Will this function always have a fixed point?

Question

$f:[0,\infty) \to [0,\infty)$ where $|f(x)-f(y)| \le \frac{1}{2}|x-y|$

Does this function always have a fixed point?

My attempt:

The function is continuous. If it becomes differentiable then it will have a fixed point.

So I was looking for some non differentiable continuous function which doesn't have a fixed point.

What I could think of is example of the form $|x-a|/2$ but I am unable to construct one. Some hints please.

Answer 1

$f$ is a contraction mapping $($meaning that it is K-Lipschitz with $K<1$(in your case $K =\dfrac12))$,and $[0,\infty)$ is a complete metric space.
By Banach Fixed Point Theorem,$f$ admits a unique fixed point in $[0,\infty)$

Answer 2

Here, $f$ is a contractive funtion, i.e., it is a K-Lipschitz function with K<1.

This means f is also continuous.

Now consider a real sequence $(x_n)_{n \in \mathbb{N}}$ with $x_{n+1} = f(x_n)$

Then $|x_{n+1}-x_n| = |f(x_n)-f(x_{n-1})| \leq \frac12 |x_n-x_{n-1}|$

So, $(x_n)$ is a contractive real sequence $\implies$ $(x_n)$ is convergent sequence.

Every contractive real sequence is convergent.

Let $(x_n)$ converges to $x$. Then $f(x_n)$ converges to $f(x)$ because f is continuous. But $f(x_n) = x_{n-1}$ and $(x_{n-1})$ converges to $x$, so $f(x_n)$ converges to $x$. This implies $f(x)=x$.

Thus, f has a fixed point.

This method is true for any contractive mapping on a complete metric space and is known as Banach fixed-point Theorem.

Answer 3

This answer follows Izaak van Dongen's argument based on his first comment. It assumes you have already correctly proved that the function is continuous (which you say you have), because it relies on the Intermediate Value Theorem.

If $f(0) = 0, $ then $x=0$ is a fixed point and we are done. So suppose the only other option:

$f(0) > 0.$ Then $\vert f(3f(0)) - f(0) \vert \leq \frac{1}{2} \vert 3f(0) - 0 \vert = \frac{3}{2}\vert f(0)\vert \overset{*}{=} \frac{3}{2} f(0),\ $ *since $\ f(0) > 0\ $ by supposition.

Now by considering the fact that if $\ u>0,\ $ then $\ \vert z - u \vert \leq \frac{3}{2} u \implies -\frac{1}{2}u\leq z \leq \frac{5}{2} u,\ $

by setting $\ z=f(3f(0))\ $ and $\ u = f(0),\ $ we see that $\ f(3f(0)) \leq \frac{5}{2}f(0)< 3f(0).\ $

Finally, let $\ g(x):= x - f(x).\ $ We have: $\ g(0) <0\ $ and $\ g(3f(0)) > 0,\ $ so by IVT,

$\ \exists\ \alpha \in (0, 3f(0) )\ $ such that $\ g(\alpha) = 0,\ $ i.e. $\ f(\alpha) = \alpha.$