Forgive me if this question does not meet the bar for this forum. But i would really appreciated some help.

I'm trying to construct a function according to some conditions in the frequency domain of the Fourier transformation. I want the function to be analytic and real when I transform it back to the time domain. The Fourier transformation of $f$ has of course some symmetry criteria to make $f$ real. But what about the Analytic property. As an analytic function imply some convergent power series expansion, and the Fourier transform of a polynomial is a sum of derivatives of Delta functions, I assume that there is a corresponding criteria of the Fourier transformation.

So the question is: If a function $f:\mathbb{R}\rightarrow \mathbb{R}$ is assumed to be analytic, what is the corresponding criteria for the Fourier transform of the function $\mathcal{F}[f] (k)$?

Edit: what I am trying to construct is probability distribution with the following condition

$f(x/\mu)/\mu=\frac{2}{3} f(x) + \frac{1}{3} (f\ast f)(x)\quad$ where $\ast$ mark the convolution, and $\mu=\frac{4}{3}$. $f$ is positive and real for $x\in [0,\infty)$

Taking the fourier transformation make the condition simpler:

$\tilde f(\mu k) = \frac{2}{3}\tilde f(k) + \frac{1}{3}\tilde f^2(k)$

So my problem is to construct $f$ (I am in particular interested in the tail behavior) and I try to use the properties of $\tilde f$. I posted a similar problem a while ago (see here). Julián Aguirre answered how to construct $\tilde f$ if it is analytic. But the inverse transformation of the power expansion is an infinite sum of derivatives of Delta functions, and is of little help.

  • $\begingroup$ Fourier inversion doesn't make sense in general. You need the function to be Schwartz (or at least L^2). $\endgroup$ Commented May 6, 2010 at 10:10
  • $\begingroup$ But I guess there exist functions which are not Schwartz, but has a well defined Fourier transform (?). In either case, is there a general way to express a Schwartz function. Like a series expansion? $\endgroup$
    – jonalm
    Commented May 6, 2010 at 10:16
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    $\begingroup$ Not true, Harry. You can take the Fourier transform of quite a lot of things. Not quite arbitrary distributions, it's true, but tempered distributions are okay. $\endgroup$ Commented May 6, 2010 at 10:17
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    $\begingroup$ Harry, I think Andrew is alluding to the fact that the Fourier transform is an automorphism of the space of tempered distributions (just as it is on L^2). So Fourier inversion applies for that whole space. This includes plenty of functions that are not in L^2, not to mention plenty of things that aren't functions. $\endgroup$ Commented May 6, 2010 at 13:17
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    $\begingroup$ Dear Harry, One can tell from the question that the OP is willing to entertain constructions based on delta functions and other distributions. This should already be enough to show that your comment is off-base. $\endgroup$
    – Emerton
    Commented May 6, 2010 at 14:52

4 Answers 4


What is sufficient (though not necessary) is that the Fourier transform decays exponentially at $\infty$ (if you want just analyticity on the line) or faster than any exponent (if you want your original function to be entire). In particular, anything with compact support will do. If this is too restrictive for your construction, you'd better just tell what exactly you are trying to construct.

  • $\begingroup$ Thanks for your interest fedja, I edited my question and explained what I'm trying to construct. $\endgroup$
    – jonalm
    Commented May 6, 2010 at 19:17

Heh. I grabbed my copy of Reed and Simon, volume 2, and by chance opened it to exactly the right page. The short answer is "exponential decay". Section IX.3 has several relevant theorems. For instance:

Theorem IX.13. Let $f$ be in $L^2(\mathbb{R}^n)$. Then $e^{b|x|} f \in L^2(\mathbb{R}^n)$ for all $b < a$ if and only if $\hat{f}$ has an analytic continuation to the set $\{\zeta : |\mathrm{Im} \zeta| < a\}$ with the property that for each $\eta \in \mathbb{R}^n$ with $|\eta| < a$, $\hat{f}(\cdot + i\eta) \in L^2(\mathbb{R}^n)$ and for any $b < a$, $\sup_{|\eta|\le b} \lVert \hat{f}(\cdot + i\eta)\rVert_2 < \infty$.

