Re: weak* topology of X* is metrizable

jane <jane1806@xxxxxxxxxx> writes:

It is a known fact that if X is a separable TVS, and K is a compact of X*
in weak* topology, then K is metrizable.

I'm interested in a more general version:

Let X be a TVS, such that the whole weak* topology of X* is metrizable. Are
there neccessary and sufficient conditions on X in this case. And if there
are i would be grateful if someone could mention the idea of the proof or a
hint.

I'll assume X is locally convex, otherwise there's no guarantee X* contains
anything but 0. Then a necessary and sufficient condition is that X has
countable dimension, i.e. there is a countable basis for X.

Suppose 0 has a countable base of neighbourhoods. Then there is a sequence
{x_n} in X such that for every y in X there is N such that for all f in
X*, |f(x_n)| < 1 for n = 1...N implies |f(y)| < 1. Now by the Hahn-Banach
separation theorem this would imply that y is in the span of x_1,...,x_n.
So X* can't be metrizable unless X has countable dimension. On the other
hand, it seems to me that if X has basis {x_n}, a suitable metric is
d(f,g) = sum_{n=1}^infty 2^(-n) min(|f(x_n)-g(x_n)|, 1).
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

Quantcast