Assume is locally Lipschitz on
, that is, for any
, there exists
(depending on
) such that
for all
.
Then, for any compact set , there exists a constant
(depending on
) such that
for all
. That is,
is Lipschitz on
.
Proof
Suppose to the contrary is not Lipschitz on
, so that for all
, there exists
such that
Then there exists two sequences such that
Since is locally Lipschitz implies
is continuous, so
is bounded on
by Extreme Value Theorem. Hence
.
By sequential compactness of , there exists a convergent subsequence
, and thus
.
Then for any , there exists
such that
but
which contradicts that
is locally Lipschitz.