= Solution
In <polar coordinates>, set
$$
u(x,y)=\left(\log\frac1{\sqrt{x^2+y^2}}\right)^{1/4}
$$
away from the origin, assigning any value at the origin. This is unbounded as $r=\sqrt{x^2+y^2}\downarrow0$. It belongs to $L^2(U)$ because
$$
\int_0^{1/2}r\left(\log\frac1r\right)^{1/2},dr<\infty.
$$
Moreover
$$
|\nabla u|^2=\frac1{16r^2}\left(\log\frac1r\right)^{-3/2},
$$
and hence
$$
\int_U|\nabla u|^2
=\frac{\pi}{8}\int_0^{1/2}\frac{dr}{r(\log(1/r))^{3/2}}<\infty.
$$
Thus $u\in H^1(U)$ but $u\notin L^\infty(U)$, exhibiting the <failure of first-order Sobolev embedding into Linfinity in two dimensions>.
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