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6 results · all verified · 5 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 1 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Harmonic Functions and Mean Values in Rn — Examples

1 · Prerequisites

2 · Summary

Examples distinguish global mean-value structure from isolated identities and classical smoothness from distributional harmonicity.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Affine functions and mixed quadratic monomials are harmonic

Example

Let n1, aR, and bRn. Every affine function xa+bx is harmonic on Rn. If n2 and i,j{0,,n1} are distinct, then xxixj is harmonic on Rn.

Verification

Given: the displayed dimension, coefficients, coordinate indices, and the classical Laplacian The Laplacian of a C2 function and of a C2 vector field.

1.1

All second partial derivatives of a+bx vanish [given].

2.1

For ij, each diagonal second derivative of xixj vanishes [given]. ∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-07Open item page →

Real and imaginary parts of holomorphic monomials

Example

For z=x+iy, the components Rez2=x2y2 and Imz2=2xy, and likewise those of every zm, are harmonic on R2.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Radial harmonic functions away from the origin

Example

Let n1 and UC2((0,)). The radial function u:Rn{0}R given by u(x)=U(x) is harmonic precisely when U(r)+n1rU(r)=0(r>0). Thus the families are U(r)=a+br for n=1, U(r)=a+blogr for n=2, and U(r)=a+br2n for n3.

Verification

Given: n1, UC2((0,)), and r=x>0.

1.1

Direct differentiation gives ΔU(r)=U(r)+(n1)U(r)/r [given].

2.1

Multiplying by rn1 gives (rn1U)=0, whose integrations give the listed cases [given, algebra]. ∎

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Harmonic on a punctured domain need not extend

Statement refuted

A function harmonic on a punctured domain need not extend harmonically across the puncture.

Counterexample

Given: n2.

1.1

On R2{0} take logx; for n3 take x2n, both harmonic by Radial harmonic functions away from the origin [given].

2.1

Each is unbounded as x0, so it has no continuous, hence no harmonic, extension [given]. ∎

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

One centred ball-mean identity does not force harmonicity

Statement refuted

One ball-mean identity at one centre forces harmonicity.

Counterexample

Given: n1, R>0, and u:RnR defined by u(x)=x2(x2n+2n+4R2).

1.1

Polar coordinates Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma give 1BRBRx2=nn+2R2 and 1BRBRx4=nn+4R4, hence 1BRBRu=0=u(0) [given, algebra].

2.1

Yet Δu=4(n+2)x22nn+2n+4R2, which is not identically zero [given, algebra]. ∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Distributional harmonicity removes an apparent interior corner

Example

Let n1 and define v:RnR by v(x)=x1. This apparent corner is not distributionally harmonic on Rn; indeed Δv=2δ{x1=0}. Thus a distributionally harmonic locally integrable function has a unique smooth harmonic representative; in particular, the actual corner x1 cannot be distributionally harmonic on any open set meeting the hyperplane {x1=0}.

Verification

Given: n1, the domain Rn, and the distributional derivative convention Distributional harmonicity and Poisson's equation on an open subset of Rn.

1.1

In one variable, integrating by parts twice gives (t)=2δ0; tensoring with the remaining variables gives the stated hyperplane term [given].

2.1

Conversely Weyl's lemma for the Laplacian gives every distributionally harmonic distribution a smooth representative [step 1.1]. ∎

Sources