Difference quotient (pde) Let $u: U\subset\mathbb{R}^n\rightarrow\mathbb{R}$. The Difference quotient of $u$ is defined by
$D_k^hu(x)=\dfrac{u(x+he_k)-u(x)}{h}$ with $h\in\mathbb{R}$, $0<|h|<\textrm{dist}(V,\partial U)$.
Here $x\in V\subset\!\subset U$ and $e_k$ is the canonical vector in $\mathbb{R}^n$.
How can I prove that $\displaystyle\int_U vD_k^{-h}w\, dx=-\int_U wD_k^h v\,dx$ 

Let $w,v$ be defined on an open subset $U\subset \mathbb R^n$ with $\text{supp } v = V \subset \subset U$. We have 
$$
\int_{U} w(x)D^h_k(v(x)) dx = \int_{U} w(x)\frac{v(x+he_k)-v(x)}{h}dx 
$$
We split the integral as sum of integrals: 
$$
\frac{1}{h}\left[\int_{U} w(x)v(x+he_k)dx- \int_{U}w(x)v(x)dx\right] = \frac{1}{h}\left[\int_{V-he_k} w(x)v(x+he_k)dx- \int_{V}w(x)v(x)dx\right]
$$
Now we change variables in the first integral by putting $y=x+he_k$: we get 
$$
\frac{1}{h}\left[\int_{V} w(x-he_k)v(x)dx- \int_{V}w(x)v(x)dx\right]
$$
which is indeed
$$
-\int_{U} v(x)D^{-h}_kw(x)dx.
$$
P.S. Another way to do this is to prove a version of Leibniz rule for the difference quotients, i.e. compute $D^h_k(uv)$ in terms of $D^h_ku$ and $D^h_kv$. From that you can easily get this sort of "integration by parts".
