Romania NMO 2023 Grade 12 P4

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Romania NMO 2023 Grade 12 P4

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monotone functions

differentiable function

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Source: Romania National Olympiad 2023

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196 posts

#1 h Apr 14, 2023, 2:44 AM

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Let $f:[0,1] \rightarrow \mathbb{R}$ a non-decreasing function, $f \in C^1,$ for which $f(0) = 0.$ Let $g:[0,1] \rightarrow \mathbb{R}$ a function defined by

$g(x) = f(x) + (x - 1) f'(x), \forall x \in [0,1].$
a) Show that

$\int_{0}^{1} g(x) \text{dx} = 0.$
b) Prove that for all functions $\phi :[0,1] \rightarrow [0,1],$ convex and differentiable with $\phi(0) = 0$ and $\phi(1) = 1,$ the inequality holds

$\int_{0}^{1} g( \phi(t)) \text{dt} \leq 0.$

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#2 h Apr 14, 2023, 12:47 PM

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I will provide a solution only for b), as a) is trivial.

We want to prove that: $\int_{0}^{1} f(\phi(x)) \text{dx} + \int_{0}^{1} \phi(x)f'(\phi(x)) \text{dx} \leq \int_{0}^{1} f'(\phi(x)) \text{dx}$
It s easier to work with $f'$ than with $f$ , as $f'$ is a function that is just nonnegative, so we will write the first integral in the L.H.S as:

$\int_{0}^{1} f(\phi(x)) \text{dx} = \int_{0}^{1} x' f(\phi(x)) \text{dx} = f(1) - \int_{0}^{1} xf'(\phi(x)) \phi'(x) \text{dx}=\int_{0}^{1} f'(x) \text{dx} - \int_{0}^{1} xf'(\phi(x)) \phi'(x) \text{dx}=\int_{0}^{1} f'(\phi(x))\phi'(x) \text{dx} - \int_{0}^{1} xf'(\phi(x)) \phi'(x) \text{dx}= \int_{0}^{1} f'(\phi(x))\phi'(x)(1-x) \text{dx}$
where the second equality is from integrating by parts, and the fourth one is from making the change of variables $x=\phi(t)$ in $\int_{0}^{1} f'(x) \text{dx}$

Now it remains to prove that: $\int_{0}^{1} f'(\phi(x)) \phi'(x)(1-x) \text{dx} + \int_{0}^{1} \phi(x)f'(\phi(x)) \text{dx} \leq \int_{0}^{1} f'(\phi(x)) \text{dx}$ , which is equivalent with

$\int_{0}^{1} f'(\phi(x))(\phi'(x)(1-x)+\phi(x)-1) \text{dx} \leq 0$
But, $\phi$ is convex, so we have $\frac{1-\phi(x)}{1-x}=\frac{\phi(1)-\phi(x)}{1-x}\geq \phi'(x) \ \forall x \in [0,1)$ , so $\phi'(x)(1-x)+\phi(x)\leq 1$ , and as $f'(\phi(x))\geq 0 \ \forall x \in [0,1]$ , we have that, $\int_{0}^{1} f'(\phi(x))(\phi'(x)(1-x)+\phi(x)-1) \text{dx} \leq 0$ . $\square$

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#3 h Apr 14, 2023, 11:18 PM

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Here is a solution for a)

Note that $g(x) =\frac{d}{dx} (x-1)f(x)$ . Thus, $\int_0^1 g(x) = [(x-1)f(x)]_0^1 =0$

This post has been edited 1 time. Last edited by CANBANKAN, Apr 14, 2023, 11:44 PM

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