= Solution
For the <Volterra operator>, the identity $K^\dagger f=f'$ 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 $\mathcal D(K^\dagger)$. 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 $|a-b|^2\leq2|a|^2+2|b|^2$ separately to the two halves defining <differentiation by one-sided difference quotients>:
$$
\|D_he\|_2^2\leq\frac2{h^2}\left[\|e\|_2^2+\int_h^{1/2+h}|e(t)|^2dt+\int_{1/2-h}^{1-h}|e(t)|^2dt\right]\leq\frac6{h^2}\|e\|_2^2.
$$
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 $h^{-1}\int_0^h f'(x+t)dt$ and the backward difference as $h^{-1}\int_0^h f'(x-t)dt$. If $M=\|f''\|_\infty$, either differs from $f'(x)$ by at most $h^{-1}\int_0^h Mt\,dt=Mh/2$. 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>:
$$
\boxed{\|D_hf^\delta-f'\|_2\leq\frac{\sqrt6\delta}{h}+\frac M2h.}
$$
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