rho_1}X_1
7  \xrightarrow{\rho_0}X_0.
8 \]
9 For sets, a point is a sequence \(x_n\in X_n\) with \(\rho_n(x_{n+1})=x_n\). Composite restrictions turn the tower into a view system on \(\N\), with an arrow \(i\to j\) when \(i\le j\). Its section space is the same coherent point type, with the same coordinates.
10 
11 A candidate transport from \(X\) to \(Y\) consists of maps \(f_n:X_n\to Y_n\). It is exact when
12 \begin{equation}\label{eq:exacttower}
13  \rho^Y_n f_{n+1}=f_n\rho^X_n\quad\text{for all }n.
14 \end{equation}
15 
16 \begin{proposition}[The exact tower category]\label{KS16}
17 Identity maps and composites of exact transports are exact. An exact transport induces a map of coherent point spaces by \(x\mapsto(f_nx_n)_n\), preserving identities and composition.
18 \end{proposition}
19 \begin{proof}
20 For two exact transports, substitute \(\rho^Z_ng_{n+1}=g_n\rho^Y_n\) and \(\rho^Y_nf_{n+1}=f_n\rho^X_n\) into the composite square. For a coherent point,
21 \[
22  \rho^Y_n f_{n+1}x_{n+1}
23  =f_n\rho^X_nx_{n+1}=f_nx_n.
24 \]
25 Identity, associativity, and composition on points are checked coordinatewise.
26 \end{proof}
27 \source{KS16}
28 This is a category on a fixed schedule. A full procategory allows additional reindexing and equivalence relations on maps. Those constructions must be supplied rather than identified with the fixed schedule by terminology.
29 
30 \subsection{The actual residual of a transport}
31 Assume \(\mathcal C\) is preadditive, so differences of maps make sense. The tower differential is
32 \[
33  (\partial f)_n=\rho^Y_nf_{n+1}-f_n\rho^X_n.
34 \]
35 Its target has components \(\Hom(X_{n+1},Y_n)\), not \(\Hom(X_n,Y_n)\).
36 
37 \begin{proposition}[The transport Leibniz identity]\label{KS17}
38 For composable candidates,
39 \[
40  \partial(gf)_n=(\partial g)_nf_{n+1}+g_n(\partial f)_n.
41 \]
42 In particular exactness is equivalent to vanishing of the differential.
43 \end{proposition}
44 \begin{proof}
45 Expand the right side. The terms \(-g_n\rho^Y_nf_{n+1}\) and \(+g_n\rho^Y_nf_{n+1}\) cancel, leaving \(\rho^Z_ng_{n+1}f_{n+1}-g_nf_n\rho^X_n\). This is the differential of the composite. The exactness statement is \eqref{eq:exacttower}.
46 \end{proof}
47 \source{KS17}
48 There is no additional mixed defect term in this identity. Such a term would require a different definition of the defect or a different expansion. Typed cancellation is a useful safeguard against importing an attractive but incorrect error formula.
49 
50 \subsection{Cofinal views preserve points, not every measurement}
51 Let \(\phi:\N\to\N\) be monotone and unbounded. The sampled tower uses levels \(X_{\phi(n)}\) and their composite restrictions.
52 
53 \begin{theorem}[Cofinal reconstruction of points]\label{KS18}
54 Sampling induces an equivalence between the coherent points of \(X\) and of its sampled tower.
55 \end{theorem}
56 \begin{proof}
57 Write $\rho_{i,j}:X_j\to X_i$ for the composite restriction when $i\le j$, with $\rho_{i,i}=I$. These maps satisfy $\rho_{i,j}\rho_{j,k}=\rho_{i,k}$. A coherent sampled point obeys
58 \[
59  \rho_{\phi(n),\phi(m)}y_m=y_n\qquad(n\le m),
60 \]
61 by iterating its adjacent coherence equations.
62 
63 For each $k$, unboundedness gives an index $n$ with $k\le\phi(n)$. Set $x_k=\rho_{k,\phi(n)}y_n$. If $m$ is another permissible index, choose $\ell\ge n,m$. Monotonicity permits both composite restrictions, and sampled coherence gives
64 \[
65  \rho_{k,\phi(n)}y_n
66  =\rho_{k,\phi(n)}\rho_{\phi(n),\phi(\ell)}y_\ell
67  =\rho_{k,\phi(\ell)}y_\ell
68  =\rho_{k,\phi(m)}y_m.
