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ISL functional equation problem
Mhremath   4
N Yesterday at 10:18 AM by pco
2. (ISL 01/A4) Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
\[
f(xy)f(x)-f(y) = (x-y)f(x)f(y)
\]for all $x,y$.
4 replies
Mhremath
Nov 14, 2024
pco
Yesterday at 10:18 AM
ISL functional equation problem
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Mhremath
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2. (ISL 01/A4) Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
\[
f(xy)f(x)-f(y) = (x-y)f(x)f(y)
\]for all $x,y$.
This post has been edited 1 time. Last edited by Mhremath, Yesterday at 7:28 AM
Reason: mistake
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UglyScientist
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It's $f(xy)(f(x)-f(y))=(x-y)f(x)f(y)$. I'l post my solution later
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UglyScientist
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Actually $P(x, 1)$ finishes.
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pco
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Mhremath wrote:
2. (ISL 01/A4) Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying
\[
f(xy)f(x)-f(y) = (x-y)f(x)f(y)
\]for all $x,y$.
Let $P(x,y)$ be the assertion $f(xy)f(x)-f(y)=(x-y)f(x)f(y)$
Let $c=f(1)$

$P(1,1)$ $\implies$ $c\in\{0,1\}$

If $c=0$, then $P(x,1)$ $\implies$ $\boxed{f(x)=0\quad\forall x}$, which indeed fits
If $c=1$ :
$P(x,1)$ implies $f(x)\ne 0$ $\forall x$
$P(x,x)$ implies $f(x^2)=1$ and then $P(x,1)$ again implies $(x-1)f(x)=0$ and so contradiction
And so no other solution
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pco
23055 posts
#6 • 1 Y
Y by Mhremath
UglyScientist wrote:
It's $f(xy)(f(x)-f(y))=(x-y)f(x)f(y)$. I'l post my solution later
Let $P(x,y)$ be the assertion $f(xy)(f(x)-f(y))=(x-y)f(x)f(y)$
Let $c=f(1)$

If $c=0$ $P(x,1)$ $\implies$ $\boxed{\text{S1 : }f(x)=0\quad\forall x}$, which indeed fits
If $c\ne 0$

$P(x,1)$ $\implies$ $f(x)(f(x)-c)=(x-1)cf(x)$ and so $f(x)\in\{0,cx\}$ $\forall x$ (note that this implies $f(0)=0$)
Let $U=\{x$ such that $f(x)=cx\}$. $0\in U$ and $1\in U$
Let $V=\{x$ such that $f(x)=0\}$. $0\in V$

Let $u,u_1\in U$ and $v,v_1\in V$
$P(u,u_1)$ $\implies$ $uu_1\in U$ $\forall u\ne u_1$
$P(v,v_1)$ is always true
$P(u,v)$ $\implies$ $uv\in V$ $\forall u\ne 0$, still true when $u=0$ and so $\forall u\in U$, $v\in V$

And so general solution :
Let $c\ne 0$
Let $U,V$ a split of $\mathbb R_{\ne 0}$ in two disjoints subsets with property :
$\forall$ distinct $u,u_1\in U$ : $uu_1\in U$
$\forall u\in U,v\in V$ : $uv\in V$
Then define : $\boxed{\text{S2 : }f(0)=0\text{  and  }f(x)=cx\quad\forall x\in U\text{  and  }f(x)=0\quad\forall x\in V}$, which indeed fits

Note that infinitely many such split $U,V$ exist. For example :

$(U,V)=(\{1\},\mathbb R\setminus\{0,1\})$ which gives solution $\boxed{f(1)=c\text{  and  }f(x)=0\quad\forall x\ne 1}$

$(U,V)=(\mathbb R_{>0},\mathbb R_{<0})$ which gives solution $\boxed{f(x)=a(|x|+x)}$

$(U,V)=(\mathbb Q_{\ne 0},\mathbb R\setminus\mathbb Q)$ which gives solution $\boxed{f(x)=cx\times 1_{\mathbb Q}(x)\quad\forall x\in\mathbb R}$

and a lot lot of other solutions
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