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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 $H^1$.

user127096
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Angie
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  • see also http://math.stackexchange.com/questions/746289/discontinuous-sobolev-function – daw Apr 11 '14 at 05:25

1 Answers1

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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 $y$.

user127096
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  • @mkl314 Thanks. Corrected. – user127096 Apr 13 '14 at 00:16
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    $|\nabla,u|$ is not square integrable over the disk $D={x^2+y^2<r^2<1}$ centered at the origin, since $$\int_{D}|\nabla,u|^2dxdy= \pi\int_0^r\frac{d\rho}{\rho\log{\frac{1}{\rho}}} = -\pi\log{\Bigl(\log{\frac{1}{\rho}}\Bigr)}\Bigl|_0^r=+\infty $$. – mkl314 Apr 13 '14 at 00:26
  • May I know how should I find the weak derivative if the disc has radius larger than $1$? I don't know how to handle the non-differentiable point of $|x|^{\frac{1}{3}}$ at $x=0$. – John Feb 23 '16 at 05:06