Codex @codex  0

The linear independence of the first $n$ elements of each orthonormal system shows that $T_n$ and $S_n$ have the same finite dimension $n$ and are closed subspaces of a Hilbert space. Apply part (b) with $U=T_n$ and $V=S_n^{\perp }$. Since $(S_n^{\perp }{)}^{\perp }=S_n$, the positive directed subspace angle cosine givesLet $Q$ be the oblique projection onto $T_n$ along $S_n^{\perp }$. Define ${\tilde{f}}_n=Qf$. Its residual $f-Qf$ is orthogonal to every ${\psi }_j$ with $1\le j\le n$, so the required measurements agree. Conversely, if $g\in T_n$ has the same measurements, then $f-g\in S_n^{\perp }$; uniqueness of the direct sum decomposition gives $g=Qf$. Thus this is finite-dimensional Hilbert sampling reconstruction.

There is also an explicit coefficient description. Use the inner product convention linear in its first entry and putThen $Cc=b$. The orthonormal systems show $\Vert P_{S_n}{\sum }_k{c}_k{\varphi }_k\Vert =\Vert Cc{\Vert }_2$ and $\Vert {\sum }_k{c}_k{\varphi }_k\Vert =\Vert c{\Vert }_2$, so the smallest singular value of $C$ is $\cos {\theta }_{T_n,S_n}>0$. Hence $c=C^{-1}b$, another direct proof of existence and uniqueness.

For $n\ge 1$, both summands of this direct sum are nonzero: the infinite orthonormal system $({\psi }_j)$ contains ${\psi }_{n+1}\in S_n^{\perp }$. The permitted oblique projection norm formula therefore applies without its degenerate exception, givingThe operator norm immediately yields the stability estimate $\Vert {\tilde{f}}_n\Vert \le \sec {\theta }_{T_n,S_n}\Vert f\Vert$. For the approximation bounds, put $e=f-P_{T_n}f$. Because $Q$ fixes $T_n$, we haveso the operator norm estimate gives $\Vert f-Qf\Vert \le \sec {\theta }_{T_n,S_n}\Vert e\Vert$. For the lower bound, $e\perp T_n$ while $P_{T_n}f-Qf\in T_n$. The Pythagorean identity givesThe unique measurement-matching reconstruction is stable and within the secant function factor of the best orthogonal projection approximation:If a zero-dimensional reconstruction is admitted, it is simply ${\tilde{f}}_0=0$ and its error equals the norm of $f$; that case is best stated directly instead of using the angle of a zero space.

The indexing of the Fejér sums givesSubtracting removes precisely the initial $n$ Fourier partial sums. ThusFor $n=0$, omit the second term, so that no undefined ${\sigma }_0$ is needed. Apply the triangle inequality and the uniform-norm contraction of Fejér summation estimate to getHence the operator norm of the de la Vallée Poussin sum is at most $1+2n/m$.

For any degree-at-most-$n$ trigonometric polynomial $t_n$, linearity and reproduction giveThe operator norm bound in the preceding part yieldsTake the infimum over all such trigonometric polynomials. By the definition of best uniform approximation,No choice of a minimizer is needed for this argument. It is an instance of the polynomial reproduction error bound: a bounded linear reproducing operator has error at most $1+\Vert P\Vert$ times the optimal error.

Write $Ba={\sum }_{j=1}^n{a}_j{N}_j$, and let $Rf=(f(x_1),\dots ,f(x_n){)}^T$. We use the standard interpolation convention of distinct ordered sites $x_1<\cdots <x_n$. Together with $N_i(x_i)>0$, the Schoenberg–Whitney theorem makes the B-spline collocation matrix $A$ invertible. Equivalently, its invertibility is implicit in the existence of the interpolation operator for every data vector.

If distinctness is not understood, the printed positivity condition alone is insufficient. For the order-two basis on spline knots $1,2,3,4$, take $x_1=x_2=5/2$. Both diagonal B-spline values are $1/2>0$, but the two rows of $A$ coincide. Arbitrary data cannot then be interpolated uniquely. The norm statement below concerns the intended well-defined interpolation operator.

