# Brezis (Variational Formulation for Boundary Value Problems)

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 $$t\in [0,1]$$.

• Just be clear, you are asking why $K$ is independent of the choice of $\tilde{g}$? Commented Aug 21, 2019 at 7:41
• That is correct. Commented Aug 21, 2019 at 8:00
• 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). Commented Aug 21, 2019 at 8:09
• 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). Commented Aug 21, 2019 at 8:21
• Yes, I have created a new question for this now. Commented Aug 21, 2019 at 8:24

## 1 Answer

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 $$K=\hat{K}$$.

• 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. Commented Aug 21, 2019 at 8:20
• 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 $\tilde{g}$". Commented Aug 21, 2019 at 8:35
• 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). Commented Aug 21, 2019 at 8:43
• OK. I think I see what is going on now. Commented Aug 21, 2019 at 8:45
• 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 $\mathbb{R}^{N}$). Commented Aug 21, 2019 at 8:47