Volterra operator

In mathematics, in the area of functional analysis and operator theory, the Volterra operator, named after Vito Volterra, is a bounded linear operator on the space L²[0,1] of complex-valued square-integrable functions on the interval [0,1]. On the subspace C[0,1] of continuous functions it represents indefinite integration. It is the operator corresponding to the Volterra integral equations.

The Volterra operator, V, may be defined for a function f ∈ L²[0,1] and a value t ∈ [0,1], as¹

${\displaystyle V(f)(t)=\int _{0}^{t}f(s)\,ds.}$

• V is a bounded linear operator between Hilbert spaces, with kernel form$${\displaystyle Vf(x)=\int _{0}^{1}1_{y\leq x}f(y)dy}$$ proven by exchanging the integral sign.
• V is a Hilbert–Schmidt operator with norm ${\displaystyle \|V\|_{HS}^{2}=1/2}$, hence in particular is compact.
• Its Hermitian adjoint has kernel form$${\displaystyle V^{*}(f)(x)=\int _{x}^{1}f(y)dy=\int _{0}^{1}1_{y\geq x}f(y)dy}$$
• The positive-definite integral operator ${\displaystyle K:=V^{*}V}$ has kernel form$${\displaystyle Kf(x)=\int _{0}^{1}\min(1-x,1-y)f(y)dy}$$proven by exchanging the integral sign. Similarly, ${\displaystyle VV^{*}}$ has kernel ${\displaystyle \min(x,y)}$. They are unitarily equivalent via ${\displaystyle Uf(x)=f(1-x)}$, so both have the same spectrum.
• The eigenfunctions of ${\displaystyle VV^{*}}$ satisfy ${\displaystyle {\begin{cases}f(1)&=0\\f'(0)&=0\\f''(x)&=-\lambda ^{-1}f\end{cases}}}$ with solution $${\displaystyle f(x)=\sin((k+1/2)\pi x),\lambda =\left({\frac {1}{(k+1/2)\pi }}\right)^{2}}$$with ${\displaystyle k=0,1,2,\dots }$.
• The singular values of V are ${\displaystyle ((k+1/2)\pi )^{-1}}$ with ${\displaystyle k=0,1,2,\dots }$.
• The operator norm of V is ${\displaystyle 2/\pi }$.
• V is not trace class.
• V has no eigenvalues and therefore, by the spectral theory of compact operators, its spectrum σ(V) = {0}.²³
• V is a quasinilpotent operator (that is, the spectral radius, ρ(V), is zero), but it is not nilpotent operator.