Re: Compactness in Hilbert spaces
- From: Ronald Bruck <bruck@xxxxxxxxxxxx>
- Date: Mon, 30 Jan 2006 18:55:28 -0800
In article
<27211130.1138663449498.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
brittanyUK <johanassenbrittany33@xxxxxxxxx> wrote:
> >> Hello
> >>
> >> I am just beginning to learn Hilbert space theory, and have come across
> >> this
> >> problem that has been troubling me for days:
> >>
> >> Let {f_n} be an orthonormal system in H, a Hilbert space, and {a_n} a
> >> sequence in l^2. Then one of the following is compact and one is not.
> >>
> >>
> >> First Set={Sum(b_n)(f_n) where Sum|b_n|^2 =< Sum|a_n|^2 }
....
> >The first is a closed ball, so yes, it's closed and >bounded. The
> >center is the origin, the radius is sqrt(\sum |b_n|^2).
....
>
> Also, how is the first set a closed ball if a_n can be any l^2 sequence?
{a_n} is fixed at the start of the definition of the set. Put
r = sqrt(\sum_n |a_n|^2).
Then you're looking at the set of x = \sum_n b_n f_n whose norms satisfy
|x| = sqrt(\sum_n |b_n|^2) <= r.
(Since the {f_n} are orthonormal.)
--Ron Bruck
.
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