Why this function has a fixed point?

Question

My professor gave us the following:

Let $f:\mathbb{R} \to \mathbb{R}$ be monotonically increasing (perhabs discontinous). Suppose $0<f(0)$ and $f(100)<100$. Prove $f(x)=x$ for some $x\in \mathbb{R}$.

And I have $2$ questions:

1. The fact that it can be discontinous is false? I am thinking on something like:

Sorry for the bad drawing, but is like $f(x)=x+1$ for a while and at some point is constant but being discontinous at the point where it should be $f(x)=x$. This is ok?

2. I am not sure how to prove that this have a fixed point, in class we studiated the Banach fixed point theorem, but with this function I don't know if it can be a contraction to apply the theorem. Any idea would help.

Answer 1

Let $A=\{x\in[0,100]:x\le f(x)\}$. Since $A$ is bounded and contains $100$, we can write $a=\sup A$.

If $a<f(a)$, then there exists $x\in(a,f(a))$. Since $f$ is increasing, $a<x$ implies $f(a)\le f(x)$. Since $a$ is an upper bound for $A$ and $x>a$, we also have $f(x)<x$. Together this forces $f(a)\le f(x)<x<f(a)$ (contradiction).

Now suppose $f(a)< a$ and take any $x\in(f(a),a)$. Since $f$ is increasing, $x<a$ implies $f(x)\le f(a)$. Since $f(a)<x$, we have $f(x)<x$, which means $x\notin A$. But now we have shown $(f(a),a]\cap A=\emptyset$, contradicting the fact that $a=\sup A$.

Answer 2

Let $g(x)=f(x)-x$; $g(0)>0$, and $g(100)<0$. Let

$$A=\{a\in[0,100]:g(x)>0\text{ whenever }0\le x<a\}\,.$$

Clearly $100$ is an upper bound for $A$, so let $a_0=\sup A$. If $g(a_0)=0$, we’re done. If $g(a_0)>0$, there is an $x\in\big(a_0,a_0+g(a_0)\big)$ such that $g(x)<0$, and it follows that

$$f(x)=g(x)+x<x<g(a_0)=f(a_0)-a_0<f(a_0)\,,$$

contradicting the monotonicity of $f$. If $g(a_0)<0$, let $x=a_0+\frac12g(a_0)$; then $x<a_0$, but

$$f(a_0)=g(a_0)+a_0<a_0+\frac12g(a_0)=x<f(x)\,,$$

again contradicting the monotonicity of $f$.

Answer 3

Hint: Suppose there is no fixed point. Let $x_0 = \sup \{x \le 100: f(x) > x\}$. We know $x_0 \ge 0$ and $x_0 < 100$. Consider the two cases $f(x_0) < x_0$ and $f(x_0) > x_0$, and get a contradiction in each case.