A hypothesis: Integral of $f^n(x)$ is monotonic as $n$ reaches infinity. When I'm solving some analysis questions I discovered some interesting phenomenons. I can't prove it or give any counterexamples so I'm seeking help.
If $f(x)$ is a continuous, monotonic, non-negative function on $[a,b]$, then sequence
$$\left\{\int_a^bf^n(x)\mathrm{d}x\right\}_{n=1}^{\infty}$$
is monotonic, or at least starting from some $N\in\mathbb{N}$ it is.
If someone can prove it or give a counterexample. Thanks!
($f^n(x)$ means $[f(x)]^n$.)
 A: I think your hypothesis is correct.
First if $\max\limits_{x\in[a,b]}f(x)\leqslant1$, then $f^{n}\geqslant f^{n+1}$ hence $\displaystyle\int_{a}^{b}f^{n}(x)\,dx\geqslant\int_{a}^{b}f^{n+1}(x)\,dx$.
Now we assume $\max\limits_{x\in[a,b]}f(x)=M>1$, let $a_{n}=\displaystyle\int_{a}^{b}f^{n}(x)\,dx$. So we have $\lim\limits_{n\rightarrow\infty}\dfrac{a_{n+1}}{a_{n}}=M>1$, which means $a_{n}$ is eventually monotonic.
By Cauchy-Schwarz,$$a_{n}=\int_{a}^{b}f^{n}(x)\,dx=\int_{a}^{b}f^{\frac{n-1}{2}}(x)f^{\frac{n+1}{2}}(x)\,dx\leqslant\left( \int_{a}^{b}f^{n-1}(x)\,dx\right) ^{\frac{1}{2}}\left( \int_{a}^{b}f^{n+1}(x)\,dx\right) ^{\frac{1}{2}}=\sqrt{a_{n-1}a_{n+1}}.$$Thus $\dfrac{a_{n+1}}{a_{n}}$ is increasing and $$\dfrac{a_{n+1}}{a_{n}}=\dfrac{\int_{a}^{b}f^{n+1}(x)\,dx}{\int_{a}^{b}f^{n}(x)\,dx}\leqslant M\dfrac{\int_{a}^{b}f^{n}(x)\,dx}{\int_{a}^{b}f^{n}(x)\,dx}=M.$$So the limit exists.
$\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{a_{n}}=M$ is because$$\lim_{n\rightarrow\infty}a_{n}^{\frac{1}{n}}=M=\lim_{n\rightarrow\infty}e^{\dfrac{\log\frac{a_{2}}{a_{1}}+\ldots+\log\frac{a_{n}}{a_{n-1}}}{n}}\cdot e^{\dfrac{\log a_{1}}{n}}=e^{\lim_{n\rightarrow\infty}\log\frac{a_{n}}{a_{n-1}}}=\lim\limits_{n\rightarrow\infty}\dfrac{a_{n+1}}{a_{n}}.$$
A: The monotonicity is not essential, but may help writing the answer. Let $f(b)>1>f(a).$ For $c$ such that  $f(c)=1$ we have $$a_{n+1}-a_n=-\int\limits_a^c[1-f(x)][f(x)]^n\,dx+\int\limits_c^b[f(x)-1][f(x)]^n\,dx\to \infty$$ as the first term tends to $0$ and the second one to $\infty.$
