# Projection exists => Uniformly convex?

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,

• 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. Nov 27, 2011 at 9:44
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.