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

complex-analysisseveral-complex-variables

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.

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 $$z=w_1$$.