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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}.$

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    Thanks for hints. Actually I suppose that I mistaked and inequality holds but with RHS equal $2\frac{f(-\pi)-f(\pi)}{n}$ instead $\frac{f(-\pi)-f(\pi)}{n}$. It follows also with the second mean theorem for integrals. But your hints are interesting. I check that inequality holds when $f$ is of $C^1$, but don't know how to use hints 2. How to approximate $f$ by decreasing functions of class $C^1$ ?2011-12-05

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I apologize; I missed a factor of $2$ in the integration by parts argument. The optimal bound is, indeed, $2 (f(-\pi) -f(\pi))/n$. Take $n=1$ and let $f(x)$ be $1$ for $- \pi \leq x < 0$ and $0 \leq x \leq \pi$. Then $\int f(x) \sin x dx = -2 = 2 (f(-1) - f(1))$.

As Thomas points out below, for all $n$ odd, this same $f$ gives $\int f(x) \sin(nx) dx = 2 (f(-\pi) -f(\pi))/n$.

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    Sorry, hope that everything is right now.2011-12-06