$|f(x)-f(y)|<|x-y|$ on a non-empty closed bounded set of real numbers

Question

Let $A$ be a non-empty closed bounded set of real numbers and $f: A \to A $ be a function such that $|f(x)-f(y)|<|x-y| , \forall x,y\in A$ , then how to show that $f$ has a unique fixed point ?

Answer 1

• For the unicity assume that $a$ and $a'$ are two different fixed points for $f$ then

$$|a-a'|=|f(a)-f(a')|<|a-a'|$$ which's a contradiction so we proved the unicity.

• For the existence let $\psi\colon x \mapsto |x-f(x)|$ which's clearly continuous (since $f$ is lipschitzian and then continuous) on the compact $A$ so it attains its minimum on say $a$. Assume that $a\ne f(a)$ we have $$|f(f(a))-f(a)|<|f(a)-a|=\min_A\psi$$ which's a contradiction. We conclude the desired result.

Answer 2

Existance:

Note that $Im (f) \subsetneq A$

Consider $a \in A$ then consider the sequence $a_n = f(a_{n-1})$ there $a_0 = a$ note that $|a_{n-1} - a_{n}| > |f(a_{n-1} - f(a_n)| = |a_n - a_{n+1}|$ so the sequence is convergent (the distance between two points is decreasing and since $A$ is bounded, the distance must tend to $0$). Since $Im (f) \subsetneq A$ then $a':=\displaystyle\lim_{n\to\infty} a_n\in A$. So $|a' - f(a')| = 0 \Rightarrow a' = f(a')$.

Uniqueness:

Assume there are two fixed points $x_0$ and $y_0$. Then

$$|f(x_0) - f(y_0)| = |x_0 - y_0|$$

Wich contradicts the initial hypothesis.