Is the following solution correct? Show that $f(x)=0$ for all real numbers $x$.

Question

I have a calculus 1 exam soon, and I just read a solution to a problem from last year's exam. I'm pretty sure the solution is almost correct, but not correct. Here's the problem:
$f:\mathbb{R} \rightarrow \mathbb{R}$ is a function. There is a polynomial $P(x)=a_n \cdot x^n+...+a_0$, $n$ is odd, $a_n \neq 0$, and $P$ satisfies that for any natural number $n$ and a real number $x$, $|f^{(n)}(x)| \leq |P(x)|$.
Show that for all real numbers $x$, $f(x)=0$. Hint: Show that there exists $x_0$ such that $f^{(n)}(x_0)=0$ for all values of $n$.
Here's the (wrong) solution: First we'll prove the hint: WLOG $a_n>0$, and then $\lim_{x \to \infty} P(x) = \infty$, and $\lim_{x \to -\infty} P(x) = -\infty$, so since $P(x)$ is continuous there exists $x_0$ such that $P(x_0)=0$. Now it is easy to show that $x_0$ is the needed point in the hint. Now, look at Taylor Polynomial of order $n$ of $f$ around $x_0$, call it $P_n(x)$. It holds that $P_n(x)=0$, So the remainder $R_n(x)$ is just $f(x)$. Take a real number $x$. We'll show that $f(x)=0$ According to lagrange remainder theorem, there exists $c_x$ between $x$ and $x_0$ such that $f(x)=R_n(x)=\frac{f^{(n+1)}(c_x)}{(n+1)!} \cdot(x-x_0)^{n+1}$. Now, $|f^{n+1}(c_x)| \leq |P(c_x)|$, and that means that $|R_n(x)| \leq |P(c_x)| \cdot \frac{(x-x_0)^{n+1}}{(n+1)!}$. Since $P(c_x)$ is constant, $\lim_{n \to \infty}|R_n(x)|=0$, but $\lim_{n \to \infty}|R_n(x)|=f(x)$, so $f(x)=0$.
I think that the wrong part here is the use of the wrong fact that $P(c_x)$ is a constant, because $c_x$ also depends on $n$, so it needs to be $c_{n,x}$. Of course this can be easily fixed because since $c_{n,x}$ is between $x$ and $x_0$ for all $n$, it's easy to find a constant $M$ such that $P(c_{n,x})<M$ for all $n$. I just want to make sure that I understand it. Am I right, and the solution is not precisely accurate?Thank you.

Answer 1

I don't think the claim is correct unless you know more about $f$, like it being analytic. Here is a counterexample. Let $P(x)=x$. Then let $$f(x)=\begin {cases} 0& |x=0 \\ e^{-1/x^2} &x\neq 0\end {cases}$$
A plot is below. Blue is $|P(x)|=|x|$ and orange is $f(x)$. $f(x)$ is a standard example of a function that has all derivatives at every point, all derivatives are $0$ at $x=0$, but is not constant. I have not checked that all derivatives are less than $|x|$ at all points, but I believe it to be true.