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

Question

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?

Answer 1

In addition to the other answer, here is a more general example to illustrate the vague-and-not-to-be-taken-too-seriously principle that

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.

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.

Answer 2

Take $$f_n(x)=(-1)^n$$

This has atleast two uniformly convergent subsequences.