Re: (Functional Analysis) Please help me understand this proof

In article <1168984555.591650.113710@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Snis Pilbor <snispilbor@xxxxxxxxx> wrote:
Hi,

Trying to understand the proof that (l^infty)^* is not isomorphic to
l^1. The proof uses the Hahn-Banach theorem to create a linear
functional f on l^infty such that f(s) = lim_{n->oo} s_n whenever this
limit exists. That's all well and good. But then it *immediately*
jumps to the conclusion that this shows (l^infty)^* is not isomorphic
to l^1. I don't see how that follows.

The function f constructed, would be mapped by an isometry to a
sequence in l^1. My guess is that we are somehow supposed to say that
since f is so nicely behaved, it can't possibly be "mimicked" by
something in l^1. But since this is a proof of nonexistence of an
isomorphism, we don't know anything about the behavior of the image of
f in l^1: is doesn't have to be in any way canonically related to the
behavior of f, a priori. Afterall an isomorphism is just a bijection
which preserves norm: we don't know anything about it "preserving the
property of mapping things with limits to their limits", or what have
you. The way I see it, all this "proof" shows is that one particularly
nice map which might have been a candidate for an isomorphism, doesn't
work.

Depending on which "the proof" you're referring to, you may be right.
A better way to prove (l^infty)^* is not isomorphic to l^1 would be
to show that (l^infty)^* is not separable. This can be done as follows.

There is a map g from the interval (0,1) to infinite sets of natural
numbers such that g(s) intersect g(t) is finite whenever s <> t.
For example, this can be done by encoding the convergents of the
continued fraction as natural numbers [plus a little fiddling to
deal with rationals, whose continued fraction is finite].
Take e(s) in l^infty such that e(s)_j = 1 for j in g(s), 0 otherwise.
If y is a linear combination of e(r) for some finite set F of r's that
does not contain s, then ||e(s) - y||_infty >= 1, because there will be
some j that is in g(s) but not in g(r) for r in F. Thus the
distance from e(s) to the linear subspace of all such y is at least 1.
Use Hahn-Banach to produce f(s) in (l_infty)^* such that f(s)(e(s)) = 1
but f(s)(e(r)) = 0 for all r <> s. Note that ||f(s) - f(r)|| >= 1 for
r <> s ...

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

.

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