69 \]
70 Thus $x_k$ is independent of the choice, including when the schedule repeats a level.
71 
72 For the adjacent equation at $k$, choose one $n$ with $k+1\le\phi(n)$. Then
73 \[
74  \rho_{k,k+1}x_{k+1}
75  =\rho_{k,k+1}\rho_{k+1,\phi(n)}y_n
76  =\rho_{k,\phi(n)}y_n=x_k.
77 \]
78 The recovered family is therefore a point of the original tower.
79 
80 Sampling a full point and recovering its $k$th coordinate gives $\rho_{k,\phi(n)}x_{\phi(n)}=x_k$ by its original coherence. Recovering a sampled point and evaluating at $k=\phi(n)$ allows the index $n$ itself, giving $\rho_{\phi(n),\phi(n)}y_n=y_n$. These are both inverse identities. No surjectivity of the bonding maps or of the point evaluations was used.
81 \end{proof}
82 \source{KS18}
83 This theorem is about exact coherent points. Raw residual size is a different quantity. On identity scalar towers, \(f_n=(-1)^n\) has nonzero fine differences, but its even subsequence is constant. The coarse residual vanishes because adjacent fine defects cancel. A cofinal equivalence of limits does not imply equality of residual norms or an absence of aliasing in uncorrected data.
84 
85 \plate{03_tower}{Exact transport commutes with every adjacent restriction. A cofinal subsequence retains coherent points, but a raw defect can cancel during coarsening. The retained coarse value alone need not record the omitted fine discrepancy.}{fig:tower}
86 
87 \input{chapters/03a_observation}
88 
89 \chap{Banach repair and constructed split coordinates}\label{sec:split}
90 \subsection{An actual complete candidate space}
91 Let \(X_n,Y_n\) be Banach spaces over a complete nontrivially normed field, with continuous linear bonding maps and uniform bounds \(\norm{\rho^X_n}\le B_X\), \(\norm{\rho^Y_n}\le B_Y\). Use the spaces of bounded operator families
92 \[
93  E=\ell^\infty_n\mathcal L(X_n,Y_n),\qquad
94  R=\ell^\infty_n\mathcal L(X_{n+1},Y_n).
95 \]
96 The source formalization derives the required completeness from the operator spaces and their bounded product. The same differential satisfies
97 \[
98  \norm{\partial f}_\infty\le(B_X+B_Y)\norm f_\infty.
99 \]
100 Thus it is a continuous linear map, and evaluation at a fixed level is continuous. These facts connect algebraic compatibility to an analytic limit argument.
101 
102 \subsection{From a correction operator to a projection}
103 The next statement is general. Let \(E\) be Banach, let \(d:E\to R\), \(p:E\to S\), and \(A:R\to E\) be continuous linear maps. Suppose \(pA=0\) and, for \(0\le q<1\),
104 \begin{equation}\label{eq:contraction}
105  \norm{df-dA(df)}\le q\norm{df}\qquad(f\in E).
106 \end{equation}
107 The hypothesis concerns residuals of actual candidates. It need not hold for every independent vector in \(R\).
108 
109 \begin{theorem}[Constructed geometric repair]\label{KS19}
110 Starting from \(f^{(0)}=f\), define
111 \[
112  f^{(m+1)}=f^{(m)}-A(df^{(m)}).
113 \]
114 There is a bounded linear projection \(P_A:E\to\ker d\) with
115 \[
116  f^{(m)}\longrightarrow P_Af,\qquad p(P_Af)=p(f),
117 \]
118 and
119 \[
120  \norm{f^{(m)}-P_Af}
121  \le\frac{\norm A}{1-q}\norm{df^{(m)}}.
122 \]
123 It fixes every exact candidate and satisfies
124 \[
125  \norm{P_A}\le1+\frac{\norm A\norm d}{1-q}.
126 \]
127 \end{theorem}
128 \begin{proof}
129 Put $r_m=df^{(m)}$. Additivity of $d$ gives
130 \[
131  r_{m+1}=r_m-dAr_m.
132 \]
133 Each $r_m$ is the residual of an actual candidate, so the stated hypothesis applies at every iteration. Induction yields, for $j\ge0$,
134 \[
135  \norm{r_{m+j}}\le q^j\norm{r_m},\qquad
136  \norm{f^{(m+j+1)}-f^{(m+j)}}
137  \le\norm A\,q^j\norm{r_m}.
