A bounded monotonic function on an closed interval has Fourier coefficient decaying in O(1/|n|) On interval $[-\pi,\pi]$, $|f|\leq B$ for $B\geq 0$, and $f$ is assumed to be monotonically increasing. 
We want to prove that $\hat f(n)=O(1/|n|)$ for $|n|$ large enough, that is for all large $|n|$,  $|\hat f(n)|\leq C/|n|$ for some $C>0$.
My thought:
$f$ is Riemann integrable, so once we proved the Riemann sum of $2\pi g(\phi)=(f(\phi)e^{-in\phi})$  on $[-\pi,\pi]$ is bounded by $C/|n|$ for $|n|$ large, we are done.
My problem:
I failed to bounded the Riemann sum of $f(\phi)e^{-in\phi}$ with some constant divided by $|n|$. Hence, I can only prove that $\hat f(n)$ is bounded.
 A: Hints:
WLOG, we assume $f$ be increasing, bounded by $M$.


*

*First check Fourier series for $\chi_{[a,b]}$ satisfies $O(1/|n|)$.

*Show that $$\sum_{k=1}^N \alpha_k\chi_{[a_k,a_{k+1}]}(x)$$ with $-\pi=a_1<a_2<\cdots<a_N<a_{N+1}=\pi$, and $-M\leq\alpha_1<\alpha_2<\cdots<\alpha_N\leq M$ has Fourier coefficient
$$\sum_{k=1}^N \alpha_k\frac{e^{-inx_{k}}-e^{-inx_{k+1}}}{in}=\frac{\alpha_1 e^{-inx_1}-\alpha_{N} e^{-inx_{N+1}}+\sum_{k=2}^{N-1} (\alpha_{k+1}-\alpha_{k})e^{-inx_{k+1}}}{in}$$
Hence $$|\sum_{k=1}^N \alpha_k\frac{e^{-inx_{k}}-e^{-inx_{k+1}}}{in}|=|\frac{\alpha_1 e^{-inx_1}-\alpha_{N} e^{-inx_{N+1}}+\sum_{k=2}^{N-1} (\alpha_{k+1}-\alpha_{k})e^{-inx_{k+1}}}{in}|\\ \leq \frac{|\alpha_1 e^{-inx_1}-\alpha_{N} e^{-inx_{N+1}}|+|\sum_{k=2}^{N-1} (\alpha_{k+1}-\alpha_{k})e^{-inx_{k+1}}|}{|n|}\\=\frac{-\alpha_1 +\alpha_{N} +\sum_{k=2}^{N-1} (\alpha_{k+1}-\alpha_{k})}{|n|}\leq \frac{4M}{|n|}$$
wihci is $O(1/|n|)$ uniformly in $N$. 


*Finally show $f$ can be approximated by this type of function due to Riemann integrability and monotonicity.

A: The typical argument ends up reproving the integration-by-parts formula in some limited form. For the Riemann-Stieljes integral, $\int_{a}^{b}f\,dg$ exists iff $\int_{a}^{b}g\,df$ exists, and, in that case,
$$
              \left.\int_{a}^{b}f\,dg = fg\right|_{a}^{b}-\int_{a}^{b}g\,df.
$$
These integrals exist if $f$ is monotone and $g$ is continuous on $[a,b]$, which applies to this case. Hence,
$$
\begin{align}
     \int_{-\pi}^{\pi}e^{-inx}f(x)\,dx & = \int_{-\pi}^{\pi}f(x)\frac{d}{dx}\frac{e^{-inx}}{-in}\,dx \\
   & = \frac{1}{-in}\int_{-\pi}^{\pi}f(x)d(e^{-inx})\\
   & = \left.\frac{e^{-inx}}{-in}f(x)\right|_{x=-\pi}^{\pi}-\int_{-\pi}^{\pi}\frac{e^{-inx}}{-in}df(x) \\
   & = -\frac{f(\pi)-f(-\pi)}{in}+\frac{1}{in}\int_{-\pi}^{\pi}e^{-inx}df(x).
\end{align}
$$
Because $f$ is non-decreasing, '$df$' is non-negative and $|e^{-inx}|=1$ gives
$$
      |\hat{f}(n)| \le \frac{f(\pi)-f(-\pi)}{n}+\frac{1}{n}\int_{-\pi}^{\pi}df(x) \\
                     = \frac{2(f(\pi)-f(-\pi))}{n}.
$$
A: Let say $$a_n=\int_{-\pi}^\pi f(x)e^{inx}dx$$
and we want to prove that $a_n\in O(\frac{1}{n})$.
Consider 
$$na_n=\int_{-\pi}^\pi f(x)e^{inx}ndx=\int_{-n\pi}^{n\pi}f(\frac{x}{n})e^{ix}dx$$
$$\operatorname{Im}(na_n)=\int_{-n\pi}^{(-n+1)\pi}f(\frac{x}{n})\sin xdx+\int_{(-n+1)\pi}^{(-n+2)\pi}f(\frac{x}{n})\sin xdx+\dots+\int_{(n-1)\pi}^{n\pi}f(\frac{x}{n})\sin xdx$$
$$=f(\frac{x_1}{n})\int_{-n\pi}^{(-n+1)\pi}\sin xdx+f(\frac{x_2}{n})\int_{(-n+1)\pi}^{(-n+2)\pi}\sin xdx+\dots+f(\frac{x_{2n}}{n})\int_{(n-1)\pi}^{n\pi}\sin xdx$$
$$=(-1)^{n}\left[f(\frac{x_1}{n})-f(\frac{x_2}{n})+f(\frac{x_3}{n})-\dots-f(\frac{x_{2n}}{n})\right]$$
since $f$ is monotonically increasing we have
$$f(-\pi)-f(\pi)\le f(\frac{x_1}{n})-f(\frac{x_{2n}}{n})\le f(\frac{x_1}{n})-f(\frac{x_2}{n})+f(\frac{x_3}{n})-\dots-f(\frac{x_{2n}}{n})\le 0$$
therefore
$$|\operatorname{Im(na_n)}|\le f(\pi)-f(-\pi)$$
Similarly we can prove that $|\operatorname{Re}(na_n)|$ is bounded. Therefore $|na_n|$ is bounded, or $a_n\in O(\frac{1}{n})$
