Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 4

Assume first that $f$ and $g$ are integrable functions, so that Fubini's theorem justifies changing the order of the two integrals. In the defining Fourier transform of the convolution, put $y=x-x^{\prime }$:This proves the convolution theorem with the stated normalization:

Direct evaluation of the Fourier transform gives, for $k\ne 0$,At $k=0$ the defining integral is the interval length, so $\boxed{\tilde{g}(0)=2}$. This also equals the continuous limit of the displayed formula.

The convolution $(g\ast g)(x)$ is the length of the overlap of $(-1,1)$ with $(x-1,x+1)$. This overlap has length $2-|x|$ when $|x|<2$ and zero otherwise, so $u=g\ast g$. The convolution theorem now givesThe value at zero is also the area of the triangular function, providing a normalization check on its Fourier transform.

The Translation property of the Fourier transform givesFor the remaining integral, the same Translation property of the Fourier transform identifies $e^{20i{k}^{\prime }}\tilde{g}(k^{\prime })$ as the Fourier transform of $g(x+20)$. By the product form of the convolution theorem, with inverse normalization $1/(2\pi )$,But $g(x+20)$ is supported in $[-21,-19]$, whereas the four translated triangular functions making up $r$ are all supported in $[-8,8]$. Their product is identically zero. Therefore the requested integral is $\boxed{0}$ for every real $k$.