Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 23I (Analysis of Functions)

Let ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ be the space of tempered distributions. For $s\in \mathbb{R}$, the Sobolev space $H^s({\mathbb{R}}^n)$ iswith normChanging the normalization of the Fourier transform changes this norm only by a fixed constant.

Let $\alpha$ be a multi-index with $|\alpha |\le k$. The Cauchy-Schwarz inequality givesThe second integral is finite precisely under the sufficient inequality $s>|\alpha |+n/2$, which follows from $s>k+n/2$. Hence $(i\xi {)}^{\alpha }\hat{u}$ is Lebesgue integrable for every $|\alpha |\le k$.

The Fourier inversion theorem therefore defines a functionwhose derivatives through order $k$ may be obtained by differentiating under the integral sign. Those derivatives are continuous, so $u^{\ast }\in C^k({\mathbb{R}}^n)$. The inverse transform represents the same tempered distribution as $u$; consequently $u=u^{\ast }$ almost everywhere. This is the Sobolev embedding theorem $H^s({\mathbb{R}}^n)\hookrightarrow C^k({\mathbb{R}}^n)$ in this range.

Taking the Fourier transform of the equation in the sense of distributions givesThe Fourier multiplieris everywhere positive. Define $\hat{u}=\hat{f}/m$. Sincewe haveThus $u\in H^{s+4}$ exists and obeys the required estimate. If two solutions existed, their difference would have $m\hat{u}=0$; positivity of $m$ proves uniqueness.

A classical solution needs continuous derivatives through order four. Applying the Sobolev embedding theorem with $k=4$ to $u\in H^{s+4}$ shows that this is guaranteed whenor equivalently