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In Hilbert space every non empty, closed, convex subset contains a unique element of smallest norm. Why is that true also in Banach space which is uniformly convex?

(normed space which is uniformly convex is a space in which for all sequences $\{x_n\}$, $\{y_n\}$ s.t $||x_n||,||y_n||\leq 1$ exists: if $\lim_n||x_n+y_n||=2$ then $\lim_n||x_n-y_n||=0$.)

I thought of defining $a=\inf\{||x|| : x\in X\}$, so exists a sequence $\{x_n\}$ s.t $\lim_n||x_n||=\inf||x_n||=a$ and then to show that $\{x_n\}$ is a Cauchy sequence and it's limit is the element we are looking for, but i didn't manage to prove that it is Cauchy.

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That's what you do... Here's a link to a complete proof: <a>… – David Mitra Nov 9 '11 at 19:27
The idea is that uniformly convex implies that if $\|x_n\|$ and $\|x_m\|$ are close enough to $a$ and $\|(x_n + x_m)/2\| \ge a$, $\|x_n - x_m\|$ must be close to 0. – Robert Israel Nov 9 '11 at 19:41

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