Let $f: [-\pi, \pi] \rightarrow \mathbb{R}$ be nonincreasing. Is it true that $ \left| \int_{-\pi}^\pi f(x) \sin (nx) dx \right| \leq \frac{f(-\pi)-f(\pi)}{n}.$ (Please, without Stieltjes integrals.)
I obtain something similar by using the second mean value theorem for integrals but with right hand side equal $2 \frac{f(-\pi)-f(\pi)}{n}$.
Thanks.
Added.
Sorry, I mistaked and this inequality is generally not true.
It holds the following inequality
$ \left| \int_{-\pi}^\pi f(x) \sin (nx) dx \right| \leq 2 \frac{f(-\pi)-f(\pi)}{n}.$
The proof goes in the followig way. By the second mean value theorem there exists a $c\in [-\pi,\pi]$ such that $\int_{-\pi}^\pi f(x) \sin nxdx=f(-\pi)\int_{-\pi}^c \sin nx dx+f(\pi) \int_c^\pi \sin nx dx$ $=-\frac{f(-\pi)}{n} (\cos nc-\cos n\pi)-\frac{f(\pi)}{n} (\cos n\pi-\cos nc)$ $=\frac{f(-\pi)-f(\pi)}{n} (\cos n\pi-\cos nc).$ Since $|(\cos n\pi-\cos nc)| \leq 2$ we obtain $ \left| \int_{-\pi}^\pi f(x) \sin (nx) dx \right| \leq 2 \frac{f(-\pi)-f(\pi)}{n}.$