Banach's Fixed Point Theorem application

Question

I'm trying to solve the following question:

Consider $M$ a complete metric space, $k > 1$ and $f: M \to M$ a surjective function, satisfying $d(f(x),f(y)) \geq k d(x,y)$, for every x,y $\in M$. Show that exists only one $a \in M$ such as $f(a) = a$.

Clearly $f$ is not a contraction, so i can't use Banach's Theorem. Any idea in how to solve the problem? Thanks in advance.

Answer 1

Just a sketch of what I think you need to do:

1. Multiply your inequality by the inverse multiplicative of $k$, so: \begin{align} d(f(x),f(y)) \geq k d(x,y) & \Rightarrow d(x,y) \leq \frac{1}{k} d(f(x),f(y)) \quad \forall x,y \in M \end{align}

you almost have the looks of a contraction, the only difference is that in the left side of the inequality you have points $x,y$ without being evaluated and in the right the evaluation of those points (we want exactly the opposite in order to assure $f$ is a contraction).

2. Use the fact that $f$ is surjective. Surjectivity assures that for any $x,y \in M$ (as codomain), there is some $w,z \in M$ (as domain) such that $f(w) = x$ and $f(z) = y$. Also note that $f(x) = f(f(w))$ and $f(y) = f(f(z))$, so you can say that

$$d(x,y) \leq \frac{1}{k} d(f(x),f(y)) \iff d(f(w),f(z)) \leq \frac{1}{k} d(f(f(w)),f(f(z))) \quad \forall ¿?$$

all is left is to argue why that this last inequality happens for all $f(w),f(z) \in M$, and then you can use the Banach Theorem, and this would conclude your question.