Past exam of the fourier analysis course of the University of Oxford / 2023 / 1

Use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-ix\xi }dx$, with complex-linear distribution pairings throughout. The Riemann-Lebesgue lemma says that, for an integrable function $f\in L^1(\mathbb{R})$, its Fourier transform is continuous and tends to zero as $|\xi |\to \infty$. It also satisfies $\Vert \hat{f}{\Vert }_{\infty }\le \Vert f{\Vert }_1$. The product rule, or convolution theorem, isHere $(f\ast g)(x)=\int f(y)g(x-y)dy$ belongs to $L^1(\mathbb{R})$ by Tonelli's theorem, and Fubini's theorem proves the displayed identity. The companion Fourier product formula, for Schwartz functions, is $\widehat{fg}=(2\pi {)}^{-1}\hat{f}\ast \hat{g}$.

A Schwartz function is a smooth function $\varphi :\mathbb{R}\to \mathbb{C}$ for which every Schwartz seminormis finite. These seminorms define the topology of the Schwartz space $\mathcal{S}(\mathbb{R})$. In particular,Thus every Schwartz function is an integrable function.

The Fourier bounds for Schwartz functions assert that each Schwartz seminorm of $\hat{\varphi }$ is bounded by finitely many Schwartz seminorms of $\varphi$. Explicitly, differentiation under the integral sign and integration by parts giveso the generalized Leibniz rule and the preceding integrability estimate giveIn particular the Fourier transform maps the Schwartz space continuously into itself.

For $\epsilon >0$, insert a Gaussian function into the proposed inverse Fourier transform:The Fourier bounds for Schwartz functions show that $\hat{\varphi }$ is an integrable function, so dominated convergence givesOn the other hand, Fubini's theorem applies because $\Vert \varphi {\Vert }_1\int e^{-\epsilon {\xi }^2/2}d\xi <\infty$. The given Fourier transform of a Gaussian, rescaled by $\sqrt{\epsilon }$, yieldsThe last integrand is bounded by $\Vert \varphi {\Vert }_{\infty }{e}^{-z^2/2}/\sqrt{2\pi }$. Since $\varphi$ is continuous, another application of dominated convergence makes this tend to $\varphi (x)$. Equating the limits proves the Fourier inversion theorem:The Gaussian functions above form an approximate identity; no Fourier inversion was used in evaluating their integrals.

A tempered distribution is a continuous linear functional $u:\mathcal{S}(\mathbb{R})\to \mathbb{C}$. Equivalently, for some integer $N$ and constant $C$,The Fourier transform of a tempered distribution is defined byThe Fourier bounds for Schwartz functions make the right side a continuous linear functional on the Schwartz space, so $\hat{u}$ is again a tempered distribution.

For an integrable function $f$, the regular distribution $u_f$ acts by $⟨u_f,\varphi ⟩=\int f\varphi$. The estimate $|⟨u_f,\varphi ⟩|\le \Vert f{\Vert }_1{p}_{0,0}(\varphi )$ proves it is a tempered distribution. Since $\iint |f(x)\varphi (\xi )|dxd\xi <\infty$, Fubini's theorem givesThus the distributional Fourier transform agrees with the ordinary Fourier transform.

Since $0<f_{\alpha }\le 1$, its regular distribution satisfiesso $f_{\alpha }$ is a tempered distribution for every $\alpha >0$.

Put $a=\alpha /2$. The defining integral of the gamma function, with $r=(1+x^2)t$, givesThe Fourier transform of a Gaussian suggests the following representative for the Fourier transform of an inverse power of one plus x squared:All factors are positive. Tonelli's theorem evaluates its total integral, allowing an infinite value at the single point zero:Assign any finite value at zero if necessary. This gives an integrable function with $\Vert h_{\alpha }{\Vert }_1=2\pi$.

To verify equality as tempered distributions, take any Schwartz function $\varphi$. Substituting the mixture into $⟨f_{\alpha },\hat{\varphi }⟩$ is justified byFor each $t>0$, Fubini's theorem and the Fourier transform of a Gaussian identify the inner pairing with $\int \sqrt{\pi /t}{e}^{-{\xi }^2/(4t)}\varphi (\xi )d\xi$. Interchanging the resulting integrals is valid because their absolute value is bounded by $2\pi \Vert \varphi {\Vert }_{\infty }$ after integrating the positive kernel. ThereforeFinally, on a compact set with $|\xi |\ge \delta >0$, the mixture integrand is bounded by $t^{a-3/2}{e}^{-t-{\delta }^2/(4t)}$, which is integrable both at zero and at infinity. Dominated convergence proves continuity there. Hence $\boxed{\widehat{f_{\alpha }}=h_{\alpha }\in L^1(\mathbb{R}),h_{\alpha }(\xi )>0\text{ and continuous for }\xi \ne 0}$.

If $\alpha >1$, then $f_{\alpha }$ is an integrable function, so the Riemann-Lebesgue lemma supplies a continuous representative of its Fourier transform. Equivalently, in the mixture for $h_{\alpha }$ the majorant $t^{(\alpha -3)/2}{e}^{-t}$ is integrable at zero precisely when $\alpha >1$. In this case dominated convergence also givesIf $0<\alpha \le 1$, Fatou's lemma applied along any sequence of nonzero $\xi$ tending to zero givesThe divergence comes from zero, where the exponent is at most $-1$. A continuous representative $H$ would agree with $h_{\alpha }$ almost everywhere. Since both are continuous on each open half-line, they must agree everywhere on those half-lines: a nonzero continuous difference would be nonzero on an interval of positive Lebesgue measure. The displayed blow-up then contradicts continuity of $H$ at zero. Changing a value at one point cannot remove this singularity. Consequently