Prove a function is constant. (complex analysis)

Suppose function $f$ is holomorphic on $\mathbb C-\{0\}$ and satisfies $$|f(z)| \le \sqrt {|z|} + \frac{1}{\sqrt {|z|}}.$$ Prove that $f$ is a constant function.

I think it's related to Laurent series representation and ML inequality but I have no idea how to prove it. Can anyone give me some hints? Thanks in advance!

Solution 1:

Let $$ f(z) = \sum_{n=-\infty}^\infty a_n z^n $$ be the Laurent series for $f$ in $\Bbb C \setminus \{ 0 \}$. The coefficients $a_n, n \in \Bbb Z,$ can be computed as an integral (compare https://en.wikipedia.org/wiki/Laurent_series) $$ a_n = \frac{1}{2 \pi i} \int_\gamma \frac{f(z)}{z^{n+1}} \, dz $$ where $\gamma$ is any closed curve which surrounds $z=0$ exactly once. Taking $\gamma$ as a circle with radius $r > 0$ gives the estimate $$ |a_n| \le \frac{1}{r^n} \max \{ |f(z)| : |z| = r \} \le \frac{ \sqrt {r} + \frac{1}{\sqrt {r}}}{r^n} \, . $$ For fixed $n \ge 1$ and $r \to \infty$ the RHS tends to zero, and the same is true if $n \le -1$ and $r \to 0$. Therefore $a_n = 0$ for all $n \ne 0$.

Alternatively, you can argue as suggested in the comments. $g(z) = z f(z)$ has a removable singularity at $z=0$ and is therefore an entire function. From $$ |g(z)| \le r^{1/2} + r^{3/2} $$ the usual estimates of the derivative with the Cauchy integral formula show that $g$ is a polynomial of degree at most one.

So we have $$ f(z) = \frac{a z + b}{z} $$ for some constants $a, b$, and from the growth restriction for $r \to 0$ it follows that $b = 0$ must hold.