Home > Complex questions > Complex #26: polynomial properties

Complex #26: polynomial properties

Problem #26: If f(z) is an entire function that is not a polynomial, prove that, given arbitrary C > 0, R > 0, and integer m > 0, there exists a z such that  |z|> R and |f(z)| > C|z|^m.

The way this is worded makes my head hurt, so let’s formulate it equivalently as Not B implies Not A instead of A implies B.
Reformulation: Let f(z) be an entire function. Suppose that there exists a C>0R > 0, and an integer m>0  such that |f(z)|\le C|z|^m whenever |z|>R. Prove that f is a polynomial.

Proof: By the given and a limiting process, |f(z)|\le CR^m for |z|=R. By the maximum modulus principle, the same fact is true for |z|\le R. By the Cauchy-Lagrange inequalities if |z|=\frac{R}{2} , we have |f^{(m)}(z)|\le \frac{m! M}{(R/2)^m}, where M is the maximum value of |f(w)| for |w-z|=\frac{R}{2}. Since this circle is completely contained in the disk \{ w:|w|\le R\}, M\le CR^m, and so we have |f^{(m)}(z)|\le (m!) 2^m C for |z|=\frac{R}{2} and thus for |z|\le \frac{R}{2}. By Liouville’s theorem, f^{(m)}(z) is a constant, and thus f is a polynomial.

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