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 $a.$

Solutions

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])

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