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.

  • 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

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.

  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.