Verification of an example on domain of holomorphy Let $D_{a,b}=\{(z_1,z_2)\in \mathbb{C}^2:|z_1|<a, |z_2|<b\}$, $a$ and $b$ are positive reals, and let $\Omega=D_{1,2}\bigcup D_{2,1}$. Let $\tilde{\Omega}=\{(z_1,z_2)\in \mathbb{C}^2:|z_1|<2, |z_2|<2, |z_1 z_2|<2\}$. Clearly $\Omega\subset \tilde{\Omega}$.
In the book, Function theory in the unit ball of $\mathbb{C}^n$ Rudin (page 7) said that every function that is holomorphic in $\Omega$ extends to a holomorphic function in $\tilde\Omega$. His claim is that any power series that converges in $\Omega$ also converges in $\tilde{\Omega}$. I cannot verify this statement. 
Any help will be appreciated. Thank you.
 A: The domain of convergence of a power series is a logarithmically convex Reinhardt domain. This is for example proved in Theorem 2.4.3. of Lars Hörmander's An introduction to complex analysis in several variables.
Since $\tilde{\Omega}$ is the smallest logarithmically convex Reinhardt domain containing $\Omega$, the result follows.
In this specific situation, for $(z_1,z_2) \in \tilde{\Omega}$ with $z_1z_2 \neq 0$, let $r_i = \lvert z_i\rvert$, and $p = \sqrt{\frac{1}{2}\lvert z_1z_2\rvert}$. With
$$\alpha_i = \frac{\log r_i - \log p}{\log 2}$$
we have $r_i = 2^{\alpha_i}\cdot p$ and
$$\alpha_1 + \alpha_2 = \frac{\log (r_1r_2) - 2\log p}{\log 2} = \frac{\log (2p^2) - \log p^2}{\log 2} = 1,$$
so by the convexity of the exponential function
$$r_1^{\mu}r_2^{\nu} = \bigl(p^{\alpha_2}(2p)^{\alpha_1}\bigr)^{\mu}\cdot\bigl((2p)^{\alpha_2}p^{\alpha_1}\bigr)^{\nu} = \bigl((2p)^{\mu}p^{\nu}\bigr)^{\alpha_1}\cdot \bigl(p^{\mu}(2p)^{\nu}\bigr)^{\alpha_2} \leqslant \alpha_1\bigl((2p)^{\mu}p^{\nu}\bigr) + \alpha_2\bigl(p^{\mu}(2p)^{\nu}\bigr),$$
and since the power series
$$\sum_{(\mu,\nu) \in \mathbb{N}^2} a_{\mu\nu} w_1^{\mu} w_2^{\nu}$$
converges absolutely at $(2p,p) \in D_{2,1}$ and at $(p,2p) \in D_{1,2}$, the above inequality shows
$$\sum_{(\mu,\nu) \in \mathbb{N}^2} \lvert a_{\mu\nu}\rvert r_1^{\mu} r_2^{\nu} < +\infty,$$
so the power series converges absolutely at $(z_1,z_2)$.
