A question on the Banach fixed point theorem.

Question

Suppose $f:(X,\tilde{d})\rightarrow(X,d)$ be a continuous function satisfying \begin{eqnarray}d(f(x),f(y))\leq \lambda d(x,y),\end{eqnarray} $\lambda > 1$. Let $\tilde{d}(x,y)=\lambda d(x,y)$. I observed that the topologies due to $d$ and $\tilde{d}$ are equivalent. Hence the "contraction" condition now reads as \begin{eqnarray}d(f(x),f(y))\leq \tilde{d}(x,y).\end{eqnarray} Can it concluded form here that a fixed point exists for $f$?. In other words I was wondering whether the Banach fixed point theorem holds if the metrics in the domain and the range spaces are equivalent and not exactly the same?.

Answer 1

You certainly have to account for the factor occuring in the equivalence. That is, if you merely have $d\circ f\le \tilde d$, then you do don't have a contraction unless you know that $\tilde d\le \lambda d $with some $\lambda<1$. For example $f\colon \mathbb R\to \mathbb R$, $x\mapsto x+1$ has no fixed points, but you can certainly find that the inequality holds if you use different metrics (i.e., different multiples of $|x-y|$).