If $s>d/2$, then $H^s(\mathbb R^d)\hookrightarrow L^\infty$. Indeed Fourier inversion and Cauchy-Schwarz give $|u(x)|\le C\|(1+|\xi|^2)^{s/2}\hat u\|_2\|(1+|\xi|^2)^{-s/2}\|_2$, and the second factor is finite exactly when $s>d/2$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 2 / 23H / b / Solution

 Solution (source code)