unique fixed point problem

Question

Let $f: \mathbb{R}_{\ge0} \to \mathbb{R} $ where $f$ is continuous and derivable in $\mathbb{R}_{\ge0}$ such that $f(0)=1$ and $|f'(x)| \le \frac{1}{2}$.

Prove that there exist only one $ x_{0}$ such that $f(x_0)=x_0$.

Answer 1

Consider $g(x)=f(x)-x$. We have that $g'(x)=f'(x)-1<-1/2<0$ and that $g(0)=1$. So $g(x)$ is strictly decreasing and $g(0)>0$, so there must me a unique $x_0$ such that $g(x_0)=0=f(x_0)-x_0$.

Remark: of course we have that $\lim_{x \to +\infty}g(x) = -\infty$

Answer 2

If $f(x)=x$ and $f(y)=y$ for $y>x$, then $$y-x=|f(y)-f(x)|=|\int_{x}^y f'(t)\mathrm{d}t|\leq \int_x^y |f'(t)|\mathrm{d}t<\frac{1}{2}(y-x),$$ a contradiction.

Answer 3

Note that $1-{1 \over 2} x \le f(x) \le 1+ {1 \over 2}x$ and so $f(x) -x \le 1-{1 \over 2}x$.

Hence $f(0) -0 = 1$ but for $x \ge 2$ we have $f(x) -x \le 0$. Hence there must be some $x \in [0,2]$ such that $f(x) -x = 0$.

Now suppose $f(x_1) = x_1, f(x_2) = x_2$. Then $|x_1-x_2| = |f(x_1)-f(x_2)| \le {1 \over 2} |x_1-x_2|$ by the mean value theorem, so we must have $x_1 = x_2$.