Theorem 5 (Weak Maximum Principle)
Suppose that $\Omega$ is bounded and that $L$ is strictly elliptic with $c \leq 0$.
If $u \in C^2(\Omega) \cap C(\bar{\Omega})$ and $L u \geq 0$ in $\Omega$, then a nonnegative maximum is attained at the boundary.Proof.
Suppose that $\Omega \subset \{|x_1| < d\}$.
Consider $w(x) = u(x) + \varepsilon e^{\alpha x_1}$ with $\varepsilon > 0$.
Then
\begin{align*}
L w
&= Lu
+ \varepsilon
\left( \alpha^2 a_{11}(x) + \alpha b_1(x) + c(x) \right)
e^{\alpha x_1} \\
&\geq \varepsilon
\left(
\alpha^2 \lambda – \alpha \| b_1 \|_\infty – \| c \|_\infty
\right)
e^{\alpha x_1}.
\end{align*}
One chooses $\alpha$ large enough to find $L w > 0$.
By the previous lemma $w$ cannot have a nonnegative maximum $\Omega$.
Hence
$$
\sup_\Omega u
\leq \sup_\Omega w
\leq \sup_\Omega w^+
= \sup_{\partial \Omega} w^+
\leq \sup_{\partial \Omega} u^+ + \varepsilon e^{\alpha d}.
$$(Original image here.)
I'm reading this proof of the weak maximum principle. The operator $L$ is this one:
$$L = \sum a_{jk}(x)\partial_j\partial_k + \sum b_j\partial_j$$
but I don't think it's relevant for what I'll ask. I understood everything except for the $\sup$ arguments.
It's natural that $\sup_{\Omega}u\le \sup_{\Omega}w$ since $w = u + \epsilon\mbox{ something positive}$. Now, what does $\sup_{\Omega}w^{+}$ mean? What does the plus sign mean and why $\sup_{\Omega}w^{}\le \sup_{\Omega}w^{+}$? Why, also, $\sup_{\Omega}w^{+} = \sup_{\partial\Omega}w^{+}$.
Why, also, the argument concludes if we let $\varepsilon\to 0$?
Where is used the hypothesis of maximum not existing in $\Omega$?
Best Answer
The theorem is stated incorrectly. In particular, consider the simple counter-example $u = -1$. In this case, $Lu = -c \geq 0$ and clearly $u$ does not attain a nonnegative maximum.
Below is the correct statement with a very detailed proof that should answer all of your questions.
Theorem 5 (Weak Maximum Principle) Suppose that $\Omega$ is bounded and that $L$ is strictly elliptic with $c\leq0$. If $u\in C^{2}(\Omega)\cup C(\overline{\Omega})$, $Lu\geq0$ in $\Omega$, and $\color{blue}{u\geq0 \text{ on } \partial \Omega}$, then a nonnegative maximum is attained at the boundary.
In the proof, we use the notation $a^{+}=\max\{a,0\}$.
Proof. Let $\epsilon>0$ and $w(x)=\epsilon e^{\alpha x_{1}}+u(x)$. Then, \begin{multline*} Lw(x)=\epsilon L[e^{\alpha x_{1}}](x)+Lu(x)\geq\epsilon L[e^{\alpha x_{1}}](x)\\ =\epsilon\left(\alpha^{2}a_{11}(x)+\alpha b_{1}(x)+c(x)\right)e^{\alpha x_{1}}\geq\epsilon\left(\alpha^{2}\lambda-\alpha\left\Vert b_{1}\right\Vert _{\infty}-\left\Vert c\right\Vert _{\infty}\right)e^{\alpha x_{1}}. \end{multline*} Since $\lambda>0$, we can always pick $\alpha$ large enough for $Lw>0$ to hold on all of $\Omega$.
Applying Lemma 4 of http://www.mi.uni-koeln.de/~gsweers/pdf/maxprinc.pdf, $w$ cannot attain a nonnegative maximum in $\Omega$. As a result of this, $w^{+}$ cannot attain a maximum in $\Omega$ since then $w$ would attain a nonnegative maximum in $\Omega$. We summarize this by writing $\sup_{\Omega}w^{+}=\sup_{\partial\Omega}w^{+}$.
Next, note that $$ w^{+}=(u+\epsilon e^{\alpha x_{1}})^{+}\leq u^{+}+(\epsilon e^{\alpha x_{1}})^{+}=u^{+}+\epsilon e^{\alpha x_{1}}. $$ Using the boundedness of $\Omega$, this implies that $\sup_{\partial\Omega}w^{+}\leq\sup_{\partial\Omega}u^{+}+\epsilon c$ for some large enough $c$. Letting $\epsilon\rightarrow0$, we have $\sup_{\partial\Omega}w^{+}\leq\sup_{\partial\Omega}u^{+}$.
Lastly, since $u\leq w\leq w^{+}$, we can combine the inequalities in the previous two paragraphs to get $\sup_{\Omega}u\leq\sup_{\partial\Omega}u^{+}$. Since $u\geq0$ on $\partial\Omega$, we have $\sup_{\partial\Omega}u^{+}=\sup_{\partial\Omega}u$. Putting this all together, $$ \sup_{\Omega}u\leq\sup_{\partial\Omega}u $$ as desired. ∎