Re: When the Gilbert space with measure is separable

On Wed, 28 Mar 2007 16:29:42 -0500, Robert Israel <israel@xxxxxxxxxxx>
wrote:

"Kostyantyn Yusenko" <kay.math@xxxxxxxxx> writes:

We are given a Gilbert space H and measure m. The question is for

[ Hilbert space, of course ]

which measures m the spaces L_2(H,m) are separable?

Is this true that when m is propabilistic measure then space L_2(H,m)
is separable?

I'm not sure what you mean by L_2(H,m). I can think of two plausible
1) m is a measure on H, and you're dealing with scalar-valued functions
f on H where ||f||^2 = int_H |f(x)|^2 dm(x) < infty
2) m is a measure on some measurable space X, and you're dealing with
H-valued functions f on X where ||f||^2 = int_X ||f(x)||^2 dm(x) < infty.

In either case, you certainly want H to be separable. Whether m is
a probability measure is not really relevant, but you do want it to be
sigma-finite.

I assumed he meant (2). Assuming so, it's not clear to me whether
you're suggesting that m sigma-finite implies that L^2(m) is
separable or not.

In fact not - whether this is for your benefit, his benefit or
nobody's benefit is not clear to me at present, but for example,
say S is an uncountable set, and let X = {0,1}^S. Give each
"factor" {0,1} a measure that assigns 1/2 to each point, and
let m be the product measure on X.

Saying x in X says that x is a function from S to {0,1}.
For a in S let X_a be the set of all x in X with x(a) = 0,
and let f_a be the characteristic function of X_a. Then
||f_a - f_b||_2 = 1/sqrt(2) for a <> b, so L^2(m) is
not separable.


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

David C. Ullrich
.

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