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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 nonpolynomial 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. 

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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 nonpolynomial 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 

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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 nonpolynomial 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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