3
$\begingroup$

Let $\Omega\subset\mathbb{R}^n$ be an open set. Let $H^{m,p}(\Omega)$ denote the Sobolev-space of at most $m$ times weakly differentiable functions whose weak derivatives are all $L^p$. Further let $C^{\infty}(\Omega)$ and $C_0^{\infty}(\Omega)$ denote respectively the smooth functions on $\Omega$ in the former case, and those who have compact support in the latter case.

There is a theorem that states that $C^{\infty}(\Omega)\cap H^{m,p}(\Omega)$ is dense in $H^{m,p}(\Omega)$. I'm wondering now if $C_0^{\infty}(\Omega)\cap H^{m,p}(\Omega)$ is dense in $H^{m,p}(\Omega)$.

Is this true or false? Could you please give me hints how to proove it or a link to literature on this?

Thank you in advance.

$\endgroup$

2 Answers 2

2
$\begingroup$

In general, it is not. For $\Omega = (0,1)$, how can you approximate the constant function $u=1$ by elements with compact support?

$\endgroup$
4
  • $\begingroup$ I don't know how. Do you have a proof that it is not possible to do so? $\endgroup$
    – Sh4pe
    Sep 24, 2012 at 9:58
  • $\begingroup$ Well, in dimension one, $H^1$-functions are continuous. $\endgroup$
    – Siminore
    Sep 24, 2012 at 10:01
  • $\begingroup$ I don't see how this helps... $\endgroup$
    – Sh4pe
    Sep 24, 2012 at 10:03
  • $\begingroup$ If a sequence converges in $H^1(a,b)$ then it converges uniformly in $(a,b)$. $\endgroup$
    – Siminore
    Sep 24, 2012 at 10:05
2
$\begingroup$

It can't happen in any bounded open set $\Omega$ as the constant function equal to $1$ is in $H^{m,p}(\Omega)\cap C^{\infty}(\Omega)$ for $m,p$, but cannot be approach by a sequence of functions $\{\phi_n\}$ of $C_0^{\infty}(\Omega)$ for the norm $W^{m,p},m\geq 1$. Otherwise, we would have $\nabla\phi_n\to 0$ in $L^p$ so by Poincaré's inequality $\phi_n\to 0$ in $L^p$, a contradiction.

$\endgroup$

You must log in to answer this question.

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