1. I was going through the paper of Brezis and Merle [1]. In theorem 3 step 2 it's written that the boundedness of $f\_n=V\_ne^{u\_n}$ in $L\_{loc}^p(\Omega)$ implies $\mu\in L^1(\Omega)\cap L\_{loc}^p(\Omega)$. Let $v\_n$ be a solution of $-\Delta v\_n=f\_n$ in $\Omega$ and $v\_n=0$ in $\partial\Omega$. Then $v\_n\to v$ uniformly on compact subsets of $\Omega$ where $v$ is solution of $-\Delta v=\mu$ in $\Omega$ and $v=0$ in $\partial\Omega$. I am not getting how this conclusion follows i.e. why this implies the uniform convergence of $v\_n\to v$ on compact subsets of $\Omega $.


2. At the very beginning they prove a basic inequality which they use later. In the last line of the proof they mention that for $y\in B\_R$ we have
$$\DeclareMathOperator{\dmu}{d\!}
\int\limits\_{B\_R}\left(\frac{2R}{|x-y|}\right)^{2-\frac{\delta}{2\pi}} \dmu x\leq\int\_{B\_R}\left(\frac{2R}{|x|}\right)^{2-\frac{\delta}{2\pi}}\dmu x.
$$
Does this result follows from translation? Otherwise $|x-y|\ngeq |x|$ in general. And also why
$$\DeclareMathOperator{\diam}{diam}
\int\limits\_{B\_R}\left(\frac{2R}{|x|}\right)^{2-\frac{\delta}{2\pi}}\dmu x=\frac{4\pi^2}{\delta}\big(\diam(\Omega)\big)^2\;?
$$
Also this is not coming from radial integration as one term $2^{2-\frac{\delta}{2\pi}}$ survives.


Any idea would be very much helpful.


**Reference**


[1] Haïm Brézis, Frank Merle, "Uniform estimates and blow-up behavior for solutions of $−Δu=V(x)e^u$
in two dimensions" Communications in Partial Differential Equations 16, No. 8-9, 1223-1253 (1991), [MR1132783](https://mathscinet.ams.org/mathscinet-getitem?mr=MR1132783), [Zbl 0746.35006](https://zbmath.org/0746.35006).



Two doubts in the paper of Brezis Merle in blow up analysis of the equation $-\Delta u=Ve^u$
$\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}$
I realize this is a bit late but I'd like to elaborate on Professor Metafune's very interesting and helpful comments. I'll prove how your claim 1. follows from the following:


**Lemma 1**
Let $\Omega$ be a $C^{1,1}$ bounded domain in $\mathbb{R}^d$, $d \geq 2$. Suppose $f \in L^1(\Omega)$ and $ u \in W\_{0}^{1,1}(\Omega)$ is a weak solution to the equation $$ -\Delta u = f .$$ Then if $ 1\leq q < \frac{d}{d-1} $there is a constant $C=C(q,d)$ such that $$ \norm u\_{W\_{0}^{1,q}} \leq C \norm f\_{L^1}.$$


Then I'll provide a proof of this Lemma that I found in Prof. Luigi Orsina's notes [Elliptic equations
with measure data]


[1]: [https://www.ugr.es/~edp/Talks/Luigi-Abril\_2009/Curso.pdf](https://www.ugr.es/%7Eedp/Talks/Luigi-Abril_2009/Curso.pdf). I'm not sure who the proof is originally due to but I suspect it is Stampacchia.