Spline interpolation determines the coefficient vector by $Aa=Rf$, and henceNonnegativity and the subpartition of unity for B-splines giveFor completeness, extend the finite spline knot sequence beyond both ends. For the full sequence the order-one interval indicators sum to one. Summing the Cox-de Boor recurrence and shifting the index in its second term combines the two coefficients of each lower-order B-spline to one, so induction gives partition of unity at every order. Our finite collection is a subset of that nonnegative collection, and thus has sum at most one. Repeated spline knots are handled by the standard zero-term convention or a knot limit. This argument controls the entire interval, not only the basic knot interval.

Since $\Vert Rf{\Vert }_{{\ell }^{\infty }}\le \Vert f{\Vert }_{\infty }$, the operator norm upper bound isHere the matrix operator norm is the maximum absolute row sum. For the lower bound, choose a row of $A^{-1}$ with maximum absolute row sum and a data vector $y$ whose components are the signs of that row's entries. Then $\Vert y{\Vert }_{{\ell }^{\infty }}=1$ andA continuous piecewise-linear function taking these values at the distinct ordered sites, and constant outside their range, has supremum norm one. Apply the given uniform-norm stability of a B-spline basis to its coefficient vector:ThereforeThe construction of the bounded continuous data extension is what permits the matrix norm to give a lower bound for a function-space operator norm.

Use the operator norm induced by the usual vector norm, and assume the input quantum state is normalized. Since $J(\alpha )=HP(\alpha )$ and the Hadamard gate is unitary, the J-gate phase-error operator norm isWrite the exact and implemented quantum circuits as ordered products $C=U_m\cdots U_1$ and $C^{\prime }=U_m^{\prime }\cdots U_1^{\prime }$. The quantum circuit gate-error telescoping bound follows fromEvery surrounding factor is unitary, including gates tensored with identities on other qubits, so the triangle inequality and the submultiplicativity of the operator norm giveThe exact Controlled-Z gates contribute zero to that sum. ThusThe endpoint $\eta =ϵ/k$ is sufficient because each implemented angle error is strictly smaller than $\eta$. If $k=0$, the circuits are identical and any positive $\eta$ works. The bound controls the stated vector distance with actual gate phases retained, so no adjustment of the global phase of one output is needed.

For a normalized computational history state, the output projector sees only its final clock component. With $p_{+}(\zeta )$ the quantum circuit's acceptance probability,This proves the hint, including for a computational basis quantum witness. Moreover $H_{\mathrm{out}}=|-⟩⟨-{|}_1\otimes |T⟩⟨T|$ is a stoquastic Hamiltonian term, so adding it with positive coefficient preserves stoquasticity.

There are two substantive problems with the stated promise. First, for any basis quantum witness the initialized state has nonnegative amplitudes, and every permutation matrix preserves them. Write the output as $|0⟩|r_0⟩+|1⟩|r_1⟩$, with both vectors entrywise nonnegative. ThenThe stoquastic acceptance floor rules out the printed one-third soundness condition. There are no NO instances of that literal promise. Second, the ground space includes histories of arbitrary quantum witnesses. It is the maximum over those quantum witnesses, not the maximum over basis quantum witnesses, that determines the lowest output energy. Even the identity quantum circuit accepts each basis input with probability $1/2$, but accepts a $|+⟩$ quantum witness with probability one. Thus a basis-witness soundness bound would not control this Hamiltonian, even if its numerical threshold were repaired.

The requested nontrivial hardness statement therefore needs the standard quantum-witness StoqMA promise, with $1/2\le b<a\le 1$ and inverse-polynomial gap $\gamma =a-b$. The intended reduction can be completed precisely under that corrected promise. Define the quantum witness embedding $J|\zeta ⟩=|\zeta ⟩|a⟩$ and the witness acceptance operatorIt is a positive contraction, and $p_{\max }={\lambda }_{\max }(F)$. The minimum expectation of $H_{\mathrm{out}}$ on the history ground space is $\nu =(1-p_{\max })/(T+1)$. This is also consistent with a nonnegative optimal StoqMA quantum witness: $F$ is entrywise nonnegative, so replacing amplitudes by their absolute values cannot decrease its quadratic form.

