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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:

 Quote: In article 500370.1152975018551.JavaMail.jakarta@nitrogen.mathforum.org>, James james545@gmail.com> wrote: 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.

It would if that were a period at the end of your quotation. But
it's a comma,

************************

David C. Ullrich
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.
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

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,.
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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