# Proof of a concentration compactness lemma

Hi I'm stuck with the proof of a concentration-compactness lemma. We have the following equation in $\mathbb{R}^N, N \ge 3$: $$-\Delta u +u=|u|^{p-2}u,$$ where $2 < p < 2^{*}$.

The functional associated to that equation is given by $$J(u) = \frac{1}{2}\|u\|^2_{H^1}-\frac{1}{p}\|u\|^p_{L^p}.$$ Because the Sobolev imbedding $H^1 (\mathbb{R}^N) \subset L^q (\mathbb{R}^N), 2 < q < 2^{*}$ is not compact, $J$ does not satisfy the Palais-Smale condition. So one considers the family $J_k$ of functionals defined by $$J_k(u) = \frac{1}{2}\|u\|^2_{H^1 (B_k)}-\frac{1}{p}\|u\|^p_{L^p(B_k)},$$ where $(B_k)_k$ is an open cover of $R^N$; $k$ positive integer.

Each $J_k$ now satisfies the PS condition.

The "lemma" says the following: Let $u_k \in H^1_0(B_k)$ be uniformly bounded in $H^1(\mathbb{R}^n)$, i.e. $\|u_k\| \le \Lambda$, with $\Lambda>0$ independent of $k$, and such that $J'(u_k) \to 0$ as $k \to \infty$. Then, along a subsequence, one of the following holds true:

1. either $u_k \to 0$ in $H^1(\mathbb{R}^N)$,
2. or, there exist $r,\delta>0$, and a sequence $a_k$ in $R^N$ such that $$\liminf_k \int_{B_{r} (a_k)}u^2_k \ge \delta.$$ I know how to prove that if 2. does not hold then 1. holds. I need a hint on how to prove that if 1. does not hold then 2. holds.

Thanks

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What are the sets $B_k$? When you say $u_k\to 0$ in $H^1$, what is the domain on which you are discussing $H^1$? If $H^1(R^N)$, how are you extending $u_k\in H^1(B_k)$ to $H^1(R^N)$? –  Willie Wong Apr 29 '11 at 23:04
Sorry! some information are missing: $B_k$ has non-empty boundary, and it is of course $H^1_0(B_k)$ instead of $H^1(B_k)$. –  Mercy Apr 30 '11 at 7:39
think of $B_k$ as $\mathbb{R}^{N-1}\times (-k,k)$. –  Mercy Apr 30 '11 at 7:51

The concentration-compactness lemma of P.L. Lions is independent of the fact your sequence is Palais-Smale. It holds for general sequences of bounded positive measure, in your case $u_k^2\; dx$.

First normalize your sequence such that $\Vert u_k^2\Vert_2 =1$.

Hence the idea is to consider the concentration function

$$Q_k(r)=\sup_{x\in R^n} \left( \int_{B(x,r)} u_k^2 \;dx\right).$$

Note that $Q_k$ is non-decreasing and non-negative bounded function on $[0,+\infty[$, with $\displaystyle \lim_{r\rightarrow +\infty }Q_k(r)=1$.

Then there exists a subsequence of $Q_k$, still denote $Q_k$, and a non-decreasing and non-negative bounded function $Q$ on $[0,+\infty[$ such that

$$Q_k(r) \rightarrow Q(r),$$

for almost all $r$. We can assume that $Q$ is left-continuous and moreover since $Q_k$ is non-decreasing and bounded, we have $$Q(r)\leq \liminf_{k\rightarrow +\infty} Q_k(r) \hbox{ for all } r.$$

Let

$$\lambda =\lim_{r\rightarrow +\infty} Q(r).$$

If $\lambda=0$, this is the vanishing case, else let $r_0$ such that

$$Q(r_0)\geq \frac{\lambda}{2}.$$

For any $k$ there exists $x_k$ such that

$$Q_k(r_0)\leq \int_{B(x_k,r_0)} u_k^2 \;dx + \frac{1}{k}.$$

Hence for $k$ big enough, you have

$$\int_{B(x_k,r_0)} u_k^2 \;dx\geq \frac{\lambda}{3}.$$

In fact, you can refine this case in two

2.1) For all $\epsilon$ there exits $R$ such that $$\int_{B(x_k,R)} u_k^2 \;dx\geq 1-\epsilon.$$ This is the compactness-case.

2.2) the measure split on disjoint set arbitrary far, this is the dichotomy case.

You will find all the details in the two papers of Lions.

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Still, I can't see why "vanishing" in $L^2$ would imply "vanishing" in $H^1$ by just considering a bounded sequence in $H^1$. –  Mercy May 4 '11 at 7:08
I never say that imply "vanishing" in $H^1$, i just give you a correct formulation of the concentration compactness. You can applicate it with the measure $(\vert \nabla u_k\vert^2 +u^2_k)ds$, which will give you the a "vanishing" case in $H^1$... But i agree that doesn't answer to the question you post again: mathoverflow.net/questions/63842/… –  Raphael May 4 '11 at 8:22
Thanks to all for nice answer. Could you please prove the following assertion? We can assume that Q(r) is left-continuous and moreover since Q_k (r) is nondecreasing and bounded, w e have Q(r)≤ limit inf Q_k (r) for all r. I will be very grateful. –  Farhad Jan 13 at 7:10
@Mercy: you probably have figured it out by now (this question is 2+ years old), maybe you could close that question by answering it yourself. If you have not, let me know. –  Jung Wen Chen Jan 14 at 12:36