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.
 A: 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.)
