Is polynomial hull contained in Runge domain? Suppose $Y$ is a compact set contained in $\Omega$ - a Runge domain, biholomorphic to $\mathbb{C}^n$. I would like to prove that polynomial hull of $Y$, $$\bar Y = \{z \in \mathbb{C}^n: |P(z)| \le ||P||_Y, \: \forall P \in \mathcal{P}(\mathbb{C} ^n) \}$$ where $\mathcal{P}(\mathbb{C} ^n)$ is the space of complex polynomials on $\mathbb{C}^n$, is contained in $\Omega$. Unfortunately I am not able to do that but at the same time I feel that assumption on being biholomorphic with $\mathbb{C}^n$ may be artificial. 
 A: The assumption that $\Omega$ is biholomorphically equivalent to $\mathbb{C}^n$ is not superfluous. By Hartogs' Kugelsatz, every holomorphic function on a spherical shell $U = \{z \in \mathbb{C}^n : 1 < \lVert z\rVert < 2\}$ has a holomorphic extension to the ball $B = \{ z \in \mathbb{C}^n : \lVert z\rVert < 2\}$ when $n \geqslant 2$. So the restriction $\rho_U \colon \mathscr{O}(B) \to \mathscr{O}(U)$ is a bijection. Since $B$ is a Runge domain and "$\mathscr{O}(U) =  \mathscr{O}(B)$", so is $U$, but the polynomial hull of $Y = \{ z \in U : \lVert z\rVert = 3/2\}$ is the ball $\{ z : \lVert z\rVert \leqslant 3/2\}$ is not contained in $U$.
The short form of the argument is:


*

*Since $\Omega$ is a Runge domain, we have $\overline{Y} \cap \Omega = \hat{Y}$, where $\hat{Y}$ is the holomorph-convex hull of $Y$, $$\hat{Y} = \{z \in \Omega : f\in \mathscr{O}(\Omega) \implies \lvert f(z)\rvert \leqslant \lVert f\rVert_Y\}.$$

*Since $\Omega$ is biholomorphically equivalent to $\mathbb{C}^n$, it is a domain of holomorphy, hence $\Omega$ is holomorphically convex, therefore $\hat{Y}$ is a compact subset of $\Omega$.

*The polynomial hull $\overline{Y}$ cannot have components contained in $\mathbb{C}^n\setminus \Omega$.
Part 1 is straightforward. Since [the restriction of] every polynomial $P$ belongs to $\mathscr{O}(\Omega)$ we have $\hat{Y} \subset \overline{Y}\cap \Omega$. And if $z_0 \in \Omega \setminus \hat{Y}$, there is an $f\in \mathscr{O}(\Omega)$ with $\lvert f(z_0)\rvert > \lVert f\rVert_Y$. Let
$$\varepsilon = \frac{\lvert f(z_0)\rvert - \lVert f\rVert_Y}{2},$$
and $P$ a polynomial with $\lVert f - P\rVert_{Y \cup \{z_0\}} < \varepsilon$. Such a polynomial exists since $\Omega$ is a Runge domain. Then $\lvert P(z_0)\rvert - \lVert P\rVert_Y > 0$, i.e. $z_0 \notin \overline{Y}$. Thus $\Omega\setminus \hat{Y} \subset \Omega \setminus \overline{Y}$ and the proof of 1 is complete.
Part 2 is not trivial, but a well-known fact about domains of holomorphy, the Cartan-Thullen theorem.
Part 3 isn't trivial either. It's an easy consequence of the Oka-Weil theorem - if $K \subset \mathbb{C}^n$ is compact and polynomially convex, then every holomorphic function on [a neighbourhood of] $K$ can be uniformly approximated [on $K$] by polynomials.
In our situation, $K = \overline{Y}$ is a compact polynomially convex set, and $K = \hat{Y} \cup (\overline{Y}\setminus \Omega)$ is a decomposition into two disjoint compact sets. A polynomial approximation of
$$f(z) = \begin{cases} 0 &, z \in \hat{Y} \\ 1 &, z \in \overline{Y}\setminus \Omega \end{cases}$$
with an error less than $\frac{1}{2}$ shows $\overline{Y}\setminus \Omega = \varnothing$.
The Oka-Weil theorem is not trivial, however. (It's theorem 1.5 in chapter VI of R. M. Range's Holomorphic Functions and Integral Representaions in Several Complex Variables for example, where you can find a proof.)
