MathLabs

Problem 5

Find all functions f:R+→R+f:\mathbb{R}^+\to\mathbb{R}^+ such that (z+1)f(x+y)=f(xf(z)+y)+f(yf(z)+x)(z+1)f(x+y)=f(xf(z)+y)+f(yf(z)+x) for all positive real numbers x,y,zx,y,z.
Step 1 of 8: ff is unbounded above
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x=y=1:2f(f(z)+1)=(z+1)f(2)x=y=1:\quad 2f(f(z)+1)=(z+1)f(2)
Detailed analysis

The identity f(x)=xf(x)=x clearly satisfies the equation. For a general solution ff, setting x=y=1x=y=1 gives 2f(f(z)+1)=(z+1)f(2)2f(f(z)+1)=(z+1)f(2) for all z>0z>0. As zz ranges over all positive reals the right-hand side is unbounded, so ff is unbounded above.