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I'm sorry and a little ashamed to ask this very simple question. The problem is that I'm not very familiar with functional equations (I just know that they can be tricky). The question is: which are the possible solutions of the following functional equation? $$[f(x)]^{2}=f(2x)$$ I assume that the solutions (in the real field $f:\mathbb{R}\to\mathbb{R}$) are just the constant functions: $$f(x)=1, \quad f(x)=0$$ Are there other solutions? How can one prove that these are the only possible solutions?

Thanks.

Mohsen Shahriari
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Luca M
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    What about $e^x$? – Bulberage Mar 23 '14 at 20:48
  • Let $f(a)=b$. It is clear that $f(2^k a)$ is determined for positive integers $k$. Then quickly we can show that this is also true for negative integers $k$. However, that's all. So there is a tremendous amount of freedom even for the values of $f$ on the rationals. – André Nicolas Mar 23 '14 at 20:55
  • Certainly $f$ must be nonnegative for this to work, and $0$ and $1$ are the only constant function solutions. Now if $f$ is a solution, let

    $$g(x) = \frac{f(x)}{e^x}$$ Then we have

    $$g^2(x) = \frac{f^2(x)}{e^{2x}} = \frac{f(2x)}{e^{2x}} = g(2x)$$

    Hence $f$ is a solution if and only if $f(x) / e^x$ is also a solution. Likewise $f(x) / e^{2x}$ is a solution, and so on: So there are, in fact, infinitely many different solutions to the equation.

     –  Mar 23 '14 at 20:56
  • Sure! what a fool am I! $f(x)=a^{x}$ is another possible solution... Thanks! Any other? – Luca M Mar 23 '14 at 20:59
  • Thanks Bulberage and @T. Bongers. It seems that (considering T.Bongers's argument) all the solutions could be written as $f(x)=a^{kx}$ – Luca M Mar 23 '14 at 21:13
  • Must the solutions be continuous? Suppose $f$ and $g$ are solutions, what about the function that is equal to $f$ on the rational numbers and $g$ on the irrationals? As André says, it looks like there is a lot of freedom here. – Ben Millwood Mar 23 '14 at 21:23
  • @Ben Millwood: Yes I'm assuming the function is continuous... – Luca M Mar 23 '14 at 21:29

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Let $f(x)=a^{g(x)}$ , where $a\in\mathbb{R}^+$ and $a\neq1$ ,

Then $a^{2g(x)}=a^{g(2x)}$

$g(2x)=2g(x)$

Let $x=2^t$ ,

Then $g(2^{t+1})=2g(2^t)$

$g(2^t)=\Theta(t)2^t$ , where $\Theta(t)$ is an arbitrary periodic function with unit period

$g(x)=\Theta(\log_2x)x$ , where $\Theta(x)$ is an arbitrary periodic function with unit period

$f(x)=a^{\Theta(\log_2x)x}$ , where $\Theta(x)$ is an arbitrary periodic function with unit period, $a\in\mathbb{R}^+$ and $a\neq1$

doraemonpaul
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  • Thanks @doraemonpaul. I knew that functional equations were tricky... Can we say that yours is the most general solution? – Luca M Mar 25 '14 at 12:45