Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 3 / 20F / iii

For $n\ge 1$, the convolution equals the length of the overlap of two intervals:It is continuous and compactly supported, with $\Vert g_n{\Vert }_{\infty }=2$. Inverse Fourier transform turns this convolution into a product, soThis function is bounded near zero and $O(x^{-2})$ at infinity, hence lies in $L^1$. Fourier inversion, or the convolution theorem applied first to the integrable interval indicators, gives ${\hat{h}}_n=g_n$.