Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function and $(f_n)_{n\geq 0}$ a sequence of functions such that $f_0=f$ and $f_{n+1}$ is an antiderivative (primitive) of $f_n$ for each $n\geq 0$, with the property that for each $x\in \mathbb{R}$ there exists $n\in \mathbb{N}$ such that $f_n(x)=0$. Prove that $f$ is identically 0.
Honestly, I do not have a viable starting point; but the problem looks very nice. Obviously, $f_n^{(n)}=f$ and one can use the general formula for the solutions of this differential equation of order $n$, but…
Best Answer
Assume that $f(x_0) \ne 0$ for some $x_0$. Then $f(x) \ne 0$ for all $x$ in an interval $[a, b]$ with $a < b$. Now $$ [a, b] = \bigcup_{n \in \Bbb N} \{ x \in [a, b]\mid f_n(x)=0\} $$ is the countable union of closed sets. The Baire category theorem implies that one of these sets has non-empty interior, i.e. $$ f_n(x) = 0 $$ for some $n$ and all $x$ in an open interval $I \subset [a, b]$. But then $f_n^{(n)}=f$ implies that $f$ is zero is on $I$, a contradiction.