Laplace transform on a convex cone

Remarks

If $y \in \Omega$, there is $c > 0$ with $\ip{y}{x} \ge c \abs{x}$ for all $x \in \Gamma$, by compactness of $\Gamma \cap \set{\abs{x} = 1}$. So if $\abs{f(x)} \le C (1 + \abs{x})^N$, the transform converges absolutely on the tube $\Omega + i \R^n$ and is holomorphic there.

Example: $\Gamma = [0, \infty)^n$ is self-dual, and $\mathcal{L}_\Gamma$ is the multidimensional Laplace transform. For $f = 1$ one gets $\mathcal{L}_\Gamma f(z) = 1/(z_1 \cdots z_n)$ on $\operatorname{Re} z_j > 0$.

Example: let $\Gamma$ be the cone of positive semidefinite real symmetric $m \times m$ matrices, with $\ip{y}{x} = \tr(yx)$. Let $dx$ be the Lebesgue measure in the entries $x_{ij}$, $i \le j$. The cone is self-dual, and $\Omega$ is the cone of positive definite matrices. For real $s > (m - 1)/2$ and $y \in \Omega$, $$\int_\Omega e^{-\tr(yx)} (\det x)^{s - (m+1)/2}\, dx = \Gamma_m(s)\, (\det y)^{-s}, \qquad \Gamma_m(s) = \pi^{m(m-1)/4} \prod_{j=0}^{m-1} \Gamma\left(s - \frac{j}{2}\right).$$

Non-example: let $\Gamma$ contain a line $\R v$. Then $\Gamma^*$ lies in the hyperplane $v^\perp$ and $\Omega$ is empty. Since $\Gamma + \R v = \Gamma$, Fubini's theorem along the direction $v$ shows that even $f = 1$ has a divergent transform at every $z$.