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In a telescoping local-shell decomposition, split the spectral filter integral at a time proportional to the shell distance. A Lieb-Robinson bound controls short times, while the two-sided almost-exponential spectral filter tail controls long times. Polynomial shell-weight growth then gives the displayed decay. The estimate is asymptotic at large distance; small shells retain the elementary operator-norm bound. The evenization of a nonnegative bandlimited filter can supply the two-sided tail from a one-sided existence statement.

For $\alpha >0$, Tikhonov regularization minimizes the strictly convex quadraticDifferentiating in every direction gives the Tikhonov normal equation $(A^{T}A+\alpha I)v=A^T{y}_{\delta }$. Symmetry of $A$ therefore givesThe spectral filter suppresses unstable division by small eigenvalues. Put $e_i=(y_{\delta }-y,u_i)$ and $x_i=(x,u_i)$. The exact noise-bias decomposition for linear regularization isBy orthonormality and the triangle inequality,The scalar gain satisfies $\lambda /({\lambda }^2+\alpha )\le 1/(2\sqrt{\alpha })$, because $(\lambda -\sqrt{\alpha }{)}^2\ge 0$. Hence the Tikhonov stability bound givesThis explicitly separates data noise from regularization bias. For a known bound $X\ge \Vert x\Vert >0$, minimizing this last bound gives an a priori regularization parameter choiceOne data-based bound is $X=(\Vert y_{\delta }\Vert +\delta )/{\lambda }_1$. More generally, $\alpha (\delta )\to 0$ and $\delta /\sqrt{\alpha (\delta )}\to 0$ make the robust estimate converge to zero.

For this fixed finite matrix, the sharper gain bound ${\max }_i{\lambda }_i/({\lambda }_i^2+\alpha )\le 1/{\lambda }_1$ is also available. Thus the ${\delta }^{2/3}$ rate is a consequence of the robust bound, not an optimality claim: choosing $\alpha X/{\lambda }_1^2\le C\delta /{\lambda }_1$ gives an $O(\delta /{\lambda }_1)$ estimate. No unique statistically optimal parameter is specified without further assumptions on the signal or noise.

For an excited eigenstate, $E_i-E_0\ge \Delta$. The matrix-element identity in condition (i), together with the two-sided Fourier cutoff, givesThe reverse matrix element has frequency $E_0-E_i\le -\Delta$ and also vanishes. Equivalently, $A^{(Z)}$ is Hermitian because $h_Z$ is Hermitian and $w$ is real, so the two elements are conjugate. ThusThe spectral gap eliminates precisely the couplings needed to make the unique ground state an eigenvector of every filtered term. Couplings between excited states with smaller energy differences can remain; the spectral filter need not diagonalize the whole operator.

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.

The filtered shell integral uses both signs of time, while the stated decay hypothesis controls only positive time. Here is a way to choose an even admissible spectral filter from the given one, rather than silently assume that its negative tail is controlled. Call the supplied spectral filter $f$. Reality and its Fourier cutoff imply that $\hat{f}$ is supported in $[-\Delta ,\Delta ]$. It is bounded and integrable in frequency, so Fourier inversion supplies a bounded continuous representative of $f$. That representative is real analytic, since its Fourier support is compact, and is not identically zero.

Define the evenization of a nonnegative bandlimited filter byThe normalizing integral is finite and positive: boundedness and integrability give finiteness, while a nonzero real-analytic nonnegative function cannot vanish on an interval, so the product is positive on some interval. The new spectral filter is even, nonnegative and normalized. Each factor has Fourier support in $[-\Delta /2,\Delta /2]$, and the convolution rule for the product therefore gives support in $[-\Delta ,\Delta ]$, including vanishing at the outer endpoints. Its positive tail is bounded byIt has the same required tail form, with a rescaled positive constant $\beta$. Since $w$ is even, its two-sided tail is twice its positive tail. The preceding parts use this chosen $w$ consistently.

