Assume that $f$ is holomorphic in $B(0,R)$ and that $|f(z)|\leq e^{-\frac{1}{|z|}} $ for all $0<|z|Assume that $f$ is holomorphic in $B(0,R)$ and that $|f(z)|\leq e^{-\frac{1}{|z|}} $ for all $0<|z|<R$ then $f=0$
Not sure how to proceed, usually we try Liouville but $ e^{-\frac{1}{|z|}}$ has an essential singularity, so that will not work. Another method uses Cauchy inequality, but that is certainly not helpful here due to Little Picard theorem. Any hints/solutions appreciated.
 A: The given estimate implies that $f(0) = 0$. The idea is that for $z \to 0$, the upper bound $e^{-1/|z|}$ decreases so fast to zero that $f$ cannot have a zero of finitely multiplicity at $z=0$.
Concretely: If $f$ is not identically zero then $f(0) = 0$ with some multiplicity $k \ge 0$, i.e.
$$
 a_k = \lim_{z \to 0} \frac{f(z)}{z^k}
$$
exists and is not zero. But
$$
\left|\frac{f(z)}{z^k} \right| \le \frac{e^{-1/|z|}}{|z|^k} \to 0
$$
for $z \to 0$ and any non-negative integer $k$, which is a contradiction.
The last limit is zero because
$$
 \lim_{r \to 0 } \frac{e^{-1/r}}{r^k} = \lim_{x \to \infty }\frac{x^k}{e^x} \le \lim_{x\to \infty }\frac{x^k}{x^{k+1}/(k+1)!} = 0 \, ,
$$
i.e. because the exponential function  growth faster than any polynomial.
One can also use the Cauchy estimates for the derivatives: If $f(z) = \sum_{n=0}^\infty a_n z^n$ is the Taylor series of $f$ in $B(0, R)$ then
$$
 |a_n| \le \frac{1}{r^n} \max_{|z|=r} |f(z)| \le \frac{e^{-1/r}}{r^n}
$$
for all $n$ and $0 < r < R$. As above, the right-hand side tends to zero for $r \to 0$, so that all $a_n = 0$.
