text:
\section{Introduction}
The inhomogeneous Muskat problem models the dynamics of two incompressible, immiscible fluids in a non-uniform, porous medium. This scenario occurs naturally when, for example, oil and water flows meet in a sand and loam media. The physical principle governing the porous media flow is Darcy's law \cite{D}, given here in the two dimensional setting:
\begin{equation}\label{darcylaw}
  \frac{\mu}{\kappa} u = -\nabla p - g \begin{bmatrix} 0 \\ \rho \end{bmatrix}
\end{equation}
where $u(x,t)$ is the fluid velocity, $\mu(x,t)$ is the fluid viscosity, $\kappa(x,t)$ is the permeability of the porous media, $p(x,t)$ is the pressure, $g$ is the gravitational constant and $\rho(x,t)$ is the fluid density. The incompressibility condition in each fluid domain is given by $\nabla \cdot u = 0$.

Given non-intersecting soil and fluid interfaces, we divide our domain into three time-dependent disjoint open regions $D_{i}(t)$ such that $D_{3}(t) = D_{3}$ is unchanging.
\begin{figure}[h]
\begin{center}
    \begin{overpic}[width=0.5\textwidth]{interfaces.png}
        \put (15, 60) {\huge$\displaystyle D_1(t)$}
        \put (65, 25) {\huge$\displaystyle D_2(t)$}
        \put (10, 10)  {\huge$\displaystyle D_3$}
        \put (40, 35) {$z(\alpha, t)$}
        \put (63, 5)  {$h(\alpha)$}
    \end{overpic}
\caption{Inhomogenous Muskat Problem.}
\end{center}
\end{figure}

Denoting the smooth, simple boundaries as $\partial D_{i}(t)$ for $i=1,2,3$,
$$D_{1}(t)\cup D_{2}(t)\cup D_{3}\cup \partial D_{1}(t) \cup \partial D_{3} = \mathbb{R}^{2}.$$
The two fluids occupy domains $D_{1}(t)$ and $D_{2}(t)\cup D_{3}$ respectively, and the two different soils occupy domains $D_{1}(t)\cup D_{2}(t)$ and $D_{3}$:
\[
    (\mu, \rho)(x,t) =
    \begin{cases}
        (\mu_1, \rho_1) & \text{if } x \in D_1(t)\\
        (\mu_2, \rho_2) & \text{if } x \in D_2(t) \cup D_3
    \end{cases}
\]
and
\[
    \kappa(x) =
    \begin{cases}
    \kappa_1 & \text{if } x \in  D_1(t) \cup D_2(t)\\
    \kappa_2 & \text{if } x \in D_3
    \end{cases}
\]
We will study the evolution of the fluid interface $\partial D_{1}(t)$. This setting is a variation of the classical Muskat problem in which the porous medium is uniform and the permeability constant is often normalized to $\kappa =1$. Note that in the case of uniform permeability, the distinction between the regions $D_{2}(t)$ and $D_{3}$ vanishes.

We obtain a contour equation for the fluid interface from \eqref{darcylaw} and the incompressibility condition on the fluid velocity by first parametrizing the fluid interface $\partial D_{1}(t)$ as
\[
    z(\alpha,t) = (z_1(\alpha,t), z_2(\alpha,t)) \qquad \text{for } \alpha\in \R,\, t \in \R_{\geq 0},
\]
and the soil interface $\partial D_{3}$ as
\[
    h(\alpha) = (h_1(\alpha), h_2(\alpha)) \qquad \text{for } \alpha \in \R
\]
as depicted in Figure 1. We assume that $z(\alpha, 0) \neq h(\alpha)$ for all $\alpha\in\mathbb{R}$.
\begin{comment}
\begin{figure}[h]
\begin{center}
    \begin{overpic}[width=0.5\textwidth]{interfaces.png}
        \put (15, 60) {\huge$\displaystyle\Omega_1(t)$}
        \put (65, 25) {\huge$\displaystyle\Omega_2(t)$}
        \put (10, 10)  {\huge$\displaystyle\Omega_3$}
        \put (40, 35) {$z(\alpha, t)$}
        \put (63, 5)  {$h(\alpha)$}
    \end{overpic}
\caption{Inhomogenous Muskat Problem.}
\end{center}
\end{figure}
\end{comment}

