### Thinking about a notorious Putnam problem

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 $n$th 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.