# Proving this Corollary regarding Fourier Series

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.

• 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 $m$. – Andrew D. Hwang Jun 14 '14 at 1:47
• 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! – user30625 Jun 14 '14 at 2:04
• 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... – C-Star-W-Star Jun 14 '14 at 2:42
• By the way I guess there should be squares everywhere in the assumption on the series... – C-Star-W-Star Jun 14 '14 at 2:47
• What do you mean by "squares everywhere in the assumption on the series"? Is this in regards to what you wrote or the Corollary? – 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.)
• What does $a.e$ and $L^2$ signify? – user30625 Jun 14 '14 at 3:01
• 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? – DisintegratingByParts Jun 14 '14 at 3:18
• 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 $[0,2\pi]$. – DisintegratingByParts Jun 14 '14 at 3:30
• 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? – user30625 Jun 14 '14 at 3:35