= Solution
The <Volterra integration operator> is injective and has dense range. Its <range of the Volterra integration operator> is $\{f\in H^1(0,1):f(0)=0\}$, and its <Moore–Penrose inverse of an operator> is differentiation on this domain. Thus the error estimate for $K^\dagger f$ implicitly requires $f(0)=0$, in addition to the printed $H^2$ assumption. For an arbitrary $H^2$ function the same calculation bounds error against $f'$, but that function need not be in $\mathcal D(K^\dagger)$.
Put $a=(1-\alpha)/2$, $b=(1+\alpha)/2$ and $h=\alpha/2$. The three stencils are forward, <central finite differences>, and backward on $[0,a)$, $[a,b)$ and $[b,1]$, respectively. All shifted evaluation points lie in $[0,1]$. For $g\in L^2(0,1)$, apply $|A-B|^2\leq2(|A|^2+|B|^2)$ and change variables in each of the six evaluation terms. The resulting integration intervals are
$$
[\alpha,b],\quad[0,a],\quad[1/2,1/2+\alpha],\quad[1/2-\alpha,1/2],\quad[b,1],\quad[a,1-\alpha].
$$
Pair $[0,a]$ with $[a,1-\alpha]$, $[\alpha,b]$ with $[b,1]$, and the two middle intervals. Each pair has disjoint interiors, so almost every point is counted at most three times. This <overlap multiplicity bound for piecewise difference operators> proves
$$
\|R_\alpha g\|_2^2\leq\frac6{\alpha^2}\|g\|_2^2,\qquad \|R_\alpha\|\leq\frac{\sqrt6}{\alpha}.
$$
For the bias on the middle interval, write the <central finite difference> as an average of $f'$. The <fundamental theorem of calculus> gives
$$
R_\alpha f(x)-f'(x)=\int_{-h}^h k_\alpha(s)f''(x+s)\,ds,\qquad k_\alpha(s)=\frac{\operatorname{sgn}(s)(h-|s|)}{\alpha},
$$
where $\|k_\alpha\|_1=\alpha/4$. Extend $f''$ by zero to the <real line>. <Young's convolution inequality> bounds the middle-interval $L^2$ error by $\alpha c/4$. Combining its squared error with the supplied outer-interval estimate gives the <central-difference noise-bias bound>
$$
\|R_\alpha f-f'\|_2\leq\sqrt{\alpha^2c^2+\alpha^2c^2/16}=\frac{\sqrt{17}}4\alpha c.
$$
The <triangle inequality> and the <operator norm> estimate now give
$$
\boxed{\|K^\dagger f-R_\alpha f^\delta\|_2\leq\frac{\sqrt6\,\delta}{\alpha}+\frac{\sqrt{17}}4\alpha c.}
$$
For $c>0$, <balancing noise and approximation bias> minimizes this upper bound at
$$
\boxed{\alpha(\delta)=2(6/17)^{1/4}\sqrt{\delta/c},\qquad \|K^\dagger f-R_{\alpha(\delta)}f^\delta\|_2\leq102^{1/4}\sqrt{\delta c},}
$$
for sufficiently small $\delta$ that $\alpha<1/2$. Capping the parameter at $1/4$ supplies an admissible rule for all positive noise levels. More generally, $\alpha\to0$ and $\delta/\alpha\to0$ suffice; $\alpha=\sqrt\delta$ avoids needing $c$ or dividing by $c=0$.
To prove convergence for every admissible datum, not just $H^2$ data, let $f=Ku$ with $u\in L^2(0,1)$ and extend $u$ by zero. Each stencil is a forward, centred or backward average of $u$. <Translation continuity in Lp> makes each averaged difference from $u$ tend to zero in $L^2(\mathbb R)$. Restricting those three bounds to their respective intervals proves $R_\alpha Ku\to u$. The <noise-bias decomposition for linear regularization>, with $\delta/\alpha\to0$, finishes the <convergent regularization of an inverse problem>.
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