Thinking about a notorious Putnam problem

by ayushkhaitan3437

Consider the following Putnam question from the 2018 exam:

Consider a smooth function f:\Bbb{R}\to\Bbb{R} such that f\geq 0, and f(0)=0 and f(1)=1. Prove that there exists a point x and a positive integer n such that f^{(n)}(x)<0.

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 x and n such that f^{(n)}(x)<0. We will compare f with functions of the form x^n in [0,1]. We will prove that f\leq x^n on [0,1]. Because x^n\to 0 on [0,1) as n\to\infty, we will have proven that f=0 on [0,1) and f(1)=1. Hence, f cannot be smooth.

Why is f\leq x^n? Let us first analyze what f looks like. It is easy to see that f(x)=0 for x\leq 0. This is because as f\geq 0, if f(x)>0 for x<0, when f will have to decrease to 0 at x=0. Hence, there will be a negative derivative involved, which is a contradiction. Hence, f(x)=0 for x\leq 0, and by continuity of derivatives for smooth functions, all derivatives at x=0 are also 0.

Now consider the functions x^n, which are 0 at x=0 and 1 at x=1. These are the same endpoints for f(x) in [0,1]. If f(x) ever crosses x^n in [0,1), then it will have a higher nth derivative than x^n at the point of intersection. As its (n+1)th derivative is also non-negative, f will just keep shooting above x^n, and hence never “return” to x^n at x=1. This contradicts the fact that f(1)=1. Hence, f will always be bounded above by x^n in [0,1]. As this is true for all n, f=0 on [0,1) and f(1)=1. This contradicts the fact that f is continuous.