Functional Equation( Japan final round 2008)

[mutasimmim]

Find all functions $ f : R\rightarrow R$ such that for any $x,y$, the relation $f(x+y)f(f(x)-y)=xf(x)-yf(y)$ satisfies.

[mutasimmim]

Hint:

Right side is (almost) symmetric, but the left side isn't.

[SANZEED]

Let us denote the given statement by $P(x,y)$. Then,
$P(x,f(x))\Rightarrow f(x+f(x))f(0)=xf(x)-f(x)f(f(x))$.
$P(x,0)\Rightarrow f(x)f(f(x))=xf(x)$.
So, $f(x+f(x))f(0)=0 \forall x\in \mathbb{R}$.
So we have two cases here.

$\textbf{Case 1:}$ When $f(x+f(x))=0 \forall x\in \mathbb{R}$.
$P(x+f(x),-x)\Rightarrow f(f(x))f(x)=xf(-x)$. But $f(f(x))f(x)=xf(x)$. So, $f(x)=f(-x)\forall x\in \mathbb{R}$.
$P(x,0)\Rightarrow f(0)f(x+f(x))=0=2xf(x)$. This yields us the solution:
$f(x)=0\forall x\neq 0$ and $f(0)=c\in \mathbb{R}$. In particular, if $c=0$, then it is the constant solution to the given equation.

$\textbf{Case 2:}$ When $f(x+f(x))$ is non-zero for all non-zero $x$.
Then $f(0)=0$. The only constant solution belongs to the family of solution in case $1$. So let $f$ be non-constant.
$P(x,-x)\Rightarrow f(x)=-f(-x)\forall x\in \mathbb{R}$.
$P(0,y)\Rightarrow f(y)(f(y)-y)=0 \forall y\in \mathbb{R}.....(*)$
Now since $f$ is non-constant, there exists a non-zero $d$ such that $f(d)$ is non-zero. Then from equation$(*)$, $f(d)=d$.
If there is a non-zero $k\in \mathbb{R}$ such that $f(k)=0$, then,
$P(k,d-k)\Rightarrow df(k-d)=(d-k)f(k-d)\Rightarrow kf(k-d)=0\Rightarrow f(k-d)=0$. So,
$P(k-d,d)\Rightarrow 0=-df(d)\Rightarrow d^{2}=0\Rightarrow d=0$ which is a contradiction.
So, $r=0$ if and only if $f(r)=0$. Thus from $(*)$ and from $f(0)=0$, we have that $f(y)=y\forall y\in \mathbb{R}$.

So finally the solutions to the given equations are:
$f(x)=0\forall x\neq 0$ and $f(0)=c\in \mathbb{R}$
or,
$f(y)=y\forall y\in \mathbb{R}$.

It is easy to check that these are indeed the solutions.

[Sazid Akhter Turzo]

Let us denote the given statement by $P(x,y)$. Then,
$P(x,f(x))\Rightarrow f(x+f(x))f(0)=xf(x)-f(x)f(f(x))$.
$P(x,0)\Rightarrow f(x)f(f(x))=xf(x)$.
So, $f(x+f(x))f(0)=0 \forall x\in \mathbb{R}$.
So we have two cases here.

$\textbf{Case 1:}$ When $f(x+f(x))=0\forall x\in \mathbb{R}$.
$P(x+f(x),-x)\Rightarrow f(f(x))f(x)=xf(-x)$. But $f(f(x))f(x)=xf(x)$. So, $f(x)=f(-x)\forall x\in \mathbb{R}$.
$P(x,0)\Rightarrow f(0)f(x+f(x))=0=2xf(x)$. This yields us the solution:
$f(x)=0\forall x\neq 0$ and $f(0)=c\in \mathbb{R}$. In particular, if $c=0$, then it is the constant solution to the given equation.

$\textbf{Case 2:}$ When $f(x+f(x))\neq 0\forall x\in \mathbb{R}$.
Then $f(00=0$. The only constant solution belongs to the family of solution in case $1$. So let $f$ be non-constant.
$P(x,-x)\Rightarrow f(x)=-f(-x)\forall x\in \mathbb{R}$.
$P(0,y)\Rightarrow f(y)(f(y)-y)=0 \forall y\in \mathbb{R}.....(*)$
Now since $f$ is non-constant, $\exists d\neq 0$ such that $f(d)\neq 0$. Then from equation$(*)$, $f(d)=d$.
If $\exists k\neq 0$ such that $f(k)=0$, then,
$P(k,d-k)\Rightarrow df(k-d)=(d-k)f(k-d)\Rightarrow kf(k-d)=0\Rightarrow f(k-d)=0$. So,
$P(k-d,d)\Rightarrow 0=-df(d)\Rightarrow d^{2}=0\Rightarrow d=0$ which is a contradiction.
So, $r=0$ if and only if $f(r)=0$. Thus from $(*)$ and from $f(0)=0$, we have that $f(y)=y\forall y\in \mathbb{R}$.

