Okay so here's the the problem:

Let $k \in \mathbb{N}$. If $f$ is periodic, with Fourier coefficients $a_n,b_n$ and the series $\sum_{n=1}^\infty{(|a_n| + |b_n|)n^k}$ converges for some $k$, then for $m \in [0,k]$ $$(1):f^{(m)}(x)=\frac{d^m}{dx^m}\sum_{n=0}^\infty[(a_n(cos(nx)+b_n(sin(nx)] $$

Now I'm not sure really sure how to approach this. It's seems quite obvious that the equality would be true, but trying to construct a proof for the past few hours has proven fruitless. My professor said to "use the Differentiation Theorem of a function" from Rudin's text (but we're not even using a rigorous textbook; our book is all application). Here's my thought process:

If we look at the right hand side of (1), we can let the inside of the summation be $\frac{d^m}{dx^m}(f_n(x))$. Then we have to prove that $$\frac{d^m}{dx^m}(f(x))=\sum_{n=1}^\infty \frac{d^m}{dx^m}f_n(x).$$ We know that $\exists k \in \mathbb{N}$ s.t for $m \in [0,k]$ $$\sum_{n=0}^\infty \frac{d^m}{dx^m}f_n(x) \leq |\sum_{n=0}^\infty \frac{d^m}{dx^m}f_n(x)| \leq \sum_{n=0}^\infty |\frac{d^m}{dx^m}f_n(x)| \leq \sum_{n=0}^\infty{(|a_n| + |b_n|)n^m}$$ $\implies$ uniform convergence by the Weierstrass M-Test since $\sum_{n=0}^\infty{(|a_n| + |b_n|)n^m}$ is convergent. We know that Weierstrass M-Test originates from the Cauchy Criterion, so it follows that we also have that $\sum_{n=1}^\infty \frac{d^m}{dx^m}f_n(x)$ is uniformly Cauchy $\forall x \in [-\pi,\pi]$.

There are two Theorems that could help here. The first one states that "If $f_n$ is uniformly Cauchy, then $\exists$ a function $f$ on S such that $f_n \to f$ uniformly on S". The second Theorem states that "If each $f_n$ is continuous on S and the series converges uniformly on S, then the series we're working with represents a continuous function. Thus, we're almost on the way to proving (1).

We know $\forall x \in [-\pi,\pi]$ the Fouier Series of $f$ is $f(x)=a_0+\sum_{n=1}^\infty [(a_ncos(nx)+b_nsin(nx)]$, so again, I felt like it would follow naturally that we could show the LHS of (1), especially after showing Uniform convergence of the RHS of (1). If anyone could give me a hand, I'd be much obliged.

  • $\begingroup$ Rather than passing the derivative through the summation (which requires strong hypotheses on $f$), compute the Fourier coefficients of $f'$ directly by integrating against $\cos(nx)$ and $\sin(nx)$; then integrate by parts, obtaining a factor of $n$ from the chain rule. Now argue inductively on $m$. $\endgroup$ – Andrew D. Hwang Jun 14 '14 at 1:47
  • $\begingroup$ I ended up with $\sum_{n=1}^\infty [-n^{2}b_nsin(nx)-n^{2}a_ncos(nx)]$. Based on the problem, I feel as though this is not correct, but the math adds up! $\endgroup$ – user30625 Jun 14 '14 at 2:04
  • $\begingroup$ The suggestion given in the first comment unfortunately doesn't apply since there's two different theorems: either prove that: If $f$ is $k$ times differentiable then the $k$-th derivative has the form$\sum\ldots$. If the expression $\sum\ldots$ is convergent then $f$ was $k$ times differentiable (and besides then of course is given by the series). You have to prove both and, unfortunately, proving that it is differentiable resambles in fighting with series... $\endgroup$ – C-Star-W-Star Jun 14 '14 at 2:42
  • $\begingroup$ By the way I guess there should be squares everywhere in the assumption on the series... $\endgroup$ – C-Star-W-Star Jun 14 '14 at 2:47
  • $\begingroup$ What do you mean by "squares everywhere in the assumption on the series"? Is this in regards to what you wrote or the Corollary? $\endgroup$ – user30625 Jun 14 '14 at 2:59

