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Assume f(x) is convex in $[-\pi,\pi ]$, $f^{'}(x)$ is bounded, Prove:

$$\frac{1}{\pi}\int^{\pi}_{-\pi}f(x) \cos 2nx \geq 0$$

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What is the source of the problems you have been posting? Are these homework or are you doing self study and plugging through some problem book? – Aryabhata Mar 16 '12 at 2:39
@Aryabhata It's truly some questions in a problem book. – 89085731 Mar 16 '12 at 2:41
Then please mention the source (including the problem number). btw, did you modify the problem statement? – Aryabhata Mar 16 '12 at 2:42
Also, this question is very similar: – Aryabhata Mar 16 '12 at 2:42
Why no $dx$ in integration? – FiniteA Mar 16 '12 at 8:49
up vote 1 down vote accepted

I think I've made an answer. Please correct it if anything is wrong. $$\int^{\pi}_{-\pi}f(x) \cos 2nx=-\frac{1}{2n}\int_{-\pi}^{\pi}f^{'}(x)sin2nx$$,then I will prove$\int_{-\pi}^{\pi}f^{'}(x)sin2nx \leq0$. Divide the interval into $4n$ parts.So the integral becomes

$$\Sigma \int_{-\pi+\frac{k\pi}{2n}}^{-\pi+\frac{k+1\pi}{2n}}f^{'}(x)sin2nx$$,in every interval,$sin2nx$ doesn't change the sign,so use Integral Mean Theorem,and it becomes $$\Sigma f^{'}(\xi_i)(-1)^{i-1}$$

With f(x) is convex,so $f^{'}(x)$ is increasing.Thus $f^{'}(\xi_i)-f^{'}(\xi_{i+1})\leq0$.

Then we can complete the proof

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