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Suppose $f \in L^1$, $2\pi$ 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 I'm not sure how to formalize my argument for this question.

jack
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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 04:04
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    @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 06:04
  • @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

1 Answers1

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

A rough'n'ready argument would be:

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.

Giuseppe Negro
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