# What is the growth of sum of absolute values of Fourier coefficients

For a periodic BV function $f$ which has jump discontinuties, is there any theorem in Fourier analysis which gives like $$\sum_{k=0}^n\left|c_k\right|\sim C\log\left(n\right)$$ where $C$ is a constant and $c_k$ are Fourier coefficients.

From google search, and refering to Trigonometric Series Vol.1 by Zygmund, I came to know that $$\left|c_k\right|\le \frac{V}{\pi k}$$, where $V$ is the total variation of $f$. But I could not find any theorem on the asymptotics of the summation, atleast on google.

I am making an attempt in this direction and something here

• Some hope : Assume $f$ is even symmetric, and the fact $\int_0^{\lambda}\left|\frac{\sin t}t\right|dt \sim C\log(\lambda)$ Jan 31, 2017 at 5:50
• Let me understand. If you allow just one jump it is easy to produce examples where $c_k\sim 1/k$, right? Precisely, what do you want more than this? Jan 31, 2017 at 11:52
• @PieroD'Ancona : Thats not proof yet. I want to know if is actually $\sim C\log(n)$ for any jumps, and also what is $C$ if the limit exists? Jan 31, 2017 at 12:11
• So you want to prove this asymptotic for all BV functions with jumps? Jan 31, 2017 at 14:54
• @PieroD'Ancona : Yes, hopefully. Jan 31, 2017 at 15:28

Well, if $\left|c_k\right|\leq \frac{V}{\pi k}$ then $$\sum_{k=1}^{N} \left|c_k\right|\leq \frac{V}{\pi} \sum_{k=1}^{n}\frac{1}{k}\leq \frac{V}{\pi} \log n$$.
Editing to add: Regarding the lower bound, the piecewise linear function with jumps $f_j$ at locations $x_j$ has Fourier coefficients
$$c_k = \frac{1}{4\pi ik} \sum_j f_je^{2\pi ikx_j} + O\left(\frac{1}{k^2}\right)\,.$$
So you can break the question into two parts: one part is finding $C$ for which $$\frac{1}{4\pi}\sum_{k=1}^n\left|\sum_j f_je^{2\pi ikx_j}\right|\frac{1}{k} \sim C\log n\,.$$ The other is the case of a continuous BV function.
• I know that its bounded above, its evident. I am looking for the asymptotic relation : $$\sum_{k=0}^n\left|c_k\right|\sim C\log\left(n\right)$$, to sttle the matter. Looks like there might be some problem as no literature has covered it. So looks like its a bit tricky. Jan 31, 2017 at 8:06
• What do you mean by partial sum unbounded? it converges to $f$ as $f$ is BV, except at points of jumps where they converge to $\frac{f(x+0)+f(x-0)}{2}$. Ofcourse conjugate partial sum blows up logarithmically (Its called Lukacs theorem), but I dont know how to use this result to arrive at what I want. Jan 31, 2017 at 8:08
• oops Looks like I have the answer if I apply Lukacs theorem, to get a lower bound. But I still need to determine $C$, the limit. (Hint I got : Conjugate partial sum goes to $-\frac{D}{\pi}\log(n)$, so I think I have proof for being positive about the $\sim C\log(n)$, I need to determine $C$ now. Jan 31, 2017 at 12:01