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.

$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.
