Problem 2

Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that for each $x \in \mathbb{R}^+$, there is exactly one $y \in \mathbb{R}^+$ satisfying $$xf(y)+yf(x) \leq 2$$

Solution

The answer is only $\boxed{f(x) = \frac{1}{x}}$. This works, as for $y = x$, we have $2xf(x) = 2$ and for $y \not= x$, by AM-GM, $xf(y) + yf(x) > 2$. We'll now prove that there are no other solutions.
Claim 01. For all $x \not= y$, we have
\[ \frac{2}{xy} < \frac{f(x)}{x} + \frac{f(y)}{y} \le \frac{1}{x^2} + \frac{1}{y^2} \]

Proof. Let us suppose that the unique $y \in \mathbb{R}^+$ satisfying $xf(y) + yf(x) \le 2$ is not $x$. We therefore conclude that $2xf(x) > 2$ and $2yf(y) > 2$, i.e. $f(x) > \frac{1}{x}$ and $f(y) > \frac{1}{y}$. However, this implies
\[ \frac{2}{xy} \ge \frac{f(x)}{x} + \frac{f(y)}{y} > \frac{1}{x^2} + \frac{1}{y^2} \ge \frac{2}{xy} \]

which is a contradiction. Therefore, we must have $xf(x) \le 1$ for all $x \in \mathbb{R}^+$; and furthermore for all $x \not= y$, we must have
\[ \frac{f(x)}{x} + \frac{f(y)}{y} > \frac{2}{xy} \]

as claimed.

Now, consider the function $g: \mathbb{R}^+ \to \mathbb{R}_{\ge 0}$ such that $g(x) = x^2 - xf \left( \frac{1}{x} \right) \ge 0$ for all $x \in \mathbb{R}^+$. Therefore the condition above rewrites to
\[ g(a) + g(b) = a^2 + b^2 - a f \left( \frac{1}{a} \right) - b f \left( \frac{1}{b} \right) < (a - b)^2 \]f

or any $a \not= b$. Let us write this as $Q(a,b) : g(a) + g(b) < (a - b)^2$.
Claim 02. $g \equiv 0$.
Proof. We have $0 \le g(a) \le g(a) + g(b) < (a - b)^2$  for all $b \not= a$. Thus,


\[ 0 \le g(a) \le \lim_{b \to a} (a - b)^2 = 0 \implies g(a) = 0 \]f

or any $a \in \mathbb{R}^+$, as desired.

This forces $f(x) = \frac{1}{x}$, as desired.