Iranian Mathematical Olympiad

We shall firstly prove following lemmas;

Lemma 1. For any real number $r$, the length of the interval $f(r)$ is at most 2. Proof. If $x,y\in f(r)$ we have $r\in f(x)\cap f(y)$ hence we have $|x-y|\le 2$.

Lemma 2. If the interval $I$ has length 2, then there is a real number $r$ such that $f(r)=I$. Proof. If $I=[x,x+2]$ then by condition 2 there is $r\in f(x)\cap f(x+2)$ so we have $x,x+2\in f(r)$. But the length of $f(r)$ is at most 2 hence we have $f(r)=[x,x+2]$, this completes our proof.

Lemma 3. f is an injective function and for any r the length of f(r) is 2. Proof. Using preceding lemmas it is enough to show that if $f(x)\subset f(y)$ then $x=y$. Assume the contrary, let $I$ be an interval of length 2 such that $x\in I,y\notin I$ then by lemma 2 there exists $r$ such that $I=f(r)$. Now we have $r\in f(x),r\notin f(y)$ which contradicts $f(x)\subset f(y)$.

Define a new function $g:\mathbb{R}\to \mathbb{R}$ such that for all $r$, $g(r)$ is the midpoint of $f(r)$. Since we know the length of $f(r)$ we can forget about $f$ and work with $g$. We apt to find all bijective functions $g$ such that: $$i.|x-g(y)|\le 1⟺|y-g(x)|\le 1.$$ $$ii.|g(x)-g(y)|\le 2⟺|x-y|\le 2.$$ $$iii.g(r)=r^2\text{ for all }0\le r\le 1.$$

Lemma 4. g is strictly increasing. Proof. It suffices to prove if $x<y,y-x<1$ then $g(x)<g(y)$. We firstly consider the case $y=0$. For each $-2\le x\le 0$ by condition 2 and $g(0)=0$ we have $$|g(x)-g(2)|>2,|g(x)|\le 2,|g(2)|\le 2$$ so we can deduce that $g(x)$ and $g(2)$ have different signs, similarly we can deduce that $g(2),g(1)=1$ have the same sign (they are both opposite to the sign of $g(-2)$). Hence we have $g(2)>0,g(x)<0$. We prove by induction on $\lfloor x\rfloor$ (We only prove the case that $0\le x<y$ the other case is similar). For the base if $0\le x<y<1$ then $x^2<y^2$. We know that $y-x-2>2$ hence by the second condition we have $$|g(y)-g(x-2)|>2,|g(y)-g(x)|\le 2,|g(x)-g(x-2)|\le 2.$$ We know by induction that $g(x)-g(x-2)>0$ hence we have $g(y)>g(x)$.

Lemma 5. we have $g(x+1)=g^{-1}(x)+1$. Proof. We know that $g$ is bijective so it suffices to prove that $g(g(x)+1)=x+1$. Assume that $g(r)=x+1$ then we have $r\le g(x)+1$. Now if $r^{\prime }>r$ we have $g(r^{\prime })>x+1$ hence by condition 1; $r^{\prime }>g(x)+1$ so we get $r=g(x)+1$.

By the last lemma we obtain that there is only one function that satisfies the condition of the problem and that function is: $$g(x)=\{\begin{array}{ll}(x-\lfloor x\rfloor {)}^2+\lfloor x\rfloor & \lfloor x\rfloor \equiv 0(\mathrm{mod}2);\\ \sqrt{x-\lfloor x\rfloor }+\lfloor x\rfloor & \lfloor x\rfloor \equiv 1(\mathrm{mod}2).\end{array}$$ It is easy to check that this function actually works. ■