138 \]
139 For $n>m$, the finite triangle inequality therefore gives
140 \[
141  \norm{f^{(n)}-f^{(m)}}
142  \le\norm A\,\norm{r_m}\sum_{j=0}^{n-m-1}q^j
143  \le\frac{\norm A}{1-q}\norm{r_m}.
144 \]
145 Since $\norm{r_m}\le q^m\norm{df}$ tends to zero, this is a Cauchy estimate. Completeness of $E$ supplies a limit, denoted $P_Af$. Passing $n\to\infty$ in the same norm inequality proves the required tail estimate.
146 
147 Continuity of $d$ implies $d(P_Af)=\lim_m r_m=0$. Protection holds before taking a limit: $pA=0$ gives $pf^{(m+1)}=pf^{(m)}$, hence $pf^{(m)}=pf$ for every $m$. Continuity of $p$ gives $p(P_Af)=pf$. This explains why continuity of the observable belongs to this analytic statement.
148 
149 For linearity, write $L=I-Ad$, so $f^{(m)}=L^mf$. For fixed $f,g$ and scalars $\lambda,\mu$,
150 \[
151  L^m(\lambda f+\mu g)=\lambda L^mf+\mu L^mg.
152 \]
153 Continuity of addition and scalar multiplication allows passage to the norm limit and proves linearity of $P_A$. The estimate with $m=0$ gives
154 \[
155  \norm{P_Af}\le\norm f+\frac{\norm A}{1-q}\norm{df}
156  \le\left(1+\frac{\norm A\norm d}{1-q}\right)\norm f,
157 \]
158 so the linear map is bounded. If $df=0$, every iterate equals $f$, hence $P_Af=f$. Its image is contained in $\ker d$ and it fixes that kernel, proving surjectivity onto its stated image and $P_A^2=P_A$. For $q=0$ the residual vanishes after the first update and the same inequalities remain valid.
159 \end{proof}
160 \source{KS19}
161 This theorem constructs the dynamics from \(A\). It does not construct a suitable \(A\) for every residual problem. That missing input can, however, be derived for a useful class of actual towers.
162 
163 \subsection{A right inverse from target splittings}
164 Suppose each target bonding map has a continuous linear section
165 \[
166  s_n:Y_n\to Y_{n+1},\qquad \rho^Y_ns_n=I,\qquad \norm{s_n}\le S.
167 \]
168 Assume the propagation gap
169 \begin{equation}\label{eq:splitgap}
170  SB_X<1,\qquad K=\frac{S}{1-SB_X}.
171 \end{equation}
172 For \(r\in R\), recursively set
173 \begin{equation}\label{eq:greenrecurrence}
174  (Ar)_0=0,\qquad
175  (Ar)_{n+1}=s_n\bigl(r_n+(Ar)_n\rho^X_n\bigr).
176 \end{equation}
177 
178 \begin{theorem}[Split tower correction]\label{KS20}
179 The recurrence defines a continuous linear map \(A:R\to E\) with
180 \[
181  \partial A=I_R,\qquad \norm A\le K.
182 \]
183 The operator \(P=I-A\partial\) is an exact projection and
184 \[
185  (Pf)_0=f_0,\qquad
186  \norm{f-Pf}_\infty\le K\norm{\partial f}_\infty.
187 \]
188 \end{theorem}
189 \begin{proof}
190 Let $\theta=SB_X<1$. The number $K=S/(1-\theta)$ is nonnegative and satisfies
191 \[
192  S+SB_XK=K.
193 \]
194 Fix $r\in R$ and define its coordinates recursively as in \eqref{eq:greenrecurrence}. The initial coordinate has norm zero. If $\norm{(Ar)_n}\le K\norm r_\infty$, submultiplicativity of operator norms gives
195 \[
196  \begin{aligned}
197  \norm{(Ar)_{n+1}}
198  &\le\norm{s_n}\bigl(\norm{r_n}+\norm{(Ar)_n}\norm{\rho^X_n}\bigr)\\
199  &\le S(1+KB_X)\norm r_\infty=K\norm r_\infty.
200  \end{aligned}
201 \]
202 Induction proves the estimate at every coordinate, so the recursively defined family actually belongs to the bounded product $E$.
203 
204 The coordinate recurrence preserves addition and scalar multiplication, starting from the zero initial coordinate. Induction therefore proves these laws for the family map $A$. Its uniform estimate proves $\norm A\le K$, giving a continuous linear map with the claimed domain and codomain.
