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Suppose a function $f : \mathbb{R} \to \mathbb{R} $ satisfies the relation $$f(x^2) = xf(x) \ \ \forall x$$ Does this imply $f$ must be a straight line, $f(x) = mx$? If so, why? If not, are there other such functions?

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    $\begingroup$ Solutions reduce to linear form if continuity (at $x=1$) is assumed, but not otherwise. For example, let $f(1)=1$, $f(-1)=-1$ and $f(x)=0$ otherwise. $\endgroup$ – Sangchul Lee Feb 15 '12 at 1:18
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    $\begingroup$ If you add continuity to the mix, then you ought be able to prove it. Let $g(x)=\frac{f(x)}{x}$ for $x\neq 0$. Then $g(x)=g(x^2)$. For $z>0$, let $z_0=z$ and let $z_{n+1}=\sqrt{z_n}$. Then $g(z_0)=g(z_1)=g(z_2)...$ and $\lim z_n = 1$, so $g(z)=g(1)$. So $f(z)=g(1)z$. For $z<0$, $g(z)=g(z^2)=g(-z)=g(1)$. Indeed, you only need continuity at $1$ and $0$. $\endgroup$ – Thomas Andrews Feb 15 '12 at 1:24
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    $\begingroup$ Related: math.stackexchange.com/questions/19257/… $\endgroup$ – Aryabhata Feb 15 '12 at 1:32
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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 $x$.

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  • $\begingroup$ Of course. Thank you. $\endgroup$ – Simon S Feb 15 '12 at 1:20
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Let $f(x)=x$ if $x$ is an algebraic number, and $f(x)=0$ if $x$ is transcendental.

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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 $\ln2$

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  • $\begingroup$ Nice answer +1 ! $\endgroup$ – mick Jan 14 '15 at 21:27

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