[proofplan]
We prove that a pointwise contraction estimate is equivalent to the same estimate on all extended diameters. The forward implication follows by bounding every pairwise distance inside $f(A)$ and then taking the extended supremum, with the case $c=0$ handled separately because of the convention $0\cdot\infty=0$. The reverse implication is obtained by applying the diameter inequality to the two-point set $\{x,y\}$ and computing the two relevant diameters exactly.
[/proofplan]
[step:Derive the image diameter estimate from the pointwise contraction estimate]
Assume that for every $x,y \in X$,
\begin{align*}
d(f(x),f(y)) \le c\,d(x,y).
\end{align*}
Let $A \subset X$ be nonempty.
First suppose $c=0$. For every $x,y \in A$, the contraction estimate gives
\begin{align*}
d(f(x),f(y)) \le 0.
\end{align*}
Since distances in a [metric space](/page/Metric%20Space) are nonnegative, $d(f(x),f(y))=0$ for all $x,y \in A$. Hence every pairwise distance in $f(A)$ is $0$, so
\begin{align*}
\operatorname{diam}(f(A))=0.
\end{align*}
By the stated convention, $0\cdot \operatorname{diam}(A)=0$, including the case $\operatorname{diam}(A)=\infty$. Therefore
\begin{align*}
\operatorname{diam}(f(A)) \le 0\cdot \operatorname{diam}(A).
\end{align*}
Now suppose $0<c<1$. For every $x,y \in A$, the definition of $\operatorname{diam}(A)$ gives
\begin{align*}
d(x,y) \le \operatorname{diam}(A).
\end{align*}
Multiplying this extended-real inequality by the positive scalar $c$ preserves the order under the stated convention. Combining it with the pointwise contraction estimate yields
\begin{align*}
d(f(x),f(y)) \le c\,\operatorname{diam}(A)
\end{align*}
for every $x,y \in A$. Since every element of $f(A)$ has the form $f(x)$ for some $x \in A$, every pairwise distance in $f(A)$ is bounded above by $c\,\operatorname{diam}(A)$. Taking the extended supremum over all such pairwise distances gives
\begin{align*}
\operatorname{diam}(f(A)) \le c\,\operatorname{diam}(A).
\end{align*}
[guided]
Assume that $f:X \to X$ satisfies the pointwise estimate
\begin{align*}
d(f(x),f(y)) \le c\,d(x,y)
\end{align*}
for every $x,y \in X$. We want to prove that the same constant controls the diameter of the image of an arbitrary nonempty subset. Fix a nonempty subset $A \subset X$.
The delicate case is $c=0$, because the statement has explicitly chosen the convention $0\cdot\infty=0$. Under the hypothesis with $c=0$, for every $x,y \in A$ we have
\begin{align*}
d(f(x),f(y)) \le 0.
\end{align*}
A metric takes values in $[0,\infty)$, so the same quantity also satisfies $0 \le d(f(x),f(y))$. Hence
\begin{align*}
d(f(x),f(y))=0
\end{align*}
for all $x,y \in A$. Thus all pairwise distances inside $f(A)$ are equal to $0$, and the supremum of those distances is
\begin{align*}
\operatorname{diam}(f(A))=0.
\end{align*}
The convention in the theorem says that $0\cdot\operatorname{diam}(A)=0$ even if $\operatorname{diam}(A)=\infty$. Therefore
\begin{align*}
\operatorname{diam}(f(A)) \le 0\cdot\operatorname{diam}(A).
\end{align*}
It remains to handle $0<c<1$. For any two points $x,y \in A$, the number $d(x,y)$ belongs to the set whose extended supremum defines $\operatorname{diam}(A)$. Therefore
\begin{align*}
d(x,y) \le \operatorname{diam}(A).
\end{align*}
Because $c>0$, multiplying by $c$ preserves the order in the extended nonnegative real line, with the convention $c\infty=\infty$. Thus
\begin{align*}
c\,d(x,y) \le c\,\operatorname{diam}(A).
\end{align*}
Combining this inequality with the contraction estimate gives
\begin{align*}
d(f(x),f(y)) \le c\,\operatorname{diam}(A).
\end{align*}
This holds for every pair $x,y \in A$. Since the points of $f(A)$ are exactly the points $f(x)$ with $x \in A$, the displayed inequality bounds every pairwise distance occurring in $f(A)$. Taking the extended supremum of those pairwise distances gives
\begin{align*}
\operatorname{diam}(f(A)) \le c\,\operatorname{diam}(A).
\end{align*}
[/guided]
[/step]
[step:Recover the pointwise contraction estimate from two-point subsets]
Assume that for every nonempty subset $A \subset X$,
\begin{align*}
\operatorname{diam}(f(A)) \le c\,\operatorname{diam}(A).
\end{align*}
Let $x,y \in X$, and define the nonempty subset $A=\{x,y\}\subset X$.
If $x=y$, then
\begin{align*}
d(f(x),f(y))=0 \le c\,d(x,y).
\end{align*}
Assume now that $x \ne y$. The pairwise distances among points of $A$ are $0$ and $d(x,y)$, so
\begin{align*}
\operatorname{diam}(A)=d(x,y).
\end{align*}
Similarly, the pairwise distances among points of $f(A)=\{f(x),f(y)\}$ are $0$ and $d(f(x),f(y))$, allowing the possibility that $f(x)=f(y)$; hence
\begin{align*}
\operatorname{diam}(f(A))=d(f(x),f(y)).
\end{align*}
Applying the assumed diameter inequality to this set $A$ gives
\begin{align*}
d(f(x),f(y))=\operatorname{diam}(f(A)) \le c\,\operatorname{diam}(A)=c\,d(x,y).
\end{align*}
Since $x,y \in X$ were arbitrary, the pointwise contraction estimate holds for all $x,y \in X$.
[/step]
[step:Conclude the equivalence]
The first step proves that the pointwise estimate implies the diameter estimate for every nonempty subset of $X$. The second step proves that the diameter estimate for every nonempty subset of $X$ implies the pointwise estimate. Therefore the two stated conditions are equivalent.
[/step]
