Levi pseudoconvex domain

Remarks

The condition does not depend on the choice of $\rho$, because another defining function multiplies the Levi form by a positive number. Example: every convex domain with $C^2$ boundary is Levi pseudoconvex. Indeed, for a convex domain the real Hessian of $\rho$ at $p$ is positive semidefinite on the real tangent space of $\partial\Omega$ at $p$. For $w \in H_p$, both $w$ and $i w$ lie in this tangent space, and $L_\rho(p; w)$ is one quarter of the sum of the real Hessian at $w$ and at $i w$. Non-example: the domain $\set{z \in \C^2 : 1 < \abs{z_1}^2 + \abs{z_2}^2 < 4}$ fails at the inner sphere. There $\rho(z) = 1 - \abs{z}^2$ is a defining function, and $L_\rho(p; w) = -\abs{w}^2 < 0$ for $w \in H_p$, $w \ne 0$. When $n = 1$ the space $H_p$ is $\set{0}$, so every planar domain with $C^2$ boundary is Levi pseudoconvex.