What are all the functions that satisfy $f(x)/f(y) = f(kx)/f(ky)$?

Find all continuous functions defined over real numbers that satisfy

$$\frac{f(x)}{f(y)} = \frac{f(kx)}{f(ky)}$$,

for any $$x$$ and $$y$$. It is possible to show that the above condition holds for $$f(x) = ax^b$$ since

$$\frac{ax^b}{ay^b} = \frac{ak^bx^b}{ak^by^b}$$.

Do functions that satisfy this property have a specific name?

• Called power functions (up to scaling anyway). – runway44 Aug 7 at 6:58
• Does this have to be true for any pair of real numbers $x, y$? Is the function defined on specific intervals? – Niki Di Giano Aug 7 at 7:13
• @NikiDiGiano Yes, for any pair of real numbers $x$ and $y$. The function is defined over $\mathbb{R}$ – KRL Aug 7 at 7:51

I will assume that $$f$$ never vanishes and $$k>0$$. Let $$g(x)=\frac {f(kx)} {f(x)}$$. The given equation becomes $$g(x)=g(y)$$ for all $$x,y$$ so $$g$$ is a constant $$c$$. Thus $$f(kx)=cf(x)$$ for all $$x$$. Put $$x=0$$ to see that $$c=1$$. Let $$h(x)=f(e^{x})$$. Then $$h(x+p)=h(x)$$ for all $$x$$ where $$p=\log \, k$$. You can retrace these steps and show that any function $$h$$ with $$h(x+p)=h(x)$$ (i.e. any periodic function $$h$$ with period $$p$$ which never vanishes) gives a solution to the given problem.
If $$f$$ is strictly monotonic (see comment by OP below) then $$f(kx) (for $$x >0 0$$) if $$k<1$$ and $$f(kx)>f(x)$$ (for $$x > 0$$) if $$k>1$$. Hence there is no solution unless $$k=1$$. Of course, the question is trivial when $$k=1$$.
• You mean $g(x) = \frac {f(kx)}{f(x)}$, I presume? – k.stm Aug 7 at 8:10
• Also, don’t you assume $c = 1$ to conclude $h(x + p) = h(x)$? – k.stm Aug 7 at 8:13
• Since $f$ never vanishes we can get $c=1$ by putting $x=0$. – Kabo Murphy Aug 7 at 8:19
• Perhaps you could ask a new question by changing the equation to $f(x)f(ky)=f(y)f(kx)$ for all $x$ and $y$. – Kabo Murphy Aug 7 at 23:38