A question about a possible solution to A5 from Putnam 2018

Question

Can the following function exist: a smooth function $f:\Bbb{R}\to\Bbb{R}$ satisfies the following condition: $f^{(n)}(0)=0$ for all $n$, but there exists a sequence of points $x_n\to 0$ such that $f^{(n)}(x_n)\to\infty$. This is in reference to problem A5 of Putnam 2018. If I can prove that such a function cannot exist, I will have proved the Putnam problem.

Answer 1

Take any $C^\infty$ function $f$ such that all $f^{(n)}(0) = 0$ but $f(x) > 0$ for $x > 0$; a standard example is

$$ f(x) = \cases{\exp(-1/x) & if $x \ge 0$\cr 0 & if $x \le 0$} $$

Note that for $x > 0$ and all positive integers $n$,

$$ f(x) = \int_0^x \frac{(x-s)^n}{n!} f^{(n+1)}(s)\; ds $$ as can be seen using integration by parts. Now $|x-s|^n/n! \le x^n/n! \to 0$ as $n \to \infty$, so there must be $x_n \in (0,x)$ with $f^{(n+1)}(x_n) \ge f(x) n!/x^n$.

For each $n$, take $y_n > 0$ so that $f(y_n) n!/y_n^n = n f(1)$. We have $y_n \to 0+$ as $n \to \infty$, because $f(x) n!/x^n \to \infty$ for any fixed $x$. Then taking $x = y_n$ in the previous paragraph, we get $x_n \in (0, y_n)$ with $f^{(n+1)}(x_n) \ge n f(1)$.