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I want to prove the following

If $N\ge 3$ there exists a constant $c_0=c_0(\Omega)$ such that for all $\alpha\ge 1$ and $z\in H^1(\Omega)$ \begin{align} \left(\int_{\Omega}|z|^{2^*}\right)^{\frac{2}{2^*}}&\le c_0\left(\int_{\Omega}|\nabla z|^2\,dx+\left(\frac{2}{|\Omega|}\int_{\Omega}|z|^{2/\alpha}dx\right)^{\alpha}\right), \end{align} with $2^*:=\frac{2N}{N-2}$.

Proof: (i) Let us prove by contradition. Suppose for all $\lambda>0$ there exists $z\in H^1(\Omega)$ and $\alpha\ge 1$ such that \begin{align*}\tag{A.1} \left(\int_{\Omega}|z|^{2^*}\right)^{\frac{2}{2^*}}>\lambda\left(\int_{\Omega}|\nabla z|^2+\left(\frac{2}{|\Omega|}\int_{\Omega}|z|^{\frac{2}{\alpha}}\right)^{\alpha}\right) \end{align*} We define a sequence $\{\lambda_n\}$ such that $\lambda_n\to\infty$ as $n\to\infty$. It follows from (A.1) that these exist $z_n\in H^1(\Omega)$ and $\alpha_n\ge 1$ such that \begin{align*}\tag{A.2} \left(\int_{\Omega}|\nabla z_n|^{2^*}\right)^{\frac{2}{2*}}>\lambda_n\left(\int_{\Omega}|\nabla z_n|^{2}+\left(\frac{2}{|\Omega|}\int_{\Omega}|z_n|^{\frac{2}{\alpha_n}}\right)^{\alpha_n}\right) \end{align*} which implies in particular that $z_n\ne 0$. We divide inequality (A.2) by $\|z_n\|_{L^{2^*}(\Omega)}^2$ to obtain \begin{align*}\tag{A.3} \int_{\Omega}\left|\nabla\frac{z_n}{\|z_n\|_{L^{2^*}(\Omega)}}\right|^2+\left(\frac{2}{|\Omega|}\int_{\Omega}\left|\nabla\frac{z_n}{\|z_n\|_{L^{2^*}(\Omega)}}\right|^{2/\alpha_n}\right)^{\alpha_n}<\frac{1}{\lambda_n} \end{align*} Setting \begin{equation*} w_n=\frac{z_n}{\|z_n\|_{L^{2^*}(\Omega)}}, \end{equation*} We deduce that \begin{align*}\tag{A.4} \int_{\Omega}\left|\nabla w_n\right|^2+\left(\frac{2}{|\Omega|}\int_{\Omega}|w_n|^{2/\alpha_n}\right)^{\alpha_n}<\frac{1}{\lambda_n} \end{align*} It follows from ($A.3$) and ($A.4$) that

\begin{align*}\tag{A.5} \|w_n\|_{L^{2^*}(\Omega)}&=1\qquad (a)\\ \|\nabla w_n\|_{L^{2}(\Omega)}&\to 0\quad\text{as}\quad n\to\infty\quad (b) \end{align*}

so that in particular there exists $w\in H^1(\Omega)$ such that as $n\to\infty$, $w_n\rightharpoonup w$ weakly in $H^1(\Omega)$ and $w_n\to w$ strongly in $L^2(\Omega)$ along a subsequence. It also follows from (A.5) and the weak lower semicontinuity of $z\mapsto\int_{\Omega}|\nabla z|^2$ that $\nabla w=0$ in $L^2(\Omega)$. Thus there exists a constant $l$ such that $w=l$ and $w_n\to l$ strongly in $H^1(\Omega)$. Since the embedding from $H^1(\Omega)$ into $L^{2^*}(\Omega)$ is continuous we have that \begin{equation} w_n\to l\quad\text{strongly in }L^{2^*}(\Omega)\,\,\text{as}\,\,n\to\infty \end{equation} and therefore it follows from (A.5) that \begin{equation} |l|=|\Omega|^{-1/2^*} \end{equation} and we may suppose that $l>0$. Furthermore, one could also deduce from (A.4) that \begin{align} \left(\frac{2}{|\Omega|}\int_{\Omega}|w_n|^{2/\alpha_n}\,dx\right)^{\alpha_n}\to 0\quad\text{as}\quad n\to\infty \end{align}



Now, here is where I confuses about 1. " It also follows from (A.5) and the weak lower semicontinuity of $z\mapsto\int_{\Omega}|\nabla z|^2$ that $\nabla w=0$ in $L^2(\Omega)$. Thus there exists a constant $l$ such that $w=l$ and $w_n\to l$ strongly in $H^1(\Omega)$. "

What results from functional analysis do I need to conclude the existence of this $l$? Then what is the reference I can look up?

  1. How does (A.5) leads to $$|l|=|\Omega|^{-1/2^*}$$
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For (b). Since $\lambda_n \to +\infty$, we have $\frac{1}{\lambda_n} \to 0$. Hence, from (A.4) and due to the fact that $$ \frac{2}{|\Omega|}\int_{\Omega}|w_n|^{2/\alpha_n} \geq 0, $$ we get $$ 0 \leq \int_{\Omega}\left|\nabla w_n\right|^2 \leq \int_{\Omega}\left|\nabla w_n\right|^2+\left(\frac{2}{|\Omega|}\int_{\Omega}|w_n|^{2/\alpha_n}\right)^{\alpha_n}<\frac{1}{\lambda_n} \to 0. $$ Thus, $$ ||\nabla w_n||_{L^2(\Omega)}^2 = \int_{\Omega}\left|\nabla w_n\right|^2 < \frac{1}{\lambda_n} \to 0 $$ as $n \to \infty$.

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  • $\begingroup$ @math101 $|\nabla w_n| \geq 0$ by definition of absolute value (or Euclidean norm if we are in $\mathbb{R}^n, n > 1$). Therefore, integral of it will be also nonnegative. $\endgroup$
    – Voliar
    Sep 21, 2014 at 11:42
  • $\begingroup$ thx for that, i deleted the silly question asked. $\endgroup$
    – math101
    Sep 21, 2014 at 13:29

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