205 
206 For every $n$, apply the target restriction to the recurrence. The equation $\rho^Y_ns_n=I$ gives
207 \[
208  \rho^Y_n(Ar)_{n+1}=r_n+(Ar)_n\rho^X_n,
209  \qquad (\partial Ar)_n=r_n.
210 \]
211 Coordinate equality yields the operator identity $\partial A=I_R$. In particular this identity holds for every bounded residual, not just residuals already known to lie in the image.
212 
213 Set $P=I-A\partial$. Then
214 \[
215  \partial P=\partial-\partial A\partial=0,\qquad
216  P^2=I-2A\partial+A(\partial A)\partial=I-A\partial=P.
217 \]
218 The expression $2A\partial$ abbreviates addition of the same operator twice and uses no division by two. Every exact $f$ is fixed because $\partial f=0$. Since $(Ar)_0=0$, the seed of $Pf$ is $f_0$. Finally $f-Pf=A\partial f$, and the bound on $A$ gives the asserted residual estimate. The denominators above are justified exactly by the strict propagation gap.
219 \end{proof}
220 \source{KS20}
221 The surjectivity is onto all bounded residual families. It is stronger than the assertion that some exact output can be assigned to each candidate. The gap and the chosen splittings are genuine inputs.
222 
223 \subsection{Protected information beyond the seed}
224 When the target grows, its new kernel directions are independent information. Define
225 \[
226  N_n(f)=(I-s_n\rho^Y_n)f_{n+1},\qquad
227  (p_Lf)_0=f_0,\quad(p_Lf)_{n+1}=N_n(f).
228 \]
229 Each \(N_n(f)\) lands in \(\ker\rho^Y_n\). The observable \(p_L\) retains both the seed and every new output component, relative to the chosen sections.
230 
231 \begin{theorem}[Split coordinate theorem]\label{KS21}
232 The observable is a bounded projection with \(p_LA=0\) and \(p_LP=p_L\). It separates exact candidates. If
233 \[
234  \mathcal S_L=\{s\in E:p_Ls=s\},
235 \]
236 then
237 \[
238  \Phi:E\simeq_L\mathcal S_L\times R,
239  \qquad \Phi(f)=(p_Lf,\partial f),
240  \qquad \Phi^{-1}(s,r)=Ps+Ar.
241 \]
242 Repair becomes \((s,r)\mapsto(s,0)\). It is the unique exact repair preserving \(p_Lf\).
243 \end{theorem}
244 \begin{proof}
245 Put \(B_n=I-s_n\rho^Y_n\). Since \(\rho^Y_ns_n=I\), direct multiplication gives
246 \[
247  B_n^2=B_n,\qquad \rho^Y_nB_n=0,\qquad B_ns_n=0.
248 \]
249 Moreover, \(B_ny=y\) whenever \(\rho^Y_ny=0\). Thus \(B_n\) really projects onto the target kernel. Its norm is at most \(1+SB_Y\). Coordinate zero of \(p_L\) is the identity, and every later coordinate applies \(B_n\), so \(p_L^2=p_L\) and \(\norm{p_L}\le1+SB_Y\). In particular the formula defines a map on the bounded product, not just on arbitrary coordinate families.
250 
251 The recurrence for \(Ar\) gives zero seed and places \((Ar)_{n+1}\) in the image of \(s_n\). The identities just proved make all its innovations zero. Therefore \(p_LA=0\), and
252 \[
253  p_LP=p_L(I-A\partial)=p_L.
254 \]
255 We next prove separation on the exact space. If \(f\) is exact, its restriction equation rewrites as
256 \[
257  f_{n+1}=s_n\rho^Y_nf_{n+1}+B_nf_{n+1}
258           =s_nf_n\rho^X_n+N_n(f).
259 \]
260 Two exact families with the same seed and innovations agree first at level zero and then at every later level by this recurrence. More generally, if two arbitrary families have the same observable and differential, their difference is exact with zero observable, so separation on the exact space makes their difference zero.
261 
262 The proposed chart has the claimed target because \(p_L(p_Lf)=p_Lf\). Given \(s\in\mathcal S_L\) and \(r\in R\), set \(f=Ps+Ar\). The preceding identities and \(\partial A=I\) give
263 \[
264  p_Lf=p_Ls=s,\qquad \partial f=r.
265 \]
266 This proves one round trip. For the other, \(f\) and \(P(p_Lf)+A\partial f\) have the same observable and residual, so the joint separation just proved identifies them. All maps are linear and bounded on their indicated subspaces and finite products, establishing a continuous linear equivalence.
