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").