Suppose that $f$ is analytic in the punctured plane $z\neq 0$ and satisfies $|f(z)|\leq\sqrt{|z|}+1/\sqrt{|z|}$. Prove $f$ is constant.

I think I will approach it this way:

1) Show that the isolated singularity is removable.

2) Use Liouville's Theorems to finish the proof.

But I am only asking for a hint as to how to show that the singularity is removable. The only thing we know in my class right now is Riemann's Principle, so that's how I would have to do it.



Hint: what can you say about the function $g(z) = zf(z)$ near $z=0$?

  • $\begingroup$ Well, that's what I've been trying to figure out. The function $f$ is bounded by $\sqrt{|z|}+1/\sqrt{|z|}=\frac{|z|+1}{\sqrt{|z|}}$, but this expression seems to me as if it will approach infinity as z approaches 0. $\endgroup$ – nonremovable Apr 1 '15 at 23:34
  • $\begingroup$ @nonremovable read my hint again. Don't look at $f$ near $0$, but at $zf$. $\endgroup$ – mrf Apr 2 '15 at 5:06

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