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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

Robert Z
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Jamaico
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1 Answers1

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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?

Robert Z
  • 145,942
  • Thanks @Robert Z, I've managed to get to this myself but my problem is how to use the |f''x| <=1 thing correctly to get this inequality..I've tried but I can't get the 4/n^2 I've tried using Triangle Inequality with Absolute Value but I don't think I'm doing it right. – Jamaico Nov 05 '17 at 13:27
  • @Jamaico Note that the remaining integral is $2n\int_0^{\pi/n} \sin(nt)dt$. – Robert Z Nov 05 '17 at 13:34
  • Thank you very much @Robert Z – Jamaico Nov 06 '17 at 12:50