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>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 105 / 2 / a / i / Solution

{title2=$H^1(U)\not\hookrightarrow L^\infty(U)$}

On a two-dimensional domain containing the origin, a logarithmic singularity such as $u(r)=(\log(1/r))^\alpha$ with $0<\alpha<1/2$ belongs locally to $H^1$ but is unbounded. Thus the critical first-order <Sobolev space> does not embed into $L^\infty$.

= Sobolev embedding
{synonym}

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