Is $\left\{\sqrt{n}\sin^{\circ n}\left(\dfrac{z}{\sqrt{n}}\right)\right\}$ locally uniformly convergent over $\mathbb{C}-\{y{\rm i}:|y|\ge\sqrt{3}\}$?

Question

Note $\sin^{\circ n}x$ as the $n$-fold iteration of $\sin x$. It is shown in this great answer that $$ \lim_{n\to\infty}\sqrt{n}\sin^{\circ n}\left(\dfrac{x}{\sqrt{n}}\right)=\dfrac{\sqrt{3}x}{\sqrt{x^2+3}}, \quad\forall x\in\mathbb{R}. $$ I was wondering about the case where $x\in\mathbb{C}$. Based on some numerical tests, it seems that this relation holds for all $x\in\mathbb{C}\setminus\{y{\rm i}:|y|\ge\sqrt{3}\}$ (where $\sqrt{z}$ is the principal square root defined over $\mathbb{C}-(-\infty,0]$). I would like to conjecture that

The function sequence $\left\{\sqrt{n}\sin^{\circ n}\left(\dfrac{z}{\sqrt{n}}\right)\right\}$ is locally uniformly convergent over $\mathbb{C}\setminus\{y{\rm i}:|y|\ge\sqrt{3}\}$.

If this is true, then we know that the limit is holomophic, so it must be $\dfrac{\sqrt{3}z}{\sqrt{z^2+3}}$ by uniqueness of holomorphic functions.

I was thinking that to imitate the proof in the answer to the original question, we need to find a uniform bound for $\left\{\sqrt{n}\sin^{\circ k}\left(\dfrac{z}{\sqrt{n}}\right):k\le n\right\}$, but I have no idea of doing this for complex $z$. So is this result of convergence correct? Any help appreciated.

Answer 1

Obviously, this answer is basically a refinement of Sangchul Lee’s linked answer.

So the answer is yes.

Lemma. Let $0< r < \sqrt{3}$ and $R>0$. Then there exists constants $C,N_0>0$ such that, for all complex numbers $z$ with $|\Im{z}| \leq r$, $|z| \leq R$, and for all $0 \leq k \leq n$ with $n \geq N_0$, one has $|\sin^{\circ k}(z/\sqrt{n})| \leq C/\sqrt{n} < 1$.

Proof of the main claim: let $K$ be a compact contained in our subspace of interest. Let $0 \leq r < \sqrt{3}$ be greater than all the $|\operatorname{Im}{z}|$ for $z \in K$, and let $R>1$ be greater than all the $|z|$ for $z \in K$. Consider the associated $N_0,C>0$.

Let, for every $z \in U:= \{t \in \mathbb{C},\, |\operatorname{Im}{z}| < r,\, |z| < R$, and every $0 \leq k \leq n$, $g_{n,k}(z)=\sin^{\circ k}{z/\sqrt{n}}$, and $f_n=g_{n,n}$.

If $g_{n,k}(z)$ vanishes with $1 \leq k \leq n$, then since $|g_{n,k-1}(z)| < 1$, $g_{n,k-1}(z) = 0$. By induction, we find that $z=0$.

In particular, we find that the $g_{n,k}$ converge uniformly (and also with respect to $k$) to zero on $U’ =U \backslash \{0\}$, so that $g_{n,k}^{-2}(z) -(\sin{g_{n,k}(z)})^{-2}$ converges uniformly in $(z,k)$ to $-1/3$.

In other words, we have $g_{n,k+1}^{-2}(z)-g_{n,k}^{-2}(z) \rightarrow 1/3$ uniformly in $(k,z)$.

By taking a Cesaro average, we find that $1/f_n^2(z) - 1/z^2 \rightarrow 1/3$ uniformly on $U’$.

Because $f_n$ is a bounded sequence of holomorphic functions on $U$ (and we remain at positive distance from $\pm 3i$), we find that $f_n^2$ converges uniformly on $U’$ to $\frac{3z^2}{z^2+3}$ (hence on $U$).

By Montel, it follows that $f_n$ is the reunion of at most two subsequences, one converging to $\lambda_+=\frac{z\sqrt{3}}{\sqrt{z^2+3}}$, and the other one to $\lambda_-=-\frac{z\sqrt{3}}{\sqrt{z^2+3}}$.

But $f_n(1) \rightarrow \frac{\sqrt{3}}{2}$ (look at the sign), so $d(f_n,\lambda_-)$ is bounded below. Therefore $f_n \rightarrow \lambda_+$ (locally uniformly).

Proof of Lemma: note that it’s enough to allow $C,N_0$ to depend on $z$ if the function $z \longmapsto C(z)+N_0(z)$ is locally bounded.

Write $z=x+iy$.

There is a constant $D$ such that for all complex numbers $z$ with $0<|z| \leq 1$, $|(\sin{z})^{-2}-z^{-2}-1/3| \leq D|z|^2$.

Choose $C>0$ (we’ll specify it later) and let $n \geq (10+C+R+D)^4$.

Let $m$ be the first integer such that $x_m > C/\sqrt{n}$ (if it doesn’t exist, we’re done). For $0 \leq k < m$, $|(x_{k+1}^{-2} -x_k^{-2}) -1/3| \leq DC^2/n$. Then $x_m^{-2}=n/z^2+m/3+u$ with $|u| \leq mDC^2/n$.

Suppose that $x^2 \geq y^2$, then take $C=R^2+1$.

Because $\operatorname{Re}(z^{-2}) \geq 0$, if $n \geq m$, we get $n/C^2 \geq |x_m^{-2}| \geq n/R^2-DC^2$. We get a contradiction for large enough $n$ (and the bound is explicit).

Suppose that $\rho=\frac{y^2-x^2}{(x^2+y^2)^2} > 1/3$. Then choose $C$ such that $\rho-1/C > 1/3$.

Again, we get, assuming that $m \leq n$, $n/C^2 \geq |x_m^{-2}| \geq n/|z|^2-m/3-DC^2 \geq n(\rho-1/3)-DC^2$. This is again a contradiction for (explicitly) large enough $n$.

So we’re left in the case where $x=ty$, $|t| < 1$ and $\frac{1-t^2}{(1+t^2)^2} \leq y^2/3$.

This implies in particular that $\frac{2|t|}{(1+t^2)^2} \geq sy^2/3$, for some explicit constant $0 < s< 1$.

In particular, we get, assuming that $n \geq m$, $n/C^2 \geq |x_m^{-2}| \geq |\operatorname{Im}(n/z^2)|-DC^2 \geq n\cdot s -DC^2$. Take $C$ such that $1/C^2 < s/2$, and we get a contradiction when $n$ is (explicitly) large enough.