how to show that f has only one fixed points?

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$$$d(f^ k (x), f^ k (y)) = 0 im getting uniques fixed points Is my proof is coorect or not correct . Pliz tell me if not correct . PLiz help me 2 Answers I propose a straightforward proof. By assumption, f^k is a contraction, so that it has a unique fixed point x \in X. From f^k(x)=x we deduce$$ 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). $$Hence x=f(x) is a fixed point of f^k, which is unique. • The last sentence doesn't mean anything. Or at best, there are steps missing (e.g. showing that a fixed point of f gives a fixed point of f^k; you've only shown the reverse). – Najib Idrissi Feb 26 '18 at 10:07 • @NajibIdrissi Well, I thought that the details could be rather easy to add... – Siminore Feb 26 '18 at 10:21 • Of course... if you know there are details to add. As it was originally phrased there was a leap in logic. Now it's fine. – Najib Idrissi Feb 26 '18 at 11:13 By Contraction mapping Theorem we could see the existence of a fixed point. Now suppose there are two distinct ones, say x,y, then$$d(x,y)=d(f^k(x),f^k(y))<c\cdot d(x,y)<d(x,y)$\$

Which is a contradiction.