SHORTLISTED PROBLEMS FOR THE 68th NMO

Let $A=\{x\ge 0:f(x)=0\}$. By (i), $0\in A$.

Let $x\in A$. Then $f(x)=0$. By (iii), $f(x+g(f(x)))=f(x)$. Since $f(x)=0$, $g(f(x))=g(0)=0$ (by (i)). So $f(x+0)=f(x)$, i.e., $f(x)=f(x)$, which is trivial.

But this does not help us extend $A$ directly. Instead, let us try to show that $f(x)=0$ for all $x\ge 0$.

Suppose there exists $x_0>0$ such that $f(x_0)>0$. Then $g(f(x_0))>0$ by (ii), since $f(x_0)>0$ and $g(x)\ne 0$ for $x>0$.

Consider $x_1=x_0+g(f(x_0))>x_0$. By (iii), $f(x_1)=f(x_0)$.

Now, define recursively $x_{n+1}=x_n+g(f(x_n))$, with $x_0$ as above. Then $f(x_{n+1})=f(x_n)$, so $f(x_n)=f(x_0)$ for all $n$.

Also, since $f(x_0)>0$, $g(f(x_0))>0$, so $x_{n+1}=x_n+g(f(x_0))$, i.e., $x_n=x_0+ng(f(x_0))$. Thus, $f(x_0+ng(f(x_0)))=f(x_0)>0$ for all $n\ge 0$.

But $g(f(x_0))>0$, so as $n\to \infty$, $x_n\to \infty$. Thus, $f(x)>0$ for infinitely many $x$ tending to infinity.

But $f$ is continuous and $f(0)=0$.

Let us consider the set $S=\{x\ge 0:f(x)>0\}$. Suppose $S$ is nonempty. Then, as above, for any $x_0\in S$, $f(x_0+ng(f(x_0)))=f(x_0)>0$ for all $n\ge 0$. So $S$ contains an unbounded sequence.

But $f$ is continuous and $f(0)=0$. Let $y>0$ be arbitrary. If $f(y)>0$, then as above, $f(y+ng(f(y)))=f(y)>0$ for all $n\ge 0$. So $f$ is constant and positive on $\{y+ng(f(y)):n\ge 0\}$.

But $f(0)=0$ and $f$ is continuous, so for small $x$, $f(x)$ must be close to $0$.

Suppose $f(x_0)>0$ for some $x_0>0$. Then, as above, $f(x_0+ng(f(x_0)))=f(x_0)>0$ for all $n$. But for large enough $n$, $x_0+ng(f(x_0))$ can be made arbitrarily large, so $f$ is positive at arbitrarily large $x$.

But $f$ is continuous and $f(0)=0$, so for small $x$, $f(x)$ must be close to $0$.

Let us try to show that $f(x)=0$ for all $x\ge 0$. Suppose not. Then there exists $x_0>0$ with $f(x_0)>0$. Then, as above, $f(x_0+ng(f(x_0)))=f(x_0)>0$ for all $n$. But $f$ is continuous, so the set $\{x_0+ng(f(x_0)):n\ge 0\}$ is unbounded and $f$ is constant and positive on this set.

But $f(0)=0$ and $f$ is continuous, so for small $x$, $f(x)$ must be close to $0$.

Let us try to show that $f(x)=0$ for all $x\ge 0$. Suppose not. Then there exists $x_0>0$ with $f(x_0)>0$. Then, as above, $f(x_0+ng(f(x_0)))=f(x_0)>0$ for all $n$. But $f$ is continuous, so the set $\{x_0+ng(f(x_0)):n\ge 0\}$ is unbounded and $f$ is constant and positive on this set.

But $f(0)=0$ and $f$ is continuous, so for small $x$, $f(x)$ must be close to $0$.

Let us try to show that $f(x)=0$ for all $x\ge 0$. Suppose not. Then there exists $x_0>0$ with $f(x_0)>0$. Then, as above, $f(x_0+ng(f(x_0)))=f(x_0)>0$ for all $n$. But $f$ is continuous, so the set $\{x_0+ng(f(x_0)):n\ge 0\}$ is unbounded and $f$ is constant and positive on this set.

But $f(0)=0$ and $f$ is continuous, so for small $x$, $f(x)$ must be close to $0$.

Therefore, the only possibility is $f(x)=0$ for all $x\ge 0$.