# Entire, $|f(z)|\le1+\sqrt{|z|}$ implies $f$ is constant

I am stuck on the following question.

Given that $f$ is an entire function with $|f(z)|\le1+\sqrt{|z|}$ for all $z\in \mathbb{C}$, show that $f$ is constant.

Can anyone give me a hint to get me started?

OK, based on one of the hints below, I define: $$g(z)=\frac{f(z)-f(0)}{z}$$ Now, let me assume that this is entire by defining $g(0)=f'(0)$. Is that fair, and if so, why?

Continuing, \begin{align*} \left|g(z)\right| &=\left|\frac{f(z)-f(0)}{z}\right|\\ &\le \left|\frac{f(z)}{z}\right|+\left|\frac{f(0)}{z}\right|\\ &=\frac{|f(z)|}{|z|}+\frac{|f(0)|}{|z|}\\ &\le\frac{1+\sqrt{|z|}}{|z|}+\frac{|f(0)|}{|z|}\\ &=\frac{1}{|z|}+\frac{\sqrt{|z|}}{|z|}+\frac{|f(0)|}{|z|}\\ &=\frac{1}{|z|}+\frac{1}{\sqrt{|z|}}+\frac{|f(0)|}{|z|}\\ &\le 1+1+|f(0)|\\ &=2+|f(0)|, \end{align*} provided $|z|\ge 1$.

Next, $g$ is entire on the disk $D=\{z:\,|z|\le 1\}$, which is compact, so it must assume a maximum value on this disk, say $|g(z)|\le M$ on $D$. Taking $M_s$ as the smaller of $2+|f(0)|$ and $M$, we have $|g(z)\le M_s$ for all $z\in C$.

Now, by Liouville's Theorem, $g$ is constant.

However, going back to the beginning of the argument, my only worry is whether $g$ is made entire by defining $g(0)=f'(0)$. I am not sure that is valid. Can someone comment on this and also comment on the above if there are errors?

Thanks.

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I guess you have to use Picard's little theorem. –  user32240 Mar 10 '13 at 22:07
What type of singularity can this have at $\infty$? –  GEdgar Mar 10 '13 at 22:08
Hint: write $f(z)-f(0)=zg(z)$ with $g$ entire and try to apply Liouville. –  julien Mar 10 '13 at 22:44
You can prove $g(z)$ is entire by examining the Taylor expansion of $f(z)-f(0)$. –  JSchlather Mar 11 '13 at 0:57
And you're not done. Since $g$ is constant $f(z)=az+b$. It remains to show that $a=0$. –  julien Mar 11 '13 at 1:07

Hint: Apply Cauchy's estimates theorem which says that if $\left|f(z)\right| \le M$ for all $z \in D(z_0, R)$ then:

$$\left|f^{(n)}(z_0)\right| \le \frac{n! M}{R^n} \quad n \in \Bbb N, n \ge 1$$

To elaborate, apply the theorem to $z_0 = 0$. For all $z \in D(0, R)$, we have $\left|f(z)\right| \le 1 + \sqrt{|z|} \le 1 + \sqrt{R}$. Thus:

$$\left|f^{(n)}(0)\right| \le \frac{n! \left(1 + \sqrt{R}\right)}{R^n} \quad n \in \Bbb N, n \ge 1$$

The RHS goes to $0$ as $R \to \infty$ (remember, $f$ is entire). This forces $f^{(n)}(0)$ to be zero for all $n \ge 1$. It follows that $f$ is constant.

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I'm sorry, I've just not made enough progress to see how this relates to the original question above. Any suggestions? –  David Mar 11 '13 at 0:42
@David I added the full solution above. –  Ayman Hourieh Mar 11 '13 at 0:47
Now I realize that it's faster to prove a stronger form of Liouville than to try to use it. +1. –  julien Mar 11 '13 at 1:59
Show that if $f$ does not have a zero it is constant.
If $f$ has a zero write $f(z)=(z-a)g(z)$ and deduce that $g$ is constant by applying your inequality.
This phrase "Show that if f does not have a zero it is constant." Is that true in general, or only true with the added requirement that $|f(z)|\le 1+\sqrt{|z|}$? And if it is true in general without the added inequality, how would one go about proving it? –  David Mar 11 '13 at 0:41
@David Consider the function $f(z)=e^z$. In general we do have Picard's little theorem, which says that if an entire function misses two points it is constant. –  JSchlather Mar 11 '13 at 0:55