# 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

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