Solutions of $f(x)\cdot f(y)=f(x\cdot y)$

Question

Can anyone give me a classification of the real functions of one variable such that $f(x)f(y)=f(xy)$? I have searched the web, but I haven't found any text that discusses my question. Answers and/or references to answers would be appreciated.

Answer 1

Let $g\colon \mathbb R\to\mathbb R$ be additive, i.e. $g(x+y)=g(x)+g(y)$ for all $x,y$. Then both \begin{align} f(x)=\begin{cases} \exp(g(\ln x)),&x>0,\\ 0,&x=0,\\ \exp(g(\ln (-x))),&x<0 \end{cases} \end{align} and \begin{align} f(x)=\begin{cases} \exp(g(\ln x)),&x>0,\\ 0,&x=0,\\ -\exp(g(\ln (-x))),&x<0 \end{cases} \end{align} are multiplicative. Except for the function which is $1$ everywhere, these should be all. I have not checked this carefully.

Now, how many additive functions are there? Well, these are exactly the $\mathbb Q$-linear functions. So choose a basis of the $\mathbb Q$-vector space $\mathbb R$ and define the images of the basis elements arbitrarily. In ZFC the existence of such a basis can be proved.

For more, follow the link given by alexjo.

The existence of additive functions which are not continuous has first been proved (using the well-ordering theorem) by Hamel. (The well-ordering theorem was new then, today this can be first semester stuff.) The article can be read (modulo possible language problems) here.

Answer 2

The functional equation $f(xy)=f(x)f(y)$ is the Power-law Cauchy equation with solution $$f(x) = x^{\gamma}$$ where $\gamma$ is an arbitrary constant. Furthermore, the function $f(x) ≡ 0$ is also a solution.

See also If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t

See:

Aczel, J. and Dhombres, J., Functional Equations in Several Variables, Cambridge Univ. Press, Cambridge, 1989

Answer 3

There is a classification of the functions $f:\mathbb R\to\mathbb R$ satisfying $$ f(x+y)=f(x)+f(y), \quad\text{for all $x,y\in\mathbb R$}. \qquad (\star) $$ These are the linear transformations of the linear space $\mathbb R$ over the field $\mathbb Q$ to itself. They are fully determined once known on a Hamel basis of this linear space (i.e., the linear space $\mathbb R$ over the field $\mathbb Q$).

This in turn provides a classification of all the functions $g:\mathbb R^+\to\mathbb R^+$ satisfying $$ g(xy)=g(x)g(y), \quad\text{for all $x,y\in\mathbb R^+$}, $$ as they have to be form $g(x)=\mathrm{e}^{f(\log x)}$, where $f$ satisfies $(\star)$. Note that $g(1)=1$, for all such $g$.

Next, we can achieve characterization of functions $g:\mathbb R\to\mathbb R^+$ satisfying $$ g(xy)=g(x)g(y), \quad\text{for all $x,y\in\mathbb R$}, $$ as $g(-x)=g(-1)g(x)$, which means that the values of $g$ at the negative numbers are determined once $g(-1)$ is known, and as $g(-1)g(-1)=g(1)=1$, it has to be $g(-1)=1$. Also, it is not hard to see that only acceptable value of $g(0)$ is $0$.

Finally, if we are looking for $g:\mathbb R\to\mathbb R$, we observe that, if $g\not\equiv 0$, and $x>0$, then $g(x)=g(\sqrt{x})g(\sqrt{x})>0$. Thus $g$ is fully determined once we specify whether $g(-1)$ is equal to $1$ or $-1$.

Note that if $g: \mathbb R\to\mathbb R$ is continuous, then either $g\equiv 0$ or $g(x)=|x|^r$ or $g(x)=|x|^r\mathrm{sgn}\, x $, for some $r>0$.

Answer 4

$f(x+h)-f(x) =f(x(1+h/x))-f(x) =f(x)f(1+h/x)-f(x) =f(x)(f(1+h/x)-1) $.

If there is an $x$ such that $f(x) \ne 0$, then $f(1) = 1$.

Therefore

$\begin{align} \frac{f(x+h)-f(x)}{h} &=\frac{f(x)(f(1+h/x)-1)}{h}\\ &=\frac{f(x)(f(1+h/x)-f(1))}{x(h/x)}\\ \end{align} $

If $f$ is differentiable at $1$, letting $h \to 0$, $f'(x) =\frac{f(x)}{x}f'(1) $ or $\frac{f'(x)}{f(x)} = \frac{f'(1)}{x} $.

Integrating, $\ln(f(x)) =f'(1)\ln(x)+c $ or $f(x) = e^c x^{f'(1)}$.

Letting $e^c=u$ and $f'(1) = v$, $f(x) = u x^v$. Putting this in the original equation, $f(xy) = u(xy)^v$ and $f(x)f(y) = ux^v uy^v =u^2(xy)^v$, so $u^2 = 1$. If we require $f(x) > 0$, $u = 1$, so the solution is $f(x) = x^v$.

This technique of converting a functional equation to a differential equation also works for $f(xy)=f(x)+f(y)$, the addition formula for $\tan^{-1}$, and others.

Of course differentiability is a stronger assumption than continuity, but non-differentiable solutions to functional equations are rarely needed except by those who need them.

Answer 5

Maybe you are interested in completely multiplicative arithmetic functions...

Answer 6

Without requiring continuity, there is no hope of a classification. If you do require continuity, take logs for simplicity, to get $l(x) + l(y) = l(x+y),$ note that $l(0) = 0,$ then if you let $l(1) = k,$ you will get $l(p/q) = k(p/q),$ then by continuity, $l(x) = k x.$ Now, exponentiate, and voila.