Volterra operator ($(Af)(x)={\int }_0^{x}f(t)dt$)

The Volterra integration operator on $L^2([0,1])$ is compact and injective, but its inverse is differentiation and is unbounded. Its singular values $2/((2n-1)\pi )$ tend to zero, making recovery of a function from noisy indefinite-integral data an ill-posed inverse problem.

On $L^2(-a,a)$, the adjoint operator is $T^{\ast }g(s)={\int }_s^{a}g(t)dt$ for $s>0$, and $T^{\ast }g(s)=-{\int }_{-a}^{s}g(t)dt$ for $s<0$. These formulas follow by reversing the triangular integration order separately on the two half-intervals. Consequently $(T+T^{\ast })g$ is constant on each half, with values ${\int }_0^{a}g$ and $-{\int }_{-a}^{0}g$. Applying this adjoint to a real weight transfers a weighted integral of an anchored function to an integral against its derivative.

For nonnegative Lebesgue measurable $f$ on $(0,1)$, the displayed Volterra power-kernel integral is defined for every $0<x<1$ as a nonnegative extended Lebesgue integral. The factor is finite and continuous on the open integration interval $(0,x)$, and an endpoint value can be assigned arbitrarily. The result may initially be infinite; for example $f(y)=1/y$ makes every value infinite. At $\beta =1$, it reduces to the ordinary Volterra operator.

For joint integration, use the integral kernel $(x-y{)}^{\beta -1}$ on $0<y<x<1$ and zero elsewhere. This piecewise prescription avoids the singular diagonal and never raises a negative number to a noninteger power. The L1 bound for a Volterra power-kernel integral gives an integrable output, finite almost everywhere, whenever $f$ is integrable.

For nonnegative integrable $f$, Tonelli's theorem applied on the triangle $0<y<x<1$ gives the exact identityFor signed or complex integrable $f$, first apply the same calculation to $|f|$ and use the triangle inequality for an integral. The defining integral then exists absolutely for almost every $x$, and the same L1 norm bound holds. The constant $1/\beta$ is sharp: use $f_n=n{1}_{(0,1/n)}$, whose L1 norm is one and whose weighted integrals above tend to $1/\beta$.

Finiteness at every point is stronger than this L1 space assertion. For $0<\beta <1$, take $x_0=1/2$, $\gamma =(1+\beta )/2$, and $f(y)=(x_0-y{)}^{-\gamma }$ on $(0,x_0)$, zero otherwise. It is integrable because $\gamma <1$, but $(V_{\beta }f)(x_0)=\infty$ because the power near $y=x_0$ is $\beta -1-\gamma =(\beta -3)/2<-1$. This is compatible with an integrable output after changing values on a null set.

The Volterra integration operator $J$ has spectrum $\{0\}$ and is injective. Hence $(\lambda I-i{J}^{-1}{)}^{-1}=J(\lambda J-iI{)}^{-1}$ exists and is bounded for every complex $\lambda$. Empty spectrum is possible for closed unbounded operators, unlike bounded operators on a nonzero complex Banach space.

// Target: analysis.bigb

For the Volterra integration operator $Af(x)={\int }_0^{x}f(t)dt$, its adjoint is $A^{\ast }g(x)={\int }_x^{1}g(t)dt$. With $a_n=(n-\frac{1}{2})\pi$, the input singular vectors are $\sqrt{2}\cos (a_{n}x)$ and the output singular vectors $\sqrt{2}\sin (a_{n}x)$, with singular values $1/a_n$. The associated Sturm-Liouville problem has Neumann condition at zero and Dirichlet condition at one for the cosine basis; the sine basis has the reversed conditions. These complete orthonormal bases give a singular value system for an injective compact operator with dense range.

The Tikhonov regularization of the Volterra integration operator replaces the unbounded differentiation gain $a_n$ by $a_n/(1+\alpha a_n^2)$. The series converges in $L^2$ for every datum and $\alpha >0$. On the range of the Volterra integration operator, it converges to the weak derivative of the datum as $\alpha \downarrow 0$. No pointwise differentiation of arbitrary noisy L2 space data is required.

On $(0,1/2)$ use the forward difference and on $(1/2,1)$ the backward difference, with $0<h<1/2$. This defines a bounded linear operator on $L^2(0,1)$ and $\Vert D_h\Vert \le \sqrt{6}/h$. For $f$ in the range of the Volterra integration operator, these differences are local averages of $f^{\prime }$ and converge to $f^{\prime }$ in $L^2$ by continuity of translations. Hence $D_h$ gives a linear regularization of differentiation. No pointwise evaluation of an arbitrary $L^2$ equivalence class is needed: translated functions are defined almost everywhere.

For $f\in C^2[0,1]$ and $\Vert f^{\delta }-f{\Vert }_2\le \delta$, two translated half-interval integrals have overlap multiplicity at most two. The inequality $|a-b{|}^2\le 2|a{|}^2+2|b{|}^2$ gives $\Vert D_h\Vert \le \sqrt{6}/h$. Averaging $f^{\prime }$ over the relevant interval and applying $|f^{\prime }(x+t)-f^{\prime }(x)|\le \Vert f^{\prime \prime }{\Vert }_{\infty }|t|$ gives the second term. For this to estimate a Moore–Penrose inverse of an operator, also require $f(0)=0$; otherwise the derivative is still approximated but $K^{†}f$ is not defined.

For $Ku(y)={\int }_0^{y}u(x)dx$ on the Hilbert space $L^2(0,1)$, every image has a weak derivative $f^{\prime }=u$ and zero trace at zero. Conversely every such Sobolev space element is this integral of its weak derivative. The range is dense because it contains smooth functions supported in $(0,1)$, but it is not closed: a step function can be approximated in $L^2$ by continuous ramps and is not itself in $H^1$. The Moore–Penrose inverse of an operator has this range as its domain and differentiates there.