Paley Wiener Theorem on sinc function

Question

Use the Paley-Wiener theorem to argue that, although ${\rm sinc}\left(t\right)$ is bandlimited, ${\rm sinc}\left(t^{3}\right)$ is not. Explain how the above result allows reconstruction of some non-bandlimited signals from nonuniformly-spaced samples.

My Attempt I know that using the Paley Wiener Theorem that \begin{equation} {\rm sinc}\left(t\right) = \frac{\sin\left(t\right)}{t}= \frac{1}{2}\int_{-1}^{1}{\rm e}^{{\rm i}xt}\,{\rm d}t \end{equation} and that \begin{equation} \left\vert\frac{\sin\left(t\right)}{t}\right\vert \leq Ce^{\left\vert t\right\vert} \end{equation} I found this result on another question but am not exactly sure how it was derived or how I would apply it to ${\rm sinc}\left(t^{3}\right)$

Answer 1

The hint about using the Paley-Wiener theorem is intended for the part about $\operatorname{sinc} (t^3)$. You did not need it for $\operatorname{sinc} t$. Indeed, the compactly supported transform is exhibited explicitly, and the exponential bound follows directly from $$|\sin z| \le \frac12( |e^{iz}|+|e^{-iz}|) \le e^{|z|}$$ (The denominator $t$ only makes this smaller when $t$ is large, but at any rate polynomial terms are negligible here.)

But once we plug a higher power of $t$ into $\operatorname{sinc} $, the exponential type is lost. For $y>0$ we have $$ |\operatorname{sinc} ((iy)^3) | \ge \frac{1}{2 y ^3} (e^{i(iy)^3}-e^{-i(iy)^3}) = \frac{1}{2 y ^3} (e^{y^3}-e^{-y^3}) $$ Clearly, this grows faster than $e^{Cy}$ for any $y$. The Paley-Wiener theorem (compact case) implies that this function does not have compactly supported Fourier transform.