Uniqueness of a multivariate exponential transform

If $k=0$, then ${\mathbb{R}}^0$ is a singleton and the open neighborhood in the hypothesis contains its only point; $L(0)=\nu ({\mathbb{R}}^0)=0$, so $\nu =0$. If $k\ge 1$, set $D=\{z\in {\mathbb{C}}^k:\mathrm{\mid }\mathrm{Re}{z}_j\mathrm{\mid }<\delta \mathrm{/}2\text{ for all }j\}$. Fix $z\in D$ and a coordinate $j$. Choose a neighborhood of $z$ on which every real coordinate has absolute value at most $3\delta \mathrm{/}4$. The moment assumption makes $e^{\delta \mathrm{\parallel }x{\mathrm{\parallel }}_1}$ integrable with respect to the positive measure [F1], and all integrands below belong to $L^1(\mathrm{\mid }\nu \mathrm{\mid })$ by [F2]. On this neighborhood, $\mathrm{\mid }{e}^{z\cdot x}\mathrm{\mid }\le e^{(3\delta \mathrm{/}4)\mathrm{\parallel }x{\mathrm{\parallel }}_1}$, while for small complex $h$ the $j$th difference quotient is bounded by $\mathrm{\mid }{x}_j\mathrm{\mid }{e}^{(3\delta \mathrm{/}4)\mathrm{\parallel }x{\mathrm{\parallel }}_1}\le C{e}^{\delta \mathrm{\parallel }x{\mathrm{\parallel }}_1}$. For every sequence $h\to 0$, [F3] gives convergence of the difference quotients in $L^1(\mathrm{\mid }\nu \mathrm{\mid })$ to $x_j{e}^{z\cdot x}$; the bound in [F2] transfers this convergence to their integrals against $\nu$. Thus $L$ is complex differentiable in coordinate $j$ at $z$, and, since $z$ and $j$ were arbitrary, $L$ is holomorphic in each coordinate on $D$.

Assume $k\ge 1$. The real neighborhood contains a product of real intervals around zero. Fix real values of coordinates $2,\dots ,k$ in those intervals. As a function of $z_1$, $L$ is holomorphic on the connected strip and vanishes on a real interval; [F4] makes it zero for every $z_1$ in that strip. Now fix such a complex $z_1$ and real coordinates $3,\dots ,k$ in their intervals and apply [F4] to coordinate $z_2$. Repeating this finite process for coordinates $1$ through $k$ proves $L(z)=0$ for every $z\in D$.

For $k\ge 1$ and each $\xi \in {\mathbb{R}}^k$, the library Fourier transform is $\widehat{\nu }(\xi )=\int e^{-2\pi i\xi \cdot x}\nu (dx)=L(-2\pi i\xi )=0$, since $-2\pi i\xi \in D$. The Fourier uniqueness result [F5] then gives $\nu =0$.