Prove there exist constant limiting integral Let $f:[0,2\pi] \to \mathbb{R}$ be lipschitz. Prove there exist contant $C>0$ such that for every $k=1,2,..$ we have : 
$\displaystyle \int_0^{2\pi} f(x)\sin(kx)dx \le \frac{C}{k}$
I tried to take $g(x)=-\frac{1}{k}\cos(kx)$ 

$$\int_{0}^{2\pi}f(x)\sin(kx)\,dx = \sum_{j=0}^{k-1}\int_{\frac{2\pi j}{k}}^{\frac{2\pi(j+1)}{k}}f(x)\sin(kx)\,dx = \frac{1}{k}\sum_{j=0}^{k-1}\int_{2\pi j}^{2\pi(j+1)}f(x/k)\sin(x)\,dx$$
but:
$$\begin{eqnarray*} \int_{2\pi j}^{2\pi(j+1)}f(x/k)\sin(x)\,dx &=& \int_{0}^{2\pi}f\left(\frac{x+2\pi j}{k}\right)\sin(x)\,dx\\&=&\int_{0}^{\pi}\left[f\left(\frac{x+2j\pi}{k}\right)-f\left(\frac{x+(2j+1)\pi}{k}\right)\right] \sin x\,dx\end{eqnarray*}$$
so by assuming that the Lipshitz constant of $f$ is $M$ and by taking absolute values we get:
$$\left|\int_{0}^{2\pi}f(x)\sin(kx)\,dx\right|\leq \frac{M\pi}{k^2}\sum_{j=0}^{k-1}\int_{0}^{\pi}\sin x\,dx = \color{red}{\frac{2\pi M}{k}}. $$
