How to show operator is compact

Question

Currently I'm self studying functional analysis, namely compact operators. In the text, the author gives the following example:

Example 1: Let $C_1$ and $C_2$ be positive constants and let $$ M:=\left\{x(t)\in C[a,b]:|x(t)|<C_1\text{ and }|x'(t)|<C_2\right\}.\tag{1} $$ Then $M$ is relatively compact.

I completely understand Example 1, it makes use of Arzela's theorem. Then the author gives the following example:

Example 2: The operator $$ Ax:=\int_0^tx(\tau)d\tau.\tag{2} $$ on $C[0,1]$ is a compact operator (use the previous example).

My question is fairly straightforward: using Example 1, how is this operator compact on $C[0,1]$?

As of now I have the following two definitions of compact operators:

Definition 1: A linear operator $A\colon X\to Y$ is called a compact operator if and only if for every bounded sequence $x_n\in X$ the sequence $Ax_n$ has a Cauchy subsequence.

Definition 2: An linear operator $A\colon X\to Y$ is compact if and only if the image of the unit ball of $X$ is a relatively compact set in $Y$.

Just going off terminology, I would guess that Definition 2 should be used to solve Exercise 2, but I'm not seeing it through all the way. Any help is appreciated!

Answer 1

I think you just need to notice that $|Ax|\leq \max_\limits{t\in[0,1]}x(t)$ (which comes immediately since $x$ is continous on a compact and the integral function is monotonic). Such maximum actually controlls also the derivative of $Ax$, thus if you take a set $M$ of functions of the form $Ax$ where $x\in B(0,1)$ you obtain a precompact space by example $1$ and with the second definition. By $B(0,1)$ I denoted the unit ball of center $0$ in $\mathcal{C}[0,1]$.