**Proof of 1.**


Set $ w\_{n} = v\_{n} -v $, so that by Lemma 1, for all $ 1< q < 2$, we have $$ \|w\_{n}\|\_{W^{1,q}(\Omega)} \leq C=C(q) .$$ Now by the Sobolev embedding theorem, $$ \|w\_{n}\|\_{L^p(\Omega)} \leq C=C(p) .$$ Indeed, if $p >2$ this follows directly from above, and otherwise it follows directly from above and Hölder's inequality. Fix a compact subset $ K$ of $\Omega$. By the Calderón–Zygmund estimate, combined with the above, $$ \|w\_{n}\|\_{W^{2,p}(K)} $$ $$\leq C (\| \Delta w\_{n} \|\_{L^p(K)} + \|w\_{n}\|\_{L^p(\Omega)}) \leq C=C(K,p),$$ since $f\_{n}$ are uniformly bounded in $L^p\_{loc}.$ By the Sobolev embedding again , there exists $\alpha > 0 $ such that $$\|w\_{n}\|\_{C^{0,\alpha}(K)} \leq C .$$ Thus by Arzela-Ascoli, $w\_{n}$ must converge uniformly on $K$ after passing to a subsequence. The uniform limit must be $0$ by the weak convergence to $0$. This proves 1.


