Re: Is Hol(D) a nice subset of B(H)?



Lee Rudolph wrote:
> Ali Taghavi <alitghv@xxxxxxxxx> writes:
>
>
> >> By "a neighborhood", I take you to mean that the
> >> neighborhood can vary
> >> with the map, so that more precisely Hol(D) is the
> >> set of all *germs*
> >> of holomorphic maps from D to C. If that is the
> >> case,
> >
> >The analytic continutity implies we do not need to use "germ"

I'm not sure what that means, but...

>
> In the only sense in what you write is correct, you
> are indeed *interpreting* elements of Hol(D) as certain
> continuous functions on D, so that for you Hol(D) is
> squeezed between two well-known and well-loved Banach
> algebras: it is a subset of C(D), the Banach algebra
> of all continuous functions on D, and a superset of
> the subalgebra of C(D) containing just those continuous
> functions on D that are holomorphic on Int D. You are
> no closer to telling us what topology you want to put
> on Hol(D). Do you want to give it the incomplete metric
> induced by its inclusion in C(D)? If not, what?

Since he is viewing these functions as elements of B(H) there is no
need to consider another topology: use that of B(H).

However, he has not said in what sense they belong to B(H) when H is
l^2. There are two common ways:

1. Identify f(z) with the diagonal operator which has the power series
coefficients of the power series for f on the diagonal. This is often
called a coefficient multiplier.

2. Identify elements of l^2 with power series (the Hardy space H^2) and
f(z) with the operator of multiplication by itself. (This is often
called an analytic Toeplitz operator.)

Probably not the first is intended, as then f corresponds to a compact
operator and is never Fredholm (if f has radius of convergence greater
than 1). Possibly the second is meant, in which case f corresponds to a
Fredholm operator if and only if it has no zeros on the unit circle.

As to whether this is a retract in Fred(H), I have no idea, though it
seems possible.


Dan

.

Relevant Pages

  • Re: Is Hol(D) a nice subset of B(H)?
    ... >> continuous functions on D, so that for you Holis ... >Probably not the first is intended, as then f corresponds to a compact ... >Fredholm operator if and only if it has no zeros on the unit circle. ... Lee Rudolph ...
    (sci.math.research)
  • Re: Is Hol(D) a nice subset of B(H)?
    ... >> of holomorphic maps from D to C. ... continuous functions on D, so that for you Holis ... squeezed between two well-known and well-loved Banach ... it is a subset of C, the Banach algebra ...
    (sci.math.research)