Darcy's law implies the vorticity ($\omega = \nabla \times v$) can only be supported on the boundaries,
\[
    \omega(x, t) = \omega_1(\alpha, t) \delta(x-z(\alpha, t)) + \omega_2(\alpha, t) \delta(x-h(\alpha)).
\]
The Biot-Savart law gives a solution for the fluid velocity in terms of the vorticity
\begin{align*}
    u(x,t) &= BR(\omega_1, z)(x, t) + BR(\omega_2, h)(x,t)\\
    &\coloneqq \frac{1}{2\pi} \pv \int_\R \frac{(x-z(\beta, t))^\perp}{\abs{x-z(\beta, t)}^2}\, \omega_1(\beta, t)\; d\beta + \frac{1}{2\pi} \pv \int_\R \frac{(x - h(\beta))^\perp}{\abs{x-h(\beta)}^2}\, \omega_2(\beta, t)\; d\beta
\end{align*}
where $BR$ stands for the Birkoff-Rott integral defined above. Taking limits in the normal direction to the boundaries, one obtains
\begin{align}
    \partial_t z(\alpha, t) &= BR(\omega_1, z)(z(\alpha, t),t) + BR(\omega_2, h)(z(\alpha, t),t) + c(\alpha, t)\ \partial_\alpha z(\alpha, t), \label{zevolution}\\
    \omega_1(\alpha, t) &= 2 A_\mu (BR(\omega_1, z)(z(\alpha, t), t) + BR(\omega_2, h)(z(\alpha, t),t))\cdot \partial_\alpha z(\alpha, t) - 2A_\rho\, \partial_\alpha z_2(\alpha, t),\label{omega1z}\\
    \omega_2(\alpha, t) &= -2A_\kappa(BR(\omega_1, z)(h(\alpha), t) + BR(\omega_2, h)(h(\alpha), t))\cdot \partial_\alpha h(\alpha)\label{omega2z},
\end{align}
where the constants are given by
\begin{equation}\label{constants}
    A_\kappa = \frac{\kappa_1 - \kappa_2}{\kappa_1 + \kappa_2}, \qquad  A_\mu = \frac{\mu_1 - \mu_2}{\mu_1 + \mu_2}, \qquad A_\rho = -\kappa_1 \frac{\rho_1 - \rho_2}{\mu_1 + \mu_2}g.
\end{equation}
In \eqref{zevolution}, the tangential term $c(\alpha,t)\partial_{\alpha}z(\alpha,t)$ is determined by the choice of parametrization of the curve and vanishes when considering the velocity normal to the interface. See \cite{P-C} for more details. If there is no jump in permeability, the constant $A_{\kappa} = 0$ and the $\omega_{2}$ terms disappear from the evolution equation for $z(\alpha,t)$.

