IMO 2007 Short list problem


Find all functions  \large f:\mathbb{R+} \rightarrow \mathbb{R+} such that

f\left(x + f\left(y\right)\right) = f\left(x + y\right) + f\left(y\right)

for all x,y \large \in  \large \mathbb{R+}

Solution :

For any positive real numbers z, we have that

f(x+f(y))+z=f(x+y)+f(y)+z
f(f(x+f(y))+z)=f(f(x+y)+f(y)+z)
f(x+f(y)+z)+f(x+f(y))=f(x+y+f(y)+z)+f(x+y)
f(x+y+z)+f(y)+f(x+y)+f(y)=f(x+2y+z)+f(y)+f(x+y)
f(x+y+z)+f(y)=f(x+2y+z)
f(a)+f(b)=f(a+b)

and by Cauchy in positive reals, then f(x)=\alpha x for all x\in (0,\infty)
Now it’s easy to see that \alpha=2 , then f(x)=2x    for all positive real numbers x

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About shivang1729

I am a young student who loves math . I like number theory and inequalities part the most , and preparing for Math Olympiads :)

Posted on Friday,July 6, 2012, in Functional equations, Problems and tagged , , , . Bookmark the permalink. 7 Comments.

  1. good question !
    f(x+f(y)) = f(x+y)+f(y)
    because f:R+ -> R+
    there is function : f(0) = p and f(q) = 0
    first substitute : x=0
    f(f(y)) = f(y)+f(y)
    f(f(y)) = 2f(y)
    let u=f(y)
    f(u) = 2u
    then f(x) = 2x 🙂

  2. Read the question carefully !! 😀
    its given that function is only for positive Real numbers
    so x can’t have value as 0 😀

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