# Are there any real-valued functions, besides logarithms, for which $f(x*y) = f(x) + f(y)$?

Is there any real-valued function, $$f$$, which is not a logarithm, such that $$∀ x,y$$ in $$ℝ$$ , $$f(x*y) = f(x) + f(y)$$?

So far, all I can think of is $$z$$ where $$z(x) = 0$$ $$∀ x$$ in $$ℝ$$

EDIT:

Functions having a domain of $$ℝ^+$$ or a domain of $$ℝ$$/{0} are acceptable as well.

What are examples of functions, $$f$$, from $$ℝ$$/{0} to $$ℝ$$ which are not logarithms, such that
$$∀ x,y$$ in $$ℝ$$, $$f(x*y) = f(x) + f(y)$$?

• If you take a closer look, the constant zero function is also the only function $f:\Bbb R\to\Bbb R$ such that $f(xy)=f(x)+f(y)$, period. Logarithms do not satisfy that property. – Saucy O'Path Sep 22 '18 at 11:41
• Probably you want functions $\mathbb R_{>0} \rightarrow \mathbb R$. – lisyarus Sep 22 '18 at 11:43
• See Overview of basic facts about Cauchy functional equation under the related equations section. – dxiv Sep 23 '18 at 19:46

$$f(0\times 0)=f(0)+f(0)$$ so $$f(0)=0$$. Put $$x=0$$ to get $$f(0)=f(0)+f(y)$$. Hence $$f \equiv 0$$. Note that this is the answer when the domain is the whole real line.

• I already gave that example in the original posting. I wrote, "So far, all I can think of is $z$ where$z(x)=0$ $∀x$ in $R$" – IdleCustard Sep 23 '18 at 19:31
• I am not giving an example. I am proving that there is no other solution if the equation is to hold for all $x,y \in \mathbb R$. – Kavi Rama Murthy Sep 23 '18 at 23:26

Yes, there are, at least if you assume the axiom of choice. Then there are functions $$g\colon\mathbb{R}\longrightarrow\mathbb{R}$$ which are not linear but which satisfy Cauchy's functional equation:

$$g(x+y)=g(x)+g(y)$$.

Now, define

$$f\colon(0,+\infty)\longrightarrow\mathbb R$$
by
$$f(x)=g\bigl(\log(x)\bigr)$$.

• OP clearly says 'for all $x,y \in \mathbb R$', so the domain is not $(0,\infty)$. – Kavi Rama Murthy Sep 22 '18 at 11:47
• @KaviRamaMurthy But the OP also clearly mentions logarithms, so he probably may have mistaken saying "for all $\in \mathbb R$". – lisyarus Sep 22 '18 at 12:01
• Quite possible. Let us see how he reacts. – Kavi Rama Murthy Sep 22 '18 at 12:03