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Regular normalized multiplicative Cauchy equations characterize the exponential

Statement

The exponential function is the unique continuous $F : R \to (0, \infty)$ satisfying $F (x + y) = F (x) F (y)$ and $F (1) = e$ . It is also the unique function differentiable at $0$ satisfying the functional equation, $F (0) = 1$ , and $F^{'} (0) = 1$ .

Facts & Assumptions

Given: A function $F$ satisfying one of the two normalizations.

[L1]

The exponential satisfies $\exp (x + y) = \exp (x) \exp (y)$ , is continuous and positive, obeys $\exp (0) = 1$ , $\exp (1) = e$ , and $\exp^{'} = \exp$ , and is the unique normalized solution of $y^{'} = y$ (The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ , The exponential function is strictly increasing, The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ , The real exponential function and the number $e$ by a power series, The exponential function is smooth and $(\exp)^{'} = \exp$ , The exponential is the unique solution of $y^{'} = y$ with $y (0) = 1$ ).

[L2]

Positive $n$ -th roots exist uniquely (Existence and uniqueness of $n$ -th roots: a unique $a^{1 / n} \geq 0$ with $(a^{1 / n})^{n} = a$ ), rational powers are Rational powers $a^{r}$ of a positive base, and rationals are dense (The rationals embed densely in the reals).

Proof

technique · cases

1.1

Under continuity and $F (1) = e$ , the equation gives $F (n) = e^{n}$ , $F (- n) = e^{- n}$ , and uniqueness of positive roots gives $F (m / n) = e^{m / n}$ for rationals $m / n$ . Density and continuity then give $F (x) = \exp (x)$ for every real $x$ .

assume-case continuousL1L2given

1.2

Under differentiability at $0$ , $\frac{F (x + h) - F (x)}{h} = F (x) \frac{F (h) - 1}{h}$ , so $F^{'} (x) = F (x) F^{'} (0) = F (x)$ . With $F (0) = 1$ , [L1] gives $F = \exp$ .

assume-case differentiablegivenL1algebra

2.1

The exponential itself satisfies both normalizations, so both uniqueness assertions follow.

step 1.1step 1.2L1cases-exhaustive∎

Depends on

Used by

A discontinuous positive solution of F(x+y)=F(x)F(y) Counterexample
FALSE: every entire function with an antiderivative is a polynomial False statement
The power-series, product-limit, IVP, functional-equation, and Picard definitions agree Theorem

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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)
Y. Vorobets, Texas A&M MATH 409 Lecture 2-07 (standard reference, not scraped)
S. G. Johnson, Exponential Functions (standard reference, not scraped)