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

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



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



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.






functional-equations







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  • You could set $f(x):=g(x)+x+1$. Then - if I did not make any mistakes - the equation will change into: $gleft(xgleft(yright)+gleft(xright)+2x+xy+1right)+xgleft(yright)=gleft(xright)$ 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 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 at 9:34















up vote
5
down vote

favorite
6












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



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



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.






functional-equations







share|cite|improve this question





















  • You could set $f(x):=g(x)+x+1$. Then - if I did not make any mistakes - the equation will change into: $gleft(xgleft(yright)+gleft(xright)+2x+xy+1right)+xgleft(yright)=gleft(xright)$ 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 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 at 9:34













up vote
5
down vote

favorite
6









up vote
5
down vote

favorite
6






6





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



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



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.






functional-equations







share|cite|improve this question













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



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



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.






functional-equations




functional-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 3 at 7:46









Wuberdall

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  • You could set $f(x):=g(x)+x+1$. Then - if I did not make any mistakes - the equation will change into: $gleft(xgleft(yright)+gleft(xright)+2x+xy+1right)+xgleft(yright)=gleft(xright)$ 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 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 at 9:34

















  • You could set $f(x):=g(x)+x+1$. Then - if I did not make any mistakes - the equation will change into: $gleft(xgleft(yright)+gleft(xright)+2x+xy+1right)+xgleft(yright)=gleft(xright)$ 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 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 at 9:34
















You could set $f(x):=g(x)+x+1$. Then - if I did not make any mistakes - the equation will change into: $gleft(xgleft(yright)+gleft(xright)+2x+xy+1right)+xgleft(yright)=gleft(xright)$ 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 at 8:04





You could set $f(x):=g(x)+x+1$. Then - if I did not make any mistakes - the equation will change into: $gleft(xgleft(yright)+gleft(xright)+2x+xy+1right)+xgleft(yright)=gleft(xright)$ 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 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 at 9:34





Thanks, the idea is great. But this new equation for g(x) seems to be almost as complicated as the original.
– Wuberdall
Sep 3 at 9:34
















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