4
$\begingroup$

The problem is as follows:

Suppose $f$ entire satisfying $ |f(z)| \leq A + B |z|^{3/2} $ for some fixed $A,B > 0$. Prove that $f$ is a linear polynomial.

I know I want to reduce it to a function where I can use a Cauchy bound, but I'm not sure how.

  • 3
    possible duplicate of [if $f$ is entire and $|f(z)| \leq 1+|z|^{1/2}$, why must $f$ be constant?](http://math.stackexchange.com/questions/151700/if-f-is-entire-and-fz-leq-1z1-2-why-must-f-be-constant)2012-11-02

2 Answers 2

4

The first technique that comes to mind is the Cauchy integral formula. $f^{(n)}(0) = \oint_{\Gamma} \dfrac{f(z)}{z^{n+1}} dz$ Hence, we have that $\left \vert f^{(n)}(0) \right \vert = \left \vert \oint_{\Gamma} \dfrac{f(z)}{z^{n+1}} dz \right \vert \leq \oint_{\Gamma} \dfrac{\left \vert f(z) \right \vert}{\left \vert z \right \vert^{n+1}} dz \leq \oint_{\Gamma} \dfrac{A + B \vert z \vert^{3/2}}{\left \vert z \right \vert^{n+1}} dz$ Take $\Gamma$ to be a circle of a very large radius $R$. We then have that$\left \vert f^{(n)}(0) \right \vert \leq \dfrac{A_1}{R^n} + \dfrac{B_1}{R^{n-3/2}}$ We can let $R$ to be arbitrarily large and hence for $n \geq 2$, we have that $f^{(n)}(0) = 0 \,\,\,\, \forall n \geq 2$ Hence, $f(z) = f(0) + f'(0) z$

1

Instead of using Cauchy estimate as shown in the comments, alternatively you may apply the maximum modulus principle to the function $\frac{f(z)-f(0)-f'(0)z}{z^2}$ to show that it is constantly $0$.