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david petry
science forum Guru

Joined: 18 May 2005
Posts: 503

Posted: Fri Jul 21, 2006 12:26 am    Post subject: Re: Entire functions, polynomial bounds

David C. Ullrich wrote:
 Quote: On 20 Jul 2006 16:09:46 -0700, "david petry" david_lawrence_petry@yahoo.com> wrote: I believe the following is true, but I don't know how to prove it. Let 'f' be a non-polynomial entire function. Then there does not exist a polynomial 'p' such that for all z, -oo < Re(f(z)) < p(|z|) Does anyone know how to prove it? I sketched a clumsy version and buggy version of a proof in one of those threads the other day. Today's version may come out better: Say f = u + iv for convenience. Also assume f(0) = 0. Suppose that u(z) <= c |z|^n for large z. Say u^+ is the positive part of u and u^- is the negative part (so u = u^+ - u^- and |u| = u^+ + u^-.) Now the mean of u^+ on the circle of radius r is <= c r^n for large r. But the mean of u itself is zero, and it follows that the mean of u^- is the same as the mean of u^+, so we actually have that the mean of |u| on a circle of radius r is <= c r^n for large r (here "c" varies from line to line in the proof). Now look at the integral formula that expresses u + iv inside a disk in terms of the boundary values of u: it shows that the sup of |v| on the circle of radius r/2 is <= c r^n. So |f| has polynomial growth, hence f is a polynomial.

Cool. That works. Thanks.
David C. Ullrich
science forum Guru

Joined: 28 Apr 2005
Posts: 2250

Posted: Fri Jul 21, 2006 12:07 am    Post subject: Re: Entire functions, polynomial bounds

On 20 Jul 2006 16:09:46 -0700, "david petry"
<david_lawrence_petry@yahoo.com> wrote:

 Quote: I believe the following is true, but I don't know how to prove it. Let 'f' be a non-polynomial entire function. Then there does not exist a polynomial 'p' such that for all z, -oo < Re(f(z)) < p(|z|) Does anyone know how to prove it?

I sketched a clumsy version and buggy version of a proof in
one of those threads the other day. Today's version may come
out better:

Say f = u + iv for convenience. Also assume f(0) = 0. Suppose that

u(z) <= c |z|^n

for large z. Say u^+ is the positive part of u and u^-
is the negative part (so u = u^+ - u^- and |u| = u^+ + u^-.)
Now the mean of u^+ on the circle of radius r is <=
c r^n for large r. But the mean of u itself is zero, and
it follows that the mean of u^- is the same as the mean
of u^+, so we actually have that the mean of |u| on a
circle of radius r is <= c r^n for large r (here "c" varies
from line to line in the proof).

Now look at the integral formula that expresses u + iv
inside a disk in terms of the boundary values of u:
it shows that the sup of |v| on the circle of radius
r/2 is <= c r^n. So |f| has polynomial growth, hence
f is a polynomial.

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

David C. Ullrich
david petry
science forum Guru

Joined: 18 May 2005
Posts: 503

 Posted: Thu Jul 20, 2006 11:09 pm    Post subject: Entire functions, polynomial bounds I believe the following is true, but I don't know how to prove it. Let 'f' be a non-polynomial entire function. Then there does not exist a polynomial 'p' such that for all z, -oo < Re(f(z)) < p(|z|) Does anyone know how to prove it?

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