Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let X be a Hilbert space, A a convex, closed subset of X. Then there exists for every x in X exactly one best approximation in A, that is there exists a y in A such that || x-y || = d(x,A) = inf { || x-z || : z in A}.

Why isn't that theorem true for Banach spaces?

share|improve this question
1  
Can you give a counterexample? –  Martijn Aug 13 '10 at 8:33
8  
Non-uniqueness of the best approximation doesn't require a complicated or unusual Banach space: take X to be \mathbb{R}^2 equipped with a norm whose unit ball is not strictly convex, and play around with some examples. (I assume you meant to require that y should be unique, otherwise the two statements which you give are not equivalent.) –  Ian Morris Aug 13 '10 at 8:34
    
In fact, a normed vector space $X$ is strictly convex if and only if for every $x$ in $X$ and every affinely convex closed set $A$ there is a unique projection of $x$ onto $A$. (see e.g. "Metric spaces, convexity and nonpositive curvature" by Athanase Papadopoulos, IRMA lectures in mathematics and theoretical physics). All Hilbert spaces are strictly convex; the space $l_\infty$ which appears in Andrew Stacey's answer is not. –  Margaret Friedland Aug 24 '11 at 18:49

3 Answers 3

Happened to have this slide in one of my lectures. The green circle and green dot correspond to the Euclidean norm, the purple square and purple line to the sup norm. With the Euclidean norm, the green dot is the closest point in the square to the blue dot; with the sup norm, the purple line is the set of closest points in the square to the blue dot.

closest point

share|improve this answer
    
Just in case it's not clear, this is (purely coincidentally) a picture of Andreas' specific example based on Ian's general example. –  Loop Space Aug 13 '10 at 12:20

A subset in a Banach space for which every point in the space has a unique element of best approximation is called a Chebyshev set. In a Banach space every closed convex set is a Chebyshev set if and only if your space is strictly convex and reflexive (so any space that fails to have these properties is a counterexample). See here for some references.

share|improve this answer
    
I got distracted by the mention of Chebyshev's sets; I am sorry for basically doubling your answer in my comments. –  Margaret Friedland Aug 24 '11 at 18:59
    
Except that I need to add that a strictly Banach convex space does not have to be reflexive; please see ams.org/journals/bull/1941-47-04/S0002-9904-1941-07451-3/… –  Margaret Friedland Aug 25 '11 at 16:48

EDIT: since I'm not an expert on Banach spaces, I feel I shouldn't say anything more, but anyway; an essential ingredient is an exact formula in Hilbert spaces for $\|x + y\|^2$. Just an idea (perhaps I am being stupid): maybe if you have a Banach space where $\| x + y \| = F(\|x\|, \|y \|, g(x,y))$ for some reasonably simple functions $F$, $g$ then something can be said.

If you examine the proof for Hilbert spaces, it makes essential use of the scalar product; so it's not really surprising that it doesn't work for general Banach spaces. The norm is a nice enough structure to do a lot of things, but not that nice.

It also demonstrates that Banach spaces have far more detailed structure than just ordinary vector spaces, but that Hilbert spaces have even more structure again. In fact, there are many properties Hilbert spaces have which general Banach spaces don't (and many which even characterise Hilbert space uniquely).

share|improve this answer
    
I think I asked the wrong question. Of course the proof for Hilbert spaces makes essential use of the scalar product, but that doesn't explain why the proof doesn't work for Banach spaces. I think a counter example could be useful. –  Linda Raabe Aug 13 '10 at 9:39
3  
Ian Morris has already pointed out a counterexample in his comment to your question. Take $V= {\mathbb R}^2$ with the $\infty$-norm $\|(x,y)\| = \max \lbrace |x|,|y|\rbrace$. Now set $B = \lbrace \xi \in V \mid \|\xi\| \leq 1 \rbrace= \lbrace(x,y) \mid -1 \leq x \leq 1, -1 \leq y \leq 1 \rbrace$ and observe that there is no unique point in $B$ which is closest to the vector $(2,0)$. Indeed, the points $(1,\alpha)$ have distance $1$ to the vector $(2,0)$ for all $\alpha \in [-1,1]$. –  Andreas Thom Aug 13 '10 at 10:33
    
@Zen: strict convexity (which is the crucial property here) is indeed defined as a property of norms; one of equivalent statements is the following: A Banach space $(X, \|\cdot\|)$ is strictly convex if and only if $x \neq y$ and $\| x \| = \| y \| = 1$ together imply that $\| x + y \| < 2$. In other words, balls are round in a strictly convex space. –  Margaret Friedland Aug 24 '11 at 18:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.