Difficult problem in Riemann integrals Could anyone help me with the following problem? Because i have stuck.

Let $f:[a,b]\rightarrow [0,\infty)$ be continuous and not the zero function. Prove that $$\lim_{n\to \infty} \frac{\int\limits_a^b f^{n+1}(x)\, \mathrm{d}x}{\int\limits_a^b f^n(x)\, \mathrm{d}x}=\sup_{x\in[a,b]}f(x)$$

Some of my thoughts:
Let $\displaystyle M=\sup_{x\in[a,b]}f(x)$ and $\displaystyle m=\inf_{x\in[a,b]}f(x)$.
Moreover define the sequences $\{I_n\}_{n=0}^{\infty}$ and $\{a_n\}_{n=0}^{\infty}$,
So, $\displaystyle I_n=\int\limits_a^b f^{n}(x)\, \mathrm{d}x$
and $\displaystyle a_n=\frac{I_{n+1}}{I_n}$.
It easy to prove that $$m(b-a)\leq I_n \leq M(b-a)$$
and $$a_n \leq M\left(\frac{M}{m}\right)^n.$$
First I tried to use the above inequalities to prove it by definition or somehow with the squeeze theorem, but unfortunately $\displaystyle M\left(\frac{M}{m}\right)^n \xrightarrow[ n \to \infty]{} \infty$
Then I thought about the integral mean value theorem, but again nothing new came up.
So do you have any suggestions, thoughts, hints or\and solutions?
 A: Thanks to Motyla Noga Tomka Mazura for part of what's below - this is easier than I've thought for years, because of the monotonicity that he pointed out. (I think this post is about half mine and half his. I'm posting it because I need the points a little more than he does - is there a way I can donate half of them to him?)
Define $M$, $I_n$ and $\alpha_n$ as you did. As @Motyla Noga Tomka Mazura points out, the Cauchy-Schwarz inequality shows that $$I_n\le\left(I_{n-1}I_{n+1}\right)^{1/2},$$ hence the sequence $\alpha_n$ is increasing. Hence $\lim\alpha_n$ exists, and hence it's enough to show that $$\lim_{n\to\infty}I_n^{1/n}=M.$$An this is easier than I thought.
First, it's clear that $f^n\le M^n$, hence $I_n\le(b-a)M^n$, so $$I_n^{1/n}\le(b-a)^{1/n}M\to M.$$ So $\limsup I_n^{1/n}\le M$.
Now let $\epsilon>0$. Since $M=\sup f$ and $f$ is continuous, there exists an open interval $E$ such that $$f(t)>(1-\epsilon)M\quad(t\in E).$$If $L$ is the length of $E$ then $$I_n\ge\int_Ef^n\ge L((1-\epsilon)M)^n,$$ so $$I_n^{1/n}\ge L^{1/n}(1-\epsilon)M\to(1-\epsilon)M.$$
Hence $\liminf I_n^{1/n}\ge(1-\epsilon)M$, and hence $\liminf I_n^{1/n}\ge M$.
A: Use the fact $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n } =g \Longrightarrow \lim_{n\to\infty} \sqrt[n]{a_n } =g$$
A: WLOG, assume the maximum value of $f$ is $1.$ Then $f^{n+1} \le f^n,$ so the ratio of integrals is $\le 1$ for all $n.$ On the other hand, Holder shows
$$\int_a^b f^n \le (\int_a^b f^{n+1})^{n/(n+1)}\cdot(b-a)^{1/(n+1)}$$
by Holder. Use this and simplify to see
$$(1)\,\,\,\,\frac{(\int_a^b f^{n+1})^{1/(n+1)}}{(b-a)^{1/(n+1)}}\le \frac{\int_a^b f^{n+1}}{\int_a^b f^n} \le 1.$$
But as is well known, $\lim_{p\to \infty}(\int_a^b f^p)^{1/p} = \sup_{[a,b]}|f| = 1.$ It follows that the left side of $(1) \to 1,$ and we're done.
