Are logarithms the only continuous function on $(0, \infty)$ that has this property?

$$ f(xy) = f(x) + f(y) $$

If so, how would we show that? If not, what else would we need to show that a function $f$ that satisfies this property is some function $\log_a$ for some $a$?

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    $\begingroup$ Of course, logarithm functions $\log_a$ are not defined (and hence not continuous) on $\mathbb{R}$. $\endgroup$ – Travis Jan 4 '15 at 1:44
  • $\begingroup$ @Travis is right. They are defined on $(0, \infty)$. $\endgroup$ – Clarinetist Jan 4 '15 at 1:44
  • $\begingroup$ I am well aware of that, but I couldn't think of the right way to go about asking it. Maybe I should specify "strictly positive" reals. $\endgroup$ – Axoren Jan 4 '15 at 1:45

Other than $f(x)=0$ for all $x$, logarithms are the only continuous functions with that property. We show this in steps.

$f(1)=f(1\cdot 1)=f(1)+f(1)$, so $f(x^0)=f(1)=0=0\cdot f(x)$. Obviously $f(x^1)=1\cdot f(x)$.

$f(x^2)=f(x\cdot x)=f(x)+f(x)=2f(x)$, and by induction we can show that $f(x^n)=nf(x)$ for all natural numbers $n$ (including zero).

For rational number $y=\frac ab$,

$$f(x^y)=f(x^{a/b})=\frac 1b\cdot bf(x^{a/b})=\frac 1bf((x^{a/b})^b)=\frac 1bf(x^a)=\frac abf(x)=yf(x)$$

So $f(x^y)=yf(x)$ for all rational numbers $y$. By continuity we can extend this to all real values $y$.

If $f(x)$ is ever non-zero, we can find $b>0$ and $c\ne 0$ such that $f(b)=c$. Then

$$0\ne c=f(b)=f(e^{\ln b})=\ln b\cdot f(e)$$

so $f(e)\ne 0$. Let $a=e^{1/f(e)}$, so $f(e)=\frac 1{\ln a}$. Then for all $x>0$,

$$f(x)=f(e^{\ln x})=\ln x\cdot f(e)=\frac{\ln x}{\ln a}=\log_a x$$

Therefore, $f(x)$ truly is a logarithm function.

If we give up the continuity restriction, I believe that the axiom of choice will show other such functions are possible. I think this is done by setting a well-order on the real numbers and using transfinite induction. But the details are probably beyond me.

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    $\begingroup$ why does $f(x^y) = yf(x)$ imply a logarithm specifically? $\endgroup$ – GFauxPas Jan 4 '15 at 1:53
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    $\begingroup$ @GFauxPas Suppose $f(a) = 1$. Then for any $b\in (0,\infty)$, we can write $b = a^{\text{log}_a(b)}$. So $f(b) = f(a^{\text{log}_a(b)}) = \text{log}_a(b)f(a) = \text{log}_a(b)$. $\endgroup$ – Alex Kruckman Jan 4 '15 at 1:56
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    $\begingroup$ One specific non-continuous function would be $\mu(n)$ from analytic number theory. The class of such functions is called the set of "multiplicative functions", with a distinction made for those which are "completely multiplicative". $\endgroup$ – abiessu Jan 4 '15 at 1:58
  • $\begingroup$ @AlexKruckman I think the only thing missing from Rory's explanation was your comment. I think that's what he was alluding to in his 4th line, but it wasn't explicit enough. Thank you for that comment. $\endgroup$ – Axoren Jan 4 '15 at 2:02
  • $\begingroup$ @Abiessu That function is only defined on $\mathbb{Z}_+$, no? Or do I misinterpret which function you mean? $\endgroup$ – Travis Jan 4 '15 at 5:52

$f(x)=0$ satisfies this property.

  • $\begingroup$ That is indeed another function that satisfies that property. Is that the only other one? $\endgroup$ – Axoren Jan 4 '15 at 1:47
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    $\begingroup$ Actually, in fact. This would be the function $\log_\infty$. $\endgroup$ – Axoren Jan 4 '15 at 1:52
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    $\begingroup$ Actually,$ 0\log x$ $\endgroup$ – Cyclohexanol. Jan 4 '15 at 2:01
  • $\begingroup$ $\log_\infty(x) = \frac{\log(x)}{\log(\infty)} = 0 \log(x)$, so yeah. $\endgroup$ – Axoren Jan 4 '15 at 2:03

Well this is a partial answer: It is the only such function differentiable at $x=1$, since we have $\displaystyle \frac{f(x+h)-f(x)}{h}=\frac{1}{x}\frac{f(1+h/x)}{h/x}\to \frac{f'(1)}{x}$ as $h\to 0\,,$ and the fundamental theorem of calculus (combined with $f(xy)=f(x)+f(y))$ gives that $f(x)=f'(1)\ln{x}\,,$ so they are all of the form $c\log{x}\,.$ So now it would be enough to show that any such function is differentiable at $x=1\,.$

Edit: Actually one can derive $\displaystyle \frac{f(1+h)}{h}=y\frac{f(y+hy)-f(y)}{hy}\,,$ so it is enough to show that there exists one point at which $f$ is differentiable.


some hints for an approach to the other part of your question.

you can show $$f(1)=0$$ and $$f(x)=-f(\frac1{x})$$

also, for a set of distinct primes $p_k$, and integers $a_k$ $$ f(\prod_{k=1}^n p_k^{a_k})=\sum_{k=1}^n a_kf(p_k) $$ so that $f$ is defined on $\mathbb{Q}$ and can be extended to $\mathbb{R}$ by continuity.

for integers $a,b$ we have $$ f(n^{\frac{b}{a}})=\frac{b}{a}f(n) $$

using continuity and rational approximation $f(a^s)=sf(a)$ for $a$ and $s \in (0,\infty)$

if $f$ is not identically zero then there is a $b$ for which $f(b)=1$


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