# Weak convergence of sequence in Sobolev space implies uniform convergence

Does weak convergence in a Sobolev space implies uniform convergence?

In the above question, a proof by contradiction is given from which it follows that if a sequence in $$W^{1,p}$$ over a nice, open and bounded domain in $$\mathbb{R}^n$$ ($$p>n$$) is weakly convergent to $$f$$ in $$W^{1,p}$$, this sequence is uniformly convergent to $$f$$.

However, I don't really get the proof. Does the fact that $$f_{n_k} \rightarrow g$$ in $$C$$ follow from the compact embedding? If yes, why is this necessarily the same subsequence as the subsequence that doesn't converge?

• Do you have problems understanding points 1 and 2 in the linked question or is only the answer unclear? – MaoWao Jan 24 at 10:51
• Only the answer is unclear to me. – Catemathics Jan 24 at 10:55

## 1 Answer

You can use the following, very useful little lemma:

A sequence $$(x_n)$$ converges to $$x$$ if and only if every subsequence has a subsequence converging to $$x$$.

It has nothing to do with Sobolev spaces or continuous functions, but holds in any topological space. Try to prove it yourself (if you are not familiar with topological spaces, you can just prove it for normed spaces).

The application to the problem as hand works as follows. By point 2 from the linked question, $$(f_n)$$ has a subsequence converging uniformly to $$f$$. Of course, instead of starting with $$(f_n)$$, you can start with an arbitrary subsequence $$(f_{n_k})$$ and extract a subsequence converging uniformly to $$f$$. The lemma above then tells you that $$f_n\to f$$ uniformly.

Note that this is just a reformulation of the linked answer, which also contains a proof of the lemma above in the case of normed spaces.