# Real Analysis

Given function $$f(x):\mathbb{R}\rightarrow \mathbb{R}$$ such that $$\gamma \in[0,1]$$ where \begin{align} f(\gamma x+(1-\gamma)y) \leq \gamma f(x)+(1-\gamma)f(y) \end{align}

for all $$x,y \in \mathbb{R}$$. Prove that

$$\displaystyle \int_{0}^{2\pi} f(x) \cos{x} \mathrm{d}x \geq 0$$

Note by Pebrudal Zanu
3 years, 9 months ago

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The given condition says that $$f$$ is convex, i.e., its derivative (assuming it to be differentiable) is non-decreasing. Integrating by parts, we have $$I=\int_{0}^{2\pi} f(x) \cos x dx = -\int_{0}^{2\pi}f'(x)\sin(x) dx = -\int_{0}^{\pi}f'(x)\sin(x) dx -\int_{\pi}^{2\pi}f'(x)\sin(x) dx$$ . Substituting $$z=x-\pi$$ in the second integral, we have $$I=-\int_{0}^{\pi}f'(x)\sin(x) dx + \int_{0}^{\pi}f'(z+\pi)\sin(z) dz$$ = $$\int_{0}^{\pi}\big(f'(x+\pi)-f'(x)\big)\sin(x) dx$$. Now since the derivative is non-decreasing, we have $$f'(x+\pi)\geq f'(x), \forall x\in \mathbb{R}$$. Also the sine function is non-negative in the interval of integration. Hence the integrand is non-negative. Hence $$I\geq 0. \hspace{15pt}\blacksquare$$

- 3 years, 9 months ago

$$f(x)$$ not differentiable.

- 3 years, 9 months ago

The proof is generalizable, because there are only a countable number points of non-differentiability (hence of Lebesgue measure zero) for a convex function $$f$$ in the open interval $$(0,2\pi)$$. See corollary 6.3 of this note.

- 3 years, 9 months ago

Hey are u Indian?

- 3 years, 9 months ago

Thank you.. I don't know about that the collorally... First, I use your step for my solution, but I don't know corollary 6.3... so, I say it's not generality. But, now I was understand, your solution :))

- 3 years, 9 months ago