The Series $f_{m}(x)=\sum_{n=0}^{\infty}a_{n}\frac{d^{m}}{dx^{m}}\cos(nx)+b_{n}\frac{d^{m}}{dx^{m}}\sin(nx)$ converges absolutely and uniformly for all $0 \le m \le k$. So $f_{m}$ is a continuous periodic function on on $[0,2\pi]$. Because these sums converge uniformly, then it possible to interchange the order of summation and integration in order to obtain the following for $1 \le m < k$: $$ \int_{0}^{x}f_{m+1}(t)\,dt = f_{m}(t)-f_{m}(0). $$ Because $f_{m+1}$ is continuous and periodic for $0 \le m < k$, then the Fundamental Theorem of Calculus gives $f_{m}'=f_{m+1}$ for $0 \le m < k$. By finite-induction, $f_{0}$ has $k$ continuous periodic derivatives with $f^{(m)}=f_{m}$ for $0 \le m \le k$.

The only part that remains to be shown is that $f_{0}=f$. I don't know what results you are allowed to use to show such a fact, but there are a variety of possibilities. If $a_{n}',b_{n}'$ are the Fourier coefficients of $f_{0}$, then you may interchange integration and summation to show that $a_{n}'=a_{n},b_{n}'=b_{n}$. At this point you need some result of the following type: If $f$ and $g$ are $L^{2}[0,2\pi]$ functions with the same Fourier coefficients, then $f=g$ a.e.. That will allow you to conclude that $f$ is equal a.e. to an $m$-times continuously differentiable function--namely $f=f_{0}$ a.e. where $f_{0}=\sum_{n}a_{n}\cos(nx)+b_{n}\sin(nx)$ on $[0,2\pi]$. (You cannot conclude $f=f_{0}$ everywhere unless you know $f$ is continuous, for example, because changing $f$ on a set of measure $0$ doesn't change its Fourier coefficients.)

  • $\begingroup$ What does $a.e$ and $L^2$ signify? $\endgroup$ – user30625 Jun 14 '14 at 3:01
  • $\begingroup$ The term a.e. means "except possibly for a set whose Lebesgue measure is 0". I assume that you have not studied Lebesgue integration because that $L$ stands for Lebesgue, too. If that is the case, then you should probably start off assuming that $f$ is continuous and periodic on $[0,2\pi]$. Then what you need to know is this: If $g$ is a periodic continuous function on $[0,2\pi]$, then $g\equiv 0$ iff all the Fourier coefficients of $g$ are $0$. Do you have such a Theorem that you can use? $\endgroup$ – DisintegratingByParts Jun 14 '14 at 3:18
  • $\begingroup$ Haha I actually went to Rudin's Principles of Mathematical Analysis to see if that was how he denoted anything regarding Lebesgue right after reading your post. Sadly, I've not gotten to Lebesgue Theory yet. And I'm going to try and find a theorem like that but currently I don't think I have it. $\endgroup$ – user30625 Jun 14 '14 at 3:27
  • $\begingroup$ A result due to Fejer that will work and that you may have learned: If $S_{n}(f)$ is the truncated Fourier Series for $f$ up to the $\cos(nx)$ $\sin(nx)$ terms, and if $f$ is continuous and periodic on $[0,2\pi]$, then average $\frac{1}{k+1}\sum_{n=0}^{k}S_{n}$ converges point-wise everywhere to $f$. Conclusion: if $f$ is continuous and periodic on $[0,2\pi]$ and has all $0$ Fourier coefficients, then $f\equiv 0$ on $[0,2\pi]$. $\endgroup$ – DisintegratingByParts Jun 14 '14 at 3:30
  • $\begingroup$ I mean we know $f$ is periodic and I showed that it was uniform convergent (which also means that it's uniformly Cauchy and that it represents a continuous function on its periodic domain). We have to show that $f^{(m)}(x)=\frac{d^m}{dx^m}\sum_{n=1}^\infty [a_ncos(nx)+b_nsin(nx)]$. If $f \equiv 0$ wouldn't that defeat the purpose? $\endgroup$ – user30625 Jun 14 '14 at 3:35

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