# Brezis-Kato regularity argument - Some questions about Struwe's proof Part II

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 goes 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 $$\int_\Omega |\nabla u|^2 \min\{|u|^{2s}, L^2\} \eta^2 \ dx + \frac s2 \int_{\{|u|^s\leq L \}} |\nabla(|u|^2)|^2 |u|^{2s - 2} \eta ^2 \ dx \\ \leq -2 \int_\Omega \nabla u u \min \{|u|^{2s}, L^2\} \nabla \eta \eta \ dx + \int_\Omega a(1 + 2|u|^2)\min \{|u|^{2s}, L^2\}\eta^2 \ dx \\ (*) \quad {\leq} \frac 12 \int_\Omega |\nabla u|^2 \min\{|u|^{2s}, L^2\}\eta^2 \ dx + c \int_\Omega |u|^2 \min\{|u|^{2s}, L^2\} |\nabla \eta|^2 \ dx \\ \quad + 3 \int_\Omega a|u|^2 \min\{|u|^{2s}, L^2\} \eta^2 \ dx + \int_\Omega |a|\eta^2 \ dx$$

Why does $$(*)$$ hold?

Thanks in advance and kind regards.

• Have you tried to apply Young's inequality, i.e. $xy\leq \frac{\epsilon^2}{2}x^2+\frac 1{\epsilon^2}y^2$? That's what it looks like at first glance, though it might take some time to figure out what $x$, $y$ and $\epsilon$ should be. – MaoWao Jul 23 at 9:57
• @MaoWao Thank you very much (for this comment and the other answer). I didn't think about Young's inequality here yet, will surely give it a thought. – Danilo Gregorin Jul 24 at 17:40

Recall Young's Inequality with an epsilon: $$ab \leq \frac 1{2\varepsilon} a^2 + \frac \varepsilon2 b^2.$$ We apply this inequality to the first term, which gives \begin{align*} - 2 \int_\Omega \nabla u u \min\{|u|^{2s}, L^2\} \nabla \eta \eta \ dx \leq & \frac 12 \int_\Omega |\nabla u|^2 \min \{|u|^{2s}, L^2\} \eta^2 \ dx \\ & + c \int_\Omega u^2 \min\{|u|^{2s}, L^2\} |\nabla \eta|^2 \ dx. \end{align*} As for the second term, we can estimate it as \begin{align*} \int_\Omega a(1+|u|^2 \min\{|u|^{2s}, L^2\} \eta^2 \ dx = & \int_\Omega a \min\{|u|^{2s}, L^2\} \eta^2 \ dx \\ & + 2 \int_\Omega a |u|^2 \min\{|u|^{2s}, L^2\} \eta^2 \ dx \\ = & 3 \int_\Omega a|u|^2 \min\{|u|^{2s}, L^2\} \eta^2 \ dx \\ & + \int_\Omega a\min\{|u|^{2s}, L^2\} \eta^2 (1 - |u|^2) \ dx \\ \leq & 3 \int_\Omega a|u|^2 \min\{|u|^{2s}, L^2\} \eta^2 \ dx \\ & + \int_\Omega a \eta^2 \ dx. \end{align*} Therefore, \begin{align} \begin{split} & \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 \\ & \qquad \leq \frac 12 \int_\Omega |\nabla u|^2 \min \{|u|^{2s}, L^2\} \eta^2 \ dx + c \int_\Omega u^2 \min\{|u|^{2s}, L^2\} |\nabla \eta|^2 \ dx \\ & \qquad \quad + 3 \int_\Omega a|u|^2 \min\{|u|^{2s}, L^2\} \eta^2 \ dx + \int_\Omega a \eta^2 \ dx. \end{split} \end{align}