Are logarithms the only continuous function on $(0, \infty)$ such that $f(xy) = f(x) + f(y)$?

Question

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 $a$?

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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 $f(b)=1$