Let $K$ be the space of infinitely differentiable functions $\mathbb{R}\to\mathbb{C}$ with compact support. I read the unproved statement in Kolmogorov-Fomin's Элементы теории функций и функционального анализа (p. 454 here) that the Fourier transform defines a bijective operator $F:K\to Z$ where $Z$ is the space of analytical entire functions $\psi:\mathbb{C}\to\mathbb{C}$ such that $$\forall z\in\mathbb{C}\quad|z|^q|\psi(z)|\le C_q e^{a|\text{ Im}z|}$$ I have been able to prove to myself that $F[\varphi](z):=\int_{\mathbb{R}}\varphi(x)e^{-izx}d\mu_x$ is entire using an analogous argument to this. I have also been able to understand why $F[\varphi]$ satisfies the above inequality $|z|^q|F[\varphi](z)|\le C_q e^{a\text{ Im}z}$ thanks to Tom and the very interesting resource he has linked.
I understand that $\frac{1}{2\pi}\int_{\mathbb{R}}\psi(\lambda)e^{-ix\lambda}d\mu_{\lambda}$ is defined for all $\psi \in Z$, but I do not see why it must be in $K$. Could anybody explain that? I $\infty$-ly thank you!
Suppose $\psi$ is an entire function, and that there is a positive integer $q$ and real $a > 0$ such that $$ |s|^{q}|\psi(s)| \le e^{a|\Im s|},\;\;\; s \in \mathbb{C}. $$ Let $\phi(s) = e^{ixs}\psi(s)$ for any $x > a$. Then $\phi$ satisfies $$ |s|^{q}|\phi(s)| \le e^{-|a-x|\Im s},\;\;\; \Im s \ge 0. $$ Then, by Cauchy's Theorem, the integral of $\phi$ over $[-R,R]$ equals the negative of the integral over the semicircular contour $|s|=R$ with $\Im s \ge 0$. That is, $$ \int_{-R}^{R}\phi(s)\,ds = -\int_{0}^{\pi}\phi(Re^{i\theta})Re^{i\theta}id\theta. $$ The integral on the right is bounded by $$ \int_{0}^{\pi}\frac{1}{R^{q}}e^{-|a-x|R\sin\theta}R\,d\theta = \frac{1}{R^{q-1}}\int_{0}^{\pi}e^{-|a-x|R\sin\theta}\,d\theta. $$ For any $q \ge 1$, the right side converges to $0$ as $R\rightarrow \infty$ by the Lebesgue bounded convergence theorem. Therefore, $$ \lim_{R\uparrow\infty}\int_{-R}^{R}e^{isx}\psi(s)\,ds =0,\;\;\; x > a. $$ You can close the contour in the lower half-plane for $x < -a$ in order to conclude that the inverse Fourier transform of $\psi$ is $0$ for $|x| > a$.