Q:

Showing an operator is compact

Can someone please explain to me how I can show the following operator is compact?
$$ Af(x)=\int_0^xf(t)dt$$
where $A:L_2[0,1]\rightarrow L_2[0,1]$.
There are two definitions of compactness that I've encountered:
Def1. An operator $L$ is compact if for a bounded sequence $\{ x_n\}$, there exists a convergent subsequence $\{ x_{n_k}\}_k$.
Def2. An operator $L$ is compact if it's continuous and maps a bounded set to a sequentially compact set.
What is the easier/standard method to prove compactness for integral operators?
I also know that "If $L$ is a bounded linear operator on a separable Hilbert space, with orthonormal basis $\{e_n\}$, and $\sum_n\| L e_n\|^2<\infty$, then $L$ is compact. "

A:

To show that $A$ is bounded, Cauchy-Schwarz works well
$$
    \|Af\|^2 = \int_0^1\left|\int_0^x f(t)dt\right|^2dx \\
    \le \int_0^1 \int_0^x|f(t)|^2dt\int_0^xdt\,dx \\
     = \int_0^1 \int_0^x|f(t)|^2dt xdx \\
     \le \int_0^1 \int_0^1 |f(t)|^2dt \, 1 dx = \|f\|^2.
$$
Then apply your theorem using the orthonormal basis $\{ e_n(t)= e^{2\pi int} \}_{n=-\infty}^{\infty}$:
$$
          Ae_n = \int_0^x e_n(t)dt = \frac{1}{2\pi in}(e_n(t)-1), \; n\ne 0, \\
            \|Ae_n\|\le \frac{1}{\pi|n|},\;\; n\ne 0.
$$

