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∈Rx,y \in \mathbb{R}. Prove that

∫02πf(x)cos⁡xdx≥0\displaystyle \int_{0}^{2\pi} f(x) \cos{x} \mathrm{d}x \geq 0

#Calculus

Note by Pebrudal Zanu
6 years, 9 months ago

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Comments

The given condition says that ff is convex, i.e., its derivative (assuming it to be differentiable) is non-decreasing. Integrating by parts, we have I=∫02πf(x)cos⁡xdx=−∫02πf′(x)sin⁡(x)dx=−∫0πf′(x)sin⁡(x)dx−∫π2πf′(x)sin⁡(x)dxI=\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−πz=x-\pi in the second integral, we have I=−∫0πf′(x)sin⁡(x)dx+∫0πf′(z+π)sin⁡(z)dzI=-\int_{0}^{\pi}f'(x)\sin(x) dx + \int_{0}^{\pi}f'(z+\pi)\sin(z) dz = ∫0π(f′(x+π)−f′(x))sin⁡(x)dx\int_{0}^{\pi}\big(f'(x+\pi)-f'(x)\big)\sin(x) dx . Now since the derivative is non-decreasing, we have f′(x+π)≥f′(x),∀x∈Rf'(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≥0.■I\geq 0. \hspace{15pt}\blacksquare

Abhishek Sinha - 6 years, 9 months ago

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f(x)f(x) not differentiable.

pebrudal zanu - 6 years, 9 months ago

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

Abhishek Sinha - 6 years, 9 months ago

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@Abhishek Sinha – Hey are u Indian?

Archiet Dev - 6 years, 8 months ago

@Abhishek Sinha – 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 :))

pebrudal zanu - 6 years, 9 months ago
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