In section 9.5 of Brezis' text, Brezis goes through the example of the homogeneous Dirichlet problem. My issue is with the final step where he recovers a classical solutions from the weak solutions. The argument is as follows:
Assume that the weak solution $u\in H^{1}_{0}(\Omega)$ belongs to $C^{2}(\overline{\Omega})$, and assume that $\Omega$ is of class $C^{1}$. Then $u=0$ on $\partial\Omega$ (by Theorem 9.17). On the other hand, we have, \begin{align} \int_{\Omega}(-\Delta u+u)v=\int_{\Omega}fv\quad\forall v\in C_{c}^{1}(\Omega), \end{align} and thus $-\Delta u+u=f$ a.e. on $\Omega$. In fact, $-\Delta u+u=f$ everywhere on $\Omega$, since $u\in C^{2}(\Omega)$; thus $u$ is a classical solution.
Can someone explain the reasoning here. What is the logical process from initially assuming $u\in C^{2}(\overline{\Omega})$ and $\Omega$ being of class $C^{1}$? How does this assumption, implying $u=0$ on $\partial\Omega$, allow us to conclude that the weak solution is in fact zero on $\partial\Omega$?
For the second part why do we not require $f$ to be continuous in order to conclude $-\Delta u+u=f$ everywhere on $\Omega$?
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}
(1)
We are assuming $u$ is a weak solution (of the homogeneous Dirichlet problem), $u\in C^2(\bar{\Omega})$, and $\Omega$ is of $C^1$ class.
Since $u\in H^1_0(\Omega)\cap C(\bar{\Omega})$ and $\Omega$ is of class $C^1$, Theorem 9.17 implies $u$, the weak solution, vanishes on $\partial \Omega$.
(2)
The assumption does not require $f$ to be continuous, but it turns out (in this case) $f$ can be redefined (on a null set) as a continuous function:
$$-\Delta u+u=f$$ a.e., which means $f$ can be redefined (on a null set) so that the equation holds everywhere. Then $f$ has to be continuous since the left hand side is continuous.