# Convergence in Sobolev spaces implies classical convergence?

Let $$p>1$$ and $$\Omega$$ be an open bounded domain in $$\mathbb{R}^N$$. Take $$(u_n)\subset W_0^{1, p}(\Omega)\cap L^{\infty}(\Omega)$$ and $$u\in W_0^{1, p}(\Omega)\cap L^{\infty}(\Omega)$$.

If $$u_n\longrightarrow u \quad\mbox{ in } W_0^{1, p}(\Omega),$$ it it true that $$u_n\longrightarrow u$$ in the classical sense (up to subsequences)?

• Any thoughts on why you expect this to be true? Can you fix the $n$ notation clash? Have you tried anything? Why did you restrict $p>1$? Commented Nov 7, 2020 at 13:37
• @CalvinKhor thanks for the comment, I edited the question fixing the right $n$. So, it is a small part of a bigger problem in which I need $p>1$ and it justifies mi choice. I didn't try anything because I don't know where to start. If you please give me a hint, I will try. Commented Nov 7, 2020 at 13:43
If $$u_n\rightarrow u$$ in $$W^{1,p}_0,$$ then (in particular) it converges in $$L^p$$. If $$u_n\rightarrow u$$ in $$L^p$$, then there exists a subsequence $$(u_{n_j})$$ of $$(u_n)$$ with the property that $$u_{n_j}\rightarrow u$$ pointwise almost everywhere. The proof of the latter fact comes in when demonstrating that $$L^p$$ is complete.