267 
268 Finally, \(Pf\) has coordinates \((p_Lf,0)\). Any exact answer with the same observable has those same two coordinates, hence equals \(Pf\). This uniqueness is relative to the complete observable of seed and innovations. It is not obtained by assuming that the seed alone determines an exact growing-target family.
269 \end{proof}
270 \source{KS21}
271 This is an actual analytic coordinate equivalence, not generally an isometry. The protected coordinate depends on the declared splitting. It should not be called an intrinsic semantic coordinate without a further comparison theorem.
272 
273 \fixedplate{04_split}{In the split coordinate chart, a candidate separates into retained information and compatibility residual. The exact subspace is the zero residual slice. The chart and its protected coordinate depend on the chosen target sections.}{fig:split}
274 
275 \subsection{A growing example and the gap boundary}
276 Use \(X_n=Y_n=\R^{n+1}\) with maximum norms. Let \(\pi_n\) delete the last coordinate, take \(\rho^X_n=\tfrac12\pi_n\), \(\rho^Y_n=\pi_n\), and let \(s_n\) append zero. Then \(S=B_Y=1\), \(B_X=1/2\), and \(K=2\).
277 
278 \begin{example}[Repair of the identity family]\label{KS22}
279 The bounded candidate \(f_n=I\) is not exact. Its repair is
280 \[
281  (Pf)_n=\operatorname{diag}(2^{-n},2^{-(n-1)},\ldots,2^{-1},1).
282 \]
283 The new last output row is retained at every stage while inherited rows are propagated by the factor \(1/2\).
284 \end{example}
285 \begin{proof}
286 At level zero the value is one. Exactness forces every inherited row at the next level to be one half of its previous value. Preservation of the innovation forces the new row to remain the corresponding identity row. Induction gives the displayed diagonal.
287 \end{proof}
288 \source{KS22}
289 For constant real scalar identity towers, by contrast, \((\partial f)_n=f_{n+1}-f_n\). The constant residual \(r_n=1\) has no bounded primitive, since any primitive satisfies \(f_n=f_0+n\). Here \(SB_X=1\). This counterexample excludes discarding the gap indiscriminately. It does not prove that the gap is necessary for every individual tower. Exact repair preserving the initial coordinate still exists in this constant example, by replacing a candidate with its initial constant family. Surjectivity onto all bounded residuals does not follow from that repair.
290 \source{KS23}
291 
292 \subsection{What is canonical}
293 For fixed splitting and protected observable, the repair above is unique. Changing the splitting can change both the observable and the repaired answer. In the growing example, replace \(s_n(y)=(y,0)\) by \(s_n'(y)=(y,y_0)\). At the first positive level, repair of the identity becomes
294 \[
295  \begin{pmatrix}1/2&0\\-1/2&1\end{pmatrix}
296  \quad\text{instead of}\quad
297  \begin{pmatrix}1/2&0\\0&1\end{pmatrix}.
298 \]
299 Both are exact continuations for their respective innovation profiles and have the same seed. Thus a unique answer for a declared preservation problem is not an answer independent of all preservation choices.
300 \source{KS24}
301 
302 \input{chapters/03b_repair_transport}
303 
304 \chap{Multiscale residuals and the limits of reconstruction}\label{sec:cascade}
305 \subsection{A split coarsener retains its missing detail}
306 Let \(C:R\to\overline R\) and \(J:\overline R\to R\) be linear maps with \(CJ=I\). Put \(H=I-JC\). Then
307 \[
308  r=JCr+Hr,\qquad CH=0,\qquad H^2=H.
309 \]
310 The two coordinates \((Cr,Hr)\) retain all information. The coarse coordinate alone generally does not.
311 
312 The actual dyadic tower coarsener has the typed formula
313 \begin{equation}\label{eq:dyadic}
314  (Cr)_n=\rho^Y_{2n}r_{2n+1}+r_{2n}\rho^X_{2n+1}.
315 \end{equation}
316 It is obtained by telescoping two adjacent compatibility squares. With target sections, a lift is
317 \[
318  (Jc)_{2n}=0,\qquad (Jc)_{2n+1}=s_{2n}c_n.
319 \]
320 The section identity gives \(CJ=I\), so the missing detail is explicitly \(Hr\), not an unspecified reconstruction error. The normed construction imposes the bounds needed to make these maps bounded on its operator family spaces.
