**Note from the answerer** : this question stems from [this article](https://www.sciencedirect.com/science/article/abs/pii/S0362546X96000223).




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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



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


