3
$\begingroup$

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?

$\endgroup$

marked as duplicate by Martin Sleziak, Ali Caglayan, Davide Giraudo functional-analysis 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.

6
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.