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Let $f,g$ be two continuous, distinct functions from $[0,1]$ to $(0, +\infty)$ such that $\int_{0}^{1} g = \int_{0}^{1} f $.
Given $n\in \mathbb{N},$ let $y_n = \int_{0}^{1} \frac{f^{(n+1)}}{g^{(n)}} $ How do I show that $(y_n)$ is increasing and divergent?

Appreciate all advice. Thank you.

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    $\begingroup$ Please credit your sources. This is from the 2005 Brazil Undergrad Math Olympiad, problem 2. I added the contest-math tag. $\endgroup$ – Nate Eldredge Oct 5 '13 at 14:38
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    $\begingroup$ Are those powers or derivatives? $\endgroup$ – Javier Oct 5 '13 at 14:39
  • $\begingroup$ Thanks for reminder. I got this question from a facebook group. $\endgroup$ – Alexy Vincenzo Oct 5 '13 at 14:41
  • $\begingroup$ Yes thanks, just saw that. $\endgroup$ – Sean Eberhard Oct 5 '13 at 14:43
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Let $A = \{ x \in [0,1] : f(x) > g(x)\}$, $B = \{ x \in [0,1] : f(x) < g(x)\}$ and $C = [0,1] \setminus (A\cup B)$. Then since $f \neq g$ and $\int_0^1 f(x)\,dx = \int_0^1 g(x)\,dx$, we have

$$\int_A f\,dx + \int_B f\,dx = \int_A g\,dx + \int_B g\,dx \iff \int_A (f-g)\,dx = \int_B (g-f)\,dx,$$

and $\int_A (f-g)\,dx > 0$. Now,

$$\begin{align} y_{n} - y_{n-1} &= \int_0^1 \frac{f^n(f-g)}{g^n}\,dx\\ &= \underbrace{\int_A \left(\frac{f}{g}\right)^n(f-g)\,dx}_{a_n} - \underbrace{\int_B \left(\frac{f}{g}\right)^n(g-f)\, dx}_{b_n}. \end{align}$$

On $A$, we have $\frac{f}{g} > 1$, and on $B$, we have $\frac{f}{g} < 1$, so $a_n > a_0 = b_0 > b_n$ for $n > 0$. Hence $y_n > y_{n-1}$. By the monotone convergence theorem. $\lim\limits_{n\to \infty} b_n = 0$ and $\lim\limits_{n\to\infty} a_n = \infty$, hence $\lim\limits_{n\to\infty} (y_n-y_{n-1}) = \infty$.

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