The following is in Appendix B of Struwe's Variational Methods
Let $u$ be a solution of $-\Delta u = g(x, u(x))$ in a domain $\Omega \subset \mathbb R^N$, $N \geq 3$, where $g$ is a Carathéodory function with subcritical superlinear growth.
Theorem:
Let $\Omega \subset \mathbb R^N$ be a smooth open set and let $g: \Omega \times \mathbb R \to \mathbb R$ be a Carathéodory function such that
$$
|g(x, u(x))| \leq a(x)(1 + |u(x)|) \quad \text{ a.e. in } \Omega
$$
for some $0 \leq a \in L_{loc}^{N/2}(\Omega)$. Let $u \in H^1_{loc}(\Omega)$ be a weak solution to $-\Delta u = g(x, u)$. Then $u \in L^q_{loc}(\Omega)$ for all $1 < q < \infty$. If $u \in H_0^1(\Omega)$ and $a \in L^{N/2}(\Omega)$, then $u \in L^q(\Omega)$ for all $1 < q < \infty$.
The proof begins as follows:
Take $\eta \in C_c^\infty(\Omega)$, $s \geq 0$ and $L \geq 0$ and let
$$
\varphi = u \min \{|u|^{2s}, L^2\} \eta^2 \in H_0^1(\Omega)
$$
Testing the equation against $\varphi$ yields
$$
\int_\Omega |\nabla u|^2 \min\{|u|^{2s}, L^2\} \eta^2 \ dx + \frac s2 \int_{\{|u|^s\leq L \}} |\nabla(|u|^2)|^2 |u|^{2s – 2} \eta ^2 \ dx \leq \\
-2 \int_\Omega \nabla u u \min \{|u|^{2s}, L^2\} \nabla \eta \eta \ dx + \int_\Omega a(1 + 2|u|^2)\min \{|u|^{2s}, L^2\}\eta^2 \ dx.
$$
Why is $\varphi \in H_0^1(\Omega)$? How do the second term in the left-hand side of the inequality arises?
I tried the following: We want to compute
$$
\int_{\{|u|^s \leq L\}} \nabla u u \nabla |u|^{2s} \eta^2 \ dx .
$$
But
$$
\nabla |u|^{2s} = \nabla(u^+ – u^-)^{2s} = 2s |u|^{2s – 1} \nabla |u|
$$
so we get
$$
\int_{\{|u|^s \leq L\}} \nabla u u \nabla |u|^{2s} \eta^2 \ dx =
2s \int_{\{|u|^s \leq L\}} (\nabla u \nabla |u|) u |u|^{2s – 1} \eta ^2 \ dx
$$
On the other hand,
$$
\frac s2 \int_{\{|u|^s \leq L\}} |\nabla |u|^2|^2 |u|^{2s – 2} \eta^2 \ dx = \frac s2 \int_{\{|u|^s \leq L\}} |2 |u| \nabla |u||^2 |u|^{2s – 2} \eta^2 \ dx \\
= 2s \int_{\{|u|^s \leq L\}} |\nabla |u||^2 |u|^{2s} \eta ^2 \ dx.
$$
How to conclude that these two expressions are the same?
Also, what is the intuition for the proof of this theorem? It is seeming like just a lot of calculations.
Thanks in advance and kind regards.
Best Answer
The function $f\colon \mathbb{R}\to\mathbb{R},\,\lambda\mapsto \lambda \min\{\lvert \lambda\rvert^{2s},L^2\}$ is Lipschitz, thus $f\circ u\in H^1_{loc}(\Omega)$. It's not super fun to prove that; it's clear for smooth $f$ with bounded derivative, and then you have to do some approximation. Anyway, this implies $\phi=\eta^2 (f\circ u)\in H^1_0(\Omega)$ -- just use the product rule and use that $\phi$ has compact support.
As for the second part, your almost there: $\nabla u=\mathrm{sgn}(u)\nabla \lvert u\rvert$, you just have to pull $\mathrm{sgn}(u)$ from the gradient to the factor $u$ to get $\lvert u\rvert$ there.