Weak convergence in $W^{1,p}_0$

Question

Note from the answerer : this question stems from this article.

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I ask this question in https://math.stackexchange.com/questions/1206617

I have a bounded sequence $(u_n)$ from $W^{1,p}_0(\Omega)$ so it weakly converge to $u\in W^{1,p}_0(\Omega)$ and strongly converge to $u$ in $L^p(\Omega).$ We define a function $f:\Omega\times \mathbb{R}\rightarrow \mathbb{R}$ a bounded Caratheodory function such that $\lim_{s\rightarrow+\infty} f(x,s)=f^{+\infty}(x)$

My question is why $$\lim_{n\rightarrow +\infty} \int_{\Omega}f(x,u_n)(u_n-u) dx=0$$ and $$\lim_{n\rightarrow +\infty}\int_{\Omega} |u_n|^{p-2} u_n(u_n-u) dx=0$$

For the first integral, I'm trying to apply Lebesgue dominated convergence, but I have no idea.

For the second integral, when $p=2$ I have no problems, because in this case we have not $|u_n|^{p-2}$ it is equal to 1 and then I just have to do $u_n(u_n-u)=(u_n-u+u)(u_n-u)$ and I do the Cauchy–Schwarz inequality, but when $p$ is not equal to 2 I have no idea.

Thank you

Answer 1

Let's call $I_1$ and $I_2$ your two integrals respectively.

You know that $u_n \to u$ strongly in $L^p$. Because $\Omega$ is bounded, $u_n$ also converges to $u$ strongly in $L^1$. As you assume that $f$ is bounded in its two arguments,

$$|I_1| \leq \|f\|_{L^{\infty}(\Omega \times \mathbb{R})} \|u_n-u\|_{L^1(\Omega)} \to 0.$$

Regarding the second integral, by Hölder inequality, you have

$$|I_2| \leq \| |u_n|^{p-2} u_n \|_{L^q(\Omega)} \|u_n-u\|_{L^p(\Omega)} $$

where $\frac 1p + \frac 1q = 1$, i.e. $q = \frac{p}{p-1}$. Thus

$$|I_2| \leq \|u_n\|_{L^p(\Omega)}^{p-1} \|u_n-u\|_{L^p(\Omega)} \to 0 $$

thanks to the strong convergence of $u_n$ in $L^p$ and its boundedness in $L^p$.