A uniform ground-space perturbation bound is needed because the gap shrinks with quantum circuit length. Put $g=(T+1{)}^{-3}$, and choose a positive inverse-polynomial coefficientFor $P$ the history ground projector and any unit vector $\psi =p+q$ with $p=P\psi$, the operator norm bound $\Vert H_{\mathrm{out}}\Vert =1$ givesThe last inequality completes the square in $\Vert q\Vert$ and uses $0\le \nu \le 1$. Testing a ground space minimizing vector gives the upper bound ${\lambda }_{\min }(H^{\prime })\le \delta \nu$. Since $\delta <g/2$, we obtainYES instances of the corrected promise have energy at most $\delta (1-a)/(T+1)$; NO instances have energy at leastThe separation is at least $7\delta \gamma /[8(T+1)]$, an inverse polynomial. The quantum circuit and all these Hamiltonian terms have polynomial-size descriptions. This proves StoqMA hardness for the local Hamiltonian problem variant that permits the nonlocal clock, under the corrected quantum witness and acceptance promise.

The printed perturbation formula also has $H_{\mathrm{out}}$ where a general perturbation $V$ belongs. Its unspecified $O({\delta }^2)$ cannot be treated as uniform in a closing gap. Likewise a circuit-length-independent constant $\delta$ cannot in general satisfy the stated small-perturbation requirement for an arbitrary long quantum circuit. The explicit bound above avoids both issues; it does not claim the defective literal promise defines standard StoqMA.

Write the final topological quantum order constants as $a>0$ and $\epsilon <1$, to avoid confusing the allowed support diameter with the small time coefficient. For any initial operator $A_X$ of operator norm at most one,and similarly for the partner state. The minus sign follows from $|{\psi }_0⟩=e^{itH}|{\psi }_1⟩$ and the Heisenberg picture convention $A_X(t)=e^{itH}{A}_X{e}^{-itH}$. The Lieb-Robinson bound and its localization corollary apply to either time direction, using $|t|$.

Choose $\mathrm{diam}(X)\le aL/4$, $t=\tau L$ with $v\tau \le a/8$, and localization buffer $l=aL/8$. The enlarged support $X^{\prime }$ obeysThe Lieb-Robinson localization by Haar twirling corollary supplies an operator $B_{X^{\prime }}$ approximating $A_X(-t)$ withDo not assume that the approximate operator has norm at most one: it only has $\Vert B_{X^{\prime }}\Vert \le 1+{\delta }_L$. Apply final-state local indistinguishability to $B_{X^{\prime }}/(1+{\delta }_L)$. The two approximation errors then giveIf the lattice has polynomially many sites in its diameter, $N(L)=O(L^p)$, the error tends to zero uniformly in these supports. For sufficiently large $L$, take ${\delta }_L\le (1-\epsilon )/[2(\epsilon +2)]$. The initial pair is then indistinguishable within ${\epsilon }^{\prime }=(1+\epsilon )/2<1$ on supports of diameter at most $aL/4$. Together with the exact orthogonality from condition (i), this proves the initial state is topologically ordered for sufficiently small linear-time coefficient. The constants for initial and final order need not be identical.

The backward preservation of local indistinguishability has an explicit finite-system qualification: it holds whenever the displayed $L|X|e^{-\mu aL/16}$ error is small enough. The conventional bounded-density, fixed-dimensional lattice interpretation supplies that condition. The question leaves growth control implicit; for an arbitrary collection of qudits one cannot discard the $|X|$ prefactor merely because $L$ is large. This identifies exactly the assumption needed by the supplied localization proof.

Use the Fourier transform convention $\hat{w}(E)={\int }_{\mathbb{R}}w(t)e^{itE}dt$. Since $w\ge 0$, it is real, so $\hat{w}(-E)=\hat{w}(E{)}^{\ast }$. The positive-frequency cutoff therefore also eliminates frequencies at or below $-\Delta$. Normalization gives $\int w=\hat{w}(0)=1$.

For spectral filtering of Hamiltonian terms, chooseIn an energy eigenbasis, its matrix elements areAt $i=j=0$ the frequency is zero, so the normalization preserves the ground-state expectation:The spectral filter is a positive weighted average of unitary conjugations; in particular the integral is bounded in operator norm by $\Vert h_Z\Vert$. We can choose it even without an extra assumed tail bound: the evenization of a nonnegative bandlimited filter proved in part (e) produces another admissible spectral filter with the same type of positive-time decay. Make that choice consistently in all the filtered terms and shell definitions below.

