I am taking Real Analysis and we recently went over the Banach Fixed-point Theorem, also commonly known as the Contraction Mapping Theorem which states:
If $(X,d)$ is a complete metric space, and $f:X\to X$ is a contraction, that is $f$ satisfies $d(f(x),f(y)) \leq L d(x,y)$ for every $x,y \in X$ and some fixed $L<1$, then $f$ has exactly one fixed point, i.e. there exists a unique $z \in X$ such that $f(z)=z$.
I was thinking about a similar statement for what I can only guess are called expansion mappings.
Suppose $(X,d)$ is a complete metric space and $f:X\to X$ satisfies $ d(f(x),f(y)) \geq L d(x,y)$ for some fixed $L>1$ and every $x,y \in X$ with $x\neq y$. Does $f$ necessarily have exactly one fixed point?
I could not come up with a counterexample using real functions, though I haven't really had time (due to homework) to put much more thought into it.
I googled "expansion mapping" and some other similar terms but there does not seem to be any useful source on the topic that I could find. I think this notion of expansion mapping is a natural one to consider after considering contraction mappings, so I don't know why there doesn't seem to be any available research on the topic.
I have a few good ideas for how I would go about trying to prove this that might help.
Firstly we note that $f$ must be injective, otherwise we would have two distinct points which get closer (distance zero) after applying the mapping which would be a contradiction.
Thus $f$ is left invertible. If I had to guess, I would say that the left inverse of an expansion mapping must be a contraction mapping (with reciprocal constant $\frac{1}{L}$?). Then that contraction must have a fixed point by the Banach Fixed-point Theorem. Perhaps it can be shown that this must also be a fixed point of $f$ itself, I haven't taken the time to see whether fixed points are preserved using only one-sided invertibility.
Any thoughts, ideas, research, or proofs on the topic of expansion mappings are welcome.