# Let $f,g$ be two distinct functions from $[0,1]$ to $(0, +\infty)$ such that $\int_{0}^{1} g = \int_{0}^{1} f$.

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?

• Please credit your sources. This is from the 2005 Brazil Undergrad Math Olympiad, problem 2. I added the contest-math tag. – Nate Eldredge Oct 5 '13 at 14:38
• Are those powers or derivatives? – Javier Oct 5 '13 at 14:39
• Thanks for reminder. I got this question from a facebook group. – Alexy Vincenzo Oct 5 '13 at 14:41
• Yes thanks, just saw that. – Sean Eberhard Oct 5 '13 at 14:43

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 $\lim\limits_{n\to\infty} (y_n-y_{n-1}) = \infty$.