Hi all,

I'm reading up on C* and Von-Neumann algebras. Jumping ahead, I have a
question. To fix ideas, fix a probability space X, a Hilbert space H
and an X-spectral measure on H. This triple gives an isometric
*-representation p of L^{\infty}(X) in the C*-algebra B(H) of bounded
linear operators H->H; this is the Borel functional calculus.

My question is the following: suppose we take the induced map
p:L^{1}(X)->B(H) and renorm the image according to the induced L^1
norm. My intuition tells me that the range is the ideal of trace class
operators and then the norm is Tr(|p(s)|) where |p(s)| is the modulus
of p(s). This is surely true for integrable elementary functions. Is
my intuition correct?

In case the answer is yes, can you provide any reference for this,
plz? Preferably online because, at the moment, I have difficulty
reaching a library.

TIA, best regards,
G. Rodrigues

On Tue, 20 Jun 2006 22:51:40 +0100, Gonçalo Rodrigues
<op73418@mail.telepac.pt> wrote:


I don't know anything about your actual question, but:


are you certain that there _is_ such an induced map? With
range contained in B(H)? I would have thought that at best
an L^1 function corresponded to some unbounded operator.



************************

David C. Ullrich

On Tue, 20 Jun 2006 22:51:40 +0100, Gonçalo Rodrigues
<op73418@mail.telepac.pt> fed this fish to the penguins:


Forget it, this is clearly wrong. Trace-class operators are conmpact,
but since X is a probability space the constant function 1 is sent
into the identity, which is not compact. Besides, on a simple function
s = \sum_n \chi_{E_n}k_n , its L^1 norm is

\sum_n \chi_{E_n}|k_n|

which is different from \sum |k_n| (the trace).

I have to stop posting questions too hastily... (my appologies). Best
regards,
G. Rodrigues

On Wed, 21 Jun 2006 07:04:59 -0500, David C. Ullrich
<ullrich@math.okstate.edu> fed this fish to the penguins:


Since X is a probability space f is integrable iff it is a.e. bounded
so such a linear map exists trivially. It is not bounded and that is
why the next sentence is "renorm the range with the induced L^1 norm"
(or some such). My bad here, with my sloppy use of language.

Of course, viewed in this way, my question has a trivial answer: the
image is the same and the norm is a "weighted trace", with weights
given by the measure on X. I really do have to stop posting questions
too hastily...

Best regards,
G. Rodrigues

On Wed, 21 Jun 2006 14:27:45 +0100, Gonçalo Rodrigues
<op73418@mail.telepac.pt> wrote:


Huh? What does it mean to say that a function is "a.e. bounded"?

(I can only think of two things this might mean:

(i) for almost every x, {f(x)} is bounded.)

That seems like a literal translation of "a.e. bounded",
but that can't be what you mean, because any function
satisfies that condition.)

Or maybe you mean

(ii) There exists c such that |f| <= c a.e.

That makes more sense, but if you mean (ii) then it's
simply not true that every L^1(X) function is a.e. bounded.

I really don't believe that f in L^1(X) induces an element
of B(H) in a natural way. And I can't make much sense of
your explanation, since I can't figure out what you mean
by "a.e. bounded".

But consider the following example: X is a probability
space, and H = L^2(X); take the spectral measure so that
the projection associated with E subset X is just the
orthogonal projection onto functions supported on E.
The for f in L^infinity(X), p(f) is just the operator
of multiplication by f. And this operator is certainly
_not_ bounded on H for f in L^1(X).


Too many "bounded"'s here. You mean that p:L^1(X)->B(H)
exists but p is not bounded? I don't believe it exists.
I really don't.



************************

David C. Ullrich

On Wed, 21 Jun 2006 08:54:00 -0500, David C. Ullrich
<ullrich@math.okstate.edu> fed this fish to the penguins:


Sorry, too many confusions on my part, so let me try to clear them up.
As you point out, of course if f in L^1 doesn't imply it is in
L^\infty (that's what I mean by a.e. bounded - your ii). So we really
should be thinking of the induced map in the *reverse* direction, e.g.

{Trace-class operators} -> L^1

whose adjoint is then our original *-representation. I suppose this
can be made to work, but I would be much more interested if there was
L^1 -> {some class of bounded operators} that was a sort of integrable
functional calculus. But as you point out, in the naive way I was
thinking this cannot be made to work - the range falls in the dual of
B(H) or some such.

Thanks for all (and appologies for my compounded confusions), best
regards,
G. Rodrigues

On Wed, 21 Jun 2006 17:14:51 +0100, Gonçalo Rodrigues
<op73418@mail.telepac.pt> wrote:


I should start by saying that I know nothing about any of this
- the extent of my knowledge of anything to do with the spectral
theorem is that it shows normal operators are equivalent to
multiplication operators on an L^2 space.

But, having clarified that I don't know what I'm talking about,
it seems to me you may be giving up too soon. My objection was
to the idea that p maps L^1(X) into B(H); it simply doesn't.
But it seems to me that p, or some extension of p which we
may as well denote again by p, does map f in L^1(X) to
an unbounded densely-defined operator. This is clear in
the example where we take H to be L^2(X), for example.

What if anything that has to do with trace-class operators
I wouldn't know.



************************

David C. Ullrich