Calculus Proof Problem Let $f$ be a continuous function whose domain includes $[0,1]$, such that $0 \le f(x) \le 1$ for all $x \in [0,1]$, and such that $f(f(x)) = 1$ for all $x \in [0,1]$. Prove that $\int_0^1 f(x)\,dx > \frac34$.
I've been stuck on this question for a while now without any idea on how to get started. Is it possible for someone to help me?
Thanks!
 A: While not a general proof of the proposition, it can be seen as true of a certain class of functions satisfying the conditions.
Let $f(x)$ be a continuous non-decreasing function defined on the interval $[0,1]$ with $a\in[0,1]$ as shown in the diagram and satisfying


*

*$f(0)=a$

*$f(a)=1$ for $x\ge a$


Then the integral $\int_0^1 f(x)\,dx$ must be greater than $a^2+(1-a)$, the sum of the two shaded rectangles.
But the minimum area of the two rectangles is $\frac{3}{4}$ and occurs when $a=\frac{1}{2}$. 
It remains to be determined whether there is any other class of function satisfying the conditions.
Addendum: Actually we can drop the non-decreasing part and merely require that $f(x)\ge a$ on $[0,a)$.

A: First, let us prove $f(1) = 1$. Since $f(f(x))=1$ for all $x \in [0, 1]$ then there must exist $x^\ast$ such that $f(x^\ast) = 1$. In particular, we see that $f(f(x^\ast))= f(1) = 1$. 
Next, let us show that $f$ has a unique fixed point. Suppose $f(x) = x$, then $f(f(x))= f(x) = 1$. Since $f(x) = x$ and $f(x) =1$ then $x= 1$. Thus, $x=1$ is the unique fixed point of $f$.
Next, let $a = \min f(x)=f(x_m)$ which exists since $f$ is continuous on a compact set. Assume $a<1$, otherwise we are done. Let us show that $f(x) \equiv 1$ for all $x \in [a, 1]$. Suppose $y \in (a, 1)$ such that $f(y)<1$. By the intermediate value theorem there exists $z \in (x_m, a)$ such that $f(z) = y$, but this contradicts the fact that $f(f(z))=1$ since $f(f(z)) = f(y)<1$. Thus $f\equiv 1$ for all $[a, 1]$. Hence $f(x)>x$ for all $x \neq 1$ and $f$ must be greater than $g$ given by
\begin{align}
g(x)=
\begin{cases}
a & \text{ if }  0\leq x \leq a\\
1 & \text{ otherwise}. 
\end{cases}
\end{align}
which means
\begin{align}
\int^1_0 f(x)\ dx>\int^1_0 g(x)\ dx = 1-a(1-a) \geq \frac{3}{4}
\end{align}
since 
\begin{align}
\frac{1}{4}-a+a^2=(\frac{1}{2}-a)^2\geq 0. 
\end{align}
A: First of all, suppose $f(1)=x$, then $f(x)=f(f(1))=1$ so $f(1)=f(f(x))=1$.
The set $[0,1]$ is a compact subset of $\mathbb{R}$ and the function $f$ is continuous, then by the extreme value theorem, $f$ must attain a minimum. We can then say that there exists $c\in [0,1]$ such that for all $x\in [0,1]$, $f(c)\leq f(x)$. 
Let $b=f(c)$ be this minimum value. By the intermediate value theorem, for any value $y$ between $f(c)=b$ and $f(1)=1$, there is a point $x$ in the interval $[c,1]$ such that $f(y)=x$ because $f$ is continuous. Let $x$ be a point in the interval $[b,1]$, then there exists a point $y\in [c,1]$ such that $f(y)=x$, so $f(x)=f(f(y))=1$. Then for all $x\in [b,1]$, $f(x)=1$.
We defined $b$ as the minimum of the image of $f$. Then for every $x\in [0,b]$, $f(x)\geq b$. We can then evaluate the integral:
$$\int_0^1\!f(x)\,\mathrm{d}x=\int_0^b\!f(x)\,\mathrm{d}x+\int_b^1\!1\,\mathrm{d}x\geq \int_0^b\!b\,\mathrm{d}x+\int_b^1\!1\,\mathrm{d}x=b^2+(1-b)$$
The point $b$ is a value between $0$ and $1$. The quadratic function $b^2-b+1$ is then at its minimum when $b=\frac{1}{2}$. So whatever $b$ is, the integral $\int_0^1\!f(x)\,\mathrm{d}x\geq \frac{3}{4}$. 
Finally, the integral cannot equal $\frac{3}{4}$, because if we suppose that the integral is $\frac{3}{4}$, then we know for sure that $b=\frac{1}{2}$. The function $f(x)=\frac{1}{2}$ when $x\in [0,\frac{1}{2}]$ and $f(x)=1$ when $x\in [\frac{1}{2},1]$ will then be discontinuous at $\frac{1}{2}$.
We can then conclude that a continuous function $f:[0,1]\to [0,1]$ such that $f(f(x))=1$ for every $x\in [0,1]$, then 
$$\int_0^1\!f(x)\,\mathrm{d}x>\frac{3}{4}$$
A: Edit: This proof does not work if f (x) does not have range [0,1] for all x in [0,1]. The simple way to prove this:
We have function f(x) which is continuous along the interval [0,1]. 
f (x) is defined such that                                                              f(x) has domain [0,1] , and range [0,1].                  (a)
f(f (x)) has domain [0,1] and range [1].                 (b)
--Note: The domain of f(f(x)) comes from the range of f(x) along the interval [0,1].--
Now we want to prove that the integral of f(x) over [0,1] is > 3/4
In order to prove this we can break the problem into two parts. First we can show that f(f(x)) and f(x) have the same range for domain [0,1]. Second we can show that because the function f's range is [1] for all x in [0,1] that the integral over [0,1] must be greater than 1.
First Part:
Since f is a function it cannot have two different outputs for one input, so (a) and (b) cannot have contradictory ranges for the same domain. From (a) we know that the range of f over domain [0,1] is 0<=f(x)<=1, from (b) we also know the range over domain [0,1] is f(x)=1. So (a) and (b)'s range do not contradict each other  in so much as f(f(x))=1 is within the range of f(x). But since f is a function it must be the case that f(x)=1 for all x in the interval [0,1]. Otherwise the function f would have multiple outputs for the same inputs and would not be a function. Therefore it must be the case that for all input for f in the interval [0,1], f has an output of 1, and thus for the function f the domain [0,1] has range [1].
The Second Part:
Since the function f on domain [0,1] has range [1], then any integral within [0,1] of inputs must be greater than or equal to 1. Therefore since the integral is greater >=1 and 1 is > 3/4, the integral is > 3/4.
Proof complete.
