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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,

1

# 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.

Best regards,