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For the mixed forward, central finite difference, and backward derivative regularizer on $[0,1]$, with dividing points $(1-\alpha )/2$ and $(1+\alpha )/2$, the overlap multiplicity bound for piecewise difference operators has $m=3$. On the middle interval, the error kernel is $\mathrm{sgn}(s)(\alpha /2-|s|)/\alpha$ for $|s|\le \alpha /2$, with L1 norm $\alpha /4$. Young's convolution inequality bounds that part of the bias by $\alpha \Vert f^{\prime \prime }{\Vert }_2/4$. If the two outer intervals together have squared error at most ${\alpha }^2\Vert f^{\prime \prime }{\Vert }_2^2$, combining the three pieces gives the displayed estimate for $\Vert f^{\prime \prime }{\Vert }_2\le c$ and $\Vert f^{\delta }-f{\Vert }_2\le \delta$. Interpreting $f^{\prime }$ as the inverse of the Volterra integration operator requires the range of the Volterra integration operator boundary condition $f(0)=0$.

For fixed finite matrices, the matrix exponential and a Taylor expansion showThis agrees through degree two with $e^{h(A+B)}$, even when $A$ and $B$ do not commute. On a fixed time interval, $\Vert S(h)\Vert \le e^{h(\Vert A\Vert +\Vert B\Vert )}$ and the telescoping identityturns the $O(h^3)$ one-step defect into $O(N{h}^3)=O(h^2)$ global error for $Nh\le T$. Hence Strang splitting is second order in general; for commuting matrices it is exact.

For the spatial discretization, use $d=2/(M+1)$, $x_m=-1+md$ and the interior vector $U=(u_1,\dots ,u_M{)}^T$, with $u_0=u_{M+1}=0$. The central finite differences giveHere $\mathrm{tridiag}$ lists lower diagonal, main diagonal and upper diagonal, in that order. Thus $A^T=A$ and $B^T=-B$. Discrete summation by parts givesThe matrix exponential $e^{hA/2}$ is consequently a contraction in the Euclidean norm, while $e^{hB}$ is an orthogonal matrix for real $\alpha$. ThereforeRepeated split steps are contractive in the mesh-weighted norm $d\sum |u_m{|}^2$, uniformly in the spatial mesh and without a Courant–Friedrichs–Lewy condition. The split semidiscretization is unconditionally stable. No commutativity or shared eigenvectors are needed for this Strang splitting contraction for symmetric diffusion and skew advection. Its time-order proof is for fixed spatial matrices; as $d\to 0$, commutators can grow, so this stability result alone is not a uniform-in-mesh second-order error estimate.

Divide the second-order difference equation by $h^2$ and set $x=nh$. The first quotient is a central finite difference approximation to $y^{\prime \prime }(x)$; the quotient $(y_{n+1}-y_{n-1})/(2h)$ approximates $y^{\prime }(x)$. Their Taylor series errors are $O(h^2)$ for a sufficiently smooth function. This is therefore a finite difference method for the second-order linear differential equation in part (a).

At a fixed $x\ge 0$, take $h\to 0$ with $n=x/h$ integral. Since $\log r_{-}=-2h+O(h^3)$, the exact bounded discrete solution satisfiesLikewise $(1-2h{)}^{x/h}\to e^{-2x}$. The decay selected by boundedness in the linear recurrence relation becomes exactly the decay selected by boundedness in the ordinary differential equation.

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 $K^{†}f$ 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 $f^{\prime }$, but that function need not be in $\mathcal{D}(K^{†})$.

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 $|A-B{|}^2\le 2(|A{|}^2+|B{|}^2)$ and change variables in each of the six evaluation terms. The resulting integration intervals arePair $[0,a]$ 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 provesFor the bias on the middle interval, write the central finite difference as an average of $f^{\prime }$. The fundamental theorem of calculus giveswhere $\Vert k_{\alpha }{\Vert }_1=\alpha /4$. Extend $f^{\prime \prime }$ 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 boundThe triangle inequality and the operator norm estimate now giveFor $c>0$, balancing noise and approximation bias minimizes this upper bound atfor sufficiently small $\delta$ that $\alpha <1/2$. Capping the parameter at $1/4$ supplies an admissible rule for all positive noise levels. More generally, $\alpha \to 0$ and $\delta /\alpha \to 0$ 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 $\delta /\alpha \to 0$, finishes the convergent regularization of an inverse problem.

If the bound is to hold for all $f\in C^3[-1,1]$, the error functional must annihilate every polynomial of degree at most two, because such a polynomial has $f^{\prime \prime \prime }=0$. Applying it to $1,x,x^2$ givesThereforeso $\eta (f)=f(-1)-2f(0)+f(1)$ is the central finite difference approximation.