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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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    Why no $dx$ in integration?2012-03-16

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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