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CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-01
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A discontinuous positive solution of F(x+y)=F(x)F(y)F(x+y)=F(x)F(y)

Statement refuted

Every positive F:R(0,)F:\mathbb R\to(0,\infty) satisfying F(x+y)=F(x)F(y)F(x+y)=F(x)F(y) is continuous and equals an ordinary exponential.

Facts & Assumptions

Counterexample

technique · constructive
1.1

Extend {1,b}\{1,b\} to a Hamel basis. Define the Q\mathbb Q-linear map AA by A(1)=1A(1)=1, A(b)=0A(b)=0, and A(v)=vA(v)=v on the remaining chosen basis elements. Then AA is additive but not the identity.

L1L2construct
2.1

Put F(x)=exp(A(x))F(x)=\exp(A(x)). Positivity and additivity give F(x+y)=F(x)F(y)F(x+y)=F(x)F(y), and F(1)=eF(1)=e.

step 1.1L3
3.1

If FF were continuous, Regular normalized multiplicative Cauchy equations characterize the exponential would give F=expF=\exp; injectivity of exp\exp would then give A(x)=xA(x)=x, contradicting A(b)=0bA(b)=0\ne b.

step 1.1step 2.1L3given
4.1

Thus FF is a discontinuous positive multiplicative solution. The construction uses Choice exactly in the basis extension.

step 3.1givendischarge-construct

Depends on

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Dependency tree · next 3 levels

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Sources