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Regular normalized multiplicative Cauchy equations characterize the exponential

Statement

The exponential function is the unique continuous F:R→(0,∞)F:\mathbb R\to(0,\infty) satisfying F(x+y)=F(x)F(y)F(x+y)=F(x)F(y) and F(1)=eF(1)=e. It is also the unique function differentiable at 00 satisfying the functional equation, F(0)=1F(0)=1, and F′(0)=1F'(0)=1.

Facts & Assumptions

Given: A function FF satisfying one of the two normalizations.

[L1]

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

[L2]

Positive nn-th roots exist uniquely (Existence and uniqueness of nn-th roots: a unique a1/n≥0a^{1/n} \ge 0 with (a1/n)n=a(a^{1/n})^n = a), rational powers are Rational powers ara^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)=eF(1)=e, the equation gives F(n)=enF(n)=e^n, F(−n)=e−nF(-n)=e^{-n}, and uniqueness of positive roots gives F(m/n)=em/nF(m/n)=e^{m/n} for rationals m/nm/n. Density and continuity then give F(x)=exp⁡(x)F(x)=\exp(x) for every real xx.

assume-case continuousL1L2given
1.2

Under differentiability at 00, F(x+h)−F(x)h=F(x)F(h)−1h\frac{F(x+h)-F(x)}h=F(x)\frac{F(h)-1}h, so F′(x)=F(x)F′(0)=F(x)F'(x)=F(x)F'(0)=F(x). With F(0)=1F(0)=1, [L1] gives F=exp⁡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

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Sources