The power of powers

In this post we make use of Taylor polynomial expansions to efficiently solve one of 2018 Putnam competition problems.

Here is the problem statement.

Let $f:\Bbb R \to \Bbb R$ be an infinitely differentiable function satisfying $f(0) = 0$ and $f(1) = 1$, and $f(x) \geq 0$ for all $x \in \Bbb R$. Show that there exist a positive integer $n$ and a real number $x$ such that $f^{(n)}(x)< 0$.

We proceed by contradiction, thus assuming that $f^{(n)}(x) \geq 0$ for all real $x$ and all positive integers $n$.

1. Show that if there exists $c < 0$ such that $f(c)>0$, then the Mean Value Theorem implies the existence of a point $x \in (c,0)$ such that $f'(x) < 0$. Conclude that it must be $f(x) = 0$ for $x \leq 0$.
2. Use the above fact, and infinite differentiability, to conclude that $$f^{(n)}(0) = 0$$ for all $n \in \Bbb Z^+$.
3. Apply Taylor’s Theorem in the interval $[0,1]$ to show that $$f(1) = \frac{f^{(n)}(\eta_n)}{n!},$$ for some $\eta_n \in (0,1)$. Conclude that for each positive integer $n$ there is a point $\eta_n \in (0,1)$ such that $$f^{(n)}(\eta_n) = n!$$
4. Use 3. and the fact that, for all real $x$, $f^{(n+1)}(x) \geq 0$ to prove that $$f^{(n)}(1) \geq n!$$
5. Apply again Taylor’s Theorem, in the interval $[1,2]$ this time, to write $$f(2) = \sum_{k=0}^n \frac{f^{(k)}(1)}{k!} + \frac{f^{(n+1)}(\xi_n)}{(n+1)!}$$ where we defined $f^{(0)}(x) = f(x)$ and where $\xi_n \in (1,2)$. By 4. and our hypothesis, we get the contradiction $$f(2) \geq n, \ \ \ \forall n \in \Bbb Z^+.$$

You might have noticed that there is nothing special in the hypothesis $f(1) = 1$. As a further exercise, you could replace this hypothesis with the weaker requirement for $f$ to be non-constant. (Hint: show that there must exist $a>0$ such that $f(a) >0$. Use the auxiliary function $g(x) = \frac{f(ax)}{f(a)}$…)