Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 1 / 23H / d

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 $L^1$.