If a sequences of functions is not pointwise convergent then it has no subsequence that is uniformly convergent

I am trying to show that $f_n(x)= \cos(nx)$ has not uniformly convergent subsequence.

$\cos(nx)$ isn't pointwise convergent on $\mathbb{R}$ so it seems that would me it can't have a subsequence which is uniform.

it is not pointwise because we can consider $\pi$ then we get $f_n(\pi) = (-1)^n$

So in general can a sequence of functions ever be not-pointwise convergent yet have a uniform convergent subsequence?

• Yes, trivially. Take $f_n(x) = (-1)^n$ as constant functions. – user296602 Aug 29 '17 at 20:30
• In general, pretty much anything that you would like to say of the form "if a sequence does ... then no subsequence does ..." is going to be difficult or impossible. You can start with a subsequence that does ... and then insert a bunch of bad terms to change the overall behavior. – user296602 Aug 29 '17 at 20:33
• @user296602 I didn't see your comment. Sorry. – tattwamasi amrutam Aug 29 '17 at 20:33
• @tattwamasiamrutam No problem. – user296602 Aug 29 '17 at 20:34

Choose a uniformly convergent sequence $f_n$ with limit $f$, and any function $g \ne f$. Then the sequence $$h_n = \left\{\begin{array}{cl} f_n & n \text{ even} \\ g & n \text{ odd}\end{array}\right.$$ is not pointwise convergent.
Take $$f_n(x)=(-1)^n$$