Find all real numbers $a$ such that exists function $\mathbb {R} \rightarrow \mathbb {R} $ satisfying the following conditions: 1) $f(f(x)) =xf(x)-ax$ for all real $x$ 2) $f$ is not constant 3) $f$ takes the value $a$
Problem
Source: 2016 Belarus TST 2.2
Tags: functional equation, functional, algebra
EmilXM
05.11.2020 00:10
Let $P(x)$ be assertion of $f(f(x))=xf(x)-ax$, and $f(t)=a$.
$P(t) \Longrightarrow f(a)=0$.
$P(0) \Longrightarrow f(f(0)) = 0$
$P(f(0)) \Longrightarrow f(0)=-af(0) \Longrightarrow f(0)(a+1)=0$
Case 1: $f(0)=0$. Assume $f(b)=0$. Then $P(b) \Longrightarrow 0=-a\cdot b \Longrightarrow b=0$.
But we also have $f(a)=0$. So $a=0$.
Case 2: $a=-1$.
Now it is sufficies to find functions for $a=0, -1$.
If $a=0$, take $f(x) = \begin{cases} 0, & \mbox{if } x\neq 1 \\ 1, & \mbox{if } x=1 \end{cases}$.
If $a=-1$, take $f(x) = \begin{cases} -1, & \mbox{if } x \neq -1 \\ 0, & \mbox{if } x=-1 \end{cases}$.
So all real values of $a$, that satisfies conditions are $\boxed{a=\{0,-1\}}$.
Keith50
08.05.2021 12:23
$\color{red}{\textit{Ans:}}$ $a=\{0,-1\}$. $\color{blue}{\textit{Proof:}}$ Let $F(x)$ be the given assertion, given that there exists $u\in \mathbb{R}$ such that $f(u)=a$, \[F(u): f(a)=au-au=0\]\[F(a): f(0)=-a^2\]\[F(0): f(-a^2)=0\]\[F(-a^2): -a^2=a^3 \implies a^2(a+1)=0 \implies a=\{0,-1\}.\]When $a=0$, take $f(x)=\begin{cases} 0 \quad \quad \textrm{if} \quad x\ne 1, \\ 1 \quad \quad \textrm{if} \quad x=1.\end{cases}$ When $a=-1$, take $f(x)=\begin{cases} -1 \quad \quad \textrm{if} \quad x\ne -1, \\ 0 \quad \quad \ \ \ \textrm{if} \quad x=-1.\end{cases}$