Ok, the sequence $f_n(x) = \sin(nx)$ - is it enough to say that $\displaystyle\lim_{n \to \infty} f_n(x)$ doesn't exist so it can't possibly be uniformly convergent?

I want to try showing it from the definition of uniform convergence on a sequence of functions, but due to $\displaystyle\lim_{n \to \infty} f_n(x)$ not existing I don't think it's really possible.

Any suggestions would be greatly appreciated :)

  • $\begingroup$ Would it be best to show this via the Cauchy Criterion. For any $0 < \epsilon < 1$ there doesn't exist an $N \in \mathbb{N}$ such that $|f_n(x) - f_m(x)| < \epsilon$ for $0 < \epsilon < 1$ and $\forall m,n > N$ $\endgroup$ – Noble. Feb 26 '13 at 10:12
  • $\begingroup$ that's enough! if a sequence of functions converges uniformly to some function $f$, then it converges simply to that same function (that is pointwise). $\endgroup$ – Olivier Bégassat Feb 26 '13 at 10:12

If a sequence of functions $f_n$ is uniformly convergent then it also converges pointwise, so it suffices to find some $x$ such that $\lim\limits_{n\to\infty}f_n(x)$ does not exist, as you suspect. One easy example to prove is $x=\pi/4$, as $\sin(2k\cdot \pi/4)$ is $0$ if $k$ is even and $\pm 1$ if it is odd.


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