Let $\;A\;$ be a $\;2\times 2\;$ matrix with entries $\;A_{ij}=f_{x_i}f_{x_j}-\delta(\frac{1}{2} {\vert \nabla f\vert}^2 -G(f))\;$ where $\;f:\mathbb R^2 \to \mathbb R^m\;,\;G:\mathbb R^m \to \mathbb R_{+}\;,\;f_{x_i}=(\frac{\partial f_1}{\partial x_i}, \dots,\frac{\partial f_m}{\partial x_i})\;$ and $\;\vert \cdot \vert\;$ stands for the Frobenius norm.
$\;A\;$ is invariant under translations. Consider $\;x,y\in \mathbb R^2\;$ two arbitary vectors and take the Euclidean inner product: $\;\langle Ax,y\rangle\;$.
If $\;\bar x\;$ is the translation of $\;x\;$, i.e. $\;\bar x =z+x\;$ for fixed $\;z\in \mathbb R^2\;$, then $\;\langle Ax,y\rangle=\langle A(\bar x -z),y \rangle=\langle A\bar x,y\rangle\;(*)$?
Is $\;(*)\;$ the appropriate way to use the translation invariance of $\;A\;$?
NOTE: I have proven that $\;\nabla f(\bar x)\;=\nabla f(x)\;$
Please correct me if it's wrong. I've been stuck to this since Saturday... Any help or hints would be valuable.
Thanks in advance!