# Discrete uniqueness sets for the two-sided Laplace transform?

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, $$\inf_{a \neq b, a,b \in A} > 0$$.

• A rather obvious comment: $T f(n) = 0$ for all natural $n$ is not enough. Indeed: if we write $Tf(s) = \int_0^\infty f(-\log t) t^{s - 1} dt$, then $T f(n) = 0$ corresponds to the Stieltjes moment problem. Jan 24 at 11:42
• In the same vein, it is not enough to assume that $T f(n) = 0$ for all integer $n$: the strong Stieltjes moment problem can be indeterminate, too. Jan 24 at 11:50

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.