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Hello,

I know that: Let X be a uniformly convex Banach-Space, $a\in X$ and $C\subset X$ closed and convex, then there is a unique $b\in C$ with $\left\Vert a-b\right\Vert=\inf_{x\in C}\left\Vert a-x \right\Vert$.

Moreover I know that: Let X be a Banach-Space, s.t. for every $a\in X$ and $C\subset X$ closed and convex there is at most one $b\in C$ with $\left\Vert a-b\right\Vert=\inf_{x\in C}\left\Vert a-x \right\Vert$. Then X is striclty convex.

So I wonder, if the following statement is true: Let X be a Banach-Space, s.t. for every $a\in X$ and $C\subset X$ closed and convex, there is a unique $b\in C$ with $\left\Vert a-b\right\Vert=\inf_{x\in C}\left\Vert a-x \right\Vert$. Then X is uniformly convex.

EDIT: This statement is false, see Hsueh-Yung Lin's comment. So I should ask: Let X be a Banach-Space, s.t. for every $a\in X$ and $C\subset X$ closed and convex, there is a unique $b\in C$ with $\left\Vert a-b\right\Vert=\inf_{x\in C}\left\Vert a-x \right\Vert$. Then every bounded sequence has a weakly convergent subsequence.

Best regards,

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  • $\begingroup$ Well, take a finite-dimensional strictly convex (but not uniformly convex Banach space). Then compactness will ensure the existence of a point of best approximation. For an infinite-dimensional example, use a reflexive strictly (but not uniformly) convex space and argue via weak compactness. I'm not convinced this is a great question for this site: math.stackexchange.com might be better. $\endgroup$ – Matthew Daws Nov 27 '11 at 9:44
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    $\begingroup$ The following article might help you: ams.org/journals/bull/1941-47-04/S0002-9904-1941-07451-3/… $\endgroup$ – HYL Nov 27 '11 at 10:27
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    $\begingroup$ @Matthew: a compactness argument actually shows that a finite dimensional strictly convex Banach space is uniformly convex. $\endgroup$ – Mark Meckes Nov 27 '11 at 12:32
  • $\begingroup$ @Hsueh-Yung Lin: interesting paper. M. Day provides a example for a reflexive strictly convex space, which is not isomorphic to a uniformly convex space. So we get the unique best approximation, because every bounded sequence admits a weakly-convergent subsequence, so I have to modify my question. Is it true, that, if every closed convex set admits a best approximation, then every bounded sequence admits a weakly-convergent subsequence. $\endgroup$ – Thomas Kuhn Nov 27 '11 at 17:23
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    $\begingroup$ @Mark: Ah, yes! Very silly... $\endgroup$ – Matthew Daws Nov 27 '11 at 18:40
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Your modified question has an affirmative answer. An equivalent form, in view of the Eberlein-Smulian theorem, is whether the Banach space $X$ must be reflexive if every closed bounded non empty set admits best approximations. If $X$ is not reflexive, then by R. C. James' famous characterization of reflexivity, there is a norm one linear functional $F$ on $X$ s.t. $F$ does not achieve its norm on the closed unit ball $B_X$. Let $C:= [F=1]\cap 2B_X$. This closed bounded non empty convex set contains no point of minimal norm.

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