Question about Lipschitz continuity

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SUMMARY

The discussion centers on proving that if \( f: X \rightarrow X \) is locally Lipschitz continuous on a normed linear space \( X \) and \( K \) is a compact set, then there exists an open set \( U \) containing \( K \) where \( f \) is Lipschitz continuous. The approach involves taking an open cover of \( K \) with sets where \( f \) is Lipschitz, extracting a finite subcover, and ensuring the resulting open set is convex. The solution confirms that replacing \( K \) with its convex hull maintains compactness, allowing for the construction of an appropriate open set.

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quasar987
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Homework Statement


This should be easy but I'm stomped.

Let K be a compact set in a normed linear space X and let f:X-->X be locally Lipschitz continuous on X. Show that there is an open set U containing K on which f is Lipschitz continuous.

Homework Equations


locally Lipschitz means that for every x in X, there is a nbdh around x on which f is Lipschitz.

The Attempt at a Solution


The obvious thing to do it seems it take an open cover of K by sets on which f is Lipschitz continuous and extract a finite subcover. But then what?!?
 
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Wouldn't it be nice if you could make sure that the final open set containing K and covered by the subcover, were convex? Replace K by it's convex hull, it's still compact, right? Then take the finite subcover. Then pick a open set just a 'little' bigger than K, but still convex and contained in the subcover. That works, doesn't it?
 
Last edited:
Of course, yes! Thanks Dick.
 

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