So finally the solutions to the given equations are:
$f(x)=0\forall x\neq 0$ and $f(0)=c\in \mathbb{R}$
or,
$f(y)=y\forall y\in \mathbb{R}$.

It is easy to check that these are indeed the solutions.

Though it seems obvious, but there must be a Case 3: $f(x+f(x)) \neq 0$ for some $x$ and $f(x+f(x))=0$ for some $x$; otherwise you'll get partial marks in any contest. Be aware of this mistakes!

[mutasimmim]

Of course $f(x)=0$ is the unique constant solution. We assume $f$ isn't constant and denote by $P(X,Y)$ the assertion that $f(x+y)f(f(x)-y)=xf(x)-yf(y)$. $P(x,0)$ implies $f(x)f((f(x))=xf(x)$ and so $f(f(x))=x$. $P(0,y)$ implies $f(y)f(f(0)-y)=-yf(y)$ and so $f(f(0)-y)=-y$, so $f(f(f(0)-y))=f(-y)$ and $f(0)-y=f(-y)$ and $f(y)=y+a$, letting $f(0)=a$. Find the value of $a$ from anywhere you want and get $a=0$. Thus $f(0)=o$ and $f(x)=x$ are the only two solutions.

PS. I leave the small argument which may arrive from getting $f(f(0)-y)=-y$ from $f(y)f(f(0)-y)=-yf(y)$ in case of $f(y)=0$.

[SANZEED]

@Turzo, sorry, actually I didn't have time to post this case, and because it's a bit obvious, I left it while posting. Anyway, here is the third case:

$\textbf{Case 3:} \exists a,b$ such that $f(a+f(a))=0$ and $f(b+f(b))$ is non-zero, where both $a,b$ are non-zero reals.
According to our previous results, $f(b+f(b))f(0)=0$. This means that $f(0)=0$. So,
$P(0,y)\Rightarrow f(y)[f(y)-y]=0$. Set $y=b+f(b)$, then we have $f(b+f(b))=b+f(b)$.
Let $d=b+f(b)$ and $k=a+f(a)$. Now unless $k=0$, we will have $d=0$, i.e. $f(b+f(b))=b+f(b)=0$ which contradicts our assumption. So, $k=f(a)+a=0$. Thus,
$P(a+f(a),-a)\Rightarrow f(a)=f(-a)$. Then $f(-a)=f(a)=-a$. So,
$P(-a,a)\Rightarrow 0=-af(-a)-af(a)=2a^{2}\Rightarrow a=0$ which is also a contradiction. So no such $a,b$ exist.
Thus either $f(x+f(x))=0\forall x\in \mathbb{R}$, or $f(x+f(x))$ is non-zero for all non-zero $x$.

[SANZEED]

Of course $f(x)=0$ is the unique constant solution. We assume $f$ isn't constant and denote by $P(X,Y)$ the assertion that $f(x+y)f(f(x)-y)=xf(x)-yf(y)$. $P(x,0)$ implies $f(x)f((f(x))=xf(x)$ and so $f(f(x))=x$. $P(0,y)$ implies $f(y)f(f(0)-y)=-yf(y)$ and so $f(f(0)-y)=-y$, so $f(f(f(0)-y))=f(-y)$ and $f(0)-y=f(-y)$ and $f(y)=y+a$, letting $f(0)=a$. Find the value of $a$ from anywhere you want and get $a=0$. Thus $f(0)=o$ and $f(x)=x$ are the only two solutions.

PS. I leave the small argument which may arrive from getting $f(f(0)-y)=-y$ from $f(y)f(f(0)-y)=-yf(y)$ in case of $f(y)=0$.

You can't declare that $f(f(x))=x$ from $f(x)f((f(x))=xf(x)$. Because you didn't prove that $f$ is non-zero for all non-zero $x$.