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As part of one problem I need to find the widest subset of $\mathbb{R}$ on which the obtained Fourier series can be integrated and derived term by term.

I found that it has something to do with uniform convergence, but I can't really apply anything I've read about it so far.

The Fourier series is: $$f(t)=\frac{2}{\pi}+\frac{4}{\pi} \sum_{n=1}^{+\infty}\left ( \frac{(-1)^{n-1}}{4n^2-1} \right )\cos(2nt)$$

Note: I can prove absolute convergence (from which uniform convergence follows) but I don't see how I can apply it to certain interval $[a,b]$ where I can discuss the bounds $a$ and $b$.

tyr
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  • The series converges absolutely and uniformly on all of $\mathbb R$, by the Weierstrass M test, because the series $\sum \left(\frac{(-1)^{n-1}}{4n^2-1}\right)$ converges absolutely. –  May 29 '16 at 18:03
  • A series of functions that converges may be differentiated term by term if the derived series converges uniformly and is of continuous functions. That is, if $f_n$ is continuously differentiable and $\sum f_n$ converges and $\sum f_n'$ converges uniformly then if $f$ is the sum of the former series we have $f'=\sum f_n'$. – Nap D. Lover May 29 '16 at 20:11
  • @LoveTooNap29: Yes, and have you tested whether in this case $\sum f_n '$ converges uniformly? – Alex M. May 29 '16 at 20:40

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Notice that for all $t \in \Bbb R$,

$$\left| \left ( \frac{(-1)^{n-1}}{4n^2-1} \right )\cos(2nt) \right| \le \left| \frac{(-1)^{n-1}}{4n^2-1} \right| \underbrace {|\cos (2nt)|} _{\le 1} \le \left| \frac{(-1)^{n-1}}{4n^2-1} \right| = \frac 1 {4n^2-1}$$

and $\sum \frac 1 {4n^2-1}$ is convergent (use the limit comparison test and compare it with $\sum \frac 1 {n^2}$, which is known to be convergent), so by Weierstrass's M-test the series $\sum \limits _{n=1}^\infty \left ( \frac{(-1)^{n-1}}{4n^2-1} \right )\cos(2nt)$ converges absolutely and uniformly on $\Bbb R$, therefore it may be integrated term by term on any interval $[a,b] \subset \Bbb R$.

Alex M.
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  • Is uniform convergence sufficient to conclude term by term differentiability? I'm not so sure. The Fourier series of a periodic "triangle" function converges uniformly, but the function is not differentiable everywhere. I think the function in the OP will also have "corners" where it is not differentiable. –  May 29 '16 at 18:15
  • I think it needs to be continuous – tyr May 29 '16 at 18:25
  • @Bungo: Uniform convergence is enough if each function in the series (in our case $\cos (2nt)$) is integrable / differentiable. – Alex M. May 29 '16 at 18:30
  • @tyr Continuity is not sufficient, as the example of the triangle function shows. A sufficient condition for termwise differentiability is: (1) the series converges uniformly, (2) the derivative of each term is continuous, and (3) the series of termwise derivatives converges uniformly. However, I don't believe this is a necessary condition. –  May 29 '16 at 18:30
  • @AlexM. But there are Fourier series which converge uniformly to functions which are not everywhere differentiable (as I believe is the case here), so termwise differentiation may not work at every point in the domain. –  May 29 '16 at 18:32
  • @Bungo: Give me a bit of time, I'm thinking about it. – Alex M. May 29 '16 at 18:34
  • @Bungo: What bothers me is the square in $4n^2-1$; had there been any other power $>2$, the result would have been easy to obtain. As a side-note, $f$ is derivable on $\Bbb R$, I've computed the sum with Mathematica and it's a finite sum of terms involving $\arctan$ of some imaginary exponentials. – Alex M. May 29 '16 at 18:56
  • @AlexM. Interesting, I don't have Mathematica on this PC so I just plotted the sum of the first 50 terms, and it looked like there were going to be corners at $t=\pm \pi/2$. (Partial sums can be misleading!) This seemed plausible given the similarity of these coefficients to those of the triangle function. –  May 29 '16 at 19:01
  • @Bungo: $f= \frac 2 \pi + \frac 1 \pi \textrm e ^{- \textrm i t} (-2 \textrm e ^{- \textrm i t} + \arctan \textrm e ^{- \textrm i t} + \textrm e ^{2 \textrm i t} \arctan \textrm e ^{- \textrm i t} + \arctan \textrm e ^{\textrm i t} + \textrm e ^{2 \textrm i t} \arctan \textrm e ^{\textrm i t})$. If Mathematica is right, then $f$ is even analytic. – Alex M. May 29 '16 at 19:05
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    It turns out that $f(x) = |\cos(x)|$. See this answer for a derivation. Consequently, $f$ is not differentiable at $x = \pm \pi/2$. –  May 29 '16 at 19:33
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    @Bungo: Nice find! This shows that Mathematica was wrong - it wouldn't be the first time. On the other hand, knowing what function that series sums up to is not a nice proof. I was thinking about using Sobolev's embedding theorem, but that approach fails too - again, because of the square in $4n^2 - 1$ (any power $>2$ would have done the job). I ask the OP to unaccept my answer, since it is obviously incomplete (and missing the more difficult half). – Alex M. May 29 '16 at 20:00
  • Some context from the OP would be good. In the first sentence, he mentions that this is "part of one problem" and that this is "the obtained Fourier series" (emphasis mine). Perhaps the problem started by defining $f(x) = |\cos(x)|$ and calculating the Fourier series was an earlier step. If not, I agree that this is not a nice proof. Unfortunately I can't see how to prove it directly from the series. –  May 30 '16 at 00:22
  • @Bungo: Exactly. You see, there are several results that give sufficient conditions for derivability, but I know of no result giving points where a function is not derivable. The problem seems thus so intractable, that I suspect its statement to be incomplete. On the other hand, if it is given that $f = |\cos|$, then the issue of its integrability / derivability is trivial and does not require its Fourier series. – Alex M. May 30 '16 at 07:03