Let
$$a:\mathbb{Z}^{d}\times\mathbb{Z}^{d}\to\mathbb{R}:(x,y)\mapsto a(x,y)$$
such that ($0<\alpha\le\beta<\infty$):
$$a(x,y)=\begin{cases} 0 &\text{if }\Vert x-y\Vert\neq 1\\ a(y,x)\in[\alpha,\beta]&\text{if }\Vert x-y\Vert\neq 1 \end{cases}$$
Define $y\mapsto A(y):=\text{diag}\{a(y,y+e_{1}),\dots,a(y,y+e_{d})\}$ where $(e_{1},\dots,e_{d})$ is the canonical basis of $\mathbb{Z}^{d}$. Note that, for any fixed $\xi\in\mathbb{R}^{d}$, $A(y)\xi$ is uniformly bounded for all $y\in\mathbb{Z}^{d}$.
Let $G(\cdot,y;a)\in L^{2}(\mathbb{Z}^{d})$ for any $y\in\mathbb{Z}^{d}$. Note that $G$ depends on $a$. In the sequel, we simply write $G(\cdot,y)$ Let's define the following function
$$f:\mathbb{Z}^{d}\to\mathbb{R}:x\mapsto f(x):=\int_{\mathbb{Z}^{d}}\nabla_{y} G(x,y)\cdot A(y)\xi\text{d}y$$
where $\int_{\mathbb{Z}^{d}}(\cdots)\text{d}y$ denotes the integral over $\mathbb{Z}^{d}$ with respect to the counting measure or, equivalently, the sum over all the elements $y$ in $\mathbb{Z}^{d}$; $\nabla_{y} G\cdot A\xi$ denotes the scalar product between $\nabla_{y} G$ and $A\xi$. The gradient is the discrete one, i.e.
$$\nabla_{y} G(x,y)={}^{t}(G(x,y+e_{1}))-G(x,y)),\dots,G(x,y+e_{d})-G(x,y))$$
We are interested in the differentiability of $f$ with respect to $a(e):=a(z,z+e_{i})$ for $i\in\{1,\dots,d\}$ fixed and $z\in\mathbb{Z}^{d}$. It is said that, given the following properties, we can conclude that $f$ is differentiable in the previous sense:
- $G(x,y;a)$ is differentiable with respect to $a(e)$. Its derivative is the following: $$\frac{\partial}{\partial a(e)}G(x,y;a)=-\nabla_{z_{i}}G(x,z;a)\nabla_{z_{i}}G(z,y;a)$$ where $\nabla_{z_{i}} u(z) := u(z+e_{i})-u(z)$
- $A(y)$ is differentiable with respect to $a(e):=a(z,z+e_{i})$ and its derivative is given by $e_{i}\otimes e_{i}$ if $y=z$ and $0_{d}\in\mathbb{R}^{d\times d}$ otherwise (if I am not mistaken).
- There exists a bounded radially symmetric function $h\in L^{1}(\mathbb{Z}^{d})$ such that $G(x,y;a)\le h(x-y)$ for any $x,y\in\mathbb{Z}^{d}$ and any $a$ as previously defined.
I haven't generalized the question because I have problems to understand why the very unhabitual condition 3. has an importance (but I suspect it is the domination of $G$ by an integrable function). However, I would like to know what are sufficient conditions for such an expression to be differentiable (references would be great)!