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 $\overline{TB}\backslash TB$?