Let \(f,g\in H^1(\mathbb{Z}^d)\). Then,
\[ \langle \partial_i f, g\rangle_{\ell^2(\mathbb{Z}^d)} = - \langle f, \partial_i g\rangle_{\ell^2(\mathbb{Z}^d)}. \]
Proof
It is
\begin{align*} \langle \partial_i f, g\rangle_{\ell^2(\mathbb{Z}^d)} &= \sum_{x\in \mathbb{Z}^d} \partial_i f(x) g(x)\\ &= \sum_{x\in \mathbb{Z}^d} (f(x+e_i)-f(x-e_i)) g(x)\\ &= \sum_{x\in \mathbb{Z}^d} f(x+e_i) g(x)- \sum_{x\in \mathbb{Z}^d} f(x-e_i) g(x)\\ &= \sum_{x\in \mathbb{Z}^d} f(x) g(x-e_i)- \sum_{x\in \mathbb{Z}^d} f(x) g(x+e_i)\\ &= -\sum_{x\in \mathbb{Z}^d} f(x) (g(x+e_i)- g(x-e_i))\\ &=-\langle f, \partial_i g\rangle_{\ell^2(\mathbb{Z}^d)}, \end{align*}where we used an index shift.