MHB Is $f$ Identically Zero on the Punctured Unit Disc?

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Here's this week's problem!

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Let $f$ be a holomorphic function on the punctured unit disc $\Bbb D \setminus \{0\}$ such that

$$|f(z)| \le \log \frac{1}{|z|} \quad \text{for all} \quad z \in \Bbb D \setminus \{0\}.$$

Prove that $f \equiv 0$.

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No one answered this week's problem. Here is my solution.

Spoiler

Since $f$ is holomorphic on $\Bbb D\setminus\{0\}$ with $|zf(z)| \le -|z|\log|z| \to 0$ as $z\to 0$, it has a removable singularity at $z = 0$. Let $a_n$ denote the $n$th coefficient of the power series expansion of $f$ near $z = 0$. By the Cauchy estimates, for all $r\in (0,1)$,

$$|a_n| \le r^{-n} \max_{|z| = r} |f(z)| \le r^{-n}\log \frac{1}{r} \quad (n = 0, 1, 2, 3,\ldots)$$

Letting $r \to 1^-$, the result follows.