Functional equations

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hanzo.ei

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hanzo.ei

#1 h Mar 29, 2025, 4:33 PM

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Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$

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GreekIdiot

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GreekIdiot

#2 h Mar 29, 2025, 5:16 PM

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Not my problem

Havent made such a beautiful FE myself yet.

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Mathzeus1024

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Mathzeus1024

#3 h Mar 30, 2025, 10:12 AM

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It works for $\textcolor{red}{f(x)=\frac{1}{x}}$ .

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GreekIdiot

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GreekIdiot

#4 h Mar 30, 2025, 10:23 AM

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Yeah we established that in another post

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hanzo.ei

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hanzo.ei

#5 h Mar 30, 2025, 10:42 AM

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pco, can you solve it :omighty:

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GreekIdiot

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GreekIdiot

#6 h Mar 30, 2025, 11:27 AM

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$f$ seems to be an involution. I wonder if we are able to prove that. Then we can eliminate other solutions to the assertion very easily.

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hanzo.ei

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hanzo.ei

#7 h Apr 2, 2025, 3:28 PM

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bump!!!!

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MR.1

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MR.1

#8 h Apr 3, 2025, 4:08 PM • 3 Y

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solved with GioOrnikapa if you guys want solution please give me $10$ likes :-D

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GreekIdiot

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GreekIdiot

#9 h Apr 3, 2025, 4:17 PM

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lol aint no way

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GioOrnikapa

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GioOrnikapa

#10 h Apr 3, 2025, 4:33 PM • 1 Y

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A lot of liars nowadays smh

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truongphatt2668

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truongphatt2668

#11 h Apr 3, 2025, 4:34 PM

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hanzo.ei wrote:

Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$

I can prove $f(x)$ is injective and $f(1)=1$ anyone continue please?

This post has been edited 3 times. Last edited by truongphatt2668, Apr 3, 2025, 4:52 PM

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GreekIdiot

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GreekIdiot

#12 h Apr 3, 2025, 5:55 PM

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truongphatt2668 wrote:

hanzo.ei wrote:

Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$

I can prove $f(x)$ is injective and $f(1)=1$ anyone continue please?

I noticed that there exists some homogenous-like function by isolating $y$ on the $RHS$ . Can you post the claims you made with proof so that we can create a complete solution?

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truongphatt2668

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truongphatt2668

#13 h Apr 4, 2025, 1:42 AM

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GreekIdiot wrote:

truongphatt2668 wrote:

hanzo.ei wrote:

Find all $f: \mathbb R_+ \rightarrow \mathbb R_+$ such that $f(xf(y)+f(x))=yf(x+yf(x)) \: \forall \: x,y \in \mathbb R_+$

I can prove $f(x)$ is injective and $f(1)=1$ anyone continue please?

I noticed that there exists some homogenous-like function by isolating $y$ on the $RHS$ . Can you post the claims you made with proof so that we can create a complete solution?

Just do it, and I will give a complete solution

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GreekIdiot

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GreekIdiot

#14 h Apr 4, 2025, 3:29 PM

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$f(x)>0$ for all $x \in \mathbb R_+$ so all $y \in \mathbb R_+$ can be written as $\dfrac {f(m)}{f(n)}$ for some $m,n \in \mathbb R_+$
Then there exists some homogenous-kinda function (lets call it $g$ ) such that $f(xf(y)+f(x))=y^{\ell +1} \cdot g(x)$ and also $f(x+yf(x))=y^{\ell} \cdot g(x)$ thats what I meant to say. Correct me if wrong lol. :oops:

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jasperE3

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jasperE3

#15 h Apr 4, 2025, 4:36 PM

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GreekIdiot wrote:

$f(x)>0$ for all $x \in \mathbb R_+$ so all $y \in \mathbb R_+$ can be written as $\dfrac {f(m)}{f(n)}$ for some $m,n \in \mathbb R_+$
Then there exists some homogenous-kinda function (lets call it $g$ ) such that $f(xf(y)+f(x))=y^{\ell +1} \cdot g(x)$ and also $f(x+yf(x))=y^{\ell} \cdot g(x)$ thats what I meant to say. Correct me if wrong lol. :oops:

What's a homogenous kinda function

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GreekIdiot

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GreekIdiot

#16 h Apr 4, 2025, 8:05 PM

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I am not sure how to call it in english or even what it is. Hope you can understand what I am saying from the symbols

Thats the important part anyways, not some random math definition.

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jasperE3

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jasperE3

#17 h Apr 4, 2025, 8:07 PM

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GreekIdiot wrote:

I am not sure how to call it in english or even what it is. Hope you can understand what I am saying from the symbols

Thats the important part anyways, not some random math definition.

I can't, can you explain?

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GreekIdiot

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GreekIdiot

#18 h Apr 4, 2025, 8:12 PM

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So basically I am trying to define a second function, g, which exists and satisfies both relations above. Then proving g must be constant will help in proving that the only sol we have found so far is unique. Hope that clears things up.

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jasperE3

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jasperE3

#19 h Apr 4, 2025, 8:15 PM

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GreekIdiot wrote:

So basically I am trying to define a second function, g, which exists and satisfies both relations above. Then proving g must be constant will help in proving that the only sol we have found so far is unique. Hope that clears things up.

How is $g$ defined

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GreekIdiot

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GreekIdiot

#20 h Apr 4, 2025, 8:18 PM

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$g(x) > 0$ is a must for all positive $x$ . Then it could be any function but we may be able to narrow it down. Just brainstorming, nothing rigorous. This FE has been unsolved for some time, I doubt that I of all people will be the one to solve.

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hanzo.ei

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hanzo.ei

#21 h Apr 17, 2025, 2:09 PM

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