Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 327 / 2

The space of smooth functions $\mathcal{E}(X)=C^{\infty }(X)$ has the topology of uniform convergence of every derivative on every compact set $K⋐X$. Thus $f_j\to f$ exactly whenfor every $K$ and every multi-index $\alpha$. Its continuous dual ${\mathcal{E}}^{\prime }(X)$ is the compactly supported distribution space; convergence in the weak dual topology means pointwise convergence on every $f\in \mathcal{E}(X)$.

Let $K$ contain the support of $u\in {\mathcal{E}}^{\prime }({\mathbb{R}}^n)$. Choose a cutoff function that equals one near $K$. It makeswell-defined, and differentiating the parameter under the pairing givesThus the Fourier transform of a compactly supported distribution is a smooth function. Since a compactly supported distribution has finite order, some $N$ and $C$ satisfyApplying this estimate to the exponential yields $|\hat{u}(\lambda )|\le C^{\prime }⟨\lambda {⟩}^N$.

Now take $v\in {\mathcal{E}}^{\prime }(X)$, multiply by a cutoff function supported in $X$ and equal to one near $\mathrm{supp}v$, and regard the result as an element of ${\mathcal{E}}^{\prime }({\mathbb{R}}^n)$. Choose $m$ so large thatis Lebesgue integrable. The inverse Fourier transform $f={\mathcal{F}}^{-1}g$ is a bounded continuous function, and the Fourier transform of a derivative givesas a distributional identity. This is the Bessel potential proof of the structure theorem for compactly supported distributions.

The function $f$ itself need not have compact support. Choose another cutoff $\rho \in \mathcal{D}(X)$ equal to one near $\mathrm{supp}v$. Then $v=\rho (1-\Delta {)}^{m}f$. Repeatedly usingexpresses $v$ as a finite sum ${\sum }_{\beta }{\partial }^{\beta }{f}_{\beta }$, where every coefficient $f_{\beta }$ is continuous and compactly supported in $X$.

By multiplication of a distribution by a smooth function, for every test function $\psi$,Hence $\rho {\delta }_0={\delta }_0$.

The distributional derivative of the Heaviside step function is $H^{\prime }={\delta }_0$. The Leibniz rule and $x{\delta }_0=0$ therefore give

Applying the distributional Leibniz rule twice givesEliminating the middle term provesChoose $\rho \in \mathcal{D}(\mathbb{R})$ with $\rho (0)=1$. Parts i and ii and the identity just proved, with $\phi =\rho$ and $u=xH$, yieldThus explicit continuous functions of compact support are