# Function $f$ with $f(x_1\cdot x_2)=f(x_1)+f(x_2)$ that is not $\log$?

Is the log-function the only function that enables the transformation of a product to a sum: $$f(x_1\cdot x_2)=f(x_1)+f(x_2)\,?$$

Yes, I can approximate the log function by a Taylor Series, but are there different functions that fulfill this property?
In order to extend the question, is there a (bijective) function (which is also defined for negative values) $$f:\mathbb{R}\setminus\{0\} \to \mathbb{R}$$ with this property? (This excludes $$\log(|x|)$$.)

Edit: I never realized how much we can do with prime numbers. @ECL is answering my original question, so I will accept it, but out of curiosity and maybe for the sake of completeness: Can we conclude that there aren't any other functions besides $$log$$ and $$exp$$ in $$\mathcal{C}^k(D) , D \subseteq \mathbb{R}^+$$ and $$k\geq0$$ with the property?

• @MartinHansen That's $f(x_1)f(x_2) = f(x_1+x_2)$. It is easy to show that this definition is satisfied by the inverse function of the one the question is asking about (even without knowing one is a log, the inverse an exponential). Just substitute $x_1 = f^{-1}(y_1), x_2 = f^{-1}(y_2)$ into the original. Jul 14, 2020 at 17:02
• @Deepak Thanks ! Jul 14, 2020 at 17:03
• @MaStbeil Might want to specify that both arguments are real and positive. $f:(\mathbb {R^+}, \mathbb{R^+} ) \to \mathbb {R}$. Jul 14, 2020 at 17:12
• In your extension, if you are allowing negative values, you should still exclude $0$. Jul 14, 2020 at 17:18
• To answer your Edit: if $f$ is continuous on $\mathbb{R}^+$ then $f=\alpha \log$ for some real $\alpha$. See the Edit to my answer for details...
– ECL
Jul 15, 2020 at 8:43

No you can't find a bijection on $$\mathbb{R}$$ with this property, the only function is $$f\equiv 0$$.

Indeed you have $$f(0) = f(0\times x) = f(0) + f(x)$$ so that $$f(x) = 0$$. This is true for all $$x\in\mathbb{R}$$.

If you exclude the $$\{0\}$$ again you cannot have a bijection since you have that $$f$$ must be a even function. Indeed $$f(x)=\frac{1}{2}f(x^2)=f(-x)$$.

However, forgetting about the bijection and looking just at $$\mathbb{R}^+$$ as a domain, it's true that $$\log$$ and its multiples are not the only possible functions. Indeed you may find everywhere discontinuous functions with the required property.

For instance you can build the function $$f$$ as follows. For each prime $$p$$ you fix $$f(p) = k_p$$, with $$k_p$$ any arbitrary number. Then for any $$q\in\mathbb{Q}$$ you have that $$f(p^q)=qf(p)=qk_p$$. The value of $$f$$ is fixed for all the $$x$$ which can be written as finite product of this terms, i.e. such that there exist a finite family of prime numbers $$\{p_i\}_{i=1\dots N}$$ and rational coefficients $$\{q_i\}_{i=1\dots N}$$ and $$x = \prod_{i=1}^N {p_i}^{q_i}\,.$$ You have indeed $$f(x) = \sum_{i=1}^N q_i k_{p_i}$$. You can show that each of these $$x$$ has a unique representation in this form, so that $$f$$ is well defined for them. For the $$x$$ which are not representable as such a finite product you define $$f(x) = 0$$. Then $$f$$ is well defined on $$\mathbb{R}^+$$ and your property holds.

All the continuous functions on $$\mathbb{R}^+$$ which respect the property are in the form $$\alpha \log$$ for some real $$\alpha$$. Actually, if $$f$$ is continuous at $$x=1$$ then it's in the form $$\alpha\log$$.
Indeed assume that f is continuous at $$1$$. First notice that $$f(1) = 0$$ since $$f(x) = f(1\times x) = f(1) + f(x)$$. Now you can prove that $$f$$ is continuous everywhere since $$\lim_{h\to 0} f(x+h) = \lim_{h\to 0}f(x(1+h/x)) = f(x)+\lim_{h\to 0}f(1+h/x) = f(x)+f(1) = f(x)\,.$$ Now you know that for any rational $$q$$, $$f(e^q) = qf(e)$$. By continuity you have that for any $$r\in\mathbb{R}$$ $$f(e^r) = r f(e)\,.$$ So $$f(x) = f(e^{\log x}) = f(e)\log x = \alpha \log x\,,$$ letting $$\alpha=f(e)$$.
For the sake of completeness, I don't think it is of much sense to speak of the property for a function defined on a generic $$D\subset \mathbb{R}^+$$. For instance for a open interval $$D=(10,11)$$, you have no $$x_1,x_2\in(10,11)$$ such that $$x_1\cdot x_2 \in (10,11)$$, which means that any function satisfies the property. However, every time that you have a neighbourhood of $$1$$ in $$D$$, and $$f$$ is continuous at $$1$$, then in each open connected component $$f$$ must be in the form $$x\mapsto k+\alpha \log x$$, with $$k$$ which can possibly vary in different connected components.
• It took me a while to understand that $f$ must be even, but that is in fact the answer I was looking for. Jul 14, 2020 at 19:14