Does there exist a holomorphic function $f$ such that $|f(z)|\geq \frac{1}{\sqrt|z|}$ for all $z \in \mathbb{C} \backslash \{ 0 \} $

Question

This is a question that was asked a few years ago at Does there exist an holomorphic function such that $|f(z)|\geq \frac{1}{\sqrt|z|}$?.

Does there exist a function $f$ holomorphic on $\mathbb{C} \setminus \{ 0 \}$ such that $|f(z)|\geq \frac{1}{\sqrt|z|}$ for all $z\in\mathbb C \setminus \{0\}$?

While i have understood the answer mentioned in the link, I wanted to know whether we can apply the Maximum Modulus Theorem(MMT) to solve this. I tried as follows:

Let the domain of $f$ be the punctured open disc, $|z| < 1, z \neq 0$. The boundary for this domain is $|z| = 1$ and $z = 0$. By MMT, if $f$ did indeed exist satisfying the given conditions, then the maximum of $f$, if it exists, must exist at the boundary of the domain : in this case at $|z|=1$ as $z=0$ is not a part of the domain. But we see that $|f|$ blows up as we get arbitrarily close to $0$. Therefore, the maximum doesn't occur at $|z|=1$ for sure. My question is, does a maximum occur anywhere at all, that I may then conclude that, by MMT, such an $f$ canNOT exist?

Answer 1

No, it is not correct. By the same argument, there would be no holomorphic function $f\colon D(0,1)\setminus\{0\}\longrightarrow\mathbb C$ such that$$\bigl(\forall z\in D(0,1)\setminus\{0\}\bigr):\bigl\lvert f(z)\bigr\rvert\geqslant\frac1{\lvert z\rvert}.$$But there is! You just take $f(z)=\frac1z$. The problem lies in assuming that the maximum is located at the boundary. Why? This only has to be truefor compact subsets of the domain.