connectedness holomorphically convex hull

Let $U\subset\mathbb{C}^n$ be a domain and $K\subset U$ compact. If $K$ is connected then $\hat{K}_U$ is also connected?

$\hat{K}_U= \{z \in U: |f(z)| \leq \sup_K |f|, \forall f\in \mathcal{O}(U)\}$: holomorphically convex hull of $K$.

Any hint would be appreciated.

• This corollary (from Shabat) holds for domains of holomorphy, but fails otherwise. You can see, that to apply Theorem 2, one needs that $\hat{K}_U$ is compact subset of $U$. If it true for any compact subset of $U$, then $U$ is holomorphically convex (by definition). – Oleg Eroshkin Sep 12 '15 at 3:36

The answer is no, unless you assume that $U$ is pseudoconvex. For example, take $K$ a unit circle. Then $\hat{K}_U$ will be the intersection of the unit disk with $U$. You can make it disconnected.
If $U$ is pseudoconvex, then the answer is yes, and it follows from Shilov's idempotents theorem.
• But from Corollary (Shabat), it seems to show that $\hat{K}_U$ is connected if $K$ is connected. – felipeuni Sep 12 '15 at 2:01
• I don't think this is right. If $U$ is an annulus like $\{ z : 1/2 < z < 2 \}$, then $\widehat{K}_U$ is $K$ itself, not the intersection of $U$ with the unit disc. (Because $\sup_{z \in K} |z|=1$ and $\sum_{z \in K} z^{-1}=1$.) To give a more pertinent example, if $U$ is disconnected, the components of $U$ which don't meet $K$ will not be in $\widehat{K}_U$. – David E Speyer Sep 12 '15 at 2:35
• @DavidSpeyer Of course it is possible in $\mathbb{C}^n$ for $n>1$. Take a complex line, a circle in that line, and a domain containing a circle. The intersection of a disk (in that line) with the domain can be disconnected. – Oleg Eroshkin Sep 12 '15 at 3:14