Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 24J (Probability and Measure)

For $f\in L^1({\mathbb{R}}^d)$, use the Fourier transform conventionThe integral is absolutely convergent for every $\xi$, and $|\hat{f}(\xi )|\le \Vert f{\Vert }_1$.

Here Fourier inversion means the absolute-integral version: $\hat{f}\in L^1({\mathbb{R}}^d)$ andThis convention is necessary for the bounded continuous version in part (d). Conditional inversion by symmetric truncation is a different statement and need not have that conclusion.

First ${\int }_{\mathbb{R}}{e}^{-x^2/2}dx=\sqrt{2\pi }$: square the positive integral and use polar coordinates to obtain $2\pi {\int }_0^{\infty }r{e}^{-r^2/2}dr=2\pi$. Scaling therefore normalizes the one-dimensional Gaussian function $g_t$.

Let $I(\xi )={\int }_{\mathbb{R}}(2\pi t{)}^{-1/2}{e}^{-x^2/(2t)}{e}^{-i\xi x}dx$. Differentiation under the integral is justified by the integrable majorant $|x|g_t(x)$. Since $g_t^{\prime }(x)=-x{g}_t(x)/t$, integration by parts, with zero endpoint terms, gives $I^{\prime }(\xi )=-t\xi I(\xi )$. With $I(0)=1$, this yields $I(\xi )=e^{-t{\xi }^2/2}$. Multiplication of the one-dimensional integrals givesApplying the same computation with reciprocal scale $1/t$ evaluates its inverse integral:For $f\in L^1$, Fubini's theorem gives $\widehat{f\ast g_t}=\hat{f}{e}^{-t|\xi {|}^2/2}$, which is integrable because $\hat{f}$ is bounded. Its inverse integral isAbsolute integrability of $|f(y)|e^{-t|\xi {|}^2/2}$ justifies this second application of Fubini's theorem. Thus Fourier inversion holds for every Gaussian convolution, in fact pointwise for this continuous representative.

Define $g$ by the inverse integral. Its absolute convergence gives $|g(x)|\le (2\pi {)}^{-d}\Vert \hat{f}{\Vert }_1$. If $x_n\to x$, the exponential factors converge pointwise and their differences are bounded by $2|\hat{f}|$. The dominated convergence theorem proves $g(x_n)\to g(x)$. Therefore $g$ is bounded and continuous and equals $f$ almost everywhere.

No, under the absolute-integral convention. Direct integration givesOn a fixed positive-length portion of every period, the numerator is bounded below. Summing the resulting lower bounds gives a divergent harmonic series, so the Fourier transform is not in $L^1$. Also there is no continuous almost-everywhere representative: continuity would force it to be $1$ on $(0,1)$ and $0$ on $(-\infty ,0)$, giving incompatible limits at $0$. Symmetrically truncated conditional inversion does hold at interior continuity points and gives half the jump value at the endpoints; this does not contradict the answer.