Consequence of Uniform Ellipticity

Question

In Evans' PDE book, the author claims (p.352) that the uniform ellipticity condition \begin{equation} \label{e:1} \sum_{i,j=1}^n a^{ij}(x)\xi_i \xi_j \ge \theta|\xi|^2 \end{equation}

implies that $$ \sum_{i,j,k,\ell=1}^n a^{ij} a^{k\ell}v_{x_i x_k}v_{x_j x_\ell} \ge \theta^2 |D^2 u|^2. $$ This is very believable to me, but I cannot write down a proof. Why is it true? It seems to me that the most obvious thing to try would be to rearrange the sum $$ \sum_{i,j,k,\ell=1}^n a^{ij} a^{k\ell}v_{x_i x_k}v_{x_j x_\ell} = \sum_{i,j=1}^n a^{ij} \sum_{k,\ell=1}^n a^{k\ell}v_{x_i x_k}v_{x_j x_\ell} $$ and try to apply uniform ellipticity to the inner sum first, followed by the outer sum. However, the inner sum here is not quite in the right form. (What would $\xi$ be? $v_{x_i x_k}$ and $v_{x_j x_\ell}$ are the $k$th and $\ell$th components of two different vectors, namely $Du_{x_i}$ and $Du_{x_j}$, respectively.) Am I missing something easy here?

Answer 1

Suppose we fix a value of $x$. Since $a^{ij}$ (evaluated at our fixed choice of $x$) is a real symmetric matrix, is it not the case that we can rotate the coordinate axes so that $a^{ij}$ becomes a diagonal matrix? Of course, our choice of rotation will vary from one value of $x$ to another, but as far as I can tell, this doesn't matter here...

So suppose we've done this rotation, and suppose the diagonal entries of $a^{ij}(x)$ are $(\theta_1, \dots, \theta_n)$. By the ellipticity condition, we must have $\theta_i \geq \theta > 0$ for each $i$. Then it is clear that $$ \sum_{i,j,k,l}a^{ij}a^{kl}v_{ik}v_{jl} = \sum_{ik} \theta_i \theta_k v_{ik}v_{ik} \geq \theta^2 \sum_{ik} v_{ik} v_{ik} .$$

Finally, notice that $\sum_{ik} v_{ik} v_{ik}$ is invariant under rotations of coordinate axes, so this quantity is the same as the $|D^2 v|^2$ evaluated in the original coordinates.