# Contraction and Fixed Point [duplicate]

This question already has an answer here:

How do I show that for $T: X \rightarrow X$ where X is complete and $T^m$ is a contraction that T has a unique fixed point $x_0 \in X$.

I know there exists $\lambda_1 \in (0,1)$ for $x, y \in X$ such that $d(T^mx, T^my) \leq \lambda_1 d(x, y)$ and I need to show that T is a contraction and then apply the fixed point theorem but how do I do that?

## marked as duplicate by Martin Sleziak, Ali Caglayan, Davide Giraudo functional-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 13 '15 at 11:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• I think you can argue by contradiction. – user99914 Oct 14 '13 at 4:00
• You don't - and in general can't - show that $T$ is a contraction. You use that $T^m$ is a contraction, hence has a unique fixed point, and show that that is also a fixed point of $T$. – Daniel Fischer Oct 14 '13 at 4:02
• – Martin Sleziak Feb 13 '15 at 11:18

## 2 Answers

You don't need to show that $T$ is a contraction. That might be false. (E.g., $X=\mathbb R^2$, $T(x,y)=(0,2x)$, $m=2$.)

then apply the fixed point theorem

So you know a fixed point theorem that would apply if $T$ were a contraction. That means that you know a fixed point theorem that does apply to $T^m$. Hence, you know that $T^m$ has a unique fixed point $x_0\in X$.

• $T$ cannot have any other fixed points, because every fixed point of $T$ is a fixed point for all powers of $T$.
• Thus the remaining work is to show that $x_0$ is in fact also a fixed point for $T$.

Note that $x_0=T^m(x_0)$ and $T(x_0)=T(T^m(x_0))=T^m(T(x_0))$, so

• $d(T(x_0),x_0)=d(T^m(T(x_0)),T^m(x_0))\leq \lambda_1 d(T(x_0),x_0)\implies d(T(x_0),x_0)=0.$
• Alternatively, as noted, $T^m(T(x_0))=T(x_0)$, which shows that $T(x_0)$ is a fixed point for $T^m$, hence $T(x_0)=x_0$ by uniqueness.

HINT: $T^k[X]\supseteq T^{k+1}[X]$ for each $k\in\Bbb N$. Using the fact that $T^m$ is a contraction and $X$ is complete, what can you say about

$$\bigcap_{k\ge 0}T^k[X]\;?$$

If that’s not quite enough, I’ve extended the hint a little in the spoiler-protected region below.

How does it compare with $\bigcap_{k\ge 0}T^{km}[X]$?