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.
 A: Sketch 


*

*You are to prove that the image of the unit ball $B \subset C[0,1]$ is precompact in $C[0,1]$. 

*Because of the fundamental theorem of calculus, you have a nice description of $T(B)$.

*Use said description to show that $T(B)$ is equicontinuous. 

*Apply Arzelà–Ascoli.
A: 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.
A: There are already two good answers but I would like to write another one, even though it does not contain any new information (it merely presents the given information differently):
To figure out whether this operator is compact you can apply the following version of the Arzelà-Ascoli theorem:
Let $X$ be a compact Hausdorff space and let $C(X)$ denote the space of continuous functions $f: X \to \mathbb R$ endowed with the sup norm $\|\cdot \|_\infty$. Then $S \subseteq C(X)$ is relatively compact if and only if it is pointwise bounded and equicontinuous.
$S$ is pointwise bounded if for every $x \in X$: $\sup_{f \in S}|f(x)|<\infty$. $S$ is equicontinuous if for every $\varepsilon > 0$ and every $x \in X$ there exists a $\delta_x$ such that $|x-y|<\delta_x$ implies $|f(x)-f(y)|<\varepsilon$ for all $f \in S$. A set $S$ is uniformly equicontinuous if for every $\varepsilon>0$ there exists $\delta$ such that $|x-y|<\delta$ implies that $|f(x)-f(y)|<\varepsilon$ for all $f \in S$. Note that uniform equicontinuity implies equicontinuity.
Now that we have set up the definitions and theorems we're good to go:
An operator $T$ is compact if $T(\overline{B(0,1)})$ is relatively compact (here $\overline{B(0,1)}$ is the closed unit ball). By Arzelà-Ascoli it is enough to show that $T(\overline{B(0,1)})$ is pointwise bounded and (uniformly) equicontinuous. 
To show pointwise boundedness of $S=T(\overline{B(0,1)})$ let $t \in [0,1]$. Then for all $y \in T(\overline{B(0,1)})$: 
$$ |y(t)|= \left |\int_0^t x(s) ds \right | \le \left |\int_0^t \|x\|_\infty ds\right | \le \int_0^t ds \le 1 $$
Hence also
$$ \sup_{y \in S}|y(t)| \le 1$$
hence  $S=T(\overline{B(0,1)})$ is pointwise bounded. 
We conclude by showing that $S$ is also uniformly equicontinuous. To this end, let $\varepsilon> 0$. Then for $t,t' \in [0,1]$ (without loss of generality assume $t \ge t'$) and $y \in S$ :
$$ \begin{align}|
y(t)-y(t')| &= \left | \int_0^t x(s) ds - \int_0^{t'}x(s) ds  \right | \\
&= \left | \int_{t'}^t x(s)  ds  \right | \le \int_{t'}^t |x(s)|  ds \le \int_{t'}^t \|x\|  ds \le |t-t'|\end{align}$$
Hence for $\delta = \varepsilon$, $ |y(t)-y(t')|<\varepsilon$ and we have shown that $S$ is uniformly equicontinuous. 
