# Function in $H^1(\Omega)$ which cannot be extended to a greater Sobolev Space

The problem is like this:

Consider the open set $\Omega \in \Bbb{R}^2$ by $\Omega=\{(x,y) : 0<x<1, 0<y<x^2 \}$

1. Is $\Omega$ with Lipschitz boundary? (i.e. the boundary is locally the graph of a Lipschitz function)

2. Prove that the function $v(x,y)=x^{1-\beta}$ with $\beta <3/2$ satisfies $v \in H^1(\Omega)$.

3. Consider any open ball $B$ containing $\Omega$. Prove that there is no function in $H^1(B)$ which extends $v$.

I solved the first two parts, but the real objective is the third: to show that if the boundary of $\Omega$ is not Lipschitz, then some functions from $H^1(\Omega)$ cannot be extended to functions in $H^1(B)$ where $B$ is a ball containig $\Omega$.

First part I solved using the fact that an open set has Lipschitz boundary if and only if it has the $\varepsilon$-cone property (i.e. for every $x \in \partial \Omega$ there exists a unit vector $\xi \in \Bbb{R}^2$ such that for all $y \in \overline{\Omega}\cap B(x,\varepsilon)$ we have that $C(y,\xi,\varepsilon) \subset \Omega$, where $C(y,\xi,\varepsilon)=\{z \in \Bbb{R}^2,\ \langle z-y ,\xi \rangle \geq \cos(\varepsilon)|z-y|$ and $0<|z-y|<\varepsilon\}$ ) If you pick $(0,0)$ which is on the boundary, then there is no $\varepsilon$-cone with vertex in $(0,0)$ contained in $\Omega$.

I would be interested how could I prove directly that $\Omega$ does not have Lipschitz boundary, using only the definition?

For the second part, $v \in L^2$ is almost smooth, and its derivatives are also in $L^2$.

I can't find the idea to solve third part. I was thinking to extend $v$ by $0$ outside $\Omega$. Why is this wrong?

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Since such a ball $B$ is a $C^1$ bounded open, we can use the fact that $H^1(B)\subset L^q(B)$ with $q\geq 2$ and a continuous embedding. If $w\in H^1(B)$ extends $v$, then we should have that $w\in L^q(B)$ for all $q\geq 2$, hence $\int_{\Omega}g^2d\lambda_2<\infty$. Therefore, $-2+(\alpha-1)q<1$ for all $q\geq 2$ and $\alpha\leq 1$. Of course, it works if $\alpha\leq 1$ since $v$ is continuous, but it's not true for $1<\alpha<\frac 32$. – Davide Giraudo Oct 6 '11 at 19:53
@DavideGiraudo: Thank you for your comment. I'll try and work on these ideas. – Beni Bogosel Oct 6 '11 at 20:23

Since such a ball $B$ is a $C^1$ bounded open, we can use the fact that $H^1(B)\subset L^q(B)$ with $q\geq 2$ and a continuous embedding (I think this result in Brezis' book, it's in the French version but I have not the English one).
If $w\in H^1(B)$ extends $v$, then we should have that $w\in L^q(B)$ for all $q\geq 2$, hence $\int_{\Omega}g^2d\lambda_2<\infty$. Therefore, $-2+(\alpha-1)q<1$ for all $q\geq 2$ and $\alpha\leq 1$. Of course, it works if $\alpha\leq 1$ since $v$ is continuous, but it's not true for $1<\alpha<\frac 32$.