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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). – t.b. Jul 30 '12 at 14:35
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. – copper.hat Jul 30 '12 at 15:00
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). – t.b. Jul 30 '12 at 16:15
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 $k_0 = k_1$. – t.b. Jul 30 '12 at 16:36
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. – t.b. Jul 30 '12 at 18:33

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