# Integral convergence and weak convergence

Given that $\Omega \subset \mathbb{R}^{n}$ is a connected bounded Lipshitz domain and $u_{k} \rightharpoonup u$ in $W^{1,p}(\Omega)$. We denote $\Gamma$ as the boundary of the domain. We have the following additional results:

$a(u_{k},\nabla v) \rightarrow a(u, \nabla v)$ in $L^{p'}(\Omega;\mathbb{R}^{n})$, $b(u_{k}) \rightarrow b(u)$ in $L^{q'}(\Gamma)$, $\nabla (u_{k}-u) \rightharpoonup 0$ in $L^{p}(\Omega;\mathbb{R}^{n})$ and $(u_{k}-u)|_{\Gamma} \rightharpoonup 0$ in $L^{q}(\Gamma)$.

How does it follow then that:

$\int_{\Omega}a(u_{k},\nabla v)\cdot\nabla(u_{k}-u)dx + \int_{\Gamma}b(u_{k})(u_{k}-u)dS \rightarrow 0$

My idea is that it seems that Holder's Inequality should be used so for the first integral term:

$\int_{\Omega}a(u_{k},\nabla v)\cdot\nabla(u_{k}-u)dx \leq (\int_{\Omega}|a(u_{k},\nabla v)|^{p'})^{\frac{1}{p'}}(\int_{\Omega}|\nabla(u_{k}-u)|^{p})^{\frac{1}{p}}$ and since $a(u_{k},\nabla v)$ is convergent it is bounded, I am just not sure we can state that $\int_{\Omega}|\nabla(u_{k}-u)^{p}|^{\frac{1}{p}} \rightarrow 0$ simply from the assumption $\nabla (u_{k}-u) \rightharpoonup 0$ in $L^{p}(\Omega;\mathbb{R}^{n})$?

Any ideas of how this would be resolved? I can add information if some important info is missing. Thanks.

## 1 Answer

You are almost there:

For the first term it is convenient to subtract and add $a(u,\nabla v)\nabla(u_k-u)$, so you get to terms: $$A = \int_\Omega (a(u_k,\nabla v) - a(u,\nabla v))\nabla(u_k-u)$$ and $$B = \int_\Omega a(u, \nabla v) \nabla(u_k-u).$$

$B$ clearly tends to $0$ be the hypothesis of weak convergence. For the $A$ term note that weakly convergent sequences are uniformly bounded in $L^p$ (this is a consequence of the uniform boundedness principle and the duality of $L^p$). So as you said, by Holder's inequality you can estimate $|A|$ by $$\|a(u_k,\nabla v)-a(u,\nabla v)\|_{L^{p'}} \|\nabla(u_k-u)\|_{L^{p}},$$ the first factor tends to $0$ by the strong convergence hypothesis, while the second factor remains uniformly bounded by the above remark.

The second integral is estimated similarly.

The basic argument using the principle of uniform boundedness above goes like this: if $f_n \rightharpoonup 0$ in $L^p$ then you can define the functionals $\Lambda_n(\psi) = \int f_n \psi$. By weak convergence, for all $\psi \in L^{p'}$ we have $\Lambda_n(\psi) \to 0$, in particular $\sup_n |\Lambda_n(\psi)| < \infty$, so by the uniform boundedness principle we must have $\sup_n \|\Lambda_n\|_{L^{p'} \to \mathbb{C}} < \infty$. Now by duality, $\|\Lambda_n\|_{L^{p'} \to \mathbb{C}} = \|f_n\|_{L^p}$, and you are done.

• Thanks for your response. Could you refer me to a good text explaining why weakly convergent sequences are uniformly bounded in $L^{p}$ due to the uniform boundedness principle and duality of $L^{p}$? – user116403 Jan 18 '14 at 15:41
• @JohnJack I have updated my answer with a proof of this fact. But this is probably in Brezis' book: amazon.com/…. – i like xkcd Jan 18 '14 at 15:49
• Okay great, thanks. – user116403 Jan 18 '14 at 16:02
• Can you see how using this it would follow that: for any $z \in W^{1,p}(\Omega)$ we also have $\int_{\Omega}a(u_{k},\nabla v)\cdot \nabla z dx + \int_{\Gamma}b(u_{k})zdS \rightarrow \int_{\omega}a(u,\nabla v)z dx + \int_{\Gamma}b(u)z dS$ – user116403 Jan 20 '14 at 16:47
• @JohnJack It's the same argument: $\nabla z \in L^p(\Omega)$, so the first term converges by the weak convergence hypothesis. – i like xkcd Jan 20 '14 at 18:56