From: bruck@pacificnet.net (Ronald Bruck)
Subject: Re: Does anyone have an example like this???
Date: Wed, 31 Mar 1999 15:17:30 -0800
Newsgroups: sci.math
Keywords: Characterization of nonreflexive Banach spaces
In article <7dtu7h$d5s$1@nnrp1.dejanews.com>, ullrich@math.okstate.edu wrote:
> In article <37019A53.4EF6BA80@math.brown.edu>,
> Munju Kim wrote:
> > X : an infinite dimensional Banach space.
> >
> > S : a convex closed subset of X which does not have an element of
> > minimal norm.
> >
> > Please tell me.
>
> Homework? Consider l_1, the space of absolutely summable
> sequences. Let e_1, e_2, ... be the standard basis vectors.
> The closed convex hull of the vectors (1 + 1/n)*e_n does it.
Actually, any nonreflexive space has this property. In fact, this is a
necessary AND sufficient condition for a space to be nonreflexive.
But that's quite a bit harder than the homework problem... (Which might
NOT be a homework problem. This approximation problem is so pervasive
it's probably popped in the work of hundreds.)
--Ron Bruck
==============================================================================
From: a.visitor@the.asylum (a.visitor)
Subject: Re: Does anyone have an example like this???
Date: Wed, 31 Mar 1999 20:38:44 +0000
Newsgroups: sci.math
In article <37019A53.4EF6BA80@math.brown.edu>, Munju Kim
wrote:
> X : an infinite dimensional Banach space.
>
> S : a convex closed subset of X which does not have an element of
> minimal norm.
>
> Please tell me.
Let
X = { f:[0,1] \to R | f is continuous and f(1)=0 }
and equip X with the uniform norm || ||_{\infty}.
Let
S = { f \in X | \int_{0}^{1}tf(t)dt = 1 }
Now prove that
(1) X is Banach;
(2) S is closed and convex;
(3) for all f\in S we have
||f||_{\infty} > inf_{g\in S} ||g||_\infty},
i.e. there is no vector in S closest to 0 (or using your terminology:
S does not have a "minimal element").