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Brezis considers the inhomogeneous Dirichlet problem, \begin{align} -\Delta u+u=f\quad &\text{in }\Omega,\\ u=g\quad &\text{on }\partial\Omega, \end{align} where $f$ is given on $\Omega$ and $g$ is given on $\partial\Omega$. In order to build up to the proposition which proves a unique weak solution exists for this problem Brezis first wishes to construct a closed convex set in $H^{1}(\Omega)$. He does this as follows:

Suppose that there exists a function $\tilde{g}\in H^{1}(\Omega)\cap C(\overline{\Omega})$ such that $\tilde{g}=g$ on $\partial\Omega$. and consider the set, \begin{align} K=\{v\in H^{1}(\Omega)\,|\,v-\tilde{g}\in H^{1}_{0}(\Omega)\}. \end{align} It follows from Theorem 9.17 that $K$ is independent of the choice of $\tilde{g}$ and depends only on $g$.

Where I require assistance: Theorem 9.17 states,

Theorem 9.17: Suppose $\Omega$ is of class $C^{1}$. Let, \begin{align} u\in W^{1,\,p}(\Omega)\cap C(\overline{\Omega})\quad\text{with }1\leq p<\infty. \end{align} Then the following properties are equivalent: \begin{align} \text{(i)}&\,u=0\text{ on }\partial\Omega,\\ \text{(ii)}&\, u\in W^{1,\,p}_{0}(\Omega). \end{align}

The way I see this working is that to show that $v-\tilde{g}\in H^{1}_{0}(\Omega)$ we can show that $v-\tilde{g}=0$ on $\partial\Omega$ and property (ii) from Theorem 9.17 follows (since the proof of (i)$\implies$(ii) does not require any assumptions on the smoothness of $\Omega$). Then from earlier we have $v=\tilde{g}=g$ on $\partial\Omega$, and so we drop the dependence on $\tilde{g}$. However, we do not know that $v-\tilde{g}\in H^{1}(\Omega)\cap C(\overline{\Omega})$ so how are we reaching this conclusion?

Showing $K$ is convex:

Suppose $u,v\in K$. Consider $t\in\mathbb{R}$, then, \begin{align} tv+(1-t)u=t(v-u)+u. \end{align} Take $\tilde{g}$ as before, then \begin{align} t(v-u)+u-\tilde{g}=t(v-\tilde{g})-t(u-\tilde{g})+(u-\tilde{g})\in H_{0}^{1}(\Omega), \end{align} since $H^{1}_{0}(\Omega)$ is a linear space. Hence $tv+(1-t)u\in K$ for $t\in [0,1]$.

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  • $\begingroup$ Just be clear, you are asking why $K$ is independent of the choice of $\tilde{g}$? $\endgroup$
    – StarBug
    Commented Aug 21, 2019 at 7:41
  • $\begingroup$ That is correct. $\endgroup$ Commented Aug 21, 2019 at 8:00
  • $\begingroup$ OK. I guess the problem in your argument is that you want to apply Thm 9.17 to $v-\tilde{g}$, which is not a continuous function and Thm 9.17 therefore not applicable. But you only need to apply Thm 9.17 to the difference of two continuous extensions (see my answer below, hope it helps). $\endgroup$
    – StarBug
    Commented Aug 21, 2019 at 8:09
  • $\begingroup$ You probably should ask the Extra Question as a new question. It doesn't really have anything to do with the first one. The comment you quote from Brezis is definitely not sufficient to understand why a weak solution is a classical solution when the data is sufficiently smooth. Most textbooks have a whole chapter devoted to this question (regularity). $\endgroup$
    – StarBug
    Commented Aug 21, 2019 at 8:21
  • $\begingroup$ Yes, I have created a new question for this now. $\endgroup$ Commented Aug 21, 2019 at 8:24

1 Answer 1

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To show that $K$ is independent of the choice of extension $\tilde{g}$, consider another extension $\hat{g}\in H^1(\Omega)\cap C(\overline{\Omega})$ with $\hat{g}=g$ on $\partial\Omega$. Put $$ \hat{K}:=\{v \in H^1(\Omega)\ |\ v-\hat{g}\in H^1_0(\Omega) \}. $$

We have to verify that $K=\hat{K}$. To this end consider $v\in K$. Then $v-\tilde{g}\in H^1_0(\Omega)$. From Theorem 9.17 we deduce that also $\hat{g}-\tilde{g}\in H^1_0(\Omega)$. Since $H^1_0(\Omega)$ is a linear space, we obtain $$ v-\hat{g} = v- \tilde{g} + (\tilde{g}-\hat{g}) \in H^1_0(\Omega). $$ Consequently, $v\in\hat{K}$. We conclude $K\subset\hat{K}$. The opposite inclusion follows by a similar argument, and we thus obtain $K=\hat{K}$.

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  • $\begingroup$ While what you have here is fine, does it really show the dependence of $K$ on $g$? I believe the dependence on $g$ Brezis talks about is that for every $v\in K$ $v=g$ on $\partial\Omega$, however this does not come across, at least for me, from your argument. $\endgroup$ Commented Aug 21, 2019 at 8:20
  • $\begingroup$ Just to be absolutely clear: $K$ depends on $g$. However, $g$ does not appear explicitly in the definition of $K$, but only implicitly via the extension $\tilde{g}$. Since such an extension is not unique, it is only natural to ask whether a different extension might yield a different set $K$. This is what is typically behind the phrase "$K$ is independent of the choice of $\tilde{g}$". $\endgroup$
    – StarBug
    Commented Aug 21, 2019 at 8:35
  • $\begingroup$ The question "is $v=g$ on $\partial\Omega$ for every $v\in K$" from your comment above is an entirely different question. In order to answer it, you first have to define what the value/meaning of $v$ is on the boundary $\partial\Omega$, which is by no means trivial since you only know $v\in H^1(\Omega)$. For this pupose you need the notion of the trace of $v$. It seems to me, however, that Brezis is trying to avoid introducing this notion (it is complicated). $\endgroup$
    – StarBug
    Commented Aug 21, 2019 at 8:43
  • $\begingroup$ OK. I think I see what is going on now. $\endgroup$ Commented Aug 21, 2019 at 8:45
  • $\begingroup$ One extra question regarding $K$. Brezis mentions it is nonempty, is this just because $\tilde{g}\in K$ always? (I should mention for this example $\Omega$ is an open bounded set in $\mathbb{R}^{N}$). $\endgroup$ Commented Aug 21, 2019 at 8:47

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