|
|
| Author |
Message |
James
science forum Guru Wannabe
Joined: 04 Feb 2005
Posts: 154
|
 Posted: Sat Jul 15, 2006 2:49 pm Post subject:
Analytic function on open disk
|
|
|
Let f be analytic in the open unit disk B(0;1) = {z : |z| < 1} and continuous on the closed unit disk. Assume that |f(z)| <= 2 for |z| = 1, Im(z) >= 0, and |f(z)| <= 8 for |z| = 1, Im(z) < 0. I need to show that |f(0)| <= 4.
I'm not sure how to approach this. The most naive thing to do would be to say that |f(z)| <= 8 on the whole disk, and then use Cauchy's integral formula to say that |f(0)| <= 8.
It's not true that f(0) = int_{|z|=1} f(w)/w dw, is it? What is true is f(0) = int_{|z|=r} f(w)/w dw, and then if f is bounded you get an estimate involving r, then by sending r to 1 you can get a bound, which is how I got |f(z)| <= 8 on the whole disk.
How should I approach this?
James |
|
| Back to top |
|
 |
eugene
science forum Guru
Joined: 24 Nov 2005
Posts: 331
|
| Posted: Sat Jul 15, 2006 3:57 pm Post subject:
Re: Analytic function on open disk
|
|
|
James wrote:
| Quote: | Let f be analytic in the open unit disk B(0;1) = {z : |z| < 1} and continuous on the closed unit disk. Assume that |f(z)| <= 2 for |z| = 1, Im(z) >= 0, and |f(z)| <= 8 for |z| = 1, Im(z) < 0. I need to show that |f(0)| <= 4.
I'm not sure how to approach this. The most naive thing to do would be to say that |f(z)| <= 8 on the whole disk, and then use Cauchy's integral formula to say that |f(0)| <= 8.
It's not true that f(0) = int_{|z|=1} f(w)/w dw, is it? What is true is f(0) = int_{|z|=r} f(w)/w dw, and then if f is bounded you get an estimate involving r, then by sending r to 1 you can get a bound, which is how I got |f(z)| <= 8 on the whole disk.
How should I approach this?
James
|
Consider, say, h(z) = f(z) * f(-z) an apply Maximum Modulus Principle.
For other useful auxiliary functions h you may look here
http://groups.google.ca/group/sci.math/browse_thread/thread/d91d7f4153e06bfe/d4888af7ba847bf2?q=inequality+and+analytical&rnum=1#d4888af7ba847bf2
where the same problem was discussed
* In that thread there was a condition of analyticity of f on the whole
unit disk, but it was superfluous : in order to apply Maximum Modulus
Principle it is enough to have analyticity inside disk and continuity
on it's boundary as in you case,. |
|
| Back to top |
|
|
A N Niel
science forum Guru
Joined: 28 Apr 2005
Posts: 475
|
| Posted: Sat Jul 15, 2006 4:04 pm Post subject:
Re: Analytic function on open disk
|
|
|
In article
<500370.1152975018551.JavaMail.jakarta@nitrogen.mathforum.org>, James
<james545@gmail.com> wrote:
| Quote: | Let f be analytic in the open unit disk B(0;1) = {z : |z| < 1} and continuous
on the closed unit disk. Assume that |f(z)| <= 2 for |z| = 1,
|
It follows that |f(z)| <= 2 on the whole disk. |
|
| Back to top |
|
|
David C. Ullrich
science forum Guru
Joined: 28 Apr 2005
Posts: 2250
|
| Posted: Sat Jul 15, 2006 4:46 pm Post subject:
Re: Analytic function on open disk
|
|
|
On Sat, 15 Jul 2006 12:04:33 -0400, A N Niel
<anniel@nym.alias.net.invalid> wrote:
It would if that were a period at the end of your quotation. But
it's a comma,
************************
David C. Ullrich |
|
| Back to top |
|
|
Google
|
|
| Back to top |
|
|
|
|
The time now is Fri Jul 30, 2010 5:43 am | All times are GMT
|
|
Copyright © 2004-2005 DeniX Solutions SRL
|
|
Other DeniX Solutions sites:
Electronics forum |
Medicine forum |
Unix/Linux blog |
Unix/Linux documentation |
Unix/Linux forums
|
Powered by phpBB © 2001, 2005 phpBB Group
|
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
Forum index
»
Science and Technology
»
Math
