# Complex Analysis Modulus Inequality

I am working on a proof, and I'm stuck with one inequality. I am given that for all $z$ (and large $z$ in particular) and $g$ an entire function, we have $$|g(z)|\leq\sqrt{|z|}+1/\sqrt{|z|}$$

Now I've been working on a contradiction in part of my proof and have arrived at this fact. There exists a $c>0$ such that

$$|g(z)|>c|z|$$

I suspect this is a contradiction when $z$ is large (the $1/\sqrt{|z|}$ term becomes negligible) but I can't seem to find a proof. Any help would be much appreciated. Thank you.

This is a nice case of Liouville's theorem. One take-away is that if $g$ is entire and bounded (for large enough $z$) by a polynomial of degree $k$, then $g$ is a polynomial of degree at most $k$. In this case you are bounded by a linear function (for large enough $z$).
• Thank you. I actually already discovered that, but now I am going further to prove that my function $g$ is, in fact, constant. I am assuming by way of contradiction that $g$ itself is a polynomial of degree 1 and that is why I am looking for a contradiction in those two inequalities that I wrote. – nonremovable Apr 3 '15 at 0:30
Cauchy's estimates: Let $M(r)$ be the maximum of $|g(z)|$ for $|z|\le r.$ Then $|g^{(n)}(0)| \le n!\frac{M(r)}{r^n}\le n!\frac{\sqrt r +1/\sqrt r}{r^n}.$ Let $r\to \infty$ to see $g^{(n)}(0)=0$ for $n>0.$ Thus $g$ is a constant (of modulus $\le 2$).