wu :: forums
wu :: forums - Sometimes unbounded entire implies constant

Welcome, Guest. Please Login or Register.
Mar 28th, 2023, 4:40am

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   general
   complex analysis
(Moderators: SMQ, ThudnBlunder, william wu, Icarus, Grimbal, towr, Eigenray)
   Sometimes unbounded entire implies constant
« Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: Sometimes unbounded entire implies constant  (Read 2794 times)
Aryabhatta
Uberpuzzler
*****






   


Gender: male
Posts: 1321
Sometimes unbounded entire implies constant  
« on: Apr 15th, 2007, 8:32pm »
Quote Quote Modify Modify

Let f be an entire function such that |f(z)| <= 1 + sqrt(|z|) for all z.
 
Show that f must be constant.
IP Logged
Obob
Senior Riddler
****





   


Gender: male
Posts: 489
Re: Sometimes unbounded entire implies constant  
« Reply #1 on: Apr 19th, 2007, 9:46am »
Quote Quote Modify Modify

Apply Cauchy's inequality to f on a disk of radius r.  Let r go to infinity to see that f' = 0.
 
What can you say about f if it is entire and |f(z)| <= (1+|z|^k)?
IP Logged
Aryabhatta
Uberpuzzler
*****






   


Gender: male
Posts: 1321
Re: Sometimes unbounded entire implies constant  
« Reply #2 on: Apr 19th, 2007, 10:20am »
Quote Quote Modify Modify

if k < 1, then your proof works in showing that f is constant.
 
if k >=1, then we have  a counterexample, f(z) = z.
IP Logged
Eigenray
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 1948
Re: Sometimes unbounded entire implies constant  
« Reply #3 on: Apr 19th, 2007, 10:24am »
Quote Quote Modify Modify

More generally, if an entire function f is everywhere bounded by a polynomial (of degree k), then f is itself a polynomial (of degree no more than k).
IP Logged
Aryabhatta
Uberpuzzler
*****






   


Gender: male
Posts: 1321
Re: Sometimes unbounded entire implies constant  
« Reply #4 on: Apr 20th, 2007, 5:06pm »
Quote Quote Modify Modify

Yes. I guess it can be proved using Cauchy's estimate for the nth derivative and the fact that the polynomial is bounded by it a finite power of z...
IP Logged
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright 2000-2004 Yet another Bulletin Board