Is there another differentiable monotone increasing (or decreasing) function $ f:\mathbb{R} \rightarrow \mathbb{R} $ with a property that $ f(xy) = f(x) + f(y) $, like the log-function has it?

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    $\begingroup$ Trivially $f(x) = 0$, $\forall x \in \mathbb{R}$ has this property. $\endgroup$ Aug 25, 2015 at 9:27
  • $\begingroup$ Well, sorry, I didn't mean some trivial functions. I'll edit my question. $\endgroup$
    – DDCh
    Aug 25, 2015 at 9:28
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    $\begingroup$ mathoverflow.net/questions/22706/… $\endgroup$ Aug 25, 2015 at 9:29
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    $\begingroup$ For positive $x$, $f(x) = c \log_e (x)$ is the general solution which includes the zero function and logarithms to other bases. $\endgroup$
    – Henry
    Aug 25, 2015 at 9:30
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    $\begingroup$ There is no such nontrivial function, including the $\log$. Such a function cannot be well defined at $0$, since $f(0\cdot x)=f(0) + f(x)$ for all $x$, meaning that $f=0$. $\endgroup$
    – 5xum
    Aug 25, 2015 at 9:38

1 Answer 1


Here is the standard derivation of the solution to the functional equation in question:

Suppose $f:(0,\infty)\to\mathbb R$ satisfies $f(xy)=f(x)+f(y)$ for all $x,y\in(0,\infty)$ and $f$ is monotone.

Now, define $g:\mathbb R\to\mathbb R$ and $g(x)=f(\exp(x))$. $g$ satisfies $g(x+y)=f(\exp(x+y))=f(\exp(x)\cdot\exp(y))=f(\exp(x))+f(\exp(y))=g(x)+g(y)$ and $g$ is monotone too.

Let $P(x,y)$ be the assertion $g(x+y)=g(x)+g(y)$. From this we can deduce: $$ P(x,0): g(x)=g(x)+g(0)\iff g(0)=0 \\ P(x,-x): g(0)=g(x)+g(-x)\iff g(-x)=-g(x) $$ Furthermore, we can prove inductively $g(x_1+x_2+…+x_n)=g(x_1)+g(x_2)+…+g(x_n)$ and therefore: $$ g(nx)=g(\underbrace{x+x+…+x}_{n})=\underbrace{g(x)+g(x)+…+g(x)}_{n}=n\cdot g(x) $$ and $$ g(x)=g\left(n\cdot\frac{x}{n}\right)=n\cdot g\left(\frac{x}{n}\right) \iff g\left(\frac{x}{n}\right)=\frac{g(x)}{n} $$ for $n\in\mathbb N$.

Consequently, we have $g\left(\frac{p}{q}\right)=p\cdot g\left(\frac{1}{q}\right)=\frac{p}{q}\cdot g(1)$ for $p,q\in\mathbb N$ and thus $g(r)=r\cdot g(1)$ for all $r\in\mathbb{Q_+}$. Together with $g(-x)=-g(x)$ we can deduce that $g(r)=r\cdot g(1)$ for all $r\in\mathbb{Q}$.

Now suppose $g$ is monotone increasing and there exists a $x\in\mathbb{R}$ for which $g(x)>x\cdot g(1)$. Because the rationals are dense, we can choose an $r\in\mathbb{Q}$ with $g(x)>g(r)=r\cdot g(1)>x\cdot g(1)$. This implies $x≥r$ and $r>x$, contradiction (note that $g(1)$ has to be positive).

The same idea works if $g(x)<x\cdot g(1)$ and also if $f$ is monotone decreasing.

So we conclude that $g(x)=c\cdot x$ for a fixed $c\in\mathbb R$ and for all $x\in\mathbb R$.

This implies $f(\exp(x))=c\cdot x \iff f(x)=c\cdot\ln(x)$. Here, we have either $c=0$ or $c=\log_a(e)$ for some $a\in\mathbb{R_{>0}}$ so either $f(x)=0$ or $f(x)=\log_a(x)$ for all $x\in(0,\infty)$.


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