Here's the theorem:

Theorem: If $f$ is periodic with Fourier coefficients $a_n,b_n$ and if the series $$\sum_{n=1}^\infty (|n^{k}a_n|+|n^{k}b_n|)$$ converges for some integer $k \geq 1$, then f has continuous derivatives $f',...,f^{(k)}$ whose Fourier series are differentiated series of $f$. $\blacksquare$

So we know that $f(x)=\sum_{n=0}^\infty a_ncos(nx)+b_nsin(nx)$ on the interval $[-\pi,\pi]$. What needs to be done next is proving that the derivative of our series $f$ converges uniformly on our interval. I know how to prove the inequality $$\sum_{n=1}^\infty [a_n(cos(nx))^{(h)}+b_n(sin(nx))^{(h)}] \leq \sum_{n=1}^\infty (|n^{h}a_n|+|n^{h}b_n|) \leq \sum_{n=1}^\infty (|n^{k}a_n|+|n^{k}b_n|)$$ for all $ h \in [1,k]$ via the Weierstrass M-Test. Using induction, is it sufficient to show $f'$ is uniformly convergent and work my way up from there to the $f^{(k)}$ case, where assume the convergence of the $(k-1)^{th}$ derivative of our series to be equal to $f^{(k-1)}$? Will that ultimately show that $$f^{(k)}(x)=\sum_{n=1}^\infty [a_n(cos(nx))^{(k)}+b_n(sin(nx))^{(k)}]?$$

If this is the case (which I'm almost positive it is based on the way uniform convergence is defined), then here is my attempt at a proof of this. If anyone could tell me if I did this even remotely correctly, I would quite grateful!

Attempted proof: Consider $$f(x)=a_0 + \sum_{n=1}^\infty [a_ncos(nx)+b_nsin(nx)]$$ which is $2\pi$-periodic. Differentiating $f$, we obtain $$f'(x) \sim \frac{d}{dx}(\sum_{n=1}^\infty [a_ncos(nx)+b_nsin(nx)]).$$ Recalling that $a_n=\int_{-\pi}^{\pi}f(x)cos(nx)dx$ and $b_n=\int_{-\pi}^{\pi}f(x)sin(nx)dx$, it using integration by parts that $$a_n'=\int_{-\pi}^{\pi}f'(x)cos(nx)dx=nb_n$$ and $$b_n'=\int_{-\pi}^{\pi}f'(x)sin(nx)dx=-na_n$$ $$\implies f'(x) \sim \frac{d}{dx}(\sum_{n=1}^\infty [a_ncos(nx)+b_nsin(nx)])$$ $$=\sum_{n=1}^\infty [-na_nsin(nx)+nb_ncos(nx)]$$ $$\equiv \sum_{n=1}^\infty [a_n\frac{d}{dx}(cos(nx))+b_n\frac{d}{dx}(sin(nx))].$$ Let's show that $\sum_{n=1}^\infty [-na_nsin(nx)+nb_ncos(nx)]$ is uniformly convergent on $[-\pi,\pi]$, which will show that that our differentiated Fourier series converges to $f'$ on $[-\pi,\pi]$. Recall that $|cos(x)| \leq 1$ which extends to $|cos(nx)| \leq 1$ and $|sin(x)| \leq 1$ which extends to $|sin(nx)| \leq 1$. Thus, $$\sum_{n=1}^\infty [-na_nsin(nx)+nb_ncos(nx)]$$ $$\leq |\sum_{n=1}^\infty [-na_nsin(nx)+nb_ncos(nx)]|$$ $$\leq \sum_{n=1}^\infty |-na_nsin(nx)+nb_ncos(nx)|$$ $$\leq \sum_{n=1}^\infty [|-na_nsin(nx)|+|nb_ncos(nx)|]$$ $$\leq \sum_{n=1}^\infty (|na_n|+|nb_n|)$$ $$ \equiv \sum_{n=0}^\infty (|n^1a_n|+|n^1b_n|)]$$ so we see our series is uniformly convergent on $[-\pi,\pi]$ by the Weierstrass M-Test, which implies that $f'$ represents our series on $[-\pi,\pi]$ and we have proved the case of $k=1$.

