This is actually a step in a bigger proof about regualarity properties of elliptic systems, but I'll write only what I don't understand. Let $L: \mathbb{R}^{m \times n} \to \mathbb{R}$ be $C^2$ and uniformely $\lambda$ quasi-convex, that is for all $\Omega \in \mathbb{R}^{n}$ open and bounded, we have:
$$ \int_{\Omega} (L(A + \nabla \phi)- L(A))dx \geq \lambda \int_{\Omega} |\nabla \phi|^2 dx \qquad \forall \phi \in C^{\infty}_0 (\Omega, \mathbb{R}^m) \; \forall A \in \mathbb{R}^{m \times n}$$
Moreover, suppose (even though I don't think it is needed here) that $|\nabla^2 L| \leq C$.
Suppose that we have a point of minimum $u$ for the functional $p \mapsto \int_{B(0,1)} L(p)dx$, and suppose that the mean of $u$ on $B(0,1)$ is zero. I define now $m= (\nabla u)_{0,1}$ to be the mean value of the gradient of $u$ over $B(0,1)$, and I define also $v(x) = u(x)-mx$. This way, I still have that the mean value of $v$ is equal to 0, but also the mean value of $\nabla v$ vanishes by construction.
Here comes the part that I don't understand. It says that given these assumptions, $v$ is a local minimizer for the integral functional associated to:
$$p \mapsto L(p+m) -L(m) - \nabla L(m)p$$
First of all, I am assuming that the last term is a sort of scalar product, that is:
$$\nabla L(m) \nabla v= \sum_{\alpha=1} ^{n} \sum_{i=1}^{m} \partial_{p^{\alpha}_i}L(m) \partial_{x_\alpha} v^i$$
but this is only my interpretation (I can't see how it could make sense, otherwise). Given that my interpretation is correct, how do you prove that $v$ is a minimizer?
In the proof it is just said without further explanations, so I guess it is somethinng very simple that I am not seeing.
P.S. For those who are wondering, this is part of a proof regarding regularity of elliptic systems, and in particular is part of a theorem which proves some decay property for the so called "excess"
Ok, I'm going to answer my own question 'cause I think I got it. The point is simply that $v$ is a LOCAL minimum. Now, it obviously is a minimum if you consider the lagrangian $L(p +m)$ because you just translate the argument, but adding the term $L(m) + \nabla L(m)p$ does not change anything because, if I look at things locally (and here's the trick) I can see all other competitors as $w= v + \rho$ with $\rho \in C^{\infty}_0$. Now it is easy to conclude.