# Is this $f(x) = x+1$ the only solution to this functional equation.

I am considering the problem of finding all functions $$f:(0,\infty)\rightarrow(0,\infty)$$ satisfying the functional equation:

$$f\big(xf(y)+f(x)\big) = 2f(x)+xy.$$

I have been able to prove the following three results/properties:

1. $$f$$ is not surjective.
2. $$f$$ does not have any fixed points.
3. $$f(x)=x+1$$ is a solution.

My intuition tells me that the solution in (3) is the only solution, but I have not been successful in proving or disproving this claim.

Any ideas on how I can make further progress is appreciated.

• You could set $f(x):=g(x)+x+1$. Then - if I did not make any mistakes - the equation will change into: $g\left(xg\left(y\right)+g\left(x\right)+2x+xy+1\right)+xg\left(y\right)=g\left(x\right)$ and the question is now: is $g(x)=0$ the only solution? I really don't know whether that will help, though. – drhab Sep 3 '18 at 8:04
• Thanks, the idea is great. But this new equation for g(x) seems to be almost as complicated as the original. – Wuberdall Sep 3 '18 at 9:34
• Injectivity can be easily established also. – Sil Apr 7 at 10:19
• The problem is not stated well, since the quantity $xf(y)+f(x)$ need not belong to the domain of $f$. If the functional equation is interpreted as applying only to those $x,y\in (0,\infty)$ for which $xf(y)+f(x)\in (0,\infty)$, then there are other solutions, for instance $f$ can be the zero function which satisfies the functional equation vacuously (since there are no $x,y$ satisfying $xf(y)+f(x)\in (0,\infty)$). The simplest way to fix the problem is by modifying the codomain to be $(0,\infty)$ instead of $\mathbb R$, which I believe is the intended question. – pre-kidney Jun 17 at 9:04