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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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    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).2012-07-30
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    @t.b. Hm... thanks. Let me rephrase. Need to check what I'm missing.2012-07-30
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    @t.b. Actually no. It's getting late here, I'll do this tomorrow.2012-07-30
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    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.2012-07-30
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    @t.b. What I want to show is that if the minimal distance to zero of every point in the set is one then there is only one point in the set that has norm one. Basically, what copper.hat said. So I get $\geq$ for free.2012-07-30
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    @copper.hat Yes thanks. I found out already : )2012-07-30
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    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).2012-07-30
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    @t.b. Thanks for the hint, I'm already sitting here trying to do exactly that. : )2012-07-30
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    (Once I have it I was planning to correct the question and post my own answer)2012-07-30
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    @t.b. Actually, the argument you're outlining shows existence but I'm still stuck trying to do uniqueness.2012-07-30
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    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 $k_0 = k_1$.2012-07-30
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    @t.b. Oh that is so obvious. The way it's written in the notes is misleading. Thanks a lot! (they seriously write "By uniform convexity this implies that $w_1 = w_2$." and I've been trying to see how.)2012-07-30
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    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.2012-07-30
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    @t.b. Thanks a lot. I'll look at this again tomorrow, I'm not doing anything anymore for today. (ok, maybe except for sitting around in chat)2012-07-30
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    You should fix the problem statement...2012-07-30
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    @t.b. You're lovely. With this comment you saved me a few hours of frustration. Thanks.2012-07-31
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    @t.b. And I think your comment with "And I really think you should" prevents anyone from answering. : ( But don't delete it please, because it contains information.2012-08-01
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    See [here](http://books.google.com/books?id=5IAWL8MUGtQC&pg=PA28) for a solution.2012-08-01

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