|
|
 | | From: | Albert | | Subject: | Complex: convergence in norm for f hol. | | Date: | Mon, 24 Jan 2005 04:08:40 +0000 (UTC) |
|
|
 |
Could someone please help me prove that if f_n->f
uniformly, then |f_n|^2->|f| also uniformly?
I can only think of something like:
1)|f_n-f|
2)|f_n-f|^-=|(f_n^-)-f^-|
with conjugation)
We then multiply 1),2) above, and get:
3)
||f_n|^2-2ref_n*f+|f|^2|
so that eps^2
Then, as n->00 f_n->f , and substitute in #3 , get:
||f_n|^2-2|f|^2+|f|^2 |=||f_n|^2-|f|^2 |
Does this work?
Thanks for any help.
|
|
| | From: | Jannick Asmus | | Subject: | Re: Complex: convergence in norm for f hol. | | Date: | Mon, 24 Jan 2005 05:35:18 +0100 |
|
|
| On 1/24/2005 5:08 AM, Albert wrote:
> Could someone please help me prove that if f_n->f
> uniformly, then |f_n|^2->|f| also uniformly?
>
> I can only think of something like:
>
> 1)|f_n-f|>
> 2)|f_n-f|^-=|(f_n^-)-f^-|>
> with conjugation)
>
> We then multiply 1),2) above, and get:
>
>
> 3)
> ||f_n|^2-2ref_n*f+|f|^2|>
> so that eps^2>
> Then, as n->00 f_n->f , and substitute in #3 , get:
>
>
> ||f_n|^2-2|f|^2+|f|^2 |=||f_n|^2-|f|^2 |>
Why that? I doubt that it is valid.
>
> Does this work?
>
> Thanks for any help.
>
Here a hint if you assume normal convergence, i.e. that f_n converges to
f uniformly on *every* compact subset:
On an arbitrary compact subset K f_n and f are bounded by a common C>0.
Then g(z)=abs(z)^2 is Lipschitz continuous on the closed disc with
centre 0 and radius C. Then g(f_n) converges uniformly to g(f) on K.
|
|
|
