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As the title, my question is: Are there any analytic functions satisfying the relation $f(x)g(y)=h(xy)$ besides $f(x)=g(x)=h(x)=x^t$ with t is a real number?

I know that when the relation is $f(x)f(y)=f(xy)$, $x^t$ is the only analytic solution for this relation. But I'm wondering if we relax the relation to $f(x)g(y)=h(xy)$ (or even the weaker version $f(x)f(y)=h(xy)$) is $x^t$ still the only possible analytical function to satisfy this relation (or the weaker version)? (Based on the comment, I should have mentioned that $f(x)$, $g(x)$ and $h(x)$ are equal to $x^t$ with a real $t$ up to a multiplicative constant)

I also wondering if this question is discussed in any of the references. Thanks!

daw
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wei chen Lin
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  • $x^{t}$ is not even uniquely defined. How can it be considered analytic? – geetha290krm May 25 '23 at 08:50
  • Sort of trivially you can take $f(x)=ax^t$, $g(x)=bx^t$ and $h(x)=abx^t$. Otherwise a quick back-of-the-envelope calculation suggests that other than the constants you can reduce it to the original problem so there should be no more solutions, but I might be wrong (first show all three functions are equal up to a multiplicative constant) – Michal Adamaszek May 25 '23 at 08:53
  • You have used Compex Analysis tag. Are $x$ and $y$ complex numbers? BTW the question has nothing to do with Functional Analysis. – geetha290krm May 25 '23 at 08:57
  • Strictly speaking zero is also the solution, but that's a tiny detail. – Michal Adamaszek May 25 '23 at 09:04
  • Sorry about the incorrect tags. I saw some similar questions, $()()=() $ using the functional analysis tag and another similar question using the complex analysis tag: https://math.stackexchange.com/questions/52255/multiplicative-analytic-functions – wei chen Lin May 25 '23 at 09:04
  • Analytic on what domain? If $t$ is not a whole number, $x^t$ is not analytic at $x=0$. – coiso May 25 '23 at 09:13
  • I think the only solutions are constant multiples of $x^{t}$ where $t$ is a non-negtiave integer. – geetha290krm May 25 '23 at 09:19

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$$f(x)g(y) = h(xy)$$. $$f'(x)g(y) = y h'(xy)$$. For a fixed $x_0$, assuming $f(x_0) \neq 0$ and $f'(x_0)$ exists, $$\frac{f'(x)}{f(x)} = \frac{yh'(xy)}{h(xy)} \implies h'(x_0y) = C\frac{h(x_0y)}{y}$$

$$\frac{h'(x_0y)}{h(x_0y)} = \frac{C}{y}$$ $$\log(h(x_0y)) = C_1\log(y) + C_2$$

$$h(x_0y) = C_2' y^{C_1} $$ Similarly, $$h(xy_0) = B_2' x^{B_1} $$

So $$f(x_0) g(y) = h(x_0y) = C_2' y^{C_1} \implies g(y) = C_2'' y^{C_1}$$ Similarly, $$f(x) = B_2'' x^{B_1}$$

So if $f(x) \neq 0$ and $f(x)$ is differentiable and similarly $g(y)$, we have only polynomial solutions.

Balaji sb
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