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In a remark to the projection theorem for Hilbert spaces I read this conjecture of a more general projection theorem:

Let $X$ be a reflexive Banach space and $K\subset X$ closed and convex. Then for every $x\in X$ there exists $y\in K$ such that $$\| x-y\|=d(x,K)=\inf_{z\in K} \|x-z\|$$

Now I tried showing this similarly to the projection theorem for Hilbert spaces, but this didn't get me far as the proof I used makes use of the parallelogram equation heavily, which does not hold in Banach spaces. So how can I show this?

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  • $\begingroup$ Martin's answer here may help. $\endgroup$ – David Mitra Jun 26 '14 at 19:12
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Recall that:

  1. Since $X$ is reflexive, any closed ball is compact in the weak topology.
  2. Every convex closed subset of $X$ (in particular $K$) is closed in the weak topology

Fix any point $w \in K$. If $||x-w|| \leq d(x, K)$, we are ok.

Otherwise consider $K' = \{ y \in K : ||x-y|| \leq ||x-w|| \} = K \cap \bar B(x, ||x-w||)$.

$K'$ is convex and closed since it is the intersection of two convex and closed sets, so it is closed for the weak topology. Moreover it is compact in the weak topology, since it is a closed subset of $\bar B(x, ||x-w||)$ which is compact.

Now, if $X$ is endowed with the weak topology, the map $y \mapsto ||x-y||$ is lower semi-continuous, hence it has a minimum in $K'$, because $K'$ is compact. But clearly a minimum in $K'$ is a minimum in $K$.

So we showed that there exists $y \in K$ such that for every $z \in K$ holds $||x-y|| \leq ||x-z||$.

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  • 1
    $\begingroup$ thanks, I totally forgot about using weak topology. Thanks! $\endgroup$ – dinosaur Jun 27 '14 at 13:07

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