# 2010 IMO Problems/Problem 1

## Problem

Find all function $f:\mathbb{R}\rightarrow\mathbb{R}$ 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 $a.$

## Solution 1

Put $x=y=0$. Then $f(0)=0$ or $\lfloor f(0) \rfloor=1$.

$\bullet$ If $\lfloor f(0) \rfloor=1$, putting $y=0$ we get $f(x)=f(0)$, that is f is constant. Substituing in the original equation we find $f(x)=0, \ \forall x \in \mathbb{R}$ or $f(x)=a, \ \forall x \in \mathbb{R}$, where $a \in [1,2)$.

$\bullet$ If $f(0)=0$, putting $x=y=1$ we get $f(1)=0$ or $\lfloor f(1) \rfloor=1$.

For $f(1)=0$, we set $x=1$ to find $f(y)=0 \ \forall y$, which is a solution.

For $\lfloor f(1) \rfloor=1$, setting $y=1$ yields $f(\lfloor x \rfloor)=f(x), \ (*)$.

Putting $x=2, y=\frac{1}{2}$ to the original we get $f(1)=f(2)\lfloor f(\frac{1}{2}) \rfloor$. However, from $(*)$ we have $f(\frac{1}{2})=f(0)=0$, so $f(1)=0$ which contradicts the fact $\lfloor f(1) \rfloor=1$.

So, $f(x)=0, \ \forall x$ or $f(x)=a, \ \forall x, \ a \in [1,2)$. ( By socrates[1])

## Solution 2

Substituting $y=0$ we have $f(0) = f(x) [f(0)]$. If $[f0)] \ne 0$ then $f(x) = \frac{f(0)}{[f(0)]}$. Then $f(x)$ is constant. Let $f(x)=c$. Then substituting that in (1) we have $c=c[c] \Rightarrow c(1-[c])=0 \Rightarrow c=0$, or $[c]=1$. Therefore $f(x)=c$ where $c=0$ or $c \in [1,2)$

If $[f(0)] = 0$ then $f(0)=0$. Now substituting $x=1$ we have $f(y)=f(1)[f(y)]$. If $f(1) \ne 0$ then $[f(y)] = \frac{f(y)}{f(1)}$ and substituting this in (1) we have $f([x]y)=\frac{f(x)f(y)}{f(1)}$. Then $f([x]y)=f(x[y])$. Substituting $x=1/2, y=2$ we get $f(0)=f(1)$. Then $f(1)=0$, which is a contradiction Therefore $f(1)=0$. and then $f(y)=0$ for all $R$

Then the only solutions are $f(x)=0$ or $f(x)=c$ where $c \in [1,2)$.( By m.candales [2])

## Solution 3

Let $y=0$, then $f(0)=f(x)\left\lfloor f(0)\right\rfloor$.

Case 1: $\left\lfloor f(0)\right\rfloor\neq 0$

Then $f(x)=\frac{f(0)}{\left\lfloor f(0)\right\rfloor}$ is a constant. Let $f(x)=k$, then $k=k\left\lfloor k \right\rfloor \Leftrightarrow k=0 \vee 1\leq k<2$. It is easy to check that this are solutions.

Case 2: $\left\lfloor f(0)\right\rfloor= 0$

In this case we conclude that $\left\lfloor f(0)\right\rfloor= 0\Rightarrow f(0)=0$

Lemma:If $y$ is such that $0\leq f(y)<1$, $f(y)=0$

Proof of the Lemma: If $x=1$ we have that $f(\left\lfloor x\right\rfloor y)=f(y)=f(x)\left\lfloor f(y)\right\rfloor =0$, as desired.

Let $0\leq x<1$, so that we have: $0=f(0)=f(\left\lfloor x\right\rfloor y)=f(x)\left\lfloor f(y)\right\rfloor\Rightarrow$ $\Rightarrow f(x)=0 \vee 0\leq f(y)<1 \Rightarrow f(x)=0 \vee f(y)=0$, using the lemma.

If $f$ is not constant and equal to $0$, letting $y$ be such that $f(y)\neq 0$ implies that $f(x)=0, \forall 0\leq x<1$.

Now it's enough to notice that any real number $x$ is equal to $ky$, where $k\in \mathbb{Z}$ and $0\leq y< 1$, so that $f(x)=f(ky)=f(\left\lfloor k\right\rfloor y)=f(k)\left\lfloor f(y)\right\rfloor=0$. Since $x$ was arbitrary, we have that $f$ is constant and equal to $0$.

We conclude that the solutions are $f(x)=k$, where $k=0 \vee 1\leq k<2$.( By Jorge Miranda [3] )

## Solution 4

Clearly $f(\left\lfloor x\right\rfloor y) = f(\left\lfloor \lfloor x \rfloor \right\rfloor y) = f(\lfloor x \rfloor)\left\lfloor f(y)\right\rfloor$, so $(f(x) - f(\lfloor x \rfloor))\left\lfloor f(y)\right\rfloor = 0$ for all $x,y\in\mathbb{R}$.

If $\left\lfloor f(y)\right\rfloor = 0$ for all $y \in \mathbb{R}$, then by taking $x=1$ we get $f(y)=f(1)\left\lfloor f(y)\right\rfloor = 0$, so $f$ is identically null (which checks).

If, contrariwise, $\left\lfloor f(y_0)\right\rfloor \neq 0$ for some $y_0 \in \mathbb{R}$, it follows $f(x) = f(\lfloor x \rfloor)$ for all $x \in \mathbb{R}$.

Now it immediately follows $f(x) = f(\lfloor x \rfloor \cdot 1) = f(x)\lfloor f(1) \rfloor$, hence $f(x)(1 - \lfloor f(1) \rfloor) = 0$.

For $x=y_0$ this implies $\lfloor f(1) \rfloor = 1$. Assume $\lfloor f(0) \rfloor=0$; then $1 \leq f(1) = f\left ( 2\cdot \dfrac {1} {2} \right ) = f(2)\left \lfloor f \left ( \dfrac {1} {2} \right ) \right \rfloor = f(2)\left \lfloor f \left ( \left \lfloor \dfrac {1} {2} \right \rfloor \right ) \right \rfloor = f(2)\left \lfloor f(0) \right \rfloor = 0$, absurd.

Therefore $\lfloor f(0) \rfloor \neq 0$, and now $y=0$ in the given functional equation yields $f(0) = f(x)\lfloor f(0) \rfloor$ for all $x \in \mathbb{R}$, therefore $f(x) = c \neq 0$ constant, with $\lfloor c \rfloor = \lfloor f(1) \rfloor = 1$, i.e. $c \in [1,2)$ (which obviously checks).( By mavropnevma [4])