Why is $H^1 \neq H_0^1$?

Question

I've been doing some reading on Sobolev-spaces and one remark said that $H_0^1$, i.e. the space of $H^1$-functions with zero-boundary values, is not the same as $H^1$. This seems clear to me, but when I tried thinking of a proof for this I had no idea how to proof this.

Answer 1

Consider $\Omega$ the unit ball of $\Bbb R^d$ and the constant function equal to $1$: it's clearly an element of $H^1(\Omega)$. However, it's not in $H^1_0(\Omega)$, otherwise we would contradict Poincaré's inequality ($\lVert u\rVert_{L^2}\leqslant C\lVert \nabla u\rVert_{L^2}$).

In particular, this proves that these space are not equal when such an inequality applies.

A notion which can help to understand the difference is the notion of trace. With good conditions of regularity of the boundary of the considered open set, functions of $H^1_0(\Omega)$ are the elements of $H^1(\Omega)$ whose trace is $0$. Intuitively, this can be explained by the fact that "with approximation by functions with compact support we can't expect to get anything of non-zero on the boundary", and to male this rigorous we need charts.