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suppose $\Omega$ is an open set in $\mathbb R^n$, for any $m\ge 0$, consider the norm on $C_0^m(\Omega)$:

$$||u||_m:=\left (\sum_{|\alpha|\le m}\ \int_{\Omega}|\partial_{\alpha} u(x)|^2 d x\right)^{1/2}$$

Let the completion of this norm be $H_0^m(\Omega)$.

My question is: why elements in $H_0^m(\Omega)$ are functions:$\Omega\to \mathbb R$? Because we cannot deduce pointwise convergence by the convergence of this norm.

I can see that a Cauchy sequence with respect to this norm is convergent in measure, hence has a subsequence convergent almost everywhere, but there is still a zero measure set that we can not define the value of the function.

Further, if we do know they are functions, then what properties can we deduce from the completion?

Richard
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1 Answers1

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Elements of $H_0^m(\Omega)$ are functions in the $L^2$ sense (i.e. they are actually equivalence classes of functions that are equal almost everywhere).

To see that you cannot do better than this in general consider the case $m = 0$. $\|\cdot\|_0$ is nothing but the $L^2$-norm on $C_0(\Omega)$ and so by a standard density result we can identify $H_0^0(\Omega) = L^2(\Omega)$.

For $m > 0$, note that convergence for $\|\cdot\|_m$ implies convergence for the $L^2(\Omega)$ norm so that $H_0^m(\Omega) \subseteq L^2(\Omega)$.

Rhys Steele
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  • You can do better for certain combinations of the parameters by Sobolev embedding. E.g. $H^1(\mathbb{R})$ can indeed be identified with a space of continuous functions. – Klaus Sep 15 '22 at 13:13
  • @Klaus Hence the wording "you cannot do better than this in general ". Note that the question does not restrict to any setting where we can guarantee such an embedding. – Rhys Steele Sep 15 '22 at 13:18
  • Thanks for your answer. I have one more question: can we embed $H_0^1(\Omega)$ into $C(\Omega)$? I know the case for $H^1$ is correct. But how about the $H_0^1$? – Richard Sep 20 '22 at 02:10
  • @Richard $H^1$ embeds into a space of continuous functions in the one-dimensional case but not in higher integer dimensions. See for example this question to see that in 2 dimensions that result already fails. However, it is straightforward to see that $H_0^1$ can be considered as a subset of $H^1$ since $C_0^m \subseteq H^1$ and $H_0^1$ is the completion with respect to the $H^1$ norm. This means that in the one dimensional case $H_0^1(\Omega) \subseteq C(\Omega)$. – Rhys Steele Sep 20 '22 at 09:30
  • @RhysSteele Thank you very much! – Richard Sep 21 '22 at 15:40