0
$\begingroup$

Find all functions $f$ defined over real numbers to real numbers such that $f(f(y))+f(x-y)=f(xf(y)-x)$

My attempt: Set $x=y=0$ to get $f(f(0))=0$. It will be very helpful if I will able to find $f(0)$ but I failed to find it. I tried to check the injectivity of $f$ but wasn't able to check it.

Please help me.

$\endgroup$
  • 1
    $\begingroup$ Brief observation: Suppose $f(y) = 0$. Then by setting $x=0$, we get $f(0) + f(-y) = f(0)$. So $y$ is a root iff $-y$ is a root. Now instead suppose $f(x) = 0$. Then by setting $x=y$, we obtain $2f(0)=f(-x)$; but we already know $f(x) = 0 = f(-x)$, so $2f(0) = f(0)$ and hence (if there is any root at all) $f(0) = 0$. $\endgroup$ – Patrick Stevens Mar 22 '19 at 4:07
  • $\begingroup$ Okay, so $f(0)=0$.Now what to do next?We can also conclude that $f(x)=f(-x)$ $\endgroup$ – Sufaid Saleel Mar 22 '19 at 4:47
  • $\begingroup$ $f(0) = 0$ if there is a root. $\endgroup$ – Patrick Stevens Mar 22 '19 at 4:51
1
$\begingroup$

Let $P(x,y)$ be the assertion that $f(f(y))+f(x-y)=f(xf(y)-x)$. Now,

Observation 1: $P(0,0)\implies f(f(0))=0$.

Observation 2: Considering $P(0,f(0))$ and $P(f(0),f(0))$ we get $f(0)=0$.

Observation 3: Considering $P(x,0)$ we get $f$ is even function, i.e;$f(x)=f(-x)$ and finally $P(x,-x)$ and $P(x,x)$ implies $f(x)=0$.

So, $f(x)=0,\forall x\in\mathbb{R}$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

I tried to solve this problem with taking help from the above comments.Please check it. enter image description here

enter image description here

I am tooo lazy to LaTeX it .Please don't mind!

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.