Minimum area bound by the function Let $f:[0,1]->[0,1]$ be a continous function such that $$f(f(x))=1$$ for all $x$ in the domain. Required is the minimum and maximum possible area bound by this function from $x=0$ to $x=1$
As the function is bounded , I start by taking some $\alpha$ as the $x$ corresponding to the max value. Then $f(f(\alpha))=1$ as well, and $1$ is the maximum possible value in itself, $$f(\alpha)=f(1)=1$$
But from here I could not proceed to find the minimum value of $\int_{0}^{1}f(x)$. A hint on how to proceed from here is highly appreciated. Thank you.
 A: Credits: @Ryszard Szwarc.
Let $f(x)$ have a minimum value of $m$. Let $f(\alpha)=m$. Now $$f(f(\alpha))=1$$
or $f(m)=1$. (Now $m$ is the input)
If $m=1$ ,then the function is constant ,giving maximum area as $1$.
For $m<1$,
By intermediate value theorem,$f(x)$ takes values between $(m,1)$.
So for all $x\geq m$, $f(x)=1$.(See EDIT)
Again for $$\int_{0}^{1} f(x)=\int_{0}^{m} f(x) + \int_{m}^{1} f(x)$$
The last term coming out to be $1-m$.
For $\int_{0}^{m} f(x)$
Since $f(x)\geq m$ , $$\int_{0}^{m} f(x) >m^2$$ A noteworthy point here is that the equality is gone , because $f(x)$ is continous and cannot be $m$ for $x<m$ and then $1$ at $m$ ($m<1$ is the case here).
Using these ,$$\int_{0}^{1} f(x) > m^2-m-1\geq 3/4$$
So 3/4 should be the 'nonstrict' minimum.
EDIT - To clear up the IVT part a bit,
Take the function $f(f(x))$,which is a constant $1$. The input to $fof$ is $f(x)$. As $f(x)\geq m$,rewrite $fof$ as
$f($value greater than m$)=1$.
And now this result is used for $f$ instead of $fof$.
