Liouville's theorem - (probably) easy question

[jinsing]

Homework Statement

If f is an entire function and |f(z)|\leq C|z|^(1/2) for all complex numbers z, where C is a positive constant, show that f is constant.

Homework Equations

All bounded and entire functions are constant.

The Attempt at a Solution

I'm 99% sure this can be easily proven using Liouville's theorem, I'm just having trouble proving that f is bounded above by a constant. What should I do with the |z|^(1/2) term?

Thanks for the help!

 

[Dick]

You can't prove it directly with Liouville's theorem, since f(z) isn't bounded. You could look at how Liouville's theorem is proved and adapt the proof.

 

[jinsing]

Would that involve looking at the Cauchy differentiation formula and using the maximum*length principle? Is it because f is entire we know we can use CDF? In other words, do we know there exists a z_0 inside a simple closed curve gamma such that CDF holds because f is entire?

 

[Dick]

Would that involve looking at the Cauchy differentiation formula and using the maximum*length principle? Is it because f is entire we know we can use CDF? In other words, do we know there exists a z_0 inside a simple closed curve gamma such that CDF holds because f is entire?

Yes, it would. If |f(z)| is bounded you can use the Cauchy integral formula to show all of the ak for k>0 in the power series expansion are equal to zero. |f(z)|<C|z^(1/2)| is also good enough. Basically the same proof.

 

[jinsing]

Ah, got it! Thank you so much!