= Solution
Choose an even positive $J(q)$. Each independent complex <Fourier mode> of the real field has density $(J(q)/\pi)e^{-J(q)|\phi_q|^2}$, so
$$
\langle\phi_q\phi_p\rangle_0=\frac{\delta_{p,-q}}{J(q)}.
$$
The quadratic part therefore has expectation $\sum_q^+G(q)/J(q)$. For the quartic part, apply <Isserlis theorem> to its four jointly centered <Gaussian random variables>, or to their real and imaginary components. Pairing the two differentiated fields together imposes $q_2=-q_1$, and its contribution is
$$
\frac BV\left(\sum_q\frac{q^2}{J(q)}\right)\left(\sum_p\frac1{J(p)}\right).
$$
Each of the two other pairings instead gives $-(B/V)|\sum_q q/J(q)|^2=0$, because the summand is odd under $q\mapsto-q$. Thus only the displayed contribution remains. Each full sum is twice its <positive-wavevector sum for a real field>; putting $S_0=\sum_q^+1/J(q)$ and $S_2=\sum_q^+q^2/J(q)$ gives $\langle H_4\rangle_0=4BS_2S_0/V$.
Substitute these expectations, $F_0$ and $\langle H_0\rangle_0$ into the <Feynman-Bogoliubov inequality>:
$$
\boxed{F\leq\mathcal F[J]=\sum_q^+\left[\log\frac{J(q)}\pi-1+\frac{G(q)}{J(q)}\right]+\frac{4B}{V}S_2S_0.}
$$
The zero mode is absent by the composition constraint. The counting uses one representative of every nonzero $\{q,-q\}$ pair, so no extra factor of two is assigned to an independent complex amplitude.
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