Show the Volterra Operator is compact using only the definition of compact The Volterra operator $V:L^{2}[0,1]\rightarrow L^{2}[0,1]$ is defined by $(Vf)(x)=\int_0^xf(t)dt$.  
I am wondering if it can be shown that $V$ is compact by definition - that is, either that $V$ maps bounded sets to precompact sets, or equivalently, that for any bounded sequence $(f_n)$ in the domain, $\{Vf_n\}$ has a convergent subsequence.
I have seen an elegant proof of the compactness of $V$ using the Arzela-Ascoli theorem.  Also, I have come across proofs using the notion of Hilbert-Schmidt operators.  However, both of these notions were foreign to me when I was assigned this problem, and I am curious if I can show $V$ is compact without using these ideas; unfortunately, I am not sure how to proceed directly.
Thank you.
 A: To directly show that $V$ maps bounded sets to precompact sets, the only idea I have would be to replicate most of the proof of Ascoli's theorem.
We can show it semi-directly, though, by exposing $V$ as the norm-limit of operators with finite-dimensional range.
For $1 \leqslant k < n$, let
$$\lambda_{n,k}(f) = \int_0^{k/n} f(t)\,dt,$$
and
$$\chi_{n,k}(x) = \begin{cases} 1 &, \frac{k}{n} \leqslant x < \frac{k+1}{n}\\ 0 &, \text{ otherwise}. \end{cases}$$
Define
$$V_n(f) = \sum_{k=1}^{n-1} \lambda_{n,k}(f)\cdot \chi_{n,k}$$
for $n \geqslant 1$. Then $V_n$ is a continuous operator with finite-dimensional range. For $0 \leqslant k < n$ and $\frac{k}{n} \leqslant x < \frac{k+1}{n}$, we have
$$\lvert V(f)(x) - V_n(f)(x)\rvert = \left\lvert\int_{k/n}^{x} f(t)\,dt \right\rvert
\leqslant \sqrt{x-\frac{k}{n}}\cdot \lVert f\rVert_{L^2} \leqslant \frac{1}{\sqrt{n}}\lVert f\rVert_{L^2},$$
and hence $\lVert V - V_n\rVert \leqslant \frac{1}{\sqrt{n}}$. Thus $V$ is the norm-limit of operators with finite-dimensional rank, therefore compact.
