Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

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 "desymetrised" 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 s.t. the continuation on each slice of the strip is of rapid decrease.

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 reasonnable 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" 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

Your Answer


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

Browse other questions tagged or ask your own question.