From: israel@math.ubc.ca (Robert Israel)
Subject: Re: No best approximation newby question
Date: 3 May 2000 17:00:02 -0500
Newsgroups: sci.math.research
Summary: [missing]

In article <390c9ed1@news.ua.pt>,
 "Delfim F. Marado Torres" <delfim@mat.ua.pt> writes:

> It is a well known fact from the theory of approximation
> that if V is a normed linear space and W a FINITE-dimensional
> subspace of V, then, given v \in V, there exists w* \in W such that
> 
> || v - w* || <= || v - w || for all w \in W.
> 
> My question is:
> 
> If W is an INFINITE-dimensional subspace of V, under
> what conditions there are no such w*?

1) If W is not a closed subspace, then no such w* exists when v is in
the closure of W but not in W.

2) Suppose V is a Banach space, and there is a bounded linear operator T 
from  V onto a Banach space Y such that the image of the closed unit ball
B = {v in V: ||v|| <= 1} is not closed.  Let W = N(T) = {w in V: T(w) = 0}.
Then for any v such that T(v) is in the closure of T(B) but not in T(B), no 
such w* exists.  
Proof: For any w in W, ||v-w||>1  since T(v-w) = T(v) is not in T(B).  
However, for any epsilon > 0 there is x in B with  
||T(v - x)|| = ||T(v) - T(x)|| < epsilon, and therefore (for some fixed
constant C, using the Open Mapping Theorem) there is w in W with 
||v - x - w|| < C epsilon, so ||v - w|| <= ||v - x - w|| + ||x|| < 1 + C epsilon.
Thus the infimum of ||v-w|| for w in W is 1, but is not attained at any w*.

For example, consider V = L^1[0,1] with T the linear functional
T(f) = int_0^1 x f(x) dx, and W = {f: int_0^1 x f(x) dx = 0}.  Note
that T(B) = {z: |z| < 1}.  Thus in this case w* never exists unless 
v is in W.

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2