Find all analytic functions $f(z)$ on the open unit disk that satisfy $|f(z)|\leq2^{-\frac{1}{|z|}}$ I am required to find all analytic functions $f(z)$ defined on the open unit disk $D=\{z\in\mathbb{C}\mid |z|<1\}$ that satisfy the following inequality:
$$\forall z\in D\setminus\{0\}:|f(z)|\leq2^{-\frac{1}{|z|}}$$
I found that the constant function $f(z)\equiv0$ is an option. I suspect this is the only option. I noticed that $f(z)$ is bounded on $D$ according to the inequality; However this is actually not an additional information since I already know that $f(z)$ is bounded on $\bar{D}$, as it is analytic (thus continuous). Something about the inequality seems odd to me. I feel that there's nothing special about the function on the RHS and this is merely a specific way to state something more general about $f(z)$. I may be wrong.
Usually I solve this kind of problems using Liouville's Theorem, however $f(z)$ is not entire here so this is not an option.
Thank you!
 A: Nevermind, I solved it eventually!
If $\alpha\in D$ is an infinite order zero of $f$ which is analytic on $D$, then $f(z)\equiv0$. Meaning, if $f(z)\neq0$ for every $z\in D$, then $\alpha\in D$ is a finite order zero of $f$ (or not a zero at all).
Therefore, we can conclude that $z=0$ is a finite order zero of $f$. Meaning, there exists $k\in\mathbb{N}$ such that for every $m\in\mathbb{N}$ that satisfies $0\leq m<k:f^{(m)}(0)=0$ and $f^{(k)}(0)\neq0$.
$f$ is analytic on $D$, thus we can find its Taylor series:
$$f(z)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n=\sum_{n=k}^{\infty}\frac{f^{(n)}(0)}{n!}z^n=z^k\sum_{n=0}^{\infty}\frac{f^{(n+k)}(0)}{(n+k)!}z^n\triangleq z^kg(z)$$
$g(z)$ is analytic on $D$, since it is a convergent power series, such that $f(z)=z^kg(z)$. Notice that $g(0)=\frac{f^{(k)}(0)}{k!}\neq0$.
But we know according to the inequality:
$$0\leq |g(z)|=\left|\frac{f(z)}{z^k}\right|\leq\frac{2^{-\frac{1}{|z|}}}{|z|^k}$$
The limit of the RHS when $z\to 0$ is $0$, thus by the Squeeze Theorem $\lim_\limits{z\to0} |g(z)|=0$, thus $\lim_\limits{z\to0} g(z)=0$. But $g(z)$ is analytic, thus continuous. Therefore:
$$0=\lim_\limits{z\to0} g(z)=g(0)\neq0$$
An absurdity.
A: Cauchy's integral formula for the derivatives gives
$$
| f^{(n)}(0) | \le \frac{n!}{r^n} \sup_{|z| = r} |f(z)| \le \frac{n! 2^{-1/r}}{r^n}
 = \frac{n!}{r^n 2^{1/r}}
$$
for $0 < r < 1$. Since
$$ \tag{*}
 \lim_{r \to 0} r^n 2^{1/r} = \infty
$$
it follows that $f^{(n)}(0) = 0$ for all $n$, i.e. $f$ is identically zero.
Proof of $(*)$:
$$
2^{1/r} = e^{(\log 2)/r} \ge \frac{(\log 2)^{n+1}}{(n+1)! r^{n+1}}
$$ follows directly from the Taylor series of the exponential function, so that
$$
r^n 2^{1/r}  \ge \frac{(\log 2)^{n+1}}{(n+1)!} \cdot \frac 1r \, .
$$
