Home > Partial Differential Equations, Sobolev Spaces > Sobolev space impossible extension

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.

Advertisements
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: