Examples of Compactness

A Volterra kernel is a kernel $K$ in $L^{2}\left(\mu \times \mu \right)$ (where $\mu$ is Lebesgue measure on the unit interval) such that $K(x,y) = 0$ when $x < y$; equivalently, a Volterra kernel is a Hilbert--Schmidt kernel that vanishes above the diagonal ($x = y$) of the unit square. The integral operator $A$ (a Volterra operator) induced by a Volterra kernel $K$ is given by

"Bounded" refers to the kernel, not the operator: the assumption is that $K$ is bounded almost everywhere in the unit square. An operator $A$ is quasinilpotent if $\lim_n \left\| A^n\right\|^{1/n} = 0$.

A Volterra kernel is a kernel $K$ in $L^{2}\left(\mu \times \mu \right)$ (where $\mu$ is Lebesgue measure on the unit interval) such that $K(x,y) = 0$ when $x < y$. The integral operator induced by a Volterra kernel $K$ is a Volterra operator, given by $(Af)(x) = \int_0^x K(x,y) f(y) \, dy$. An operator $A$ is quasinilpotent if $\lim_n \left\| A^n\right\|^{1/n} = 0$.

An integral operator $A$ induced by a Volterra kernel $K$ (a kernel $K \in L^{2}\left(\mu \times \mu \right)$ on the unit square with $K(x,y) = 0$ when $x < y$) is called a Volterra operator; $A$ is given by $(Af)(x) = \int_0^x K(x,y) f(y) \, dy$. The simplest non-trivial Volterra operator is the one whose kernel is the characteristic function of the triangle $\left\{ \left\langle x, y \right\rangle : 0 \leq y \leq x \leq 1\right\}$: this is the Volterra integration operator $V$, defined on $L^{2}\left((0,1)\right)$ by

In other words, $V$ is indefinite integration, with the constant of integration adjusted so that every function in the range of $V$ vanishes at $0$.

An integral operator induced by a Volterra kernel (a kernel $K \in L^{2}\left(\mu \times \mu \right)$ on the unit square with $K(x,y) = 0$ when $x < y$) is called a Volterra operator. The Volterra integration operator $V$ is the one defined on $L^{2}\left((0,1)\right)$ by $(Vf)(x) = \int_0^x f(y) \, dy$.

There is an operator $V_0$ on $L^{2}\left((-1,+1)\right)$ (Lebesgue measure) that bears a faint formal resemblance to $V$; by definition

Note that $V_0$ is the integral operator induced by the kernel that is the characteristic function of the butterfly $\left\{ \left\langle x, y \right\rangle : 0 \leq \left|y\right| \leq \left|x\right| \leq 1\right\}$.

The Volterra integration operator $V$ is the operator defined on $L^{2}\left((0,1)\right)$ by $(Vf)(x) = \int_0^x f(y) \, dy$.

A subspace is invariant under an operator if the operator maps it into itself; "non-trivial" means different from both $0$ and the whole space.

Suppose that $\left\{ \alpha_n\right\}$ is a monotone sequence ($\alpha_n \geq \alpha_{n+1}$, $n = 0, 1, 2, \ldots$) of positive numbers ($\alpha_n > 0$) such that $\sum_{n=0}^{\infty } \alpha_n^2 < \infty$. The unilateral weighted shift with the weight sequence $\left\{ \alpha_n\right\}$ is called a monotone $l^2$ shift. The span of the basis vectors $e_n, e_{n+1}, e_{n+2}, \ldots$ is invariant under such a shift, $n = 0, 1, 2, \ldots$. The orthogonal complement, i.e., the span $\mathbf{M}_n$ of $e_0, \ldots , e_{n-1}$, is invariant under the adjoint, $n = 1, 2, 3, \ldots$; the principal result is that every invariant subspace of that adjoint is one of these orthogonal complements.