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Find all function $f$: $\Bbb R^+ \rightarrow \Bbb R^+$ such that $$f\left(\frac{f(x)}{y}\right) = yf(y)*f(f(x))$$ for all $x, y\in\Bbb R^+$

My attempt:-

Put $y = 1$,
$f(f(x)) = f(1)*f(f(x))$ $\implies f(1) = 1 \hspace{1cm} \{\because f(x) > 0\}$

Put $y = f(x)$,
$f(1) = f(x)*f(f(x))^2$
$\implies f(f(x)) = \cfrac{1}{\sqrt{f(x)}}$ $\implies f(y) = \cfrac{1}{\sqrt{y}}$ and finished.
But when I saw the solution it says "We cannot conclude from it". Is this because I assume $y = f(x)$ and I need to do more steps? Or something else?

EDIT:-
There are so many confusions in comments. This is the solution of the question. I didn't understand from last $3^{rd}$ line of LHS

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Edit again:- After so many confusion what I understood is, I proved $f(y) = \cfrac{1}{\sqrt{y}}$ for $y \in$ {all possible values of $f(x)$ for $x \in \Bbb R^+\}$ but I have to prove for $y \in \Bbb R^+$ ($x$ and $y \in \Bbb R^+$ is given but not $f(x) \in \Bbb R^+$). Since we don't know yet if $f(x) \in \Bbb R^+$, the way we prove $f(x) \in \Bbb R^+$ is using $f(\frac{u}{v})$ where $\frac{u}{v} = y\in \Bbb R^+$ as stated by Anne in the answer, Thanks Anne for helping me in understanding the solution.

user1150809
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  • your answer seems ok. But I think you also have to verify your solution by substituting your solution in the original equation and see if is satisfied for every positive x and y – NivMan Apr 10 '23 at 09:18
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    @NivMan No it is not ok, for the reason detected by the OP himself: not every $y$ is $f(x)$ for some $x.$ – Anne Bauval Apr 10 '23 at 09:25
  • @AnneBauval. The solution does seem to be right. In Samar's solution he assumed y=f(x). So the equation holds under this assumption. But it may hold also in the general case (for every positive y and x). To verify that the solution $f(y)=1/\sqrt{y}$ holds for arbitrary positive x,y we should substitute the solution $f(y)=1/\sqrt{y}$ in tne given equation. it gives $x^{1/4}y^{1/2}$ in both sides of the equation which proves that the solution is valid for every positive y,x – NivMan Apr 10 '23 at 11:05
  • @Gonçalo His proof of $f(1)=1$ is correct (in that part, his choice of $y=1$ does not assume that $y=f(x)$ for some $x.$ But the rest of his proof is not correct (see next comment). – Anne Bauval Apr 10 '23 at 11:35
  • @NivMan (See previous comment) But the rest of his proof is not correct because, as you said yourself: "the equation holds under this assumption", i.e. only on ${\rm im}(f)$ and not necessarily on the whole domain. The proof of the converse (i.e. what you detail in your comment) is a formality, that is not the point. – Anne Bauval Apr 10 '23 at 11:36
  • so to sum up, if there is a solution it must be $f(y)=1/\sqrt{y}$. so there is at most one solution. to find that there is at least one solution we should substitute $f(y)=1/\sqrt{y}$ in both sides of the given equation and see if the equation holds for all positive x,y. it gives $x^{1/4}y^{1/2}$ in both sides of the equation which proves that this solution is valid for every positive y,x. – NivMan Apr 10 '23 at 11:38
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    @NivMan Again: 1) no: "if there is a solution it must be $f(y)=1/\sqrt y$" is clear for $y\in{\rm im}(f)$ but still not proved for all $y\in\Bbb R^+$. This is in that sense that Samar's answer is not ok, as himself suspected. 2) The rest of your comment is "a formality" i.e. it is obvious, it is useless to repeat it, and is not the point. – Anne Bauval Apr 10 '23 at 11:59
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    I added the solution, please refer it and try to help me in understanding what it says. – user1150809 Apr 10 '23 at 14:32
  • The end of your last edit surprises me. We knew of course (given $f:\Bbb R^+\to\Bbb R^+$) that $f(x)\in\Bbb R^+.$ It is the converse that we didn't know, i.e. we did not know that every element $y\in\Bbb R^+$ is of the form $f(x)$ for some $x.$ And actually we never know it, untill we ultimately know exactly which function $f$ is. What we prove instead is the (weaker but sufficient) fact that every element $y\in\Bbb R^+$ is of the form $\frac{f(s)}{f(t)}.$ – Anne Bauval Apr 11 '23 at 09:36
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    I am leaving the question. May be I need more time and more questions of these types to really understand deeply. – user1150809 Apr 11 '23 at 10:29

1 Answers1

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Thank you for having added the solution. Let us read it together, comparing it with your attempt.

You proved that $f(1)=1$ (so do they), and (instantiating the hypothesis with $y=f(x)$) $$\forall y\in{\rm im}(f),\quad f(y)=\frac1{\sqrt y},$$ and you could not conclude that the same holds for every $y\in\Bbb R^+,$ because you did not know yet that $f$ is onto.

What they proved instead (instantiating with $y\in{\rm im}(f)$ but not related to $x$) is: $$\forall u,v\in{\rm im}(f),\quad f(u/v)=\frac1{\sqrt{u/v}}.$$ In order to conclude that $$\forall y\in\Bbb R^+,\quad f(y)=\frac1{\sqrt y},$$ they managed to prove that every $y\in\Bbb R^+$ can be written $\frac uv$ for some $u,v\in{\rm im}(f),$ by the following trick (which I simplify without losing its efficiency, instantiating the hypothesis with $x=1$ and using thrice the fact that $f(1)=1$): $$y=\frac{f(1/y)}{f(y)}.$$ This ends the proof (the converse fact that $y\mapsto\frac1{\sqrt y}$ is a solution goes without saying).

FShrike
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Anne Bauval
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  • Thanks for the explanation. I am still trying to understand it, idk why I am unable to process this even after so many comments and solution. – user1150809 Apr 11 '23 at 03:13
  • Do you need me to detail why $y=\frac{f(1/y)}{f(y)}$ is true? @SamarKamboj or why $\forall u,v\in{\rm im}(f),\quad f(u/v)=\frac1{\sqrt{u/v}}?$ – Anne Bauval Apr 11 '23 at 06:43
  • I understand why $y = \frac{f(1/y)}{f(y)}$ is true, just by putting $x = 1$ and $f(1) = 1$ as you mentioned but still how does it prove that it is surjective. I am unable to understand how to prove it is onto. – user1150809 Apr 11 '23 at 06:57
  • The surjectivity they are proving is not that of $f$ but that of $g:(s,t)\mapsto\frac{f(s)}{f(t)}$ (from $(\Bbb R^+)^2$ to $\Bbb R^+$). Thanks to this surjectivity, the formula $ f(y)=\frac1{\sqrt y},$ which was proved for all $y\in{\rm im}(g),$ is astually true for all $y\in\Bbb R^+.$ – Anne Bauval Apr 11 '23 at 07:10