Evaluating a limit involving integral Let $f: [0,\infty)\to [0,\infty)$ be bounded and continuous. Find if exists :
$$\displaystyle \lim_{n\to \infty}n\left(\sqrt[n]{\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x} \;-\;\sqrt[n]{\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x}\right)$$
Note : $f^n(x)=(f(x))^n$
This problem was given in a facebook group and everyone claimed it was right. I have made zero progress so far. Can someone solve it? Thanks a lot. 
 A: Notice that
$$ n\left(\sqrt[n]{\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x} \;-\;\sqrt[n]{\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x}\right) = n \left[ \exp \left( \frac1{n}\ln\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x\right)- \exp\left( \frac1{n}\ln\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x\right) \right]$$
Using the mean value theorem
$$ n\left(\sqrt[n]{\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x} \;-\;\sqrt[n]{\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x}\right) = e^{\xi_n}\ln \left(\frac{\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x}{\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x}\right)$$
where $\xi_n$ is some number between the two arguments of the $\exp$ function in the first set of equations.
Because $f$ is bounded, the integrals converge
$$\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x \leq (\sup f(x))^n\int_{0}^{\infty}e^{-x}\,\mathrm{d}x=(\sup f(x))^n$$
Also
$$\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x \leq \sup f(x)\int_{0}^{\infty}f^n(x)e^{-x}\,\mathrm{d}x$$
and
$$\frac{\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x}{\int_{0}^{\infty}f^n(x)e^{-x}\,\mathrm{d}x}  \leq \sup f(x)$$
which provides an upper bound for the limit of the LHS.
To get a lower bound first change variables to $u=e^{-x}$ and apply Holder's inequality to obtain
$$\int_{0}^{1}f^{n}(-\ln u)\,\mathrm{d}u \leq \left(\int_{0}^{1}f^{n+1}(-\ln u)\,\mathrm{d}u\right)^{n/(n+1)}\left(\int_{0}^{1}\,\mathrm{d}u\right)^{1/(n+1)}=\left(\int_{0}^{1}f^{n+1}(-\ln u)\,\mathrm{d}u\right)^{n/(n+1)}$$
Rearranging and changing variables back to x we find
$$\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x \leq \left(\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x\right)\left(\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x\right)^{-1/(n+1)}$$
and
$$\left(\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x\right)^{1/(n+1)} \leq \frac{\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x}{\int_{0}^{\infty}f^{n}(x)e^{-x}\,\mathrm{d}x}$$
It can be shown that the limit of the LHS is $\sup f(x)$ -- providing a lower bound for the limit of the RHS, that is identical to the upper bound.
Therefore,
$$ \lim_{n \rightarrow \infty}\frac{\int_{0}^{\infty}f^{n+1}(x)e^{-x}\,\mathrm{d}x}{\int_{0}^{\infty}f^n(x)e^{-x}\,\mathrm{d}x} = \sup f(x)$$
We can also show $e^{\xi_n}$ converges to $\sup f(x)$ and the limit of the original expression is $\sup f(x) \ln(\sup f(x)).$
