MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is known that smooth functions with exponential decay at $\pm\infty$ are functions whose Fourier transform have analytic continuation in some suited complex strip. I was wondering what happens if we ask for exponential decay from only one side. More precisely :

The context : In Fourier Analysis, Self-Adjointness by Reed and Simon (Methods of Modern Mathematical Physics, Vol. 2), Theorem IX.14 tells us that (I'll take dimension 1 for simplicity) :

if $T$ is a tempered distribution on $\mathbb R$ such that :

  • $\hat T$ has an analytic continuation to $|\Im z| < a$ for some $a > 0$
  • on each slice $\mathbb R + i\eta$ with $|\eta| < a$, $\hat T$ is integrable
  • for each $0 < b < a$, the supremum of the integrals on the slices $\mathbb R + i\eta$ with $|\eta| < b$ is finite

then $T$ is a bounded continuous function and for any $0 < b < a$, there exists $C_b \geq 0$ s.t. $$|T(x)| \leq C_b e^{-b|x|}$$

This result can be "desymmetrised" easily : by assuming analytic continuation in $-a_- < |\Im z| < a_+$ with $a_-,a_+ > 0$ we get decay faster than $e^{-a_+x}$ as $x\to -\infty$ and than $e^{a_-x}$ as $x\to+\infty$.

In, authors mention a similar result which is an equivalence in $\mathcal S(\mathbb R)$ morally saying

Functions with such kind of decay are functions whose Fourier transform are of rapid decrease and admit a holomorphic continuation in the associated strip such that the continuation on each slice of the strip is of rapid decrease.

The question : I was wondering if anyone would knew what happens if we only ask for exp. decay at $-\infty$ : for instance for a $L^\infty$ function, is there some reasonable equivalent condition on the Fourier transform (in the sense of tempered distributions) for it to have $e^{ax}$ decay as $x\to-\infty$ ?

Not assuming exp. decay at $+\infty$ makes that such functions can have "ugly" Fourier transforms (since they're not in $L^1$ or $L^2$ in general), for instance $e^x H(-x) + H(x)$ where $H$ denotes the Heaviside step function yields $\delta_0 + i \text{pv}\frac{1}{\xi} - \frac{i}{\xi-i}$. Nonetheless, the "functional part" (in some kind of bad and undefined sense...) of this transform indeed has holomorphic continuation in the half-plane $\Im z < 1$ (but is not $L^1$ on any slice). This is the kind of observations that decided me to ask this question here :)

share|cite|improve this question
Since one can decompose any f as $\chi_{|(-\infty,0]}f+\chi_{|[0,\infty)}f$ one can apply the twosided version to $\chi_{|(-\infty,0]} f$ (which has exponential decay in both directions) and has reduced the question to "What can be said about Fouriertransformations of functions with $f(x)=0$ for $x<0$." Therefore I would lookup Laplace transformations of distributions, maybe the answer to that new question is already known. – Johannes Hahn Oct 30 '13 at 14:53
up vote 6 down vote accepted

Let me summarize what has been already said. If $f$ is a locally integrable function, you can break it into two parts: $f=f_1+f_2$, where $f_1$ is supported on the positive ray, and $f_2$ on the negative ray. Then you consider two "halves" of the Fourier transform: $$F^-(z)=\int_0^\infty e^{-izt}f_1(t)dt$$ and $$F^+(z)=\int_{-\infty}^0 e^{-itz}f_2(t)dt.$$ The first function is analytic in the lower half-plane, and the second is analytic in the upper half-plane. This pair of analytic functions is called "generalized Fourier transform in the sense of Carleman", or "Fourier transform in the sense of hyperfunctions". If "usual" Fourier transform exists in some sense, like when $f\in L^2$ or $f\in L^1$, or $f$ is a temperate distribution, then it is recovered as the sum of boundary values of these two analytic functions. Boundary values must be understood in the appropriate sense, dependng on your conditions on $f$. Now if $f$ decreases exponentially, say on the negative ray, this means that $F^+$ has an analytic continuation from the upper half-plane to a larger half-plane $\Im z>-a$. That is Fourier transform on the real line is a sum of of a function analytic in the lower half-plane and another one analytic in the larger upper half-plane $\Im z>-a$. This is the characterization you asked for.

In general, a temperate distribution on the real line has a canonical representation as sum of the boundary values of two analytic functions, one in the upper half-plane, another in the lower half-plane. So this property (analytic continuation of one of these functions to a larger half-plane) is an intrinsic property which exactly characterizes the exponential decrease of the original on one of the semiaxes.


  1. Carleman, L'Intégrale de Fourier et Questions que s'y Rattachent, Publications Scientifiques de l'Institut Mittag-Leffler, 1. Uppsala, 1944.

  2. A. Kaneko, Introduction to hyperfunctions, Kluwer, Dordrecht, Tokyo, 1988.

share|cite|improve this answer
Thank you very much, it is indeed what I was looking for. It is nice to see that a characterization has to use hyperfunctions, so in some sense I've naturally come across them. I'll try to read more about that, this seems very interesting. – Laurent Oct 31 '13 at 0:12

The short answer is: if you assume that $f$ decays exponentially only on one side (and increases say polynomially on the other side) then its Fourier transform is the boundary value of a holomorphic function, defined on the upper (or lower) complex plane. This can be made much more precise and the rate of increase as $y\to0$ is connected with the degree of smoothness of $\hat f$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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