Past exam of the fourier analysis course of the University of Oxford / 2023 / 2

Let $R\varphi (x)=\varphi (-x)$, and define reflection of a tempered distribution by $⟨Ru,\varphi ⟩=⟨u,R\varphi ⟩$. The Fourier inversion theorem on the Schwartz space says $\hat{\hat{\varphi }}=2\pi R\varphi$. ThusReflection commutes with the Fourier transform, since $\widehat{R\varphi }=R\hat{\varphi }$ and the same relation follows by duality for tempered distributions. Define the inverse Fourier transform on tempered distributions byIt follows thatIn particular $u=(2\pi {)}^{-1}R\hat{\hat{u}}$ is the general Fourier inversion formula. When $\hat{u}$ is an integrable function, this is the regular distribution of $(2\pi {)}^{-1}\int \hat{u}(\xi )e^{ix\xi }d\xi$; an arbitrary tempered distribution need not have an ordinary integral representation.

Set $\tilde{\varphi }(x)=\overline{\varphi (-x)}$. This is a Schwartz function and its Fourier transform is $\hat{\tilde{\varphi }}(\xi )=\overline{\hat{\varphi }(\xi )}$. The convolution theorem givesThe convolution of two Schwartz functions is again a Schwartz function: differentiating under the integral and expanding $x^m=(y+(x-y){)}^m$ bounds every weighted derivative by products of integrable and bounded weighted derivatives. Evaluate Fourier inversion for this convolution at zero:The Schwartz space is dense in the L2 space. For $f\in L^2(\mathbb{R})$ choose Schwartz functions ${\varphi }_j\to f$ in the L2 norm. The identity makes ${\hat{\varphi }}_j$ a Cauchy sequence in the L2 space, and its limit defines $\mathcal{F}f$, independently of the approximation. It agrees with the distributional Fourier transform, because Cauchy-Schwarz inequality makes convergence in the L2 norm imply convergence when paired with every Schwartz function. For $f\in L^1\cap L^2$ it therefore agrees almost everywhere with the ordinary Fourier transform.

The extension is onto: its image is closed by the norm identity and contains the Schwartz space, since Fourier inversion maps that space to itself. This proves Plancherel theorem:Its inverse is the extension of the inverse Fourier transform, with limits understood in the L2 norm.

For real $s$, the Sobolev space $H^s(\mathbb{R})$ consists of those tempered distributions $u$ whose Fourier transform is represented by a locally square-integrable function and for whichThis is the Sobolev norm used here; inserting the fixed factor $(2\pi {)}^{-1}$ gives an equivalent convention.

For the constant function $c$, Fourier inversion at zero givesA nonzero Dirac delta distribution cannot be represented by a locally square-integrable function. Indeed, if $G$ represented $2\pi c{\delta }_0$, choose a test function $\rho$ with $\rho (0)=1$, and put ${\rho }_{\epsilon }(\xi )=\rho (\xi /\epsilon )$. On a fixed compact interval containing their supports, Cauchy-Schwarz inequality givesThus $c=0$. Every Sobolev space requires local square integrability of the Fourier transform, regardless of how negative $s$ is. ThereforeThe zero constant function belongs to every Sobolev space.

The distributional derivative is defined by $⟨u^{\prime },\varphi ⟩=-⟨u,{\varphi }^{\prime }⟩$. Since differentiation under the integral sign gives $(\hat{\varphi }{)}^{\prime }=\widehat{-i\xi \varphi }$, we obtainThis proves the Fourier transform of a derivative identity $\boxed{\widehat{u^{\prime }}=i\xi \hat{u}}$ for all tempered distributions.

If $u,u^{\prime }\in H^s(\mathbb{R})$, their Fourier transforms are represented by locally square-integrable functions, and the identity identifies $\widehat{u^{\prime }}$ with $i\xi \hat{u}$ almost everywhere. Hence the Sobolev norms satisfyso $\boxed{u\in H^{s+1}(\mathbb{R}),\Vert u{\Vert }_{H^{s+1}}^2=\Vert u{\Vert }_{H^s}^2+\Vert u^{\prime }{\Vert }_{H^s}^2}$ for every real $s$.

Write $U=\hat{u}$, $V=\hat{v}$, and $⟨\xi ⟩=(1+{\xi }^2{)}^{1/2}$. By the Sobolev space hypotheses, $A=⟨\xi {⟩}^{s}U$ and $B=⟨\xi {⟩}^{t}V$ belong to the L2 space. Their product is an integrable function by Cauchy-Schwarz inequality. Thus $UV=⟨\xi {⟩}^{-s-t}AB$ is a locally integrable function defining a tempered distribution, with estimateThe supremum is controlled by finitely many Schwartz seminorms. Define the convolution of Sobolev distributions byIt is independent of the representatives chosen for $U,V$, and the convolution theorem shows that it agrees with ordinary convolution for Schwartz functions. More generally, approximate $u,v$ by Schwartz functions in their respective Sobolev norms; the displayed estimate shows convergence of the products in tempered distributions, so their ordinary convolutions converge to this definition. For $u,v\in L^1$ that also satisfy the stated Sobolev space hypotheses, the ordinary convolution theorem gives the same result.

There need not be any real $r$ with $u\ast v\in H^r(\mathbb{R})$. ChooseHere $1_{(0,1)}$ is an indicator function. Since ${\int }_0^1{\xi }^{-2/3}d\xi =3$ and every weight $(1+{\xi }^2{)}^q$ is bounded above and below by positive constants on $[0,1]$, both $u$ and $v$ belong to every Sobolev space. But their convolution has Fourier transform $UV={\xi }^{-2/3}{1}_{(0,1)}$, and for every real $r$,Thus $\boxed{u,v\in \bigcap _{q\in \mathbb{R}}{H}^q(\mathbb{R})\text{but}u\ast v\notin \bigcup _{r\in \mathbb{R}}{H}^r(\mathbb{R})}$. The obstruction is failure of local square integrability at zero; a weight controlling large frequencies cannot fix it.