4

Let $f:(0,\infty)\to \Bbb R$ be a continuous function and $\alpha,\beta>0$ be two fixed real numbers.

Under what conditions on $\alpha$ and $\beta$ can we deduce that $f$ is a constant function if we know $$ f(x)=f(\alpha x) = f(\beta x)? $$

I noticed that if $\alpha$ is an integer power of $\beta$ then the statement is not true. Indeed, for $\alpha=4,\beta=2$ the function $$ f(x) = \sin(2\pi \log_2(x)) $$ provides a counterexample.

I believe that if $\frac{\log\alpha}{\log\beta}\in \Bbb R\backslash\Bbb Q$ then the statement should hold. For example, I think it is likely that $f(x)=f(2x)=f(3x)$ implies $f$ is constant but I don't know how to prove it.

BigbearZzz
  • 15,530
  • I think it is easier if you put $x=e^y, \alpha=e^a, \beta=e^b$. The property turns into $g(y)=g(y+a)=g(y+b)$. And now this looks like a standard question about periodic functions; of course, I haven't said that the answer is easy, only that now it should be easier to look for it somewhere. – Giuseppe Negro Jun 10 '19 at 09:26
  • The correct condition is $\log\alpha / \log\beta \in\mathbb{R}\backslash\mathbb{Q}$ . – JWL Jun 10 '19 at 09:27
  • 1
    https://math.stackexchange.com/questions/775718/a-real-continuous-periodic-function-with-two-incommensurate-periods-is-constant?r=SearchResults – Martin R Jun 10 '19 at 09:29
  • @JWL That's actually what I meant to write, thanks! In my short note I tried some quick calculation on this I use variables that look similar i.e. $a,b$ and $\alpha,beta$ where $\alpha = \log a$ and such, hence the mistake. – BigbearZzz Jun 10 '19 at 09:35

1 Answers1

3

Writing $g(x)=f(e^{x})$ we can reduce the problem to the following: if $a$ and $b$ (non-zero) are periods of a continuous function $g$ under what conditions can we say $g$ is a constant. The answer is $a/b$ is irrational. Hence the answer to the present question is $f$ is a constant if $\frac {\log \, \alpha} {\log\, \beta}$ is irrational.

Kavi Rama Murthy
  • 359,332