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Let $(E, | \cdot|)$ be a Banach space and $F$ a closed subset of $E$. Fix $a \in E$. In general, there does not necessarily exist $b \in F$ such that $$ |b-a| = \inf_{x\in F}|x-a|. $$

Are there some conditions on the set $F$ to guarantee that such projection exists for every $a\in E$?

Update: I have recently solved this question. It suggests that in case of the dual space $E'$, if $M \subseteq E'$ is closed in the weak$^\star$ topology $\sigma(E', E)$, then the projection into $M$ exists for any $f\in E'$.

Analyst
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    If $F$ is compact then $b$ exists necessarily. – Kavi Rama Murthy Feb 22 '22 at 11:30
  • @KaviRamaMurthy Fix $a\in E$, then the map $x \mapsto |x-a|$ is continuous and thus attains both minimum and maximum on the compact set $F$. Compactness is usually a strong condition. Are there weaker ones? – Analyst Feb 22 '22 at 11:46
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    There is no general weaker condition that I can think of. If $E$ is finite dimensional then any closed set will do. – Kavi Rama Murthy Feb 22 '22 at 11:50
  • @KaviRamaMurthy Thank you so much for your help! – Analyst Feb 22 '22 at 11:51
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    weak compactness should work as well, as the norm is weakly lower semicontinuous – daw Feb 22 '22 at 13:54
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    Such sets that admit best approximations are often called proximinal. It can be shown that if $E$ is reflexive, and $F$ is closed and convex then $F$ is proximinal. Furthermore, if $E$ is reflexive and strictly convex, then there exists exactly one best approximation. In that case, $F$ is said to be Chebyshev, and the (nonlinear) map that sends each $x \in E$ to its best approximation in $F$ is called the metric projection of $F$. – Evangelopoulos Foivos Feb 23 '22 at 11:48
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    @EvangelopoulosF. Could you post your comment as an answer so that I can accept it? – Analyst Feb 23 '22 at 12:25

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Such sets that admit best approximations are often called proximinal. It can be shown that if $E$ is reflexive, and $F$ is closed and convex then $F$ is proximinal. (The converse is also true; namely, if every closed and convex subset of $E$ is proximinal then $E$ is reflexive). Furthermore, if $E$ is reflexive and strictly convex, then there exists exactly one best approximation. In that case, $F$ is said to be Chebyshev. (In fact, $E$ is strictly convex and reflexive iff each of its closed and convex subsets is Chebyshev). The (nonlinear) map that sends each $x \in E$ to its best approximation in $F$ is called the metric projection of $F$.

Evangelopoulos Foivos
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