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?