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 $t\in [0,1]$.
Best 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 $K=\hat{K}$.