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

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