LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A real-linear functional on $C^{m}$ is complex linear exactly when its antiholomorphic part vanishes

Statement

Fix $m \geq 1$ and let $T : C^{m} \to C$ be $R$ -linear, that is additive with $T (λ h) = λ T (h)$ for every real $λ$ . Then there are unique $c_{0}, \dots, c_{m - 1}$ and $d_{0}, \dots, d_{m - 1}$ in $C$ with

$T (h) = \sum_{k < m} c_{k} h_{k} + \sum_{k < m} d_{k} \overline{h_{k}} (h \in C^{m}),$

namely $c_{k} = \frac{1}{2} (T (e_{k}) - i T (i e_{k}))$ and $d_{k} = \frac{1}{2} (T (e_{k}) + i T (i e_{k}))$ . Moreover $T$ is $C$ -linear if and only if $d_{k} = 0$ for every $k < m$ .

Facts & Assumptions

Given: An $R$ -linear $T : C^{m} \to C$ , with $C^{m}$ read through Complex $m$ -space and its real coordinate dictionary.

[L1]

A map between Euclidean spaces is linear when it preserves real linear combinations (A linear map $L : R^{m} \to R^{n}$ in Euclidean coordinates); $C$ -linear additionally requires $T (λ h) = λ T (h)$ for every complex $λ$ (Holomorphic functions on an open subset of $C^{m}$ ).

[L2]

$C^{m}$ is a vector space over $C$ (Vector space over a field, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ) with standard basis $e_{0}, \dots, e_{m - 1}$ , and every $h$ 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 $0$ ).

[L3]

For $z = a + b i$ with $a, b$ real, $Re z = a$ , $Im z = b$ and $\overline{z} = a - b i$ (Real and imaginary parts, complex conjugation, and modulus).

[L5]

The Wirtinger operators of a real totally differentiable $f$ satisfy $D f (a) h = \sum_{k < m} (\partial_{z_{k}} f (a)) h_{k} + \sum_{k < m} (\partial_{{\overset{ˉ}{z}}_{k}} f (a)) \overline{h_{k}}$ (Wirtinger operators in $C^{m}$ ).

Proof

technique · direct

1.1givenL1L2L3algebra

Write $h_{k} = ξ_{k} + i η_{k}$ with $ξ_{k}, η_{k}$ real, as in [L3]. By [L2] and $R$ -linearity, $h = \sum_{k < m} (ξ_{k} e_{k} + η_{k} (i e_{k}))$ and hence $T (h) = \sum_{k < m} (ξ_{k} T (e_{k}) + η_{k} T (i e_{k}))$ .

1.2givenalgebra

Put $c_{k} = \frac{1}{2} (T (e_{k}) - i T (i e_{k}))$ and $d_{k} = \frac{1}{2} (T (e_{k}) + i T (i e_{k}))$ ; then $c_{k} + d_{k} = T (e_{k})$ and $i (c_{k} - d_{k}) = T (i e_{k})$ .

2.1step 1.1step 1.2L3algebra

Substituting $ξ_{k} = \frac{1}{2} (h_{k} + \overline{h_{k}})$ and $η_{k} = \frac{1}{2 i} (h_{k} - \overline{h_{k}})$ from [L3] into step 1.1 and collecting, the coefficient of $h_{k}$ is $\frac{1}{2} T (e_{k}) + \frac{1}{2 i} T (i e_{k}) = c_{k}$ and the coefficient of $\overline{h_{k}}$ is $\frac{1}{2} T (e_{k}) - \frac{1}{2 i} T (i e_{k}) = d_{k}$ , so $T (h) = \sum_{k < m} c_{k} h_{k} + \sum_{k < m} d_{k} \overline{h_{k}}$ .

2.2step 1.2L2L3algebra

The coefficients are unique: if $\sum_{k} c_{k}^{'} h_{k} + \sum_{k} d_{k}^{'} \overline{h_{k}}$ represents $T$ as well, evaluating at $h = e_{k}$ gives $c_{k}^{'} + d_{k}^{'} = T (e_{k})$ and at $h = i e_{k}$ gives $i (c_{k}^{'} - d_{k}^{'}) = T (i e_{k})$ , a system whose only solution is the pair of step 1.2.

3.1step 2.1L1algebra

If every $d_{k} = 0$ then $T (h) = \sum_{k} c_{k} h_{k}$ , which satisfies $T (λ h) = λ T (h)$ for every complex $λ$ , so $T$ is $C$ -linear in the sense of [L1].

3.2step 2.1L1L3algebra

Conversely, suppose $T$ is $C$ -linear. Taking $λ = i$ in [L1] and using step 2.1 gives $\sum_{k} c_{k} (i h_{k}) + \sum_{k} d_{k} \overline{i h_{k}} = i \sum_{k} c_{k} h_{k} + i \sum_{k} d_{k} \overline{h_{k}}$ ; since $\overline{i h_{k}} = - i \overline{h_{k}}$ by [L3], the left side is $i \sum_{k} c_{k} h_{k} - i \sum_{k} d_{k} \overline{h_{k}}$ , so $2 i \sum_{k} d_{k} \overline{h_{k}} = 0$ for every $h$ . Evaluating at $h = e_{k}$ gives $d_{k} = 0$ for each $k < m$ .

4.1step 2.1step 2.2step 3.1step 3.2L5∎

Steps 2.1, 2.2, 3.1 and 3.2 prove the representation, its uniqueness, and the stated equivalence; the case $m = 1$ and the zero functional, for which every $c_{k}$ and $d_{k}$ vanishes, are included with no separate argument. By [L5] the representation applied to $T = D f (a)$ has $c_{k} = \partial_{z_{k}} f (a)$ and $d_{k} = \partial_{{\overset{ˉ}{z}}_{k}} f (a)$ , so the criterion reads: the real differential is $C$ -linear exactly when every $\partial_{{\overset{ˉ}{z}}_{k}} f (a)$ vanishes.

Depends on

Dependency tree · 2 levels

Sources

J. Lebl, Tasty Bits of Several Complex Variables, §1.3

A real-linear functional on CmCm is complex linear exactly when its antiholomorphic part vanishes