Finite rank volterra operator

Question

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

Answer 1

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.