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.
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.