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

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-Schwartz inequality, but when $p$ is not equal to 2 i have no idea.

Thank you

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 $L^p$.

• I juste don't understand why $|||u_n|^{p-2}u_n||_{L^q}=||u_n||^{p-1}_{L^p}$ ? and i asked the question to a professor he told me that $f$ is bounded means that it sends bounded sets to bounded sets – Vrouvrou Mar 28 '15 at 10:08
• @Vrouvrou : By definition, the $L^q$ norm is equal to $(\int | |u_n|^{p-2}u_n|^q)^{\frac 1q}$. But as $q = \frac{p}{p-1}$, the integrand is equal to $|u_n|^{p}$ and the integral is to the power of $\frac{p-1}{p}$, so that's a mere rewriting. Regarding $f$, okay, there's something more to do. – Hachino Mar 28 '15 at 10:10
• @Vrouvrou : Also, could you please tell me what are your assumptions on the limit $f^{\infty}$ ? – Hachino Mar 28 '15 at 10:15
• no assumptions on $f^{\infty}$, if $f$ sends bounded sets to bounded sets $||f||_{L^\infty}<\infty$ ? – Vrouvrou Mar 28 '15 at 10:18
• @Vrouvrou : Sadly no, since any continuous function on $\mathbb{R}$ sends bounded sets to bounded sets... But this implies that, for $s$ in a, bounded set $f(x,s)$ remains bounded, since $x$ lies in $\Omega$, bounded. – Hachino Mar 28 '15 at 10:22