I'm reading 'The Schwarz Lemma' by Sean Dineen, and was wondering if anyone could expand on an assertion he makes. On page 59, he considers the following domain in two complex variables:
Fix $r, s$ such that $0<r<1$ and $0<s<1$. Let $\mathcal D = \mathbb D^2 \setminus \left\{ (z, w); |z|<r \text{ and } |w|>s \right\}$.
Then says: 'Using the distinguished boundary of the bidisc $\mathbb D^2$, the Cauchy integral formula in two variables, and the maximum modulus theorem we see that each $f\in H(\mathcal D, \mathbb D)$ can be extended to an element of $H(\mathbb D^2, \mathbb D).$'
Now I'm trying to go through all these steps. Obviously the distinguished boundary of $\mathbb D^2$ is $\left\{(z,w): |z|=1,|w|=1 \right\}$, so is it just a question of specifying the function that extends from $\mathcal D$ to $\mathbb D^2$, ie. the function given by the Cauchy integral formula over the distinguished boundary? That is, $$f(\zeta) = \frac{1}{(2 \pi i)^2} \int_{|z|=1} \int_{|w|=1} \frac{f(z, w)}{(z-\zeta_1)(w-\zeta_2)} dz dw$$
That's all I've come up with so far, but it seems there's more work involved. Any help would be great. Thanks in advance.
The distinguished boundary is right, but you don't really want to use it as you stated, as the function is not even defined there. You could use the distinguished boundary of a slightly smaller bidisk. That is $$ g(\zeta) = \frac{1}{(2 \pi i)^2} \int_{|z|=r_0} \int_{|w|=s_0} \frac{f(z, w)}{(z-\zeta_1)(w-\zeta_2)} dw dz $$
As long as $r < r_0 < 1$ and $0 < s_0 < 1$.
You really need show
1) that the function $g$ you get by the integral formula has anything to do with $f$. You can do that by considering the inside integral (The one in $dw$ first. You can see that this is just the one dimensional integral and for any fixed $z$ with $|z|=r_0 > r$, for all $\zeta_2$ with $|\zeta_2| < s_0$ $$ f(z,\zeta_2) = \frac{1}{2\pi i} \int_{|w|=s_0} \frac{f(z, w)}{(w-\zeta_2)} dw $$ Therefore $$ g(\zeta) = \frac{1}{2 \pi i} \int_{|z|=r_0} \frac{f(z, \zeta_2)}{(z-\zeta_1)} dz $$ Now $f$ is holomorphic for all $(z,\zeta_2)$ when $|\zeta_2| < s$. So for small $\zeta_2$ which are small, then $f$ is holomorphic for all $$ g(\zeta) = \frac{1}{2 \pi i} \int_{|z|=r_0} \frac{f(z, \zeta_2)}{(z-\zeta_1)} dz = f(\zeta) $$ The function $g$ is holomorphic, so by identity theorem it has to equal to $f$ for all $\zeta$.
2) Take $r_0$ and $s_0$ to 1 to get it for the entire bidisc.
3) The maximum principle is used to show that you still have a function to a disc.