Why is $|m(\xi)| < C_k\left(1 + \sum_{|\xi_j| < R^\delta} \xi_j^2\right)^{-\frac{k}{2}}R^{\delta k}$? Let $$m(\xi) = \prod_{j=1}^n \frac{\sin \pi \xi_j}{\pi \xi_j}$$ where $\xi = (\xi_1,\dots,\xi_n) \in \mathbb{R}^n$.
In a paper I am reading, it is claimed that

Claim. For all $\delta > 0$ and $k \ge 1$,
$$ |m(\xi)| < C_k\left(1 + \sum_{|\xi_j| < R^\delta} \xi_j^2\right)^{-\frac{k}{2}}R^{\delta k}, \label{ineq}\tag{1}$$ where $R > 1$ and $C_k$ is a constant depending only on $k$.

The proof given was quite terse:

Denote $I_0 = \{ j : 1 \le j \le n, |\xi_j| > 1\}$. We have
$$
\prod_{j \notin I_0} \left|\frac{\sin \pi \xi_j}{\pi \xi_j}\right| \le e^{-c\sum_{j\notin I_0}\xi_j^2} \label{2}\tag{2}
$$ and
$$
\prod_{j \in I_0} \left|\frac{\sin \pi \xi_j}{\pi \xi_j}\right| \le e^{-c|I_0|}. \label{3}\tag{3}
$$
Estimating
$$
\sum_{|\xi_j| < R^\delta} \xi_j^2 < R^{2\delta}|I_0| + \sum_{j\notin I_0}\xi_j^2, \label{4}\tag{4}
$$
$\eqref{ineq}$ follows.

I see how to deduce $\eqref{2}$ using the Taylor series for $\sin x \over x$ for $|x| \le 1$, and $\eqref{3}$ using simple inequalities. I also see $\eqref{4}$ easily. My confusion is how it implies $\eqref{ineq}$.
Any comments, hints or clarifications are welcome.
 A: Let $\xi = (\xi_1,\cdots,\xi_n) \in \mathbb{R}^n$ be given. By (2) and (3), there exists a constant $c>0$ such that, adopting all your notations:
\begin{equation*}
\begin{split}
|m(\xi)| &= \left|\prod_{j \in I_0}\frac{\sin \pi \xi_j}{\pi \xi_j}\prod_{j \notin I_0}\frac{\sin \pi \xi_j}{\pi \xi_j}\right|  \\ &\leqslant e^{-c(\sum_{j \notin I_0}\xi_j^2+|I_0|)}\\ &< \exp{\{-c (-R^{2\delta}|I_0|+|I_0|+\sum_{|\xi_j|<R^{\delta}}\xi_j^2)\}} 
\\&= \exp{\{-c(\sum_{|\xi_j|<R^{\delta}}\xi_j^2+1)+c+c|I_0|R^{2\delta}-c|I_0|\}}\\ &= d \exp{\{-c(\sum_{|\xi_j|<R^{\delta}}\xi_j^2+1)\}}\exp{\{c|I_0|R^{2\delta}\}}
\end{split}
\end{equation*}
Here $d$ is a constant depending on $c$ and $|I_0|$. Now let $k \in [1,\infty)$ be given (no need to be integer). If $x=\sum_{|\xi_j|<R^{\delta}}\xi_j^2+1$, we can use the limit:
\begin{equation*}
\lim_{x \to \infty} \frac{x^{\frac{k}{2}}}{e^{cx}} = 0
\end{equation*}
That is, the left side is positive and bounded for $x >0$. There exists a constant $C_k>0$ dependent only on $k$ such that $\frac{x^{\frac{k}{2}}}{e^{cx}}<C_k$, or equivalently $e^{-cx} < C_k x^{-\frac{k}{2}}$. I will leave the rest to you.
