Realization Calculus: twin edition

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1 \chap{A complete regularization diagnostic for the escaping model}\label{sec:diagonal-regularization}
2 The abstract alternatives become explicit for the diagonal model. They also answer a natural objection to the hard prefix corrections: perhaps a smoother correction could fit increasingly well without escaping. It can fit more efficiently, but it cannot avoid the obstruction.
3 
4 \subsection{The entire regularized family is explicit}
5 Use $Tu=(u_n/n)$ and $r=w=(1/n)$ as in Section~\ref{sec:escape}. Choose $\alpha=m^{-2}$ for an integer $m\ge1$. The normal equation in each coordinate gives
6 \[
7  a^{(m)}_n=\frac{m^2}{m^2+n^2},\qquad
8  v^{(m)}_n=r_n-(Ta^{(m)})_n=\frac{n}{m^2+n^2}.
9 \]
10 The repaired ambient candidate is $(-a^{(m)},1)$. Its protected coordinate is one for every $m$.
11 
12 \begin{proposition}[Energy, residual, and certified tails]\label{WR10}
13 The displayed $a^{(m)}$ belongs to $\ell^2$ and is the unique regularized correction at $\alpha=m^{-2}$. Set
14 \[
15  A_m=\norm{a^{(m)}}^2=\sum_{n\ge1}\frac{m^4}{(m^2+n^2)^2},\qquad
16  B_m=\norm{v^{(m)}}^2=\sum_{n\ge1}\frac{n^2}{(m^2+n^2)^2}.
17 \]
18 Then
19 \begin{equation}\label{eq:soft-bounds}
20  \frac m4\le A_m\le\frac{4m}{3},\qquad
21  \frac1{4(m+1)}\le B_m\le\frac2m.
22 \end{equation}
23 For the sums $A_{m,K},B_{m,K}$ through any integer $K\ge m$,
24 \begin{equation}\label{eq:soft-tails}
25  0\le A_m-A_{m,K}\le\frac{m^4}{3K^3},\qquad
26  0\le B_m-B_{m,K}\le\frac1K.
27 \end{equation}
28 \end{proposition}
29 \begin{proof}
30 For $n>m$, one has $(a_n^{(m)})^2\le m^4/n^4$, so the correction sequence is square-summable. The diagonal operator is bounded with norm one, and $w=(1/n)\in\ell^2$, hence the residual sequence is also in $\ell^2$. The operator is self-adjoint, and the coordinate equality
31 \[
32  \frac1n\frac{n}{m^2+n^2}
33  =m^{-2}\frac{m^2}{m^2+n^2}
34 \]
35 proves $T^*v^{(m)}=\alpha a^{(m)}$ as an equality of Hilbert vectors. Theorem~\ref{WR06} therefore identifies this actual infinite vector as the unique minimizer, rather than just a coordinatewise candidate.
36 
37 For $1\le n\le m$, $m^2/(m^2+n^2)\ge1/2$. There are $m$ such coordinates, so $A_m\ge m/4$. For the upper bound, the first $m$ squares are at most one, and the decreasing function $x^{-4}$ gives
38 \[
39  \sum_{n>m}\frac1{n^4}\le\int_m^\infty x^{-4}\,dx=\frac1{3m^3}.
40 \]
41 Thus $A_m\le m+m^4/(3m^3)=4m/3$.
42 
43 For $n\le m$, the residual square is at most $n^2/m^4$. The sum-of-squares estimate
44 \[
45  \sum_{n=1}^m n^2=\frac{m(m+1)(2m+1)}6\le m^3
46 \]
47 holds for $m\ge1$, so this part contributes at most $1/m$. For $n>m$, each residual square is at most $1/n^2$, whose tail is at most $1/m$. This proves $B_m\le2/m$.
48 
49 For the lower bound, $n>m$ implies $m^2+n^2\le2n^2$, hence
50 \[
51  \frac{n^2}{(m^2+n^2)^2}\ge\frac1{4n^2}.
52 \]
53 The other direction of the decreasing-function integral comparison gives
54 \[
55  \sum_{n>m}\frac1{n^2}\ge\int_{m+1}^\infty x^{-2}\,dx=\frac1{m+1}.
56 \]
57 This proves $B_m\ge1/(4(m+1))$.
58 
59 Finally, for $K\ge m$, both omitted tails consist of nonnegative terms. Bound the first by $m^4\sum_{n>K}n^{-4}$ and the second by $\sum_{n>K}n^{-2}$. The same upper integral comparisons give the two certified remainders in \eqref{eq:soft-tails}. No finite truncation is substituted for an equality of infinite energies.
60 \end{proof}
61 \written{WR10}
62 
63 Thus the correction norm grows on the order of $\sqrt m$, the residual norm decays on the order of $m^{-1/2}$, and the penalty $\alpha A_m=A_m/m^2$ tends to zero. The smooth correction avoids the hard discontinuity at a prefix boundary, but it still approaches the forbidden constant sequence coordinate by coordinate.
