How to figure whether it is a compact operator How to figure whether it is a compact operator:
$$T:C[0,1]\rightarrow C[0,1] $$ $C[0,1]$:the space of all continous function on [0,1] with supremum norm
$$(Tx)(t)=\int^t_0 x(s)ds, \ \ \forall t\in[0,1]$$
Could you please help with this question.

Fix $\varepsilon>0$ and choose $\delta=\varepsilon>0$. Consider arbitrary $y\in T(B_{C([0,1])})$ and $t_1,t_2\in[0,1]$ such that $|t_2-t_1|<\delta$. Since $y\in T(B_{C([0,1])})$ there exist $x\in B_{C([0,1])}$ such that $T(x)=y$. Since $x\in B_{C([0,1])}$, then for all $s\in [0,1]$ we have $|x(s)|\leq\sup_{s\in[0,1]}|x(s)|=\Vert x\Vert\leq 1$. Now we have the following inequality
$$
|y(t_2)-y(t_1)|
=|T(x)(t_2)-T(x)(t_1)|
=\left|\int_{t_1}^{t_2} x(s)ds\right|
\leq\int_{t_1}^{t_2} |x(s)|ds
$$
$$
\leq\int_{t_1}^{t_2}ds
=t_2-t_1
<\delta
=\varepsilon
$$
Thus we proved that
$$
\forall \varepsilon>0\quad\exists\delta>0\quad\forall t_1,t_2\in[0,1]\quad\forall y\in T(B_{C([0,1])})\quad (|t_2-t_1|<\delta\implies|y(t_2)-y(t_1)|<\varepsilon)
$$
By Arzela-Ascoli theorem this means that $T(B_{C([0,1])})$ is precompact, i.e. $\operatorname{cl}_{C([0,1])} (T(B_{C([0,1])}))$ is compact. This means that $T$ is compact.
