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Homework Statement
Let M and N be two metric spaces. Let f:M \to N. Prove that a function that is locally Lipschitz on a compact subset W of a metric space M is Lipschitz on W.
A similar question was asked here
https://www.physicsforums.com/showthread.php?t=325759&highlight=locally+lipschitz+compact,
but it didn't really address my question.
Homework Equations
Definition: a function f:M \to N is said to be Lipschitz on a set S if there exists a positive constant L \in \Re^+ such that \forall x,y \in S, d_N(f(x),f(y)) \leq L*d_M(x,y)
Definition: a function f:M \to N is said to be locally Lipschitz on a set S if for every point x_i \in S, there exists an open ball B_{r_i}(x_i) of radius r_i centered at x_i such that f is Lipschitz on B_{r_i}(x_i) with Lipschitz constant L_i.
The Attempt at a Solution
Because we are given that f is locally Lipschitz, we know that \forall x_i \in W: \exists r_i \in \Re^+: f is Lipschitz on B_{r_i}(x_i). The set \{B_{r_i}(x_i) | x_i \in W \} is an open cover of W. Since W is compact, we can extract a finite subcover so that (after a possible re-ordering of indices of balls) W \subseteq \bigcup_{i=1}^{N}B_{r_i}(x_i).
Now, for any two x,y \in W there are two possibilities:
i) \exists i \in \{1,...,N\}: x,y \in B_{r_i}(x_i). In this case, we have d_N(f(x),f(y)) \leq L_i*d_M(x,y). If this were true for all pairs of points, I could simply choose a Lipschitz constant for W to be \max_i(L_i).
ii) \forall i \in \{1,...,N\}: x, y are not in the same ball B_{r_i}(x_i). In this case, I'm really stuck regarding what to do. I have tried using the triangle inequality with little success.
This actually isn't a homework assignment. The ODE textbook I'm using stated this fact, but relegated its proof to the exercises. Any help would be appreciated
