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Let $B$ be a uniformly convex Banach space. Let $K \subset B$ be a closed convex subset.

Edit Thank you for your helpful comments.

I am trying to show that if $K$ is a closed convex subset of $B$ and $b_0$ is any point in $B$ then there exists a unique point $k_0$ in $K$ such that $\|k_0 - b_0 \| = \inf_{k \in K} \|k - b_0 \|$.

With your comments I've managed to show uniqueness. Now for existence, after shifting $K$ and $b_0$ by $-b_0$ and scaling the whole space by $\frac{1}{ \inf_{k \in K} \|k\|}$, to get $b_0=0$ and $\inf_{k \in K} \|k\| = 1$, I'm wondering why I can't argue as follows:

Since $\inf_{k \in K} \|k\| = 1$, there is a sequence in $K$ converging to a point $b$ in $B$ with $\|b\|=1$. But $K$ is closed so $b \in K$.

Of course this has to be wrong, since it doesn't use uniform convexity of $B$. Perhaps where I claim that there is a sequence in $K$ converging to $b$ but I don't see why this is wrong.

Thanks for your help!

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    $\begingroup$ Sorry: what are you trying to show? There's something missing here (you're probably assuming that $0 \notin K$ but that's still not enough). Note also that you haven't used uniform convexity so far (maybe you should state the definition and do some more work from there). $\endgroup$ – t.b. Jul 30 '12 at 14:35
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    $\begingroup$ Is $k$ meant to be a point of minimal distance to some point outside $K$? If you take $K$ to be the closed unit ball, then obviously there are lots of points with unit norm. $\endgroup$ – copper.hat Jul 30 '12 at 15:00
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    $\begingroup$ Then you should add that to your question because as it stands it makes no sense. And I really think you should try to do that yourself before asking it here (take a sequence approximating the infimum and use uniform convexity to show that it is a Cauchy sequence). It's essential that you work in a complete space here: so (banach-spaces), not (normed-spaces). $\endgroup$ – t.b. Jul 30 '12 at 16:15
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    $\begingroup$ Once you've shown a minimum exists, take two minima $k_0$, $k_1$. Then your argument will also show that the sequence given by $k_{2n} = k_0$ and $k_{2n+1} = k_1$ is Cauchy, hence $k_0 = k_1$. $\endgroup$ – t.b. Jul 30 '12 at 16:36
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    $\begingroup$ You can of course apply uniform convexity directly: If you have two minima $k_0$ and $k_1$ then, assuming $\varepsilon = \lVert k_0 - k_1\rVert \gt 0$, there is $\delta \gt 0$ such that $\lVert\frac{k_0+k_1}{2}\rVert \lt 1-\delta$ but by convexity of $K$ we have $(k_0 - k_1)/2 \in K$, so $\lVert (k_0 - k_1) \rVert \geq 1$, a contradiction. $\endgroup$ – t.b. Jul 30 '12 at 18:33
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The problem of the proof is that although you can find a minimizing sequence $k_n \in K$ such that $||k_n|| \to 1$ as $n \to \infty$, it is not true to claim that such sequence is converging to some specific point.

Hence uniform convexity is necessary in the proof for existence. By uniform convexity one can show that such sequence $k_n$ is in fact a Cauchy sequence in the Banach space $B$. Therefore we can safely conclude the convergence of $k_n$.

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