Why weak convergence and a.e. convergence imply the convergence of this integral?

Question

In the proof of the Brezis-Nirenberg theorem on the book "Nonlinear Analysis and Semilinear Elliptic Problems" by Ambrosetti and Malchiodi, it is used lemma 11.11 at pag. 183, in whose proof we have a sequence $(u_n)_{n\in\mathbb{N}}$ that converges to $\bar u$ in the following senses:

• weakly in $H^1_0(\Omega)$;
• almost everywhere;

where $\Omega$ is a non-empty bounded open subset of $\mathbb{R}^N$ with $N\ge3$ and so, by Sobolev embedding theorem, $(u_n)_{n\in\mathbb{N}}$ converges weakly in $L^{2^*}(\Omega)$ to $\bar u$, where $2^*=\frac{2N}{N-2}$. Now, it is clamed that: $$\forall v\in H^1_0(\Omega), \int_\Omega |u_n(x)|^{2^*-2}u_n(x) v(x)\operatorname{d}x\to \int_\Omega |\bar u(x)|^{2^*-2}\bar u(x) v(x)\operatorname{d}x, n\to \infty.$$ I tried to figure out why it is so, using Hölder inequality or trying to find a dominating function like in the proof of Riesz-Fischer theorem, but failed in both cases... it seems to me that what I actually need to prove such a claim using that instruments it is that $\|u_n-u\|_{2^*}\rightarrow0,n\rightarrow\infty$, a thing that we don't have here. Obviously I'm missing something... can anyone help me to figure out what I'm missing?

Answer 1

Edit The proof originally incorrectly applied a theorem concerning linear operators to a non-linear one, which has been fixed in the edit.

You need to exploit the weak convergence of $u_n.$ Observe that since $|(|u_n|^{2^*-2}u_n)| = |u_n|^{2^*-1},$ for each $n$ we get, $$|u_n|^{2^*-2}u_n \in L^{\frac{2^*}{2^*-1}}(\Omega) = L^{\frac{2N}{N+2}}.$$ Moreover the sequence $|u_n|^{2^*-2}u_n$ is uniformly bounded in $L^p$ where $p = \frac{2N}{N+2} > 1,$ so by reflexivity we get a weakly convergent subsequence $$|u_{n_k}|^{2^*-2}u_{n_k} \rightharpoonup v \in L^p(\Omega).$$ Since it also convergences a.e., we get $v = |\overline u|^{2^*-2}\overline u.$ As the above holds for every subsequence, we conclude that the entire sequence $|u_n|^{2^*-2}u_n$ converges weakly to $|\overline u|^{2^*-2}\overline u$ in $L^p(\Omega).$

As the conjugate dual of $p=\frac{2N}{N+2}$ is $2^*,$ for all $v \in L^{2^*}(\Omega)$ we have, $$ \int_{\Omega} |u_n|^{2^*-2}u_n v \,\mathrm{d} x \rightarrow \int_{\Omega} |\overline u|^{2^*-2}\overline u v \,\mathrm{d}x. $$ As $H^1_0(\Omega) \hookrightarrow L^{2^*}(\Omega)$ by Sobolev embedding, the result follows.

Answer 2

Let's use $\Vert . \Vert_{\alpha}$ to denote the usual norm on $L^{\alpha}(\Omega)$ for every $1\leq \alpha \leq \infty $. It is sufficient to show that: $$ \Vert (u_n|u_n|^{{2^*-2}} -\bar{u}|\bar{u}|^{{2^*-2}})v\Vert_1\to 0 \text{ as } n \to \infty $$ Now, for any $\phi\in C^{\infty}_0(\Omega)$: $$ \Vert (u_n|u_n|^{{2^*-2}} -\bar{u}|\bar{u}|^{{2^*-2}})v\Vert_1\\ \leq \Vert (u_n|u_n|^{{2^*-2}} -\bar{u}|\bar{u}|^{{2^*-2}})\phi\Vert_1 + \Vert (u_n|u_n|^{{2^*-2}} -\bar{u}|\bar{u}|^{{2^*-2}}) (v - \phi)\Vert_1\\ $$

