The functional equation $f(xy)=f(x)f(y)$

Question

Let $f(x)$ be a function that satisfies this functional equation, $f(xy)=f(x)f(y)$.

With a little bit of intuition and luck one may come to a conclusion that these are perhaps the solutions of $f(x)$,

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

However, these solutions are family solutions of $f(x)=x^n$. What I meant by this is that, when $n=1$ you get the function $f(x)=x$. When $n=0$ you get $f(x)=1$ and when $x=0$ well you get $f(x)=0$. So, it seems $f(x)=x^n$ is the genuine solution to that functional equation and when you're taking different values for $x$ and $n$ you're getting bunch of other functions of the same family.

Getting excited by this I tried to take different values for $x$, for instance when $x=2$, $x^n$ becomes $2^n$. So, now I'm expecting the function $f(x)=2^n$ to satisfy this functional equation $f(xy)=f(x)f(y)$. However, it doesn't. I don't know why it's not satisfying. May I get your explanation?

In the expression $f(x)=x^n$ the argument of the function is $x$, not $n$. $n$ is fixed (for any specific function of the type you like). you can't get a new solution by picking a value for $x$. — lulu, Aug 04 '18 at 10:31
Also, there are other solutions to your functional equation, though not any other nice ones. See, for example, this question and its answers. — lulu, Aug 04 '18 at 10:31
It simply isn't true that every solution is of the form $x^n$. for some $n$. $0$ isn't of that form and, as I mentioned, there are discontinuous solutions as well. — lulu, Aug 04 '18 at 10:35
I don't think this is a duplicate of @greedoid's proposed original. The original functional equation is the same, but the answers there don't (for good reason) address the particular confusion the OP has here. — hmakholm left over Monica, Aug 04 '18 at 11:02
What is the domain of $x$? There are quite a few completely multiplicative functions out there ... — rtybase, Aug 04 '18 at 11:19

score 0 · Answer 1 · answered Aug 04 '18 at 11:15

If you take the equation $$ \tag{*} f(x) = x^n $$ and "set $x=0$", what you get is not $f(x)=0$, but $$ f(0) = 0^n $$ which does not define a function -- it only says what the function value must be at $0$ (and doesn't even do that until you decide what $n$ is).

As @lulu pointed out in comments, your assumption that every solution of the functional equation must have the form $\text{(*)}$ is simply not true. $f(x)=0$ does not have this form, and there is an infinity of "wild" discontinuous solutions too (at least if we assume the Axiom of Choice).

mengdie1982 · Answer 2 · 2018-08-04T12:06:52.330

-1

We may prove that the non-identical function $f(x)$ which is continuous over $(0,+\infty)$ and satisfies $$f(xy)=f(x)f(y)$$ is $$f(x)=x^\alpha$$ only.

edited Aug 04 '18 at 12:06

answered Aug 04 '18 at 12:01

mengdie1982

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The functional equation $f(xy)=f(x)f(y)$

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