# Construct locally lipschitz map from a bounded one

Let $X$ be a Banach space and $BC(X)$ the space of all bounded closed subsets in $X$. It can be shown that $(BC(X),d_H)$ is a complete metric space (see this page for a definition of $d_H$). If $f:X\to X$ bounded. Does this imply that $F:BC(X)\to BC(X)$ given by $F(A)=\overline{\{f(a):a\in A\}}$ is locally Lipschitz continuous?

And if the map $f: X\to X$ is locally Lipschitz countinuous, is it automaticaly bounded?

• Does this hold for $f:\mathbb R\to\mathbb R$ given by $f(x)=\sqrt{x}$, which has arbitrarily large derivative near $0$? – Alex Becker May 10 '12 at 17:28

Take $X=\mathbb R$ and $f$ any continuous function on $\mathbb R$. Consider $A$, $B$ given by single points $A= \{a\}$, $B=\{b\}$. Then $F(A) = \{f(a)\}$, $F(B) = \{f(b)\}$, and $d_H(F(A),F(B)) = |f(a) - f(b)|$ while $d_H(A,B) = |a-b|$. So if $F$ is locally Lipschitz, $f$ must also be locally Lipschitz; if $F$ is bounded, $f$ must also be bounded. Any $f$ that is bounded but not locally Lipschitz is a counterexample to your first question. Any $f$ that is locally Lipschitz but not bounded is a counterexample to your second question (assuming the "it" there refers to $F$).