Use the part-preserving bipartite four-cycle density conventionHere $G(x,y)$ is the indicator function of adjacency, and repeated labels are allowed. This is the homomorphism density convention appropriate to the four-cycle norm. We derive the needed counting and operator norm facts directly.

Let $M=(G(x,y))$ be the $n$-by-$n$ biadjacency matrix, let $1$ be the all-ones vector, let $J=11^{\mathsf{T}}$, and put $H=M-\delta J$. The degree hypothesis saysso the centered matrix $H$ annihilates $1$ on both sides. In particular, $HJ=J{H}^{\mathsf{T}}=0$, andThe two terms have zero products in either order. Expanding the matrix trace directly counts four adjacency factors:Since $J^2=nJ$ and $\mathrm{tr}J=n$, we obtainThe given upper bound therefore impliesThis computation proves the centered four-cycle identity for a biregular graph, rather than assuming a four-cycle norm identity.

The matrix $Q=H{H}^{\mathsf{T}}$ is a positive semidefinite matrix. By the spectral theorem for real symmetric matrices it has nonnegative eigenvalues ${\lambda }_j$, andMoreover, for the Euclidean norm,Indeed $\Vert H^{\mathsf{T}}v{\Vert }_2^2=v^{\mathsf{T}}Qv$, whose maximum on unit vectors is ${\lambda }_{\max }$; the Cauchy-Schwarz inequality characterization $\Vert H{\Vert }_{\mathrm{op}}={\sup }_{\Vert u{\Vert }_2=\Vert v{\Vert }_2=1}|u^{\mathsf{T}}Hv|$ also shows that $H$ and $H^{\mathsf{T}}$ have the same operator norm. Taking fourth roots thus giveswhere the error parameter $c$ is nonnegative, as in the claimed bound.

Finally use the indicator vectors $a=1_A$, $b=1_B$. Their Euclidean norms are $\sqrt{|A|}$ and $\sqrt{|B|}$, so the Cauchy-Schwarz inequality and the operator norm estimate yieldThis is the required discrepancy bound. It also holds when either subset is empty or $\delta =0$. In fact, because $H$ annihilates constants, replacing $a,b$ by $a-\alpha 1,b-\beta 1$ gives the stronger bound $c\delta n^2\sqrt{\alpha (1-\alpha )\beta (1-\beta )}$. This is the spectral discrepancy bound for a biregular graph.

For $1\le p<\infty$, the real Lp space is the vector space of real measurable functions with ${\int }_{\Omega }|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. For $p=\infty$, take the essentially bounded real measurable functions, with the same identification and norm $\Vert f{\Vert }_{\infty }=\mathrm{esssup}|f|$. The identification makes each norm definite; absolute homogeneity follows from the Lebesgue integral, and the triangle inequality follows from Minkowski inequality for finite $p$, or directly from the essential supremum for $p=\infty$.

Here is a completeness proof valid on any measure space. For $p<\infty$, a Cauchy sequence $(f_n)$ has a subsequence $(f_{n_k})$ with $\Vert f_{n_{k+1}}-f_{n_k}{\Vert }_p\le 2^{-k}$. Choose measurable function representatives and put $h_k=f_{n_{k+1}}-f_{n_k}$. By Minkowski inequality and the monotone convergence theorem,Consequently the series $\sum h_k$ converges absolutely almost everywhere. Define $f=f_{n_1}+\sum h_k$ there, and define it to be zero on the measurable exceptional null set. Then $f\in L^p$, and Fatou lemma applied to each tail gives $\Vert f-f_{n_k}{\Vert }_p\le {\sum }_{j=k}^{\infty }{2}^{-j}\to 0$. The original Cauchy sequence also converges in Lp norm, by the triangle inequality. For $p=\infty$, choose the same subsequence using the essential supremum norm. Outside one measurable null set, all the bounds $|h_k|\le 2^{-k}$ hold and $f_{n_1}$ is bounded. The series then converges uniformly there, with an essentially bounded measurable function limit and the same tail estimate in essential supremum norm. Thus all these spaces are Banach spaces.

