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.

  • $\begingroup$ 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

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.

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