Assume f(x) is convex in $[-\pi,\pi ]$, $f^{'}(x)$ is bounded, Prove:
$$\frac{1}{\pi}\int^{\pi}_{-\pi}f(x) \cos 2nx \geq 0$$
Assume f(x) is convex in $[-\pi,\pi ]$, $f^{'}(x)$ is bounded, Prove:
$$\frac{1}{\pi}\int^{\pi}_{-\pi}f(x) \cos 2nx \geq 0$$
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