I know this is simple, and I see all the pieces of the puzzle are there, but I can't seem to get it. Let $u$ be a solution of $\Delta u = f \;\;\; x \in B_4 $ Then if we can bound $\int_{B_4} |D^2 u|^2 \leq 1,\;\;\; \int_{B_4} |f|^2 \leq \delta^2$ then $\int_{B_4} |\nabla u - \overline{\nabla u}_{B_4}|^2 \leq C_1.$ Let $v$ be the solution to $\begin{cases} \Delta v = 0 & \\ v = u - (\overline{\nabla u})_{B_4}\cdot \vec{x} - \overline{u}_{B_4} & \partial B_4 \end{cases}.$ Then by minimality of harmonic function with respect to energy in $B_4$, $\int_{B_4} |\nabla v|^2 \leq \int_{B_4} |\nabla u - \overline{\nabla u}_{B_4}|^2 \leq C_1.$
I understand how one uses minimality of the solution in the last part, but I'm not sure how you get the $C_1$ bound on $\nabla u - \overline{\nabla u}_{B_4}$. In particular, I'm not sure what the overline means. Restricted to the boundary?