# Fixed point for expansion mapping.

Let $f$ be a continuous mapping of a complete metric space $M$ onto itself satisfying the following condition for any $x, y$ $\in$ M: $d(f(x), f(y))$ is greater than or equal to $\alpha d(x; y)$, $\alpha$> 1 (greater than 1). Prove that the mapping f has a unique fixed point.

I know the Fixed point theorem for contraction mapping and I understand the proof of this theorem. But how can you the similar proof to prove the statement above?

Note that $f$ is injective and surjective, hence has an inverse on its range. Let $Y=f(M)$, note that $Y=M$ and let $\rho(y_1,y_2) = d(f(y_1),f(y_2))$.
Show that $\rho$ is a distance and $(M,\rho)$ is complete. Note that $f^{-1}$ is a contraction on $M$ (with rank ${1 \over \alpha} < 1$). Hence $f^{-1}$ has a unique fixed point. Hence $f$ has a unique fixed point.
• The question states that $f$ is onto $M$, hence surjective. If $x \neq y$ then $d(f(x),f(y)) \ge \alpha d(x,y) >0$ hence injective. – copper.hat Feb 19 '16 at 9:02