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 must be weakly increasing is obvious from the statement. Suppose that . Then , whence from the hypothesis on . Since is non-negative on , the result follows.
The second conclusion is a bit more interesting. We first show that for all integers . The case is from the hypothesis. From there, we see that , whence . The desired conclusion then follows from induction.
Now suppose that there exists a positive number such that . Then we have , by the non-negativity of and our hypothesis on . Hence, we have restricted our to the interval . Iterating this argument, we find that , so in fact . Iterating this and recalling the previous paragraphing, we find that for all positive integers $k$, so that . But from the inequality and the non-negativity of , it follows that , so this is a contradiction. Hence no such exists.