# Proof of a special case of Banach's fixed point theorem

I have to prove the following special case of the theorem:

Let $f : I \to I$ be Lipschitz continuous on the closed (not bounded) interval $I=[0,\infty)$ with Lipschitz constant $L \lt 1$. Then $f$ has a unique fixed point in $I$ i.e. $\exists^{1} x \in I : f(x) = x$

The steps that I'm supposed to take are:

(i) Show that $f$ can only have one fixed point
(ii) Fix $x_0 \in I$ and define $x_n = f(x_{n-1})$ for $n \in \mathbb{N}$; prove that |$x_{n+1} - x_n$| $\leq L^{n} |x_1 - x_0$|
(iii) Show that ($x_n$) is a Cauchy sequence (hint: geometric sequences)
(iv)Show that x is the limit of $x_n$ as $n$ tends to infinity and that $x \in I$
(v)Show that $x$ is a fixed point of $f$

I suppose it seems straightforward but I've fallen at the first hurdle (i.e. part (i)). Any help would be appreciated.

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First hurdle. Assume $f(x_0)=x_0$ and $f(y_0)=y_0$. If $x_0\neq y_0$. Then $|x_0-y_0|=|f(x_0)-f(y_0)|<|x_0-y_0|$. Impossible. – 1015 Apr 15 '13 at 16:52
Why does that last inequality have to hold? – Mathlete Apr 15 '13 at 16:54
Do you know what Lipschitz mean? – 1015 Apr 15 '13 at 16:57
Yes I do, sorry, I just had a total mind blank. How would I continue? – Mathlete Apr 15 '13 at 16:58
(ii) induction (iii) geometric sum formula (iv) $I$ is complete (v) uniqueness of limit and sequential continuity of $f$: $f(x)=\lim f(x_n)=\lim x_{n+1}=\lim x_n=x$. – 1015 Apr 15 '13 at 17:01

If $f(x)=x$ and $f(y)=y$ for some $x\ne y$ with $x,y\in I$, what can you say about $|f(x)-f(y)|$ in comparison to $|x-y|$? Is this possible?

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Of course, because its Lipschitz continuous. How would I go on from there though? – Mathlete Apr 15 '13 at 16:57
There's nothing more to show. You assumed that $f$ had more than one fixed point, and derived a contradiction. Thus, $f$ has at most one fixed point. You'll show that $f$ has exactly one fixed point in part (v). – Cameron Buie Apr 15 '13 at 17:07
Oh right, I see. thank you very much. – Mathlete Apr 15 '13 at 17:08