On a step in Brezis-Merle's inequality

Question

I am reading the proof of theorem 1 from Brezis and Merle's paper "Uniform estimates and blow–up behavior for solutions of −δ(u)=v(x)eu in two dimensions". We have a bounded domain $\Omega \subset \mathbb{R}^2$, a function $f \in L^1(\Omega) $, and $u$ a solution to $$ -\Delta u(x) = f(x) $$ in $\Omega$ with $u$ being $0$ on the boundary of $\Omega$. We'd like to show that for each $0< \delta < 4\pi $, one has $$ \int_{\Omega}\exp(\frac{(4\pi-\delta)u(x)}{\int_{\Omega}|f|})dx \leq \frac{4\pi^2diam(\Omega)^2}{\delta}.$$ Brezis and Merle then take $R = 1/2 diam(\Omega)$ so that $\Omega \subset B_{R}$ for some ball, and extend f to be $0$ oustide $\Omega$. They then define $$\bar u(x) = \int_{B_R}\ \log(\frac{2R}{|x-y|})|f(y)|dy,$$ and then claim that $$(*) -\Delta \bar u = |f| $$ in $\mathbb{R}^2$. Now, it is clear that $(*)$ holds when $x$ is not in $B_{R}$ since then there is no singularity in the integrand and one can just pull the Laplacian inside the integrand. However, when $x$ is in $B_{R}$ this becomes much more subtle. I suspect that since this cannot be done for generic integrable $f$ (but can be done for $f \in L^p, 1<p<\infty$ by Calderon-Zygmund theory), we either must interpret (*) weakly, or are using something about the domain and the factor of $R$ in the logarithm.

Answer 1

When $f \in L^1(B_R)$ we may verify that the suggested function $$u(x) = - \frac{1}{2\pi}\int_{B_R} f(y) \log |x-y|\,dy$$ is a distributional solution at the very least. Note that $u \in L^1_{\text{loc}}(\mathbb{R}^2)$, indeed for any $B_r(x_0) \subset \mathbb{R}^2$ by splitting the integral into $B_R\setminus B_{r+1}(x_0)$ and $B_{r+1}(x_0)$ we have \begin{align} \int_{B_r(x_0)} |u(x)| \,dx & \le \frac{1}{2\pi}\int_{B_R \setminus B_{r+1}(x_0)} \left(\int_{B_r(x_0)} \log |x-y| \,dx\right) |f(y)| \,dy \\& \quad + \frac{1}{2\pi} \int_{B_{r+1}(x_0)} \left(\int_{B_r(x_0)} |\log |x-y|| \,dx\right) |f(y)|\,dy = I_1 + I_2.\end{align} For the first integral $I_1$ note that for $y \not\in B_{r+1}(x_0)$ the function $x \mapsto \log |x-y|$ is a positive harmonic in $B_r(x_0)$ therefore by mean value property we have $$\int_{B_r(x_0)} \log |x-y| \,dx = |B_r|\log |x_0 - y| < |B_r|\log (|x_0| + R).$$ As for the second integral $I_2$ for $y \in B_{2r+1}(x_0)$ we have $B_{r}(x_0) \subset B_{|x_0 - y| + r}(y) \subset B_{2r+1}(y)$ so that $$\int_{B_r(x_0)}|\log |x-y||\,dx \le \int_{B_{|x_0 - y| + r}(y)} |\log |x - y||\,dx \le \int_{B_{2r+1}} |\log |x||\,dx < +\infty.$$

Now it's straightforward to verify that $u$ is a distributional solution, by Fubini's theorem we may interchange the order of integration in the usual way to show that $$\int_{\mathbb{R}^2} \left(-\int_{B_R} \frac{1}{2\pi}\log |x-y| f(y) \,dy\right) \Delta \varphi (x) \,dx = \int_{\mathbb{R}^2} f \varphi \,dx$$ for all $\varphi \in C_c^\infty(\mathbb{R}^2)$.

N.B.: The $\log 2R$ is just there to make $\overline{u}$ positive with $2R \ge |x-y|$ for $x,y \in B_R$. You can refer to some standard Potential theory book for the fact that distributional sub/super-harmonic functions actually equal a.e. to a u.s.c. (resp. l.s.c.) sub/super-harmonic function so that the stated maximum principle in the paper may apply (otherwise let me know, I'll try to dig up some reference).