= Solution
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 $\widehat f$ 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> by
$$
w(t)=C f(t/2)f(-t/2),\qquad C^{-1}=\int_{\mathbb R}f(t/2)f(-t/2)dt.
$$
The 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 by
$$
\int_T^\infty w(t)dt\leq2C\|f\|_\infty\int_{T/2}^\infty f(u)du.
$$
It 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 $\int w=1$ give
$$
\left\|\int_{|t|\leq T_d}w(t)\bigl(\tau_t^{H_d}(h_Z)-\tau_t^{H_{d-1}}(h_Z)\bigr)dt\right\|\leq C\|h_Z\|T_d d^\alpha e^{-\mu d+\lambda T_d}=O(\|h_Z\|d^{\alpha+1}e^{-\mu d/2}).
$$
For the long-time part, each conjugated operator has norm $\|h_Z\|$, so their difference has norm at most $2\|h_Z\|$. The two-sided <spectral filter> tail gives
$$
\left\|\int_{|t|>T_d}\cdots dt\right\|=O\left(\|h_Z\|(\log(\beta T_d))^2 e^{-\beta T_d/(\log(\beta T_d))^2}\right).
$$
Let $c=\beta\mu/(2\lambda)>0$, 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. Therefore
$$
\boxed{\|A^{(Z,d)}\|=O\left(\|h_Z\|d^{\alpha+1}\exp\left[-\frac{cd}{(\log(cd))^2}\right]\right).}
$$
This is an asymptotic statement for large $d$; small shells have the elementary bound $2\|h_Z\|$, avoiding the meaningless substitution $cd=1$ into the logarithmic expression. If $ks=0$, there is no dynamical spreading and the shell increments vanish. \b[Filtering gives almost-exponentially decaying shells despite the filtered operator's potentially global support.]
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