Problem 1751

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 $g(x) \neq 0$ 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., $x_n = x_0 + n g(f(x_0))$.
Thus, $f(x_0 + n g(f(x_0))) = f(x_0) > 0$ 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 $x_0 \in S$, $f(x_0 + n g(f(x_0))) = f(x_0) > 0$ 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, $f(y + n g(f(y))) = f(y) > 0$ for all $n \ge 0$.
So $f$ is constant and positive on $\{y + n g(f(y)) : n \ge 0\}$.

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, $f(x_0 + n g(f(x_0))) = f(x_0) > 0$ for all $n$.
But for large enough $n$, $x_0 + n g(f(x_0))$ 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, $f(x_0 + n g(f(x_0))) = f(x_0) > 0$ for all $n$.
But $f$ is continuous, so the set $\{x_0 + n g(f(x_0)) : n \ge 0\}$ 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, $f(x_0 + n g(f(x_0))) = f(x_0) > 0$ for all $n$.
But $f$ is continuous, so the set $\{x_0 + n g(f(x_0)) : n \ge 0\}$ 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, $f(x_0 + n g(f(x_0))) = f(x_0) > 0$ for all $n$.
But $f$ is continuous, so the set $\{x_0 + n g(f(x_0)) : n \ge 0\}$ 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 $x \ge 0$.