I have a function $f(x)$ in $[0,2\pi]$ for which $f(0) = f(2\pi)$ and for which $|f''(x)| \le 1$. Show that $$\left|\int_0^{2\pi} f(t)\sin(nt)dt\right| \le \frac{4}{n^2}$$ for every natural $n$.
How do I prove this? I manage to get to the integral using the substitution method twice but I can't get the inequality.
Thanks
For any positive integer $n$, by integration by parts, \begin{align*}\int_0^{2\pi} f(t)\sin(nt)dt&=\left[\frac{-f(t)\cos(nt)}{n}\right]_0^{2\pi}+\frac{1}{n}\int_0^{2\pi} f'(t)\cos(nt)dt\\&=\left[\frac{f'(t)\sin(nt)}{n^2}\right]_0^{2\pi}-\frac{1}{n^2}\int_0^{2\pi} f''(t)\sin(nt)dt\\ &=-\frac{1}{n^2}\int_0^{2\pi} f''(t)\sin(nt)dt. \end{align*} Hence, by $|f''(x)|\leq 1$, it follows $$\left|\int_0^{2\pi} f(t)\sin(nt)dt\right| \leq \frac{1}{n^2}\int_0^{2\pi} |\sin(nt)|dt.$$ Can you take it from here?