$f(x^2) = xf(x)$ implies that $ f(x) = mx$?

Solution 1:

No, there are other such functions. For example, define $f\colon\mathbb{R}\to\mathbb{R}$ by $$ f(x) \;=\; \begin{cases} 2x & \text{if }x\text{ is algebraic,} \\[6pt] 3x & \text{if }x\text{ is transcendental.}\end{cases} $$ Both algebraic and transcendental numbers are closed under the operation of squaring, and therefore $f(x^2) = x\,f(x)$ for all $x$.

Solution 2:

Let $f(x)=x$ if $x$ is an algebraic number, and $f(x)=0$ if $x$ is transcendental.

Solution 3:

Of course not.

To find the general solution of $f(x^2)=xf(x)$ :

First, this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe2308.pdf

Let $x_1=\ln x$ , $f_1(x_1)=f(x)$ ,

Then $f_1(2x_1)=e^{x_1}f_1(x_1)$

Then, this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe2303.pdf

Let $x_2=\ln x_1$ , $f_2(x_2)=f_1(x_1)$ ,

Then $f_2(x_2+\ln2)=e^{e^{x_2}}f_2(x_2)$

$x_2\to x_2\ln2$ :

$f_2(x_2\ln2+\ln2)=e^{e^{x_2\ln2}}f_2(x_2\ln2)$

$f_2((x_2+1)\ln2)=e^{2^{x_2}}f_2(x_2\ln2)$

$f_2(x_2\ln2)=\Theta_1(x_2)\prod_{x_2}e^{2^{x_2}}$, where $\Theta_1(x_2)$ is an arbitrary periodic function with unit period

According to http://en.wikipedia.org/wiki/Indefinite_product#Rules ,

$f_2(x_2\ln2)=\Theta_1(x_2)e^{\sum_{x_2}2^{x_2}}$, where $\Theta_1(x_2)$ is an arbitrary periodic function with unit period

According to http://en.wikipedia.org/wiki/Indefinite_sum#Antidifferences_of_exponential_functions ,

$f_2(x_2\ln2)=\Theta_1(x_2)e^{2^{x_2}}$, where $\Theta_1(x_2)$ is an arbitrary periodic function with unit period

$f_2(x_2\ln2)=\Theta_1(x_2)e^{e^{x_2\ln2}}$, where $\Theta_1(x_2)$ is an arbitrary periodic function with unit period

$f_2(x_2)=\Theta(x_2)e^{e^{x_2}}$, where $\Theta(x_2)$ is an arbitrary periodic function with period $\ln2$

$f(x)=\Theta(\ln\ln x)x$, where $\Theta(x)$ is an arbitrary periodic function with period $\ln2$