Function in $H^1$, but not continuous

By the Sobolev embedding theorem, if $\Omega$ is bounded, $H^s(\Omega)\subset C(\Omega)$ for $s>1$, in $\mathbb R^2$. Where I can find a counterexample (if one exists) for the case $s=1$? I mean a discontinuous function ($n=2$), that belongs to $H^1$.

Solution 1:

Let $\Omega$ be the disk of radius $r<1$ centered at the origin. Define $u(x,y) = (-\log(x^2+y^2))^{1/3}$. Clearly, this is not a continuous function.

But on every line not passing through the origin, this function is Lipschitz, therefore absolutely continuous. It remains to check that its partial derivatives are in $L^2$. By the chain rule, $$u_x(x,y) = \frac13 (-\log(x^2+y^2))^{-2/3} (x^2+y^2)^{-1} (2x)$$ Since $|x|\le (x^2+y^2)^{1/2}$, it follows that $$|u_x(x,y)| \le (-\log(x^2+y^2))^{-2/3} (x^2+y^2)^{-1/2} $$ This function is square integrable, as one can see by integration in polar coordinates. The same applies to $u_y$; indeed we can simply relabel $x$ and $y$.