Convergence of this finite difference sum on a small lattice Consider the unit box $Q_1=[0,1]^3$ and the associated network of size $\epsilon$ : $\epsilon \mathbb{Z}^3 \cap Q_1$.
Let $W_\epsilon$ the set of functions $v : \mathbb{R}^3 \rightarrow \mathbb{R}$ such that for every $i \in \mathbb{Z}^3$, there exists $v_i \in \mathbb{R}$ with
$$v(i \epsilon +y \epsilon)= v_i \text{ for all } y \in \mathring{Q_1}$$
(understand that $v$ is piece-wise constant on all cubic tiles of the network $\epsilon \mathbb{Z}^3$).
We then define the following space by restricting functions of $W_\epsilon$ to $Q_1$:
$$V_\epsilon=\{ v|_{Q_1}, \ v \in W_\epsilon\}$$
I'm considering the following sequence $(u_\epsilon)_\epsilon$ such that :

*

*$u_\epsilon \in V_\epsilon$ for all $\epsilon > 0$

*$u_\epsilon$ converges toward a linear profile $x \mapsto \xi \cdot x$ in $L^2(Q_1)$ as $\epsilon$ goes to zero, where $\xi$ is a given vector in $\mathbb{R}^3$.

I would like to understand the following limit if it exists :
$$\underset{\epsilon \rightarrow 0}{\lim} \ \epsilon \sum_{(x,y)\ \in (\epsilon \mathbb{Z}^3 \cap Q_1)^2 \\ \quad |x-y|=\epsilon} |u_\epsilon(x)-u_\epsilon(y)|^2$$
which is just a sum of finite differences over all neighboring couple $(x,y)$ on the lattice.
With a quick formal analysis, we have $|u_\epsilon(x)-u_\epsilon(y)|^2 \approx |\xi \cdot(x-y)|^2 \approx |\xi|^2 \epsilon^2$ and the sum becomes
$$\epsilon^3  \sum_{(x,y)\ \in (\epsilon \mathbb{Z}^3 \cap Q_1)^2 \\ \quad |x-y|=\epsilon} |\xi|^2 \approx \epsilon^3 \frac{|Q_1|}{\epsilon^3} |\xi|^2 \approx |\xi|^2 |Q_1|,$$
so much that I expect the limit to be $|\xi|^2 |Q_1| = |\xi|^2$. However, I'm struggling to prove rigorously the result and it might be false. Do you have any clue on how to compute this limit ?
For what matters, I'm also fine having just a lower estimate for the liminf of this quantity, something in this flavor:
$$\underset{\epsilon \rightarrow 0}{\liminf} \ \epsilon \sum_{(x,y)\ \in (\epsilon \mathbb{Z}^3 \cap Q_1)^2 \\ \quad |x-y|=\epsilon} |u_\epsilon(x)-u_\epsilon(y)|^2 \geq C |\xi^2| |Q_1|.$$
Thanks for your help.
 A: Here's a lower bound in the case that the functions $u_\epsilon$ converge uniformly to $x\mapsto \xi\cdot x$. I imagine that one can generalize this to the non-uniform case as well.
Let $\xi^\vee$ represent the function $x\mapsto \xi\cdot x$ on $Q_1$. For some $\epsilon$, let $\delta$ be such that $\|u_\epsilon-\xi^\vee\|_\infty\leq\delta$. Then, letting $n=\lfloor 1/\epsilon\rfloor$,
\begin{align*}
\epsilon \sum_{\substack{(x,y)\ \in (\epsilon \mathbb{Z}^3 \cap Q_1)^2 \\ \quad |x-y|=\epsilon}} |u_\epsilon(x)-u_\epsilon(y)|^2
&\geq \epsilon\sum_{b=0}^n\sum_{c=0}^n\left(\sum_{a=0}^{n-1}\left|u_\epsilon\big((a+1)\epsilon,b\epsilon,c\epsilon)\big)-u_\epsilon\big((a\epsilon,b\epsilon,c\epsilon)\big)\right|^2\right)\\
&\ \ \ \ + \epsilon \sum_{c=0}^n\sum_{a=0}^n\left(\sum_{b=0}^{n-1}\left|u_\epsilon\big(a\epsilon,(b+1)\epsilon,c\epsilon)\big)-u_\epsilon\big((a\epsilon,b\epsilon,c\epsilon)\big)\right|^2\right)\\
&\ \ \ \ + \epsilon \sum_{a=0}^n\sum_{b=0}^n\left(\sum_{c=0}^{n-1}\left|u_\epsilon\big(a\epsilon,b\epsilon,(c+1)\epsilon)\big)-u_\epsilon\big((a\epsilon,b\epsilon,c\epsilon)\big)\right|^2\right)\\
\end{align*}
The inner sum of the first sum is at least, using QM-AM,
\begin{align*}
& \frac1n\left(\sum_{a=0}^{n-1}\left|u_\epsilon\big((a+1)\epsilon,b\epsilon,c\epsilon)\big)-u_\epsilon\big((a\epsilon,b\epsilon,c\epsilon)\big)\right|^2\right)
 \geq \frac1n \Big(  (\sum_{a=0}^{n-1}u_\epsilon\big((a+1)\epsilon,b\epsilon,c\epsilon)\big)-u_\epsilon\big((a\epsilon,b\epsilon,c\epsilon)\big)\Big)^2\\
&=\frac1n\left(u_\epsilon\big(n\epsilon,b\epsilon,c\epsilon\big)-u_\epsilon\big(0,b\epsilon,c\epsilon\big)\right)^2\\
&\geq \frac1n\left(\left|\xi\cdot(n\epsilon,b\epsilon,c\epsilon)-\xi\cdot(0,b\epsilon,c\epsilon)\right|-2\delta\right)^2\\
&=\frac1n\left(n\epsilon|\xi_1|-2\delta\right)^2.
\end{align*}
This means that the desired quantity is at least
$$\sum_{i=1}^3\epsilon n(\epsilon n|\xi_i|-2\delta)^2;$$
since $1-\epsilon<\epsilon n\leq 1$, this is at least
$$\sum_{i=1}^3(1-\epsilon)\big((\epsilon n)^2|\xi_i|^2-4\delta(\epsilon n)|\xi_i|\big)\geq (1-\epsilon)^3\|\xi\|_2^2-4\delta\|\xi\|_1.$$
As $\delta,\epsilon\to 0$, this tends towards $|\xi|^2$, as desired.
