Find functions R to R

[Phlembac Adib Hasan]

Find all functions $f : \mathbb {R} \longrightarrow \mathbb {R}$ such that for all $x, y \in \mathbb {R}$, the following equation holds:\[ f \left( x y + f \left( x \right) \right) = x f \left ( y \right ) + f \left ( x \right ) \]

[Phlembac Adib Hasan]

Solution:

Assume $P(x,y)\Longrightarrow f \left( x y + f \left( x \right) \right) = x f \left ( y \right ) + f \left ( x \right )$
$P(0,0)\Longrightarrow f(f(0))=f(0).......(i)$
$P(x,0)\Longrightarrow f(f(x))=xf(0)+f(x)$
So if $f(0)\not =0$, then the function is injective.But then $(i)$ will imply $f(0)=0$.So a contradiction and $f(0)=0$.Hence $f(f(x))=f(x)$
$P(f(x),x)\Longrightarrow f(f(x)(x+1))=f(x)(f(x)+1).....(ii)$
$P(x,f(x))\Longrightarrow f(f(x)(x+1))=f(x)(x+1).......(iii)$
From $(ii)$ and $(iii),f(x)(f(x)+1)=f(x)(x+1)\Rightarrow f(x)=0\; or\; x\forall x\in \mathbb R$
So there are three possible functions:
1.$f(x)=x\forall x\in \mathbb R$
2.$f(x)=0\forall x\in \mathbb R$
3.$f(x)=0\; or\; x\forall x\in \mathbb R$
Clearly the first two are correct solutions.Now we'll show the third is not correct.
Let $\exists c,d$ such that $c,d\neq 0$ and $f(c)=c$ but $f(d)=0$.
$P(c,d)\Longrightarrow f(cd+c)=c$
But $f(cd+c)=0\; or\; cd+c$.None of these is equal to RHS.So a contradiction and we are done.

[SANZEED]

All the substitutions are in the main equation.
Set $x=y=0$. Then $f(f(0))=f(0)$.
Set $y=0$. Then $f(f(x))=xf(0)+f(x)$. Now if $f(0)\ne 0$, then $f(x)$ is injective, which would imply $f(0)=0$. This contradiction implies that $f(0)=0,f(f(x))=f(x)$.
Set $x=f(x),y=x$. Then $f(f(x)x+f(f(x)))=f(f(x)x+f(x))=f(x)^{2}+f(f(x))=f(x)^{2}+f(x)$
Set $y=f(x)$. Then $f(xf(x)+f(x))=xf(f(x))+f(x)=xf(x)+f(x)$
From the last two,we have that $f(x)^{2}=xf(x)$. So the possible solutions are:
$f(x)=x \forall x\in \mathbb{R}$
$f(x)=0 \forall x\in \mathbb{R}$
$f(x)=0$ for some $x\in \mathbb{R}$ and $f(x)=x$ for the others.
The first two can be checked easily. For the third,to prove it not to be a solution,take $a,b\in \mathbb{R}$ such that $f(a)=a,f(b)=0$. Then substituting $x=c,y=d$ yields that $f(cd+c)=c$. But from our assumption it should have been $0$ or $cd+c$. This contradiction finishes the solution.

[Phlembac Adib Hasan]

All the substitutions are in the main equation.
Set $x=y=0$. Then $f(f(0))=f(0)$.
Set $y=0$. Then $f(f(x))=xf(0)+f(x)$. Now if $f(0)\ne 0$, then $f(x)$ is injective, which would imply $f(0)=0$. This contradiction implies that $f(0)=0,f(f(x))=f(x)$.
Set $x=f(x),y=x$. Then $f(f(x)x+f(f(x)))=f(f(x)x+f(x))=f(x)^{2}+f(f(x))=f(x)^{2}+f(x)$
Set $y=f(x)$. Then $f(xf(x)+f(x))=xf(f(x))+f(x)=xf(x)+f(x)$
From the last two,we have that $f(x)^{2}=xf(x)$. So the possible solutions are:
$f(x)=x \forall x\in \mathbb{R}$
$f(x)=0 \forall x\in \mathbb{R}$
$f(x)=0$ for some $x\in \mathbb{R}$ and $f(x)=x$ for the others.
The first two can be checked easily. For the third,to prove it not to be a solution,take $a,b\in \mathbb{R}$ such that $f(a)=a,f(b)=0$. Then substituting $x=c,y=d$ yields that $f(cd+c)=c$. But from our assumption it should have been $0$ or $cd+c$. This contradiction finishes the solution.

See my solution.Is it necessary to re-post it?

[SANZEED]

Sorry,but I didn't see your solution. If I had,I wouldn't have re-posted it.

[sourav das]

All the substitutions are in the main equation.
Set $x=y=0$. Then $f(f(0))=f(0)$.
Set $y=0$. Then $f(f(x))=xf(0)+f(x)$. Now if $f(0)\ne 0$, then $f(x)$ is injective, which would imply $f(0)=0$. This contradiction implies that $f(0)=0,f(f(x))=f(x)$.
Set $x=f(x),y=x$. Then $f(f(x)x+f(f(x)))=f(f(x)x+f(x))=f(x)^{2}+f(f(x))=f(x)^{2}+f(x)$
Set $y=f(x)$. Then $f(xf(x)+f(x))=xf(f(x))+f(x)=xf(x)+f(x)$
From the last two,we have that $f(x)^{2}=xf(x)$. So the possible solutions are:
$f(x)=x \forall x\in \mathbb{R}$
$f(x)=0 \forall x\in \mathbb{R}$
$f(x)=0$ for some $x\in \mathbb{R}$ and $f(x)=x$ for the others.
The first two can be checked easily. For the third,to prove it not to be a solution,take $a,b\in \mathbb{R}$ such that $f(a)=a,f(b)=0$. Then substituting $x=c,y=d$ yields that $f(cd+c)=c$. But from our assumption it should have been $0$ or $cd+c$. This contradiction finishes the solution.

See my solution.Is it necessary to re-post it?

Adib, Don't discourage anyone to re-post same solution in a different way. Because still it may give us some interesting thought or may have some faults. Everyone in this forum is trying to learn. So from me, SANZEED, Don't be shy. Re-post if you need to...