Difference-quotient calculus: integration by parts, product rule, commutation

The map $T(x)=x+h{e}_i$ is a bijection from ${\mathrm{\Omega }}_{i,h}$ onto ${\mathrm{\Omega }}_{i,-h}$: if $x\in {\mathrm{\Omega }}_{i,h}$ then $y:=x+h{e}_i$ satisfies $y\in \mathrm{\Omega }$ and $y-h{e}_i=x\in \mathrm{\Omega }$, so $y\in {\mathrm{\Omega }}_{i,-h}$; conversely, if $y\in {\mathrm{\Omega }}_{i,-h}$ then $x:=y-h{e}_i$ satisfies $x\in \mathrm{\Omega }$ and $x+h{e}_i=y\in \mathrm{\Omega }$, so $x\in {\mathrm{\Omega }}_{i,h}$, and the two passages are inverse to each other. Hence by [F2], for every $f\in L^1({\mathrm{\Omega }}_{i,-h})$, ${\int }_{{\mathrm{\Omega }}_{i,h}}f(x+h{e}_i)dx={\int }_{{\mathrm{\Omega }}_{i,-h}}f(y)dy\mathrm{.}$

For $x\in {\mathrm{\Omega }}_{i,h}$ write $u_{+}=u(x+h{e}_i)$ and $v_{+}=v(x+h{e}_i)$. The algebraic identity $u_{+}{v}_{+}-uv=(u_{+}-u)v+u_{+}(v_{+}-v)=(u_{+}-u)v_{+}+u(v_{+}-v)$ holds pointwise at every $x$ where the four values are finite, hence almost everywhere on ${\mathrm{\Omega }}_{i,h}$; dividing by $h$ and reading $u_{+}=({\tau }_{-h{e}_i}u)(x)$ and $v_{+}=({\tau }_{-h{e}_i}v)(x)$ gives both displayed forms of the product rule (ii).

Applying step 1.1 to $f:=u{\tau }_{h{e}_i}v$, whose class lies in $L^1({\mathrm{\Omega }}_{i,-h})$ under the absolute-convergence hypothesis, gives ${\int }_{{\mathrm{\Omega }}_{i,h}}u(x+h{e}_i)v(x)dx={\int }_{{\mathrm{\Omega }}_{i,-h}}u(y)v(y-h{e}_i)dy={\int }_{{\mathrm{\Omega }}_{i,-h}}u({\tau }_{h{e}_i}v)dy,$ because $f(x+h{e}_i)=u(x+h{e}_i)v(x)$ and $f(y)=u(y)v(y-h{e}_i)$. Since ${\delta }_h^{i}u=(u_{+}-u)\mathrm{/}h$ and ${\delta }_{-h}^{i}v=(v-{\tau }_{h{e}_i}v)\mathrm{/}h$, and ${\delta }_h^{i}uv=(u_{+}v-uv)\mathrm{/}h$, splitting the first integral and using the displayed identity for its translated part yields the two-domain formula ${\int }_{{\mathrm{\Omega }}_{i,h}}{\delta }_h^{i}uvdx=\frac{1}{h}({\int }_{{\mathrm{\Omega }}_{i,-h}}u{\tau }_{h{e}_i}vdy-{\int }_{{\mathrm{\Omega }}_{i,h}}uvdx)=-{\int }_{{\mathrm{\Omega }}_{i,-h}}u{\delta }_{-h}^{i}vdx+\frac{1}{h}({\int }_{{\mathrm{\Omega }}_{i,-h}}uvdx-{\int }_{{\mathrm{\Omega }}_{i,h}}uvdx)\mathrm{.}$

