3
$\begingroup$

Possible Duplicate:
If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t
Solution for exponential function's functional equation by using a definition of derivative

I can think of three functions that satisfy the condition $f(xy) = f(x)f(y)$ for all $x, y$, namely

  • $f(x) = x$
  • $f(x) = 0$
  • $f(x) = 1$

Are there more?

And is there a good way to prove that such a list is exhaustive (once expanded to include any other examples that I haven't thought of)?

  • 3
    $f(x) = |x|^a$ for any constant $a$ will work. I think these are the only *continuous* functions $f$ satisfying this condition (but there are many other "monstrous" solutions).2011-11-15
  • 0
    http://math.stackexchange.com/questions/43964/if-fxy-fxfy-then-show-that-fx-xt-for-some-t http://math.stackexchange.com/questions/64766/solution-for-exponential-functions-functional-equation-by-using-a-definition-of2011-11-15
  • 0
    Oh yes, I forgot all about the fact that $(xy)^n = x^n y^n$. So I guess the answer to the first question is "yes", but my second question remains.2011-11-16
  • 0
    There is a term for what you are asking about, 'totally multiplicative function': http://en.wikipedia.org/wiki/Completely_multiplicative_function2011-11-16
  • 1
    @Wade: That's not a good link. "Completely multiplicative function" is a term of art in number theory to refer to functions whose domain are the *positvie integers* and that are multiplicative. The very first sentence in the wikipedia page specifies that we are talking about functions "of positive integers".2011-11-16

2 Answers 2