For the almost-exponential locality of filtered Hamiltonian terms, split the integral defining $A^{(Z,d)}$ at $|t|=T_d$. Write $\lambda =2ks>0$ and choose $T_d=\mu d/(2\lambda )$. On the short-time part, part (d) and $\int w=1$ giveFor the long-time part, each conjugated operator has norm $\Vert h_Z\Vert$, so their difference has norm at most $2\Vert h_Z\Vert$. The two-sided spectral filter tail givesLet $c=\beta \mu /(2\lambda )>0$, so $\beta T_d=cd$. The logarithmic prefactor is bounded by $O(d^{\alpha +1})$, and the first, exponentially decaying contribution is asymptotically smaller than this almost-exponential contribution. ThereforeThis is an asymptotic statement for large $d$; small shells have the elementary bound $2\Vert h_Z\Vert$, avoiding the meaningless substitution $cd=1$ into the logarithmic expression. If $ks=0$, there is no dynamical spreading and the shell increments vanish. Filtering gives almost-exponentially decaying shells despite the filtered operator's potentially global support.

Fix a singular system of a compact operator, with $K{v}_j={\sigma }_j{w}_j$ and $K^{\ast }{w}_j={\sigma }_j{v}_j$, ${\sigma }_j>0$. A spectral regularization method has the formIt annihilates $\mathcal{R}(K{)}^{\perp }$. Standard sufficient spectral filter conditions are boundedness of ${\sup }_{0<\sigma \le \Vert K\Vert }|\sigma g_{\alpha }({\sigma }^2)|$ for each $\alpha$, pointwise $\lambda g_{\alpha }(\lambda )\to 1$ as $\alpha \downarrow 0$ for $\lambda >0$, and a uniform bound on $|\lambda g_{\alpha }(\lambda )|$. These give bounded linear operators and convergence on the domain of the Moore–Penrose inverse of an operator by the Picard criterion and dominated convergence theorem.

For Tikhonov regularization,This can be computed by solving the Tikhonov normal equation, without knowing the singular value decomposition.

For spectral cutoff regularization, takewith value zero at $\lambda =0$. Only singular values ${\sigma }_j\ge \sqrt{\alpha }$ are inverted. Here $g_{\alpha }$ multiplies $\sigma$; conventions calling the full coefficient $\sigma g_{\alpha }({\sigma }^2)$ the spectral filter are equivalent.

A regularization of an inverse problem replaces the unbounded generalized inverse by bounded operators $R_{\alpha }:Y\to X$ satisfying $R_{\alpha }y\to A^{†}y$ as $\alpha \downarrow 0$ on admissible exact data. A spectral regularization method can be writtenwhere $f_{\alpha }(\sigma )\to 1$ for fixed $\sigma >0$ while $f_{\alpha }(\sigma )/\sigma$ remains bounded for each fixed $\alpha$. A uniform bound on $f_{\alpha }$ supplies the usual dominated-convergence argument on exact data.

This defines $f_{\alpha }$ as the dimensionless damping factor. In the alternative convention calling the entire inverse coefficient the spectral filter, that filter is $f_{\alpha }(\sigma )/\sigma$. For data error bounded by $\delta$, a sufficient regularization parameter choice also requires $\delta \Vert R_{\alpha (\delta )}\Vert \to 0$; consistency alone does not control noise.

Let $({\sigma }_j,u_j,v_j)$ be a singular system of a compact operator, withThe Moore–Penrose inverse of an operator is the generally unbounded mapdefined when the Picard criterion holds, with the component in $\ker A^{\ast }$ sent to zero. It is the minimum-norm least-squares solution of $Af=g$.

A regularization of an inverse problem consists of bounded maps $R_{\alpha }:Y\to X$ and a parameter rule $\alpha =\alpha (\delta ,g^{(\delta )})$ such thatwhenever $\Vert g^{(\delta )}-g\Vert \le \delta$ and $g$ is in the domain of $A^{†}$.

For Tikhonov regularization, minimizinggivesThe scalar spectral filter satisfiesand consequentlyFor exact data, each filter factor ${\sigma }_j^2/({\sigma }_j^2+\alpha )$ tends to one, so $R_{\alpha }g\to A^{†}g$. Choosingtherefore makes both the approximation error and propagated data error vanish. For example, $\alpha (\delta )=\delta$ is an admissible a priori regularization parameter choice.