TITLE: Does the Legendre-Hadamard condition imply a generalized Gårding inequality?
QUESTION [14 upvotes]: For simplicity, we restrict to constant coefficients. Let $A^{ij}_{ab} \in \mathbb{R}$, $1 \le i, j \le n$ and $1 \le a, b\le m$, satisfy the Legendre-Hadamard condition:
$$
A^{ij}_{ab}\xi_i\xi_jv^av^b \ge \lambda |\xi|^2|v|^2
$$
for some $\lambda > 0$ and any $\xi \in \mathbb{R}^n$ and $v \in \mathbb{R}^m$. Let $B$ be the unit ball in $\mathbb{R}^n$.
It is straightforward to use the Fourier transform to prove that there exists $c > 0$ such that given any $u \in C^\infty_0(B,\mathbb{R}^m)$,
$$
\int_{B} A^{ij}_{ab}\partial_iu^a\partial_ju^b \ge c\int_B |\partial u|^2.
$$
If the (stronger) Legendre condition
$$
A^{ij}_{ab}p_i^ap_j^b \ge \lambda\sum_{i,a}(p_i^a)^2
$$
for any $p_i^a \in \mathbb{R}$ holds, it is easy to use an extension operator to extend the inequality to any function $u \in C^\infty(\overline{B},\mathbb{R}^m)$. 
Question: Does the Legendre-Hadamard condition imply a Garding inequality of the form
$$
\int_{B} A^{ij}_{ab}\partial_iu^a\partial_ju^b \ge \int_B c|\partial u|^2 - c'|u|^2,
$$
for any $u \in C^\infty(\overline{B},\mathbb{R}^m)$? If not, what is a counterexample?

REPLY [4 votes]: Gui-Qiang Chen suggested me to look at your question. 
I assume that your notation $\partial u$ means the gradient $\nabla u$. In the following I use $\bar\partial u$ to denote the standard first order operator for complex functions.
There is another simple counter-example in the $2\times 2$ case. Consider the convex functional
$I(u)=\int_B|\bar\partial u|^2dx$ for $u=u_1+iu_2$ and write the quadratic form in terms of
$u_1$ and $u_2$. Since $|\bar\partial u|^2=|P_{E_{\bar\partial}}\nabla u|^2$, where $P_{E_{\bar\partial}}$ is the orthogonal projection to the subspace of anti-conformal matrices and $\nabla u$ is the gradient of $(u_1,u_2)$, we have, for every rank-one matrix $\xi\otimes\eta$ that  $|P_{E_{\bar\partial}} \xi\otimes\eta|^2=|\xi|^2|\eta|^2/2$. It is easy to see that the corresponding coefficient tensor $A$ satisfies the strong L-H condition. However, if one takes any holomorphic function $u$, then $I(u)=0$. This leads to
easy counter-examples for every fixed pair of constants $c>0$ and $c^\prime>0$, e.g. $u_n=e^{nz}$. This type of constructions by using subspaces of conformal and anti-conformal matrices in the space of $2\times 2$ real matrices are commonly used in the vectorial calculus of variations related to quasiconvexity and material microstructure. 
If $A$ is not a constant tensor, say,
$A\in L^\infty$, even under the homogeneous Dirichlet boundary condition, Garding's inequality fails in general. I constructed such an example some years ago: `A counterexample in the theory of coerciveness for elliptic systems. J. Partial Differential Equations 2 (1989), no. 3, 79–82 (MR1026095)'.  I also have some recent results on the so-called universal coercivity problem: 'On coercivity and regularity for linear elliptic systems.  Calc. Var. PDEs 40 (2011), no. 1-2, 65–97 (MR2745197)'.