I attempted a Putnam problem, and I would like some verification on my proof. I did not see it listed in the solutions. This is problem A5 from the 2018 Putnam.
Let $f : \Bbb R \to \Bbb R$ be an infinitely differentiable function satisfying $f(0) = 0$, $f(1) = 1$, 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 $f^{(n)}(x) < 0$.
My attempt. Let $f:\Bbb R \to \Bbb R$ be smooth, non-negative, and $f(0)=0$ and $f(1)=1$. We use a proof by contradiction. Suppose bwoc that for all positive integers $n$, for all real numbers $x$, $f^{(n)}(x)\geq 0$.
First we show $f(x)=0$ for all $x\leq 0$. Suppose bwoc there exists $c<0$ such that $f(c)>0$. By the mean value theorem, this implies $f'(a)=-f(c)/c$ for some $a\in(c,0)$. This contradicts $f^{(1)}(x)\geq 0$ for all $x$.
Observe that $f(x)=0$ for all $x\leq 0$ implies $f^{(k)}(x)=0$ for all $x<0$, for all $k\geq0$. Now let us write $f$ as a Taylor series centered at some point $c<0$. But this Taylor series is identically zero:
$$ f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(c)}{k!}(x-c)^k = 0$$
Since $f$ is smooth, it must be equal to its own Taylor series. Therefore $f(x)=0$ for all real numbers $x$. This contradicts $f(1)=1$, completing the proof.
I didn't see this proof listed in the solutions. Is this proof
It's not a proof. You've assumed that any infinitely differentiable function is equal to its Taylor series. That's only true of analytic functions, and there are functions that are infinitely differentiable (over $\Bbb R$) that are not analytic. A classic example is: $f(x) = 0 \text{ for } x \leq 0, f(x) = e^{-\frac 1x} \text{ for } x \gt 0$.