# Is the Hausdorff semi-distance Lipschitz?

Let $X$ be Banach (with metric $d$) and let $H(X)$ be the set of closed bounded subsets of $X$. Define for $A,B\in H(X)$ $$\delta(A,B)=\sup_{a\in A}\inf_{b\in B}d(a,b)$$ be the Hausdorff semi-distance and $d_H=\max(\delta(A,B), \delta(B,A))$ denote that Hausdorff metric. The metric space $(H(X),d_H)$ is complete. How can I show that the map $$H(X)\xrightarrow{f_A}\mathbb{R}\colon f_A(B)=\delta(A,B)$$ is Lipschitz for a fixed $A\in H(X)$?

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Your definition of the "Hausdorff metric" does not give a metric. $\:$ –  Ricky Demer Mar 22 '12 at 20:29
Try using Didier's answer here –  t.b. Mar 22 '12 at 20:29