I am working on the problem:
Let $\{ f_n \}$ be a sequence of real-valued functions on $[0,1]$ defined by $f_0 = f \in C[0,1]$ and $f_n$ is an anti-derivative of $f_{n-1}.$ Suppose that for each $x \in [0, 1]$ there is $n \in \mathbb{N}$ for which $f_n(x) = 0.$ Prove that $f = 0.$
Here is my attempt:
We first note that if $f_n$ is an anti-derivative of $f_{n-1}$, this implies that $f_{n-1}$ is a derivative of $f_n$. i.e. \begin{equation} f_{n-1} = \frac{d}{dx}(f_n). \end{equation} Case 1: If $n = 1$ such that $f_n = 0$ for all $x \in [0,1]$. Here if $n = 1$ then $f_1(x) = 0$ for all $x \in [0,1]$. Now, put $n = 1$ in equation (1). We get that $$f_{1-1} = \frac{d}{dx}(f_1(x)) \implies f_0 = 0 = f \forall x \in [0,1].$$ Case 2: If $n>1$ such that $f_n(x) = 0$ for all $x \in [0,1]$. Here if $n > 1$, from equation (1) we get that $$f_{n-1} = \frac{d}{dx}(f_n) \implies f_{n-1} = 0.$$ Now take the derivative of $f_{n-1}$. $$\frac{d}{dx}(f_{n-1}) = f_{n-2} \implies \frac{d^2}{dx^2}(f_n) = \frac{d}{dx}(f_{n-1}) = f_{n-2}.$$ This then implies $$\frac{d^n}{dx^n}(f_n) = f_{n-n} \forall x \in [0,1].$$ Hence $\frac{d^n}{dx^n}(0) = f_0 = f$ for all $x \in [0,1].$ Thus $f = 0$ for all $x \in [0,1].$ $\square$
Please let me know if I am on the right track or assist me in proving this correctly
It suffices to show that $f$ is zero on a dense subset, since $f$ is continuous.
Let $y\in [0,1]$ and $\epsilon >0$ be arbitrary. For each $n\in \mathbb N$, consider
$$ \mathscr O_n = \{ x\in [0,1]\cap [y-\epsilon,y+\epsilon] \ \ f_n(x) = 0\}.$$
Note that each $\mathscr O_n$ is closed since $f_n$ is continuous and by assumption, $$ \bigcup_{n\in \mathbb N} \mathscr O_n = [0,1]\cap [y-\epsilon,y+\epsilon].$$
This implies, by the Baire Category, that one of $\mathscr O_n$ is not nowhere dense. Thus $\mathscr O_n$ is dense in some non-empty interval $I_n$ in $[0,1]\cap [y-\epsilon, y+\epsilon]$. This implies $f_n$ is zero on $I_n$ and thus (by differentiating $n$ times) $f$ is zero on $I_n$. Since $y, \epsilon$ are arbitrary, $f$ is zero on a dense subset.