Thinking about a notorious Putnam problem
Consider the following Putnam question from the 2018 exam:
Consider a smooth function such that , and and . Prove that there exists a point and a positive integer such that .
This is a problem from the 2018 Putnam, and only 10 students were able to solve it completely, making it the hardest question on the exam. I spent a day thinking about it, and my “proof” differs a lot from the official solutions, and is really a heuristic.
Proof: Assume that there does not exist any and such that . We will compare with functions of the form in . We will prove that on . Because on as , we will have proven that on and . Hence, cannot be smooth.
Why is ? Let us first analyze what looks like. It is easy to see that for . This is because as , if for , when will have to decrease to at . Hence, there will be a negative derivative involved, which is a contradiction. Hence, for , and by continuity of derivatives for smooth functions, all derivatives at are also .
Now consider the functions , which are at and at . These are the same endpoints for in . If ever crosses in , then it will have a higher th derivative than at the point of intersection. As its th derivative is also non-negative, will just keep shooting above , and hence never “return” to at . This contradicts the fact that . Hence, will always be bounded above by in . As this is true for all , on and . This contradicts the fact that is continuous.