Alphabeta Math
TheoremStatement: AI-adaptedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13
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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.

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The C2 real and imaginary parts of a holomorphic function satisfy Laplace's equation and form a harmonic-conjugate pair

Statement

Let f=u+iv be holomorphic on an open set U⊆C, and assume u,v∈C2(U). Then

uxx+uyy=0,vxx+vyy=0.

A C2 real function satisfying this equation is called harmonic. Thus u and v are harmonic, and v is a harmonic conjugate of u in the sense that u+iv is holomorphic. No automatic C2 regularity or global existence of harmonic conjugates is asserted.

Facts & Assumptions

Given: A holomorphic f=u+iv on U with u,v∈C2(U).

[F1]

A C2 function has continuous iterated partial derivatives through order two (Ck maps and multi-index derivative notation in Euclidean space).

[L2]

If a function is C2 on an open subset of Rm, then its mixed second partial derivatives agree (Clairaut--Schwarz theorem for continuous second partial derivatives).

Proof

technique · direct
1.1

Differentiating ux=vy in x and uy=−vx in y gives uxx=vyx and uyy=−vxy.

givenL1F1
1.2

Differentiating ux=vy in y and uy=−vx in x gives uxy=vyy and uyx=−vxx.

givenL1F1
2.1

By [L2], vyx=vxy, so step 1.1 gives uxx+uyy=0.

step 1.1L2algebra
3.1

By [L2], uxy=uyx, so step 1.2 gives vxx+vyy=0. The terminology in the Statement now applies to the given pair u,v.

step 1.2L2algebra∎

Depends on

Used by

Dependency tree · two levels

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Sources