# Prove that a real variable function that satisfies certain conditions has a fixed point.

Let $f: \mathbb R \to \mathbb R$ be a differentiable function such that $f(0)=1$ and $|f'(x)|<\dfrac{1}{2}$.

$i)$ Prove that there exists $x_0 \in [0,2] :f(x_0)=x_0$.

$ii)$ Let $\phi:C[\frac{1}{2},2] \to C[\frac{1}{2},2]$ defined as $\phi(g)(x)=\dfrac{(g(x))^2}{2x}$. Prove that if $x_0 \in [\frac{1}{2},2]$, then $x_0$ is also a fixed point of $D(\phi)(f)(f)$.

My attempt at a solution:

I could do almost anything of this exercise. The only thing I could get to is:

Given any two $x,y \in \mathbb R$, by the mean value theorem, we have that there is $c \in (x,y)$ $f(x)-f(y)=f'(c)(x-y)$. Then $|f(x)-f(y)|=|f'(c)||x-y|$. But $|f'(c)|<\dfrac{1}{2}$, so $|f(x)-f(y)|=|f'(c)||x-y|<\dfrac{1}{2}|x-y|$, this proves $f$ is a contraction and as $\mathbb R$ is complete, then, by the fixed point theorem, there is a unique $x \in \mathbb R: f(x)=x$. Now, if I want to prove there is a fixed point in $[0,2]$ and $x$ is the only fixed point, I must prove that $x \in [0,2]$. I don't see at all why the fact that $f(0)=1$, among the other hypothesis, would imply this.

For point $ii)$, I don't even understand well what is $D(\phi)(f)(f)$, is it the differential the function that must have the fixed point? What it means $D(\phi)(f)(f)$?, if it is the differential of the function $\phi$ evaluated at $f$, wouldn't a fixed point be in $C[\dfrac{1}{2},2]$ instead of $\mathbb R$?. Two last question: why $D(\phi)(f)(f)$ has a double $(f)(f)$? and wouldn't I have to prove first that the linear operator $\phi$ is differentiable?

• I'm afraid most of us won't know what question ii) means. – Stefan Smith Nov 28 '13 at 4:16
• Neither do I, I'll ask my classmates and if someone clarifies it, I'll write it here. – user100106 Nov 28 '13 at 20:01

Regarding part i, here is one approach: consider the function $g(x) = f(x) - x$. We note that $g(0)>0$, and (using the mean value theorem) you can show that $g(2) < 0$. We now know that there is a $c \in (0,2)$ for which $g(c) = f(c) - c = 0$. This is your (unique) fixed point.
As for the second part, would $D(\phi)(f)(f)(x)$ be $[D\phi](f(f(x)),$ or $D[\phi \circ f \circ f](x),$ or is it perhaps $D[\phi \circ f](f(x))$? Honestly, I'm not sure how to make sense of the notation you've used here.
• If you do use my proof, you would have to prove that this fixed point is unique; you can do so using the mean value theorem to show that if there are two distinct fixed points, then there is a $c$ between them for which $f'(c) = 1>\frac 12$. – Omnomnomnom Nov 28 '13 at 18:26
Brouwer's fixed-point theorem http://en.wikipedia.org/wiki/Brouwer_fixed-point_theorem tells you that any continuous map from a compact, convex subset of Euclidean plane has a fixed point. Now, consider the restriction of $f$ to the interval $[0,2]$. We show that $f([0,2] \subset [0,2]$, and then the existence of the fixed point follows. By the mean-value theorem:
$-1/2 <\frac {f(2)-f(0)}{2-0}=\frac {f(2)-1}{2} < 1/2$ , so that $f(x)<2$ on $[0,2]$; for any $x$ in $[0,2]$use: $$-1/2<\frac{f(x)-1}{x}<1/2$$ , so that $$\frac{3-x}{2}<f(x)<\frac {1+x}{2}$$ . This guarantees that the restriction of $f(x)$ to $[0,2]$ is a map from the compact space $[0,2]$ to itself, and so Brouwer's fixed point theorem guarantees the existence of a fixed point.