Show the Volterra Operator is compact using only the definition of compact

Solution 1:

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.