IMO 1986 Problem 5

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 10m14s

Problem

Find all functions $f$ defined on the non-negative reals and taking non-negative real values such that: $f(2)=0,f(x)\ne0$ for $0\le x<2$, and $f(xf(y))f(y)=f(x+y)$ for all $x,y$.

Exploration

Substituting $y=2$ into the functional equation yields

$f(xf(2))f(2)=f(x+2).$

Since $f(2)=0$, this forces $f(x+2)=0$ for all $x\ge 0$, hence $f(x)=0$ for all $x\ge 2$.

This immediately isolates the active region $[0,2)$, where $f$ is strictly positive.

Substituting $x=0$ gives

$f(0)f(y)=f(y),$

so $f(0)=1$ because $f(y)\ne 0$ on $[0,2)$.

The key structural tension appears when $x+y=2$. Then

$f(xf(y))f(y)=f(2)=0,$

and since $f(y)>0$ for $y<2$, this forces

$f(xf(y))=0.$

Because all zeros occur at inputs $\ge 2$, this creates a sharp inequality constraint linking $f(y)$ to $(2-y)$.

Testing small values suggests the function must blow up as $y\to 2^{-}$, ruling out bounded or affine behavior. This contradicts the previously assumed linear model, so a different form must be derived from the boundary interaction at $x+y=2$.

Problem Understanding

We seek all functions $f:\mathbb{R}{\ge 0}\to \mathbb{R}{\ge 0}$ such that

$f(xf(y))f(y)=f(x+y),$

with $f(2)=0$ and $f(x)\ne 0$ for $x<2$.

The condition $f(x)=0$ for $x\ge 2$ follows immediately from substituting $y=2$. Thus the equation splits into a nontrivial positive regime on $[0,2)$ and a forced zero regime beyond $2$.

The main difficulty is determining the exact functional form on $[0,2)$ under the constraint that the equation must transition consistently at $x+y=2$.

Key Observations

From $x+y=2$, we obtain for every $y<2$:

$f((2-y)f(y))=0.$

Since $f$ is nonzero on $[0,2)$, the argument must satisfy

$(2-y)f(y)\ge 2.$

This gives a lower bound:

$f(y)\ge \frac{2}{2-y}.$

If the inequality were strict for some $y$, then for sufficiently small $x>0$ with $x<2-y$, one would still have $xf(y)\ge 2$, forcing $f(xf(y))=0$ while $f(x+y)>0$, contradicting the functional equation. Hence equality must hold:

$(2-y)f(y)=2.$

Thus the function is determined pointwise on $[0,2)$.

Solution

From the equality

$(2-y)f(y)=2,$

we obtain

$f(y)=\frac{2}{2-y} \quad \text{for } 0\le y<2.$

Also $f(2)=0$, and from the earlier propagation result,

$f(x)=0 \quad \text{for } x\ge 2.$

We now verify the functional equation.

Let $x,y\ge 0$.

If $y\ge 2$, then $f(y)=0$, so

$f(xf(y))f(y)=0,$

and since $x+y\ge 2$, also $f(x+y)=0$.

Assume $y<2$, so $f(y)=\frac{2}{2-y}$.

If $x+y\ge 2$, then $f(x+y)=0$. Also

$xf(y)=\frac{2x}{2-y}\ge 2,$

so $f(xf(y))=0$, and the equation holds.

Now assume $x+y<2$. Then $xf(y)<2$, so the formula applies:

$f(xf(y))=\frac{2}{2-\frac{2x}{2-y}}.$

Compute:

$2-\frac{2x}{2-y}=\frac{2(2-y-x)}{2-y}.$

Hence

$f(xf(y))=\frac{2(2-y)}{2(2-y-x)}=\frac{2-y}{2-y-x}.$

Therefore,

$f(xf(y))f(y)=\frac{2-y}{2-y-x}\cdot \frac{2}{2-y}=\frac{2}{2-x-y}=f(x+y).$

Thus the equation holds in all cases.

The function satisfies all conditions and is uniquely determined by the boundary constraint.

Verification of Key Steps

The deduction from $x+y=2$ to $(2-y)f(y)\ge 2$ is forced by the fact that zeros occur only at inputs $\ge 2$, since $f$ is strictly positive on $[0,2)$.

The strengthening to equality follows because any strict inequality would force, for fixed $y$, a range of $x<2-y$ with $xf(y)\ge 2$, which would collapse the left-hand side to zero while the right-hand side remains positive, contradicting the functional equation.

All substitutions in the verification step stay within the correct domain splits $x+y<2$, $x+y\ge 2$, and $y\ge 2$, ensuring no hidden case omissions.

Alternative Approaches

One can instead define $g(x)=\frac{1}{f(x)}$ on $[0,2)$ and transform the equation into a Cauchy-type additive relation for $g$, which leads to linearity of $2/f(x)$ and yields the same reciprocal form.

Another approach is to analyze the transformation $x\mapsto xf(y)$ as a semigroup action and classify all invariant blow-up barriers at $2$, again forcing $f(x)$ to be proportional to $(2-x)^{-1}$.