# rate of decay of fourier coefficients vs smoothness

suppose $f \in L1$, 2pi periodic and that the fourier coefficients decay with order $|n|^{-k}, k \gt 2$

show that the derivative of f is continuous

i read that the rate of decay of fourier coefficients relates that the "smoothness" of the function. but im not sure how to formalize my argument for this question

any help would be much appreciated

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possible duplicate of Differentiability and decay of magnitude of fourier series coefficients. Prof. Blatter's hints in his answer should give enough info to start working. – t.b. Mar 31 '11 at 4:04
@Theo: On the face of it, I don't see anything in that thread (or in the one linked to in its comments) that goes in the direction from Fourier coefficients to continuity/differentiability. If any of the statements made there in the other direction can be turned around, I think this should be made explicit to answer the present question. – joriki Mar 31 '11 at 6:04
@joriki: I'm thinking about writing this stuff up once and for all. First, it's not that hard at all and second it's at least the fifth time I see it coming up in one direction or the other (some of these have been deleted in the meantime). – t.b. Mar 31 '11 at 10:02

[Notation. $a_n \lesssim b_n$ means: there exists some positive constant $C$ s.t. $a_n \le C b_n$.]

let

$$c_n=\frac{1}{2 \pi} \int_{-\pi}^{\pi}f(y)e^{-i n y}\, dy$$

and write

$$f(x)=\sum_{n\in \mathbb{Z}} c_n e^{i n x}\quad (1)$$

The decay condition on $c_n$ implies uniform convergence of this series:

$$\lvert c_n e^{i n x} \rvert \le \lvert n\rvert^{-k}\lvert n^k c_n \rvert \lesssim \lvert n\rvert^{-k}$$

and $\sum_{n \in \mathbb{Z}} \lvert n\rvert^{-k}$ is a convergent numerical series. Now differentiate (1) termwise: you get

$$\sum_{n \in \mathbb{Z}}i n c_n e^{i n x}$$

which is again a uniformly convergent series:

$$\lvert i n c_n \rvert \lesssim \lvert n\rvert^{1-k}.$$

So (1) is a uniformly convergent series whose term-by-term derivative is uniformly convergent. This implies that $f$ is differentiable and

$$f'(x)=\sum_{n \in \mathbb{Z}}i n c_n e^{i n x}$$

so that, again by uniform convergence, $f'$ is also continuous.

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You don't have to exclude $0$ in the summation; the corresponding term is $0$ anyway. – joriki Mar 31 '11 at 8:00
@joriki: Of course. Thank you. – Giuseppe Negro Mar 31 '11 at 8:23