Putnam 2022, Problem B-6

Find all continuous functions $f : \mathbb{R}^+ \to \mathbb{R}^+$ such that $f(xf(y)) + f(yf(x)) = 1 + f(x + y)$ for all $x, y > 0$.


Solution:
The only such functions are the functions \( f(x) = \frac{1}{1 + cx} \) for some \( c \geq 0 \) (the case \( c = 0 \) giving the constant function \( f(x) = 1 \)). Note that we interpret \( \mathbb{R}^+ \) in the problem statement to mean the set of positive real numbers, excluding 0.  

For convenience, we reproduce here the given equation:  
\[
f(x f(y)) + f(y f(x)) = 1 + f(x + y) \tag{1}
\]  

We first prove that  
\[
\lim_{x \to 0^+} f(x) = 1. \tag{2}
\]  

Set  
\[
L_- = \liminf_{x \to 0^+} f(x), \quad L_+ = \limsup_{x \to 0^+} f(x).
\]  

For any fixed \( y \), we have by (1)  
\[
L_+ = \limsup_{x \to 0^+} f(x f(y))
\]  
\[
\leq \limsup_{x \to 0^+} (1 + f(x + y)) = 1 + f(y) < \infty.
\]  

Consequently, \( x f(x) \to 0 \) as \( x \to 0^+ \). By (2) with \( y = x \),  
\[
2L_+ = \limsup_{x \to 0^+} 2f(x f(x))
\]  
\[
= \limsup_{x \to 0^+} (1 + f(2x)) = 1 + L_+
\]  
\[
2L_- = \liminf_{x \to 0^+} 2f(x f(x))
\]  
\[
= \liminf_{x \to 0^+} (1 + f(2x)) = 1 + L_-
\]  
and so \( L_- = L_+ = 1 \), confirming (2).  

We next confirm that  
\[
f(x) \geq 1 \text{ for all } x > 0 \implies f(x) = 1 \text{ for all } x > 0. \tag{3}
\]  

Suppose that \( f(x) \geq 1 \) for all \( x > 0 \). For \( 0 < c \leq \infty \), put \( S_c = \sup\{ f(x) : 0 < x \leq c \} \); for \( c < \infty \), (2) implies that \( S_c < \infty \). If there exists \( y > 0 \) with \( f(y) > 1 \), then from (1) we have \( f(x + y) - f(x f(y)) = f(y f(x)) - 1 \geq 0 \); hence  
\[
S_c = S_{(c - y) f(y)}
\]  
\[
\left( c \geq c_0 = \frac{y f(y)}{f(y) - 1} \right)
\]  

and (since \( (c - y) f(y) - c_0 = f(y)(c - c_0) \)) iterating this construction shows that \( S_\infty = S_c \) for any \( c > c_0 \). In any case, we deduce that  
\[
f(x) \geq 1 \text{ for all } x > 0 \implies S_\infty < \infty. \tag{4}
\]  

Still assuming that \( f(x) \geq 1 \) for all \( x > 0 \), note that from (1) with \( x = y \),  
\[
f(x f(x)) = \frac{1}{2}(1 + f(2x)).
\]  

Since \( x f(x) \to 0 \) as \( x \to 0^+ \) by (2) and \( x f(x) \to \infty \) as \( x \to \infty \), \( x f(x) \) takes all positive real values by the intermediate value theorem. We deduce that \( 2S_\infty \leq 1 + S_\infty \) and hence \( S_\infty = 1 \); this proves (3).  

We may thus assume hereafter that \( f(x) < 1 \) for some \( x > 0 \). We next check that  
\[
\lim_{x \to \infty} f(x) = 0. \tag{5}
\]  

Put \( I = \inf\{ f(x) : x > 0 \} < 1 \), choose \( \epsilon \in (0, (1 - I)/2) \), and choose \( y > 0 \) such that \( f(y) < I + \epsilon \). We then must have \( x f(x) \neq y \) for all \( x \), or else  
\[
1 + I \leq 1 + f(2x) = 2f(y) < 2I + 2\epsilon,
\]  
contradiction. Since \( x f(x) \to 0 \) as \( x \to 0^+ \) by (2), we have \( \sup\{ x f(x) : x > 0 \} < \infty \) by the intermediate value theorem, yielding (5).  

By (2) plus (5), \( f^{-1}(1/2) \) is nonempty and compact. We can now simplify by noting that if \( f(x) \) satisfies the original equation, then so does \( f(cx) \) for any \( c > 0 \); we may thus assume that the least element of \( f^{-1}(1/2) \) is 1, in which case we must show that \( f(x) = \frac{1}{1 + x} \).  

We next show that  
\[
\lim_{x \to \infty} x f(x) = 1. \tag{6}
\]  

For all \( x > 0 \), by (1) with \( y = x \),  
\[
f(x f(x)) = \frac{1}{2}(1 + f(2x)) > \frac{1}{2} = f(1),
\]  
so in particular \( x f(x) \neq 1 \). As in the proof of (5), this implies that \( x f(x) < 1 \) for all \( x > 0 \). However, by (5) and (7) we have \( f(x f(x)) \to 1/2 \) as \( x \to \infty \), yielding (6).  

By substituting \( y \mapsto xy \) in (1),  
\[
f(x f(xy)) + f(xy f(x)) = 1 + f(x + xy).
\]  

Taking the limit as \( x \to \infty \) and applying (6) yields  
\[
f(1/y) + f(y) = 1. \tag{8}
\]  

Combining (1) with (8) yields  
\[
f(x f(y)) = f(x + y) + f\left( \frac{1}{y f(x)} \right).
\]  

Multiply both sides by \( x f(y) \), then take the limit as \( x \to \infty \) to obtain  
\[
1 = \lim_{x \to \infty} x f(y) f(x + y) + \lim_{x \to \infty} x f(y) f\left( \frac{1}{y f(x)} \right)
\]  
\[
= f(y) + \lim_{x \to \infty} x f(y) y f(x)
\]  
\[
= f(y) + y f(y)
\]  
and solving for \( f(y) \) now yields \( f(y) = \frac{1}{1 + y} \), as desired.  

**Remark.** Some variants of the above approach are possible. For example, once we have (5), we can establish that \( f \) is monotone decreasing as follows. We first check that  
\[
f(x) < 1 \text{ for all } x > 0. \tag{9}
\]  

Suppose by way of contradiction that \( f(x) = 1 \) for some \( x \). By (1),  
\[
f(2x) + 1 = 2f(x f(x)) = 2f(x) = 2
\]  
and so \( f(2x) = 1 \). It follows that \( f^{-1}(1) \) is infinite, contradicting (5).  

We next check that  
\[
x < y \implies f(x) > f(y). \tag{10}
\]  

For \( x < y \), by substituting \( x \mapsto y - x \) in (1) we obtain  
\[
1 + f(y) = f((y - x) f(x)) + f((y - x) f(y - x))
\]  
\[
< 1 + f((y - x) f(x)),
\]  
whence \( f((y - x) f(x)) > f(y) \). Because \( (y - x) f(x) \to 0 \) as \( x \to y^- \) and \( (y - x) f(x) \to y \) as \( x \to 0^+ \), \( (y - x) f(x) \) takes all values in \( (0, y) \) as \( x \) varies over \( (0, y) \); this proves (10).