**Now here is the proof of Lemma 1**


We denote various function spaces without the domain $\Omega$, as $\Omega$ is understood to be the domain of all functions defined here. Roughly speaking, we truncate $f$, work with solutions to the Poisson equation with this truncated $f$, and estimate the measures of level sets using the Layer-Cake formula. Define $f\_{n} = f1\_{\{|f|<n\}}$, so that clearly $f\_{n} \in L^{\infty} $, $ \|f\_{n}\|\_{L^1} \leq \|f\|\_{L^1} $ and $ f\_{n} \to f $ in $L^1.$ Consider $u\_{n}$ to be the $H\_{0}^1$ solution to $$ (1) \hspace{1 cm} -\Delta u\_{n} = f\_{n}. $$Define $$u\_{n}^k = u\_{n}1\_{\{|u\_{n}| <k \}} + k1\_{\{|u\_{n} \geq k\}}.$$ Then $u\_{n}^k$ is a valid test function to test equation (1) against. Since $\nabla u\_{n} = \nabla u\_{n}^k$ on the support of $ \nabla u\_{n}^k, $ we get $$(2) \hspace{1 cm} \int\_{\Omega} |\nabla u\_{n}^k|^2 \leq \int\_{\Omega} f\_{n}u\_{n}^{k}\leq k\|f\|\_{L^1}. $$ Denote by $2^\* $ the Sobolev conjugate to $2$, if $d \geq 3.$ If $ d = 2, $ we simply work with some $p >2$ in what follows and the argument does not change substantially. By Sobolev's inequality and (2) , $$\bigg(\int\_{\Omega} (u\_{n}^k)^{2^\*} \bigg)^{\frac{2}{2^\*}} \leq Ck\|f\|\_{L^1} .$$ Set $A\_{n,k} = \{ |u\_{n}| > k \} . $ From the above estimate, and using that $u\_{n}^k = k$ on $A\_{n,k}$, we obtain $$(3) \hspace{1 cm} |A\_{n,k}| \leq \big(\frac{\|f\|\_{L^1}}{k}\big)^{\frac{d}{d-2}}.$$ Now fix $\lambda > 0$. Observe that $$ \{|\nabla u\_n | > \lambda \} \subset A\_{n,k} \cup \{ |\nabla u\_{n}| > \lambda \cap |u\_{n}| < k\}.$$ However using Chebyshev and (2), we get $$(4) \hspace{1 cm} |\{ |\nabla u\_{n}| > \lambda \cap |u\_{n}| < k\} | \leq |\{ |\nabla u\_{n}^k | > \lambda \}| \leq \frac{1}{\lambda ^2}\int\_{\Omega}|\nabla u\_{n}^k|^2 \leq \frac{k\|f\|\_{L^1}}{\lambda ^2} .$$ In total, using (3) and (4) we get the following upper bound : $$(5) \hspace{1 cm} | \{|\nabla u\_n | > \lambda \} | \leq |A\_{n,k}| + \frac{k \|f\|\_{L^1}}{\lambda ^2} \leq \big(\frac{\|f\|\_{L^1}}{k}\big)^{\frac{d}{d-2}} + \frac{k \|f\|\_{L^1}}{\lambda ^2} .$$ Now we pick $k$ to make the right hand side as small as possible, we choose $$ k = \lambda ^{\frac{d-2}{d-1}} \|f\|\_{L^1}^{\frac{1}{d-1}}.$$ This gives $$(6) \hspace{1 cm} | \{|\nabla u\_n | > \lambda \} | \leq C\big(\frac{\|f\|\_{L^1}}{\lambda}\big)^{\frac{d}{d-1}}.$$ Fix $1\leq q < \frac{d}{d-1},$ and $t > 0.$ Then using the Layer-Cake formula and (6) we get
\begin{equation\*} 
\begin{split}\| \nabla u\_{n}\|\_{q}^{q} &\leq \int\_{\{|\nabla u\_{n}| > t\}}|\nabla u\_{n}|^q + \int\_{\{|\nabla u\_{n}| \leq t\}} |\nabla u\_{n}|^q \leq \int\_{\{|\nabla u\_{n}| > t\}}|\nabla u\_{n}|^q + t^q|\Omega| 
\\
&\leq 2t^q|\Omega| + q\int\_{t}^{\infty}\lambda^{q-1}|\{|\nabla u\_{n} > \lambda \}|d\lambda \leq 2t^q|\Omega|^q + Cq\int\_{t}^{\infty}\lambda^{q-1}\|f\|\_{L^1}\lambda^{\frac{-d}{d-1}}.
\end{split}
\end{equation\*}
Evaluating this latter integral, we get $$ (7) \hspace{1 cm} \| \nabla u\_{n}\|\_{q}^{q} \leq 2t^q + C \frac{q\|f\|\_{L^1}^{\frac{d}{d-1}}}{\frac{d}{d-1}-q}\frac{1}{t^{ \frac{d}{d-1} -q}} .$$ Choosing $ t = \|f\|\_{L^1},$ we arrive at $$(8) \hspace{1 cm} \| \nabla u\_{n}\|\_{q} \leq C(q,d)\|f\|\_{L^1}.$$
Thus, $ u\_{n} $ is a bounded sequence in $ W\_{0}^{1,q} ,$ so by Rellich-Kondrachev up to a subsequence, we have $ u\_{n} \to u'$ in $L^q$. Moreover, $ \nabla u\_{n}$ are bounded in $L^q$, so up to a subsequence $ \nabla u\_{n} \rightharpoonup v'$ in the weak $L^q$ topology. It is easy to see that $v'$ must be the weak derivative of $u'$. Note that $ u'$ is in $W^{1,q}$, and $u'$ is the weak limit of the $u\_{n}$ in the weak $W^{1,q}$ topology. But $W\_{0}^{1,q}$ is a strongly closed and convex subset of $W^{1,q}$, so it follows that $u' \in W\_{0}^{1,q} .$ But now note that for all $\varphi \in C^{\infty}\_{c}(\Omega) $ $$ \int\_{\Omega} f\varphi = \lim \int\_{\Omega}f\_{n}\varphi = \lim \int\_{\Omega} \langle \nabla \varphi , \; \nabla u\_{n} \rangle = \int\_{\Omega} \langle \nabla \varphi , \; \nabla u' \rangle , $$ so $u'$ solves the same Dirichlet problem as $u$. By a result of Lions and Dautray, since $\partial \Omega \in C^{1,1} $, we have $u = u'.$ But then by (8) and the fact that $ \nabla u\_{n} \rightharpoonup \nabla u $, $ \| \nabla u\|\_{q} \leq \liminf \| \nabla u\_{n}\|\_{q} \leq C \|f\|\_{L^1} .$ By Poincare, we have the desired estimate $$ \|u\|\_{W\_{0}^{1,q}} \leq C \|f\|\_{L^1} $$


