Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 8 / 3 / d / Solution

Start with $u\in H_s$ for some real $s$. The equation becomesSmooth $f$ belongs to every periodic Sobolev space, and multiplication by the smooth coefficient preserves $H_s$ by part (c), so the right side lies in $H_s$. The Fourier multiplier $(1+\Delta {)}^{-1}$ has coefficient $(1+|m{|}^2{)}^{-1}$ and maps $H_s$ isometrically to $H_{s+2}$. Thus $u\in H_{s+2}$. Repeat this argument to get $u\in H_{s+2j}$ for every nonnegative integer $j$. For any derivative order $k$, choose $j$ with $s+2j>k+n/2$ and apply part (b). Therefore $\boxed{u\in C^{\infty }({\mathbb{T}}^n)}$. This is elliptic bootstrapping directly from the Fourier multiplier, with no assumption that the solution initially has nonnegative Sobolev order.