Now suppose $$f^{(k-1)}(x)=\frac{d^{(k-1)}}{dx^{(k-1)}}(\sum_{n=1}^\infty [a_ncos(nx)+b_nsin(nx)])$$ $$\equiv \sum_{n=1}^\infty [a_n\frac{d^{(k-1)}}{dx^{(k-1)}}(cos(nx))+b_n\frac{d^{(k-1)}}{dx^{(k-1)}}(sin(nx))]$$ on the interval $[-\pi,\pi]$. We have currently have, since we don't know if the k-th derivative of $f$'s Fourier series converges to $f^{(k)}$, $$f^{(k)} \sim \sum_{n=1}^\infty [a_n\frac{d^{(k)}}{dx^{(k)}}(cos(nx))+b_n\frac{d^{(k)}}{dx^{(k)}}(sin(nx))].$$ This part of the proof is actually proven in an older post on stackexchange, which I link here:

"Proving that $\sum_{n=1}^\infty{[a_ncos(nx)]^{(m)}+[b_nsin(nx)]^{(m)}} \leq \sum_{n=1}^\infty{(|a_nn^m| + |b_nn^m|)}$"

It follows, finally, that $$f^{(k)}=\sum_{n=1}^\infty [a_n\frac{d^{(k)}}{dx^{(k)}}(cos(nx))+b_n\frac{d^{(k)}}{dx^{(k)}}(sin(nx))]$$ on the interval $[-\pi,\pi]$, so by the Principle of Mathematical Induction, we see that $f$ has $k$-continuous derivatives given the conditions imposed. $$\blacksquare$$


1 Answer 1


The proof is correct but so wordy... How about this:

  1. There is a standard theorem in real analysis: if $\sum f_n'$ converges uniformly on some interval $I$ and $\sum f_n$ converges at some point of $I$, then $\sum f_n$ converges uniformly on $I$, its sum $f$ is differentiable, and $f'=\sum f_n'$.

  2. The hypotheses in 1 apply to $f_n(x)=a_n\cos nx+b_n\sin nx$, because of the estimate $$|f_n'(x)| = |-n a_n \sin nx + nb_n\cos nx| \le | n a_n \sin nx| + |nb_n\cos nx| \\ \le |n a_n|+|n b_n|$$ and the Weierstrass M-test.

  3. The case $k=1$ has been proved. The general case follows by induction.

  • $\begingroup$ Thank you for the feedback! It's a relief to know that I was correct. I know it's quite long, I just wanted to make it as detailed as possible so as to avoid any potential holes in my logic. That theorem you posted is incredibly helpful. I actually just found in Bartle's Introduction to Real Analysis, 3rd edition along with a few other useful Theorems to strengthen the proof (by citing of course, as I have no intentions of increasing the length of this already monstrous proof). $\endgroup$ Jun 25, 2014 at 5:57
  • $\begingroup$ In the morning, I will be attempting to write a shorter proof with your suggestions. My only fear is that the $f^{(k}) will still be quite lengthy. Oh well. We'll see what I work out $\endgroup$ Jun 25, 2014 at 6:09
  • $\begingroup$ Okay so I figured out a way to make it shorter: basically leave out the derivation of how to get to $f'$, as well as leaving out the summations when doing the Weierstrass M-Test. I was curious about something, if you're still willing to answer any more of my questions. If we just have $f \sim \sum ...$ then is it okay to make the assumption that $f_n$ converges for at least one point on $[-\pi,\pi]$ or does there need to be more information? If you're not sure, no worries. Just thought I'd ask. $\endgroup$ Jun 26, 2014 at 4:45
  • $\begingroup$ @SavageHenry Kolmogorov constructed an $L^1$ function whose Fourier series diverges everywhere. $\endgroup$
    – user147263
    Jun 26, 2014 at 4:54
  • $\begingroup$ LOL oh my. That doesn't seem too promising for the direction I was thinking of going... $\endgroup$ Jun 26, 2014 at 5:47

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