Mathematical Reflections 2014, Issue 5 - Problem J318

Problem:
Determine the functions $f : \mathbb{R} \longrightarrow \mathbb{R}$ satisfying $f(x-y)-xf(y) \leq 1-x$ for all real numbers $x$ and $y$.

Proposed by Marcel Chirita

Solution:
Observe that if $x=0$, we get $f(-y) \leq 1$ for all $y \in \mathbb{R}$, i.e. $f(x) \leq 1$ for all $x \in \mathbb{R}$. If $y=0$, then

$f(x)-xf(0) \leq 1-x,$                               (1)

and if $y=x$, then

$f(0)-xf(x) \leq 1-x.$                               (2)

Summing up (1) and (2), we get $(1-x)(f(x)+f(0)) \leq 1-x$ and if $x>1$, we obtain $f(x)+f(0) \geq 2$, for all $x \in (1,\infty)$, i.e. $f(x) \geq 2-f(0) \geq 1$. Therefore, $f(x)=1$ for all $x \in (1,\infty)$ and $f(0)=1$. Now, setting $x=1$ in the given relation, we have $f(1-y) \leq f(y)$ and for all $1-y>1$, we have $1 \leq f(y)$. So, $f(x)=1$ for all $x \in (-\infty,0)$. Now, setting $x=1$ in (1) and (2), we get $f(0)=f(1)$. Finally, if we set $x=1/t$ and $y=x$ in the given relation, we get $$f(0)-\dfrac{1}{t}f\left(\dfrac{1}{t}\right) \leq 1-\dfrac{1}{t},$$ i.e. $$tf(0)-f\left(\dfrac{1}{t}\right) \leq t-1.$$ This relation, $f(0)=1$ and (1) give $$f(x) \leq f\left(\dfrac{1}{x}\right),$$ so $f(x)=1$ for all $x \in (0,1)$. In conclusion, $f(x)=1$ for all $x \in \mathbb{R}$.