Logarithmically convex domain

Remarks

Example: for the unit ball $\mathbb{B}^2$, $\lambda(\mathbb{B}^{2*}) = \set{x \in \R^2 : e^{2x_1} + e^{2x_2} < 1}$, a sublevel set of the convex function $e^{2x_1} + e^{2x_2}$, hence convex. Every polydisc with centre $0$ is logarithmically convex, since its image is a product of half-lines $(-\infty, \log r_j)$. Non-example: let $D = \set{\abs{z_1} < 1, \abs{z_2} < e^{-2}} \cup \set{\abs{z_1} < e^{-2}, \abs{z_2} < 1}$, a Reinhardt domain. The points $(e^{-1}, e^{-3})$ and $(e^{-3}, e^{-1})$ lie in $D$, with images $(-1, -3)$ and $(-3, -1)$. The midpoint $(-2, -2)$ of these images is the image of $(e^{-2}, e^{-2})$, which is not in $D$, so $\lambda(D^*)$ is not convex. For a complete Reinhardt domain, logarithmic convexity is equivalent to being a domain of holomorphy.