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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
    $\begingroup$ $f(x) = |x|^a$ 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). $\endgroup$ – Srivatsan Nov 15 '11 at 22:10
  • $\begingroup$ 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. $\endgroup$ – bryn Nov 16 '11 at 7:44
  • $\begingroup$ There is a term for what you are asking about, 'totally multiplicative function': en.wikipedia.org/wiki/Completely_multiplicative_function $\endgroup$ – cobaltduck Nov 16 '11 at 16:37
  • 1
    $\begingroup$ @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". $\endgroup$ – Arturo Magidin Nov 16 '11 at 17:05

Since $f(x) = f(1)f(x)$ for all $x$, either $f(1)=1$, or else $f(x)=0$ for all $x$. Assume $f(1)=1$. If $f(a)=0$ for some $a\neq 0$, then $f(b) = f(a)f(b/a) = 0$ for all $b$, so we may assume $f(a)\neq 0$ for all $a\neq 0$.

Also, $f(1) = f(-1)^2$, so either $f(-1)=1$ or $f(-1)=-1$. If $a\gt 0$ then $f(a)=(f(\sqrt{a}))^2$, so $a\gt 0$ implies $f(a)\gt 0$. Thus, if $f(-1)=-1$ then $f$ is odd on the nonzero numbers; and if $f(-1)=1$ then $f$ is even.

As for $f(0)$, since $f(0)=f(0)^2$, either $f(0)=0$ or $f(0)=1$. If $f(0)=1$, then $1=f(0) = f(0a) = f(a)$, so $f(a)=1$ for all $a$.

So we have a couple of "degenerate" solutions: $f(a)=0$ for all $a$; $f(a)=1$ for all $a$; and $f(a)=1$ for all $a\neq 0$ and $f(0)=0$.

So now assume that $f(1)=1$, $f(0)=0$, $f(a)\gt 0$ for all $a\gt 0$.

Now consider $g(x) = \ln(f(e^x))$. Then $$g(x+y) = \ln(f(e^{x+y})) = \ln(f(e^xe^y)) = \ln (f(e^x)f(e^y)) = \ln(f(e^x))+\ln(f(e^y)) = g(x)+g(y),$$ so $g$ satisfies Cauchy's Functional Equation.

Both the "even" and the "odd" version of $f$ yield the same $g$, since $g$ only depends on the values of $f$ on the positive reals.

Conversely, given any $g\neq 0$ that satisfies Cauchy's Functional Equation, define $h$ on the positive reals by $h(x) = \exp(g(\ln(x))$. Then $h(xy) = h(x)h(y)$, and $h(x)$ is not the constant function $1$. Letting $h(0)=0$, you obtain two functions $f$ that satisfy your original equation: $f(x) = h(|x|)$, and $f(x) = \mathrm{sgn}(x)h(|x|)$.

Now, I cheated, because Cauchy's Functional Equation does not have an explicit complete list of solutions, but it's been well-studied, so you might as well look into that instead of trying to re-invent the wheel.


Here is one more: $$ f(x)= \begin{cases} 0&\mathrm{for}& x=0,\\ 1&\mathrm{for}& x\neq 0. \end{cases} $$


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