I found the following functional equation:
Find all functions f:R→R such that:
xf(x)−yf(y)=(x−y)f(x+y) for all x,y∈R
Could you please help me? I think I proved that if f(0)=0 then for each x∈Q f(kx)=kf(x), but I don't know how to continue. Thanks in advance.
Answer
It's obvious that every function of the form f(x)=ax+b satisfies the equation:
xf(x)−yf(y)=(x−y)f(x+y)(1)
It can be shown that those are the only solutions indeed. To show that, let a=f(1)−f(0) and b=f(0), and define g(x)=f(x)−ax−b. It's easy to see that by (1), g satisfies
xg(x)−yg(y)=(x−y)g(x+y)(2)
and we have g(0)=0 and g(1)=0. Letting x=1 and y=−1 in (2) we get g(−1)=0. Now, letting y=1 and y=−1 in (2), we respectively get:
xg(x)=(x−1)g(x+1)(3)
xg(x)=(x+1)g(x−1)(4)
Substituting x+1 for x in (4) we have:
(x+2)g(x)=(x+1)g(x+1)(5)
Subtracting (5) and (3) we get 2g(x)=2g(x+1) and thus g(x)=g(x+1). Hence by (5) we have (x+2)g(x)=(x+1)g(x) and therefore g is the constant zero function. So f(x)=ax+b.
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