Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 3 / 20F (Analysis of Functions)

Dominated convergence makes $\hat{f}$ continuous, and $\Vert \hat{f}{\Vert }_{\infty }\le \Vert f{\Vert }_1$. Smooth compactly supported functions are dense in $L^1$. Their Fourier transforms decay at infinity by integration by parts, so uniform approximation proves the Riemann-Lebesgue lemma: $\hat{f}\in C_0$.

For density, smooth compactly supported functions are also uniformly dense in $C_0$ by cutoff and mollification. Each such target $g$ is a Schwartz function; its inverse Fourier transform is another Schwartz function, hence belongs to $L^1$, and Fourier inversion recovers $g$. Therefore the image of a function of $L^1$ is dense in $C_0$.

The $L^1$ Fourier inversion theorem says that if $f,\hat{f}\in L^1$, then $f$ agrees almost everywhere with the inverse integral of $\hat{f}$. If $\hat{f}=0$, this applies and gives $f=0$ almost everywhere. Hence the Fourier transform is injective as a map of $L^1$ equivalence classes, with dense image in $C_0$.

Integrating the exponential gives the Fourier transform of an interval indicatorThe value at zero is the removable limit, for $a\le b$.

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 ${\hat{h}}_n=g_n$.

If the bounded injective Fourier transform $L^1\to C_0$ were surjective, the open mapping theorem for Banach spaces would make its inverse bounded, yielding $\Vert h{\Vert }_1\le C\Vert \hat{h}{\Vert }_{\infty }$. But the preceding $g_n$ have uniform norm $2$, whereas $\Vert h_n{\Vert }_1$ diverges.

For a quantitative lower bound, on $0<x\le 1/4$ we have $\sin (2\pi x)\ge 4x$. On each interval $I_j=[(j+1/12)/n,(j+5/12)/n]$, $|\sin (2\pi nx)|\ge 1/2$. For $1\le j\le \lfloor n/8\rfloor$ these intervals lie below $1/4$ for large $n$, soThis contradicts boundedness of the inverse. Therefore the Fourier transform has proper dense image in $C_0(\mathbb{R})$.