APMO 2011 problem-5

[sourav das]

Determine all functions $f:\mathbb{R}\to\mathbb{R}$, where $\mathbb{R}$ is the set of all real numbers, satisfying the following two conditions:
1) There exists a real number $M$ such that for every real number $x,f(x)<M$ is satisfied.
2) For every pair of real numbers $x$ and $y$,


\[ f(xf(y))+yf(x)=xf(y)+f(xy) \]

[*Mahi*]

Determine all functions $f:\mathbb{R}\to\mathbb{R}$, where $\mathbb{R}$ is the set of all real numbers, satisfying the following two conditions:
1) There exists a real number $M$ such that for every real number $x,f(x)<M$ is satisfied.
2) For every pair of real numbers $x$ and $y$,


\[ f(xf(y))+yf(x)=xf(y)+f(xy) \]

$f(x)=0 \forall x \in \mathbb{R}$ is the trivial solution.
Let us first set $K$ such that $\forall x , f(x) \leq K$, which means there exists $x'$ such that $f(x')=K$.
Now, we see that, setting $y=1$, $f(xf(1))=xf(1)$. So if $|f(1)|>0$ then there will always be an $x$ such that $f(x)>K$, which is impossible, so $f(1)=0$.Again, $x=0$ in the first equation gives $f(0)=0$
Setting $x=1,f(f(y))=2f(y)$...(i)
Now, let us set $x=1,y=x'$.
Then \[f(f(x'))+x' .0=1.f(x')+f(x' .1)\]
Or $f(K)=2K \leq K$
So, we get $K \leq 0$.
But we've already found $x$ such that $f(x)=0$.
So $K=0$
Now, substituting $y$ for $f(y)$,
\[ f(2xf(y))+f(y)f(x)=2xf(y)+f(xf(y)) \]
Or, $f(2xf(y))-2xf(y)=f(xf(y))-f(y)f(x) \leq 0 $
So, setting $x$ instead of $2x$, $f(xf(y)) \leq xf(y)$, and thus $yf(x) \geq f(xy)$.
Now, if $x > 0$ and $y=\frac 1x$, then $f(x) \geq 0$, and thus , for all non negative $x$, $f(x)=0$.
Now let for a negative $a, f(a)<0$.Then ,
\[f(2xf(a))+f(a)f(x)=2xf(a)+f(xf(a))\](As $f(f(a))=2f(a)$)
But, for negative $x,xf(a)>0$ and thus $f(xf(a))=0=f(2xf(a))$
So, $f(a)f(x)=2xf(a)$ or $f(x)=2x$ $\forall x<0$.
And the solution is $f(x)=0 \forall x$ $\in \mathbb{R}$ or
\[f(x)=\left\{\begin{matrix}
f(x)=0&\forall x \geq 0 , \\
f(x)=2x&\forall x<0
\end{matrix}\right.\]
And we can check these two solutions to see that both of them satisfies the conditions.

Quite classic!

[Corei13]

$f(2xf(y))-2xf(y)=f(xf(y))-f(y)f(x) \leq 0 $

So, setting $x$ instead of $2x$, $f(xf(y)) \leq xf(y)$, and thus $yf(x) \geq f(xy)$.

Why ? :-/

[*Mahi*]

$f(2xf(y))-2xf(y)=f(xf(y))-f(y)f(x) \leq 0 $

So, setting $x$ instead of $2x$, $f(xf(y)) \leq xf(y)$, and thus $yf(x) \geq f(xy)$.

Why ? :-/

$f(xf(y))+yf(x)=xf(y)+f(xy)$(given) so if $f(xf(y)) \leq xf(y)$ then to make the two sides equal, $yf(x) \geq f(xy)$ is a must.

[Corei13]

I asked why $f(2xf(y))-2xf(y)=f(xf(y))-f(y)f(x) \leq 0$? and why $x\longrightarrow 2x\Longrightarrow f(xf(y)) \leq xf(y)$ ?

[*Mahi*]

misunderstood!

I asked why $f(2xf(y))-2xf(y)=f(xf(y))-f(y)f(x) \leq 0$? and why $x\longrightarrow 2x\Longrightarrow f(xf(y)) \leq xf(y)$ ?

$f(xf(y)) \leq 0 ,-f(y)f(x) \leq 0$ as $f(x)f(y) \geq 0$ so $f(xf(y))-f(y)f(x) \leq 0$ and thus $f(2xf(y)) \leq 2xf(y)$. Here
$x\longrightarrow \frac x2 \Longrightarrow f(xf(y)) \leq xf(y)$