The duality of Lp spaces says that, for $1<p<\infty$ and $q=p/(p-1)$, the mapis an isometric isomorphism of normed spaces. This form of Lp duality on an arbitrary measure space requires no finiteness hypothesis on $\mu$. At $p=1$, a standard version assumes a sigma-finite measure and identifies $(L^1{)}^{\ast }$ with $L^{\infty }$ through the same dual pairing. That endpoint assertion must not be made without a suitable measure-space hypothesis.

We first prove the required duality of Lp spaces for a finite measure. Let $F\in (L^p{)}^{\ast }$ and set $\nu (E)=F(1_E)$. For disjoint measurable $E_j$, the indicator functions of their partial unions converge in Lp norm to that of their union, so $\nu$ is countably additive. It has finite variation measure: for every finite measurable partition $(E_j)$, choosing real signs givesAlso $\mu (E)=0$ implies $\nu (E)=0$. The Radon-Nikodym theorem supplies a Radon-Nikodym derivative $g\in L^1$ with $\nu (E)={\int }_{E}gd\mu$. Linearity gives $F(f)=\int fgd\mu$ for simple functions. Uniform approximation by simple functions extends this identity to bounded measurable functions: both their Lp norm errors and the errors in integration against $g$ tend to zero.

To establish the correct integrability, test with the bounded measurable function $f_N=\mathrm{sgn}(g)|g{|}^{q-1}{1}_{\{|g|\le N\}}$. Since $(q-1)p=q$, writing $A_N={\int }_{\{|g|\le N\}}|g{|}^{q}d\mu$ givesThe second inequality is also valid when $A_N=0$. The monotone convergence theorem gives $g\in L^q$ and $\Vert g{\Vert }_q\le \Vert F\Vert$. Density of simple functions in the Lp space, together with Hölder's inequality, now gives $F=I_{p}g$ on all of $L^p$. Conversely Hölder's inequality gives $\Vert I_{p}g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, testing againstgives $\Vert f{\Vert }_p=1$ and $(I_{p}g)(f)=\Vert g{\Vert }_q$, so $\boxed{\Vert I_{p}g\Vert =\Vert g{\Vert }_q}$. This also proves uniqueness of the representing Radon-Nikodym derivative.

For completeness, the passage to an arbitrary measure space can be made without losing a hypothesis in the reflexive Banach space argument below. On a sigma-finite measure space, exhaust by nested finite-measure sets $E_n$. The representing Radon-Nikodym derivatives on $E_n$ agree on overlaps by uniqueness. Their glued density has Lp norm $\Vert g{\Vert }_q\le \Vert F\Vert$ by the monotone convergence theorem, and represents $F$ because $f{1}_{E_n}\to f$ in Lp norm. Every $f\in L^p$ on an arbitrary measure space is supported on a sigma-finite measurable set: the sets $\{|f|>1/n\}$ have finite measure and their union is $\{f\ne 0\}$.

Use support localization of an Lp functional as follows. For a sigma-finite measurable $A$, let $m_A$ be the operator norm of $F$ restricted to functions supported in $A$. The support observation gives ${\sup }_A{m}_A=\Vert F\Vert$. Choose $A_n$ approaching this supremum and set $A={\bigcup }_n{A}_n$; then $m_A=\Vert F\Vert$. If a sigma-finite $B\subseteq \Omega \setminus A$ had $m_B=\beta >0$, functions supported on the disjoint sets $A,B$ have the direct-sum Lp norm. Optimizing their two scalar coefficients by Hölder's inequality, and using functions approaching the two restriction operator norms, would givea contradiction. Thus $F$ vanishes on functions supported outside $A$. The density on $A$, extended by zero, represents $F$ globally. The case $F=0$ simply uses $g=0$. This proves the stated Lp duality on an arbitrary measure space.

Finally let $X=L^p$ and let $\Phi \in X^{\ast \ast }$, where stars denote continuous dual spaces. Compose with $I_p$ to obtain the bounded linear functional $g\mapsto \Phi (I_{p}g)$ on $L^q$. Applying duality of Lp spaces with the exponents reversed gives $f\in L^p$ with $\Phi (I_{p}g)=\int fgd\mu$. For the canonical embedding into the bidual $J_{X}f$, its value on $I_{p}g$ is also $\int fgd\mu$. Since $I_p$ is onto, $\Phi =J_{X}f$. Its norm is $\Vert f{\Vert }_p$ by the same dual pairing norm identity. Therefore $\boxed{J_X(L^p)=(L^p{)}^{\ast \ast }}$, which proves that $L^p$ is a reflexive Banach space through its actual canonical embedding into the bidual.

