1
$\begingroup$

Let $(E,\|.\|)$ be a real normed vector space and $d$ be the distance associated to the norm. I am wondering if there exists a characterization of the subsets $F$ of $E$ such that for all $x\in E$, there exists $y\in F$ satisfying $d(x,y)=d(x,F)$. If $E$ is finite dimensional, these sets are exactly the closed sets, but in the infinite dimensional case, the closedness is only necessary. Are there additional constraints which give $F$ this property?

$\endgroup$
  • 1
    $\begingroup$ Compactness is sufficient. If the norm is nice enough, convexity (in addition to closedness, of course) suffices. $\endgroup$ – Daniel Fischer Oct 4 '13 at 15:35
1
$\begingroup$

In addition to compactness as @DanielFischer has pointed out, there is another interesting condition (due to Phelps - see this link) :

Let $M$ be a subspace of a normed linear space $E$, and $M^{\perp}$ denote its annihilator in the dual space $E^{\ast}$. Then every linear functional on $M$ has a unique norm-preserving extension to $E$ iff for every $f\in E^{\ast}$, there is a unique best approximation $g\in M^{\perp}$ such that $\|f-g\| = d(f,M^{\perp})$

I am not sure if this of use to you, but I thought this might make for interesting reading.

$\endgroup$
  • $\begingroup$ Ok, thank for the reference! $\endgroup$ – TerranDrop Oct 7 '13 at 12:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.