# A problem about regularity and mean value property in the Merle and Brezis work on $-\Delta u = V(x) \exp u$ in $\mathbb{R}^2$ plane

I'm reading the Theorem2 in UNIFORM ESTIMATES AND BLOW-UP BEHAVIOR FOR SOLUTIONS OF $$-\Delta u=V(x) e^u$$ IN TWO DIMENSIONS They prove that for the solution of $$-\Delta u= V(x)\exp u \text { in } \mathbb{R}^2$$ if $$V \in L^{p}(\mathbb{R}^2)$$, $$\exp u \in L^{p^{\prime}}\left(\mathbb{R}^2\right)$$, with $$1, then $$u \in L^{\infty}\left(\mathbb{R}^2\right)$$. (Here $$1/p + 1/p^{\prime}=1$$)

First fix $$0<\epsilon<1$$, note that for any $$1 \leq p, so they split $$V(x)\exp u$$ as $$V(x)\exp u=f_1+f_2$$ with $$\left\|f_1\right\|_{L^1\left(\mathbb{R}^2\right)}<\epsilon$$ and $$f_2 \in L^{\infty}\left(\mathbb{R}^2\right)$$. Let $$B_r$$ be the ball of radius $$r$$ centered at $$x_0$$.

They denote by $$C$$ various constants independent of $$x_0$$ (but possibly depending on $$\epsilon$$ ). Then let $$u_i$$ be the solution of $$\left\{\begin{array}{rlll} -\Delta u_i & =f_i & \text { in } & B_1 \\ u_i & =0 & \text { on } & \partial B_1 \end{array}\right.$$

By a lemma they proved before, they proved that

Assume $$\Omega \subset \mathbb{R}^2$$ is a bounded domain and let $$u$$ be a solution of

$$\left\{\begin{array}{ccc} -\Delta u=f(x) & \text { in } \quad \Omega, \\ u=0 & \text { on } \quad \partial \Omega, \end{array}\right.$$ Then for every $$\delta \in(0,4 \pi)$$ we have $$\int_{\Omega} \exp \left[\frac{(4 \pi-\delta)] u(x) \|}{\|f\|_1}\right] \mathrm{dx} \leq \frac{4 \pi^2}{\delta}(\operatorname{diam} \Omega)^2 .$$

So applied with $$\delta=4 \pi-1$$, they have $$\int_{B_1} \exp \left[\frac{1}{\epsilon}\left|u_1\right|\right] \leq \mathrm{C}$$ and in particular $$\left\|u_1\right\|_{L^1\left(B_1\right)} \leq C$$. We also have $$\left\|u_2\right\|_{L^{\infty}\left(B_1\right)} \leq C$$. Let $$u_3=u-u_1-u_2$$ so that $$\Delta u_3=0$$ on $$B_1$$. The mean value theorem for harmonic functions implies that $$$$\tag{1} \left\|u_3^{+}\right\|_{L^{\infty}\left(B_{1 / 2}\right)} \leq \mathrm{C}\left\|u_3^{+}\right\|_{L^1\left(B_1\right)}$$$$ (This can be simply verified by the fact that let $$\Omega \subset \mathbb{R}^n$$ open and let $$u$$ be a harmonic function in $$\Omega$$. If $$K \subset \Omega$$ is compact, then $$\sup _{x \in K}|u(x)| \leq \frac{n}{\omega_n \operatorname{dist}(K, \partial \Omega)^n} \int_{\Omega}|u(x)| \mathrm{d} x .)$$

On the other hand they have $$u_3^{+} \leq u^{+}+\left|u_1\right|+\left|u_2\right|$$ and since $$p^{\prime}\int_{R^2} u^{+}\mathrm{d} x \leq \int_{R^2} \exp p^{\prime}u\mathrm{d} x \leq C$$ we see that $$\left\|u_3^{+}\right\|_{L^1\left(B_1\right)} \leq \mathrm{C}$$. Combining this with (1) they find that $$\left\|u_3^{+}\right\|_{L^{\infty}\left(B_{1 / 2}\right)} \leq C.$$ Finally they write $$$$\tag{2} -\Delta u= V(x)\exp u= V(x)\exp u_1 \exp (u_2+u_3)=g$$$$ (with $$V \in L^p(B_1)$$, $$\|g\|_{L^{1+\delta}\left(B_{1 / 2}\right)} \leq C$$ for some $$\delta>0$$ (since $$e^{u_2+u_3} \in L^{\infty}\left(B_{1 / 2}\right)$$, $$\exp u_1 \in L^{1 / \epsilon}\left(B_1\right)$$ with $$\left.1 / \epsilon> p^{\prime}\right)$$.

What I'm confused is the last step, they said they use once more the mean value theorem and standard elliptic estimates they deduce from (2) that

$$\left\|u^{+}\right\|_{L^{\mathbb{\infty}}\left(B_{1 / 4}\right)} \leq C\left\|u^{+}\right\|_{L^1\left(B_{1 / 2}\right)}+C \|u\|_{L^{1+\delta}\left(B_{1 / 2}\right)} \leq C .$$ Since $$C$$ is independent of $$x_0$$ they conclude that $$u^{+} \in L^{\infty}\left(\mathbb{R}^2\right)$$.

I'm very confused about how they use the mean value property again and what kind of regularity they use ?

By Caldrron-Zygmund inequality, you may obtain : $$\|u\|_{W^{2,1+\delta}~~(B_{\frac{1}{4}})} \leq C\|u\|_{L^{1+\delta}~~(B_{\frac{1}{2}})} +C\|g\|_{L^{1+\delta}~~(B_{\frac{1}{2}})}.$$ Since the dimension of space is 2, $$\|u\|_{L^{\infty}~~(B_{\frac{1}{4}})}\leq C\|u\|_{W^{2,1+\delta}~~(B_{\frac{1}{4}})}.$$ Using the $$L^1$$- theory, see
you can obtain $$\|u_1\|_{L^{1+\delta} ~~(B_{\frac{1}{2}})}$$ is bounded. By the mean value theorem or harnack inequality you can obtain $$\|u_3\|_{L^{\infty}~~(B_{\frac{1}{2}})}\leq C$$, combine with the fact that $$\|u_2\|_{L^{\infty}~~(B_{\frac{1}{2}})}\leq C$$ you may obtain $$\|u\|_{L^{1+\delta}~~(B_{\frac{1}{2}})} \leq C,$$ and finish the proof.
• Thanks for your answer! I want to know what this $L^1$-theory is. Aug 18, 2023 at 9:29