Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 1 / 23H (Analysis of Functions)

With the convention in convolution theorem, absolute integrability permits Fubini:

For Schwartz $f,g$, Fourier inversion and Fubini giveTaking $g=f$ proves Plancherel theorem; polarization gives the full identity.

Approximate $g\in L^2$ by Schwartz $g_n$. Young's inequality gives
$\Vert f\ast (g_n-g){\Vert }_2\le \Vert f{\Vert }_1\Vert g_n-g{\Vert }_2$, while multiplication by bounded $\hat{f}$ is continuous in $L^2$. Pass to the limit in
$\widehat{f\ast g_n}=\hat{f}{\hat{g}}_n$.

If $g\in L^2$ is orthogonal to every translate of $f$, then the correlation $g\ast \tilde{f}$ vanishes. By convolution theorem,
$\hat{g}\overline{\hat{f}}=0$ almost everywhere. Since $\hat{f}$ vanishes only at one point, a null set, $\hat{g}=0$ almost everywhere. By Plancherel theorem, $g=0$. Orthogonal-complement characterization proves density in $L^2$.

The nonzero bounded functionalis continuous on $L^1$. For every translate, $\Lambda (f(\cdot -t))=e^{-2\pi it}\hat{f}(1)=0$. Thus it annihilates $Y$, and density by annihilators shows that $Y$ is not dense in $L^1$.