Find all multiplicative continuous functions on $(0,\infty)$

Question

If $x>0,y>0$ and if $f(xy)=f(x)f(y)$, then $f=\, ?$

I tried the problem. And got it as $f(x)^n=f(x^n)$. But answer is $f(x)=x^n$. How?
$f$ is a continuous function.

Answer 1

Trusting that you demand continuity, so that we have $f(x^n)=f(x)^n\;\forall n\in \mathbb R$ (and not just $\mathbb Q$) we can complete the argument as follows:

Suppose that $f(e)=\lambda$. We note that $\lambda ≥0$, since $f(\sqrt e)^2=\lambda$. If $\lambda =0$ then $f(x)$ is identically $0$, which is a valid solution. Otherwise, suppose that $\lambda > 0$. Then for any $x\in \mathbb R$ we write $x=e^{\ln x}$. Then, by what you have already proven, $f(x)=f(e^{\ln x})=f(e)^{\ln x}=\lambda^{\ln x}$. We conclude by writing $$f(x)=\lambda^{\ln x}=e^{(\ln \lambda)( \ln x)}=x^{\ln \lambda}$$

Note: absent continuity, the claim is not generally true. To construct a counterexample, start with a discontinuous function $\Psi(x)$ satisfying $\Psi(x+y)=\Psi(x)+\Psi(y)$. (The usual examples start by thinking of $\mathbb R$ as a vector space over $\mathbb Q$ and letting $\Psi$ denote projection onto one basis line.). Then, to get a discontinuous example of your type, define $f(x)=e^{\Psi(\ln x)}$.

Answer 2

4 obvious solutions spring to mind.

$f(x) = 0$ then $f(xy) = 0 = 0*0 =f(x)f(y) $.

$f(x) = 1$ then $f(xy) = 1 = 1*1 = f(x)f(y) $.

$f(x) = x$ then $f(xy) = xy = f(x)f(y)$.

And $f(x) = x^n$ for some $n \in \mathbb R$ then $f(xy) = x^ny^n = f(x)f(y)$.

These are all really the same class of answers. If $n = 1; f(x) = x = x^1$. If $n = 0; f(x) =1 = x^0$. $f(x) = 0$ can be considered an exception. However if we were to allowed $n$ in the extended reals. $n = -\infty; f(x) = 0 = x^{-\infty}$.

But are is this the only class of solutions?

let $b \ne 1; f(b) = c$. Then $c = b^{\log_b c}$. Let $n = \log_b c$. So $f(b) = b^n$.

(I am assuming $c > 0$.[footnote])

Let $x > 0$. Let $k = \log_b x$. So $x = b^k$.

$f(x) = f(b^k) = f(b)^k = b^{nk} = x^n$. And we are done

(I am assuming $f(x^k) = f(x)^k$ for all $k \in \mathbb R$. [footnote 2])

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[footnote]

If $f(b) = 0$ then $f(1) = f(b)f(1/b) = 0$. S0 $f(x) = f(x)f(1) = 0$ for all $x$ and $f(x) = 0$ is one of the cases we discussed and we can assume we are trying for other cases. So we may assume $f(b) \ne 0$. Furthermore no $x > 0$ is such $f(x) = 0$.

$f(x) = f(x)f(1)$ so $f(1) = 1 > 0$. If $f(b) < 0$ then by intermediate value theorem there exist a value $c$ between $b$ and $x$ where $f(c) = 0$ but we've shown that is not the case. So $f(x) >0$ for all $x$.

[footnote 2]

$f(x^n) = f(x*x*...) = f(x)^n$ and $f({x}^{1/n})^n = f(x)$ so $f({x}^{1/n}) = f(x)^{1/n}$. So $f(x^r)= f(x)^r; \forall r \in \mathbb Q$. As $f$ is continuous, $f(x^y) = f(x)^y$ for all $y \in \mathbb R$.