The modulus of a holomorphic function on a closed polydisc is bounded by its supremum on the distinguished boundary

The sets ${\bar{\mathrm{\Delta }}}_r(a)$ and ${\mathrm{\Gamma }}_r(a)$ are closed and bounded by [L2] and [L9], hence compact and nonempty by [L5], so $\mathrm{\mid }f\mathrm{\mid }$ attains a maximum on each by [L5] and both suprema are attained real numbers; write $P={\sup }_{{\mathrm{\Gamma }}_r(a)}\mathrm{\mid }f\mathrm{\mid }$. By [L6] the function $f$ is uniformly continuous on ${\bar{\mathrm{\Delta }}}_r(a)$.

For a polyradius $s$ with $s_k<r_k$ for every $k$, put $A_p:=\sup \{\mathrm{\mid }f(w)\mathrm{\mid }:w\in {\bar{\mathrm{\Delta }}}_s(a)\text{ and }\mathrm{\mid }{w}_k-a_k\mathrm{\mid }=s_k\text{ for every }k<p\}$ for $0\le p\le m$; these are attained maxima by the argument of step 1.1 applied to the corresponding closed bounded sets, and $A_m={\sup }_{{\mathrm{\Gamma }}_s(a)}\mathrm{\mid }f\mathrm{\mid }$.

Let $0<t<1$ and take $s=tr$. Every $w\in {\mathrm{\Gamma }}_{tr}(a)$ satisfies $a+(w-a)\mathrm{/}t\in {\mathrm{\Gamma }}_r(a)$ and $\parallel (a+(w-a)\mathrm{/}t)-w\parallel =(1\mathrm{/}t-1)\parallel w-a\parallel$ by [L9], which tends to $0$ as $t\to 1$ uniformly in $w$ because $\parallel w-a\parallel$ is bounded on ${\bar{\mathrm{\Delta }}}_r(a)$; so by the uniform continuity of step 1.1 and [L10], ${\sup }_{{\mathrm{\Gamma }}_{tr}(a)}\mathrm{\mid }f\mathrm{\mid }\le P+\epsilon$ once $t$ is close enough to $1$, for any prescribed $\epsilon >0$.

Fix such an $s$ and $p<m$, and let $w\in {\bar{\mathrm{\Delta }}}_s(a)$ have $\mathrm{\mid }{w}_k-a_k\mathrm{\mid }=s_k$ for $k<p$. Replacing the $p$th coordinate of $w$ by any $\xi$ with $\mathrm{\mid }\xi -a_p\mathrm{\mid }<r_p$ leaves the point in ${\mathrm{\Delta }}_r(a)$, because the other coordinates satisfy $\mathrm{\mid }{w}_k-a_k\mathrm{\mid }\le s_k<r_k$; so by [L3] the slice $g(\xi )$ is holomorphic on the disc $D(a_p,r_p)$ and in particular continuous on the closed disc $\bar{D}(a_p,s_p)$, which is the closure of $D(a_p,s_p)$ since every point of the circle is a limit of interior points along its radius and the closed disc is closed by [L8] and [L9].

The disc $D(a_p,s_p)$ is a bounded complex domain by [L2], [L7] and [L8], being nonempty, open, convex hence connected, and bounded, and its topological boundary is the circle $\mathrm{\mid }\xi -a_p\mathrm{\mid }=s_p$ by step 3.1. So [L1] applied to the slice gives $\xi$ on that circle with $\mathrm{\mid }f(w)\mathrm{\mid }\le \mathrm{\mid }g(\xi )\mathrm{\mid }$, and the point obtained from $w$ by putting $\xi$ in the $p$th slot lies in ${\bar{\mathrm{\Delta }}}_s(a)$ with its first $p+1$ coordinates on their circles; hence $\mathrm{\mid }f(w)\mathrm{\mid }\le A_{p+1}$. Taking the supremum over such $w$ gives $A_p\le A_{p+1}$.

By [L4] the chain of step 4.1 gives $A_0\le A_m$, that is ${\sup }_{{\bar{\mathrm{\Delta }}}_s(a)}\mathrm{\mid }f\mathrm{\mid }\le {\sup }_{{\mathrm{\Gamma }}_s(a)}\mathrm{\mid }f\mathrm{\mid }$ for every polyradius $s$ with $s_k<r_k$.

Let $z\in {\bar{\mathrm{\Delta }}}_r(a)$ and $\epsilon >0$. For $t<1$ the point $a+t(z-a)$ lies in ${\mathrm{\Delta }}_{tr}(a)\subseteq {\bar{\mathrm{\Delta }}}_{tr}(a)$ by [L2] and [L9], so steps 5.1 and 2.2 give $\mathrm{\mid }f(a+t(z-a))\mathrm{\mid }\le {\sup }_{{\mathrm{\Gamma }}_{tr}(a)}\mathrm{\mid }f\mathrm{\mid }\le P+\epsilon$ for $t$ close enough to $1$; letting $t\to 1$ and using the continuity of $f$ at $z$ gives $\mathrm{\mid }f(z)\mathrm{\mid }\le P+\epsilon$, hence $\mathrm{\mid }f(z)\mathrm{\mid }\le P$.

Step 6.1 gives ${\sup }_{{\bar{\mathrm{\Delta }}}_r(a)}\mathrm{\mid }f\mathrm{\mid }\le P$, and the reverse inequality holds because ${\mathrm{\Gamma }}_r(a)\subseteq {\bar{\mathrm{\Delta }}}_r(a)$ by [L2]; so the two suprema are equal and attained by step 1.1. For $m\ge 2$ the set ${\mathrm{\Gamma }}_r(a)$ is by [L2] a proper subset of the topological boundary, so the bound is by a strictly smaller set than in the one-variable statement.