Logarithm of Infinite Product of Complex Functions

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

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

Then the following are equivalent:

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


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

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

Let the product $\displaystyle \prod_{n \mathop = 1}^\infty f_n$ 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:

  • $f_n(x)\neq0$ for $n\geq n_0$ and $x\in U$
  • The series $\displaystyle \sum_{n \mathop = n_0}^\infty \log f_n$ converges uniformly on $U$ to $\log g + 2k\pi i$, where $g = \displaystyle \prod_{n\mathop =n_0}^\infty f_n$.


1 implies 2

It suffices to show that:

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

Real part

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

By Logarithm of Infinite Product of Real Functions $\displaystyle \sum_{n \mathop = 1}^\infty \log |f_n| = \log |f|$ 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 $\left({\theta - \pi \,.\,.\, \theta + \pi}\right]$.

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:

$\displaystyle \sum_{j \mathop = 1}^n \Im(\log f_n(x)) = \theta_n(x) + 2 k_n(x) \pi$

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

By the Triangle Inequality:

$2 \pi \sup_{x\in K}\left\vert{k_{n + 1}(x) - k_n(x)}\right\vert \le \sup_{x\in K}\left\vert{\theta_{n + 1}(x) - \theta_n(x)}\right\vert + \sup_{x\in K}\left\vert{\Im(\log f_{n + 1}(x) )}\right\vert$

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 $|f_n-1|\leq \frac12$ for $n\geq n_0$.

By Complex Logarithm is Continuous Outside Branch and Heine-Cantor Theorem, $\log$ is uniformly continuous on $\overline B(1,\frac12)$.

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

Thus $2 \pi \sup_{x\in K}\left\vert{k_{n + 1}(x) - k_n(x)}\right\vert\to 0$.

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


$\displaystyle \sum_{j \mathop = 1}^n \Im(\log f_n) \to \theta + 2 k \pi$ uniformly on $K$.


2 implies 1

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

Also see