321 \source{KS25}
322 
323 \subsection{A whole cascade}
324 At scale \(j\), let \(C_j:R_j\to R_{j+1}\) and \(J_j:R_{j+1}\to R_j\) be bounded linear maps with \(C_jJ_j=I\). For \(x\in R_0\), define
325 \[
326  x_0=x,\quad x_{j+1}=C_jx_j,\quad
327  L_0=I,\quad L_N=J_0\cdots J_{N-1},
328 \]
329 \[
330  h_j=x_j-J_jx_{j+1},\qquad Q_Nx=L_Nx_N,
331  \qquad S_N=I-Q_N.
332 \]
333 The map \(Q_N\) restores the surviving coarse remainder to the original space. It is this restored remainder, not just \(x_N\), that controls reconstruction.
334 
335 \begin{theorem}[Finite reconstruction and nested remainders]\label{KS26}
336 For every \(N\),
337 \[
338  x=Q_Nx+\sum_{j<N}L_jh_j,
339  \qquad S_Nx=\sum_{j<N}L_jh_j.
340 \]
341 Each \(Q_N\) is a projection, and for \(N\le M\),
342 \[
343  Q_NQ_M=Q_MQ_N=Q_M.
344 \]
345 The vectors with all details zero are exactly the persistent subspace
346 \[
347  \mathcal P_\infty=\{x:Q_Nx=x\text{ for every }N\}.
348 \]
349 \end{theorem}
350 \begin{proof}
351 Let $D_0=I$ and $D_N=C_{N-1}\cdots C_0:R_0\to R_N$. Thus $x_N=D_Nx$, $Q_N=L_ND_N$, and repeated use of $C_jJ_j=I$ gives
352 \[
353  D_NL_N=I_{R_N}.
354 \]
355 This cancellation takes place between adjacent maps with matching domains. No identification of different residual spaces is needed.
356 
357 Since $L_{j+1}=L_jJ_j$ and $x_{j+1}=C_jx_j$,
358 \[
359  L_jh_j=L_j(x_j-J_jx_{j+1})=Q_jx-Q_{j+1}x.
360 \]
361 Summing from $j=0$ to $N-1$ leaves $x-Q_Nx$ because $Q_0=I$. This proves both finite reconstruction identities, including $N=0$, when the detail sum is empty.
362 
363 Idempotence follows from $D_NL_N=I$:
364 \[
365  Q_N^2=L_N(D_NL_N)D_N=Q_N.
366 \]
367 For $M\ge N$, put $C_{N,M}=C_{M-1}\cdots C_N$ and $J_{N,M}=J_N\cdots J_{M-1}$, using identities for $M=N$. Then
368 \[
369  D_M=C_{N,M}D_N,\quad L_M=L_NJ_{N,M},\quad C_{N,M}J_{N,M}=I.
370 \]
371 The identities $D_NL_M=J_{N,M}$ and $D_ML_N=C_{N,M}$ now give
372 \[
373  Q_NQ_M=L_NJ_{N,M}D_M=Q_M,\qquad
374  Q_MQ_N=L_MC_{N,M}D_N=Q_M.
375 \]
376 Both orders have been checked, rather than inferred from one range inclusion.
377 
378 If all $h_j$ vanish, the finite telescope gives $Q_Nx=x$ for every $N$. Conversely, if every $Q_Nx=x$, then $L_jh_j=Q_jx-Q_{j+1}x=0$. Apply the left inverse $D_j$ of $L_j$ to obtain $h_j=0$. The injectivity of the restoration composite is the step that recovers the original detail from its transported value.
379 \end{proof}
380 \source{KS26}
381 
382 \fixedplate{05_cascade}{A split cascade retains every discarded component as a detail. The finite sum is always exact after the restored coarse remainder is included. An infinite detail expansion requires that remainder to vanish in the topology being used.}{fig:cascade}
383 
384 \subsection{Convergence has its own theorem}
385 Call \(x\) norm recoverable when \(Q_Nx\to0\). This is equivalent to \(S_Nx\to x\), but it does not follow merely from vanishing persistent subspace. Let
386 \[
387  \mathcal F=\bigcup_N\ker Q_N
388 \]
389 be the finite core.
390 
391 \begin{theorem}[Density criterion for norm recovery]\label{KS27}
392 Suppose \(\sup_N\norm{Q_N}\le K<\infty\). Then the norm recoverable vectors are exactly \(\overline{\mathcal F}\). In particular all vectors are norm recoverable if and only if the finite core is dense.
