Prove that: $\sup_{z \in \overline{D}} |f(z)|=\sup_{z \in \Gamma} |f(z)|$ Suppose $D=\Delta^n(a,r)=\Delta(a_1,r_1)\times \ldots  \times \Delta(a_n,r_n) \subset \mathbb{C}^n$
and 
$\Gamma =\partial \circ \Delta^n(a,r)=\left \{ z=(z_1, \ldots , z_n)\in \mathbb{C}^n:|z_j-a_j|=r_j,~ j=\overline{1,n}  \right \}$.
Let $f \in \mathcal{H}(D) \cap \mathcal{C}(\overline{D})$. 
Prove that: $\sup_{z \in \overline{D}} |f(z)|=\sup_{z \in \Gamma} |f(z)|$
I need your help. Thanks.

Since it's homework, a few hints:
First assume that $f$ is holomorphic on a neighbourhood of $\bar D$. Then use the maximum principle in each variable separately to conclude that it's true in this case. Finally, approximate your $f$ by functions that are holomorphic on a neihbourhood of $\bar D$.

More details I'll do it for $n=2$ under the assumption that $f$ extends to a neighbourhood of $\bar D$, and leave the general case up to you. Let $a \in \partial \Delta$ and define $\phi_a(\zeta) = (a,\zeta)$. Then $f(\phi_a(\zeta))$ is holomorphic on a neighbourhood of $\bar\Delta$, so by the maximum modulus principle in one variable applied to $f\circ \phi_a$, $\sup_{\zeta\in\bar\Delta} f(a, \zeta) = \sup_{\zeta\in\partial\Delta} f(a, \zeta)$. Do the same for $\psi_a(\zeta) = (\zeta,a)$ and take the sup over all $a \in \Delta$ to obtain
$$\sup_{z\in \partial(\Delta \times \Delta)} f(z) = \sup_{z\in\partial\Delta\times\partial\Delta} f(z)$$
since the union of the images of $\phi_a$ and $\psi_a$ cover the entire boundary of $\Delta\times\Delta$. To finish off, use the maximum modulus principle in $\mathbb{C}^n$ to see that
$$\sup_{z\in \bar\Delta \times \bar\Delta} f(z) = \sup_{z\in\partial(\Delta\times\Delta)} f(z)$$
or, if you prefer, do the same argument again with $a \in \Delta$.
