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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?

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    $\begingroup$ Yes, trivially. Take $f_n(x) = (-1)^n$ as constant functions. $\endgroup$ – user296602 Aug 29 '17 at 20:30
  • $\begingroup$ 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. $\endgroup$ – user296602 Aug 29 '17 at 20:33
  • $\begingroup$ @user296602 I didn't see your comment. Sorry. $\endgroup$ – tattwamasi amrutam Aug 29 '17 at 20:33
  • $\begingroup$ @tattwamasiamrutam No problem. $\endgroup$ – user296602 Aug 29 '17 at 20:34
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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.

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

This has atleast two uniformly convergent subsequences.

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  • $\begingroup$ oh of course, the counterexample is right there $\endgroup$ – oliverjones Aug 29 '17 at 20:32
  • $\begingroup$ Of course, this even has uncountably many uniformly convergent subsequences, not just two $\endgroup$ – Hagen von Eitzen Aug 29 '17 at 20:35

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