A complete regularization diagnostic for the escaping model

Proof

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

proves \(T^*v^{(m)}=\alpha a^{(m)}\) as an equality of Hilbert vectors. Theorem 29.1 therefore identifies this actual infinite vector as the unique minimizer, rather than just a coordinatewise candidate.

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

Thus \(A_m\le m+m^4/(3m^3)=4m/3\).

For \(n\le m\), the residual square is at most \(n^2/m^4\). The sum-of-squares estimate

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 \(B_m\le2/m\).

For the lower bound, \(n>m\) implies \(m^2+n^2\le2n^2\), hence

The other direction of the decreasing-function integral comparison gives

This proves \(B_m\ge1/(4(m+1))\).

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 (30.2). No finite truncation is substituted for an equality of infinite energies.

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Proof

Step 1: control the infinite Riemann sums uniformly. Define

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

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 \([0,L]\).

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

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.

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

Step 1 now proves the two stated asymptotics.

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

Proposition 29.2 identifies \(\beta(M_m)=\sqrt{B_m}\). Thus

For every sufficiently large \(M\), choose \(m\) with \(M_m\le M\le M_{m+1}\). Monotonicity of \(\beta\) gives

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.

Step 4: compare the hard-prefix method. That correction has norm \(\sqrt N\) and residual square \(\sum_{n>N}n^{-2}\). Integral comparison yields

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.

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Proof

The exact-data bias has coordinates \(n/(m^2+n^2)\), so its squared norm is precisely \(B_m\). By Proposition 30.1,

Theorem 29.4 at the same parameter gives

The triangle inequality proves the first bound.

Because \(0<\delta\le1\), the number \(u=\delta^{-2/3}\) is at least one. The ceiling satisfies \(u\le m<u+1\le2u\). Therefore

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.

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