Concentration-compactness Lemma

Question

Let $N\geq 3$ y $2^{*} := 2N/(N - 2)$. The space $\mathcal{D}^{1,2}(\mathbb{R}^N) := \left\lbrace u \in L^{2^{*}} (\mathbb{R}^N) : \nabla u\in L^2(\mathbb{R}^N)\right\rbrace $, with the inner product $\int_{\mathbb{R}^N} \nabla u\nabla v$, and the corresponding norm $\left( \int_{\mathbb{R}^N} |\nabla u|^2\right)^{1/2} $.

The optimal constant in the Sobolev inequality is given by $$S:= \underset{\begin{matrix}u\in\mathcal{D}^{1,2}(\mathbb{R}^N)\\ |u|_{2^*}=1\end{matrix}}{\inf} |\nabla u|_{2}^{2}>0.$$ Concentration-compactness Lemma: Let $\left\lbrace u_n\right\rbrace\subset \mathcal{D}^{1,2}(\mathbb{R}^N)$ be a sequence such that $$ \begin{matrix} u_n\rightharpoonup u & \text{in } \mathcal{D}^{1,2}(\mathbb{R}^N),\\ |\nabla (u_n - u)|^2 \rightharpoonup \mu & \text{in } \mathcal{M}(\mathbb{R}^N),\\ |u_n -u|^{2^{*}}\rightharpoonup \nu & \text{in } \mathcal{M}(\mathbb{R}^N),\\ u_n\rightarrow u & \text{a.e. on } \mathbb{R}^N, \end{matrix} $$and define $$ \mu_\infty :=\lim_{R\to\infty} \limsup_{n\to\infty} \int_{|x|\geq R} |\nabla u_n|^2\, ,\qquad \nu_\infty :=\lim_{R\to\infty} \limsup_{n\to\infty} \int_{|x|> R} | u_n|^{2^{*}}\, . $$ Then it follows that $$ ||\nu||^{2/2^{*}} \leq S^{-1}||\mu|| , $$$$ \nu_{\infty}^{2/2^{*}} \leq S^{-1} \mu_{\infty}, $$$$ \limsup_{n\to\infty} |\nabla u_n |_{2}^{2}= |\nabla u|_{2}^{2}+||\mu|| + \mu_{\infty}, $$$$ \limsup_{n\to\infty} | u_n |_{2^{*}}^{2^{*}}= |u|_{2^{*}}^{2^{*}}+||\nu|| + \nu_{\infty}. $$Moreover, if $u=0$ and $||\nu||^{2/2^{*}} = S^{-1}||\mu|| $, then $\nu$ and $\mu$ are concentrated at a single point.

Proof:

1. Assume first $u = 0$. Choosing $h \in\mathcal{D} (\mathbb{R}^N )$, we infer from the Sobolev inequality that $\{ u_n\}$ $$\left( \int |hu_n|^{2^{*}}\, dx\right) ^{2/2^{*}}\leq S^{-1} \int |\nabla (hu_n)|^{2}\, dx .$$ Then $$\mid \| \nabla (hu_n)\|_{2} -\| h\nabla u_n\|_{2}\mid \,\leq \|u_n \nabla h\|_2$$ Taking $\text{supp}(\nabla h)\subseteq K $, we have $$\int_{K} |u_n \nabla h|^2\; dx \,\leq \|\nabla h\| \int_K |u_n|^2 dx\rightarrow 0$$ since $u_n \to 0$ en $L_{loc}^2$. Therefore we can write the inequality as $$\left( \int |hu_n|^{2^{*}}\, dx\right) ^{2/2^{*}}\leq S^{-1} \int |h(\nabla u_n)|^{2}\, dx .$$ We would now like to consider $\mu$ and $\nu$. We have as hypothesis $$|\nabla (u_n - u)|^2 \rightharpoonup \mu\;\;\text{en}\;\;\mathcal{M}(\mathbb{R}^N),$$ $$|u_n -u|^{2^{*}}\rightharpoonup \nu\;\;\text{en}\;\;\mathcal{M}(\mathbb{R}^N).$$ Combining the above results, we can rewrite the equation as \begin{equation} \left( \int |h|^{2^{*}}\, d\nu \right) ^{2/2^{*}}\leq S^{-1} \int |h|^{2}\, d\mu . \end{equation} Then we have t \begin{align*} \left( \int |h|^{2^{*}}\, d\nu \right) ^{2/2^{*}}&\;\leq\; S^{-1} \int |h|^{2}\, d\mu \\ \underset{\begin{matrix} |h|_{\infty}=1 \end{matrix}}{\sup} |\left\langle \nu , h\right\rangle |^{2/2^{*}} &\;\leq\; S^{-1} \underset{\begin{matrix} |h|_{\infty}=1 \end{matrix}}{\sup} |\left\langle \mu , h\right\rangle | \\ \| \nu\|^{2/2^{*}}&\;\leq\;S^{-1}\|\mu\| \end{align*} Therefore, the first inequality is verified.

