0
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ Perhaps something like math.umn.edu/~garrett/m/fun/notes_2013-14/paley-wiener.pdf will help? $\endgroup$ – Tom Dec 1 '14 at 15:41
  • $\begingroup$ @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. $\endgroup$ – Self-teaching worker Dec 1 '14 at 17:02
  • 1
    $\begingroup$ @DavideZena : You missed an absolute value. You should have $e^{a|\Im z|}$ for the bound. Perhaps that is confusing you? $\endgroup$ – Disintegrating By Parts Dec 1 '14 at 19:19
  • 1
    $\begingroup$ @T.A.E. Oh, thank you: edited. No, I mistyped, but I had $|\Im z|$ in mind. What I don't grasp is why $\forall\psi\in Z\quad F^{-1}[\psi]\in K$... $\endgroup$ – Self-teaching worker Dec 1 '14 at 19:50
  • 1
    $\begingroup$ @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. $\endgroup$ – Disintegrating By Parts Dec 1 '14 at 21:31
1
$\begingroup$

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 $|x| > a$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – Self-teaching worker Dec 2 '14 at 10:22
  • $\begingroup$ 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! $\endgroup$ – Self-teaching worker Dec 2 '14 at 11:08
  • 1
    $\begingroup$ @DavideZena : I would guess that your assumptions in the first comment to this post are correct. Then the technique in the second is correct, except that it is almost always easier to deal with integrals instead of derivatives; the PV is not a problem because the integrals are absolutely convergent by the assumed bounds. $\endgroup$ – Disintegrating By Parts Dec 2 '14 at 12:50
  • 1
    $\begingroup$ 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! $\endgroup$ – Self-teaching worker Dec 2 '14 at 20:37
  • 1
    $\begingroup$ @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 $s$. $\endgroup$ – Disintegrating By Parts Dec 2 '14 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.