Converse of a fixed-point theorem

Question

I'm having some trouble furnishing a proof here.

Let $(E, d)$ be a metric space such that any $k$-Lipschitz function has a fixed point for $0 < k < 1$. Does it follow, then, that $E$ is complete?

Answer 1

No. Take, for example, $E=\{(x,\sin(1/x)\mid 0<x\le 1\}$.

Suppose $f\colon E\to E$ is a $k$-Lipschitz function. It can be extended to a function on the completion of $E$, $E'=E\cup\{(0,y)\mid |y|\le1\}$. Now to show that the original function has a fixed point we only need to prove that the fixed point of the extension can't lie on y axis.

The idea is, I believe, that in such case "each wave of sine maps to a wave nearer to the y axis" and then each point of $\{(0,y)\mid |y|\le1\}$ would be a limit point of some sequence of the form $f^n(a)$ ($a\in E$) -- i.e. all $E'\setminus E$ would consist of fixed points which is imposible.

For a complete proof and other examples see e.g. On a converse to Banach's Fixed Point Theorem