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LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26
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A real-linear functional on Cm is complex linear exactly when its antiholomorphic part vanishes

Statement

Fix m1 and let T:CmC be R-linear, that is additive with T(λh)=λT(h) for every real λ. Then there are unique c0,,cm1 and d0,,dm1 in C with

T(h)=k<mckhk+k<mdkhk(hCm),

namely ck=12(T(ek)iT(iek)) and dk=12(T(ek)+iT(iek)). Moreover T is C-linear if and only if dk=0 for every k<m.

Facts & Assumptions

Given: An R-linear T:CmC, with Cm 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:RmRn in Euclidean coordinates); C-linear additionally requires T(λh)=λT(h) for every complex λ (Holomorphic functions on an open subset of Cm).

[L3]

For z=a+bi with a,b real, Rez=a, Imz=b and z=abi (Real and imaginary parts, complex conjugation, and modulus).

[L5]

The Wirtinger operators of a real totally differentiable f satisfy Df(a)h=k<m(zkf(a))hk+k<m(zˉkf(a))hk (Wirtinger operators in Cm).

Proof

technique · direct
1.1

Write hk=ξk+iηk with ξk,ηk real, as in [L3]. By [L2] and R-linearity, h=k<m(ξkek+ηk(iek)) and hence T(h)=k<m(ξkT(ek)+ηkT(iek)).

givenL1L2L3algebra
1.2

Put ck=12(T(ek)iT(iek)) and dk=12(T(ek)+iT(iek)); then ck+dk=T(ek) and i(ckdk)=T(iek).

givenalgebra
2.1

Substituting ξk=12(hk+hk) and ηk=12i(hkhk) from [L3] into step 1.1 and collecting, the coefficient of hk is 12T(ek)+12iT(iek)=ck and the coefficient of hk is 12T(ek)12iT(iek)=dk, so T(h)=k<mckhk+k<mdkhk.

step 1.1step 1.2L3algebra
2.2

The coefficients are unique: if kckhk+kdkhk represents T as well, evaluating at h=ek gives ck+dk=T(ek) and at h=iek gives i(ckdk)=T(iek), a system whose only solution is the pair of step 1.2.

step 1.2L2L3algebra
3.1

If every dk=0 then T(h)=kckhk, which satisfies T(λh)=λT(h) for every complex λ, so T is C-linear in the sense of [L1].

step 2.1L1algebra
3.2

Conversely, suppose T is C-linear. Taking λ=i in [L1] and using step 2.1 gives kck(ihk)+kdkihk=ikckhk+ikdkhk; since ihk=ihk by [L3], the left side is ikckhkikdkhk, so 2ikdkhk=0 for every h. Evaluating at h=ek gives dk=0 for each k<m.

step 2.1L1L3algebra
4.1

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 ck and dk vanishes, are included with no separate argument. By [L5] the representation applied to T=Df(a) has ck=zkf(a) and dk=zˉkf(a), so the criterion reads: the real differential is C-linear exactly when every zˉkf(a) vanishes.

step 2.1step 2.2step 3.1step 3.2L5

Depends on

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