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 http://arxiv.org/abs/math/0007097, 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 :)