Convergence/Divergence speed of $u_n = \int_0^1 f(x)^n g(x) \textrm{d}x$ given $f, g$ continuous and non-negative Let be $f, g : [0, 1] \to \mathbb{R}_{+}^{*}$ continuous maps such that:
$\forall n \in \mathbb{N}, u_n = \int_0^1 f(x)^n g(x) \textrm{d}x$
I want to show that $v = \left(\dfrac{u_{n + 1}}{u_n}\right)_n$ converges to a finite limit.
What I have done so far, is:
By the second mean value theorem, I have $(c_n)_n \in ]0, 1[^{\mathbb{N}}$ such that:
\begin{equation*}
u_n = f(c_n)^n \int_0^1 g(x) \textrm{dx} = f(c_n)^n u_0
\end{equation*}
Then:
\begin{align*}
\dfrac{u_{n + 1}}{u_n} & = f(c_{n + 1}) \left(\dfrac{f(c_{n + 1})}{f(c_n)}\right)^n
\end{align*}
Now, I'm stuck, because I wish I could know something better on $(c_n)_n$, but all I know is, by Weierstrass: $(c_n)_n$ has at least one accumulation point $c_0$ such that, I know that $v$ has at least one accumulation point, that is: $f(c_0)$.
So, basically, I have to show that the only limit of $v$ is $f(c_0)$ and that will conclude, I have no idea on how to proceed. If I try to consider proof by contradiction, knowing that I have another accumulation point $c_1$ will not help me.
Bonus: I wonder if we can get an asymptotic relation of $(u_n)_n$ like: $u_n \sim f(c_0)^n M$.
 A: Note that $$\frac{u_{n+1}}{u_{n}}=\frac{\int_{0}^{1}f\left(x\right)^{n+1}g\left(x\right)dx}{\int_{0}^{1}f\left(x\right)^{n}g\left(x\right)dx}=f\left(\xi_{n}\right)\frac{\int_{0}^{1}f\left(x\right)^{n}g\left(x\right)dx}{\int_{0}^{1}f\left(x\right)^{n}g\left(x\right)dx}=f\left(\xi_{n}\right)\leq M,\,\xi_{n}\in\left(0,1\right)$$ by the mean value theorem and the extreme value theorem, so the sequence $\frac{u_{n+1}}{u_{n}}$ is bounded. Furthermore, by the Cauchy inequality, we have $$\left(\int_{0}^{1}f\left(x\right)^{n+1}g\left(x\right)dx\right)^{2}=\left(\int_{0}^{1}f\left(x\right)^{n/2+1}g\left(x\right)^{1/2}f\left(x\right)^{n/2}g\left(x\right)^{1/2}dx\right)^{2}$$ $$\leq\left(\int_{0}^{1}f\left(x\right)^{n+2}g\left(x\right)dx\right)\left(\int_{0}^{1}f\left(x\right)^{n}g\left(x\right)dx\right)$$ hence $$\frac{u_{n+1}}{u_{n}}\leq\frac{u_{n+2}}{u_{n+1}}$$ and so the sequence is monotone non-decreasing. This allow to conclude that it has a finite limit, by the monotone convergence theorem.
