Selection and Training Session

Answer : $a\in \{0\}\cup [1,\infty )$. Let function $f:{\mathbb{R}}_{\ge 0}\to {\mathbb{R}}_{\ge 0}$ satisfy the equality $$f(f(x)+f(y))=yf(1+yf(x))\text{\hspace{1em}(1)}$$ for all nonnegative real numbers $x$ and $y$. Set $f(0)=c$ and $f(1)=b$. For $y=0$ from equality (1) we get $$f(f(x)+c)=0.\text{\hspace{1em}(2)}$$ for all $x\in [0,\infty )$. For $x=f(x)+c$ from equality (1) we get $$f(f(y))=by.\text{\hspace{1em}(3)}$$ for all $y\in [0,\infty )$. Suppose $b\ne 0$. From equality (3) it follows that function $f$ is injective. Hence from equality (2) it follows that $f(x)+c=f(0)$, a contradiction.

Therefore $b=0$ and equality (3) is equivalent to $$f(f(y))=0\text{\hspace{1em}(4)}$$

For $y=0$ from equality (4) we get $f(c)=0$ hence $f(0)=f(f(c))=0$. So $c=f(0)=0$. For $y=1$ from equality (1) we get $$f(1+f(x))=0(5)$$ for all $x\in [0,\infty )$. Suppose that $f(a)=d\ne 0$ for some $a>1$. Substituting $x=a$, $y=\frac{a-1}{d}$ to equality (1) we get $$f(f(\frac{a-1}{d})+d)=a-1.$$ which leads to a contradiction with equality (5) for $x=f(\frac{a-1}{d})+d$. Hence for any $a\in \{0\}\cup [1,\infty )$ the equality $f(a)=0$ is proved. The next example shows that there exists a function satisfying equality (1) such that $f(x)\ne 0$ for any $x\in (0,1)$: $$f(x)=\{\begin{array}{ll}0,& x=0;\\ \frac{1}{x},& x\in (0,1);\\ 0,& x\ge a.\end{array}$$