I have a question regarding a proof of a lemma (I.1) in P.L. Lions "concentration-compactness principle" (locally compact case, 2nd part (See p.10 here). Here it is:
Let $1 < p \leq \infty$, $1 \leq q < \infty$ with $q \neq \frac{Np}{N-p}$ if $p<N$. Assume $u_n$ is bounded in $L^q$ and $\nabla u_n$ is bounded in $L^p$ and we have $$\sup_{y \in \mathbb{R}^n} \ \int_{y+B_R} |u_n|^q \rightarrow 0$$ for some $R > 0$. Then $u_n \rightarrow 0$ in $L^{\alpha}$ for $\alpha \in [q, \frac{Np}{N-p}]$.
Now to the proof: We first assume that $u_n$ is also bounded in $L^{\infty}$. He picks $q_1 > q$, $\infty > (q_1 - 1)p' > q$.
He remarks that by Hölder:
\begin{alignat*}{1} \underset{y \in \mathbb{R}^n}{sup}\int_{y + B_R}|u_n|^{q_1-1}||\nabla u_n| \rightarrow 0 \end{alignat*} So far so good. Then he claims that by Sobolev embeddings, if $\gamma \in ]1, \frac{N}{N-1}[$ ($N$ being the dimension) there exists a constant $c_0$ independent of y such that:
\begin{alignat*}{1} \int_{y + B_R}|u_n|^{q_1\gamma} \leq (\int_{y + B_\rho}|u_n|^{q_1} + |u_n|^{q_1-1}|\nabla u_n|dx)^{\gamma}\\ \leq \epsilon_n^{\gamma - 1}\int_{y + B_\rho}|u_n|^{q_1} + |u_n|^{q_1-1}|\nabla u_n| \end{alignat*} I am totally lost by those 2 last inequalities. It do not understand how we can introduce the $\nabla u_n$ term. If I had to guess the term $\epsilon_n$ is just $(\int_{y + B_\rho}|u_n|^{q_1} + |u_n|^{q_1-1}|\nabla u_n|dx)^{\gamma-1}$, I do not see why this quantity can be made small, a priori we only know it is bounded. Also I do not see why we have to change the radius of the ball from $R$ to $\rho$. Thanks a lot for any ideas on how to tackle this!
Let $\Omega = y + B_R$. Since $u_n \in L^{\infty}$ and the measure of $y + B_R$ is finite, $|u_n|^{q_1} \in L^1$. Moreover, $|\nabla |u_n|^{q_1}| = q_1|u_n|^{q_1-1}|\nabla u_n|$ is integrable as well. Therefore, $u_n \in W_1^1(\Omega)$. Thus, by the Sobolev embedding theorem, $$\left(\int_{\Omega}|u_n|^{q_1\gamma}\right)^{1/\gamma} = ||(|u_n|^{q_1})||_{L^\gamma} \le c_0||(|u_n|^{q_1})||_{W_1^1} = c_0\int_{\Omega}\left(|u_n|^{q_1} + q_1|u_n|^{q_1-1}|\nabla u_n|\right).$$ In other words, $$\int_{\Omega}|u_n|^{q_1\gamma} \le C\left(\int_{\Omega}|u_n|^{q_1} + |u_n|^{q_1-1}|\nabla u_n|\right)^{\gamma},$$ which, I believe, is how the first inequality should look like. Next, I think, we are going to observe that $$\epsilon_n = \sup\limits_{y \in \mathbb{R}^n}\int_{y+B_R}|u_n|^{q_1} + |u_n|^{q_1-1}|\nabla u_n| \to 0$$ as $n \to \infty$. Note that $\epsilon_n$ does not depend on $y$. It follows that $$\int_{\Omega}|u_n|^{q_1\gamma} \le C\epsilon_n^{\gamma-1}\int_{\Omega}|u_n|^{q_1} + |u_n|^{q_1-1}|\nabla u_n|$$
I am relatively sure the next step would be something like covering $\mathbb{R}^n$ with balls of radius $R$ and, choosing $\gamma$ approprietly get the estimate $$||u_n||_{L^\alpha} \le C_0\epsilon^{\gamma-1} \to 0.$$