How to prove that $\int_U vD_{k}^{-h}wdx=-\int_{U} wD_{k}^{h}vdx$

Question

How do we prove the following? $$\int_U vD_{k}^{-h}w\,dx=-\int_{U} wD_{k}^{h}v\,dx$$ Here $v$ is smooth with compact support and $U$ is open and bounded. I'm reading Partial differential equations by Evans, page 311. Here $$D_{k}^{h}u(x) = \frac{u(x+he_k)-u(x)}{h}.$$ So, my first attempt is a variable change, I mean I start with the right side so $$-\int w(x)\frac{v(x+he_k)-v(x)}{h}\,dx\quad \text{ and }\quad y=x+he_k$$ imply $dx=dy$ and $x=y-he_k$, so my new integral is $$\int w(y-he_k)\frac{v(y-he_k)-v(y)}{h}\,dy.$$ I was thinking to take the limit with $h\to 0$ and do integration by parts but its not working. Can you help me please? Any hint? Thank you.

Answer 1

Well i separeted like following. $\int vD_{k}^{h}w= \int v(x)\frac{w(x-he_k)-w(x)}{-h}= \int \frac{vw}{h}- \int \frac{v(x)w(x-he_k}{h}$ and take $y=x-he_k$ the integral above is equal $\int \frac{v(x)w(x)}{h} -\int \frac{v(x+he_k)w(x)}{h}=-\int w(x)(\frac{v(x+he_k)-v(x)}{h})=-\int wD_{k}^{h}v.$ well i believe is fine right?