How to find a bound for Fourier coeffiecients Let $ -\pi=x_{0}<x_{1}<...<x_{m}=\pi $
be a partition of the interval $ [-\pi,\pi) $
And let $ s\left(x\right)=\sum_{j=1}^{m}\alpha_{j}1_{\left(x_{j-1},x_{j}\right)}\left(x\right)$ be a step function such that $ \alpha_{1}\leq\alpha_{2}\leq...\leq\alpha_{m}\in\mathbb{R} $
Where $ 1_{\left(x_{j-1},x_{j}\right)}\left(x\right)=\begin{cases}
1 & x\in\left(x_{j-1},x_{j}\right)\\
0 & x\in[-\pi,\pi)\setminus\left(x_{j-1},x_{j}\right)
\end{cases} $
I have to find a constant $ C $ such that $ C $ does not depend on $ s $ and the following inequality holds:
$ |\hat{s}\left(k\right)|\le C\frac{\max\left\{ -\alpha_{1},\alpha_{m}\right\} }{|k|} $
I already calculated the Fourier coefficients of the indicators, and it is given by:
$ \hat{1}_{\left(a,b\right)}\left(k\right)=\begin{cases}
\frac{b-a}{2\pi} & k=0\\
\frac{1}{2\pi ik}\left(e^{-ika}-e^{-ikb}\right) & k\ne 0
\end{cases} $
I actually did find such $ C $, I managed to prove that for $ C=\frac{m}{2\pi} $ the inequality holds. Is it possible to find such $ C $ that would not depend on $ s $ or on the partition $ x_1<x_2<...<x_m $ ?
This is part of a guided exercise and I need such $ C $ for the next steps.
Thanks in advance.
 A: We want to bound $ |\sum_{j=1}^{m}\alpha_{j}\left(e^{-ikx_{j-1}}-e^{ikx_{j}}\right)| $
Let us denote $ B_{i}=\sum_{j=1}^{i}\left(e^{-ikx_{j=1}}-e^{-ikx_{j}}\right) $
So by Abel's summation formula we have:
$ |\sum_{j=1}^{m}\alpha_{j}\left(e^{-ikx_{j-1}}-e^{ikx_{j}}\right)|=|\alpha_{m}\left(e^{-ik_{m-1}}-e^{-ikx_{m}}\right)+\sum_{i=1}^{m-1}B_{i}\left(\alpha_{i+1}-\alpha _{i}\right)| $
$ \underset{a_{i}\leq a_{i+1}}{\leq}2|\alpha_{m}|+\sum_{i=1}^{m-1}|B_{i}|\left(\alpha_{i+1}-\alpha_{i}\right) $
Now notice that $ B_i $ is just a telescopic sum and thus $ |B_i| < 2 $ So:
$ \underset{a_{i}\leq a_{i+1}}{\leq}2|\alpha_{m}|+\sum_{i=1}^{m-1}|B_{i}|\left(\alpha_{i+1}-\alpha_{i}\right)\leq2|\alpha_{m}|+2\sum_{i=1}^{m-1}\left(\alpha_{i+1}-\alpha_{i}\right)\underset{telescopic\thinspace\thinspace\thinspace sum}{=}2|\alpha_{m}|+2\left(\alpha_{m}-\alpha_{1}\right)\leq4|\alpha_{m}|+4|\alpha_{1}| $
So all in all
$ |\hat{s}\left(k\right)|=|\frac{1}{2\pi}\intop_{-\pi}^{\pi}\sum_{j=1}^{m}\alpha_{j}1_{\left(x_{j-1},x_{j}\right)}e^{-ikx}dx|=|\frac{1}{2\pi}\sum_{j=1}^{m}\alpha_{j}\hat{1}_{\left(x_{j-1},x_{j}\right)}\left(k\right)|=\frac{1}{2\pi|k|}|\sum_{j=1}^{m}\alpha_{j}\left(e^{-ikx_{j-1}}-e^{ikx_{j}}\right)|\leq\frac{4}{2\pi|k|}\left(|\alpha_{1}|+|\alpha_{m}|\right) $
And since $ a_1\leq a_2 \leq ... \leq a_m $
we have
$ |\hat{s}\left(k\right)|\leq\frac{8}{2\pi|k|}\max\left\{ -a_{1},a_{m}\right\}  $
