Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 4

For the stated Fourier transform convention, Fourier inversion isFor example, the pointwise formula holds at continuity points under the usual integrability hypotheses; the same normalization applies to the $L^2$ or distributional inverse.

Assume initially that $f,g$ are integrable, so Fubini's theorem justifies interchanging the integrals. Transform the convolution and put $u=x-z$:This is the convolution theorem. No factor $2\pi$ appears in this direction because the forward transform in the question is unnormalized.

Differentiate the defining Fourier integral $n$ times with respect to $k$:For a nonnegative integer $n$, this givesIntegrability of $x^{n}f$ together with suitable lower moments justifies differentiation under the integral. The same multiplication-derivative identity holds more generally for tempered distributions.

Evenness means the negative half of the function is determined as well: this is the triangular function $q(x)=1-|x|$ on $|x|\le 1$, zero outside. Its transform is real and even:Integration by parts, or integrating the two terms separately, givesThe apparent singularity at zero is removable, and its limiting value is the area under $q$. This is the Fourier transform of a triangular function.

Let $r(z)$ be the indicator of $[-1,1]$. The finite integral is exactly the convolution $p=q\ast r$. Its additional transform factor isApplying the convolution theorem and part (d),Equivalently this is $2(\sin k/k)[\sin (k/2)/(k/2){]}^2$. The value at zero also follows from $\int p=(\int q)(\int r)=1\times 2$.