393 \end{theorem}
394 \begin{proof}
395 The nested identities first imply that the kernels increase: if $Q_My=0$ and $N\ge M$, then $Q_Ny=Q_NQ_My=0$. Thus the finite core really is an increasing union of linear subspaces.
396 
397 Suppose $Q_Nx\to0$. Put $y_N=S_Nx=x-Q_Nx$. Idempotence gives $Q_Ny_N=Q_Nx-Q_N^2x=0$, so $y_N\in\mathcal F$. Since $x-y_N=Q_Nx\to0$, these are finite-core approximants converging to $x$. Hence $x\in\overline{\mathcal F}$.
398 
399 Conversely, fix $x\in\overline{\mathcal F}$ and $\epsilon>0$. If $K=0$, the common operator bound gives $\norm{Q_Nx}=0$ for every $N$, so convergence is immediate. Suppose $K>0$. Choose $y\in\mathcal F$ satisfying $\norm{x-y}<\epsilon/(K+1)$. Some $M$ has $Q_My=0$. For every $N\ge M$, nesting and the common operator bound give
400 \[
401  \norm{Q_Nx}=\norm{Q_N(x-y)}\le K\norm{x-y}
402  <\epsilon\frac{K}{K+1}\le\epsilon.
403 \]
404 The approximation $y$ is chosen before the depth $N$, giving precisely the quantifier order in norm convergence. This proves equality of the two sets. Applying it to every $x\in R_0$ gives the density criterion.
405 \end{proof}
406 \source{KS27}
407 There are other sufficient conditions. If the transported details are absolutely summable in the Banach space \(R_0\), their sum exists and the remaining boundary lies in \(\mathcal P_\infty\). If that persistent subspace is zero, the sum equals \(x\). Alternatively,
408 \[
409  \norm{x-S_Nx}\le\left(\prod_{j<N}\norm{J_j}\right)\norm{x_N}
410 \]
411 gives a direct sufficient decay test. A decreasing coarse norm by itself is not enough when restoration gains grow.
412 
413 \subsection{The actual dyadic result}
414 For the dyadic residual hierarchy defined above, the persistent residual is zero. Zero detail at a scale reveals every even coordinate, while an odd coordinate is passed to a smaller index at the next scale. Strong induction on the coordinate index therefore proves that a residual with all details zero is zero.
415 
416 \begin{theorem}[Dyadic determination and finite coordinate recovery]\label{KS28}
417 In the actual dyadic system, all detail coordinates jointly determine the residual. Moreover,
418 \[
419  k+1<2^N\quad\Longrightarrow\quad
420  (Q_Nr)_k=0,\qquad(S_Nr)_k=r_k.
421 \]
422 Thus every fixed residual coordinate is recovered exactly at a sufficiently large finite depth. The statement does not assert convergence in the supremum norm over all coordinates.
423 \end{theorem}
424 \begin{proof}
425 Write \(R_j\) for the residual space at scale \(j\). Its lift has the actual coordinate formula
426 \[
427  (J_jz)_{2m}=0,\qquad (J_jz)_{2m+1}=s^{(j)}_{2m}z_m,
428 \]
429 where \(s^{(j)}_{2m}\) is the corresponding target section at that scale. The expression on the right is composition of the section with the residual operator. Its source and target are those of the odd coordinate at scale \(j\).
430 
431 Set \(L_{j,0}=I_{R_j}\) and \(L_{j,N+1}=J_jL_{j+1,N}\). We prove, simultaneously for all starting scales \(j\), that
432 \[
433  k+1<2^N\quad\Longrightarrow\quad (L_{j,N}z)_k=0
434  \qquad(z\in R_{j+N}).
435 \]
436 For \(N=0\), the inequality \(k+1<1\) has no solution. Suppose the assertion holds at depth \(N\). At depth \(N+1\), an even coordinate vanishes by the displayed lift formula. For an odd coordinate \(k=2m+1\), the hypothesis gives
437 \[
438  2(m+1)<2^{N+1},\qquad m+1<2^N.
439 \]
440 The induction hypothesis at starting scale \(j+1\) makes \((L_{j+1,N}z)_m\) zero. Composing that zero operator with \(s^{(j)}_{2m}\) gives zero at coordinate \(k\). This completes the induction without identifying residual spaces at different scales.
