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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-105/2/a/i/solution