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The following statement is true for Hilbert spaces $H$:

Every closed convex set ${\cal C} \subset H$ has a unique element $x$ such that for any $y \in C$, we have $|x| \leq |y|$.

Is this statement still true for Banach spaces? If not, what is a counterexample?

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$\mathcal C$ better be nonempty. –  Jonas Meyer Feb 2 '13 at 2:54
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3 Answers

This is a bit too long for a comment. The question was answered by Jonas and Giuseppe. Let me add that a result sometimes called the Day-James theorem characterizes the spaces for which the statement in your question is true:

For a normed space $X$ the following are equivalent:

  1. $X$ is strictly convex and reflexive (in particular $X$ is complete).
  2. Every non-empty closed convex set in $X$ has a unique point of minimal norm.

See the discussion after the proof of Corollary 5.1.19 on page 436 of Megginson's book An introduction to Banach space theory.

The space in Giuseppe's answer is not strictly convex and the ones in Jonas's answer are neither strictly convex nor reflexive.

The question Distance minimizers in $L^1$ and $L^{\infty}$ could also be of interest to you.

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Exercises 4 and 5 of Chapter 5 of Rudin's Real and complex analysis state:

4 Let $C$ be the space of all continuous function on $[0,1]$, with the supremum norm. Let $M$ consist of all $f\in C$ for which $$\int_0^{1/2}f(t)\,dt-\int_{1/2}^1f(t)\,dt=1.$$ Prove that $M$ is a closed convex subset of $C$ which contains no element of minimal norm.

5 Let $M$ be the set of all $f\in L^1([0,1])$, relative to Lebesgue measure, such that $$\int_0^1f(t)\,dt=1.$$ Show that $M$ is a closed convex subset of $L^1([0,1])$ which contains infinitely many elements of minimal norm. (Compare this and Exercise 4 with Theorem 4.10.)

You can guess what Theorem 4.10 is.

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The answer is no, in general. For a simple example consider the plane $\mathbb{R}^2$ equipped with the norm $$\lVert (x, y)\rVert_{\infty}=\max(\lvert x\rvert, \lvert y\rvert).$$ Then consider the vertical segment $\{(x, y)\ :\ x=1,\ y\in[-1, 1]\}$.

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