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).


  • 1
    $\begingroup$ Looking at homogeneous polynomials of degree $n$ the constant goes to infinity with the degree $\endgroup$
    – Conrad
    Mar 2 at 23:19
  • $\begingroup$ @Conrad But we assume that our degree $n$ will always be finite? Of course, if $n \to \infty$, then $C \to \infty$. But this is a case we exclude here. $\endgroup$
    – anon
    Mar 3 at 10:21
  • $\begingroup$ not sure what you mean - the problem you mention seems to specifically call for all polynomials hence the degree can increase to infinity (in other words for degree $n$ you get a finite constant $C_n$ but $C_n \to \infty$ so cannot work for all polynomials); if you bound the degree of allowed polynomials probably you are good sure $\endgroup$
    – Conrad
    Mar 3 at 16:07
  • $\begingroup$ @Conrad Yes, the degree is bound. $\endgroup$
    – anon
    Mar 3 at 17:25
  • $\begingroup$ The text of the problem doesn't specify but says all polynomials which to me means that the degree can be any finite number however high $\endgroup$
    – Conrad
    Mar 3 at 20:11

1 Answer 1


(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 $\{p_n\}_n$).

  • $\begingroup$ I really think @Conrad's comment to the Question is on the right track -- if the degrees of the allowed $p$s are unbounded, this is hopeless. (Just let $p_n$ be successively longer approximations to $1/(1 - 1/n - x)$ so an ever-less-bounded spike pushes towards the boundary of $\Bbb{D}^2$. (Again, it's late and I may hove gotten myself turned around...) $\endgroup$ Mar 3 at 3:16
  • $\begingroup$ Interesting. So this means, that for $r>1$, it's essentially hopeless? Because this is what I would actually need: Whether the supremum of $p$ over an "extended bidisc" $r \mathbb{D}^2$ is bounded from above by the supremum of $p$ over the standard bidisc times a positive constant. I know that it's possible to obtain $C$ which depends on $p$, but I need a $C$ which does not depend on $p$ (and only on $r$). $\endgroup$
    – anon
    Mar 3 at 10:16
  • $\begingroup$ @anon : In the holomorphic functions on $\Bbb{C}^1$ setting, this condition is equivalent to $|f(z)| \leq C_1z$ for $|z|>1$, so that $|f'(z)| \leq C_2$ for $|z| > 1$, forcing $f'$ constant (because $f'$ is bounded on the closed disc automatically). So the only functions that work are linear. The $\Bbb{C}^1$ setting is embedded in your setting... $\endgroup$ Mar 3 at 16:14

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