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I am reading Gunning's book Introduction to Holomorphic Functions of Several Variables, Vol. I, and I am stuck in the proof of Maximum modulus theorem: if $f$ is holomorphic in a connected open subset $D \subset \mathbb{C}^{n}$ and if there is a point $A \in D$ such that $|f(Z)| \leq |f(A)|$ for all points $Z$ in some open neighborhood of $A$, then $f(Z) = f(A)$ for all points $Z \in D$. In the proof Gunning says that for any polydisc $\Delta = \Delta(A; R) \subset D$ for which $\overline{\Delta} \subset D$ we have as a consequence of the Cauchy integral formula that

$$|\Delta| f(A) = \int_{\Delta} f(Z) dV(Z),$$ where $dV(Z)$ is the Euclidean volume element in $\mathbb{C}^{n} = \mathbb{R}^{2n}$ and $|\Delta| = \int_{\Delta} dV(Z) = \pi^{n}r_{1}^{2} \cdots r_{n}^{2}$ is the Euclidean volume of $\Delta$.

It looks very easy, but I am stuck on it for a long time. I can not see how this is a consequence of Cauchy integral formula, since the integral is on $\Delta$ and not on the product of $|\zeta_{j} - a_{j}| = r_{j}$. We can not apply Stokes, because the form is of degree $n$. Maybe the Intermediate Value Theorem for integrals solve it, but how to assure that the point giving the equality is $A$? Maybe a change of variables?

Thanks for help.

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up vote 2 down vote accepted

It is an integrated form of the Cauchy formula. The single complex variable case illustrates what's going on. For example, $$ f(0) = {1\over 2\pi i}\int_{|z|=1} {f(z)\over z}\;dz = {1\over 2\pi} \int_0^{2\pi} {f(re^{i\theta})\over r\,e^{i\theta}}\,d(re^{i\theta}) = {1\over 2\pi} \int_0^{2\pi} f(re^{i\theta})\;i\,d\theta $$ can be integrated (for example) $\int_0^1 \ldots r\,dr$: $$ f(0)\cdot \int_0^11\cdot r\,dr \;=\; {1\over 2\pi}\int_0^1 \int_0^{2\pi} f(re^{i\theta})\;r\;d\theta\,dr $$ The obvious absolute-value estimate gives $$ |f(0)|\;\le\; {1\over \pi} \int_{|z|\le 1} |f(z)|\,dV $$

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