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, such that 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 strictly convex.


So I wonder, if the following statement is true: Let $X$ be a Banach space, such that 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](https://mathoverflow.net/questions/81979/projection-exists-uniformly-convex#comment210420_81979). So I should ask: Let $X$ be a Banach space, such that 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.



Projection exists ⇒ Uniformly convex?
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.


