Fourier coefficients at infinity

Question

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 $\frac{1}{n}$?

Answer 1

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.

Answer 2

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.

Answer 3

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 $x \to +0$

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

Answer 4

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.