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## Sobolev space impossible extension

Let $\Omega=\{ (x,y) \in \Bbb{R}^2 : x \in (0,1),\ y \in (0,x^2)\}$. And for $\beta < 3/2$ (!!! Correction: $1<\beta<3/2$) consider $v: \Omega \to \Bbb{R},\ v(x,y)=x^{1-\beta}$. Prove that:

1) $\Omega$ does not have Lipschitz boundary (i.e. its boundary is not locally the graph of Lipschitz functions).

2) $v \in H^1(\Omega)$.

3) For every ball $B$ which contains $\Omega$, there is no function in $H^1(B)$ which extends $v$.

This problem gives a counter example which states that if $\Omega$ doesn’t have Lipschitz boundary, then there may be no extension to some functions in $H^1(\Omega)$ to greater Sobolev spaces.

Proof: 1) We use the fact that a set has Lipschitz boundary if and only if it has the $\varepsilon$-cone property (that is, for every $x \in \partial \Omega$ there exists a unit vector $\xi \in \Bbb{R}^2$ such that for every $y \in B(x,\varepsilon)$ the cone $C(y,\xi,\varepsilon)=\{ z \in \Bbb{R}^2 : \langle z-y,\xi \rangle \geq \cos \varepsilon |z-y| \text{ and }0<|z-y|<\varepsilon\}$ is contained in $\Omega$). Pick $(0,0)$ which is on the boundary of $\Omega$, and since the graph of $y=x^2$ is tangent to $Ox$ in the origin, if we pick some $y$ close to $(0,0)$ in $\Omega$, its $\varepsilon$-cone will not be contained in $\Omega$. Therefore $\Omega$ does not have Lipschitz boundary.

Another approach would be to argue by contradiction. Suppose $\Omega$ has Lipschitz boundary, so around the origin $(0,0)$ $\partial\Omega$  is the graph of a Lipschitz function. That is not possible, since the boundary of $\Omega$ around $(0,0)$ is not the graph of any function, no matter what coordinates we establish centered in $(0,0)$.

2) Notice that the partial derivatives with respect to $y$ are zero, and $v$ is smooth with respect to $x$ in $\Omega$. So the partial derivatives exist and we must check that they and $v$ belong to $L^2(\Omega)$. To do this, note that $\displaystyle \int_\Omega v^2=\int_0^1 \int_0^{x^2} x^{2-2\beta}dydx=\int_0^1 x^{4-2\beta}dx=\frac{1}{5-2\beta}<\infty$. The partial derivative with respect to $y$ is zero, and therefore belongs to $L^2(\Omega)$. The partial derivative with respect to $x$ is $\displaystyle \frac{\partial v}{\partial x}=(1-\beta)x^{-\beta}$. To check that this belongs to $L^2$, see that $\displaystyle \int_\Omega \left(\frac{\partial v}{\partial x}\right)^2=(1-\beta)^2 \int_0^1 \int_0^{x^2} x^{-2\beta}dydx=(1-\beta)^2 \int_0^1 x^{2-2\beta}dx=\frac{(1-\beta)^2}{3-2\beta}$ which is finite.

3) Suppose that $v$ admits an extension to $H^1(B)$. Since $B$ is a ball and therefore it is bounded and of class $C^1$, it follows by the Rellich-Kondrachov Theorem (See Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (2011) Thm 9.16) that $H^1(B) \subset L^q(B)$ for all $q \geq 2$ with compact injection. Anyway, this means that the extension $w$ of $v$ is in $L^q(B)$ for $q \geq 2$. But then $\displaystyle \int_B |w|^q \geq \int_\Omega |w|^q=\int_\Omega |v|^q=\int_0^1 x^{2+(1-\beta)q}dx$, which for $q$ large enough it is not finite since $2+(1-\beta)q$ becomes negative. Contradiction.