Spectral radius Volterra operator with an arbitrary kernel from $L^2$

Question

How to prove that the spectral radius $r \left( V_K \right)$ of a Volterra operator with an arbitrary kernel is zero.

$$V_K : L^2 \left[a,b \right] \rightarrow L^2 \left[a,b \right]$$ $$ (V_Kf)(x) = \int_a^x K \left(x,y \right) f(y) dy, $$ where $K\left(x,y \right) \in L^2 \left( \left[ a,b \right] \times \left[ a,b \right] \right)$

I know how to prove this if the kernel is bounded, but how to act in an arbitrary case?

Thank you!

Answer 1

$$ |V_K^nf|^2= \\ = \left|\int_{a}^{x}K(x,x_{n-1})\cdots\int_{a}^{x_2}K(x_2,x_1)\int_{a}^{x_1}K(x_1,x_0)f(x_0)dx_0 dx_1\cdots dx_{n-1}\right|^2 \\ \le \left[\int_{a}^{x}\cdots\int_{a}^{x_2}\int_{a}^{x_1}|K(x,x_{n-1})\cdots K(x_2,x_1)K(x_1,x_0)||f(x_0)|dx_0dx_1\cdots dx_{n-1}\right]^{2} \\ \le\left(\int_a^x\cdots\int_a^{x_1}|K(x,x_{n-1})\cdots K(x_1,x_0)|^2dx_0dx_1\cdots dx_{n-1}\right) \\ \times \left(\int_{a}^{x}\cdots\int_a^{x_1}|f(x_0)|^2dx_0dx_1\cdots dx_{n-1}\right) \\ \le \|f\|_{L^2}^2 \int_a^x\cdots\int_{a}^{x_3}\int_a^{x_2}dx_1dx_2\cdots dx_{n-1} \\ =\|f\|_{L^2}^2\frac{(x-a)^{n-1}}{(n-1)!} $$ Therefore, after integrating in $x$ on $[a,b]$, one obtains $$ \|V_K^n\| \le \sqrt{\frac{(b-a)^{n-1}}{(n-1)!}} $$ That's enough to give an infinite radius of convergence for the exterior resolvent expansion.