 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 inequalityreduces the weighted convolution formula for $\widehat{Vf}$ to an ${\ell }^1\ast {\ell }^2$ estimate. Thus $\Vert Vf{\Vert }_{H^s}\le C_{V,s}\Vert f{\Vert }_{H^s}$ 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 $k\ne 0$, the Fourier multiplier of $\Delta$ is $|k{|}^2$, andThe zero coefficient is already finite. Hence $f\in H^{s+2}$. Iterating the argument gives $f\in H^{s+2j}$ for every $j\ge 0$. 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.