extension by zero for sobolev functions.

Question

Given some function in $H^1(A)$, and if $B$ is an open subset (A also open) containing $A$, do we get an element of $H^1(B)$ if we just extend by 0? I dont think so, but what would be a simlple counterexample? Why exactly is it impossible to do this in general?

Answer 1

A simple counterexample is $A=(0,1)$, $B=(-1,1)$. Set $v=1$ on $A$. Then $v\in H^1(A)$, but the extension by zero is not in $H^1(B)$: $v$ is discontinuous on $B$, but by Sobolev embeddings $H^1(B)$ is continuously imbedded into $C(\bar B)$.

For regular domains, $H^1$ functions that can be extended by zero and still are $H^1$ are precisely the funcitons in $H^1_0$.