Problem Statement
Suppose $f:\mathbb{R}\to\mathbb{R}$ is a smooth function (infinitely differentiable) where $f(x)\geq 0$ for all $x\in\mathbb{R}$, $f(0)=0$, and $f(1)=1$. Show that there is some $n\in\mathbb{N}$ and an $x_0\in\mathbb{R}$ such that $f^{(n)}(x_0)<0$.
Proof Attempt
We can consider two cases:
Case (1): There exists a $z<0$ such that $f(z)>0$. By the MVT, there exists a point $z_0\in(z,0)$ such that $$f'(z_0)=\frac{f(z)-f(0)}{z-0}=\frac{f(z)}{z}<0 $$ so we are done.
Case (2): For all $x<0$, $f(x)=0$. In this case, $f'(x)=0$ for all $x<0$, which implies that, by left sided derivatives and the fact that $f'$ is smooth, $f'(0)=0$. It then follows that, since $f'(x)=0$ on $(-\infty,0]$, we have $f^{(n)}(x)=0$ for all $x\in(-\infty,0]$.
For $n=1$, define $z_1$ as the real number in $(0,1)$ such that $$f'(z_1)=\frac{f(1)-f(0)}{1-0}=1$$For $n\in\mathbb{N}$, define $z_{n+1}$ by the real number in $(0,z_n)$ such that $$f^{(n+1)}(z_{n+1})=\frac{f^{(n)}(z_n)-f^{(n)}(0)}{z_n-0}=\frac{f^{(n)}(z_{n})}{z_n}>0$$ whose existence is given by the mean value theorem. This defines the sequence $(z_n)$, which converges to $0$ and is strictly decreasing.
Edit 2: Continuing, we see that by definition, we have $$f^{(n+1)}(z_{n+1})=\frac{f^{(n)}(z_n)}{z_n}>f^{(n)}(z_n)$$ for any $n\in\mathbb{N}$ with $n\geq 2$, since $0<z_n<1$, thus the sequence $f^{(n)}(z_n)$ is strictly increasing and always greater than or equal to $1$. Now, $f^{(n)}(0)=0$, but since $(z_n)$ can get arbitrarily close to $0$ and $f^{(n)}(z_n)\geq 1$ for all $n\in\mathbb{N}$, we have $f^{(n)}(0)=0<1\leq f^{(n)}(z_n)$ for large enough $n$.
This is has almost shown that $f$ is not differentiable for some value of $n$, but I am not quite sure that it has exactly shown that. If it has, then we are done, as Case (2) cannot happen. I would like help either formalizing this or fixing this if it is not true.
This is where my argument, to me, seems to break down. Intuitively, the slope of the secant lines between $f^{(n)}(z_n)$ and $f^{(n)}(0)$ is increasing without bound as $n$ increases. My idea was to create a sequence of points which converged to $0$, but broke the differentiability at $0$. I would greatly appreciate any tips in proceeding, critiques of my proof outline, or alternative proof methods. Thank you in advance.
Edit 1: In Case (2), I am not explicitly making any assumptions that would lead to a contradiction, but why is this proof going in that direction? Either Case (2) cannot happen, which I find intuitively true, but my intuition often fails me, or one of my assumptions in the beginning of Case (2) is false.
Edit: Just to close this off, one of my faulty assumptions was that I assumed the sequence I constructed converged to $0$, which it doesn't have to.
This is actually a very hard question. So hard, that it appeared in the 2018 Putnam Exam. There are lots of solutions to this and I do not know which one suits your knowledge best. So just choose.