Question: Does a finite constant $C>0$ exist such that for $\forall\ p\in\mathbb{C}[z_1,z_2]$ we have:
$$\sup_{z\in r \mathbb D^2}|p(z_1,z_2)|\le C\sup_{z\in \mathbb D^2}|p(z_1,z_2)|$$
where $r>0$?
I define as $p\in\mathbb{C}[z_1,z_2]$ a finite function of the following formula:
$$p(z_1, z_2) := \sum_{i,j=0}^k \alpha_{p_{ij}} z_1^i z_2^j$$
where $\alpha_{p_{ij}} \in \mathbb{C}$.
Of course, $\mathbb D^2$ is the (open) bidisc.
Important: The constant $C$ cannot depend on $p$. It can only depend on $r$.
Is this even possible to prove? If not, is it maybe possible for a specific family of functions in $\mathbb{C}[z_1,z_2]$? Is it maybe possible for a closed bidisc?
I really need this for something I work on currently (PhD student here).
Thanks.
(I think this works. Maybe I'm tireder than I think.)
Let $p_n(z) = n z_1^{2n}$. (This is holomorphic in the first coordinate and oblivious to the second coordinate. So we know that the supremum is attained on the boundary of these scaled bidiscs (in fact at any point that is on the boundary of the disc in the first coordinate).) Then \begin{align*} C(r) &\geq \frac{\sup_{z \in r\Bbb{D}^2} \left| p_n(z_1,z_2) \right|}{\sup_{z \in \Bbb{D}^2} \left| p_n(z_1,z_2) \right|} \\ &= \frac{n r^{2n}}{n 1^{2n}} \\ &= r^{2n} \text{.} \end{align*}
So this sequence of functions satisfies your requirement only if $0 < r \leq 1$ and does not do so if $1 < r$ (since in that case, $C$ is unbounded as we proceed through the sequence $\{p_n\}_n$).