Re: Convergence on Hardy spaces
- From: David C. Ullrich <ullrich@xxxxxxxxxxxxxxxx>
- Date: Sun, 17 Apr 2005 08:12:22 -0500
On 16 Apr 2005 22:55:50 -0700, bryant_j_j@xxxxxxxxx wrote:
>
>David C. Ullrich wrote:
>> On 15 Apr 2005 18:05:24 -0700, bryant_j_j@xxxxxxxxx wrote:
>>
>> >howdy...
>> >
>> > it is trivial that if a sequnce of functions in $H^{\infty}$ on
>the
>> >unit circle converges then it also converges as a sequence of
>functions
>> >in $H_p$ on the unit circle for all $1\leq p <\infty$. are there
>> >sufficient conditions on the sequence for the converse (i.e. a
>sequence
>> >of functions converging in $H^p$ will also converge in $H^{\infty}$)
>to
>> >be true, especially for the case where $p=2$?
>>
>> What's the point to all those dollar signs? You really think that
>> $p=2$ is somehow more clear than p=2?
>>
>
> well it's a habit you know for using too much LaTeX.
>
>
>> Anyway, it's hard to imagine such a condition that's weak
>> enough to give an interesting result. The best I can do is
>> that if f_n -> f in H^2 and f_n -> g in H^infinity then
>> g = f.
>>
>
> is that really the best possible result?
How would anyone know? What other result do you have
in mind? It's the best I can think of, and nobody else
has added anything...
>> If you have some specific condition in mind someone might
>> be able to tell you whether it's sufficient.
>>
>
> i was thinking that maybe someone has done research on this and have
>proposed some generic sufficient conditions.
I can't imagine that the sort of condition you probably have in
mind is _possible_. Of course that's too vague a statement to be
provable, but I can't imagine what sort of condition would do.
Do you have some specific condition in mind?
I mean here's another obvious sufficient condition, except that
I can't imagine it's useful, since I don't see how you could
check it: If the f_n form a totally bounded set in H^infinity
then f_n -> f in H^2 implies that f_n -> f uniformly.
Can't imagine a situation where that would be useful, since
I don't see how you're going to know that {f_n} is totally
bounded...
>anyway, let's say that the
>sequence f_n are rational functions with no roots on C (the unit
>circle), and it converges in H^2 to another rational function f with no
>roots on C. are there sufficient conditions such that also f_n->f in
>H^{\infty}?
Not that I know of. Again, if you have some condition in mind
people might be able to tell you whether it's sufficient.
Do you know anything else about these functions?
>references would help. TIA.
************************
David C. Ullrich
.
- References:
- Convergence on Hardy spaces
- From: bryant_j_j
- Re: Convergence on Hardy spaces
- From: David C . Ullrich
- Re: Convergence on Hardy spaces
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- Convergence on Hardy spaces
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