Cuchta-Georgiev Fourier transform of delta derivatives
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Theorem
If $f$ is $k$-times delta differentiable and for all $\ell \in \{0,\ldots,k-1\}$, $\displaystyle\lim_{t \rightarrow \pm \infty} f^{\Delta^{\ell}}(t)e_{\ominus iz}(t,s)=0$, then $$\mathcal{F}_{\mathbb{T}}\left\{f^{\Delta^k}\right\}(z;s) = (iz)^k \mathcal{F}_{\mathbb{T}}\{f\}(z;s).$$
