Saturday, 1 June 2013

Easy functional equation



Find all functions f:RR such that:




f(2f(x)+f(y))=2x+f(y)x,yR.





If you put x=y=0, you get f(3f(0))=f(0). What deductions about f(0) can you then make?



Clearly from above f(0)=0 is a solution . . . so,



Putting x=0 gives f(2f(0)+f(y))=f(y)



f(f(y))=f(y)



So f(x)=x is a solution, but is it the only one?




I think it probably is, but how to prove?


Answer



f(2f(x)+f(y))=2x+f(y)x,yR.


Interchaning x and y you get



f(f(x)+2f(y))=f(x)+2y.



Claim 1: f(x) is 1 to 1.




Indeed, if f(x)=f(y) then



2x+f(x)=2x+f(y)=f(2f(x)+f(y))=f(f(x)+2f(y))=f(x)+2y



This implies that x=y.



Now, you can do part of what you did:



f(2f(0)+f(y))=f(y)yR.




Since f is 1 to 1 you get



2f(0)+f(y)=y.



Thus



f(y)=y2f(0).



Setting y=0 you get f(0)=0 and thus f(x)=x is the only solution.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...