The question of well-posedness of a fluid-fluid interface in porous media has been well-studied. The local well-posedness depends on the system initially satisfying the Rayleigh-Taylor condition, which requires the jump of the gradient of the pressure in the normal direction to the interface to be strictly positive. A system satisfying the Rayleigh-Taylor condition is said to be in the stable regime, see e.g. \cite{Ambrose}. In the case that there is a density jump across the interface, $A_{\rho} \neq 0$, the Rayleigh-Taylor condition requires that the denser fluid lay below the interface, meaning $\rho_{2}>\rho_{1}$. The scaling invariance of the Muskat problem gives the criticality of $H^{1+\frac{d}{2}}$ regularity, where $d$ is the dimension of the interface. Similarly, it can be seen that $W^{1,\infty}$, $C^{1}$ and $\mathcal{F}^{1,1}_{0}$ are also all scale invariant. (For the definition of $\mathcal{F}^{1,1}_{0}$, see \eqref{Fspnu} below.)
\begin{comment}
In the stable regime with uniform permeability $A_{\kappa}=0$, there are numerous works giving conditions for local and global well-posedness of the interface system. For example, in \cite{CCG2} and \cite{CCG1}, local well-posedness is shown for initial data satisfying the Rayleigh-Taylor condition in $H^{d+2}$ for interface dimension $d=1,2$. Later results prove local well-posedness for interfaces of lower regularity, with the scaling critical Sobolev regularity shown to be at the level of $W^{1,\infty}$ or $H^{s+\frac{d}{2}}$ for $d=1,2$. In two dimensions, local well-posedness of graphical interfaces in $H^{2}$ was shown in \cite{matioc}. Without a jump in viscosities, well-posedness is shown in $H^{s}$ for $s \in (\frac{3}{2},2)$ by \cite{Matioc2}. In \cite{CGSV}, local well-posedness for initial data in $W^{2,p}$ with bounded slope is proven, with global existence in the case of a small slope. A paradifferential approach is used in \cite{Nguyen} and \cite{NguyenP} to show local well-posedness in all subcritical spaces $H^{s}$ for $s> 1 + \frac{d}{2}$ for arbitrary dimension and with various characteristics such as finite depth, viscosity jump and surface tension. Another result \cite{AlazardLazar} gives a local well-posedness in $\dot{H}^{1}\cap \dot{H}^{s}$ for all $s>\frac{3}{2}$ by paralinearizing the system. More recently, for both $d=1,2$ dimensional interfaces, without viscosity jump, \cite{AlazardNguyen1} and \cite{AlazardNguyen2} show local well-posedness for Lipschitz data in the scaling critical space $H^{1+\frac{d}{2}}$ as well as global well-posedness in the case of small data of the same regularity. For arbitrary Lipschitz data but small $\dot{H}^{2}$ norm of the interface in three dimensions, global well-posedness is shown in \cite{GancedoLazar}. Using a fractional logarithmic correction of $H^{\frac{3}{2}}$, \cite{AlazardNguyen3} can drop the Lipschitz requirement for the initial data.
\end{comment}

The stable regime with uniform permeability $A_{\kappa}=0$ has been extensively studied; see for example, \cite{AbelsMatioc}, \cite{AlazardLazar}, \cite{AlazardNguyen3}, \cite{AlazardNguyen1}, \cite{AlazardNguyen2}, \cite{CCFGL}, \cite{CGS}, \cite{CGSV}, \cite{CCG2}, \cite{CCG1}, \cite{CG1}, \cite{GancedoLazar}, \cite{GGNP}, \cite{Matioc2}, \cite{matioc}, \cite{NguyenGlobal}, \cite{Nguyen}, \cite{NguyenP}, etc. and references therein. Of particular interest to this paper are results showing global well-posedness for initial data of \textit{medium size} in the case of uniform permeability. Previously, global in time results are known in both two and three dimensions for systems with uniform permeability for medium size initial data without a viscosity jump \cite{CCGRS}, \cite{CCGS}, \cite{PS} and with a viscosity jump \cite{GGPS} in the case of an infinite graph interface by using the norms \eqref{Fspnu}. Under the effects of surface tension, the stability of medium size perturbations of a gravity unstable bubble interface has also been proven in \cite{gancedo2020global}. A modulus of continuity approach in \cite{Cameron1} and \cite{Cameron2} gives a different medium size condition in both 2D and 3D without viscosity jump.

As discussed earlier, in this paper, we consider the non-uniform permeability setting, also called the inhomogeneous Muskat problem. In this setting, for a horizontal permeability jump boundary and without viscosity jump, graph interface solutions were shown to be locally well-posed using energy estimates in Sobolev spaces and there exists interfaces starting in the stable regime that become unstable in finite time \cite{BCG}. A class of graph solutions exhibiting this turning behavior was also demonstrated by a computer-assisted proof in \cite{JaviGraneroInhom}. In the same setting, \cite{inhomShkoller} demonstrated global existence and decay to the flat fluid interface for small initial data in $H^{2}$. For a general interface curve and with a viscosity jump, local well-posedness was shown in \cite{P-C} in Sobolev spaces. The existence of splash singularities was later shown in \cite{CordobaTania}, although splat singularities were ruled out. This paper will address the problem of global well-posedness in the critical regularity $\mathcal{F}^{1,1}_{0}$ for medium sized initial data in the regime that also has a viscosity jump. Moreover, the estimates proven in this paper will imply the interface is instantly analytic and decay to the flat solution as a corollary. The authors would like to note that while writing the final version of this paper, a concurrent result \cite{GraneroAlonso} in the regime without a viscosity jump, demonstrates precise conditions for $H^{3}$ solutions to show decay in the Lipschitz norm and further establishes a global existence and decay result for Lipschitz solutions with small initial data.

