from we have for some $c > 0$ (writing also $(U^x,U^y) = {\nabla}^\perp \psi$)

$$\begin{aligned}
{\left\lVert P_{\neq 0}\omega(t,x+ty,y) \right\rVert}_{H^\sigma} & \lesssim {\left\lVert \omega_{in} \right\rVert}_{H^\sigma} e^{-c \nu t^3 } \label{ineq:LinearMixing} \\ 
{\left\lVert P_{0}\omega(t,x,y) \right\rVert}_{H^\sigma} & \lesssim \frac{{\left\lVert \omega_{in} \right\rVert}_{H^\sigma} + {\left\lVert \omega_{in} \right\rVert}_1}{{\left\langle \nu t \right\rangle}^{1/4}} \\ 
{\left\lVert P_{\neq 0} U^x(t) \right\rVert}_2 + {\left\langle t \right\rangle} {\left\lVert U^y(t) \right\rVert}_2 & \lesssim \frac{{\left\lVert \omega_{in} \right\rVert}_{H^\sigma} e^{-c \nu t^3 }}{{\left\langle t \right\rangle}}. \label{ineq:UlinDamp}\end{aligned}$$

where all implicit constants are independent of $\nu$ and $t$ and we are defining the orthogonal projections to $x$-independent modes $P_0 g = \frac{1}{2\pi} \int g(x,y) dx$ and $x$-dependent modes $P_{\neq 0}g = g - \frac{1}{2\pi} \int g(x,y) dx$.

There are several details of Proposition \[prop:LinBehave\] that warrant attention. First, is the requirement of three derivatives, which is necessary to deduce . Indeed, in order to get the inviscid algebraic decay from , one requires localization of the Fourier transform of the initial data. Physically, this is to control the amount of information which is being *un-mixed* by the Couette flow (see below and [@BM13] for further discussion). Secondly, notice that the regularity estimates in are only after pulling back by the Couette flow characteristics, another evidence of the mixing 