Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 27K (Probability and Measure)

Use the conventionThe Fourier inversion theorem says that, when $f,\hat{f}\in L^1({\mathbb{R}}^d)$,for almost every $x$.

Sincethe Fubini's theorem and Fourier inversion giveThus the Plancherel theorem identity in this convention is

The inverse integral is a continuous function of $x$, since $\hat{f}\in L^1$ permits dominated convergence. If $f$ is continuous, it and the inverse integral are continuous and agree almost everywhere. They must agree everywhere: otherwise their continuous difference would be nonzero on an open set of positive measure. This is the continuous version of Fourier inversion.

PutThis function is bounded near zero and is $O(u^{-2})$ at infinity, so $H\in L^1(\mathbb{R})$ and the defining integral for $g$ converges absolutely.

For the triangular functiondirect integration givesThe Fourier inversion theorem, or equivalently the Fourier transform of a triangular function, therefore yieldsIn particular $g(x)=0$ whenever $|x|>1$. Finally,This also agrees with the Plancherel theorem applied to $h$ and $H$.