The function $t\to t|t|^{{2^*-2}}$ is Lipschitz on every subset of the form $[-M,M]$ where its Lipschitz constant is $(2^*-1)M^{{2^*-2}}$. This gives us the following estimate: $$|a|a|^{{2^*-2}} -b|b|^{{2^*-2}}| \leq (2^*-1)(|a|+|b|)^{{2^*-2}}|a-b| \qquad \forall a,b \in \mathbb{R} $$ Therefore, thanks to this relation and Hölder inequality we are able to prove: \begin{align} \Vert (u_n|u_n|^{{2^*-2}} &-\bar{u}|\bar{u}|^{{2^*-2}})\phi\Vert_1 \leq \Vert \phi\Vert_{\infty} \Vert u_n|u_n|^{{2^*-2}} -\bar{u}|\bar{u}|^{{2^*-2}}\Vert_1 \\ &\leq \Vert \phi\Vert_{\infty} \Vert (2^*-1)(|u_n| + |\bar{u}|)^{2^*-2} (u_n-\bar{u}) \Vert_1 \\ &\leq (2^*-1) \Vert \phi\Vert_{\infty} \Vert (|u_n| + |\bar{u}|)^{2^*-2} \Vert_{\frac{2^*-1}{2^*-2}} \Vert u_n-\bar{u} \Vert_{2^*-1}\\ &= (2^*-1) \Vert \phi\Vert_{\infty} \Vert|u_n| + |\bar{u}|\Vert^{2^*-2}_{2^*-1} \Vert u_n-\bar{u} \Vert_{2^*-1} \\ &\leq (2^*-1) \Vert \phi\Vert_{\infty} (\Vert u_n\Vert_{2^*-1} + \Vert \bar{u}\Vert_{2^*-1})^{2^*-2} \Vert u_n-\bar{u} \Vert_{2^*-1} \\ &\leq (2^*-1) \Vert \phi\Vert_{\infty} (2C_1)^{2^*-2} \Vert u_n-\bar{u} \Vert_{2^*-1} \end{align} where $C_1=\max \{ \sup_{n} \{ \Vert u_n \Vert_{2^*-1} \},\Vert \bar{u} \Vert_{2^*-1} \}$ is finite since $H^1_0(\Omega)$ embeds continuously in $L^{2^*-1}(\Omega)$ and $u_n$ weakly converges to $u$ in $H^1_0(\Omega)$. We have to estimate the other term now: \begin{align} \Vert (u_n|u_n|^{{2^*-2}} &-\bar{u}|\bar{u}|^{{2^*-2}})(v - \phi)\Vert_1 \leq \Vert u_n|u_n|^{{2^*-2}} -\bar{u}|\bar{u}|^{{2^*-2}}\Vert_{\frac{2^*}{2^*-1}} \Vert v - \phi\Vert_{2^*}\\ &\leq (\Vert u_n|u_n|^{{2^*-2}}\Vert_{\frac{2^*}{2^*-1}} +\Vert \bar{u}|\bar{u}|^{{2^*-2}}\Vert_{\frac{2^*}{2^*-1}}) \Vert v - \phi\Vert_{2^*}\\ &= (\Vert u_n\Vert^{2^*-1}_{2^*} +\Vert \bar{u}\Vert^{2^*-1}_{2^*}) \Vert v - \phi\Vert_{2^*}\\ &\leq (2C_2)^{2^*-1} \Vert v - \phi\Vert_{2^*} \end{align} Where $C_2:=\max\{ \sup_{n}\{\Vert u_n\Vert_{2^*}\},\Vert \bar{u}\Vert_{2^*} \}$ that is finite thanks to Sobolev embedding of $H^1_0(\Omega)$ in $L^{2^*}(\Omega)$ combined, once again, with the weak convergence of $u_n$ in $H^1_0(\Omega)$. $$ 0\leq\limsup_{n \to \infty} \Vert (u_n|u_n|^{{2^*-2}} -\bar{u}|\bar{u}|^{{2^*-2}})v\Vert_1\\ \leq \limsup_{n \to \infty} (2^*-1) \Vert \phi\Vert_{\infty} (2C_1)^{2^*-2} \Vert u_n-\bar{u} \Vert_{2^*-1} +\\ + \limsup_{n \to \infty} (2C_2)^{2^*-1} \Vert v - \phi\Vert_{2^*}\\ = (2C_2)^{2^*-1} \Vert v - \phi\Vert_{2^*} $$ using that $u_n$ converges to $\bar{u}$ stongly in $L^{2^*-1}(\Omega)$ due to Rellich-Kondrachov compact embeddings. Since $\Vert v - \phi\Vert_{2^*}$ can be chosen arbitrarily small as $C^{\infty}_0(\Omega)$ is dense in $L^{2^*}(\Omega)$, the thesis follows.