Understanding uniqueness in Banach's Fixed Point Theorem

Question

Banach's Fixed Point Theorem proves uniqueness of fixed points. Suppose the contraction map $g$ had two fixed points $x$ and $y$. Then,

$$|x - y| = |g(x) - g(y)| \leq L|x - y|,$$

which implies $(1 - L)|x - y| \leq 0$. But since $(1 - L) > 0$, we can deduce $x = y$.

---

• I get that the first equality follows from the definition of a fixed point
• Second equality follows from definition of contraction map
• How does it imply $(1-L)|x - y| \leq 0$, and how does $(1 - L) > 0$ lead us to $x = y$?

Answer 1

You have:$$\lvert x-y\rvert\leqslant L\lvert x-y\rvert.$$Subtracting $L\lvert x-y\rvert$ from both sides you get that$$(1-L)\lvert x-y\rvert\leqslant0.\tag1$$But $1-L>0$ and $\lvert x-y\rvert\geqslant0$. So, we can only have $(1)$ if $\lvert x-y\rvert=0$.

Answer 2

Another way: as $x \neq y$, $p:=|x-y| >0$. So we can divide by $p$ in $p \le Lp$ (which is your combined inequality) and we get $1 \le L$ contradicting that $L<1$.

Answer 3

You may also reason as follows:

If $x,y$ are two fixpoints, then if $g^n$ denotes the $n$-fold composition of $g$: $$0\leq|x-y| = |g^n(x) - g^n(y)| \leq L^n|x-y| \stackrel{n\to \infty}{\longrightarrow}0$$

Hence, $x=y$.