Periodic integration by parts on a flat torus

Fix $j\in \{1,\dots ,d\}$ and $h\in C_{\mathrm{p}\mathrm{e}\mathrm{r}}^1({\mathbb{T}}^d;\mathbb{R})$. Put ${\mathrm{\ell }}_n:=2\pi \mathrm{/}n$ and $x_k:={\mathrm{\ell }}_{n}k$ for $k\in \{0,\dots ,n-1{\}}^d$. The uniform-grid Riemann sum $S_n:={\mathrm{\ell }}_n^d{\sum }_k{\mathrm{\partial }}_{j}h(x_k)$ converges to ${\int }_{{\mathbb{T}}^d}{\mathrm{\partial }}_{j}hdx$: if ${\omega }_j(\delta )$ is the supremum oscillation of ${\mathrm{\partial }}_{j}h$ between points of $Q_d$ at distance at most $\delta$, the upper-minus-lower Darboux sum on this grid is at most $(2\pi {)}^d{\omega }_j(\sqrt{d}{\mathrm{\ell }}_n)$, which tends to $0$ by [F1], and $S_n$ lies between those Darboux sums.

For each grid point $x_k$, apply [F2] to the one-variable restriction of $h$ along the $j$th coordinate segment of length ${\mathrm{\ell }}_n$. Its difference quotient differs from ${\mathrm{\partial }}_{j}h(x_k)$ by at most ${\omega }_j({\mathrm{\ell }}_n)$. The sum of these difference quotients over each complete $j$-cycle telescopes to $[h(\dots ,2\pi ,\dots )-h(\dots ,0,\dots )]\mathrm{/}{\mathrm{\ell }}_n=0$ by periodicity; when $d=1$ the remaining-coordinate index is the single empty tuple. Hence $\mathrm{\mid }{S}_n\mathrm{\mid }\le (2\pi {)}^d{\omega }_j({\mathrm{\ell }}_n)\to 0$. By step 1.1, ${\int }_{{\mathbb{T}}^d}{\mathrm{\partial }}_{j}hdx=0$.

Apply step 2.1 to each component $h=F_j$. The finite sum $\mathrm{div}F={\sum }_j{\mathrm{\partial }}_j{F}_j$ is uniformly continuous; the same Darboux-gap bound from step 1.1, with its own modulus of continuity, proves it is Riemann integrable and that its uniform-grid sums converge. Each such sum is the sum of the corresponding component sums, so taking their limits gives ${\int }_{{\mathbb{T}}^d}\mathrm{div}Fdx={\sum }_{j=1}^d{\int }_{{\mathbb{T}}^d}{\mathrm{\partial }}_j{F}_{j}dx=0$.

The product field $gF$ is periodic. For each component, expanding its difference quotient as $\frac{g(x+h{e}_j)F_j(x+h{e}_j)-g(x)F_j(x)}{h}=\frac{g(x+h{e}_j)-g(x)}{h}{F}_j(x+h{e}_j)+g(x)\frac{F_j(x+h{e}_j)-F_j(x)}{h}$ and taking $h\to 0$ proves ${\mathrm{\partial }}_j(g{F}_j)=({\mathrm{\partial }}_{j}g)F_j+g{\mathrm{\partial }}_j{F}_j$. These derivatives are finite sums of products of bounded uniformly continuous fields, hence uniformly continuous; the Darboux-gap argument of step 1.1 makes them Riemann integrable, so $gF$ qualifies in step 3.1. The same argument makes $\mathrm{\nabla }g\cdot F$ and $g\mathrm{div}F$ integrable, and the finite Riemann sums are linear. Applying step 3.1 to $gF$ therefore gives $0={\int }_{{\mathbb{T}}^d}\mathrm{\nabla }g\cdot Fdx+{\int }_{{\mathbb{T}}^d}g\mathrm{div}Fdx$. Rearranging proves the second identity.