Inequality $\left| \int_{-\pi}^{\pi} \varphi(t) \sin(n t) \operatorname{d} t \right| \leq \frac{2 \pi}{n} \max_{x \in [-\pi, \pi]} |\varphi(x)|$ I have a monotonic increasing function $\varphi$ which is continious and not negative with domain $[-\pi,\pi]$. I want to show that for all $n \in \mathbb N$:
$$ \left| \int_{-\pi}^{\pi} \varphi(t) \sin(n t) \operatorname{d} t \right| \leq \frac{2 \pi}{n} M$$
Where $|\varphi(x)| \leq M < \infty$ for all $x \in [-\pi, \pi]$.
I tried many things, but I can't figure out how to prove it... Here's what I've tried.
1. Try: Using the mean value theorem for integration we obtain
$$ \left| \int_{-\pi}^{\pi} \varphi(t) \sin(n t) \operatorname{d} t \right| = 2 \pi  |\varphi(\xi) \sin(n \xi) |  \leq 2 \pi \max_{x \in [-\pi, \pi]} \varphi(x)$$ 
for some $\xi \in [-\pi,\pi]$, but this is too big. I don't know how the get the $\frac{1}{n}$ there...
2. Try: Since 
$$ \left| \int_{-\pi}^{\pi} \varphi(t) \sin(n t) \operatorname{d} t \right| \leq  \int_{-\pi}^{\pi} \left| \varphi(t) \sin(n t) \right| \operatorname{d} t $$ and $\sin(nt) = 0$ if $t = \frac{k\pi}{n}$, $k = 0, \dots, n$ in $[0,\pi]$ one can split the right hand side up in integrals from $k\pi/n$ to $(k+1)\pi/n$, but this makes it even bigger...
I also tried partial integration but I didn't get anything that I found was of interest... The problem with substitution is that sine is not always positive in $-\pi$ and $\pi$. Any ideas? Thank you!
 A: 
The key is to figure out how to translate the hypothesis that $\varphi$ is monotonous.

The function $\varphi$ is nondecreasing and continuous hence there exists some nonnegative integrable function $g$ such that, for every $t$ in $(-\pi,\pi)$, $$\varphi(t)=\varphi(-\pi)+\int_{-\pi}^tg(s)\mathrm ds.$$ 
The integral $I_n$ to be evaluated is
$$
I_n=\int_{-\pi}^{\pi} \varphi(t) \sin(n t) \operatorname{d} t=\varphi(-\pi)\int_{-\pi}^\pi \sin(n t) \operatorname{d} t+\int_{-\pi}^{\pi} g(s)\int_s^\pi \sin(n t) \operatorname{d} t \operatorname{d} s,$$ that is, $$
I_n=\int_{-\pi}^{\pi} g(s)\frac{\cos(n s)-\cos(n\pi)}n \operatorname{d} s.$$ For every $s$, $|\cos(n s)-\cos(n\pi)|\leqslant2$ and $g(s)\geqslant0$ hence $$|I_n|\leqslant\frac2n\int_{-\pi}^{\pi} g(s) \operatorname{d} s=\frac2n(\varphi(\pi)-\varphi(-\pi))\leqslant\frac2n\varphi(\pi)=\frac2n\max \varphi=\frac2n\|\varphi\|_\infty,$$ where we used the hypothesis that $\varphi$ is nonnegative, hence $\varphi(-\pi)\geqslant0$ and $\varphi(\pi)=\max \varphi=\|\varphi\|_\infty$.
One sees that the factor $2\pi$ in the statement of the problem can be replaced by $2$. Conversely, the factor $2$ is optimal.
A: Hint: let $t'=nt$, your integral becomes
$$\left|\frac1n \int_{-n\pi}^{n\pi} \varphi(t'/n) \sin t' dt'\right| \leq
\frac1n \max_{[-\pi,\pi]} \varphi \left|\int_{-n\pi}^{n\pi} \sin t'd t'\right|  $$
and compute the integral...
