# An interesting problem

Recently in my Facebook feed, I sometimes get “problems of the week” from the Canadian Math Society with respect to the Canadian Open Mathematics Challenge. This week’s problem seemed quite fun, so I tried it. The question is stated in the link above.

The fact that $f(x)$ must be weakly increasing is obvious from the statement. Suppose that $0 \leq x < y \leq 4$. Then $0 \leq y-x \leq 4$, whence $f(y) \geq f(x) + f(y-x)$ from the hypothesis on $f$. Since $f$ is non-negative on $[0,4]$, the result follows.

The second conclusion is a bit more interesting. We first show that $f(1/2^k) \leq 1/2^{k+1}$ for all integers $k \geq -2$. The case $k = -2$ is from the hypothesis. From there, we see that $f(4) \geq 2f(2)$, whence $f(2) \leq 1$. The desired conclusion then follows from induction.

Now suppose that there exists a positive number $x \in [0,4]$ such that $f(x) > x$. Then we have $2 \geq f(4-x) + f(x) > x$, by the non-negativity of $f$ and our hypothesis on $x$. Hence, we have restricted our $x$ to the interval $[0,2)$. Iterating this argument, we find that $1 = f(2) \geq f(2-x) + f(x) > x$, so in fact $x \in [0,1)$. Iterating this and recalling the previous paragraphing, we find that $x < 1/2^k$ for all positive integers $k$, so that $x = 0$. But from the inequality $f(4) \geq f(4) + f(0)$ and the non-negativity of $f$, it follows that $f(0) = 0$, so this is a contradiction. Hence no such $x$ exists.