Consider the sequence of functions defined as $ g_n =\int_{0}^{1} x^n f(x)dx$. Show that $ \frac {g_{n+1}}{g_n} \ge \frac {g_1}{g_0}$

Question

The following problem was given by my friend as a challenge, but unfortunately I was not able to solve.

Let $f$ be a continuous and non negative real valued function defined on $[0,1]$. Consider the sequence of functions defined as $ g_n =\int_{0}^{1} x^n f(x)dx$. Show that $ \frac {g_{n+1}}{g_n} \ge \frac {g_1}{g_0}$.

I think this problem is hard (well, at least for me) and i am sorry that i am not getting a way to solve this. Any hints/ideas?

Answer 1

Put $D=[0,1]^2$. We have:

$$g_{n+1}g_0-g_ng_1=\int\int_D x^{n}(x-t)f(x)f(t)dxdt=\int\int_D t^{n}(t-x)f(x)f(t)dxdt$$

Hence adding the two integrals: $$2(g_{n+1}g_0-g_ng_1)=\int\int_D (x^{n}-t^n)(x-t)f(x)f(t)dxdt$$

And as $(x^n-t^n)(x-t)\geq 0$ for all $x,t$, we get $g_{n+1}g_0-g_1g_n\geq 0$.