unique fixed point problem 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 $f(x_0)=x_0$.

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.
