Find all functions f:R→R such that:
f(2f(x)+f(y))=2x+f(y)∀x,y∈R.
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,y∈R.
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)∀y∈R.
Since f is 1 to 1 you get
2f(0)+f(y)=y.
Thus
f(y)=y−2f(0).
Setting y=0 you get f(0)=0 and thus f(x)=x is the only solution.
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