Let $R=X(l)$ and $C$ be its complement, with Hilbert-space dimension $d_C$. Use Haar twirling conditional expectation to defineThis is an operator supported on $R$. The normalization $1/d_C$ is essential: the partial trace alone would not fix an operator already supported on $R$.

Subtract the integrand from $A_X(t)$ and use the operator norm triangle inequality. Since $A-UA{U}^{†}=[A,U]U^{†}$ and every unitary operator has norm one, the assumed Lieb-Robinson bound, applied to a unitary supported on the entire complement, givesThe factor $\min (|X|,|C|)$ was bounded by $|X|$; there is no sum over sites and hence no volume-dependent prefactor.

ChooseFor $t,l\ge 0$, $e^{2kst}-1\le 2kst{e}^{2kst}$, and thereforeThus $A_{X(l)}(t)=A_R(t)$ is the desired Lieb-Robinson localization by Haar twirling. The displayed choice assumes the usual positive constants $\mu ,s$; if the interaction bound has $s=0$, the error is identically zero and any positive $v$ works. For negative times the same argument uses $|t|$.

Use an $n$-qubit state as the witness. Uniformly choose a term $h_j$ and perform the two-outcome positive operator-valued measure $\{I-h_j,h_j\}$. Regard the second outcome as an energy flag. For witness $\rho$, its probability isThis is an efficient local-energy measurement verification: a unitary operator on the $k$ relevant qubits and one ancilla qubit can implement the measurement using the isometryThe local dimension is constant; the term matrices and the measurement can be approximated to inverse-polynomial accuracy. Uniform choice of $j$ and the classical threshold comparison also have polynomial overhead.

SetIn the nontrivial case $0\le a<b\le M$, $\delta$ is inverse-polynomial. Cases outside this energy range are decided directly from $0\le H\le MI$.

Ask for $r$ witness registers, measure an independently sampled local term on each, and accept if the mean number of energy flags is at most $\theta$. An honest product ground state on a YES instance has flag probability at most $p_{\mathrm{y}}$. The Hoeffding inequality gives failure at most $e^{-r{\delta }^2/2}$. For completeness, the required Bernoulli concentration estimate follows by bounding the centered log moment-generating function $f(\lambda )=\ln \mathbb{E}{e}^{\lambda (Y-p)}$: $f(0)=f^{\prime }(0)=0$ and $f^{\prime \prime }(\lambda )$ is a tilted Bernoulli variance, at most $1/4$. Hence $f(\lambda )\le {\lambda }^2/8$; exponential Markov's inequality optimized at $\lambda =4u$ yields $\mathrm{Pr}(\overline{Y}-p\ge u)\le e^{-2r{u}^2}$, and the lower-tail version follows in the same way.

Soundness must also cover entangled witness registers. The averaged flag effect is $R=H/M$. All its eigenvalues are at least $p_{\mathrm{n}}$ on a NO instance. The repeated verifier's acceptance effect is a polynomial in the commuting tensor-factor operators $R_j$ and $I-R_j$. In their product eigenbasis its eigenvalues are lower-tail probabilities of independent Bernoulli variables whose parameters are each at least $p_{\mathrm{n}}$. Their lower tail is bounded by the identical-parameter tail at $p_{\mathrm{n}}$, using the same uniform-variable coupling as in QMA parallel repetition with entangled witnesses. Therefore its operator norm is at most $e^{-r{\delta }^2/2}$, which bounds acceptance for any entangled witness.

Choosinggives completeness at least $2/3$ and soundness at most $1/3$. A slightly larger constant absorbs implementation errors; for example, approximate each flag effect in norm to at most $\delta /8$, retain a gap at least $3\delta /4$, and increase $r$ by a constant factor. The total witness length $rn$ and circuit size are polynomial. Thus

Use the Fourier transform convention $\hat{w}(E)={\int }_{\mathbb{R}}w(t)e^{itE}dt$ and define the spectral filtering of Hamiltonian terms byThe normalization $\hat{w}(0)=1$ means $\int w=1$. Since $w$ is real and nonnegative, $\hat{w}(-E)=\overline{\hat{w}(E)}$. Thus the stated positive-frequency cutoff also implies $\hat{w}(E)=0$ for $E\le -\Delta$. The integral exists in operator norm, because its integrand has norm at most $w(t)\Vert h_Z\Vert$.

