Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 4 / 15D (Complex Methods)

For example, it suffices to assume $f,g\in L^1(\mathbb{R})$. Indeed the absolute double integral isThus Fubini's theorem permits interchange in the Fourier transform of the convolution. With $u=x-t$,Hence the convolution theorem in this normalization isThere is no extra $2\pi$ in this formula; that factor belongs to the inverse transform and Parseval identity for the convention being used.

Integrating the interval indicator givesThe convolution $f_2(x)$ is the length of overlap of two intervals of length one whose centres differ by $x$. Thus $f_2(x)=1-|x|$ for $|x|\le 1$, and zero otherwise. The convolution theorem givesFor this Fourier transform normalization, the Parseval identity says $\int |f{|}^{2}dx=(2\pi {)}^{-1}\int |\hat{f}{|}^{2}d\lambda$. Sincewe have $\int |{\hat{f}}_2{|}^{2}d\lambda =4\pi /3$. Substituting $\lambda =2y$ givesThe quotient at zero is understood by its continuous limit.