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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 $K$.

enter image description here enter image description here Any hint would be appreciated.

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  • 2
    $\begingroup$ 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). $\endgroup$ – Oleg Eroshkin Sep 12 '15 at 3:36
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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.

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  • $\begingroup$ But from Corollary (Shabat), it seems to show that $\hat{K}_U$ is connected if $K$ is connected. $\endgroup$ – felipeuni Sep 12 '15 at 2:01
  • $\begingroup$ 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 $\widehat{K}_U$. $\endgroup$ – David E Speyer Sep 12 '15 at 2:35
  • $\begingroup$ I just saw that wikipedia defines a domain to be connected. But I deny that you can make a connected open set which contains the unit circle and whose intersection with the unit disc is disconnected. $\endgroup$ – David E Speyer Sep 12 '15 at 2:46
  • $\begingroup$ @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. $\endgroup$ – Oleg Eroshkin Sep 12 '15 at 3:14
  • $\begingroup$ Yes, I see what you are getting at now. Sorry for being dumb. $\endgroup$ – David E Speyer Sep 12 '15 at 3:17

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