Given $f'(x)$ is strictly bounded by $1$, $f(x)$ has at least one fixed point Let $f:\mathbb{R}\to\mathbb{R}$ be differential function such that $f'(x)\leq r<1$ for all $x\in \mathbb{R}$. Then $f$ has at least one fixed point.
What I am considering is $h(x)=f(x)-x$ then $h(x)$ is strictly decreasing, but I did not get why it will hit the x axis???
 A: (Elaborating on the comments to the question:)
The function $h(x) = f(x) - x$ satisfies $h'(x) \le r - 1 < 0$.
So $h$ is not only decreasing, it decreases with a certain "minimum rate".
This vague formulation can be made into a precise statement
using the mean value theorem,
which states  that for $x > 0$
$$ \begin{aligned}
 h(x) &= h(0) + (x-0) \, h'(t) \quad \text{for some } t \in (0, x) \\
    &= f(0) + x \, h'(t) \\
&\le f(0) + x \, (r-1)  \, .
\end{aligned}$$
In particular, for any $x \ge a := \dfrac{|f(0)|}{1-r}$,
$$
h (x) \le f(0) - |f(0)| \le 0 \, .
$$
In a similar manner you can show that
for any $x \le -a = \dfrac{|f(0)|}{r-1}$,
$$
h (x) \ge f(0) + |f(0)| \ge 0 \, .
$$
Now it follows from the intermediate value theorem that $h(c) = 0$ 
for some $c$ in the interval  $[-a, a]$. So $f$ has a fixed point
in this interval.
A: The idea is good: you have $h'(x)=f'(x)-1<0$, so the function is strictly decreasing.
Now you have to show that $h$ assumes positive and negative values. Suppose not; either $f(x)>x$ for every $x$ or $f(x)<x$ for every $x$.
Let's do the first case; by Lagrange's theorem you have, for $x>0$,
$$
\frac{f(x)-f(0)}{x-0}=f'(c)
$$
for some $c\in(0,x)$. Thus
$$
f(x)=f'(c)x+f(0)>x
$$
Rearranging it, we can write
$$
f(0)>x(1-f'(c))
$$
so that, using $f'(c)\le r$,
$$
f(0)>x(1-r)
$$
for every $x>0$. This is of course a contradiction.
Suppose instead $f(x)<x$, for every $x$. Then, for $x<0$,
$$
\frac{f(x)-f(0)}{x-0}=f'(c)
$$
so
$$
f(x)=xf'(c)+f(0)<x
$$
that implies $f(0)<x(1-f'(c))$. Since $f'(c)\le r$, we have $1-f'(c)\ge1-r$ and so $x(1-f'(c))\le x(1-r)$ and therefore
$$
f(0)<x(1-r)
$$
for every $x<0$, a contradiction.
A: Banach fixed point theorem: the mean value theorem shows that $|f(x) - f(y)| = |f'(c)||x-y|$ for some $c$, so $|f(x) - f(y)| \le r|x-y|$. So apply Banach's theorem.
