Given $f$ such that $f''$ is continious and differentiable on $[0,2\pi]$, and that $f(0)=f(2\pi)$ and $ |f''(x)|\leq1$ for all $x\in[0,2\pi]$ I need to show

$$\left|\int_{0}^{2\pi}f(t)\sin nt\text{d}t\right|\leq\frac{4}{n^2},\ \forall n$$

I started off with integration by parts twice and got that this integral is

$-\underbrace{\frac{1}{n}\left(f(2\pi)\cos2\pi n-f(0)\cos 0\right)}_{\text{=0}}+\frac{1}{n^{2}}\left(f'(2\pi)\underbrace{\sin2\pi n}_{\text{=0}}-f'(0)\underbrace{\sin0}_{\text{=0}}\right)-\intop_{0}^{2\pi}f''(t)\cdot\frac{\sin nt}{n^{2}}\text{d}t$

I checked I have no mistakes for like an hour or so by now. However, this doesn't seem to work out now because

$$\left|\intop_{0}^{2\pi}f''(t)\cdot\frac{\sin nt}{n^{2}}\text{d}t\right|\leq\frac{1}{n^{2}}\left|\intop_{0}^{2\pi}1\cdot\sin nt\text{d}t\right|=0$$ which is problematic.

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    $\begingroup$ Move the absolute value inside: $$\left|\int_0^{2\pi}f''(t)\cdot \frac{\sin nt}{n^2}\ dt\right| \leq \frac{1}{n^2}\int_0^{2\pi}|\sin nt|\ dt$$ $\endgroup$ – Bungo Oct 5 '17 at 20:49
  • $\begingroup$ Is that not troubling though that I got that this integral must be 0? $\endgroup$ – Theorem Oct 5 '17 at 20:53
  • $\begingroup$ And with your method, am I supposed to split $[0,2\pi]$ into $2n$ equal pieces, claiming the integral over each part is $\frac{2}{n}$? $\endgroup$ – Theorem Oct 5 '17 at 20:55
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    $\begingroup$ It's not troubling that $\int_0^{2\pi} 1 \cdot \sin nt\ dt = 0$, but your inequality as written is wrong. $\endgroup$ – Bungo Oct 5 '17 at 20:55
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    $\begingroup$ Yes, that's correct, there are exactly $2n$ periods of $|\sin nt|$ in $[0,2\pi]$, so it suffices to integrate one of the periods and multiply the result by $2n$. $\endgroup$ – Bungo Oct 5 '17 at 20:57

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