6 TST for BMO

Let $f(0)=a,a\in \mathbb{R}$. We set $x=0$ and obtain $f(a)=0$. Next, $y=a$ yields $$f(f(x))=xf(a+1)(1).$$ Assume first that $f(a+1)\ne 0$. Then $f$ is an injection. Indeed, assuming $f(x_1)=f(x_2)$ for some $x_1\ne x_2$, after putting these values in (1) we get contradiction. Set $x=1$ in the initial condition. Since $f$ is injective, it follows $$f(1)+f(y)=y+1(2)$$ Setting $y=1$ yields $2f(1)=2$, hence $f(1)=1$. $f(y)=y+1$. Now, (2) gives us $f(y)=y,\forall y\in \mathbb{R}$, which indeed is solution.

Let now consider the case $f(a+1)=0$. Then $f(f(x))=0,\forall x\in \mathbb{R}.$ This means that $f(y)=0,\forall y\in \text{Im}(f)$. Assume that there exists a point $y_0$, for which $f(y_0+1)\ne 0$. Setting $y=y_0$ in the initial condition yields $$f(f(x)+xf(y_0))=xf(y_0+1)(3).$$ For any $x_0\in \mathbb{R}$ putting $x:=x_0/(f(y_0+1))$ in (3), we get, $$f(f(x)+xf(y_0))=x_0.$$ Since $x_0$ is arbitrary, it means that $f$ is surjective, which implies $\text{Im}(f)=\mathbb{R}$. Therefore, $f(y)=0,\forall y\in \mathbb{R}$, which contradicts the assumption $f(y_0+1)\ne 0$. So, in this case we get $f(x)=0,\forall x\in \mathbb{R}$, which also is solution. $\square$