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Find all functions $f: \mathbb R_{>0} \rightarrow \mathbb R_{>0}$ such that $f(f(f(x)f(y)))=f(x)f(y^2)$ for all $x, y \in \mathbb R_{>0}$.

I made this problem myself.

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    $\begingroup$ f(x) = k gives an entire family of functions. $\endgroup$ – Chris Cudmore Jul 22 '16 at 20:02
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    $\begingroup$ @ChrisCudmore That doesn't work for every $k$. $\endgroup$ – wythagoras Jul 22 '16 at 20:03
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$f(x)f(1)=f(f(f(x)f(1)))=f(f(f(1)f(x)))=f(1)f(x^2)$. Therefore $f(x)=f(x^2)$ for all $x$.

Now $f(x)^2=f(x)f(x^2)=f(f(f(x)f(x)))=f(f(f(x)))$. So $f(f(x))=f(f(x)^2)=f(f(f(f(x))))=f(f(x))^2$. Therefore $f(f(x))=1$ for all $x$. But now $f(x)^2=f(f(f(x)))=1$, so $f(x)=1$.

The only solution is the constant function $f(x)=1$.

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    $\begingroup$ That's doing my brain in... how do you get $f(f(f(f(x)))) = f(f(x))^2$? I just about get every other part though - nice solution! $\endgroup$ – Shuri2060 Jul 22 '16 at 21:21
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    $\begingroup$ With $f(z)^2=f(f(f(z)))$, substitute $z=f(x)$. $\endgroup$ – f'' Jul 22 '16 at 21:29
  • $\begingroup$ Thanks. Out of interest, is the solution to $f(x) = f(x^2)$ a constant in general (out of context of the question)? $\endgroup$ – Shuri2060 Jul 22 '16 at 21:33
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    $\begingroup$ If $f$ must be continuous over $\mathbb{R}^+$, then yes, because $\lim_{x\rightarrow1}f(x)$ would not exist if it was not constant. $\endgroup$ – f'' Jul 22 '16 at 21:41
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    $\begingroup$ most appropriate username for this solution $\endgroup$ – elias Jul 22 '16 at 22:03
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Extremely-partial initial observations:

The LHS is symmetrical in x,y so the RHS is too so $f(x)f(y^2)=f(x^2)f(y)$ so $f(x^2)/f(x)$ is constant; say it equals $k$. Now our equation is $f(f(f(x)f(y)))=kf(x)f(y)$ or $f(f(z))=kz$ when $z=f(x)f(y)$.

Handwaving,

morally the first of those observations says f looks like a logarithm and the second says it looks like a scaling, and the two together suggest that actually it'll have to be constant (hence identically 1).

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Incomplete Answer:

$f$ cannot be a polynomial with degree larger than 0.

Proof:

Let f(x) have degree $n$.

Let us consider the degrees of the $x$ and $y$ terms separately in each side of the equation.

LHS:

The largest $x$ term has degree $n^3$.

The largest $y$ term has degree $n^3$.

RHS:

The largest $x$ term has degree $n$.

The largest $y$ term has degree $2n$.

Hence it is impossible for $f$ to be a polynomial with a degree larger than 0, as we should be able to equate these degrees for both sides of the equation.

The only polynomial with degree 0 (of the form $f(x)=k$) which works is $f(x) = 1$ as it is the only solution to $f(x) = (f(x))^2$.$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ ($f(x) = 0$ doesn't work as $f(x) \not\in \mathbb{R}^+$)

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  • $\begingroup$ Well, technically f(x) = 1 is a polynomial and it satisfies the equation given by wythagoras. $\endgroup$ – C. Woods Jul 22 '16 at 20:49
  • $\begingroup$ Yes, my comment was before you edited your answer. $\endgroup$ – C. Woods Jul 22 '16 at 20:50
  • $\begingroup$ Well maybe we commented/edited at the same time. I'll also point out that f(x) = 0 is not a solution because the question specifies that the function must map into the positive reals. $\endgroup$ – C. Woods Jul 22 '16 at 20:54

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