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Let $X$ be a uniformly convex Banach space, $x\in X$ and $C\subset X$ closed and convex, then there is a unique $y\in C$ with $$\lVert x-y\rVert=\inf_{z\in C}\lVert x-z \rVert.$$

Is there a good book where I can find a proof of this theorem?

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Theorem 8.2.2 of Larsen, "Functional Analysis", Marcel Dekker, 1973

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This is Corollary 5.2.17 in the book "An introduction to Banach space theory" by Megginson, where the term "uniformly rotund" is used instead of "uniformly convex".

For completeness, I sketch the proof.

  1. We may assume $x=0$, by translation.
  2. Let $d = \|y_n\|\to\inf \{\|y\| : y\in C\}$. We may replace $C$ with $\{y\in C: \|y\|\le 2d\}$ without changing the problem. This makes $C$ bounded.
  3. Take a sequence $y_n $ such that $\|y_n\|\to d$.
  4. Uniformly convex spaces are reflexive. Therefore, $C$ is weakly compact by Alaoglu's theorem.
  5. Replace $\{y_n\}$ by a weakly convergent subsequence, and let $y$ be its limit. Note that $y\in C$ because closed convex sets are weakly closed. Hence $\|y\|\ge d$. But since closed balls around $0$ are also weakly closed, we have $\|y\|\le d$. Thus $\|y\|=d$. This proves the existence.
  6. For the uniqueness, strict convexity is enough: the midpoint of a line segment between two "nearest" points would be even closer to $0$.
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does the strict convexity mean a stronger triangle inequality, namely $\|u+v\| < \|u\|+\|v\|$ whenever $v,u$ are linearly independent ? – user1952009 May 20 at 1:16

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