How to figure whether it is a compact operator

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.