This must be widely known, but it is not to me. What is the solution of the functional equation $f(x^2)=xf(x),\,\forall x\in\mathbb R$ or for $x$ in a finite field? What is it when $f$ is continuous or differentiable when $x\in\mathbb R$? Obviously $f(x)=ax$ for some constant $a$ is a solution. It is not the unique solution to the first question while I do not know how to prove the uniqueness for the second if it is uniqueness for the second.

  • $\begingroup$ You should at least mention the domain and codomain of $f$, rather than let people guess... $\endgroup$ – WhatsUp Dec 18 '19 at 23:14
  • $\begingroup$ @WhatsUp: I left it unspecified intentionally to get the widest possible solution. But I now added the condition. I may ask another question relaxing the restriction. $\endgroup$ – Hans Dec 18 '19 at 23:19
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    $\begingroup$ @heepo Please don't make unnecessary edits. $\endgroup$ – Trevor Gunn Dec 18 '19 at 23:27
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    $\begingroup$ It's actually quite important. Think about the question over a finite field... the answer will be of very different nature. $\endgroup$ – WhatsUp Dec 18 '19 at 23:29
  • $\begingroup$ @WhatsUp That continuity and differentiability are mentioned imply we are probably on a subset of $\mathbb R$ already. $\endgroup$ – Simply Beautiful Art Dec 18 '19 at 23:31

Suppose $f(x)\ge0$ for $x>1$. Let $2^{g(x)}=f(2^{2^x})$. Then we have


Now let $g(x)=h(x)+2^x$ to get


That is, take any 1-periodic function for $h$ and you will have a solution for $f$ when $x>1$. One can construct the function on the negatives using $f(-x)=-f(x^2)/x$, and another solution for $|x|<1$ in the same manner by considering $f(2^{-2^x})$. As the cases when $|x|=1$ do not depend on other values, they can also be defined on their own.

Supposing $f(x)<0$, we can use the same procedure but with $2^{g(x)}=-f(2^{2^x})$ or likewise as stated above.

An example solution, not of the provided form:

$$h(x)=\sin(2\pi x)$$

$$g(x)=\sin(2\pi x)+2^x$$


which is clearly not linear.

Note that the continuity requirement requires $\lim_{x\to-\infty}h(x)$ to exist, and hence it must be constant in such a case.

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  • $\begingroup$ Note that this example does not contradict Trevor Gunn's answer, since the domain of this particular $f(x)$ is $(1, \infty )$. $\endgroup$ – Crostul Dec 18 '19 at 23:38
  • $\begingroup$ Note the note at the end. $\endgroup$ – Simply Beautiful Art Dec 18 '19 at 23:40

When $f$ is continuous, we consider the sequence $x_n = x^{1/2^n}$, which tends to $1$ for any positive $x$. We can show that $f(x^{2^n}) = x^{2^n - 1} f(x)$. For instance, $f(x^4) = x^2 f(x^2) = x^3f(x)$. It follows that

$$f(x) = f(x_n^{2^n}) = x_n^{2^n - 1}f(x_n) \to xf(1).$$

We also have $f(0) = 0f(0) = 0$ and $$f(-x) = \frac{1}{-x}(-xf(-x)) = \frac{1}{-x} f(x^2) = \frac{1}{-x}(xf(x)) = -f(x).$$

Therefore, if $f$ is continuous, then $f(x) = xf(1)$ so $f$ is linear.

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We have

$$ f\left(e^{\ln x^2}\right)-xf\left(e^{\ln x}\right)=0 $$


$$ F(2\ln x)-x F(\ln x) = 0 $$

now making $z = \ln x$ we have

$$ F(2z) - e^z F(z)=0 $$

and following

$$ F\left(2^{\log_2 (2z)}\right)-e^z F\left(2^{\log_2 z}\right) = 0 $$


$$ \mathbb{F}(\log_2 z+1) - e^z \mathbb{F}(\log_2 z) = 0 $$

now making $u = \log_2 z$ we have

$$ \mathbb{F}(u+1) - e^{2^u} \mathbb{F}(u) = 0 $$

which solved furnishes

$$ \mathbb{F}(u) = \Phi(u)e^{2^u} $$

where $\Phi(u)$ is a generic periodic function with period $1$ as for example $\Phi(u) = \cos(2\pi u)$ .Now going back $\mathbb{F}\to F\to f$ we arrive at

$$ f(x) = \Phi(\log_2(\ln x)) x $$

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