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Weierstrass M-test for complex-valued function series

Statement

Let $X$ be a set and $f_{n} : X \to C$ . Suppose $M_{n} \geq 0$ , $∣ f_{n} (x) ∣ \leq M_{n}$ for all $n, x$ , and the real series $\sum M_{n}$ converges. Then $\sum f_{n} (x)$ converges absolutely for every $x$ and its partial sums converge uniformly on $X$ in the sense of Uniform convergence and the uniformly Cauchy condition for complex-valued functions, with the componentwise dictionary.

Facts & Assumptions

Given: Functions $f_{n}$ and a convergent nonnegative majorant series $\sum M_{n}$ as in the Statement.

[L1]

For complex numbers, $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[L2]

A complex-valued function sequence converges uniformly if and only if it is uniformly Cauchy (A sequence of complex-valued functions converges uniformly if and only if it is uniformly Cauchy).

[L3]

The real Weierstrass M-test states that the same majorant hypotheses give absolute pointwise and uniform convergence for real-valued functions (The Weierstrass M-test gives absolute pointwise convergence and uniform convergence of a function series).

Proof

technique · direct

1.1L1algebra

For partial sums $S_{N} (x) = \sum_{n < N} f_{n} (x)$ and $q > p$ , [L1] gives $∣ S_{q} (x) - S_{p} (x) ∣ \leq \sum_{p \leq n < q} M_{n}$ for every $x$ .

2.1step 1.1L2

Since the real series $\sum M_{n}$ is Cauchy, its tails make the bound in step 1.1 uniformly small; thus $(S_{N})$ is uniformly Cauchy and converges uniformly by [L2].

3.1L3∎

For each $x$ , the nonnegative series $\sum ∣ f_{n} (x) ∣$ is bounded termwise by $\sum M_{n}$ , exactly the comparison used in [L3], and therefore converges. Zero majorants and the empty set require no separate choice.

Depends on

Dependency tree · 2 levels

Sources

L. Ahlfors, Complex Analysis, 3rd ed., Ch. 2