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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 

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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. 

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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,. 

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James1118 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 

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