2. For $R > 1$, let$\psi_R \in C^1(\mathbb{R}^ N )$ be such that $\psi_R (x) = 1$ for $|x| > R+ 1$, $\psi_R (x) = 0$ for $|x| < R$ and $0\leq\psi_R (x)\leq 1$ in $\mathbb{R}^N$. By the Sobolev inequality, we have \begin{equation} S^{1/2}\|\psi_R u_n\|_{2^{*}}\leq \|\nabla(\psi_R u_n)\|_2 \leq\|\nabla u_n \psi_R \|_2 +\| u_n\nabla \psi_R\|_2 \end{equation} or in other words $$ S^{1/2}\left( \int |\psi_R u_n|^{2^{*}}\, dx \right) ^{1/2^{*}}\leq \left( \int |\nabla u_n |^{2}|\psi_R |^{2}\, dx\right) ^{1/2} + \left( \int | u_n|^{2}|\nabla \psi_R |^{2}\, dx\right)^{1/2} .$$ Then $$\mid \| \nabla (\psi_R u_n)\|_{2} -\| \psi_R\nabla u_n\|_{2}\mid \,\leq \|u_n \nabla \psi_R \|_2$$ Taking $\text{supp}(\nabla \psi_R)\subseteq K $, we have $$\int_{K} |u_n \nabla \psi_R|^2\; dx \,\leq \|\nabla \psi_R\|_\infty \int_K |u_n|^2 dx\rightarrow 0$$ since $u_n \to 0$ en $L_{loc}^2$. Therefore we can write the inequality as $$\left( \int |\psi_R u_n|^{2^{*}}\, dx\right) ^{2/2^{*}}\leq S^{-1} \int |\psi_R(\nabla u_n)|^{2}\, dx .$$ Taking the superior limit on both sides of the inequality, the following is obtained \begin{multline} \limsup_{n\to\infty}\left( \int |\psi_R u_n|^{2^{*}}\, dx \right) ^{2/2^{*}}\leq S^{-1} \left( \limsup_{n\to\infty}\int | u_n|^{2}|\nabla\psi_R|^{2}\, dx\right) . \end{multline}

In addition, we have $$\int_{|x|> R+1} |\nabla u_n|^{2}\, dx\leq \int_{\mathbb{R}^N} |\nabla u_n|^{2}\psi_R^{2}\, dx\leq \int_{|x|>R} |\nabla u_n|^{2}\, dx$$ y $$\int_{|x|> R+1} | u_n|^{2^{*}}\, dx\leq \int_{\mathbb{R}^N} | u_n|^{2^{*}}\psi_R^{2^{*}}\, dx\leq \int_{|x|>R} |u_n|^{2^{*}}\, dx$$ Taking the limit when $R\to\infty$, we have that by hypothesis are defined as $$ \mu_\infty :=\lim_{R\to\infty} \limsup_{n\to\infty} \int_{\mathbb{R}^N} |\nabla u_n|^2\psi_R^{2}\, dx,$$ $$ \nu_\infty :=\lim_{R\to\infty} \limsup_{n\to\infty} \int_{\mathbb{R}^N} | u_n|^{2^{*}}\psi_R^{2^{*}}\, dx . $$ Combining the above results \begin{align*} \lim_{R\to\infty} \limsup_{n\to\infty} \int_{\mathbb{R}^N} | u_n|^{2^{*}}\psi_R^{2^{*}}\, dx &\leq S^{-1}\lim_{R\to\infty} \limsup_{n\to\infty} \int_{\mathbb{R}^N} |\nabla u_n|^2\psi_R^{2}\, dx\\ \nu_\infty \;\;&\leq S^{-1}\;\; \mu_\infty \end{align*}

Verifying that $\nu_{\infty}^{2/2^{*}} \leq S^{-1} \mu_{\infty}$.

