Suppose is a closed subset with the property that for all there exists a unique point such that . Then the map is continuous.

What is the most general setting where this statement holds? Must be convex in this new setting?

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# Pseudo-convexity

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4 thoughts on “Pseudo-convexity”

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Suppose is a closed subset with the property that for all there exists a unique point such that . Then the map is continuous.

What is the most general setting where this statement holds? Must be convex in this new setting?

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The following argument should work for (and more generally for spaces where closed balls are compact?).

Let and call . We must show that .

Since is cauchy, the set is bounded and thus there is a subsequence of the sequence such that for some .

Obviously we have . But also . Notice that , which is a continuous function. Then taking the limit on we get . And thus , and by uniqueness . Now notice that exactly the same argument works for any subsequence of , and thus any subsequence of has a subsequence with as its limit. This settles the result.

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Awesome

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Minor remark: all closed balls are compact iff the space has the Heine-Borel property.

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Apparently this is a well-studied problem (see this): sets with this property are called Chebsyhev sets. A brief summary of some results:

A Chebyshev set such that the map is continuous (this condition is called ‘having continuous metric projection’ in the thesis linked above) in a Banach space with strictly convex dual space (for instance, any Hilbert space or any space with finite ) is convex.

One can guarantee that the map is continuous adding some technical hypotheses on both the Chebyshev set and the ambient space (see for instance Theorem 2.6.1).

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