 Solution

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