Let $f$ and $g$ be two continuous, distinct functions from $[0,1] \rightarrow (0,+\infty)$ such that $\int_{0}^{1}f(x)dx = \int_{0}^{1}g(x)dx$. Let $y_n=\int_{0}^{1}{\frac{f^{n+1}(x)}{g^{n}(x)}dx}$ for $n\in\mathbb{N}$.
Prove that $(y_n)$ is an increasing and divergent sequence.
Here's a solution one of my colleagues proposed.
It is easy to see that there is some $\alpha\in [0,1]$ such that $f(\alpha)>g(\alpha)$.
Let us consider the function $\psi : [0,1]\to\mathbb{R_{+}}^{*}$, with $\psi (x)=\frac{f(x)}{g(x)}.$ Since $\psi(\alpha) >1$, from $\psi$ being continuous we obtain the existence of an interval $[a,b]$ in $[0,1]$ such that $\psi(x) >1$ for any $x\in [a,b].$
So there is some $\lambda >1$ satisfying the following inequality:
$ x_{n}=\int_{0}^{a} f(x) \psi ^{n} (x) dx +\int_{a}^{b} f(x) \psi ^{n} (x) dx+\int_{b}^{1} f(x) \psi ^{n} (x) dx\geq \lambda^{n}\int_{a}^{b} f(x)dx $
Also, from the mean value theorem for integrals we have some $c\in (a,b)$ such that $\int_{a}^{b} f(x)dx=(b-a)f(c) >0$
So, the given sequence diverges.
Could someone please help me figure out how exactly the above inequality has been established, I'm not able to understand that particular part. A rather detailed solution, while mentioning the important properties used at each step would be of great help.
Also, is there any other method to solve this problem? I've been thinking of induction, but haven't been able to show something like the difference of consecutive terms is of the same sign, that is positive.
Thanks a lot.
Since $f$ and $\psi$ are positive in $[0,1]$ then $$\int_{0}^{a} f(x) \psi ^{n} (x) dx\geq 0\quad\text{and}\quad \int_{b}^{1} f(x) \psi ^{n} (x) dx\geq 0.$$
Moreover $\psi(x)\geq \lambda:=\min_{t\in [a,b]}\psi(t)>1$ in $[a,b]$ (and $f>0$) implies $$\int_{a}^{b} f(x) \psi ^{n} (x) dx\geq \lambda^{n}\int_{a}^{b} f(x)dx.$$