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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.

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  • $\begingroup$ 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. $\endgroup$ – drhab Sep 3 '18 at 8:04
  • $\begingroup$ Thanks, the idea is great. But this new equation for g(x) seems to be almost as complicated as the original. $\endgroup$ – Wuberdall Sep 3 '18 at 9:34
  • $\begingroup$ Injectivity can be easily established also. $\endgroup$ – Sil Apr 7 at 10:19
  • $\begingroup$ 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. $\endgroup$ – pre-kidney Jun 17 at 9:04

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