2005 IMO Shortlist Problems/A2


(Bulgaria) Let $\mathbb{R}^+$ denote the set of all postive real numbers. Determine all functions $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that

$\displaystyle f(x)f(y) = 2f(x+yf(x))$

for all positive real numbers $\displaystyle x$ and $\displaystyle y$.


Solution 1

Lemma 1. $f$ is non-decreasing.

Proof. Suppose, on the contrary, that there exist $x> z$ such that $f(x) < f(z)$. We set $y = \frac{x-z}{f(z)-f(x)} > 0$, so $x + yf(x) = z + yf(z)$, to obtain

$f(x)f(y) = 2f(x+yf(x)) = 2f(z+yf(z)) = f(z)f(y)$,

or $f(x) = f(z)$, a contradiction.

Lemma 2. $f$ is not strictly increasing.

Proof. Assume the contrary. Then for all $x ,y$ we have

$f(x)f(y) = 2f(x + yf(x)) > 2f(x)$,

so $f(y) > 2$. Furthermore, since $f$ is injective, we have $f(x+1 \cdot f(x)) = f(x)f(1) = f(1+ x\cdot f(1))$, which implies $x+ f(x) = 1+ x \cdot f(1))$, or

$f(x) = (f(1)-1)x + 1$.

But then for $x$ arbitrarily close to 0, $f(x)$ becomes less than 2, a contradiction. Thus $f$ is not strictly increasing.

Now, let $x>z$, $f(x) = f(z)$. If $y$ is in the interval $(0, (x-z)/f(x)$, then $z< z+yf(z) \le x$, so

$f(z) \le f(z+yf(z)) \le f(x) = f(z)$,

so $f(x+yf(z)) = f(x)$. It follows that

$f(z)f(y) = 2f(z + yf(z)) = 2f(x) = 2f(z)$,

so $f(y) = 2$, for all $y$ in that interval.

But if ${} f(y_0) = 2$, then setting $x=y=y_0$ in the original equation gives $f(3^n y_0) = 2$, by induction. It follows that for any $x$, there exist $a<x$ and $c>x$ such that $f(a) = f(c) = 2$, and since $f$ is non-decreasing, we must have $f(x) = 2$ for all $x$. It is easy to see that this satisfies the given equation. Q.E.D.

Solution 2

Lemma 1. There exist distinct positive $\displaystyle a,b$ such that $\displaystyle f(a) = f(b)$.

Proof. Suppose the contrary, i.e., suppose $\displaystyle f$ is injective. Then for any $\displaystyle x$, we have

$f(x+1\cdot f(x)) = f(x)f(1) = f(1 + x\cdot f(1))$,

which implies

$\displaystyle x + f(x) = 1 + f(1) \cdot x$
$\displaystyle f(x) = (f(1)-1)x +1$.

This means that either $\displaystyle f(1)=1$ and $\displaystyle f(x) = 1$ for all $\displaystyle x$ (a contradiction, since that is not injective), or $\displaystyle f(x) = ax+1$, for some real $\displaystyle a$. But setting $\displaystyle y=x$ then gives us a quadratic in $\displaystyle x$ with a nonzero leading coefficient, which has at most two real roots, implying that $\displaystyle x$ can only assume two different values, a contradiction.

Lemma 2. There exist $\displaystyle c_0$ and infinitely many $\displaystyle c_n = 3^n \cdot c_0$ such that $\displaystyle f(c_n) = 2$, for all nonnegative integers $\displaystyle n$.

Proof. Let $\displaystyle a,b$ be the distinct positive reals of Lemma 1 such that $\displaystyle f(a)=f(b)$; without loss of generality, let $\displaystyle a>b$. Letting $\displaystyle c_0 = \frac{a-b}{f(b)}$ yields

$f(b)f(c_0) = 2f\left( b + \frac{a-b}{f(b)} \right) = 2f(a) = 2f(b)$.

Since $f(b) \in \mathbb{R}^+$, this implies $\displaystyle f(c_0) = 2$. We now prove that $\displaystyle f(c_n) = 2$ for all nonnegative $\displaystyle n$, by induction. We have just proven our base case. Now, assume $\displaystyle f(c_n) = 2$. Setting $\displaystyle x=y=c_n$ gives us

$2\cdot 2 = f(c_n)f(c_n) = 2 f(c_n + c_n f(c_n)) = 2f(3c_n) = 2f(c_{n+1})$,

so $\displaystyle f(c_{n+1}) = 2$, as desired.

Lemma 3. For all $x\in \mathbb{R}^+$, $f(x) \ge 1$.

Proof. We note that $\displaystyle y = \frac{-x}{f(x)-1}$ is the solution to the equation $\displaystyle y  = x + y\cdot f(x)$, which is positive when $\displaystyle f(x)<1$, so if this is the case, setting $\displaystyle y$ to this value gives us

$\displaystyle f(x)f(y) = 2f(y)$,

and since $\displaystyle f(y)>0$, this implies, $\displaystyle f(x) = 2 \not< 1$, a contradiction.

Lemma 4. $\displaystyle f(x) \ge 2$.

Proof. We will prove by induction that $f(x) \ge 2^{1- \frac{1}{2^n}}$, for all $\displaystyle n$. Our base case comes from Lemma 3. Now, if for all $\displaystyle x$, $\displaystyle f(x) \ge 2^{1-\frac{1}{2^n}}$, then for some $\displaystyle z$,

$[f(x)]^2 = f(x)f(x) = 2f(z) \ge 2 \cdot 2^{1-\frac{1}{2^n}} = 2^{2\left(1 - \frac{1}{2^{n+1}} \right) }$,

so $f(x) \ge 2^{\frac{1}{2^{n+1}}}$, as desired.

Now, suppose there exists some $\displaystyle f(x) < 2$. Then there exists some $\displaystyle n$ such that $2^{1-\frac{1}{2^n}} > f(x)$. But this gives us $f(x) \ge 2^{1-\frac{1}{2^n}} > f(x)$, a contradiction. Thus for all $\displaystyle x$, $\displaystyle f(x) \ge 2$.

We will now prove that $\displaystyle f(x) = 2$ is the only solution to the functional equation. Consider any value of $\displaystyle x$. By Lemma 2, there exists some $\displaystyle {} c_n > x$ such that $\displaystyle f(c_n) = 2$. Setting $\displaystyle y= \frac{c_n -x}{f(x)}$ then gives us

$f(x) f\left( \frac{c_n -x}{f(x)} \right) = 2f(c_n) = 4$.

But from Lemma 4, we know $f(x), f\left( \frac{c_n-x}{f(x)} \right) \ge 2$, so we must have $\displaystyle f(x) = f\left( \frac{c_n -x}{f(x)} \right) = 2$. This constant function clearly satisfies the given equation. Q.E.D.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.