BMO 2024 | Problem 4 | Solution - 30 April 2024

Problem 4.
Let ℝ⁺=(0,

\infty

) be the set of all positive real numbers. Find all functions f:ℝ⁺→ℝ⁺ and polynomials P(x) with non-negative real coefficients such that P(0)=0 which satisfy the equality f(f(x)+P(y))=f(x-y)+2y for all real numbers x>y>0.

Solution

We prove as before $f(x) \geq x$ . Now let $g(x)=f(x)-x$ so that :
$g(g(x)+x+P(y))+g(x)+P(y) = g(x-y)+y.$ Note that $Q(x,x-c) \Rightarrow P(x-c) \leq x-c+g(c)$ so $\deg P \leq 1$ . It is easy to show $\deg P \not = 0$ so assume $P=aX$ with $a \in \mathbb{R}^+_0$ and so :
$g(g(x)+x+ay)+g(x)+ay=g(x-y)+y.$
It is easy to show $a \leq1$ . Note that $g(x) \leq g(x-y)+y(1-a)$ and so for $\epsilon<a$ there exists $N\in \mathbb{R}^+$ such that $g(x) \leq (1-\epsilon)x$ for all $x\geq N$ .

Now assume there exists $N, b<1$ such that $g(x) \leq bx$ for all $x \geq N$ . Note that if $x \geq \frac{N}{1-b}$ then $x-g(x) \geq N$ and $x\geq N$ so :
$Q(x,g(x)) \Rightarrow ag(x)\leq g(x-g(x)) \leq bx-bg(x) \Rightarrow g(x) \leq \frac{b}{a+b}x.$ Hence we get $g(x) \leq \frac{b}{a+b}x$ whenever $x \geq \frac{N}{1-b}$ .

Consider the sequence defined by $x_1=b$ and $x_{n+1} = \frac{x}{a+x}$ . It is quite easy to show that $x_n \to 0$ when $x\to \infty$ so for any $c$ there exists $N_c$ such that $g(x) \leq cx$ whenever $x \geq N_c$ .

Finally take $x \geq N_c$ . This means :
$c^2x+cx+acy +ay+cx \geq g(g(x)+x+ay)+g(x)+ay \geq y.$ Taking $c\to 0$ and $y \to x$ gives $a \geq 1$ and so $a=1$ .

Thus we are left with :
$g(g(x)+x+y)+g(x)=g(x-y).$ Note that $g$ is decreasing and that translating $x,y$ fixes the LHS. Thus $g$ cannot decrease stricly far enough, i.e $g(x)>g(y)$ while $x<y$ when $x,y \geq N$ . Hence $g$ is constant far enough and this clearly implies that $g$ is constant.

Reciprocally we get $g(x)=0$ , hence $f(x)=x$ , which clearly works.