3. Now we want to see that $\nu$ is concentrated at a single point. For this let us further assume that $||\nu||^{2/2^{*}} = S^{-1}||\mu|| $. From Holder's inequality for $h \in\mathcal{D} (\mathbb{R}^N )$, and combined with $\left( \int |h|^{2^{*}}\, d\nu \right) ^{2/2^{*}}\leq S^{-1} \int |h|^{2}\, d\mu .$ \begin{equation*} \left( \int |h|^{2^{*}}\, d\nu \right) ^{1/2^{*}}\leq S^{-1/2}||\mu||^{1/N} \left(\int |h|^{2^{*}}\, d\mu \right) ^{1/2^{*}}. \end{equation*} It follows that $\nu \leq S^{-2^{*}/2}||\mu||^{2/N-2}\mu$. If we combine it with $||\nu||^{2/2^{*}} = S^{-1}||\mu|| $ we have $$\nu = S^{-2^{*}/2}||\mu||^{2/N-2}\mu$$ and $$\mu = S||\nu||^{-2/N}\nu.$$

From $\left( \int |h|^{2^{*}}\, d\nu \right) ^{2/2^{*}}\leq S^{-1} \int |h|^{2}\, d\mu $ we have that, for $h \in\mathcal{D} (\mathbb{R}^N )$, \begin{equation*} \left( \int |h|^{2^{*}}\, d\nu \right) ^{2/2^{*}}\leq \int |h|^{2}\, ||\nu||^{-2/N}d\nu , \end{equation*} i.e. \begin{equation*} \left( \int |h|^{2^{*}}\, d\nu \right) ^{1/2^{*}}||\nu||^{1/N}\leq \left(\int |h|^{2}\, d\nu \right) ^{1/2}, \end{equation*} and so on, for each open set $\Omega$ $$\nu(\Omega)^{1/2^{*}}\nu(\mathbb{R}^N)^{1/N}\leq \nu(\Omega)^{1/2}.$$ Then $\nu(\Omega)=0$ or $\nu(\mathbb{R}^N)\leq \nu(\Omega)$. It follows that $\nu$ is concentrated in a single point.

4. Considering now the general case, we write $v_n := u_n - u$. Since $$v_n \rightharpoonup 0, \,\, \text{en } \mathcal{D}^{1,2} (\mathbb{R}^N )$$ we have $$|\nabla u_n|^{2}\rightharpoonup \mu +|\nabla u|^{2} \,\, \text{in } \mathcal{M} (\mathbb{R}^N ).$$ According to the Brezis-Lieb Lemma, we have for every non negative $h \in\mathcal{K} (\mathbb{R}^N )$ $$\int h|u|^{2^{*}}\, = \lim_{n\to\infty} \left( \int h|u_n|^{2^{*}}-\int h|v_n|^{2^{*}}\right) .$$ Hence we obtain $$| u_n|^{2^{*}}\rightharpoonup \nu +| u|^{2^{*}} \,\, \text{in } \mathcal{M} (\mathbb{R}^N ).$$ Inequality $||\nu||^{2/2^{*}} \leq S^{-1}||\mu||$ follows from the corresponding inequality for $\{v_n\}$.

5. Since $$\limsup_{n\to\infty}\int_{|x|> R} |\nabla v_n|^{2}\,= \limsup_{n\to\infty}\int_{|x|> R} |\nabla u_n|^{2}- \int_{|x|> R} |\nabla u|^{2}$$ we obtain $$\lim_{R\to\infty}\limsup_{n\to\infty}\int_{|x|> R} |\nabla v_n|^{2}=\mu_\infty .$$ By the Brezis-Lieb Lemma, we have $$\int_{|x|> R} |u|^{2^{*}}\, = \lim_{n\to\infty} \left( \int_{|x|> R} |u_n|^{2^{*}}-\int_{|x|> R} |v_n|^{2^{*}}\right) .$$ and so $$\lim_{R\to\infty}\limsup_{n\to\infty}\int_{|x|\geq R} | v_n|^{2^{*}}=\nu_\infty .$$ Inequality $\nu_{\infty}^{2/2^{*}} \leq S^{-1} \mu_{\infty}$ follows then from the corresponding inequality for $\{v_n\}$.

