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Here is the problem.

Let $k\in \mathbb{N}$. Suppose that there is a constant $C$ such that $|c_n|<\frac{C}{|n|^{k+1}}$ ($c_n$ here is the $n$th Fourier coefficient). Prove that $f(x)=\sum_{n\in\mathbb{N}}c_ne^{inx}$ has $k$ continuous derivatives.

I have absolutely no idea where to even start doing this. Any guidance would be greatly appreciated.

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So the first k-1 continuous derivatives comes from the uniform convergence of $\sum \frac{1}{|n|^{i}}$, $i\geq 2$. The tricky part is about the k-th derivative. – Walt Mar 27 '13 at 5:17
@Yunfeng I have shown that if $f$ has a $k$-th derivative then $\frac{1}{2\pi}f(x)e^{inx}$ is bounded by $C/|n|^k$. Would that help at all? – crf Mar 27 '13 at 5:41
I could be wrong, but I don't think this is true. Consider the function $f(x) = x^2$, defined on the interval $[-\pi,\pi]$. By requiring $f(x+2\pi) = f(x)$, extend it to a function on $\mathbb{R}$. This extended function is continuous, but not differentiable at odd multiples of $\pi$. Its $n$th Fourier coefficient is a constant multiple of $1/n^2$ (with the exception of $n=0$), and these Fourier coefficients therefore satisfy your assumptions with $k=1$. But since the Fourier series converges uniformly, it converges to $f$, and in particular, it is not continuously differentiable. – Nick Strehlke Mar 27 '13 at 7:16
I asked the converse of this question; the answer provided is very good and may guide your answer:… – Ron Gordon Mar 27 '13 at 10:57

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