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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 $x_0$.

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Apply your last sentence $n$ times.

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By definition of contraction, \begin{equation} 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)). \end{equation}

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