the symmetry of $A$ and the anti-symmetry of $ik U$. Now, integrating by parts and using and , we have $$\begin{aligned}
\l ikB \psi , A(t)\omega  \r&=\sum_{j\in\J} \l ik B  \psi , \chi_j A_j(t)\chi_j \omega  \r=-\sum_{j\in\J} \l ikB  \psi , \chi_j  \Delta_k \psi_{j,A}\r\notag\\
&=k^3\sum_{j\in\J} \l i B  \psi , \chi_j  \psi_{j,A}  \r+k\sum_{j\in\J} \l i\de_y(\chi_j  B  \psi) , \de_{y} \psi_{j,A}  \r.\end{aligned}$$ Thus, we obtain $$\begin{aligned}
|\l i kB  \psi , A(t)\omega  \r|
&\leq k^3\sum_{j\in\J} \l i B  \psi , \chi_j  \psi_{j,A}  \r+k\sum_{j\in\J} \l i\de_y(\chi_j  B  \psi) , \de_{y} \psi_{j,A}  \r\notag\\
&\leq \sum_{j\in \J}\bigg[k^3\int_I |B(y)|  |\psi(y)| |\psi_{j,A}(y) |  \chi_j(y)\dd y+ k\int_I |B(y)| |\de_y (\chi_j(y) \psi(y))||\de_y \psi_{j,A}(y)|\dd y \notag\\
&\qquad\quad+ k\int_I |B'(y)| |  \psi(y)|\de_y \psi_{j,A}(y) |\chi_j(y)\dd y\bigg].
$$ We now use , and Lemma \[lem:local\_elliptic\] to deduce that $$\begin{aligned}
   \ddt \l \omega, A(t)\omega  \r &\leq \Re\l \omega,  \dot A(t)\omega\r +C \sum_{j\in\J}\int_{I_j} \left[|B(y)|+|B'(y)|\right] \left[|\nabla_k  \psi(y)|^2 +k^2|\nabla_k \psi_{j,A}(y) |^2\right]\dd y\notag \\   
   &\leq \Re\l \omega,  \dot A(t)\omega\r +C\eps_0 \sum_{j\in\J}\int_{I_j} |U'(y)| \left[|\nabla_k  \psi(y)|^2 +k^2|\nabla_k \psi_{j,A}(y) |^2\right]\dd y\notag \\   
   &\leq \Re\l \omega,  \dot A(t)\omega\r +C\eps_0 \sum_{j\in\J}\int_{I_j} |U'(y)| \left[|\nabla_k  \psi_j(y)|^2 +k^2|\nabla_k \psi_{j,A}(y) |^2\right]\dd y,
  \end{aligned}$$ and we are done.

The L2 stability theorem
------------------------

Using the reductions of Proposition \[prop:reduction\], the proof of Theorem \[thm:L2\] reduces to constructing suitable 