# When is a local contraction mapping globally contraction.

Let $A$ be a complete metric space and $T:A\rightarrow A$ is an continuous operator. There exists some $\Delta$ such that for $\forall x,y\in A$ and $d(x,y)<\Delta$, there's some $\epsilon <1$ s.t. $d(Tx,Ty)\le \epsilon d(x,y)$.

This is what I mean by the "local contraction" property and I'm wondering when is $T$ is a global contraction.

My conjecture is that when $T$ is an "uniformly local contraction", i.e. the $\Delta$ and $\epsilon$ are identical for all $x,y$. Then for arbitrarily chosen $x$ and $y$ we can always find some points between them in $A$, and the distance between each of those points are less than $\Delta$, then applying the triangle inequality gives the global contraction result.

I'm not sure whether this is an established result or whether there are some traps to be aware of.

Thank you!

Specifically, I'm interested in the fixed point property about contraction mappings. I've read the paper Remarks on metric transforms and fixed-point theorems that gives the result if any two points in $A$ can be joined by a rectifiable curve, then a local contraction acts like a global contraction in deriving fixed points. However the space $A$ I'm working on is a functional space and I'm fuzzy about the "rectifiable curve" concept and it seems I only need a weaker result than this.

• This is the 450,000th question. – wythagoras Jun 13 '15 at 18:54
• Good to know :) – iridium Jun 13 '15 at 18:58
• You need some kind of connectedness assumption. Any discrete space with any map has the "local contraction" property. – TYS Jun 13 '15 at 19:50
• @YanShuoTan, Thank you for the comment, I've made some complements concerning your comment. Is the concept in that paper equivalent as what you mean by connectedness? Thanks again! – iridium Jun 13 '15 at 22:17

Even being connected by rectifiable curves is not enough to get global contraction from local. Consider the following map, which flips the top line segment to the right:

Small distances are not increased, but $|A'D'|>|AD|$. To get a counterexample in your setting, shrink A'B'C'D' slightly.

A sufficient condition is the space being a length space, that is: for every $a,b$ the distance $d(x,y)$ is the infimum of lengths of rectifiable curves connecting $a$ to $b$. Indeed, if $\gamma$ is a curve of length $\ell$, then an $\epsilon$-local contraction maps it to a curve of length $\le \epsilon \ell$, and the conclusion follows.