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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 $|\Im(z)|â‰Īr$ for $r>0$.

Background: This arises from solving Borwein integrals via Fourier transform.

Infiniticism
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  • The naive bound fails but it is clear the solution is to show that $f(z)=\prod_{k=0}^{\infty}\text{sinc} \left(\frac{\pi z}{2 k+1}\right)$ is entire and rapidly decreasing on the horizontal strips $|\Im(z)|\le r$ from which the Cauchy integral formula implies that all its derivatives are rapidly decreasing too, so that $f(t)$ is Schwartz and so is its inverse Fourier transform. – reuns May 24 '20 at 07:44
  • @reuns Following your hint I've proved your first statement. Would you please elaborate the way of using Cauchy integral formula here? I didn't figure out how derivatives can be donimated. – Infiniticism Jun 01 '20 at 13:01
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    Did you show that $|f(x+iy)|$ is bounded by a rapidly decrasing function $h(x)$ on a strip $|y|\le r$ ? Then $$f^{(k)}(x) = \frac{k!}{2i\pi} \int_{|s-x|=r/2} \frac{f(s)}{(s-x)^{k+1}}ds\le k! (r/2)^{k+1} h(x)$$ – reuns Jun 01 '20 at 14:45

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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 $f\in S$.

Infiniticism
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