# Pseudo-convexity

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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1. gciruelos says:

Awesome

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

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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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