On p. 33 of these lecture notes, we deduce that:
$$-\Delta q=\operatorname{div}\operatorname{div}u$$
on the $n$-d torus $\mathbb T^n$. The deduction is then made that:
$$\|q\|_{L^2(\mathbb T^n\times(0,T))}\leq\|u\|_{L^2(\mathbb T^n\times(0,T))}.$$
Apart from suspecting this should be a $\lesssim$, that is there is a missing constant on the RHS, how would one prove this? I was pointed to Gilbarg-Trudinger and to Evans, but neither seems to have an esimate that produces this one in the respective section about elliptic regularity estimates. Am I missing something, is the estimate I need somewhere else in those books, or anyway, what fact guarantees that deduction holds?
The article is working on $\mathbb T^n$, a torus, so we can use Fourier serieses. Write:
\begin{align*} q={}&\sum_{\mathbb Z^n}c_ke^{ik\cdot x} \\ u={}&\sum_{i,j}\sum_{\mathbb Z^n}d_k^{(i,j)}e^{ik\cdot x}e_i\otimes e_j. \end{align*}
The equation $-\Delta q=\operatorname{div}\operatorname{div}u$ then reads:
$$-k^2c_k=\sum_{i,j}k_ik_jd_k^{(i,j)}\qquad\qquad\forall k.$$
Divide by $-k^2$ and square:
$$c_k^2=\sum_{i_1,j_1,i_2,j_2}\frac{k_{i_1}k_{j_1}k_{i_2}k_{j_2}}{k^4}d_k^{(i_1,j_1)}d_k^{(i_2,j_2)}.$$
Taking modulus on the RHS, the fraction is at most 1, then we use a Cauchy-Schwartz type inequality to get:
$$c_k^2\leq\sqrt{\sum_{i_1,j_1,i_2,j_2}d_k^{(i_1,j_1),2}}\sqrt{\sum_{i_1,j_1,i_2,j_2}d_k^{(i_2,j_2),2}}=n^4\sum_{i,j}(d_k^{i,j})^2\leq C^2n^4\|u\|_{L^2}^2.$$
where I used that, whatever matrix norm I pick for my $\|u\|_{L^2}$, there is $C$ such that $C|u|\geq\sqrt{\sum_{i,j}(d_k^{i,j})^2}$ by equivalence of norms in finite dimension. Summing over $k$, we get exactly the desired estimate.