64 
65 \fixedplate{35_regularized_escape}{At each fixed coordinate, the soft correction approaches one as $m$ tends to infinity. Its transition moves outward as $m$ increases. The chart compares three exact coordinate formulas $m^2/(m^2+n^2)$, not finite approximations to an unspecified optimization procedure. Every displayed sequence has finite energy, but the family's energies are unbounded.}{fig:regularized-escape}
66 
67 \subsection{The sharp asymptotic correction cost}
68 The preceding bounds establish divergence and decay. The leading constant can also be computed, which permits a meaningful comparison with hard prefix repair.
69 
70 \begin{theorem}[A sharp budget asymptotic for the diagonal datum]\label{WR11}
71 As $m\to\infty$,
72 \[
73  \frac{A_m}{m}\longrightarrow\frac\pi4,\qquad
74  mB_m\longrightarrow\frac\pi4.
75 \]
76 For the exact budget profile of this infinite problem,
77 \begin{equation}\label{eq:sharp-budget}
78  \lim_{M\to\infty}M\beta_{T,w}(M)=\frac\pi4.
79 \end{equation}
80 The hard prefix correction at budget $M=\sqrt N$ instead has
81 \[
82  \sqrt N\,\norm{w-T\mathbf1_N}\longrightarrow1.
83 \]
84 \end{theorem}
85 \begin{proof}
86 \emph{Step 1: control the infinite Riemann sums uniformly.} Define
87 \[
88  f(x)=\frac1{(1+x^2)^2},\qquad
89  g(x)=\frac{x^2}{(1+x^2)^2}.
90 \]
91 Then $A_m/m=m^{-1}\sum_{n\ge1}f(n/m)$ and $mB_m=m^{-1}\sum_{n\ge1}g(n/m)$. Fix $L\ge2$ and put $N=\lfloor mL\rfloor$. On $[0,L]$, both functions are uniformly continuous. If $\omega_f(h)$ is a modulus of continuity there, comparison on the $N$ intervals of length $1/m$ gives
92 \[
93  \left|\frac1m\sum_{n=1}^N f(n/m)-\int_0^{N/m}f(x)\,dx\right|
94  \le L\omega_f(1/m).
95 \]
96 The missing interval $[N/m,L]$ has length below $1/m$ and a bounded integrand, so it contributes a term tending to zero. The same argument applies to $g$. This proves convergence of each truncated sum to its integral on $[0,L]$.
97 
98 The tails require a bound independent of sufficiently fine discretization. Since $N\ge mL/2$, the inequalities $f(x)\le x^{-4}$ and $g(x)\le x^{-2}$ give
99 \[
100  \frac1m\sum_{n>N}f(n/m)
101  \le m^3\sum_{n>N}n^{-4}\le\frac{m^3}{3N^3}
102  \le\frac8{3L^3},
103 \]
104 \[
105  \frac1m\sum_{n>N}g(n/m)
106  \le m\sum_{n>N}n^{-2}\le\frac mN\le\frac2L.
107 \]
108 The integral tails are at most $1/(3L^3)$ and $1/L$. Given an error tolerance, choose $L$ first to control both discrete and integral tails, then choose $m$ to control the compact-interval sums. This justifies passing from finite-interval Riemann convergence to the full half-line sums. No uncontrolled interchange of two infinite limits is involved.
109 
110 \emph{Step 2: evaluate the integrals.} The substitution $x=\tan\theta$ gives $dx=\sec^2\theta\,d\theta$ and $1+x^2=\sec^2\theta$, so
111 \[
112  \int_0^\infty f(x)\,dx=\int_0^{\pi/2}\cos^2\theta\,d\theta=\frac\pi4,
113  \qquad
114  \int_0^\infty g(x)\,dx=\int_0^{\pi/2}\sin^2\theta\,d\theta=\frac\pi4.
115 \]
116 Step 1 now proves the two stated asymptotics.
117 
118 \emph{Step 3: move from regularization parameters to arbitrary budgets.} Set $M_m=\sqrt{A_m}$. Every positive coordinate $m^2/(m^2+n^2)$ increases strictly with $m$, so $A_m$, and hence $M_m$, increases strictly. Since $A_m/m\to c=\pi/4>0$, these budgets tend to infinity and
119 \[
120  \left(\frac{M_{m+1}}{M_m}\right)^2
121  =\frac{A_{m+1}/(m+1)}{A_m/m}\frac{m+1}{m}\longrightarrow1.
122 \]
123 Proposition~\ref{WR07} identifies $\beta(M_m)=\sqrt{B_m}$. Thus
124 \[
125  M_m\beta(M_m)=\sqrt{(A_m/m)(mB_m)}\longrightarrow c.
126 \]
127 For every sufficiently large $M$, choose $m$ with $M_m\le M\le M_{m+1}$. Monotonicity of $\beta$ gives
128 \[
129  M_m\beta(M_{m+1})\le M\beta(M)\le M_{m+1}\beta(M_m).
130 \]
131 The lower bound is $(M_m/M_{m+1})[M_{m+1}\beta(M_{m+1})]$ and the upper is $(M_{m+1}/M_m)[M_m\beta(M_m)]$. Both tend to $c$, proving the full budget limit, not merely convergence on a selected subsequence of budgets.
