[proofplan] We prove the uniform $\varepsilon$-$\delta$ condition directly from the Lipschitz estimate. The only point requiring separation is the case $c=0$, since the usual choice $\delta=\varepsilon/c$ is then unavailable. When $c>0$, the Lipschitz inequality converts the bound $d_X(x,y)<\delta$ into the required bound on $d_Y(f(x),f(y))$; when $c=0$, the map has zero distance between all image values. The contraction case follows because a contraction is, by definition, a map satisfying such a Lipschitz estimate with contraction constant less than $1$. [/proofplan]

[step:Choose a uniform radius for each $\varepsilon>0$]Let $\varepsilon>0$ be arbitrary. If $c>0$, define $\delta=\varepsilon/c$; if $c=0$, define $\delta=1$. In either case, $\delta>0$.[/step]

[guided]To prove [uniform continuity](/page/Uniform%20Continuity), we must show that for every tolerance $\varepsilon>0$ in the codomain metric $d_Y$, there is a single radius $\delta>0$ in the domain metric $d_X$ that works for all pairs of points $x,y \in X$. Fix such an arbitrary $\varepsilon>0$. The Lipschitz estimate in the hypothesis says that distances after applying $f$ are bounded by $c$ times the original distance. If $c>0$, the natural way to make $c\,d_X(x,y)$ smaller than $\varepsilon$ is to force $d_X(x,y)<\varepsilon/c$. Thus in this case we set \begin{align*} \delta=\varepsilon/c. \end{align*} This is positive because both $\varepsilon>0$ and $c>0$. If $c=0$, division by $c$ is not available. In that case the Lipschitz estimate is even stronger: it will force all image distances to be zero. Therefore any positive radius is sufficient, and we choose \begin{align*} \delta=1. \end{align*} This gives a positive number $\delta$ in every case.[/guided]

[step:Verify the uniform continuity condition using the Lipschitz bound]Let $x,y \in X$ satisfy $d_X(x,y)<\delta$. If $c>0$, then by the assumed Lipschitz estimate and the definition of $\delta$, \begin{align*} d_Y(f(x),f(y)) \le c\,d_X(x,y) < c\delta = \varepsilon. \end{align*} If $c=0$, then the Lipschitz estimate gives \begin{align*} d_Y(f(x),f(y)) \le 0\cdot d_X(x,y)=0. \end{align*} Since $d_Y$ is a metric, $d_Y(f(x),f(y))\ge 0$, hence \begin{align*} d_Y(f(x),f(y))=0<\varepsilon. \end{align*} Thus for every $\varepsilon>0$ there exists $\delta>0$ such that, for all $x,y \in X$, \begin{align*} d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\varepsilon. \end{align*} This is exactly uniform continuity of $f:X\to Y$.[/step]

[guided]Now take arbitrary points $x,y \in X$ satisfying \begin{align*} d_X(x,y)<\delta. \end{align*} We must prove \begin{align*} d_Y(f(x),f(y))<\varepsilon. \end{align*} First suppose $c>0$. The hypothesis applies to this particular pair $x,y \in X$, so \begin{align*} d_Y(f(x),f(y)) \le c\,d_X(x,y). \end{align*} Because $d_X(x,y)<\delta$ and $c>0$, multiplying the strict inequality by $c$ preserves its direction: \begin{align*} c\,d_X(x,y)<c\delta. \end{align*} By the definition $\delta=\varepsilon/c$, we have \begin{align*} c\delta=c(\varepsilon/c)=\varepsilon. \end{align*} Combining these inequalities gives \begin{align*} d_Y(f(x),f(y))<\varepsilon. \end{align*} Now suppose $c=0$. The same Lipschitz estimate gives \begin{align*} d_Y(f(x),f(y)) \le 0\cdot d_X(x,y)=0. \end{align*} Every metric takes nonnegative values, so \begin{align*} d_Y(f(x),f(y))\ge 0. \end{align*} Therefore \begin{align*} d_Y(f(x),f(y))=0. \end{align*} Since $\varepsilon>0$, this implies \begin{align*} d_Y(f(x),f(y))<\varepsilon. \end{align*} In both cases, the same $\delta$ works for all choices of $x,y \in X$ satisfying $d_X(x,y)<\delta$. This proves the uniform continuity condition: \begin{align*} \forall \varepsilon>0\ \exists \delta>0\ \forall x,y\in X,\quad d_X(x,y)<\delta \implies d_Y(f(x),f(y))<\varepsilon. \end{align*} Hence $f:X\to Y$ is uniformly continuous.[/guided]

[step:Apply the result to contraction mappings]Let $(X,d_X)$ be a [metric space](/page/Metric%20Space) and let $f:X\to X$ be a [contraction mapping](/page/Contraction%20Mapping). By the definition of a contraction mapping, there exists a constant $c\in[0,1)$ such that, for all $x,y\in X$, \begin{align*} d_X(f(x),f(y))\le c\,d_X(x,y). \end{align*} Since $c\ge 0$, the first part of the theorem applies with $Y=X$ and $d_Y=d_X$. Therefore $f$ is uniformly continuous.[/step]

[guided]Let $(X,d_X)$ be a metric space and let $f:X\to X$ be a contraction mapping. The definition of contraction mapping supplies a constant $c\in[0,1)$ such that every pair $x,y\in X$ satisfies \begin{align*} d_X(f(x),f(y))\le c\,d_X(x,y). \end{align*} This is exactly the Lipschitz hypothesis from the first part of the theorem, with the codomain metric equal to the domain metric: take $Y=X$ and $d_Y=d_X$. The interval condition $c\in[0,1)$ also gives $c\ge 0$, so all hypotheses of the first part are satisfied. Therefore the first part implies that $f:X\to X$ is uniformly continuous.[/guided]