$|f|\leq |g|$ in each $r\mathbb{T}^2$ implies on $\mathbb{D}^2$

Question: Let $f,g\in \mathcal{O}(\mathbb{D^2})$ (holomorphic functions on bidisc). Assume $|f|\leq|g|$ holds on each $r\mathbb{T}^2:=\{(z_1,z_2)\in \mathbb{D}^2: |z_1|=|z_2|=r\}$, $-\leq r<1$. And the zero set $Z(g)\subseteq Z(f)$. It is ture that $|f|\leq |g|$ on $\mathbb{D}^2$?

I try to put $h=f/g$ and use something like Riemann's removable-singularity Theorem or maximal modular principal. But I fails.

Solution 1:

We do not need the assumption $Z(g) \subseteq Z(f)$.

Let $w = (w_1,w_2) \in \mathbb{D}^2$. We want to show that $|f(w)| \leq |g(w)|$.

If $|w_1| = |w_2|$, then $w \in |w_1|\mathbb{T}^2$, and $|f(w)| \leq |g(w)|$ holds by hypothesis.

Thus, suppose $|w_1| \ne |w_2|$ and, wlog, $|w_1| < |w_2| =:r$. Let $F(z) = f(z,w_2)$ and $G(z) = g(z,w_2)$ (functions of one complex variable). For all $z \in \mathbb{C}$ with $|z|=r$, we have $(z,w_2) \in r\mathbb{T}^2$ and, hence, $|F(z)| \leq |G(z)|$. By the maximum principle, it follows that $|F(z)| \leq |G(z)|$ for all $z \in \mathbb{C}$ with $|z| \leq r$. In particular, this holds for $z=w_1$.