Could you help me to justify better and in detail the proof of the theorem? This is the theorem 1.40 of the book Minimax Theorems by M. Willem
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 $u=0$ y $||\nu||^{2/2^{*}} = S^{-1}||\mu|| $, yhen $\nu$ and $\mu$ are concentrated at a single point. Theorem: Let $\{u_n\}\subset \mathcal{D}^{1,2} (\mathbb{R}^N )$ be a minimizing sequence satisfying$$|u_n|_{2^*} =1,\quad | \nabla u_n|_{2}^{2}\rightarrow S, \quad n\rightarrow\infty.$$Then there exists a succession $\{y_n, \lambda_n\} \subset \mathbb{R}^N \times (0,\infty) $ such that $ \{ u_n^{y_n, \lambda_n} \}$ contains a convergent subsubsession. In particular there exists a minimizer for $S$.
Proof: We define the Levy concentration functions as $$Q_n (\lambda):= \sup_{y\in\mathbb{R}^N} \int_{B(y,\lambda)} |u_n|^{2^{*}}$$Since, for every $n$, $$\lim_{\lambda\to 0^+} Q_n(\lambda)=0,\quad \lim_{\lambda\to \infty} Q_n(\lambda)=1,$$there exists $\lambda_n>0$ such that $Q_n(\lambda_n)=1/2$. Moreover, there exists $y_n \in\mathbb{R}^N$ tsuch that $$\int_{B(y_n,\lambda_n)} |u_n|^{2^{*}}=Q_n(\lambda_n)=1/2$$since$$\lim_{|y|\to\infty}\int_{B(y,\lambda_n)} |u_n|^{2^{*}}=0.$$Let us define $v_n:= u_n^{y_n, \lambda_n}$. Hence $|v_n|_{2^{*}} =1$, $|\nabla v_n|_2^2\to S$ and $$ \frac{1}{2}= \int_{B(0,1)} |v_n|^{2^{*}}=\sup_{y\in\mathbb{R}^N} \int_{B(y,1)} |v_n|^{2^{*}}. \qquad\qquad(1)$$ Since $\{ v_n\}$ is bounded in $\mathcal{D}^{1,2} (\mathbb{R}^N )$, we may assume, going if necessary to a subsequence, $$ \begin{matrix} v_n\rightharpoonup v & \text{en } \mathcal{D}^{1,2}(\mathbb{R}^N),\\ |\nabla (v_n - v)|^2 \rightharpoonup \mu & \text{en } \mathcal{M}(\mathbb{R}^N),\\ |v_n -v|^{2^{*}}\rightharpoonup \nu & \text{en } \mathcal{M}(\mathbb{R}^N),\\ v_n\rightarrow v & a.e. \text{ en } \mathbb{R}^N \end{matrix} $$ By the preceding lemma, $$ S=\lim |\nabla v_n|_{2}^2 =|\nabla v|_{2}^2 + ||\mu ||+\mu_\infty , \qquad\qquad(2) $$$$ 1=| v_n|_{2^{*}}^{2^{*}} =| v|_{2^{*}}^{2^{*}} + ||\nu ||+\nu_\infty , \qquad\qquad(3) $$where $$ \mu_\infty :=\lim_{R\to\infty} \limsup_{n\to\infty} \int_{|x|> R} |\nabla v_n|^2\, ,$$$$ \nu_\infty :=\lim_{R\to\infty} \limsup_{n\to\infty} \int_{|x|> R} | v_n|^{2^{*}}\, . $$ We deduce from (2), $||\nu||^{2/2^{*}} \leq S^{-1}||\mu|| $, $\nu_{\infty}^{2/2^{*}} \leq S^{-1} \mu_{\infty}$ and Sobolev inequality, $$ S\geq S\left( (|v|_{2^{*}}^{2^{*}})^{2/2^{*}} + ||\nu ||^{2/2^{*}} + \nu_{\infty}^{2/2^{*}}\right) .$$It follows from (3) that $|v|_{2^{*}}^{2^{*}}$, $||\nu ||$ and $\nu_{\infty}$ are equal either to $0$ or to 1. By (1), $\nu_{\infty}\leq 1/2$ so that $\nu_{\infty}=0$. If $||\nu ||=1$ then $v =0$ and $||\nu ||^{2/2^{*}} \geq S^{-1}||\mu||$. The preceding lemma implies that $\nu$ is concentrated at a single point $z$. We deduce from (1) the contradiction $$\frac{1}{2}=\sup_{y\in\mathbb{R}^N} \int_{B(y,1)} |v_n|^{2^{*}}\geq \int_{B(z,1)} |v_n|^{2^{*}}\rightarrow ||\nu||=1.$$Thus $|v|_{2^{*}}^{2^{*}}=1$ and then $$|\nabla v|_{2}^{2}=S=\lim |\nabla v_n|_{2}^{2}.$$Therefore $v$ is a minimizer for $S$.