Given a fixed non-zero constant $k\in\mathbb{R}$, find all functions $f:\mathbb{R}\to\mathbb{R}$ satisfying $$f(f(x))=kx\quad\text{and}\quad f\left(x^2\right)=xf(x).$$

If $f$ is continuous, then we can show that the solutions to the second equation are of the form $f(x)=mx$ (see here for example). With the first equation, this immediately implies that $f(x)=\sqrt{k}x$ or $f(x)=-\sqrt{k}x$. However, I am struggling to extend this to non-continuous $f$. Any help would be appreciated.

  • $\begingroup$ It is not likely true for non-continuous $x$. $\endgroup$ – Thomas Andrews Jan 24 '16 at 13:35
  • $\begingroup$ Hint: the function can jump between two values and still satisfy f(f(x))=kx for example. Note that this is just a hint, not a solution. You need to carefully choose which subsets have which value. $\endgroup$ – user2469 Jan 24 '16 at 15:20
  • $\begingroup$ @barrycarter do you have a complete solution? I'm struggling to use your idea $\endgroup$ – jlammy Jan 24 '16 at 21:16
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    $\begingroup$ More of a hint: f(x) = x*Sqrt[k] for x in set S, and f(x) = x*-Sqrt[k] for x outside of S. Just choose S carefully. $\endgroup$ – user2469 Jan 24 '16 at 22:04

Let's suppose for a non-zero constant $k$ and every real number $x$, we have: $$f\big(f(x)\big)=kx\tag0\label0$$ $$f\big(x^2\big)=xf(x)\tag1\label1$$ By \eqref{0} you can find out that $f$ is injective. Now, by \eqref{0} and \eqref{1} you have: $$f\left(f(x)^2\right)=f(x)f\big(f(x)\big)=kxf(x)=kf\left(x^2\right)=f\bigg(f\Big(f\left(x^2\right)\Big)\bigg)=f\left(kx^2\right)$$ $$\therefore\quad f(x)^2=kx^2\tag2\label2$$ Therefore by \eqref{2} we have $k=f(1)^2$ and hence $k$ is positive. Again by \eqref{2}, for every real number $x$ we get $f(x)=\pm\sqrt kx$.

Now let's define $K^\pm:=\big\{x\ne0\big|f(x)=\pm\sqrt kx\big\}$. So we get $\mathbb R=\{0\}\cup K^+\cup K^-$. By \eqref{1} we have $f(0)=0$. You can reformulate \eqref{0} and \eqref{1} this way: $$x\in K^\pm\quad\text{iff}\quad\sqrt kx\in K^\pm$$ $$x\in K^\pm\quad\text{iff}\quad x^2\in K^\pm$$ It's easy to see that every function satisfying $f(0)=0$ and $f(x)=\pm\sqrt kx$ for $x\in K^\pm$ is a solution, where $K^\pm$ satisfy the above conditions.

The trivial cases happen when one of $K^\pm$ is empty, and give us the linear solutions $f(x)=\sqrt kx$ and $f(x)=-\sqrt kx$. For a nontrivial case, you can take $K^+=\{\pm k^q\vert q\in\mathbb Q\}$ and $K^-=\{\pm k^q\vert q\in\mathbb R\backslash\mathbb Q\}$.

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  • $\begingroup$ As immediate conclusion from $(2)$ it seems we get $f(x)=\pm\sqrt k x$ only for $x\ge 0$. However, $f(-x)=-\frac1x\cdot(-x)f(-x)=-\frac1xf(x^2)=-\frac1x\cdot xf(x)=-f(x)$. $\endgroup$ – Hagen von Eitzen Jan 25 '16 at 21:26
  • $\begingroup$ What if k=1 and we want a discontinuous f ? $\endgroup$ – DanielWainfleet Jan 25 '16 at 23:11
  • $\begingroup$ For $ k=1$ we can let $K^+=\{\pm (2^{2^n} ) :n\in Z\}$. $\endgroup$ – DanielWainfleet Jan 25 '16 at 23:17
  • $\begingroup$ @HagenvonEitzen Why not concluding $f(x)=\pm\sqrt kx$ for $x<0$ immediately from $(2)$? I see no assumption forcing us to do it for $x\ge0$. $\endgroup$ – Mohsen Shahriari Jan 26 '16 at 2:45

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