1
$\begingroup$

I have some trouble with proper understanding of $H_0^1$ space. I confess that I am now beginning to study functional analysis and maybe my question may seem rather trivial. However I would like to know if $$H_0^1(\Omega)$$ where $\Omega$ is an open bounded subset of $\mathbb R^N$, is a finite dimensional Hilbert space.

$\endgroup$
  • $\begingroup$ It includes $C_c^\infty (\Omgea)$, which is infinite dimensional. Hence, $H_0^1$ is also infinite dimensional. $\endgroup$ – PhoemueX Jun 3 '15 at 19:38
2
$\begingroup$

It is not finite dimensional.

One way of seeing this is if you consider the simple case $\Omega = ~ ]0,1[ \subset \Bbb R$. Then $H^1_0(\Omega)$ is the set of square integrable functions, whose derivative is also square integrable, and that the trace map on the boundary is $0$.

A subset of this is the set of continuous functions on $]0,1[$ such that vanish at $0$ and $1$. This set is infinite dimensional: think Fourier transform.

In fact, $H^1_0(\Omega)$ is a separable Hilbert space. This means that it admits a countable Hilbert basis.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.