[proofplan] The proof is a direct expansion of the definition of a contraction on the [metric space](/page/Metric%20Space) $(X,d)$. We first write the contraction inequality using the metric $d$, then substitute the defining formula $d(x,y)=\|x-y\|_X$. Since the same constant $c\in[0,1)$ works before and after this substitution, the two formulations are equivalent. [/proofplan] [step:Expand the contraction condition in the metric $d$] By definition, the map $f:X\to X$ is a contraction with respect to the metric $d$ if and only if there exists a constant $c\in[0,1)$ such that \begin{align*} d(f(x),f(y)) \le c\,d(x,y) \end{align*} for all $x,y\in X$. [guided] We begin by using only the definition of contraction in a metric space. The metric space in question is $(X,d)$, and the map under consideration is \begin{align*} f:X\to X. \end{align*} Saying that $f$ is a contraction with respect to $d$ means exactly that there is one constant $c\in[0,1)$ such that the distance between the images is bounded by $c$ times the distance between the original points. In symbols, this is \begin{align*} d(f(x),f(y)) \le c\,d(x,y) \end{align*} for every pair $x,y\in X$. The important point is that the same constant $c$ must work uniformly for all pairs of points. [/guided] [/step] [step:Substitute the norm formula for the induced metric] Since $d$ is the metric induced by the norm $\|\cdot\|_X$, its defining formula gives \begin{align*} d(f(x),f(y)) = \|f(x)-f(y)\|_X \end{align*} and \begin{align*} d(x,y)=\|x-y\|_X \end{align*} for all $x,y\in X$. Therefore the contraction inequality \begin{align*} d(f(x),f(y)) \le c\,d(x,y) \end{align*} is exactly the inequality \begin{align*} \|f(x)-f(y)\|_X \le c\|x-y\|_X \end{align*} for all $x,y\in X$. [/step] [step:Conclude the two formulations are equivalent] If $f$ is a contraction with respect to $d$, the first step gives a constant $c\in[0,1)$ satisfying the metric contraction inequality, and the second step converts it into \begin{align*} \|f(x)-f(y)\|_X \le c\|x-y\|_X \end{align*} for all $x,y\in X$. Conversely, if such a constant $c\in[0,1)$ satisfies the displayed norm inequality for all $x,y\in X$, then substituting $\|u-v\|_X=d(u,v)$ for $u,v\in X$ gives \begin{align*} d(f(x),f(y)) \le c\,d(x,y) \end{align*} for all $x,y\in X$. Hence $f$ is a contraction with respect to $d$. This proves the equivalence. [/step]