# Let $X$ be a metric space, and let $f:X \to X$ be a $k$ contraction. Prove that for all $n \in \mathbb{N}$, $d(f^n(x),f^{n+1}(x)) \leq k^nd(x,f(x))$.

Let $$X$$ be a metric space, and let $$f:X \to X$$ be a $$k$$ contraction. Prove that for all $$n \in \mathbb{N}$$, $$d(f^n(x),f^{n+1}(x)) \leq k^nd(x,f(x))$$.

By definition, because $$f(x)$$ is a $$k$$-contraction, $$d(f(x),f(y)) \leq kd(x,y)$$ for all $$x,y \in X$$.

I also have this theorem I could employ? Because $$f$$ is $$k$$-contracted, then we know $$f$$ has a unique fixed point at $$x_0$$ and for all $$x \in X$$, the iterated map on $$f$$ based at $$x$$ converges to $$x_0$$.

Apply your last sentence $$n$$ times.
By definition of contraction, $$$$d(f^{n}(x), f^{n+1}(y)) \leq kd(f^{n-1}(x), f^{n}(y)) \leq k^2 d(f^{n-2}(x), f^{n-1}(y)) \leq ... \leq k^{n-1}d(f(x), f^2(y)) \leq k^n d(x, f(y)).$$$$