= Solution
Interpret $H_{-\infty}(\mathbb T^n)=\bigcup_{s\in\mathbb R}H^s(\mathbb T^n)$, the periodic <distributions> of finite negative <Sobolev space> order. Smooth multiplication is bounded on every <periodic Sobolev space>. One direct proof uses the rapid decay of the <Fourier coefficients> of $V$: the inequality
$$
\langle k\rangle^s\leq C_s\langle k-\ell\rangle^{|s|}\langle\ell\rangle^s,\qquad\langle k\rangle=(1+|k|^2)^{1/2},
$$
reduces the weighted convolution formula for $\widehat{Vf}$ to an $\ell^1*\ell^2$ estimate. Thus $\|Vf\|_{H^s}\leq C_{V,s}\|f\|_{H^s}$ for every real $s$.
Suppose initially $f\in H^s$. The equation gives $\Delta f=u-Vf\in H^s$, since a <smooth function> belongs to every <periodic Sobolev space>. For $k\ne0$, the <Fourier multiplier> of $\Delta$ is $|k|^2$, and
$$
(1+|k|^2)^{s+2}|\widehat f(k)|^2
\leq4(1+|k|^2)^s|\widehat{\Delta f}(k)|^2.
$$
The zero coefficient is already finite. Hence $f\in H^{s+2}$. Iterating the argument gives $f\in H^{s+2j}$ for every $j\geq0$. Applying the <Sobolev embedding theorem> also to each distributional <partial derivative> produces a smooth representative. Therefore $\boxed{f\in C^\infty(\mathbb T^n)}$. This is an explicit <elliptic regularity bootstrap>, valid for complex or real $V$ and without a sign assumption on it.
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