A: Let $M=\max_{x\in[a,b]}f(x)$. By assumption, $M>0$. Choose $x_{0}\in[a,b]$
such that $M=f(x_{0})$. By continuity, there exists $\delta>0$ such
that $f(x)>\frac{M}{2}$ for whenever $x\in[x_{0}-\delta,x_{0}+\delta]$
(If $x_{0}=a$ or $x_{0}=b$, we modify the interval in an obvious
way). For each $n\in\mathbb{N}$, Observe that $\int_{a}^{b}f^{n}(x)dx\geq\int_{x_{0}-\delta}^{x_{0}+\delta}(\frac{M}{2})^{2}dx=2\delta(\frac{M}{2})^{n}>0$.
Therefore, $\frac{\int_{a}^{b}f^{n+1}(x)dx}{\int_{a}^{b}f^{n}(x)dx}$
is well-defined.
We go to show that $\frac{\int_{a}^{b}f^{n+1}(x)dx}{\int_{a}^{b}f^{n}(x)dx}\rightarrow M$
as $n\rightarrow\infty$. Firstly, we prove under the assumption that
$M=1$. Clearly, $f^{n+1}(x)\leq f^{n}(x)$, so $\int_{a}^{b}f^{n+1}\leq\int_{a}^{b}f^{n}$.
Therefore, $\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}}\leq1$.
In particular, $\limsup_{n}\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}}\leq1$.
Let $\alpha\in(0,1)$ be arbitrary. Let $A=\{x\in[a,b]\mid f(x)>\alpha\}$,
then $\mu(A)>0$ (i.e., the Lebesgue measure of $A$ is positive)
because $f$ is continuous. We have that $\int_{a}^{b}f^{n+1}\geq\int_{A}f^{n+1}\geq\int_{A}\alpha f^{n}$.
Therefore,
\begin{eqnarray*}
\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}} & \geq & \frac{\alpha\int_{A}f^{n}}{\int_{A}f^{n}+\int_{A^{c}}f^{n}}\\
 & \geq & \frac{\alpha\int_{A}f^{n}}{\int_{A}f^{n}+\alpha^{n}(b-a)}.\\
 & = & \alpha\frac{1}{1+\frac{\alpha^{n}(b-a)}{\int_{A}f^{n}}}.
\end{eqnarray*}
We assert that $\frac{\alpha^{n}}{\int_{A}f^{n}}\rightarrow0$ as
$n\rightarrow\infty$. Choose $\beta\in(\alpha,1)$.
Let $B=\{x\in[a,b]\mid f(x)>\beta\}$, then $\mu(B)>0$. Observe that
$B\subseteq A$, so $\int_{A}f^{n}\geq\int_{B}f^{n}\geq\mu(B)\beta^{n}$.
It follows that $\frac{\alpha^{n}}{\int_{A}f^{n}}\leq\frac{\alpha^{n}}{\mu(B)\beta^{n}}\rightarrow0$
as $n\rightarrow\infty$. Now, it is clear that
\begin{eqnarray*}
\liminf_{n}\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}} & \geq & \liminf_{n}\alpha\frac{1}{1+\frac{\alpha^{n}(b-a)}{\int_{A}f^{n}}}\\
 & = & \alpha.
\end{eqnarray*}
Since $\alpha\in(0,1)$ is arbitrary, we actually have $\liminf_{n}\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}}\geq1$.
Combining, we have $\liminf_{n}\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}}=\limsup_{n}\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}}=1$.
This shows that $\lim_{n}\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}}=1$.
Finally, we drop the assumption that $M=1$. Let $g(x)=\frac{1}{M}f(x)$,
then $g$ satisfies all the conditions in above with $\max g=1$.
Therefore $\lim_{n}\frac{\int_{a}^{b}g^{n+1}}{\int_{a}^{b}g^{n}}=1$
and it follows that $\lim_{n}\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}}=M.$
