Show that $T$ is a compact operator

Question

This is an old preliminary exam problem: Let $T:C([0,1])\rightarrow C([0,1])$ be defined by $Tf(x)=\int_0^x f(t)dt$. Prove that $T$ is a compact operator, i.e. the image of the unit ball is pre compact.

So, if I understand correctly, I need to show that the image of $B:=\{f\in C([0,1]) | \sup_{x\in[0,1]}|f(x)|<1\}$ has compact closure. Here is my attempt:

Note that $TB=\{f\in C[0,1]| \sup_{x\in[0,1]} \int_0^x f(t)dt<1\}$

My first thought is to rewrite this as $TB=\{F'(x) | \sup_{x\in[0,1]} F(x)<1\}$

I think this is justified because if $F(x)=\int_0^x f(t)dt$ then $F'(x)=f(x)$ (since w.l.o.g. $F(0)=0$) for almost all $x$.

So the next step is to figure out what the closure of $TB$ looks like, but this is where I'm stuck. Any help would be appreciated. Thanks!

Update: I understand, from the answer given by user251257 that $TB$ is uniformly bounded and equicontinuous, so that, by Arzela Ascoli, any sequence in $TB$ has a uniformly convergent subsequence. But I still don't understand why this shows that the closure of $TB$ is compact. In order for $\overline{TB}$ to be compact, don't we need for any sequence in $\overline{TB}$ to have a convergent subsequence? What if we take a sequence in $\overline{TB}\backslash TB$?

Answer 1

Hints:

1. $TB = \{ g\in C^1[0,1] \mid \|g'\| < 1, g(0) = 0 \}$
2. $TB$ is uniformly Lipschitz continuous by the Lipschitz constant $1$, in particular it is equicontinuous.
3. $TB$ is uniformly bounded, as for $g\in TB$ we have $$ |g(x)| \le \int_0^x |g'(t)| dt \le x \le 1. $$
4. By Arzela Ascoli theorem $TB$ is pre compact.
5. For $g_k \in \overline{TB}$ there exists a $h_k\in TB$ with $\|g_k-h_k\| < 1/k$. Then, $h_k$ has a convergent subsequence and the corresponding subsequence of $g_k$ has the same limit.