= Solution
\b[The filtered decomposition is not necessarily frustration-free.] Commuting with $P_0$ makes $|\phi_0\rangle$ an <eigenvector> of each term; it does not make that <eigenvector> a lowest-energy state of each term. This is the distinction expressed by <commuting with a ground projector does not imply frustration freeness>.
For an explicit positive, two-local counterexample on three <qubits>, let $P_0^{(i)}=|0\rangle\langle0|_i$ and $P_1^{(i)}=|1\rangle\langle1|_i$, and use the two distinct interaction sets $\{1,2\}$ and $\{2,3\}$:
$$
h_{12}=P_1^{(1)}+P_0^{(2)},\qquad
h_{23}=P_1^{(3)}+2P_1^{(2)}.
$$
Both are <positive semidefinite> and commute with their sum. The total energy on a computational basis vector is
$$
E(b_1,b_2,b_3)=b_1+b_3+1+b_2.
$$
Thus $H=h_{12}+h_{23}$ has unique <ground state> $|000\rangle$, energy $1$, and <spectral gap> $1$. Since $[H,h_{12}]=[H,h_{23}]=0$, <spectral filtering of Hamiltonian terms> leaves both terms unchanged for every normalized filter:
$$
g^{(12)}=h_{12},\qquad g^{(23)}=h_{23}.
$$
But $h_{12}|000\rangle=|000\rangle$, whereas $h_{12}|010\rangle=0$. The global <ground state> therefore does not minimize $g^{(12)}$, proving that the resulting decomposition is not a <frustration-free Hamiltonian>.
There is a useful positive result if the original decomposition already is a <frustration-free Hamiltonian>. If $a_Z=\lambda_{\min}(h_Z)$ and $h_Z|\phi_0\rangle=a_Z|\phi_0\rangle$, then $h_Z-a_ZI\geq0$. Averaging its unitary conjugates with nonnegative $w$ gives $g^{(Z)}-a_ZI\geq0$, while $g^{(Z)}|\phi_0\rangle=a_Z|\phi_0\rangle$. Thus <filtering preserves existing frustration freeness>; it does not create it for an arbitrary gapped Hamiltonian.
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