Proof: f(z) is Entire & |f(z)|<|z|^{n} $\Rightarrow$ f(z) is a Polynomial

[futurebird]

Homework Statement

Prove that if f(z) is entire and $$|f(z)|<|z|^{n}, \forall |z|> R$$ then f(z) must be a polynomial.

Homework Equations

Taylor Series

$$f(z) =\displaystyle\sum_{j=0}^{\infty}\frac{f^{(j)}(z_{0})}{j!}(z-z_{0})^{j}$$

Cauchy's Estimate
If f is analytic interior to and on a circle C centered at $$z_{0}$$, radius $$R = |z-z_{0}|$$. Where M is the maximum value of |f(z)|.

$$\left| f^{j}(z_{0}) \right| \leq \frac{j!M}{R^{n}}$$

The Attempt at a Solution

f(z) is entire so it has Taylor series.

$$f(z) =\displaystyle\sum_{j=0}^{\infty}\frac{f^{(j)}(z_{0})}{j!}(z-z_{0})^{j}$$

I need to show that this series has a finite number of terms.

$$=\displaystyle\sum_{j=o}^{n}\frac{f^{(j)}(z_{0})}{j!}(z-z_{0})^{j}+\displaystyle\sum_{j=n+1}^{\infty}\frac{f^{(j)}(z_{0})}{j!}(z-z_{0})^{j}$$

The first sum is the polynomial part of the series, now to show the other part is zero. Consider the coefficient of the (n+1)th term:

$$a_{n+1}=\frac{f^{(n+1)}(z_{0})}{(n+1)!}$$

Because,

$$|f^{(n+1)}(z_{0})|\leq \frac{M(n+1)!}{R^{n+1}}$$

R is the radious of the circle around z0, and M is the max value of f(z).

$$|a_{n+1}|\leq MR^{-n-1}$$

Now I'm lost... Is M the same thing as $$|z|^{n}$$?
Is this even making any sense?

If M is at most $$|z|^{n}$$ then can I say that for R large this term is zero?

 

[futurebird]

Anyone?

 

[Dick]

It making sense. Set z0 to be 0. M is the maximum of |f(z)| over the contour. Since the contour is a circle of radius R, |z|=R. |f(z)|<=|z|^n=R^n. So you have shown |f^(n+1)(0)|<=(n+1)!/R for any R. So f^(n+1)(0)=0. Now just change the argument at little to show that f^(n+j)=0 for j>=1.