Proved that the given set is not closed in the function space $\mathcal{C}([0,1])$

Question

The problem is actually taken from Davidson's Real analysis:

Prove that the set $S= \{ F:F(x) = \int_0^x f(t)dt, ||f|| \leq 1 ,\, f\in \mathcal{C}([0,1])\}$ is not closed.

This means we should find a sequence $(F_n)_{n=1}^{\infty}$ in $S$ that converges uniformly to a function $F \not \in S$. Since every $F_n$ is continuous, the function $F$ should also be continuous, by uniform convergence. However, I have totally no idea how to construct such sequence. I'd like to get some hints. Thanks!

Answer 1

Note that, by the Fundamental Theorem of Calculus, a function $F$ is in $S$ if and only if ist is continuously differentiable, $F(0)=0$ and it satisfies $|F_0^\prime(x)|\le 1\,\forall x $.

So you should try to find a function $F_0$ with $F_0(0)=0$ which is continuous but not $C^1$, and a sequence of $C^1$ functions $F_n$ with $F_n(0)=0$ and $||F_n^\prime||\le 1$ converging uniformly to $F_0$. The most nasty task will be to ensure the bound on $F_n^\prime$