132 
133 \emph{Step 4: compare the hard-prefix method.} That correction has norm $\sqrt N$ and residual square $\sum_{n>N}n^{-2}$. Integral comparison yields
134 \[
135  \frac{N}{N+1}\le N\sum_{n>N}\frac1{n^2}\le1.
136 \]
137 Taking the limit and then a square root gives its leading product one. The two constants are consequently compared at large actual correction budgets, rather than at arbitrarily equated parameter labels.
138 \end{proof}
139 \written{WR11}
140 
141 At large matched correction budgets, the optimal soft repair therefore has about $\pi/4$ times the hard-prefix residual. This improves the efficiency of approximation, not the existence result. Both residuals tend to zero only as the budget tends to infinity. The limiting constant is a calculation for this declared model, not a universal regularization efficiency factor.
142 
143 \fixedplate{36_budget_product}{The plotted intervals enclose the actual infinite quantities $\sqrt{A_mB_m}$, using finite sums plus the proved tails. The corresponding hard-prefix intervals enclose $\sqrt{N\sum_{n>N}n^{-2}}$ with $N=m$. These have different natural budget sequences. Their limits give the matched-budget asymptotic constants $\pi/4$ and one.}{fig:budget-product}
144 
145 \subsection{The same operator has an attainable datum}
146 Change only the residual to $b_n=1/n^2$. Its minimum norm preimage is $a^\dagger_n=1/n\in\ell^2$. At $\alpha=m^{-2}$, the regularized correction is
147 \[
148  c^{(m)}_n=\frac{m^2}{n(m^2+n^2)},\qquad
149  a^\dagger_n-c^{(m)}_n=\frac{n}{m^2+n^2}.
150 \]
151 The bias square is exactly the previously computed $B_m$. This links the two cases without conflating their existence hypotheses: the escaping residual of one problem is the regularization bias of an attainable problem.
152 
153 \begin{corollary}[An explicit noisy attainable repair]\label{WR12}
154 For the attainable datum $b=(1/n^2)$, suppose $\norm{\widetilde b-b}\le\delta$ with $0<\delta\le1$. Choose
155 \[
156  m=\lceil\delta^{-2/3}\rceil,\qquad\alpha=m^{-2}.
157 \]
158 Then the noisy regularized correction satisfies
159 \[
160  \norm{a_\alpha(\widetilde b)-a^\dagger}
161  \le\sqrt{\frac2m}+\frac{\delta m}{2}
162  \le(\sqrt2+1)\delta^{1/3}.
163 \]
164 \end{corollary}
165 \begin{proof}
166 The exact-data bias has coordinates $n/(m^2+n^2)$, so its squared norm is precisely $B_m$. By Proposition~\ref{WR10},
167 \[
168  \norm{a_{m^{-2}}(b)-a^\dagger}\le\sqrt{2/m}.
169 \]
170 Theorem~\ref{WR09} at the same parameter gives
171 \[
172  \norm{a_{m^{-2}}(\widetilde b)-a_{m^{-2}}(b)}
173  \le\frac{\delta}{2\sqrt{m^{-2}}}=\frac{\delta m}{2}.
174 \]
175 The triangle inequality proves the first bound.
176 
177 Because $0<\delta\le1$, the number $u=\delta^{-2/3}$ is at least one. The ceiling satisfies $u\le m<u+1\le2u$. Therefore
178 \[
179  \sqrt{2/m}\le\sqrt2\,\delta^{1/3},
180  \qquad \frac{\delta m}{2}\le\delta u=\delta^{1/3}.
181 \]
182 Adding these estimates proves the displayed rate with every ceiling factor accounted for. The positive-noise condition is needed to define this particular finite integer choice. The statement does not prescribe an infinite $m$ at $\delta=0$, nor does it use a probabilistic assumption on the perturbation.
183 \end{proof}
184 \written{WR12}
185 
186 This is a deterministic guarantee for the specified datum and Hilbert noise bound. It is not a statistical confidence statement. The identical parameter rule in the escaping problem cannot produce convergence to an exact protected Hilbert solution, because no such solution exists.
187 
188 \subsection{What the calculation permits a machine to certify}
189 The finite program can check a normal equation, exact preservation, and a quadratic optimality identity. It can return a dual witness ruling out a smaller residual at a given budget. In the infinite diagonal example it can return partial sums together with rational tail enclosures. Those are distinct certificates with distinct conclusions.
190 
191 None of them alone establishes unrestricted semantic realizability. The bounded-budget theorem needs a genuinely uniform bound. The exactification theorem needs control of candidate movement. The regularization limit needs actual attainment of the projected datum. The repeated appearance of a small residual does not replace any of those hypotheses.
192 
193 The useful next diagnostic is therefore not just ``Did the residual decrease?'' It is ``What happened to the correction norm, what quantity was preserved, and what existence theorem applies to this datum?''

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