Holomorphically convex hull

Remarks

Always $K \subseteq \hat K_D$, and $\hat K_D$ is closed in $D$. Using $f(z) = z_j$ shows that $\hat K_D$ is bounded, and using $f = e^{\ell}$ for complex linear $\ell$, with $\abs{e^{\ell}} = e^{\operatorname{Re} \ell}$, shows that $\hat K_D$ lies in the convex hull of $K$.

Example: $D = \C$ and $K = \set{\abs{z} = 1}$. By the maximum principle every entire $f$ satisfies $\abs{f(z)} \le \max_K \abs{f}$ for $\abs{z} \le 1$, and $f(z) = z$ excludes the points with $\abs{z} > 1$; so $\hat K_D$ is the closed unit disc.

With the same $K$ and $D = \C \setminus \set{0}$, the function $f(z) = 1/z$ is holomorphic on $D$ and has $\abs{f(z)} > 1 = \max_K \abs{f}$ for $0 < \abs{z} < 1$; so here $\hat K_D = K$.

When $D = \C^n$, the hull coincides with the polynomially convex hull, since entire functions are uniform limits of polynomials on compact sets.