Let $(X,d)$ be a metric space, define $F(X)$ as the set of all non-empty compact subsets of $X$, for $A,B \in F(X)$ we define $$d(A,B) = \sup_{a \in A} \, \inf_{b \in B} d(a,b)$$ We now define the Hausdorff-metric $h$ as $h(A,B) = \max \{ d(A,B),d(B,A) \}$. For any function $f:X \rightarrow X$ we define the function $f:F(X) \rightarrow F(X):A \mapsto f(A)$. It's not too hard to show that if $f$ is a contraction on $X$, then the corresponding function on $F(X)$ is also a contraction with the same Lipschitz constant. Now if $w_1, w_2, ..., w_k$ are all contractions on $X$ with Lipschitz constants $s_1,s_2,...,s_k$, show that $W:F(X) \rightarrow F(X):A \mapsto \bigcup_{n=1}^{k} w_{n}(A)$ is also a contraction with Lipschitz constant $s = \max s_{n}$.
I've been able to show that $W$ is a contraction whose Lipschitz constant is at most $s$, but I have failed to show that it actually equals $s$. How can you prove this? Is it even true?
Several articles claim it is the case, such as https://zaguan.unizar.es/record/108540/files/texto_completo.pdf, but none seem to have a proof. Hutchinson's original article (https://maths-people.anu.edu.au/~john/Assets/Research%20Papers/fractals_self-similarity.pdf) only proves the inequality and doesn't even mention that the Lipschitz constant equals $s$.