Functional equation : $ f(f(x))=bxf(x)+a$

Question

Does there exist real number $a, b$ and onto function $f : \mathbb{R} \to \mathbb{R}$ satisfying $$ f(f(x))=bxf(x)+a$$ for all real numbes $x$ ?

My attempt :

Let $f(x_1)=f(x_2)$, so $f(f(x_1))=f(f(x_2))$

so $bx_1f(x_1)=bx_2f(x_2)$

then $bx_1f(x_2)=bx_2f(x_2)$

For $b \not= 0$, we have $x_1=x_2$

Hence $f$ is one-to-one function.

Please suggest how to proceed.

Answer 1

Hint: First, consider $w$ such that $f(w)=0$ and see what happens when you iterate $f$ on $w$. Then consider $v$ such that $f(v)=-1/b$ and see what happens when you iterate $f$ on $v$.

A full solution is hidden below.

Note that $b\neq 0$, since otherwise we would have $f(f(x))=a$ for all $x$ and $f$ could not be surjective. Now let $w$ be such that $f(w)=0$. We then have $$f(0)=f(f(w))=bwf(w)+a=a$$ and therefore $$f(a)=f(f(0))=b\cdot 0\cdot f(0)+a=a$$ and therefore $$a=f(f(a))=baf(a)+a=ba^2+a.$$ Since $b\neq 0$, this implies $a=0$.

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Now let $v$ be such that $f(v)=-1/b$. We then have $$f(-1/b)=f(f(v))=bvf(v)=-v$$ and therefore $$f(-v)=f(f(-1/b))=b\cdot(-1/b)\cdot f(-1/b)=v$$ and therefore $$f(v)=f(f(-v))=b\cdot(-v)\cdot f(-v)=-bv^2.$$ But $f(v)=-1/b$ as well, so $v=\pm 1/b$. If $v=-1/b$ we have both $f(v)=v$ and $f(v)=-v$, which is a contradiction. If $v=1/b$ we have both $f(-v)=-v$ and $f(-v)=v$, again a contradiction. Thus no such $f$ exists.