# Fourier transform of compactly supported differentiable function

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!

• Perhaps something like math.umn.edu/~garrett/m/fun/notes_2013-14/paley-wiener.pdf will help? – Tom Dec 1 '14 at 15:41
• @Tom I've been able to understand the proof of the inequality even with my $\varepsilon$ of knowledge of these wonderfully interesting topics. I heartily thank you! The only thing I still doesn't grasp is how to see $F$'s bijectivity. – Self-teaching worker Dec 1 '14 at 17:02
• @DavideZena : You missed an absolute value. You should have $e^{a|\Im z|}$ for the bound. Perhaps that is confusing you? – Disintegrating By Parts Dec 1 '14 at 19:19
• @T.A.E. Oh, thank you: edited. No, I mistyped, but I had $|\Im z|$ in mind. What I don't grasp is why $\forall\psi\in Z\quad F^{-1}[\psi]\in K$... – Self-teaching worker Dec 1 '14 at 19:50
• @DavideZena : Have you seen the Paley Wiener Theorem. This is typically cast as a problem of holomorphic functions in the right half-plane, but it applies to the upper half-plane, too. – Disintegrating By Parts Dec 1 '14 at 21:31

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 $|x| > a$.
• Although Kolmogorov-Fomin's tends to be quite ambiguous and doesn't say whether $a$ and $q$ are fixed, I think, from the proof of Paley-Wiener theorem, that the correct definition of $Z$ is the set of entire functions $\psi$ such that $\exists a>0:\forall q\in\mathbb{N}^+\quad\exists C_q\in\mathbb{R}:\forall z\in\mathbb{C}\quad|z|^q|\psi(z)|\le C_q e^{a|\text{ Im}z|}$ (and this $a$ is such that $\text{supp}(\varphi)\subset B(0,a)$): am I right? That allows us to verify that $R^{1-q}\int_0^\pi e^{-|a-x|R\sin\theta}d\theta\to 0$. Thank you so much again! – Self-teaching worker Dec 2 '14 at 10:22
• One thing isn't clear to me: why $F^{-1}[\psi]\in C^\infty(\mathbb{R})$. The tools used by the paper linked to by Tom are unknown to me: cannot it be proved with more elementary means? I see that $\int_{-R}^R(ix)^Ne^{isx}\psi(s)ds=\frac{\partial^N}{\partial x^N}\int_{-R}^Re^{isx}\psi(s)ds$ by dominated convergence, but I'm not sure that it's straightforward that $\frac{\partial^N}{\partial x^N}\text{PV}\int_{-\infty}^\infty e^{isx}\psi(s)ds=\lim_{R\to\infty}\frac{\partial^N}{\partial x^N}\int_{-R}^Re^{isx}\psi(s)ds$... $\infty$ thanks! – Self-teaching worker Dec 2 '14 at 11:08
• I missed that: $\int_{\mathbb{R}}\phi d\mu$ exists and, therefore, $\int_{\mathbb{R}}\phi d\mu=\text{PV}\int_{-\infty}^\infty\phi(t)dt$, Thank you very very much! – Self-teaching worker Dec 2 '14 at 20:37
• @DavideZena : If $\phi \in L^{2}(\mathbb{R})$, it is also useful to know that $PV \int_{-\infty}^{\infty}e^{-ist}\phi(t)\,dt$ converges in $L^{2}(\mathbb{R})$, even though you may not be able to say much about pointwise convergence at any particular $s$. – Disintegrating By Parts Dec 2 '14 at 20:49