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Is there a function analytic on $\mathbb{C}-\{0\}$ such that $|f(z)|\ge \frac{1}{\sqrt{|z|}}$ for $z\ne 0$?

My attempt: $f(z)\ge \frac{1}{\sqrt{|z|}}\implies |zf(z)^2|\ge 1$ for $z\ne 0$. Thus $g(z):=\frac{1}{zf(z)^2}$ is analytic everywhere except at $0$ and $|g(z)|\le 1$. Thus we can conclude that $g$ has a removable singularity at $0$ and we will let $\tilde{g}$ denote the analytic continuation of $g$. Thus $\tilde{g}$ is a function that is analytic everywhere and bounded. Hence by Liouville's Theorem, it must be constant. Thus implies $f(z)^2=\frac{c}{z}$, but $\frac{1}{\sqrt{z}}$ is not an analytic function on it's domain of definition. Thus there can be no such function $f$.

Is this solution correct?

Bruno B
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Takamoto Yuji
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  • This solution is indeed correct. I think the only thing of note would be that I personally wouldn't use a bar for $\bar{g}$ due to conjugation being a thing in the complex numbers but that's really only a matter of preference and nitpicking. – Bruno B Jul 16 '23 at 16:04
  • @BrunoB that’s a tilde, but yea, I’m not a fan of tilde here either lol – peek-a-boo Jul 16 '23 at 16:09
  • @peek-a-boo oh wow! It looks like a bar with normal zooming out, but maybe it's just my computer. When zooming in it is a tilde though indeed. – Bruno B Jul 16 '23 at 16:11

1 Answers1

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Almost correct. You really need to do a little case work. If $c\neq 0$, then you use the fact that $\frac{1}{\sqrt{z}}$ and hence $\frac{c}{\sqrt{z}}$ cannot be entire. On the other hand, if $c=0$, then it follows that $f=0$ identically, but this would violate your initial inequality, so in both cases we get a contradiction.

Here’s another way of presenting: since $f$ is holomorphic on $\Bbb{C}\setminus\{0\}$ and is nowhere-vanishing (due to the inequality) it follows $g=\frac{1}{f}$ is holomorphic here and satisfies $|g(z)|\leq\sqrt{|z|}$, and thus has an entire extension (which I shall continue to denote $g$) with $g(0)=0$. Now, the growth condition $|g(z)|\leq \sqrt{|z|}$ already implies $g$ is constant, so $g=g(0)=0$ identically, but this is not possible since $g=\frac{1}{f}$ is the quotient of a holomorphic function. So, actually what this proof shows is that $f=\infty$ identically, and this is the only holomorphic (in the extended sense of mapping into the Riemann sphere) function which satisfies this inequality.

As for how I concluded $g$ must be constant, it is a consequence of Cauchy’s inequalities, and this is a general principle to keep in mind: whenever you have some growth estimates on a holomorphic function, by plugging that into Cauchy’s integral formula, you can obtain conclusions for certain Laurent coefficients (in particular you can prove that it is entire, or that it must be a polynomial of such and such degree etc). See here for an illustration.

peek-a-boo
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