Proof of Degree <= 1 for Entire Function f

[Silviu]

Homework Statement

Suppose f is entire and there exist constants a and b such that $|f(z)| \le a|z|+b$ for all $z \in C$. Prove that f is a polynomial of degree at most 1.

Homework Equations

The Attempt at a Solution

We have that for any $z \neq 0$, $\frac{|f(z)|}{a|z|} \le b$. So if we take the limit as $z \to \infty$ it is obvious that if f is polynomial, it can't have a degree greater than 1. However I am not sure why it must be polynomial.

 

[stevendaryl]

Homework Statement

Suppose f is entire and there exist constants a and b such that $|f(z)| \le a|z|+b$ for all $z \in C$. Prove that f is a polynomial of degree at most 1.

Homework Equations

The Attempt at a Solution

We have that for any $z \neq 0$, $\frac{|f(z)|}{a|z|} \le b$. So if we take the limit as $z \to \infty$ it is obvious that if f is polynomial, it can't have a degree greater than 1. However I am not sure why it must be polynomial.

I'm not sure what you are supposed to be able to use in your proof. One fact about entire functions is the Cauchy estimate formula, which says:

• Take a circle in the complex plane of radius R centered on z=0.
• Let M_R be the largest value of |f(z)| on the circle.
• Then |f^{(n)}(0)| \leq \frac{n! M_R}{R^n} (where f^{(n)} means the n^{th} derivative).

See if you can use this to prove f^{(n)} = 0 for n &gt; 1.

 

[Mark44]

We have that for any $z \neq 0$, $\frac{|f(z)|}{a|z|} \le b$.

This doesn't follow from $|f(z) | \le a|z| + b$