Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 3 / 20F / iv / Solution

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 $C_0(\mathbb{R})$.