441 
442 Now \(Q_Nr=L_{0,N}x_N\), so the support assertion applies to the restored remainder. Since \(S_Nr=r-Q_Nr\), it gives the reconstruction equation. Finally, if two residuals have identical details at every scale, the finite telescope makes their \(S_N\) reconstructions equal for every \(N\). At any fixed coordinate choose \(N\) with \(k+1<2^N\). The reconstruction equation then makes their original coordinates equal. Hence the residuals are equal.
443 \end{proof}
444 \source{KS28}
445 The strict inequality matters: the next coordinate, \(k=2^N-1\), can still have a nonzero remainder. There is also a sharper, nonconsecutive description of the support.
446 
447 \begin{proposition}[The dyadic support pattern]\label{KS55}
448 For the same lifts,
449 \[
450  2^N\nmid(k+1)\quad\Longrightarrow\quad(Q_Nr)_k=0.
451 \]
452 Thus coordinate \(k\) is certainly recovered once \(N>\nu_2(k+1)\), where \(\nu_2(m)\) is the largest power-of-two exponent dividing the positive integer \(m\). The only possible unrecovered coordinates at depth \(N\) are
453 \[
454  2^N-1,\quad 2\cdot2^N-1,\quad3\cdot2^N-1,\quad\ldots.
455 \]
456 \end{proposition}
457 \begin{proof}
458 Prove the assertion for \(L_{j,N}\) by induction, uniformly in \(j\). At depth zero every positive integer is divisible by one, so there is nothing to prove. At depth \(N+1\), even coordinates vanish. If \(k=2m+1\), then
459 \[
460  2^{N+1}\mid(k+1)\quad\Longleftrightarrow\quad2^N\mid(m+1).
461 \]
462 When divisibility fails, the inner coordinate in \(J_jL_{j+1,N}\) vanishes by induction. The formula for the odd lift finishes the proof. Apply the result to \(Q_Nr=L_{0,N}x_N\).
463 \end{proof}
464 \source{KS55}
465 This refinement is formalized by a scale-uniform induction on the actual varying-fiber residual system in \code{DyadicSupport.lean}. It asserts where a remainder \emph{may} survive, not that every allowed coordinate is nonzero. In the scalar identity-bonding model with \(C(a,b)=a+b\), the residual constantly equal to one has value \(2^N\) at every surviving position. That example proves sharpness of the uniform prefix cutoff.
466 
467 \plate{16_dyadic_support}{The possible support of the restored dyadic remainder is sparse and periodic. At depth \(N\), only positions whose index plus one is divisible by \(2^N\) can remain. Open cells are guaranteed recovered. Marked cells are allowed to survive, not asserted nonzero for every residual.}{fig:dyadicsupport}
468 
469 \subsection{Two boundary examples}
470 On \(\ell^\infty(\N,\R)\), let \(C\) remove the first coordinate and let \(J\) prepend zero. Then \(Q_Nx\) is the tail of \(x\), placed at its original coordinates with zeros before \(N\). Every fixed coordinate is eventually zero in the remainder, and \(\norm{Q_N}\le1\). Nevertheless the constant sequence \(\mathbf1\) satisfies \(\norm{Q_N\mathbf1}=1\) for every \(N\).
471 
472 \begin{proposition}[The recovery domain is \(c_0\)]\label{KS29}
473 For this shift cascade,
474 \[
475  S_Nx\to x\text{ in }\ell^\infty
476  \quad\Longleftrightarrow\quad x_n\to0.
477 \]
478 For \(x_n=q^n\), \(0\le q<1\), the error is exactly \(q^N\).
479 \end{proposition}
480 \begin{proof}
481 The error norm is \(\sup_{n\ge N}|x_n|\). It tends to zero precisely when the sequence tends to zero. For a geometric sequence its largest tail value is the first one, \(q^N\).
482 \end{proof}
483 \source{KS29}
484 As a second example, take scalar spaces with \(C=\tfrac12I\), \(J=2I\). The coarse value \(x_N=2^{-N}x\) tends to zero, but restoration gives \(Q_Nx=x\). Every detail is zero and the entire input is persistent. The example belongs to this scalar cascade, not to the actual dyadic hierarchy, whose persistent subspace was proved to vanish.
485 
486 \begin{insight}
487 Retained packets make omission explicit. They do not guarantee convergence. Coordinate recovery, bounded reconstruction, summability, and attainment remain different mathematical properties.
488 \end{insight}

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