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If $f(z)$ is an analytic function in a complex neighborhood (in $\mathbb C^n$) of a real point $x^0 \in \mathbb R^n \subset \mathbb C^n$, then $\log|f(x)|$ is integrable over some neighborhood $U \subset \mathbb R^n$ of $x^0$. Does anyone know a reference for this fact? I looked into a number of books on several complex variables without finding the statement.

I want a reference, because I don't think the validity of the assertion is obvious. However, here is sketch of proof. We may assume that $x^0= (0, \ldots, 0)$. Choose coordinates so that $f(0, \ldots 0, x_n)$ does not vanish identically near $x_n= 0$. Then the same is true for $x'$ in some neighborhood $V$ of $(0, \ldots, 0) \in \mathbb R^{n-1}$, hence $x_n \mapsto \log|f(x', x_n)|$ is locally integrable for every $x' \in V$. But we need to prove that the integrability is uniform with respect to $x'$. However, using Weierstrass' preparation theorem it is now easy to see that the integral $\int\log|f(x', x_n)|dx_n$ over a neighborhood of $x_n = 0$ is locally bounded as a function of $x'$ after (if needed) a choice of new coordinates.

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    $\begingroup$ I know a reference for the Mahler measure, which is a special case, namely $f(z)=P(e^{iz_1},\ldots,e^{iz_n})$ where $P$ is any complex polynomial, see Everest, Ward, "Heights of polynomials and entropy in algebraic dynamics", Section 3.2. They show in this case that locally the measure of the set $\{|f| < \varepsilon\}$ is $\ll \varepsilon^\alpha$ for some $\alpha>0$. Probably this estimate holds for arbitrary $f$. $\endgroup$ Commented Jan 21, 2023 at 8:15
  • $\begingroup$ This is very interesting. Thank you very much! $\endgroup$
    – Jan Boman
    Commented Jan 26, 2023 at 22:10

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