CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-01

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A discontinuous positive solution of $F (x + y) = F (x) F (y)$

Statement refuted

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

Facts & Assumptions

Given: An irrational real $b$ , so ${1, b}$ is linearly independent over $Q$ , and the Axiom of Choice.

[L1]

Every independent set extends to a basis (Zorn's lemma gives a basis between any linearly independent set and any spanning set containing it: if $L \subseteq S \subseteq V$ with $L$ independent and $span (S) = V$ , there is a basis $B$ of $V$ with $L \subseteq B \subseteq S$ , Vector space over a field, The Axiom of Choice).

[L2]

A $Q$ -linear map is additive (Cauchy's functional equation $f (x + y) = f (x) + f (y)$ , and the additive functions $R \to R$ ).

[L3]

The ordinary exponential is positive, injective, and multiplicative (The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ , The exponential function is strictly increasing, The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ ).

Counterexample

technique · constructive

1.1

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

L1L2construct

2.1

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

step 1.1L3

3.1

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

step 1.1step 2.1L3given

4.1

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

step 3.1givendischarge-construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

59 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)
J. Lebl, Basic Analysis, Analytic Functions (standard reference, not scraped)
S. G. Johnson, Exponential Functions (standard reference, not scraped)
E. Gselmann, habilitation thesis on functional equations (standard reference, not scraped)

A discontinuous positive solution of F(x+y)=F(x)F(y)F(x+y)=F(x)F(y)