Opposite of a contraction mapping 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.
 A: Let $X=[0,\infty)$, and consider
\begin{align*}
f:X&\rightarrow X\newline
x&\mapsto 2\cdot x+1
\end{align*}
Then $f$ is expanding, sinde given $x,y\in X$,
$$d(f(x),f(y))=|(2\cdot x+1)-(2\cdot y+1)|=2\cdot |x-y|=2\cdot d(x,y)\geq 2\cdot d(x,y)$$
But, it has no fixed point in $X$, althoug $X$ is obviously complete, since
$$2\cdot x+1=x$$
is equivalent to
$$x=-1\not\in X$$
As you can see the lack of suprajectivity is the great problem for having a fixed point, since it makes impossible to assure the good properties of the inverse. That is, if $f$ is expanding and bijective your argumentation is true.
Returning to the example, when calculating a left inverse it can be
\begin{align*}
g:X&\rightarrow X\newline
x&\mapsto \begin{cases}\frac{x-1}{2} &\text{if }x\in [1,\infty)\newline0 &\text{if }x\in [0,1)&\end{cases}
\end{align*}
Which obviously is contractible, and has fixed point zero. However,
$$0=g(0)$$
does not permitt us affirm
$$f(0)=0$$
A: $X=[1,\infty)$, $f(x)=2x$ is a counterexample.
There is at most one fixed point in general, because if $x$ and $y$ are both fixed points, then $d(x,y)=d(f(x),f(y))\geq L d(x,y)$ implies $x=y$.
For the case $X=\mathbb R$, you could take $f(x)=2x+1$ if $x\geq 0$, and $f(x)=2x$ if $x<0$.  However, there are no continuous counterexamples for $X=\mathbb R$, which can be proved using the intermediate value theorem.
A: Just to add a simple case in which your intuition is true, let us assume that $f$ is continuous and that $X=\mathbb{R}^n$. The Theorem of the invariance of the domain says $f$ is an open mapping, since it is an injection. Additionally, the function has a closed image, due to your assumption $-$ the mapping sends Cauchy sequences in the image space back to Cauchy sequences in the domain space, and thus the image is a closed subspace of $\mathbb{R}^{n}$. Since the image is closed and open, the mapping is surjective. Now, being an invertible function and having an inverse map that is a contraction, the function $f$ has a fixed point, since if $$f^{-1}(x)=x,$$ we have $$f(x)=x.$$
