Finite rank volterra operator I am wondering when a Volterra integral operator $V_K:L_2(0,1)\to L_2(0,1)$ is a finite rank operator:
$$V_Kf=\int_0^xK(x,y)f(y)dy$$
thanks in advance for your help
 A: The Volterra integral operator $V_K: L_2(0,1)\to L_2(0,1)$ which is given by
$$ (V_K f)(x)=\int_{0}^{x}K(x,y) f(y)dy\qquad (f\in L_2(0,1))$$
is of rank at most $n$ if and only if the kernel is of the form
$$ K(x,y)=g_1(x)\overline{h_1(y)}+\cdots+g_n(x)\overline{h_n(y)} \tag1$$
for some functions $g_j, h_j \in L_2(0,1)$ $(1\leq j \leq n)$ such that
$$ \chi_{[0,x]}(y)\bigl(g_1(x)\overline{h_1(y)}+\cdots+g_n(x)\overline{h_n(y)}\bigr)=g_1(x)\overline{h_1(y)}+\cdots+g_n(x)\overline{h_n(y)}, \tag2$$
i.e., the support of $g_1(x)\overline{h_1(y)}+\cdots+g_n(x)\overline{h_n(y)}$ has to be in the triangle $0\leq y \leq x \leq 1$.
For instance, let $a_1,\ldots, a_n \in (0,1)$ be arbitrary. Let, for each $j=1,\ldots, n$,
$$ g_j(x)=\left\{ \begin{array}{ccl}
0&; & 0\leq x<a_j\\
\gamma_{j}(x) &;& a_j\leq x\leq 1
\end{array} \right.\quad \text{and}\quad
h_j(y)=\left\{ \begin{array}{ccl}
\eta(y)&; & 0\leq y<a_j\\
0 &;& a_j\leq y\leq 1
\end{array} \right. \tag3$$
where $\gamma_j\in L_2(a_j,1)$ and $\eta_j\in L_2(0,a_j)$. Then these functions satisfy (2). Of course, the kernel $K$ can have several representations of the form (1) which satisfy (2), however at least one of them is (3), I guess.
