Solutions of $f(x)\cdot f(y)=f(x\cdot y)$ 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.
 A: 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. 
A: 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
A: 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 $r>0$. 
A: $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.
A: Maybe you are interested in completely multiplicative arithmetic functions...
A: 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.
