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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!

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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.

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  • $\begingroup$ 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
  • $\begingroup$ @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

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