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 4 of 8: Derive

yf(x)=f(xf(y))

Previous step Next step

(y+1)f(x)=f\!\left(\tfrac{x}2f(y)+\tfrac x2\right)+f\!\left(\tfrac{x}2f(y)+\tfrac x2\right)=f(xf(y))+f(x) \implies yf(x)=f(xf(y))

Detailed analysis

Setting $x=y=\tfrac x2$ and $z=y$ in the functional equation gives $(y+1)f(x)$ equal to twice the identical term $f(\tfrac x2f(y)+\tfrac x2)$ . The two arguments summed equal $xf(y)+x$ , which also equals the sum $xf(y)+x$ split as one term $xf(y)$ and one term $x$ ; by additivity over equal sums (step 3), twice that term equals $f(xf(y))+f(x)$ . Hence $(y+1)f(x)=f(xf(y))+f(x)$ , so $yf(x)=f(xf(y))$ for all $x,y>0$ .