Differentiation by one-sided difference quotients ($D_{h}f(x)=\frac{f(x+h)-f(x)}{h}\text{or}\frac{f(x)-f(x-h)}{h}$)

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.