Subfield: Algebra


Question: Let $f$ be any function that maps the set of real numbers into the set of real numbers. Prove that there exist real numbers $x$ and $y$ such that

$$
f(x-f(y))>y f(x)+x .
$$

Solution:
Assume that

$$
f(x-f(y)) \leq y f(x)+x \quad \text { for all real } x, y
\tag{1}
$$

Let $a=f(0)$. Setting $y=0$ in (1) gives $f(x-a) \leq x$ for all real $x$ and, equivalently,

$$
f(y) \leq y+a \text { for all real } y
\tag{2}
$$

Setting $x=f(y)$ in (1) yields in view of (2)

$$
a=f(0) \leq y f(f(y))+f(y) \leq y f(f(y))+y+a .
$$

This implies $0 \leq y(f(f(y))+1)$ and thus

$$
f(f(y)) \geq-1 \text { for all } y>0 \text {. }
\tag{3}
$$

From (2) and (3) we obtain $-1 \leq f(f(y)) \leq f(y)+a$ for all $y>0$, so

$$
f(y) \geq-a-1 \text { for all } y>0
\tag{4}
$$

Now we show that

$$
f(x) \leq 0 \text { for all real } x \text {. }
\tag{5}
$$

Assume the contrary, i.e. there is some $x$ such that $f(x)>0$. Take any $y$ such that

$$
y<x-a \quad \text { and } \quad y<\frac{-a-x-1}{f(x)}
$$

Then in view of (2)

$$
x-f(y) \geq x-(y+a)>0
$$

and with (1) and (4) we obtain

$$
y f(x)+x \geq f(x-f(y)) \geq-a-1,
$$

whence

$$
y \geq \frac{-a-x-1}{f(x)}
$$

contrary to our choice of $y$. Thereby, we have established (5).

Setting $x=0$ in (5) leads to $a=f(0) \leq 0$ and (2) then yields

$$
f(x) \leq x \quad \text { for all real } x
\tag{6}
$$

Now choose $y$ such that $y>0$ and $y>-f(-1)-1$ and set $x=f(y)-1$. From (1), (5) and



(6) we obtain

$$
f(-1)=f(x-f(y)) \leq y f(x)+x=y f(f(y)-1)+f(y)-1 \leq y(f(y)-1)-1 \leq-y-1,
$$

i.e. $y \leq-f(-1)-1$, a contradiction to the choice of $y$.
Assume that

$$
f(x-f(y)) \leq y f(x)+x \quad \text { for all real } x, y
\tag{7}
$$

Let $a=f(0)$. Setting $y=0$ in (7) gives $f(x-a) \leq x$ for all real $x$ and, equivalently,

$$
f(y) \leq y+a \text { for all real } y
\tag{8}
$$

Now we show that

$$
f(z) \geq 0 \text { for all } z \geq 1 \text {. }
\tag{9}
$$

Let $z \geq 1$ be fixed, set $b=f(z)$ and assume that $b<0$. Setting $x=w+b$ and $y=z$ in (7) gives

$$
f(w)-z f(w+b) \leq w+b \quad \text { for all real } w
\tag{10}
$$

Applying to $w, w+b, \ldots, w+(n-1) b$, where $n=1,2, \ldots$, leads to

$$
\begin{gathered}
f(w)-z^{n} f(w+n b)=(f(w)-z f(w+b))+z(f(w+b)-z f(w+2 b)) \\
+\cdots+z^{n-1}(f(w+(n-1) b)-z f(w+n b)) \\
\leq(w+b)+z(w+2 b)+\cdots+z^{n-1}(w+n b) .
\end{gathered}
$$

From (8) we obtain

$$
f(w+n b) \leq w+n b+a
$$

and, thus, we have for all positive integers $n$

$$
f(w) \leq\left(1+z+\cdots+z^{n-1}+z^{n}\right) w+\left(1+2 z+\cdots+n z^{n-1}+n z^{n}\right) b+z^{n} a .
\tag{11}
$$

With $w=0$ we get

$$
a \leq\left(1+2 z+\cdots+n z^{n-1}+n z^{n}\right) b+a z^{n} .
\tag{12}
$$

In view of the assumption $b<0$ we find some $n$ such that

$$
a>(n b+a) z^{n}
\tag{13}
$$

because the right hand side tends to $-\infty$ as $n \rightarrow \infty$. Now (12) and (13) give the desired contradiction and (9) is established. In addition, we have for $z=1$ the strict inequality

$$
f(1)>0 \text {. }
\tag{14}
$$

Indeed, assume that $f(1)=0$. Then setting $w=-1$ and $z=1$ in (11) leads to

$$
f(-1) \leq-(n+1)+a
$$

which is false if $n$ is sufficiently large.

To complete the proof we set $t=\min \{-a,-2 / f(1)\}$. Setting $x=1$ and $y=t$ in (7) gives

$$
f(1-f(t)) \leq t f(1)+1 \leq-2+1=-1 .
\tag{15}
$$



On the other hand, by (8) and the choice of $t$ we have $f(t) \leq t+a \leq 0$ and hence $1-f(t) \geq 1$. The inequality (9) yields

$$
f(1-f(t)) \geq 0
$$

which contradicts (15).