 Solution

For $s>n/2$ and $u\in H^s({\mathbb{R}}^n)$, Fourier inversion and Cauchy--Schwarz giveThe last integral is finite exactly when $s>n/2$. Thus the sobolev embedding theorem gives

To see why the endpoint relevant to $H^1({\mathbb{R}}^3)$ fails, choose a smooth cutoff $\chi$ supported near the origin and equal to one there, and setNear zero, $|u{|}^2$ contributes ${\int }_0^1{r}^{3/2}dr$, while $|\nabla u{|}^2$ contributes a constant multiple of ${\int }_0^1{r}^{-1/2}dr$. Both are finite, so $u\in H^1({\mathbb{R}}^3)$, but $u$ is unbounded.