If $|f(z)| < \sqrt{\left| z\right|}$, then $\lim_{z\to\infty} f(z)$ exists?

Question

Let $f(z)$ be a holomorphic function defined on $D=\{ z\in\mathbb{C} | \left| z \right| > 1\}$. For all $z\in D$, we have $\left| f(z) \right| \le \sqrt{\left| z \right|}$. Show that $\lim_{z\to\infty} f(z)$ exists.

How should I show this? I was thinking could I show that $f(z)$ is actually constant, but that doesn't seem to be the case. I think I can find $f(z)$ that is not constant but has limit at infinity. What is the right "picture" to think about here? Hint is greatly appreciated.

Update: So I considered $g(z) = f(\frac{1}{z})$. Take $C_R$ be circle of radius $R$. Then I got something like

$$\int_{C_R} \left| g(z) \right| dz=\int_{C_R} \left| f(\frac{1}{z})\right| dz\le \int_{C_R} \frac{1}{\sqrt{R}} dz = 2\pi \sqrt{R}$$

Then by Residue Theorem,

$$2\pi i \text{Res}_{z= 0} g(z) = 2\pi \sqrt{R}$$

Since this is true for all $R$, $\text{Res}_{z=0} f(z) = 0$

Update 2: I actually considered $\lim_{z\to 0} zg(z)$. Since $$ \left| z f\left(\frac{1}{z}\right)\right|<\left| \frac{z}{\sqrt{\left| z\right|}}\right| \to 0 \text{ as } z \to 0$$

$g(z)$ should have a removable singularity at $z=0$. So $f(z)$ should have a removable singularity at infinity?

Answer 1

You proved $f\left(\frac{1}{z}\right), |z|<1$ has a removable singularity at $0$ and by Riemann's removable singularity theorem $f\left(\frac{1}{z}\right)$ is continuously extensible over $0$ or $\lim_{z \rightarrow 0} f\left(\frac{1}{z}\right)$ exists. Similarly $f(z)$ has a removable singularity at $\infty$ or the limit exists, more details here, 8.4.

Alternatively (a longer version), the function is holomorphic on $1 < |z-0|<R$ so it can be represented by a Laurent series: $$f(z)=\sum_{n=-\infty}^{\infty}a_nz^n$$ where $$a_n=\frac{1}{2\pi i}\oint_{C_R} \frac{f(z)}{z^{n+1}}dz$$ for any $R>1$.

Case 1. $n \geq 1$ $$|a_n|=\left| \frac{1}{2\pi i}\oint_{C_R} \frac{f(z)}{z^{n+1}}dz \right|\leq \frac{1}{2\pi} \oint_{C_R} \left| \frac{f(z)}{z^{n+1}}\right| dz \leq \frac{1}{2\pi} \oint_{C_R} \left| \frac{\sqrt{|z|}}{z^{n+1}}\right| dz=$$ $$=\frac{1}{2\pi} \oint_{C_R} \frac{\sqrt{|z|}}{|z|^{n+1}} dz=\frac{1}{2\pi} \frac{1}{R^{n+\frac{1}{2}}} \oint_{C_R} dz=\frac{1}{2\pi} \frac{1}{R^{n+\frac{1}{2}}} 2\pi R = \frac{1}{R^{n-1+\frac{1}{2}}}$$ Taking limit $R \rightarrow \infty$ we obtain $|a_n|=0, n \geq 1$. So far $$f(z)=\sum_{n=-\infty}^{0}a_nz^n$$

Case 2. $n\geq 0$ $$|a_{-n}|=\left| \frac{1}{2\pi i}\oint_{C_R} f(z)z^{n-1} dz \right| \leq ... \leq \frac{1}{2 \pi} R^{n-\frac{1}{2}} 2\pi R=R^{n+\frac{1}{2}}$$ Taking limit $R \rightarrow 1$ we obtain $|a_{-n}| \leq 1, n\geq 0$.

As a result $$f(z)=a_0+\sum_{n=1}^{\infty}\frac{a_{-n}}{z^n}, |z|>1$$ Or $$|f(z)|=\left| a_0+\sum_{n=1}^{\infty}\frac{a_{-n}}{z^n} \right| \leq |a_0|+\sum_{n=1}^{\infty}\frac{|a_{-n}|}{|z|^n} \leq 1 + \sum_{n=1}^{\infty}\frac{1}{|z|^n} = \frac{1}{1-\frac{1}{|z|}} =\frac{|z|}{|z|-1} $$

Altogether, we have $|f(z)| \leq \sqrt{2} < 2$ for $1< |z| \leq 2$ and $|f(z)| \leq \frac{|z|}{|z|-1} < 2$ for $|z|>2$. Or $|f(z)| < 2$ on $D$. Same applies to $f\left(\frac{1}{z}\right), |z|<1$ and by the same Riemann's removable singularity theorem $f\left(\frac{1}{z}\right)$ has a removable singularity at $0$ and so on ...