= Solution
For an excited eigenstate, $E_i-E_0\geq\Delta$. The matrix-element identity in condition (i), together with the two-sided Fourier cutoff, gives
$$
\langle\phi_i|A^{(Z)}|\phi_0\rangle=\widehat w(E_i-E_0)\langle\phi_i|h_Z|\phi_0\rangle=0.
$$
The reverse <matrix element> has frequency $E_0-E_i\leq-\Delta$ and also vanishes. Equivalently, $A^{(Z)}$ is Hermitian because $h_Z$ is Hermitian and $w$ is real, so the two elements are conjugate. Thus
$$
\boxed{\langle\phi_0|A^{(Z)}|\phi_i\rangle=\langle\phi_i|A^{(Z)}|\phi_0\rangle=0\quad(i>0).}
$$
The <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.
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