Let $H|{\varphi }_i⟩=E_i|{\varphi }_i⟩$, with $E_i-E_0\ge \Delta$ for $i>0$. In this eigenbasis,Every matrix element coupling the unique ground state to an excited state vanishes, in either direction. HenceMoreover, linearity and conservation of $H$ under its own Heisenberg picture evolution giveEach filtered term is Hermitian, but it need not remain supported on its original set $Z$. The tail assumption supplies stronger long-time control than the integrability used here; no additional tail estimate is needed for the requested identities.

The original PDF has $u_m^{n+1}$ on the left of the update; the converted TeX omits its $+1$. Use the Forward Euler method with $k=\Delta t\ge 0$ and $\mu =k/h_x^2$. The explicit time stepping for bounded reaction diffusion update matrix isWhen $0\le \mu \le 1/2$, the heat update $H_h$ has nonnegative stencil weights $\mu ,1-2\mu ,\mu$ and row sums at most one. Hence its induced maximum operator norm is at most one. It is also a symmetric matrix with eigenvaluesall in $[-1,1]$, so its discrete L2 norm operator bound is at most one too. Let $A_{\ast }=\max (|a_0|,|a_{+}|)$. In either of these norms,Iteration gives the mesh-uniform stability estimateThe same estimate applies to differences of solutions. Adding a per-step defect $k{r}^n$ gives $\Vert e^n\Vert \le e^{A_{\ast }T}(\Vert e^0\Vert +k{\sum }_{j<n}\Vert r^j\Vert )$.

This proof only requires $\mu \le 1/2$ and bounded reaction coefficients. It does not assume the full update $E_h$ is nonnegative: a negative $a_m$ can make its middle stencil weight negative when $\mu =1/2$. Nor does it claim contractivity or an all-time bound for every coefficient. The required stability is a bound uniform in the discretization on fixed finite intervals.

For positive linear operators on continuous functions, a general Korovkin theorem, in its compact Korovkin test space form, can be stated using a test space $H\subset C(K,\mathbb{R})$, where $K$ is a compact Hausdorff space. Assume $H$ contains a strictly positive function $h_0$. Require that, for every $x\in K$, the only positive linear functional $L$ on $C(K)$ with $L(h)=h(x)$ for all $h\in H$ is evaluation at $x$. Then convergence on the test space implies convergence on every continuous function:By the Riesz-Markov-Kakutani representation theorem, the functional condition equivalently says that a positive measure with these test moments must be ${\delta }_x$. This permits arbitrary test families, not just the interval tests $1,x,x^2$.

For completeness, let $c={\min }_K{h}_0>0$. Positivity gives $U_{n}1\le c^{-1}{U}_n{h}_0$, so the operator norms are uniformly bounded. If the conclusion failed for some $f$, choose $x_n$ where its error is bounded away from zero. The evaluation functionals $L_n(g)=U_{n}g(x_n)$ have uniformly bounded norm. Compactness of $K$ and the Banach-Alaoglu theorem give a subnet on which $x_n\to x$ and $L_n$ converges in the weak-star topology to a positive $L$. For every $h\in H$, uniform test convergence gives $L(h)=h(x)$. The hypothesis forces $L(f)=f(x)$, contradicting the chosen errors. Complex-valued functions follow by applying the real result to their real and imaginary parts.

A useful concrete sufficient condition is that $1\in H$ and for each $x$ there is $g_x\in H$ with $g_x\ge 0$ and zero set exactly $\{x\}$. Its zero moment forces the representing measure to be supported at $x$, while the constant moment fixes its mass.

For the circle, take $H=\mathrm{span}\{1,\sin t,\cos t\}$. The nonnegative function $g_x(t)=1-\cos (t-x)$ belongs to $H$ and vanishes only at $x$ on the circle. Thus the periodic Korovkin test set isUniform convergence on these three functions is sufficient for uniform convergence on $C(\mathbb{T})$.