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Can you prove or disprove the following:

The Fourier coefficients of a continuous function with discontinuous derivative decay like $ 1/n^2$.

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  • $\begingroup$ Doubtful. Take any discontinuous function with slow decay, i.e. $f(x) = \sum_k \frac{e^{ikx}}{\log k}$ (ptwise limit exists) and integrate it (though not necessarily term by term)? On the positive side, if the derivative is in $L^2$, then you have a positive result. $\endgroup$ – Evan Aug 14 '13 at 20:54
  • $\begingroup$ @Evan .Thank you for the answer.Let say $f'$ belongs to $L^2$. $\endgroup$ – BigM Aug 14 '13 at 20:59
  • $\begingroup$ Hmm.. I might be wrong about the decay rate actually.. But it's possible to prove that if $f'$ belongs to $L^2$ then the Fourier series converges uniformly. Would have to work it out again, but the idea is to use the fact that the Fourier coefficients of $f'$ are in $l^2$ and use Cauchy Schwarz on $\sum \frac{|a_k|}{k}$, which bounds the $l^1$ norm of the Fourier coefficients of $f$. $\endgroup$ – Evan Aug 14 '13 at 21:37
  • $\begingroup$ This is the Fourier transform version: Relation between function discontinuities and Fourier transform at infinity. $\endgroup$ – user Aug 15 '13 at 2:03
  • $\begingroup$ @user89499 thanks for the link.mm I went through the comments,however quite honestly didnt see a crisp proof or disprove for the Fourier transform version.Actually i used this statement somewhere and my advisor said: "its baloney".lol $\endgroup$ – BigM Aug 15 '13 at 2:19
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In stated generality this is false. Take some periodic function with behaves like $f(x)=x^{1/3}$ near $0$. Since $f'(x)\notin L^2$, the Fourier coefficients of $f'$ are not in $\ell^2$. In particular, they cannot decay like $1/n$.

The following (adapted from here) is true.

Claim. Suppose $f $ is piecewise differentiable, $f'$ has bounded variation on each "piece" $(a_k,a_{k+1})$ and has nonzero jump discontinuity at some $a_k$. Then the sequence $n^2\widehat f(n)$ is bounded but does not tend to $0$.

Proof: $n^2 \widehat f(n)$ has the same size as $\widehat \mu(n)$, where $\mu$ is the signed measure such that $f'=\int \,d\mu$. (That is, $\mu$ is the weak/distributional derivative of $f'$). By the assumptions on $f'$, the measure $\mu$ consists of absolutely continuous component and some point masses. The Fourier series for the absolutely continuous part has coefficients that tend to $0$, by the Riemann-Lebesgue lemma. The Fourier series for the sum of point masses is a trigonometric polynomial; therefore it is bounded but does not tend to $0$. $\quad \Box$

With some extra machinery, Claim can be generalized:

Claim 2. Suppose $f $ is absolutely continuous, $f'$ has bounded variation and is not continuous. Then the sequence $n^2\widehat f(n)$ is bounded but does not tend to $0$.

Proof goes as above, but the sticky point is to show that $\widehat \mu(n)\not\to 0$. A measure such that $\widehat \mu(n) \to 0$ is called a Rajchman measure. From the survey Seventy years of Rajchman measures by Lyons we find that a Rajchman measure has no atoms (proved by Neder in 1920). But $f'$ is not continuous, which means the measure $\mu$ has atoms. $\quad \Box$

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  • $\begingroup$ very interesting. The function in hand is $$\ Sqrt(x) Log(1+x)$$ . It seems to me it it of BV on [0,1]. because its derivative is integrable. Would you agree with me on that? $\endgroup$ – BigM Aug 16 '13 at 0:08
  • $\begingroup$ @BigM That is certainly true. Whether you have read my answer is less certain. $\endgroup$ – user Aug 16 '13 at 1:22

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