How do I prove that the distances from boundary are equal for holomorphic convex set of $K$ and $K$? Let $\Omega(\subset \mathbb{C}^n)$ be a domain of holomorphy. Let $K$ be a compact subset of $\Omega$. Let $K_\Omega$ be the holomorphic convex hull of $K$. Then, how do I prove that $d(K,\partial\Omega)=d(K_\Omega,\partial\Omega)$?
 A: I'll do only the case $n=1$ . Consider $\eta \in \partial \Omega$. The function $f_{\eta} \colon z\mapsto \frac{1}{z-\eta}$ is holomorphic on $\Omega$.
The supremum of $|f_{\eta}|$ on a compact subset $L$ of $\Omega$ is $\frac{1}{d(\eta, L)}$. Now, for every holomorphic function $f$ on $\Omega$ we have
$$\sup_{z \in K} |f(z)| = \sup_{z \in K_{\Omega}} |f(z)|$$
Applying this to $f_{\eta}$ we conclude
$$d(\eta, K) = d(\eta, K_{\Omega})$$
A: Theorem: For a domain $ \subset \mathbb{C}^n$ to be holomorphically convex, it is both necessary and sufficient that for any $K$ compactly embedded in $D$, $$d(K, \partial D) = d(\widehat{K}, \partial D).$$ 
Proof: Sufficiency is obvious. To show necessity, observe that the right hand side of the above expression never exceeds the left hand side. If the inequality is strict, then we could find a point $\zeta \in \widehat{K}$ such that $d(\zeta, \partial D) < d(K, \partial D)$. Using the Simultaneous Extension Lemma however, we know that any function $f \in \mathscr{O}(D)$ would then extend into the polydisk which is centred at $\zeta$ and has radius $d(K, \partial D)$. That is, beyond the boundary of $D$. This implies that $D$ is not a domain of holomorphy and we have the result. 
Notice that intuitively, this theorem says that the passage from compact sets to their holomorphically convex hulls occurs without decreasing the distance to the boundary. This passage is therefore, in some sense, just filling holes and cavities. 
Reference: Shabat's Introduction to Complex Analysis -- Part II. See page 184. 
The simultaneous extension lemma can be found on page 183.
If you have any further questions, please do not hesitate to ask. 
