Calculus Proof Problem

Solution 1:

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

1. $f(0)=a$
2. $f(a)=1$ 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 $[0,a)$.

Solution 2:

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}

Solution 3:

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

$$\int_0^1\!f(x)\,\mathrm{d}x>\frac{3}{4}$$