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Is there an obvious reason why Fourier coefficients of a given function cannot decay at infinity like, say, $\frac{1}{\sqrt{n}}$? Why do they have to decay like integer powers of $\frac{1}{n}$?

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    $\begingroup$ The Fourier coefficients of a function $f\in L^2([-\pi,\pi])$ are in $\ell^2$, i.e., $\sum\limits_{n=-\infty}^\infty |\hat{f}(n)|^2=\|f\|_2^2<\infty$. $\endgroup$
    – Jochen
    Commented Jun 18, 2019 at 10:16

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There is no reason whatsoever, and it is not true that Fourier coefficients must decay like integer powers of $1/n$. In fact, given a sequence $c_k$ of positive real numbers that tends to zero, there is a function $f\in L_1(0,2\pi)$ such that $\hat{f}(k)\geq c_k$ for all $k\in\mathbb{ Z}$. In other words, given any rate of decay, there is an integrable function on $(0,2\pi)$ whose Fourier coefficients have slower rate of decay.

If however you add some smoothness assumptions to $f$, such as-- $f$ is several times differentiable, then it is possible to prove that the Fourier coefficients have decay at a rate bounded by some polynomial rate of decay.

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    $\begingroup$ Ok, so say I have more regularity, like piecewise smoothness. Why do they have to be powers of integers then? $\endgroup$
    – Karl
    Commented Jun 18, 2019 at 10:22
  • $\begingroup$ Well, they don't have to be exactly. All the estimates are inequalities, and being limited-scoped humans as we are, we find it convenient to compare growth-rates to integer powers. Usually we differentiate integer-number of times, so naturally comparisons are with integer powers. However, if you accept a notion of a fractional derivative, you will get fractional powers as rates of growth. $\endgroup$ Commented Jun 18, 2019 at 10:23
  • $\begingroup$ Wow, this is distressing. An exercise in my book, using the typical theorem $\mathcal{C}^k \leftrightarrow \sim \frac{1}{n^k}$, concludes that a piecewise smooth function with a jump discontinuity has fourier coefficients that decay like $\frac{1}{n}$. I guess it is wrong and all you can say is that it is at most like #\frac{1}{n}#? $\endgroup$
    – Karl
    Commented Jun 18, 2019 at 10:26
  • $\begingroup$ I don't see why would anyone would be distressed to learn that some functions have Fourier coefficients decaying like $1/n$, and some do not. $\endgroup$ Commented Jun 18, 2019 at 10:28
  • $\begingroup$ Because it is expressed in that form. Piecewise smooth with jump discontinuity implies decay like $\frac{1}{n}$. Everytime, not just some. $\endgroup$
    – Karl
    Commented Jun 18, 2019 at 10:30
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For any function $f$ is bounded variation $|\hat {f} (n)| \leq \frac M n$ where $M$ is the total variation of $f$. [Ref. Fourier Series by Edwards]. Any piecewise differentiable function $f$ is of bounded variation.

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They can very well decay like $\frac{1}{\sqrt{k}}$.

Look at this

$$s_{\frac{1}{2}} = \sum _{k=1}^{\infty } \frac{\sin (k x)}{\sqrt{k}} = \frac{1}{2} i \left(\text{Li}_{\frac{1}{2}}\left(e^{-i x}\right)-\text{Li}_{\frac{1}{2}}\left(e^{i x}\right)\right)$$

which gives well defined function of $x$ in terms of polylog functions.

It is antisymmetric, has period $2 \pi$ and for $x \to +0$ it behaves as $x \zeta \left(-\frac{1}{2}\right)+\frac{\sqrt{\frac{\pi }{2}}}{\sqrt{x}}$

For more general decays we have

$$s_{p}=\sum _{k=1}^{\infty } \frac{\sin (k x)}{k^p}= \frac{1}{2} i \left(\text{Li}_p\left(e^{-i x}\right)-\text{Li}_p\left(e^{i x}\right)\right)$$

In this case we have for $x \to +0$

$$s_{p} \simeq x^{p-1} \cos \left(\frac{\pi p}{2}\right) \Gamma (1-p)-\frac{1}{48} x^{p+2} \sin \left(\frac{\pi p}{2}\right) \Gamma (4-p)+x \zeta (p-1)$$

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  • $\begingroup$ Hi Wolfgang, I apologise but I was referring to a more regular context, say a piecewise smooth function on some interval, then extended by periodicity to the whole real line. $\endgroup$
    – Karl
    Commented Jun 18, 2019 at 11:13
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The Parseval's Theorem states that the average power of a signal equals the sum of the average power of its Fourier components. $$ \eqalign{ & f(t) = {{a_{\,0} } \over 2} + \sum\limits_{1\, \le \,n} {a_{\,n} \cos \left( {nt} \right) + b_{\,n} \sin \left( {nt} \right)} \quad \Rightarrow \cr & \Rightarrow \quad {1 \over {2\pi }}\int_{ - \pi }^\pi {f(t)^{\,2} dt} = \left( {{{a_{\,0} } \over 2}} \right)^{\,2} + {1 \over 2}\sum\limits_{1\, \le \,n} {\left( {a_{\,n} ^{\,2} + b_{\,n} ^{\,2} } \right)} \cr} $$

Thus, for a finite power signal, i.e. a phisycally possible signal, the sum of the squares of its components must converge.
Clearly, if the amplitudes decay as $1/ \sqrt{n}$, the overall power won't be finite.

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