Show that any polynomially bounded C-differentiable function on C must be a polynomial, of degree <= N.?

A function f: C-%26gt;C is said to be polynomially bounded if there are constants C1, C2 and a non-negative integer N such that:





|f(z)|%26lt;= C1+c2|z|^N, for all z element of C.





Show that any polynomially bounded C-differentiable function on C must be a polynomial, of degree %26lt;= N.

Show that any polynomially bounded C-differentiable function on C must be a polynomial, of degree %26lt;= N.?
We'll prove this by mathematical induction. For the case when N = 0, this means that f is a bounded, entire (ie. C-differentiable on the whole of C) function; thus, by Liouville's theorem, f is constant.





Suppose the result holds for some non-negative integer N. Given an entire function f for which





|f(z)| %26lt;= C1 + C2.|z|^(N+1) for a z in C,





define g: C --%26gt; C by





g(0) = f'(0),


g(z) = [f(z) - f(0)] / z, for non-zero z.





Then g is an entire function which satisfies





|g(z)| %26lt;= D1 + D2|z|^N





for some positive constants D1 and D2, and so is a polynomial of degree no greater than N, which, by our definition of g, implies that f is a polynomila of degree no greatern that N+1. Hence, by induction, the result holds for all non-negative integers N.
Reply:I am not sure about the previous answer but using Liouville's theorem is definitely the key. In fact, my solution is as simple as going to this page http://www.algebra.com/~pavlovd/wiki/Lio...





and repeating the steps of the proof one to one, except for the last two inequalities. In this case |f| is bounded differently and hence the net results is going to be that





|a_k| %26lt;= (C1 + C2 |r|^N)/|r|^k





which implies that for k%26gt;N when r goes to infinity (same as in Liouville's theorem proof) |a_k| %26lt;= 0 which implies all those coefficients are 0s and then the Taylor series representation for f is exactly the polynomial of degree %26lt;= N.