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

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

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TopNewestThe 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 \(I\geq 0. \hspace{15pt}\blacksquare\) – Abhishek Sinha · 2 years, 10 months ago

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– Pebrudal Zanu · 2 years, 10 months ago

\(f(x)\) not differentiable.Log in to reply

this note. – Abhishek Sinha · 2 years, 10 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 ofLog in to reply

– Archiet Dev · 2 years, 10 months ago

Hey are u Indian?Log in to reply

– Pebrudal Zanu · 2 years, 10 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 :))Log in to reply