I am currently studying Fourier Analysis on my own. In the Notes I use the following comment is made, which I unfortunately don't understand:

Given that we know the series

$f(x) = \sum c_k e^{ikx}$

converges pointwise (where the $c_k$ are the Fourier coefficients and $f$ is a periodic function), to show uniform convergence it is enough to show that

$ \sum |c_k| < \infty $

I tried to find the result regarding uniform convergence that this comment refers to but so far I wasn't successful.

Could somebody help me and give a hint as to why this is true ? Many thanks!


This is just an application of the Weierstrass M-test.

It's a rather simple result, and it is edifying to prove the uniform convergence of your series directly:

Let $m> n$, then $$ \biggl| \sum_{j=1}^m c_j e^{ijx} - \sum_{j=1}^n c_j e^{ijx} \biggr| =\biggl| \sum_{j=n+1}^m c_j e^{ijx} \biggr| \le \sum_{j=n+1}^m |c_j e^{ijx} |= \sum_{j=n+1}^m |c_j |. $$ Since $\sum\limits_{j=1}^\infty |c_j|<\infty$, we can make the right hand side of the above as small as we wish provided $n$ is sufficiently large. Thus, we can make the left hand side as small as we wish, independently of $x$, as long as $n$ is sufficiently large.

It follows that $ \sum\limits_{j=1}^\infty c_j e^{ijx} $ is uniformly Cauchy, and, thus, uniformly convergent.

  • $\begingroup$ Perhaps you should say something regarding the lower index, since complex Fourier series starts from $i=-\infty$. $\endgroup$ – AD. Feb 1 '12 at 16:12
  • $\begingroup$ But $\sum\limits_{j=1}^\infty |c_j|<\infty$ is not a given, though, is it? $\endgroup$ – Ryker Feb 15 '16 at 0:36
  • 1
    $\begingroup$ @Ryker I interpreted the OP's question as "why would assuming $\sum|c_j|<\infty$ imply $f(x)$ is uniformly convergent?" $\endgroup$ – David Mitra Feb 15 '16 at 1:20
  • $\begingroup$ @DavidMitra, I see. Yeah, taking another look at it I think you're right. $\endgroup$ – Ryker Feb 15 '16 at 1:24
  • $\begingroup$ Isn't the last equality actually an inequality? I believe you were adapting the proof itself of the Weierstrass M-Test for this sequence of functions specifically. $\endgroup$ – DrHAL Sep 1 '17 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.