= Solution
For $s>n/2$ and $u\in H^s(\mathbb R^n)$, Fourier inversion and Cauchy--Schwarz give
$$
|u(x)|\leq C\int|\widehat u(\xi)|\,d\xi
\leq C\left(\int(1+|\xi|^2)^s|\widehat u|^2d\xi\right)^{1/2}
\left(\int(1+|\xi|^2)^{-s}d\xi\right)^{1/2}.
$$
The last <integral> is finite exactly when $s>n/2$. Thus the <sobolev embedding theorem> gives
$$
\|u\|_\infty\leq C_{n,s}\|u\|_{H^s}.
$$
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 set
$$
u(x)=\chi(x)|x|^{-1/4}.
$$
Near zero, $|u|^2$ contributes $\int_0^1r^{3/2}dr$, while $|\nabla u|^2$ contributes a constant multiple of $\int_0^1r^{-1/2}dr$. Both are finite, so $u\in H^1(\mathbb R^3)$, but $u$ is unbounded.
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