Translation commutes with weak differentiation. Let $w\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathrm{\Omega })$ have $D^{\alpha }w\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathrm{\Omega })$. For $\phi \in C_c^{\mathrm{\infty }}({\mathrm{\Omega }}_{i,h})$ the function $\phi (\cdot -h{e}_i)$ lies in $C_c^{\mathrm{\infty }}({\mathrm{\Omega }}_{i,-h})$, since $\mathrm{supp}\phi +h{e}_i\subseteq {\mathrm{\Omega }}_{i,-h}$; step 1.1 applied to the integrable functions $w{\tau }_{h{e}_i}\phi$ and $D^{\alpha }w{\tau }_{h{e}_i}\phi$ gives ${\int }_{{\mathrm{\Omega }}_{i,h}}w(x+h{e}_i)D^{\alpha }\phi (x)dx={\int }_{{\mathrm{\Omega }}_{i,-h}}w(y)D^{\alpha }(\phi (\cdot -h{e}_i))(y)dy=(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}{\int }_{{\mathrm{\Omega }}_{i,-h}}{D}^{\alpha }w(y)\phi (y-h{e}_i)dy=(-1{)}^{\mathrm{\mid }\alpha \mathrm{\mid }}{\int }_{{\mathrm{\Omega }}_{i,h}}\left(D^{\alpha }w\right)(x+h{e}_i)\phi (x)dx,$ by [F4] applied on the open set ${\mathrm{\Omega }}_{i,-h}$, whose test function $\phi (\cdot -h{e}_i)$ is compactly supported there. Hence $D^{\alpha }({\tau }_{-h{e}_i}w)={\tau }_{-h{e}_i}(D^{\alpha }w)$ weakly on ${\mathrm{\Omega }}_{i,h}$.

The correction term in step 2.1 vanishes whenever ${\int }_{{\mathrm{\Omega }}_{i,h}}uv={\int }_{{\mathrm{\Omega }}_{i,-h}}uv$. If $uv=0$ a.e. outside ${\mathrm{\Omega }}_{i,h}\cap {\mathrm{\Omega }}_{i,-h}$, then both integrals equal ${\int }_{{\mathrm{\Omega }}_{i,h}\cap {\mathrm{\Omega }}_{i,-h}}uv$, so this holds; and if $uv$ has compact support in $\mathrm{\Omega }$ with $\mathrm{\mid }h\mathrm{\mid }<\mathrm{dist}(\mathrm{supp}(uv),\mathrm{\partial }\mathrm{\Omega })$, then every $x\in \mathrm{supp}(uv)$ satisfies $\mathrm{dist}(x,\mathrm{\partial }\mathrm{\Omega })>\mathrm{\mid }h\mathrm{\mid }$, hence $x+h{e}_i\in \mathrm{\Omega }$ and $x-h{e}_i\in \mathrm{\Omega }$, so $\mathrm{supp}(uv)\subseteq {\mathrm{\Omega }}_{i,h}\cap {\mathrm{\Omega }}_{i,-h}$ and again both integrals equal ${\int }_{\mathrm{\Omega }}uv$. If ${\mathrm{\Omega }}_{i,h}={\mathrm{\Omega }}_{i,-h}$ the two integrals are literally the same. For $\mathrm{\Omega }={\mathbb{R}}^n$ one has ${\mathrm{\Omega }}_{i,h}={\mathrm{\Omega }}_{i,-h}={\mathbb{R}}^n$. If $1\le p<\mathrm{\infty }$, translation invariance from step 1.1 applied to $\mathrm{\mid }u{\mathrm{\mid }}^p$ gives $\mathrm{\parallel }u(\cdot +h{e}_i){\mathrm{\parallel }}_{L^p}=\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p}$ and hence $\mathrm{\parallel }{\delta }_h^{i}u{\mathrm{\parallel }}_{L^p}\le 2\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p}\mathrm{/}\mathrm{\mid }h\mathrm{\mid }$. If $p=\mathrm{\infty }$, the measure-preserving translation in [F2] preserves null sets: applying its change-of-variables identity to indicators of null superlevel sets shows $\mathrm{\parallel }u(\cdot +h{e}_i){\mathrm{\parallel }}_{L^{\mathrm{\infty }}}=\mathrm{\parallel }u{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}$, and the triangle inequality gives the same difference-quotient bound in $L^{\mathrm{\infty }}$. In either case, Hölder's inequality [F3] makes both sides of the identity in step 2.1 absolutely convergent for $u\in L^p$ and $v\in L^{p^{\mathrm{\prime }}}$.

By step 2.2 applied to $w:=u$, and by linearity of weak differentiation [F4], $D^{\alpha }({\delta }_h^{i}u)=\frac{1}{h}(D^{\alpha }({\tau }_{-h{e}_i}u)-D^{\alpha }u)=\frac{1}{h}({\tau }_{-h{e}_i}(D^{\alpha }u)-D^{\alpha }u)={\delta }_h^i(D^{\alpha }u)\text{weakly on }{\mathrm{\Omega }}_{i,h}\mathrm{.}$ Together with steps 2.1 and 3.1 and the product rule of step 1.2 this proves (i)-(iii).