Denote
$$F(z)=\prod _{k=0}^{\infty}\text{sinc} \left(\frac{\pi z}{2 k+1}\right)=\prod _{n=1}^{\infty } \cos \left(\frac{\pi z}{2 n}\right)$$
How can we prove $F\in S(\mathbb{R})$ (Schwartz space) ? I've already shown that $F(z)$ is entire and rapidly decreasing in strip $|\Im(z)|≤r$ for $r>0$.
Background: This arises from solving Borwein integrals via Fourier transform.
Best Answer
Following reun's hint, $|f(x+iy)|$ is donimated by rapidly decreasing $h(x)$ on strip $\Im z\leq r$, $r>0$ ($\text{sinc}(z)=O(\frac{1}{x})$ on the strip as $x\to \infty$). Cauchy formula gives $|f^{(k)}|\leq k! (r/2)^{-k} h(x-r/2)$, thus all derivatives of $f$ are also rapidly decreasing, so $f\in S$.