$\newcommand\normt[1]{\left\lVert#1\right\rVert_{L^2}}$
We want to minimize the distance function or its square, $$f(\zeta_1,\zeta_2) = \normt{u(x)-\sum_{k=1,2}\sigma_kQ(x-\zeta_k)}^2 = \int \left[u(x)-\sum_{k=1,2}\sigma_kQ(x-\zeta_k)\right]^2dx$$ subject to the constraint $\textbf{C }:|\zeta_1-\zeta_2| > \frac{3}{4}|\log \gamma|.$ Here $Q$ is a function in $L^2$ and $\sigma_k=\pm 1$ for $k=1,2$ and $\gamma>0$ is some constant. Suppose that the infimum is achieved at $z_1$ and $z_2$ such that $|z_1-z_2|>\frac{3}{4}|\log \gamma|$ then we denote, $$\epsilon(x) = u(x) - \sigma_1Q(x-z_1) - \sigma_2Q(x-z_2) = \inf_{|\zeta_1-\zeta_2| > \frac{3}{4}|\log \gamma|}f(\zeta_1,\zeta_2).$$ I want to show that $$\langle\epsilon, \partial_{x_j}Q(\cdot-z_1)\rangle = \langle\epsilon, \partial_{x_j}Q(\cdot-z_2)\rangle = 0$$ where the inner product is the usual inner product of two functions in $L^2(\mathbb{R}^N)$ $$\langle u,v \rangle = \int u(x)v(x) dx.$$
My Attempt: On differentiating this expression at the point $(z_1,z_2)$ we get that, $$\frac{\partial}{\partial \zeta_1} f_{|(z_1,z_2)}= 2\int \left[u(x)-\sum_{k=1,2}\sigma_kQ(x-z_k)\right](\sigma_1 \sum_{j=1}^{N}\partial_{x_j} Q(x-z_1)) = 0$$ which implies that $$\langle \epsilon(x),\sum_{j=1}^{N}\partial_{x_j} Q(x-z_1))\rangle = 0.$$ Similarily by computing the other partial derivative we can get, $$\langle \epsilon(x),\sum_{j=1}^{N}\partial_{x_j} Q(x-z_2))\rangle = 0.$$ Is this computation correct up to this point? I am not sure how to proceed further, all I know is that $Q$ is a positive and radial function.