The problem is like this:
Consider the open set $\Omega \in \Bbb{R}^2$ by $\Omega=\{(x,y) : 0
Is $\Omega$ with Lipschitz boundary? (i.e. the boundary is locally the graph of a Lipschitz function)
Prove that the function $v(x,y)=x^{1-\beta}$ with $\beta <3/2$ satisfies $v \in H^1(\Omega)$.
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?