Suppose F\subseteq \Bbb{R} is a closed subset with the property that for all x\in \Bbb{R} there exists a unique point y(x)\in F such that d(x,F)=d(x,y(x)). Then the map y:\Bbb{R}\to F is continuous.

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


4 thoughts on “Pseudo-convexity

  1. The following argument should work for \mathbb{R}^n (and more generally for spaces where closed balls are compact?).

    Let x_n \to x and call y_n := y(x_n). We must show that y_n \to y = y(x).
    Since x_n is cauchy, the set \{y_n\} is bounded and thus there is a subsequence y_k of the sequence y_n such that y_k \to y' for some y' \in F.
    Obviously we have d(x,F) \leq d(x,y'). But also d(x,y') \leq d(x,x_k) + d(x_k, y_k) + d(y_k, y'). Notice that d(x_k,y_k) = d(x_k,F), which is a continuous function. Then taking the limit on k we get d(x,y')\leq d(x,F). And thus d(x,y') = d(x,F), and by uniqueness y' = y. Now notice that exactly the same argument works for any subsequence of y_n, and thus any subsequence of y_n has a subsequence with y as its limit. This settles the result.

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

    A Chebyshev set such that the map y 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 L_p space with finite p) is convex.

    One can guarantee that the map y 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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