how to show that f has only one fixed points?

Question

Let (X, d) be a complete metric space. Let$ f : X → X$ be a function such that for all distinct$ x, y ∈ X$ ,

$ d(f^ k (x), f^ k (y)) < c · d(x, y)$, for some real number $c < 1$ and an integer $k > 1$. Show that f has a unique fixed point.

my attempt : i take $f(x) = x $and $f(y) = y$ .now $f^k = f(f(...(x),,))))= x $ similarly $f^k =f(f(,,,,,(y))..) = y $

here im getting $ d(f^ k (x), f^ k (y)) =c · d(x, y)$ $=0$.By fixed point theorem $f(x) =f(y) = x= y$ so im getting$ $$ d(f^ k (x), f^ k (y)) = 0$

im getting uniques fixed points

Is my proof is coorect or not correct . Pliz tell me

if not correct . PLiz help me

Answer 1

I propose a straightforward proof. By assumption, $f^k$ is a contraction, so that it has a unique fixed point $x \in X$. From $f^k(x)=x$ we deduce $$ f^{k+1}(x) = f(f^k(x))=f(x), $$ so that $f(x)$ is a fixed point of $f^k$. By the uniqueness of $x$, we deduce $f(x)=x$. Hence $f$ possesses a fixed point. This point is also unique, since it coincides with the fixed point of $f^k$. Indeed, whenever $f(x)=x$, we get $$ f^2(x)=f(f(x))=f(x), \quad f^{3}(x)=f(f^2(x))=f(f(x))=f(x), \ldots f^k(f(x))=f(x). $$ Hence $x=f(x)$ is a fixed point of $f^k$, which is unique.

Answer 2

By Contraction mapping Theorem we could see the existence of a fixed point. Now suppose there are two distinct ones, say $x,y$, then

$$d(x,y)=d(f^k(x),f^k(y))<c\cdot d(x,y)<d(x,y)$$

Which is a contradiction.