6. For every $R > 1$, we have \begin{equation*} \begin{split} \limsup_{n\to\infty}\int |\nabla u_n|^{2} & = \limsup_{n\to\infty}\left( \int \psi_{R}|\nabla u_n|^{2} + \int (1-\psi_{R})|\nabla u_n|^{2}\right) \\ & = \limsup_{n\to\infty} \int \psi_{R}|\nabla u_n|^{2} \\ &\qquad\quad+ \int (1-\psi_{R})\, d\mu + \int (1-\psi_{R})|\nabla u|^{2} \end{split} \end{equation*} When $R \to\infty$, we get, by the Lebesgue dominated convergence theorem $$\limsup_{n\to\infty}\int |\nabla u_n|^{2} =\mu_{\infty} +\int d\mu + \int |\nabla u|^2 = \mu_{\infty} + ||\mu|| + |\nabla u|_{2}^{2}.$$

Now applying Brezis-Lieb's lemma, for $R>1$ \begin{equation*} \begin{split} \limsup_{n\to\infty}\int |u_n|^{2^*} & = \limsup_{n\to\infty}\left( \int \psi_{R}|u_n|^{2^*} + \int (1-\psi_{R})| u_n|^{2^*}\right) \\ & = \limsup_{n\to\infty} \int \psi_{R}|u_n|^{2^*} \int (1-\psi_{R})\, d\nu + \int (1-\psi_{R})| u|^{2^*} \end{split} \end{equation*} When $R \to\infty$, by the Lebesgue dominated convergence theorem we get $$\limsup_{n\to\infty}\int |u_n|^{2^*} = | u|_{2^*}^{2^*} + ||\nu|| +\nu_{\infty} .$$ We have therefore proved $\limsup_{n\to\infty} |\nabla u_n |_{2}^{2}= |\nabla u|_{2}^{2}+||\mu|| + \mu_{\infty}$ and $ \limsup_{n\to\infty} | u_n |_{2^{*}}^{2^{*}}= |u|_{2^{*}}^{2^{*}}+||\nu|| + \nu_{\infty}$.

There are parts I don't understand. My doubts are: In part 3 of the proof I don't realize how this was obtained: From Holder's inequality for $h \in\mathcal{D} (\mathbb{R}^N )$, and combined with $\left( \int |h|^{2^{*}}\, d\nu \right) ^{2/2^{*}}\leq S^{-1} \int |h|^{2}\, d\mu .$ \begin{equation*} \left( \int |h|^{2^{*}}\, d\nu \right) ^{1/2^{*}}\leq S^{-1/2}||\mu||^{1/N} \left(\int |h|^{2^{*}}\, d\mu \right) ^{1/2^{*}}. \end{equation*} It follows that $\nu \leq S^{-2^{*}/2}||\mu||^{2/N-2}\mu$.

The same for this part: \begin{equation*} \left( \int |h|^{2^{*}}\, d\nu \right) ^{1/2^{*}}||\nu||^{1/N}\leq \left(\int |h|^{2}\, d\nu \right) ^{1/2}, \end{equation*} and so on, for each open set $\Omega$ $$\nu(\Omega)^{1/2^{*}}\nu(\mathbb{R}^N)^{1/N}\leq \nu(\Omega)^{1/2}.$$

In part 4 of the demonstration I don't realize how this follows:

According to the Brezis-Lieb Lemma, we have for every non negative $h \in\mathcal{K} (\mathbb{R}^N )$ $$\int h|u|^{2^{*}}\, = \lim_{n\to\infty} \left( \int h|u_n|^{2^{*}}-\int h|v_n|^{2^{*}}\right) .$$ Hence we obtain $$| u_n|^{2^{*}}\rightharpoonup \nu +| u|^{2^{*}} \,\, \text{in } \mathcal{M} (\mathbb{R}^N ).$$

In part 6 of the demonstration I don't realize how this is deduced: \begin{equation} \begin{split} \limsup_{n\to\infty}\int |\nabla u_n|^{2} & = \limsup_{n\to\infty}\left( \int \psi_{R}|\nabla u_n|^{2} + \int (1-\psi_{R})|\nabla u_n|^{2}\right) \\ & = \limsup_{n\to\infty} \int \psi_{R}|\nabla u_n|^{2} \\ &\qquad\quad+ \int (1-\psi_{R})\, d\mu + \int (1-\psi_{R})|\nabla u|^{2} \end{split} \end{equation} When $R \to\infty$, we get, by the Lebesgue dominated convergence theorem $$\limsup_{n\to\infty}\int |\nabla u_n|^{2} =\mu_{\infty} +\int d\mu + \int |\nabla u|^2 = \mu_{\infty} + ||\mu|| + |\nabla u|_{2}^{2}.$$