# functional equation $f(x^2)=xf(x)$

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.

• You should at least mention the domain and codomain of $f$, rather than let people guess... – WhatsUp Dec 18 '19 at 23:14
• @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. – Hans Dec 18 '19 at 23:19
• @heepo Please don't make unnecessary edits. – Trevor Gunn Dec 18 '19 at 23:27
• It's actually quite important. Think about the question over a finite field... the answer will be of very different nature. – WhatsUp Dec 18 '19 at 23:29
• @WhatsUp That continuity and differentiability are mentioned imply we are probably on a subset of $\mathbb R$ already. – 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

$$g(x+1)=g(x)+2^x$$

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

$$h(x+1)=h(x)$$

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:

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

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

$$f(x)=2^{\sin(2\pi\log_2(\log_2(x)))}x$$

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.

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

We have

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

or

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

now making $$z = \ln x$$ we have

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

and following

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

or

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

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

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