MR0493420 (58 #12429b) Reed, Michael; Simon, Barry Methods of modern mathematical physics. II. Fourier analysis, self-adjointness. Academic Press [Harcourt Brace Jovanovich, Publishers], New York-London, 1975. xv+361 pp. (Reviewer: P. R. Chernoff) 47-02 (81.47)

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    $\begingroup$ For ease of searching for the OP, I should mention that those are all variations on the classical Paley-Wiener theorem. $\endgroup$ Commented May 6, 2010 at 15:44
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    $\begingroup$ Another place to look besides Reed and Simon maybe Hormander's Analysis of Linear Partial Differential Operators, probably either volume 1 or 2. $\endgroup$ Commented May 6, 2010 at 15:48

If you need an "if and only if" result, you should use a generalization of the Fourier transform called the FBI transform. There's a nice theorem linking the real analyticity of a function to the decay of its FBI transform. On page 137 of A Primer of Real Analytic Functions, Krantz and Parks state it this way:

Fix $x_0 \in \mathbb{R}$. An integral function $f$ is real analytic at $x_0$ if and only if $f$ satisfies condition $RA(x_0)$.

A function satisfies "condition $RA(x_0)$" if its FBI transform decays exponentially as you vary the parameters in a certain way.

I suspect the FBI transform might be overkill for what you're trying to do, but if you're interested, check out Section 5.3 of A Primer of Real Analytic Functions.

  • $\begingroup$ You can also find the answer at en.wikipedia.org/wiki/… $\endgroup$
    – S. Carnahan
    Commented May 6, 2010 at 16:45
  • $\begingroup$ Thanks for the input. But I do need the linearity of the transformation (see the edit of my question), and that is not the case according to wikipedia. $\endgroup$
    – jonalm
    Commented May 6, 2010 at 19:24

The following basic result needs to be quoted on these matters of analyticity: the Paley-Wiener-Schwartz theorem gives a characterization of distributions with compact support. Let $u$ be a tempered distribution ; $u$ is compactly supported in a ball of center $0$ and radius $R$ if an only if $\hat u$ is an entire function such that $$\exists C_0, \exists N_0,\forall \zeta\in \mathbb C^d,\quad \vert\hat u(\zeta)\vert\le C_0(1+\vert \zeta\vert)^{N_0}e^{R\vert\Im\zeta\vert}. $$ Something analogous allows a characterization of $C^\infty$ functions with compact support.

This leads to the following characterization of the analytic wave front set, due to Bros and Iagolnitzer. Let $v\in \mathcal E'(\mathbb R^d)$. We define the Fourier-Bros-Iagolnitzer transform $Tv$ of $v$ by the following formula, where the integral is in fact a bracket of duality, $$ (Tv)(z,\lambda)=\int_{\mathbb R^d} e^{-\pi\lambda(z-x)^2} v(x) dx,\qquad z\in \mathbb C^d, \lambda >0. $$ Let $\Omega$ be an open subset of $\mathbb R^d$; let us note $\Omega\times(\mathbb R^d\backslash \{0\})$ by $\dot T^*(\Omega)$ and by $dL(z)$ the Lebesgue measure on $\mathbb C^d$. Let $u\in \mathcal D'(\Omega)$. The analytic wave-front-set of $u$, denoted by $WF_{A}(u)$, is the complement in $\dot T^*(\Omega)$ of the set of points $(x_{0},\xi_{0})$ such that $$ \exists W_{0}\in \mathscr V_{x_{0}-i\xi_{0}}, \exists \chi_{0}\in C^\infty_c(\Omega), \chi_{0}(x)=1\ \text{near $x_{0}$}, \exists \epsilon_{0}>0\quad \text{with} $$ $$ \sup_{\lambda\ge 1, z\in W_{0}}e^{\epsilon_{0}\lambda} \vert{(T\chi_{0} u)(z,\lambda)}\vert e^{-\pi\lambda(\Im z)^2} <+\infty. $$ The first projection of $WF_A(u)$ is the analytic singular support.


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