# Sequence of antiderivatives of a continuous function Let f : <mrow class="MJX-TeXAtom-ORD

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