Show that $\{n^2F(\frac{1}{n})\}$ is bounded.

Question

Given that $F:\Bbb C\to \Bbb C$ is an analytic function such that $|F(\frac{1}{n})|\le \dfrac{1}{n^{\frac{3}{2}}}\forall n\in \Bbb N$.Show that $\{n^2F(\frac{1}{n})\}$ is bounded.

Though there is already an answer posted here by another user to show $\{n^2f(\frac{1}{n})\}$ is bounded. I am unable to follow it up.

Since $F$ is analytic so it is continuous and together with $|F(\frac{1}{n})|\le \dfrac{1}{n^{\frac{3}{2}}}\forall n\in \Bbb N$ we have $F(0)=0$. So $F(z)=zG(z);G(0)\ne0$ and $G$ is analytic.

As posted in the answer we have to use Schwarz Lemma which states that

If $f$ is analytic in the open unit disc and $f(0)=0$ then $|f(z)|\le |z|$ and $|f^{'}(0)|\le 1$

But how should I use the above lemma here.Please provide some help.

Answer 1

Indeed $G(0)=0$: $|\frac{1}{n}G(\frac{1}{n})|=|F(\frac{1}{n})|\le\frac{1}{n^{3/2}}$. Thus $|G(\frac{1}{n})|\le\frac{n}{n^{3/2}}=\frac{1}{n^{1/2}}$, by continuity of $G$ we get $G(0)=0$.

Now, since $G$ is analytic and $G(0)=0$, we have $|G(z)|\le|z|$ on the unit disc. Therefore $|F(z)|=|zG(z)|\le|z|^2$ on the unit disc. Taking $z=\frac{1}{n}$, done.