I've seen a similar result in complex analysis, when an entire complex function satisfies $f(z)f^{(n)}(z)=0$ for all $z \in \mathbb{C}$ implies that $f(z)$ is polynomial.
What if when $f: \mathbb{R} \rightarrow \mathbb{R}$ is a $n$ times differentiable function such that $f(x)f^{(n)}(x)=0$ for all $x\in\mathbb{R}$, does it follow that $f$ is polynomial?
thanks.
what i have tried so far: I tried induction: for $n=1$ we have f(x)f'(x)=0 which means $f(x)^2=c$ hence f'(x)=0 for all $x$. Supposing the statement is true for $nā1$, I want to prove that $f^{(n)}(x)f(x)=0$ will implies f^{(n)}(x)f'(x)=0 which implies $f^{(n)}(x)=0$ (by induction hypothesis) .. my idea is to define function $g$ such that i could get f^{(n)}(x)f'(x)=0.. i'm stuck here.. I have another try, by analysing $\bigcup A_j$, where $A_j$ is the open interval on which $f(x)$ is not zero (it can be proved it is indeed interval). I also have tried taylor theorem.