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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.

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up vote 5 down vote accepted

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 $$f(0)=0$$

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$X=[1,\infty)$, $f(x)=2x$ 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.

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What if one adds an additional hypothesis that $f$ is onto? Then the inverse is a contraction mapping on the whole space and so has a unique fixed point. – Michael Hardy Nov 9 '11 at 23:09
Michael Hardy: That's a good point, which is also at least hinted at in Iasafro Maesman's answer ("lack of surjectivity is the great problem"). It is one way to think of why is is true for the case of continuous $f:\mathbb R\to\mathbb R$, because the intermediate value theorem can be used to show that $f$ is onto. – Jonas Meyer Nov 9 '11 at 23:29
A different aspect: we may not insist that the space be compact, as the maximum distance is achieved by two points, and an expansive mapping has nowhere to put them. – Will Jagy Nov 10 '11 at 0:41
@WillJagy: That's a good point, but it's even worse than that. If $X$ has more than one point, then taking $x\neq y$, the set $\{d(x,y),d(f(x),f(y)),d(f(f(x)),f(f(y))),d(f(f(f(x))),f(f(f(y)))),\ldots\}$ is unbounded, so $X$ must be unbounded. (Your argument would also apply with the weaker condition $d(f(x),f(y))>d(x,y)$.) – Jonas Meyer Nov 10 '11 at 1:26

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