# (Non-continuous) solutions to $f(f(x))=kx$ and $f(x^2)=xf(x)$

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.

• It is not likely true for non-continuous $x$. – Thomas Andrews Jan 24 '16 at 13:35
• 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. – user2469 Jan 24 '16 at 15:20
• @barrycarter do you have a complete solution? I'm struggling to use your idea – jlammy Jan 24 '16 at 21:16
• 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. – 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 $$K^-=\{\pm k^q\vert q\in\mathbb R\backslash\mathbb Q\}$$.
• As immediate conclusion from $(2)$ it seems we get $f(x)=\pm\sqrt k x$ only for $x\ge 0$. However, $f(-x)=-\frac1x\cdot(-x)f(-x)=-\frac1xf(x^2)=-\frac1x\cdot xf(x)=-f(x)$. – Hagen von Eitzen Jan 25 '16 at 21:26
• For $k=1$ we can let $K^+=\{\pm (2^{2^n} ) :n\in Z\}$. – DanielWainfleet Jan 25 '16 at 23:17
• @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 $x\ge0$. – Mohsen Shahriari Jan 26 '16 at 2:45