Let $f : \mathbb R\_+ \to \mathbb C$ be a measurable and integrable function where $\mathbb R\_+ = [0,\infty)$. The Laplace transform of $f$ is given by
$$
Lf(s) = \int\_0^\infty f(x)e^{-sx} \, dx.
$$
A classical Theorem due to Lerch [M. Lerch, Sur un point de la théorie des fonctions génératrices d’Abel, Acta Math. 27, 339 -351 (1903)] states that if
$$
Lf(\delta n) = 0, \quad n \in \mathbb N
$$
for some $\delta>0$ then $f=0$ almost everywhere. Suppose now that we replace the Laplace transform by the two-sided Laplace transform,
$$
Tf(s) = \int\_{\mathbb R} f(x)e^{-sx} \, dx. \quad (\*)
$$
Suppose that $Tf$ defines an entire function (clearly, we need a decay assumption on the negative axis so that $Tf$ is entire).


**Question:** Are there Lerch-type results for the two-sided Laplace transform? That is, suppose that $f$ belongs to a certain function class $C \subset L^1(\mathbb R)$. Can we find discrete sets $A \subset \mathbb R$, so that $Tf(a)$ for all $a \in A$ implies $f=0$ almost everywhere? In particular, I'm interested in function classes $C$ so that $A$ can be chosen to be uniformly discrete, $\inf\_{a \neq b, a,b \in A} > 0$.



Discrete uniqueness sets for the two-sided Laplace transform?
Of course, if $A$ has accumulation points then
your statement is correct, since you assume the function $Lf$ to be entire.


Otherwise, there are no restrictions on zeros of such functions $Lf$ (unless you somehow restrict your class $C$ of functions $f$). For example, let $f$ be an infinite sum of $\delta$-functions sitting at integers. Then your integral becomes a sum, and changing the variable to $z=e^{-s}$ we obtain an arbitrary Laurent series in $\mathbf{C}\backslash\{0\}$. This can have
arbitrary sequence of zeros.


Of course a sum of delta-functions is not integrable, but it is easy to modify this example, to make it integrable by integrating by prts:
$$\sum\_{-\infty}^\infty a\_ke^{-sk}=\int\_{-\infty}^\infty e^{-sx}dn(x)=-s\int\_{-\infty}^\infty n(x)e^{-sx}dx,$$
where $n$ is a step function jumping by $a\_k$ at $k$.
The integral in the RHS have the same zeros as the LHS, except at $s=0$, and $n(x)$ is an integrable (step) function. By integrating few more times, you can make your $f$ arbitrarily smooth.


The main condition of Lerch's theorem is that $f(x)=0$ for $x<0$, which ensures that $LF$ is bounded in the right half-plane. If we allow an arbitrary support of $f$, no conclusion about zeros can be made, except, of course that this is a discrete set.


