CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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Every continuous $f$ with $f (x y) = f (x) + f (y)$ is $f (x) = c \log x$ for a unique $c$ , including $c = 0$

Statement

If $f : (0, \infty) \to R$ is continuous and $f (x y) = f (x) + f (y)$ for all $x, y > 0$ , then there is a unique $c \in R$ such that

$f (x) = c \log x (x > 0) .$

Here $c = 0$ gives the zero function.

Facts & Assumptions

Given: A continuous $f$ satisfying the product-to-sum equation.

[F1]

$E = L^{- 1}$ (The integral exponential $E : R \to (0, \infty)$ as the inverse of $L$ ) and $E (a + b) = E (a) E (b)$ (The inverse $E$ satisfies $E (a + b) = E (a) E (b)$ ).

[L1]

$E$ is differentiable (The inverse $E$ is differentiable, $E^{'} = E$ , and $E (0) = 1$ ), so it is continuous (A function differentiable at $c$ is continuous at $c$ ).

[L2]

Composites of continuous functions are continuous (A composite of continuous functions is continuous, with no side hypothesis of the kind the composition of limits needs).

[L3]

Every continuous additive real function has the form $g (t) = c t$ (Six regularity conditions each force an additive $f : R \to R$ to be $x \mapsto f (1) x$ : continuity at a single point, monotonicity on a nondegenerate interval, boundedness above on one, boundedness below on one, constancy of sign on one, and a graph that is not dense in $R^{2}$ ).

[L4]

$E = \exp$ (The integral exponential $E$ is the published exponential function), $e = \exp (1)$ (The real exponential function and the number $e$ by a power series), and $L = \log$ (The integral logarithm $L$ is the published natural logarithm).

[F2]

For $b > 0$ , $b \neq 1$ , one defines $\log_{b} x = \log x / \log b$ (The logarithm to a positive base other than one).

[L5]

The natural logarithm is continuous and satisfies $\log (x y) = \log x + \log y$ (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

Proof

technique · direct

1.1

Put $g (t) = f (E (t))$ . By [F1] and the functional equation, $g$ is additive. By [L1], [L2], and continuity of $f$ , it is continuous.

F1L1L2given

1.2

Conversely, [L5] shows that each function $x \mapsto c \log x$ is continuous and satisfies the product-to-sum equation. For $c = 0$ this is the zero function.

L5algebra

1.3

If $c \neq 0$ , put $b = E (1 / c)$ . Then $b > 0$ and $\log b = 1 / c \neq 0 = \log 1$ , so $b \neq 1$ ; [F2] gives $\log_{b} x = c \log x$ . Thus the nonzero members are exactly the constant-multiple forms underlying logarithms to bases, while the zero member requires no division.

F1L4L5F2algebra

2.1

By [L3], $g (t) = c t$ for a unique scalar $c = g (1)$ .

step 1.1L3

3.1

For $x > 0$ , $x = E (L (x))$ , so $f (x) = g (L (x)) = c L (x) = c \log x$ . By [L4], $E (1) = e$ , hence $c = g (1) = f (e)$ , which also proves uniqueness.

F1step 2.1L4

4.1

Steps 3.1, 1.2, and 1.3 prove the classification and its endpoint case.

step 3.1step 1.2step 1.3∎

Depends on

Used by

Dropping f(e)=1 leaves the whole family c log, including logarithms to other bases and the zero function Example

Dependency tree · two levels

52 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

Henry Ricardo, The Equivalence of Definitions of the Natural Logarithm Function (standard reference, not scraped)

Every continuous ff with f(xy)=f(x)+f(y)f(xy)=f(x)+f(y) is f(x)=clog⁡xf(x)=clogx for a unique cc, including c=0c=0