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Let $f,g:[0,1] \rightarrow (0,\infty)$ be two distinct, continuous functions such that $$\int_0^1 f(x)dx=\int_0^1 g(x)dx$$ Denote $$y_n=\int_0^1{\frac{f^{n+1}(x)}{g^n(x)}}dx$$ for all integer $n \geq 0$. Prove that $(y_n)_{n \geq 1}$ is an increasing and divergent sequence.

Here is the source of the problem.

My attempt:

Note that $$y_n -y_{n-1}=\int_0^1{\frac{f^{n+1}(x)}{g^n(x)}}dx-{\frac{f^{n}(x)}{g^{n-1}(x)}}dx = \int_0^1{\frac{f^{n+1}(x)-f^n(x)g(x)}{g^n(x)}}dx = \int_0^1{\frac{f^n(x)}{g^n(x)}(f(x)-g(x))}dx \geq A\int_0^1(f(x)-g(x)dx=0$$

where $A>0$.

Hence, $(y_n)_{n \geq 1}$ is increasing. However, I have trouble to show the sequence diverges. Can anyone give some hint?

Idonknow
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  • This may be obvious, but what is the reason for the last inequality? – Clement C. May 20 '15 at 14:06
  • http://math.stackexchange.com/questions/515584/let-f-g-be-two-distinct-functions-from-0-1-to-0-infty-such-that-i?rq=1 (contains spoiler). I've read it and a hint I can give you is to consider the integral as the sum of integrals $\int_A+\int_B$, where $A:={x\in [0,1] : f(x)>g(x)},B:={x\in [0,1]: g(x)>f(x)}$. – Alberto Debernardi May 20 '15 at 14:11

1 Answers1

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There is an $c \in [0,1]$ such that $f(c)>g(c)$. Using the continuity we get that there is $\lambda >1$ and $[\alpha , \beta ] \subset [0,1]$ such that $\frac{f(x)}{g(x)} \ge \lambda $ on $[ \alpha , \beta ]$ . Then $y_n \ge \int_ {\alpha }^{\beta }f(x) \lambda ^n dx$, and so $\lim_{n \to \infty }y_n= \infty$.

npatrat
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