Show that $\left\{x\in[0,1] \rightarrow F(x) = \int_{0}^{x} f(t)dt : f\in C([0,1])\text{ and }||f||_\infty \leq B\right\}$ is compact. I am trying to show that $\mathscr{F}$ is compact, where $$\mathscr{F} = \left\{F(x) = \int_{0}^{x} f(t)dt\mid x \in [0,1],f\in C([0,1])  \text{ and } ||f||_\infty \leq B\right\}$$
To do so, I was planning to prove this by showing that $\mathscr{F}$ is complete and totally bounded.
I know that $\mathscr{F}$ is pointwise bounded, and pointwise equicontinuous. And, furthermore, by the Ascoli Arzel Theorem, I know that its closure is compact. I am pretty lost on how to show $\mathscr{F}$ is compact.
Any help is appreciated.
 A: The set $\mathcal{F}$ is not closed, therefore it is not compact. It consists of differentiable functions. Let
$$f_n(x)=\begin{cases}
-2 &0\le x\le  {1\over 2}\\
4nx-2n-2 & {1\over 2}<x\le {1\over 2}+{1\over n}\\
2 & {1\over 2}+{1\over n}<x\le 1\end{cases}$$
Then $f_n$ is continuous and $|f_n(x)|\le 2.$
Moreover $$f_n(x)\nearrow f(x)=\begin{cases} -2 & 0\le x\le {1\over 2} \\ 2 & {1\over 2}<x\le 1
\end{cases}$$ The convergence is uniform on  $0\le x\le {1\over 2}$ and on ${1\over 2}+\delta\le x\le 1$ for any fixed $0<\delta<{1\over 2}.$
For $n<m$ we have
$$|F_n(x)-F_m(x)|=\left |\int\limits_{{1\over 2}+{1\over m}}^{{1\over 2}+{1\over n}}[f_n(t)-f_m(t)]\,dt\right | \\
\le \int\limits_{{1\over 2}+{1\over m}}^{{1\over 2}+{1\over n}}[|f_n(t)|+|f_m(t)|]\,dt \le 4\left ({1\over n}-{1\over m}\right )\le {4\over n}
$$
Therefore $F_n$ satisfies the Cauchy condition, hence is uniformly convergent. Let $F$ denote the uniform limit of $F_n.$
Then for any fixed $x\in [0,1]$ we get
$$F(x)=\lim_n F_n(x)=\lim_n\int\limits_0^xf_n(t)\,dt
=\int\limits_0^xf(t)\,dt\\
=\begin{cases}-2x & 0\le x\le {1\over 2}\\
2x-2 & {1\over 2}<x\le 1
\end{cases}$$
The function $F$ does not belong to $\mathcal{F}$ because it is not differentiable at $x={1\over 2}.$
