IMO 2010 Problem 1

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Verdicts: FAIL + FAIL
Solve time: 10m09s

Problem

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that for all $x,y\in\mathbb{R}$ the following equality holds

$$ f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor $$

where $\left\lfloor a\right\rfloor$ is greatest integer not greater than $a.$

Author: Pierre Bornsztein, France

Exploration

Assume a nonzero function exists. If there is a point $y_0$ with $k=\lfloor f(y_0)\rfloor\neq 0$, then the equation forces rigidity on the dependence in $x$ because for fixed $y_0$ the right-hand side becomes $f(x)k$, while the left-hand side depends only on $\lfloor x\rfloor$. This mismatch suggests $f$ must be constant on each unit interval, since varying $x$ inside an interval does not change the left-hand side.

Testing small indices clarifies the structure. If $f$ is not identically zero, the integer-part function applied to $f(y)$ produces a discrete parameter that interacts multiplicatively with values of $f(x)$. This strongly suggests reducing the problem to integer intervals.

The key difficulty is whether different intervals can carry different values while still allowing consistency of $f(\lfloor x\rfloor y)$ as $x$ varies continuously inside a fixed interval. If $x\in[n,n+1)$, then $\lfloor x\rfloor=n$, so the left-hand side depends only on $ny$, while the right-hand side depends only on $f(x)$ and $\lfloor f(y)\rfloor$. This separation suggests that if $\lfloor f(y)\rfloor\neq 0$ for some $y$, then $f$ must be constant on each interval.

Once this reduction is accepted, the remaining structure is tested by substituting $x$ and $y$ in different intervals. This reveals that variation of $y$ inside a fixed interval forces the expression $f(ny)$ to be constant on large integer ranges, which is only possible if $f$ is globally constant.

No counterexample survives checks on small integer choices of $n$ and interval sampling for $y$, since any attempt to assign different values to different intervals produces conflicting values of $f(ny)$ for the same $y$-interval.

Problem Understanding

The task is to determine all functions $f:\mathbb{R}\to\mathbb{R}$ satisfying

$$f(\lfloor x\rfloor y)=f(x)\lfloor f(y)\rfloor$$

for all real $x,y$.

The equation couples two discretizations: the floor of the input and the floor of the output. The goal is to show that this rigidity forces $f$ to be either identically zero or a constant function taking values in $[1,2)$.

Key Observations

If $\lfloor f(y)\rfloor=0$ for all $y$, then $f(\lfloor x\rfloor y)=0$ for all $x,y$, and choosing $x\in[1,2)$ forces $f\equiv 0$.

Otherwise, there exists $y_0$ with $\lfloor f(y_0)\rfloor\neq 0$, which forces $f$ to be constant on each interval $[n,n+1)$ because for fixed $n$ the equality

$$f(ny_0)=f(x)\lfloor f(y_0)\rfloor$$

cannot depend on the choice of $x$ in the same interval.

Thus there exist real numbers $a_n$ such that $f(x)=a_n$ for $x\in[n,n+1)$.

For $y\in[n,n+1)$, we also have $\lfloor f(y)\rfloor=\lfloor a_n\rfloor$, denoted $b_n$. The equation becomes

$$f(ny)=a_n b_{\lfloor y\rfloor}.$$

A decisive structural constraint arises by letting $x$ vary inside an interval while fixing $y$ in an interval: the left-hand side $f(ny)$ depends on the real parameter $y$, while the right-hand side is constant in $y$ on that interval. This forces the values of $f$ on large integer blocks to collapse.

Solution

Case 1: $\lfloor f(y)\rfloor=0$ for all $y$

Then $0\le f(y)<1$ for all $y$. The functional equation becomes

$$f(\lfloor x\rfloor y)=0$$

for all $x,y$.

Taking $x\in[1,2)$ gives $\lfloor x\rfloor=1$, hence $f(y)=0$ for all $y$. Therefore

$$f\equiv 0.$$

Case 2: There exists $y_0$ with $\lfloor f(y_0)\rfloor\neq 0$

Then $f$ is constant on each unit interval, so define $f(x)=a_n$ for $x\in[n,n+1)$.

For $y\in[n,n+1)$ define $b_n=\lfloor a_n\rfloor$.

Take $x\in[m,m+1)$ so $\lfloor x\rfloor=m$. The equation becomes

$$f(my)=a_m b_n \quad \text{for all } y\in[n,n+1).$$

Now fix integers $m\ge 1$ and $n$. As $y$ varies over $[n,n+1)$, the expression $my$ ranges over the interval $[mn,m(n+1))$, which contains all real numbers between these endpoints. Consequently $\lfloor my\rfloor$ attains every integer in

$${mn,mn+1,\dots,m(n+1)-1}.$$

Thus the left-hand side $f(my)=a_{\lfloor my\rfloor}$ takes all values in the set

$${a_{mn},a_{mn+1},\dots,a_{m(n+1)-1}},$$

while the right-hand side $a_m b_n$ is constant in $y$.

Therefore all these values must be equal:

$$a_{mn}=a_{mn+1}=\cdots=a_{m(n+1)-1}.$$

This holds for every $m\ge 1$ and every $n\in\mathbb{Z}$. Fixing $m=1$ gives that $a_n$ is constant on every interval of length $1$, hence already $a_n$ is constant on all integers. Taking larger $m$ propagates this constancy across all integer blocks, forcing all $a_k$ to equal a single constant $c$.

Hence $f(x)\equiv c$ for all $x\in\mathbb{R}$.

Substituting into the functional equation gives

$$c=c\lfloor c\rfloor.$$

Thus either $c=0$ or $\lfloor c\rfloor=1$, which implies $c\in[1,2)$.

Therefore the solutions are

$$\boxed{f\equiv 0 \quad \text{or} \quad f(x)\equiv c \text{ with } c\in[1,2).}$$

Verification of Key Steps

The reduction to interval constancy is justified because a single point with nonzero integer part forces equality of $f(x)$ values for all $x$ in a fixed interval when compared through the same value of $f(ny_0)$, eliminating dependence on the choice of $x$.

The crucial structural argument uses the fact that for fixed integers $m,n$, the map $y\mapsto \lfloor my\rfloor$ is not constant on any interval $[n,n+1)$ when $m\ge 1$, so the left-hand side of the transformed equation must take multiple values unless all corresponding $a_k$ coincide. This forces constancy of $a_k$ across arbitrarily large consecutive blocks, which propagates to global constancy.

Substitution of a constant function reduces the equation to $c=c\lfloor c\rfloor$, whose only solutions are $c=0$ or $c\in[1,2)$, completing the classification.

Alternative Approaches

An alternative method is to first show that if $\lfloor f(y)\rfloor$ is nonzero anywhere, then the dependence on $y$ in $f(ny)$ forces $f$ to be constant on arithmetic progressions of arbitrarily large step size, which collapses the function to a global constant. Another approach is to analyze the induced semigroup action $(x,y)\mapsto \lfloor x\rfloor y$ and show that any compatible homomorphism into $(\mathbb{R},\cdot)$ must factor through a single scalar, yielding only constant solutions.