PropositionStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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Sums, products and nonvanishing quotients of holomorphic functions are holomorphic

Statement

Let $m \geq 1$ , let $U \subseteq C^{m}$ be open and let $f, g : U \to C$ be holomorphic. Then $λ f + μ g$ is holomorphic on $U$ for all $λ, μ \in C$ , $f g$ is holomorphic on $U$ , and $f / g$ is holomorphic on the open set ${z \in U : g (z) \neq 0}$ , with

$D (λ f + μ g) (a) = λ D f (a) + μ D g (a), D (f g) (a) = f (a) D g (a) + g (a) D f (a),$ $D (f / g) (a) = \frac{g (a) D f (a) - f (a) D g (a)}{g (a)^{2}} (g (a) \neq 0),$

and correspondingly $\partial_{z_{k}} (f g) = f \partial_{z_{k}} g + g \partial_{z_{k}} f$ and $\partial_{z_{k}} (f / g) = (g \partial_{z_{k}} f - f \partial_{z_{k}} g) / g^{2}$ for each $k < m$ . In particular the holomorphic functions on $U$ form a commutative ring under pointwise operations, containing the constants.

Facts & Assumptions

Given: An open $U \subseteq C^{m}$ and holomorphic $f, g : U \to C$ ; $C^{m}$ is read through Complex $m$ -space and its real coordinate dictionary.

[L1]

$f$ is complex differentiable at $a$ when there is a $C$ -linear $L$ with $f (a + h) = f (a) + L (h) + r (h)$ and $∣ r (h) ∣ / ∥ h ∥ \to 0$ ; $L$ is unique and written $D f (a)$ (Holomorphic functions on an open subset of $C^{m}$ ).

[L2]

A holomorphic function of several variables is continuous, and $D f (a) h = \sum_{k < m} (\partial_{z_{k}} f (a)) h_{k}$ (A holomorphic function of several variables is continuous and separately holomorphic).

[L3]

A map $T (h) = \sum_{k < m} c_{k} h_{k}$ is $C$ -linear, and for a differentiable $f$ the coefficients are $\partial_{z_{k}} f$ (A real-linear functional on $C^{m}$ is complex linear exactly when its antiholomorphic part vanishes, Wirtinger operators in $C^{m}$ ).

[L4]

For every linear $L : R^{m} \to R^{n}$ there is $K \geq 0$ with $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for every $h$ (Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ ).

[L5]

Linear combinations, products and nonvanishing quotients of complex numbers obey the field laws ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ), and the one-variable derivative rules take the displayed forms (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L6]

A set is open exactly when each of its points admits a ball inside it (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space).

[L7]

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

[L8]

Every $h \in C^{m}$ satisfies $h = \sum_{k < m} h_{k} e_{k}$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

Proof

technique · direct

1.1givenL1L4

Fix $a \in U$ and write $f (a + h) = f (a) + D f (a) h + r_{f} (h)$ and $g (a + h) = g (a) + D g (a) h + r_{g} (h)$ as in [L1], with $∣ r_{f} (h) ∣, ∣ r_{g} (h) ∣ = o (∥ h ∥)$ ; by [L4] read through the dictionary there is $K \geq 0$ with $∣ D f (a) h ∣ \leq K ∥ h ∥$ and $∣ D g (a) h ∣ \leq K ∥ h ∥$ .

2.1step 1.1L1L3L7L8

For $λ, μ \in C$ the map $h \mapsto λ D f (a) h + μ D g (a) h$ is $C$ -linear by [L3] and [L8], and the remainder of $λ f + μ g$ at $a$ is $λ r_{f} (h) + μ r_{g} (h)$ , which is $o (∥ h ∥)$ by [L7]; so [L1] makes $λ f + μ g$ complex differentiable at $a$ with the stated differential.

2.2step 1.1L1L3L5L7

Multiplying the two expansions of step 1.1 and collecting, $f (a + h) g (a + h) = f (a) g (a) + (f (a) D g (a) h + g (a) D f (a) h) + ϱ (h)$ , where $ϱ (h) = D f (a) h D g (a) h + (f (a) + D f (a) h) r_{g} (h) + (g (a) + D g (a) h + r_{g} (h)) r_{f} (h)$ . By step 1.1 and [L7] the first term is at most $K^{2} ∥ h ∥^{2}$ and the others are bounded quantities times $o (∥ h ∥)$ , so $ϱ (h) = o (∥ h ∥)$ ; the first-order part is $C$ -linear by [L3], so [L1] gives the product rule.

2.3step 1.1L1L2L3L5L6L7

Suppose $g (a) \neq 0$ . By [L2] the function $g$ is continuous, so [L6] gives a ball $B$ about $a$ inside $U$ on which $∣ g ∣ \geq ∣ g (a) ∣ / 2 > 0$ ; in particular ${z \in U : g (z) \neq 0}$ is open by [L6]. On $B$ write $\frac{1}{g (a + h)} - \frac{1}{g (a)} = \frac{g (a) - g (a + h)}{g (a) g (a + h)} = \frac{- D g (a) h - r_{g} (h)}{g (a) g (a + h)}$ ; using $g (a + h) \to g (a)$ this equals $- \frac{D g (a) h}{g (a)^{2}} + o (∥ h ∥)$ by step 1.1 and [L7]. So $1 / g$ is complex differentiable at $a$ with differential $h \mapsto - D g (a) h / g (a)^{2}$ , which is $C$ -linear by [L3].

3.1step 2.2step 2.3L2L3L5L8

Combining steps 2.2 and 2.3 gives the quotient rule for $f / g$ at every point where $g$ does not vanish, and reading each differential at $h = e_{k}$ with [L2], [L3] and [L8] gives the displayed formulas for $\partial_{z_{k}}$ .

4.1step 2.1step 2.2L1L5∎

Steps 2.1 and 2.2 make the holomorphic functions on $U$ closed under pointwise addition and multiplication; those operations are commutative, associative and distributive because the values lie in the field $C$ ([L5]), and every constant function is holomorphic with zero differential by [L1]. So the holomorphic functions on $U$ form a commutative ring containing the constants.

Depends on

Used by

Dependency tree · two levels

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Sources

J. Lebl, Tasty Bits of Several Complex Variables, §1.2 (standard reference, not scraped)