Questions on a subset of $C[0,1]$ Problem: Let $$\mathcal{F} =\left\{F\in C[0,1]: F(x) = \int_0^x f(t) \, dt, \,\, f\in C[0,1],\, \, ||f||_\infty \leq 1\right\}. $$
Questions:
a) Show that $\mathcal{F}$ is not closed. 
b) Show that $\overline{\mathcal{F}} = \{f \in C[0,1]:f(0) = 0 , \, \, \text{Lip}(f)\leq 1\}$ (Lip represents Lipschitz constant).
I'm not quite sure how to start with these two questions. For part (a), I was trying to construct a sequence that does not converge in $\mathcal{F}$ but this seems to be difficult. For part (b), I was given a hint to find $F_n \in \mathcal{F}$ so that $F_n(\frac{k}{2^n}) = (1-\frac{1}{n})f(\frac{k}{2^n})$ but I'm not too sure how does that help with this question.
Any suggestion or hint for these two parts? Thank you.
 A: Hints: Let $g$ be any integrable function on $[0,1]$ which is not almost everywhere equal to any continuous such that $|g(x)| \leq 1$ for all $x$. (Eg: characteristic fucntion of $(0,\frac 1  2)$).  Let $f(x)=\int_0^{x} g(t)dt$. Then $|f(x)-f(y)|\leq |x-y|$. We can find  sequence $(g_n)$ of continuous functions  such that $|g_n(x)| \leq 1$ for all $n,x$ and $\int_0^{1} |g_n(x)-g(x)|dx \to 0$. Let $f_n(x)=\int_0^{x} g_n(t)dt$. Then each $f_n \in \mathcal F$ and $f_n \to f$ uniformly. Hence, $f$ belongs to the closure of $\mathcal F$ (w.r.t. the sup norm). But $f$ does not belong to $\mathcal F$.
For the second part let $f$ be any Lipschitz function with Lipschitz constant less than or equal to $1$ vanishing at $0$. Then $f$ is absolutely continuous, and we can write $f(x)=\int_0^{x} f'(t)dt$. Approximating $f'$ by continuous functions as above we see that $f$ belongs to the closure of $\mathcal F$.  The reverse inclusion is obvious.
A: For (a), if you want an explicit example: consider the sequence $$F_n(x) = \int^x_0 \frac{2}{\pi}\arctan(n(\tfrac 1 2-t))dt.$$ As $n \to \infty$, the integrand converges to the function which is $1$ on $[0,1/2)$ and $-1$ on $(1/2,1]$, so $F_n(x)$ converges to $F(x) = \frac 1 2 - \lvert x - \frac 1 2 \rvert$, which is not $C^1$, and hence cannot be written as the integral of a continuous function.
