 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 $\hat{f}$ 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.