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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!

cqfd
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math_student
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  • Well, if you alraedy suspect it's related to the Laurent series you should write that down. Then consider growth of the function as $|z|$ tends to $0$ and $\infty$. – Thomas Apr 17 '16 at 06:53
  • Im not sure but is this idea correct? as |z| tends to zero, the negative power terms in the Laurent series has something to do with 1/sqrt(|z|) part (maybe compare the magnitude?) and as |z| tends to infinity, the positive power terms has something to do with sqrt|z| part? – math_student Apr 17 '16 at 07:03
  • Oh can it be done by taking the Laurent series, and for the coefficient,( which can be obtained by integrating along a closed loop), take the loop with infinite radius? – math_student Apr 17 '16 at 07:14
  • Yes, that appears to be the idea. This is a real analysis approach to the problem, there may some complex analysis approach which (I don't know of without doing some research and which) is easier – Thomas Apr 17 '16 at 07:15
  • Ok i will try. Thank you! – math_student Apr 17 '16 at 07:18
  • Hint: $g(z)=zf(z)$ can be shown to have a removable singularity at $z=0$ with value $0$. What does that say about $f(z)$ at the origin? – Greg Martin Apr 17 '16 at 07:39
  • it means f has a removable singularity at z=0 with arbitrary complex value, but not infinity? – math_student Apr 17 '16 at 07:51
  • Anyone can provide a solution or more hints? I've tried for few hours and still im not able to prove it.. – math_student Apr 17 '16 at 12:16

1 Answers1

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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.

Riemann
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Martin R
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