$|f(z)|\leq e^{-1/|z|}$ implies that a Holomorphic function is identically 0 on upper half plane

This is a question from an exam in an undergraduate complex analysis course.

We denote $G=\{z\in \mathbb{C}|Im(z)>0\}$ - the upper half plane. Given a holomorphic function $f\in Hol(G)$ such that $|f(z)|\leq e^{-1/|z|}$ for all $z\in G$, prove that $f=0$ identically on G.

The question provides a hint, which is to take a large positive $R>0$, and a sufficiently large natural number $N$, and define the function: $$g(z)=\prod_{k=0}^{N-1}f(e^{2\pi ik/N}*z+iR)$$ on the circle $D_R=\{z\in \mathbb{C}||z|<R\}$. I tried to define a sequence of such functions $g_N(z)$ and managed to show that they converge uniformly to 0 when $N$ approaches infinity through: $$0\leq |g_N(z)|=\prod_{k=0}^{N-1}|f(e^{2\pi ik/N}*z+iR)|\leq e^{-N/|z|}\rightarrow0$$However I got stuck afterwards, and couldn't find a way to apply the identitiy theorem as I wanted.

• I don't know an answer, I guess one should figure out where it is important that $R$ and $N$ are "sufficiently big"...how did you get that that product is $\le e^{-N/|z|}$? Jul 28 '18 at 18:54
• My guess at how a solution might go: for large enough $R,N$, $g_{N,R}(z)$ satisfies some kind of bound like that, which is an analytic function on $|z|<R$ that vanishes to infinite order at 0. Hence $g=0$, which means at least one of the terms forming the product is 0. Letting $N$ run through bigger and bigger primes, we see that the zero set of $f$ is not isolated, so $f$ is identically 0. Jul 28 '18 at 19:00
• I think $\leq e^{-N/|z|}$ which I wrote might actually be wrong. Your intuition seems correct but for a finite $N$ we don't necessarily vanish, and at the limit $g$ is no longer a finite product, so I don't know if a single term needs to be $0$ on its own. Jul 28 '18 at 19:19
• If a bound like that is correct, then you /do/ vanish at $z=0$, not talking about the limit $N→ ∞$. Note $e^{1/|z|} ≥ \frac{1}{K!|z|^K}$ so $e^{-1/|z|} = o(|z|^K)$ as $z→ 0$ for every $K$ Jul 28 '18 at 19:35