Let ##\forall x,y\in\mathbb{Q}^+## and ##f(xf(y))=\dfrac{f(x)}y##

[littlemathquark]

Homework Statement

Let $\forall x,y\in\mathbb{Q}^+$ and $f(xf(y))=\dfrac{f(x)}y$ so find $f:\mathbb{Q}^+\to\mathbb{Q}^+$

Relevant Equations

Let $\forall x,y\in\mathbb{Q}^+$ and $f(xf(y))=\dfrac{f(x)}y$ so find $f:\mathbb{Q}^+\to\mathbb{Q}^+$

I have no idea for this question. Can you give me some clue, please?

 

[pasmith]

Have you tried substituting any special values for x or y?

Have you considered a power law relation f(x) = Ax^b?

 

[littlemathquark]

Yes,
İf $y=1$ then $f(xf(1))=f(x)$ and $f(1)=1$

for $x=1$ then $f(f(y))=\dfrac{f(1)}{y}=\dfrac{1}{y}$

Sİnce $f(xf(y))=\dfrac 1y\cdot f(x)=f(f(y))\cdot f(x)$ so $f$ must be multiplicative function so $f(xy)=f(x)f(y)$ and also I found $f(x/y)=f(x)/f(y)$

 

[pasmith]

Yes,
İf $y=1$ then $f(xf(1))=f(x)$ and $f(1)=1$

This doesn't follow. Have you shown that f is injective, so as to justify the assertion that f(xf(1)) = f(x) \Rightarrow xf(1) = x?

for $x=1$ then $f(f(y))=\dfrac{f(1)}{y}=\dfrac{1}{y}$

Sİnce $f(xf(y))=\dfrac 1y\cdot f(x)=f(f(y))\cdot f(x)$ so $f$ must be multiplicative function so $f(xy)=f(x)f(y)$ and also I found $f(x/y)=f(x)/f(y)$

So f(x) = x should work. Does it?

 

[littlemathquark]

$f(y_1)=f(y_2)\implies f(f(y_1))=f(f(y_2))\implies \frac{f(1)}{y_1}=\frac{f(1)}{y_2}\implies y_1=y_2$ so f injective.

 

[fresh_42]

Yes,
İf $y=1$ then $f(xf(1))=f(x)$ and $f(1)=1$

for $x=1$ then $f(f(y))=\dfrac{f(1)}{y}=\dfrac{1}{y}$

Sİnce $f(xf(y))=\dfrac 1y\cdot f(x)=f(f(y))\cdot f(x)$ so $f$ must be multiplicative function so $f(xy)=f(x)f(y)$ and also I found $f(x/y)=f(x)/f(y)$

We only have $f(xy)=f(x)f(y)$ for $x>0$ and $y\in f\left(\mathbb{Q}^+\right).$ Is $f$ surjective?

 

[littlemathquark]

Yes, f is surjective because $f(f(y)) =1/y$ and take $a/b\in Q^+$
$f(f(b/a)) =a/b$

 

[littlemathquark]

This doesn't follow. Have you shown that f is injective, so as to justify the assertion that f(xf(1)) = f(x) \Rightarrow xf(1) = x?



So f(x) = x should work. Does it?

f(x) =x doesn't work.

 

[fresh_42]

Yes, f is surjective because $f(f(y)) =1/y$ and take $a/b\in Q^+$
$f(f(b/a)) =a/b$

Hence, we are left with the question to determine all automorphisms of $\left(\mathbb{Q}^+,\cdot\right)$ that also fulfill the functional equation. $f=\operatorname{id}$ does not, so we are looking for a proper subset of automorphisms here, and there are no inner automorphisms.

We can also rule out order-preserving automorphisms:
https://www.ams.org/journals/proc/1...-1994-1195720-5/S0002-9939-1994-1195720-5.pdf

Btw., just a question: If you already proved that $f$ is bijective and multiplicative, why don't you tell us before anybody does, or ask about it like @pasmith and me, what you already did? I'm asking this before I test $f$ on its order-preserving property, in case you already did it.

 

[littlemathquark]

Hence, we are left with the question to determine all automorphisms of $\left(\mathbb{Q}^+,\cdot\right)$ that also fulfill the functional equation. $f=\operatorname{id}$ does not, so we are looking for a proper subset of automorphisms here, and there are no inner automorphisms.

We can also rule out order-preserving automorphisms:
https://www.ams.org/journals/proc/1...-1994-1195720-5/S0002-9939-1994-1195720-5.pdf

Btw., just a question: If you already proved that $f$ is bijective and multiplicative, why don't you tell us before anybody does, or ask about it like @pasmith and me, what you already did? I'm asking this before I test $f$ on its order-preserving property, in case you already did it.

When I asked question I have no idea about it, really. But I have tried to solve it all day and I found what I write to you. So I'm looking for multiplicative and 1-1 and surjective function but I couldn't construct it for now.

 

[fresh_42]

When I asked question I have no idea about it, really. But I have tried to solve it all day and I found what I write to you. So I'm looking for multiplicative and 1-1 and surjective function but I couldn't construct it for now.

The paper says it cannot be order-preserving, and $q\cdot f^2(q) =1$ for all $q\in \mathbb{Q}^+$ suggests it has to be order-reversing since this condition doesn't look as if it can be a wild permutation. Are you sure such a function exists? Maybe we need two different functions depending on whether $q<1$ or $q>1.$

 

[fresh_42]

Here is my solution:
$$
f(x\cdot f(y))=\dfrac{f(x)}{y}
$$
implies that $f^2(x)=\dfrac{1}{x}$ and $f^4(x)=x$ or $(f^4-\operatorname{id})(x)=0.$ Thus
\begin{align*}
(f-\operatorname{id})\circ (f+\operatorname{id})\circ (f^2+\operatorname{id}) =(f^2-\operatorname{id})\circ (f^2+\operatorname{id})=f^4-\operatorname{id}=0
\end{align*}
Since all functions commute with each other, we do not have to care about the order. Now you can solve
$$
(f^2-\operatorname{id})\circ (f^2+\operatorname{id})(x)=\left(\left(\dfrac{1}{\operatorname{id}}-\operatorname{id}\right)\circ \left(\dfrac{1}{\operatorname{id}}+\operatorname{id}\right)\right)(x)=0
$$
which hasn't any solution in $\mathbb{Q}^+$ let alone holds for all positive rationals.