Problem 5

Find all functions

f:\mathbb{R}^+\to\mathbb{R}^+

such that

(z+1)f(x+y)=f(xf(z)+y)+f(yf(z)+x)

for all positive real numbers

x,y,z

Step 3 of 8: Additivity over equal sums

Previous step Next step

a+b=c+d \implies f(a)+f(b)=f(c)+f(d)

Detailed analysis

Given $a,b,c,d>0$ with $a+b=c+d$ , unboundedness of $f$ lets us pick $e$ with $f(e)$ larger than $1,a/b,b/a,c/d,d/c$ . Applying the lemma with $c=f(e)$ to the pairs $(a,b)$ and $(c,d)$ produces $u,v,w,t>0$ with $f(e)u+v=a$ , $u+f(e)v=b$ , $f(e)w+t=c$ , $w+f(e)t=d$ , and one checks $u+v=w+t$ since $(u+v)(f(e)+1)=a+b$ and $(w+t)(f(e)+1)=c+d$ . Substituting $x=u,y=v,z=e$ into the functional equation gives $f(a)+f(b)=(e+1)f(u+v)$ ; similarly $f(c)+f(d)=(e+1)f(w+t)$ , and the claim follows.