7
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ a) what is the domain on which this space is considered? This is important. b) is the constant function $1$ in $H^1$? Is it in $H_0^1$? $\endgroup$
    – 75064
    May 2, 2013 at 19:39

1 Answer 1

7
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Intuitively this would mean if $u \in H_0^1$ and $f:\mathbb{R}\rightarrow \mathbb{R}$ is such that $f(0)=0$ and $f(u)\in H^1$ then $f(u)$ is zero because it must have boundary values that are zero. But how to prove that in terms of the trace operator? $\endgroup$ Dec 13, 2023 at 21:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .