= Solution
Use the <Fourier transform> convention $\widehat w(E)=\int_{\mathbb R}w(t)e^{itE}\,dt$ and define the <spectral filtering of Hamiltonian terms> by
$$
\boxed{g^{(Z)}=\int_{\mathbb R}
w(t)e^{itH}h_Ze^{-itH}\,dt.}
$$
The normalization $\widehat w(0)=1$ means $\int w=1$. Since $w$ is real and nonnegative, $\widehat w(-E)=\overline{\widehat w(E)}$. Thus the stated positive-frequency cutoff also implies $\widehat w(E)=0$ for $E\leq-\Delta$. The integral exists in <operator norm>, because its integrand has norm at most $w(t)\|h_Z\|$.
Let $H|\phi_i\rangle=E_i|\phi_i\rangle$, with $E_i-E_0\geq\Delta$ for $i>0$. In this <eigenbasis>,
$$
\langle\phi_i|g^{(Z)}|\phi_j\rangle
=\widehat w(E_i-E_j)\langle\phi_i|h_Z|\phi_j\rangle.
$$
Every matrix element coupling the unique <ground state> to an excited state vanishes, in either direction. Hence
$$
\boxed{[g^{(Z)},P_0]=0,\qquad
g^{(Z)}|\phi_0\rangle
=\langle\phi_0|h_Z|\phi_0\rangle|\phi_0\rangle.}
$$
Moreover, linearity and conservation of $H$ under its own <Heisenberg picture> evolution give
$$
\sum_Zg^{(Z)}
=\int w(t)e^{itH}\left(\sum_Zh_Z\right)e^{-itH}\,dt
=H\int w(t)\,dt
=\boxed{H.}
$$
Each 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.
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