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For the Volterra operator, the identity $K^{†}f=f^{\prime }$ requires $f(0)=0$ in addition to the printed $C^2$ hypothesis. The constant function $f=1$ is a counterexample to the unrestricted domain assertion: it is not in $\mathcal{D}(K^{†})$. The estimate below is valid for derivatives of every $C^2$ function, and estimates the Moore–Penrose reconstruction when this missing boundary condition holds.

Let $e=f^{\delta }-f$. Apply $|a-b{|}^2\le 2|a{|}^2+2|b{|}^2$ separately to the two halves defining differentiation by one-sided difference quotients:The two translated intervals cover each point at most twice; this proves the factor six. The shifts act on almost-everywhere equivalence classes, so no undefined pointwise data sampling is involved.

For exact data, write the forward difference as $h^{-1}{\int }_0^h{f}^{\prime }(x+t)dt$ and the backward difference as $h^{-1}{\int }_0^h{f}^{\prime }(x-t)dt$. If $M=\Vert f^{\prime \prime }{\Vert }_{\infty }$, either differs from $f^{\prime }(x)$ by at most $h^{-1}{\int }_0^{h}Mtdt=Mh/2$. The interval has length one, so this also bounds the $L^2$ approximation error. The triangle inequality proves the error bound for one-sided differentiation:

For $\delta >0$ and $M=\Vert f^{\prime \prime }{\Vert }_{\infty }>0$, differentiate the bound $F(h)=\sqrt{6}\delta /h+Mh/2$. Its derivative vanishes at $h_{\ast }^2=2\sqrt{6}\delta /M$, and $F^{\prime \prime }(h)>0$. Consequently, when $h_{\ast }<1/2$,If $h_{\ast }\ge 1/2$, the infimum on the open allowed interval is approached as $h\uparrow 1/2$, rather than attained at an admissible endpoint. If $M=0$ and $\delta >0$, the bound decreases throughout the interval and has the same boundary behavior. At zero noise with $M>0$, its infimum is approached as $h\downarrow 0$.

A practical regularization parameter choice independent of $M$ isIt satisfies $h\to 0$ and $\delta /h\to 0$, hence gives a convergent regularization of an inverse problem on the domain of the Volterra operator inverse. This extends beyond $C^2$ exact data: if $f=Ku$ with $u\in L^2$, the differences are local averages of $u$. Extend $u$ by zero outside the interval; continuity of translations in $L^2$ makes these averages converge to $u$ on both halves. Together with $\Vert D_h(f^{\delta }-f){\Vert }_2\le \sqrt{6}\delta /h$ this proves convergence uniformly over the noise ball for every admissible exact datum.

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.