Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 327 / 1

Using multi-index notation, the Schwartz space isA sequence ${\phi }_m$ converges to $\phi$ in this Fréchet space when $p_{\alpha ,\beta }({\phi }_m-\phi )\to 0$ for every $\alpha ,\beta$. The space ${\mathcal{S}}^{\prime }({\mathbb{R}}^n)$ of tempered distributions is the continuous dual of $\mathcal{S}({\mathbb{R}}^n)$, and $u_m\to u$ there means weak convergence of distributions, namely $⟨u_m,\phi ⟩\to ⟨u,\phi ⟩$ for every $\phi \in \mathcal{S}$.

Continuity of a linear functional immediately implies that ${\phi }_m\to 0$ entails $⟨u,{\phi }_m⟩\to 0$. Conversely, enumerate the Schwartz seminorms as $p_1,p_2,\dots$. If $u$ were not continuous, then for each $m$ one could choose ${\phi }_m$ such thatEvery fixed seminorm tends to zero along this sequence, so ${\phi }_m\to 0$ in $\mathcal{S}$, contradicting the assumed sequential property. This is the sequential continuity criterion for a linear map on a metrizable topological vector space.

The angular-frequency Fourier transform and its inverse on $\mathcal{S}({\mathbb{R}}^n)$ areIt extends to a tempered distribution by duality:

Since ${\delta }_{1/t}\phi (x)=\phi (x/t)$ maps the Schwartz space continuously to itself, the formuladefines a continuous linear functional on $\mathcal{S}$, hence a tempered distribution. For a locally integrable function $u$, the change of variables formula $y=tx$ givesso this dilation of a distribution agrees with ordinary function dilation.

The scaling property of the Fourier transform gives, first for Schwartz functions and then by duality,If the homogeneous distribution $u$ has degree $\sigma$, thenPutting $s=1/t$ yields ${\delta }_s\hat{u}=s^{-n-\sigma }\hat{u}$. Thus the Fourier transform of a homogeneous distribution has degree $\boxed{-n-\sigma }$.

Because $0<\alpha <n$, $|x{|}^{-\alpha }$ is locally integrable at the origin and has only polynomial growth at infinity, so it defines a tempered distribution. It is a homogeneous distribution of degree $-\alpha$ and is radial. Its Fourier transform is therefore radial and homogeneous of degree $\alpha -n$, so it must have the form $c_{\alpha }|\lambda {|}^{\alpha -n}$. In particular, $\boxed{\beta =n-\alpha }$.

To determine the constant, use the stated Gamma integral representation, Fubini's theorem, and the Fourier transform of a Gaussian:The change of variables formula $s=|\lambda {|}^2/(4\tau )$ then givesThis is precisely the Fourier transform of the Riesz kernel.