Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 23G (Analysis of Functions)

Taking Fourier transforms givesThusThe multiplier and all its polynomially weighted derivatives have at most polynomial growth, so this defines a tempered distribution. In fact it maps $H^s$ continuously into $H^{s+4}$. Since the multiplier never vanishes, any homogeneous tempered solution has zero Fourier transform; uniqueness follows.

A compactly supported smooth function belongs to $H^s({\mathbb{R}}^n)$ for every real $s$. Part (a) therefore puts $u$ in $H^{s+4}$ for every $s$. Taking $s$ arbitrarily large and applying the Sobolev embedding theorem to every derivative shows that $u\in C^{\infty }({\mathbb{R}}^n)$.

In three dimensions, ${\delta }_0\in H^s$ for every $s<-3/2$. Hence the solution belongs to $H^{s+4}$ for every such $s$; choosing, for example, $s=-2$ gives $u\in H^2({\mathbb{R}}^3)$. Since $2>3/2$, Sobolev embedding supplies a unique continuous representative. It solves the equation distributionally, and uniqueness follows from part (a), so this is the unique continuous solution.