## Sobolev space impossible extension

Let . And for (**!!! Correction: **) consider . Prove that:

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

2) .

3) For every ball which contains , there is no function in which extends .

This problem gives a counter example which states that if doesn’t have Lipschitz boundary, then there may be no extension to some functions in to greater Sobolev spaces.

**Proof: **1) We use the fact that a set has Lipschitz boundary if and only if it has the -cone property (that is, for every there exists a unit vector such that for every the cone is contained in ). Pick which is on the boundary of , and since the graph of is tangent to in the origin, if we pick some close to in , its -cone will not be contained in . Therefore does not have Lipschitz boundary.

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

2) Notice that the partial derivatives with respect to are zero, and is smooth with respect to in . So the partial derivatives exist and we must check that they and belong to . To do this, note that . The partial derivative with respect to is zero, and therefore belongs to . The partial derivative with respect to is . To check that this belongs to , see that which is finite.

3) Suppose that admits an extension to . Since is a ball and therefore it is bounded and of class , it follows by the **Rellich-Kondrachov** Theorem (See Brezis, *Functional Analysis, Sobolev Spaces and Partial Differential Equations* (2011) Thm 9.16) that for all with **compact injection.** Anyway, this means that the extension of is in for . But then , which for large enough it is not finite since becomes negative. Contradiction.