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.


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