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How to show that $D=\{ |z_1|<1\} \cup \{ |z_2|<1\} \subset \mathbb{C}^2$ is not a domain of holomorphy. What is the smallest domain of holomorphy $ S\supset D $ ?

I think we cannot just add the distinguished boundary (the torus $|z_1|=1, |z_2|=1$) to $D$ because it would violate analiticity. Should we use logarithmically convexity argument to extend our domain? Can anyone help?

Obrigado.

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1 Answer 1

For Reinhardt domains, recognizing domains of holomorphy is a lot simpler than for general domains.

To see that the $D$ in question is not a domain of holomorphy, the easiest is probably to use the following theorem on convergence of power series:

Theorem Let $\Omega \subset \mathbb{C}^n$ be a connected Reinhardt domain containing $0$ and assume that $f$ is holomorphic on $\Omega$. Then the power series of $f$ around the origin converges (normally) on $\tilde \Omega$ -- the logarithmically convex hull of $\Omega$.

Hence any function holomorphic on $\Omega$ extends to $\tilde\Omega$. In fact more is true, namely $\tilde\Omega$ is the smallest domain of holomorphy containing $\Omega$. So in your case, you want to find $\tilde D$.

It's also possible to see that $D$ is not a domain of holomorphy by a "pushing discs" argument, but this is a bit messy to write down explicitly.

For a reference to the above theorem, see for example Krantz, Function theory of several complex variables. In my (somewhat old) edition, it's proposition 2.3.16. The proof follows from Cauchy's integral formula on polydiscs, and is not very difficult.

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