Logarithm of Infinite Product of Complex Functions

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Let $X$ be a weakly locally compact topological space.

Let $\sequence {f_n}$ be a sequence of everywhere nonzero continuous mappings $f_n: X \to \C$.

Then the following are equivalent:

$(1): \quad$ The product $\ds \prod_{n \mathop = 1}^\infty f_n$ converges locally uniformly to $f$.
$(2): \quad$ The series $\ds \sum_{n \mathop = 1}^\infty \ln f_n$ converges locally uniformly to $\ln f + 2k\pi i$ for some mapping $k:K\to\Z$.


Let $X$ be a locally compact and locally connected metric space.

Let $\sequence {f_n}$ be a sequence of continuous mappings $f_n: X \to \C$.

Let the product $\ds \prod_{n \mathop = 1}^\infty f_n$ converge locally uniformly to $f$.

Let $x_0 \in X$.

Then there exist $n_0 \in \N$, $k \in \Z$ and a neighborhood $U$ of $x_0$ such that:

$(1): \quad \map {f_n} x \ne 0$ for $n \ge n_0$ and $x \in U$
$(2): \quad$ The series $\ds \sum_{n \mathop = n_0}^\infty \ln f_n$ converges uniformly on $U$ to $\ln g + 2 k \pi i$, where $g = \ds \prod_{n \mathop = n_0}^\infty f_n$.


1 implies 2

It suffices to show that:

$(1): \quad \ds \sum_{n \mathop = 1}^\infty \map \Re {\ln f_n} = \map \Re {\ln f}$ locally uniformly
$(2): \quad \ds \sum_{n \mathop = 1}^\infty \map \Im {\ln f_n} = \map \Im {\ln f} + 2 k \pi$ locally uniformly for some $k: K \to \Z$

Real part

By Absolute Value of Uniformly Convergent Product, $\ds \prod_{n \mathop = 1}^\infty \cmod {f_n} = \cmod f$ locally uniformly.

By Logarithm of Infinite Product of Real Functions $\ds \sum_{n \mathop = 1}^\infty \ln \cmod {f_n} = \ln \cmod f$ locally uniformly.

Hence the result.

Imaginary Part

Let $K\subset X$ be compact.

Let $P_n$ denote the $n$th partial product.

Then $P_n \to f$ uniformly on $K$.

Let $\theta$ be the argument of $f$.

Let $\theta_n = \arg P_n$ be the argument of $P_n$ in the half-open interval $\hointl {\theta - \pi} {\theta + \pi}$.

By the corollary to Uniform Convergence of Complex Functions in Polar Form:

$\theta_n \to \theta$ uniformly on $K$.

Let $k_n: K \to \Z$ be such that:

$\ds \sum_{j \mathop = 1}^n \map \Im {\ln \map {f_n} x} = \map {\theta_n} x + 2 \map {k_n} x \pi$

We show that $k_n$ is eventually equal to a fixed function $k: K \to \Z$.

By the Triangle Inequality for Complex Numbers:

$2 \pi \sup_{x \mathop \in K} \size {\map {k_{n + 1} } x - \map {k_n} x} \le \sup_{x \mathop \in K} \cmod {\map {\theta_{n + 1} } x - \map {\theta_n} x} + \sup_{x \mathop \in K} \size {\map \Im {\ln \map {f_{n + 1} } x } }$

By Factors in Uniformly Convergent Product Converge Uniformly to One, $f_n \to 1$ uniformly on $K$.

Let $n_0 \in \N$ be such that $\cmod {f_n - 1} \le \dfrac 1 2$ for $n \ge n_0$.

By Complex Logarithm is Continuous Outside Branch and Heine-Cantor Theorem, $\ln$ is uniformly continuous on $\map {\overline B} {1, \dfrac 1 2}$.

By Uniformly Continuous Function Preserves Uniform Convergence, $\ln f_n \to 0$ uniformly on $K$.

Thus $\ds 2 \pi \sup_{x \mathop \in K} \cmod {\map {k_{n + 1} } x - \map {k_n} x} \to 0$.

So the sequence $k_n$ is eventually constant, say equal to $k: K \to \Z$.


$\ds \sum_{j \mathop = 1}^n \map \Im {\ln f_n} \to \theta + 2 k \pi$ uniformly on $K$.


2 implies 1

Follows from Complex Exponential is Uniformly Continuous on Half-Planes.

Also see