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

Use ${\mathbb{T}}^n={\mathbb{R}}^n/(2\pi \mathbb{Z}{)}^n$ with normalized Haar measure and Fourier basis $e_k(x)=e^{ik\cdot x}$, $k\in {\mathbb{Z}}^n$. For any real $s$, the periodic Sobolev space isWeighted Fourier coefficients identify it isometrically with ${\ell }^2$, so it is a Hilbert space and complete. This definition includes negative orders and distributions. For nonnegative integer $s$ the norm is equivalent to the sum of squared L2 norms of weak derivatives through order $s$.

Write $⟨k⟩=(1+|k{|}^2{)}^{1/2}$. There are $O(2^{jn})$ lattice points in the dyadic shell $2^j\le |m|<2^{j+1}$, while each contributes at most a constant times $2^{-2js}$. The geometric series ${\sum }_j{2}^{j(n-2s)}$ converges when $2s>n$, proving ${\sum }_m⟨m{⟩}^{-2s}<\infty$. For $|\alpha |\le k$ and $u\in H_{s+k}$, the Cauchy-Schwarz inequality yieldsThus every derivative Fourier series through order $k$ converges absolutely and uniformly. Starting with the continuous sum for $u$, uniform convergence of successive derivative series justifies differentiation, so these are its continuous classical derivatives. $\boxed{H_{s+k}({\mathbb{T}}^n)\hookrightarrow C^k({\mathbb{T}}^n),s>n/2}$, with a continuous norm bound.

For every real $s$, the elementary weight estimatefollows from $⟨x+y⟩\le \sqrt{2}⟨x⟩⟨y⟩$, using the reverse rearrangement for negative $s$. If $a$ is smooth its Fourier coefficients decay faster than every power, so $c_j=⟨j{⟩}^{|s|}|\hat{a}(j)|$ is summable. The Fourier convolution formula and Young convolution inequality giveInitially this holds for trigonometric polynomials, and density extends it to every $u\in H_s$, agreeing with distributional multiplication. Hence smooth multiplication preserves every order.

If $s>n/2$, the stronger estimate $⟨m{⟩}^s\le C_s(⟨l{⟩}^s+⟨m-l{⟩}^s)$ givesusing the summability estimate from part (b) with $k=0$. Approximation extends the product, and the continuous embedding identifies it with the pointwise product. Completeness from part (a), this estimate and the constant unit make $H_s$ a commutative Sobolev algebra for $s>n/2$. Rescaling the norm by $C_s^{\prime }$ if necessary gives an equivalent strictly submultiplicative Banach-algebra norm.

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.