Liouville's Theorem: Prove f is Constant

In summary, Liouville's theorem states that if a function is bounded and entire, then its derivative is also bounded and entire. This means that if f(z) is a constant, then so is g(z).
  • #1
retrostate
7
2

Homework Statement


Show that if f is an entire function that satisfies

|1000i + f(z)| ≥ 1000, for all z ∈ C, then f is constant.

Homework Equations


(Hint: Consider the function g(z) = 1000/1000i+f(z) , and apply Liouville’s Theorem.)

The Attempt at a Solution


Ok, so I assume that as f is entire, then for it to be a constant, it must be bounded (Liouville’s Theorem).

Am I right in thinking that as g(z) is the reciprocal of |1000i + f(z)|, Then

1000/1000i+f(z) ≤ 1000/1000 =1

This is as far as I’ve got, I’ve sat here for hours, so any help would be very much appreciated….Thank you
 
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  • #2
retrostate said:

Homework Statement


Show that if f is an entire function that satisfies

|1000i + f(z)| ≥ 1000, for all z ∈ C, then f is constant.

Homework Equations


(Hint: Consider the function g(z) = 1000/1000i+f(z) , and apply Liouville’s Theorem.)

The Attempt at a Solution


Ok, so I assume that as f is entire, then for it to be a constant, it must be bounded (Liouville’s Theorem).

Am I right in thinking that as g(z) is the reciprocal of |1000i + f(z)|, Then

1000/1000i+f(z) ≤ 1000/1000 =1

This is as far as I’ve got, I’ve sat here for hours, so any help would be very much appreciated….Thank you

You should write that a little more carefully. Use parentheses where you need them and don't drop the absolute value. But yes,

|1000/(1000i+f(z))|<=1. So the function g(z)=1000/(1000i+f(z)) is bounded, yes? Is it entire? Why? What does that let you say about f(z)?
 
  • #3
I would say g(z) is differentiable on the whole C - {-1000i}, and so would be entire within the region {z:Imz > - 1000}. And is bounded above at g(z)=1,Therefore, by Liouvilles Theorem f(z) is a constant? i.e. if f(z)=0 then g(z)=i < 1…?
 
  • #4
retrostate said:
I would say g(z) is differentiable on the whole C - {-1000i}, and so would be entire within the region {z:Imz > - 1000}. And is bounded above at g(z)=1,Therefore, by Liouvilles Theorem f(z) is a constant? i.e. if f(z)=0 then g(z)=i < 1…?

To apply Liouville's theorem to g(z) it has to be analytic EVERYWHERE. Since g(z) is a quotient of analytic functions it's analytic everywhere that the denominator doesn't vanish. Does the denominator ever vanish? Remember you are given |1000i+f(z)|>=1000.
 
  • #5
Ahhh, no the denominator can't vanish because we are dividing 1000 by an inequality which is greater or equal to 1000, therefore, the maximum value g(z) can take is 1 if |1000i+f(z)|=1000.
So, as the denominator does not vanish it must be analytic everywhere
 
  • #6
retrostate said:
Ahhh, no the denominator can't vanish because we are dividing 1000 by an inequality which is greater or equal to 1000, therefore, the maximum value g(z) can take is 1 if |1000i+f(z)|=1000.
So, as the denominator does not vanish it must be analytic everywhere

Ok, so if g(z) is bounded and entire what does Liouville tell you about it? What does that imply for f(z)?
 
  • #7
That f(z) is constant if and only if g(z) is constant, and because g(z) is bounded and entire, by Liouvilles theorem it is constant, which implies f(z) is constant
 
  • #8
retrostate said:
That f(z) is constant if and only if g(z) is constant, and because g(z) is bounded and entire, by Liouvilles theorem it is constant, which implies f(z) is constant

Sure, that's it!
 
  • #9
Thank you Dick, I can't tell you how much I appreciate the time you have spent helping me.
 

1. What is Liouville's Theorem?

Liouville's Theorem is a mathematical theorem that states that if a function is bounded and holomorphic (analytic) in the entire complex plane, then it must be a constant function.

2. How do you prove that a function is constant using Liouville's Theorem?

In order to prove that a function is constant using Liouville's Theorem, you must show that the function is bounded and holomorphic in the entire complex plane. If this condition is met, then the theorem guarantees that the function must be a constant.

3. What does it mean for a function to be bounded?

A function is considered bounded if its values do not exceed a certain limit as the input approaches infinity. In other words, the function has a finite range and does not grow without bound.

4. How can you determine if a function is holomorphic?

A function is considered holomorphic if it is complex-differentiable at every point in its domain. This means that the function must satisfy the Cauchy-Riemann equations, which are a set of necessary conditions for complex differentiability.

5. Can Liouville's Theorem be applied to functions that are not holomorphic?

No, Liouville's Theorem only applies to functions that are holomorphic in the entire complex plane. If a function is not holomorphic, then the theorem cannot be used to prove that it is constant.

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