1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Let f be entire and |f| >= 1. prove f is constant

  1. Jun 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Let f(z) be entire and let |f(z)| ≥ 1 on the whole complex plane. Prove f is constant.


    2. Relevant equations
    Theorem 1: Let f be analytic in the domain D. If |f(z)| = k, where k is a constant, then f is constant.

    Maximum Modulus Principle: Let f be analytic and non constant in the bounded domain D. If f is continuous on the closed region R that consists of D and all of its boundary points B, then |f(z)| assumes its max value, and does so only at points on the boundary B.


    3. The attempt at a solution

    Below is my attempt. Let me know if I am even in the right ballpark.

    Proof:

    Note that from Theorem 1 if |f(z)| = 1 then f is constant and we are done. Therefore we want to show f is constant for
    |f(z)| > 1.

    To show this we use contradiction. Suppose f(z) is entire, |f(z)| > 1, and f is non constant. Let D = {z: |z| < 1}. Since f is entire it is continuous on the complex plane. Consequently it is continuous on a region R consisting of D and its boundary points B. By the max modulus principle |f(z)| assumes its max value at the boundary points B. Therefore |f(z)| ≤ 1. Which is contradicts our hypothesis. Therefore f is constant.
     
  2. jcsd
  3. Jun 7, 2014 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    What about an f where |f(z)|=1 somewhere, but not everywhere? You do not cover this case.

    Okay so far.

    Why? All you have is |z|<1.

    That approach does not work.
     
  4. Jun 7, 2014 #3
    ok can we consider 1/|f(z)|? which is analytic since we were given f is analytic and |f|≥ 1 on the whole complex plane?

    Then 1/|f(z)| ≤ 1 for all z. By liouville's theorem 1/f(z) is constant. Am i warmer?
     
    Last edited: Jun 7, 2014
  5. Jun 8, 2014 #4

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    That works.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Let f be entire and |f| >= 1. prove f is constant
Loading...