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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?

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

NOTE: There are discontinuous functions satisfying the given identity.

If $f$ is strictly monotonic (see comment by OP below) then $f(kx)<f(x)$ (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 $k=1$.

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  • $\begingroup$ You mean $g(x) = \frac {f(kx)}{f(x)}$, I presume? $\endgroup$ – k.stm Aug 7 at 8:10
  • $\begingroup$ Yes, surely. Thanks for pointing out. $\endgroup$ – Kabo Murphy Aug 7 at 8:11
  • $\begingroup$ Also, don’t you assume $c = 1$ to conclude $h(x + p) = h(x)$? $\endgroup$ – k.stm Aug 7 at 8:13
  • $\begingroup$ Since $f$ never vanishes we can get $c=1$ by putting $x=0$. $\endgroup$ – Kabo Murphy Aug 7 at 8:19
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    $\begingroup$ Perhaps you could ask a new question by changing the equation to $f(x)f(ky)=f(y)f(kx)$ for all $x$ and $y$. $\endgroup$ – Kabo Murphy Aug 7 at 23:38

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