We will consider the case of a graphical fluid interface $z(\alpha,t) = (\alpha, f(\alpha,t))$ and a fixed horizontal permeability jump interface $h(\alpha) = (\alpha, -h_2)$ for $h_2 > 0$ under the assumption $\norm{f(\alpha, 0)}_{L^\infty}  < h_2$ as in \cite{BCG}, although we will allow for a viscosity jump.
Setting
\begin{align*}
    \Delta_\beta f(\alpha) &= \frac{f(\alpha) - f(\alpha - \beta)}{\beta},
\end{align*}
the choice of a graph interface determines the tangential constant $c(\alpha,t)$ and turns the system \eqref{zevolution}-\eqref{omega1z}-\eqref{omega2z} into
\begin{align}
\label{eq: interface}
    \partial_t f(\alpha) = \frac{1}{2\pi}(I_1(\alpha) + I_2(\alpha) + I_3(\alpha) + I_4(\alpha))
\end{align}
for
\begin{align}
\label{eq: I_1}
    I_1(\alpha) &= \pv \int_\R \frac{1}{1 + \Delta_\beta f(\alpha)^2} \frac{\omega_1(\alpha - \beta)}{\beta} \; d\beta,\\
\label{eq: I_2}
    I_2(\alpha) &= \pv \int_\R \frac{ \partial_\alpha f(\alpha)\Delta_\beta f(\alpha)  }{1 + \Delta_\beta f(\alpha)^2} \frac{\omega_1(\alpha - \beta)}{\beta} \; d\beta,\\

\label{eq: I_3}
    I_3(\alpha) &=  \pv \int_\R \frac{\beta}{\beta^2 + (f(\alpha) + h_2)^2}\ \omega_2(\alpha - \beta)\; d\beta,\\
\label{eq: I_4}
    I_4(\alpha) &= \pv \int_\R \frac{\partial_\alpha f(\alpha) (f(\alpha) + h_2)}{\beta^2 + (f(\alpha) + h_2)^2}\ \omega_2(\alpha - \beta)\; d\beta,
\end{align}
in which
\begin{align}
\begin{split}
    \omega_1(\alpha) &= \frac{A_\mu}{\pi} \pv \int_\R \frac{\partial_\alpha f(\alpha) - \Delta_\beta f(\alpha) }{1 + \Delta_\beta f(\alpha)^2} \frac{\omega_1(\alpha - \beta)}{\beta}\; d\beta \\
    &\hspace{2.5em}+ \frac{A_\mu}{\pi} \pv \int_{\R} \frac{ \beta \partial_\alpha f(\alpha) -(f(\alpha) +h_2)  }{\beta^2 + (f(\alpha)+h_2)^2} \omega_2(\alpha - \beta)\; d\beta -2A_\rho \partial_\alpha f(\alpha),
    \end{split} \label{omega1equation}\\
    \omega_2(\alpha) &= -\frac{A_\kappa}{\pi} \pv \int_\R  \frac{f(\alpha - \beta) + h_2}{\beta^2 + (f(\alpha - \beta) + h_2)^2}\ \omega_1(\alpha - \beta)\; d\beta.\label{omega2equation}
\end{align}
Defining $\partial_{\alpha}\Omega_{i} = \omega_{i}$, it can be derived that
\begin{align}
    \begin{split}
        \Omega_1(\alpha) &= -\frac{A_\mu}{\pi} \int_\R \frac{\Delta_\beta f(\alpha) - \partial_\alpha f(\alpha - \beta)}{1 + \Delta_\beta f(\alpha)^2} \frac{\Omega_1(\alpha - \beta)}{\beta}\; d\beta \\
        &\hspace{2.5em}- \frac{A_\mu}{\pi} \int_\R \frac{f(\alpha) + h_2}{\beta^2 + (f(\alpha) + h_2)^2}\Omega_2(\alpha - \beta) \; d\beta - 2A_\rho f(\alpha)\label{Omega1}
    \end{split}\\
    \Omega_2(\alpha) &= -\frac{A_\kappa}{\pi} \int_\R \frac{f(\alpha - \beta) + h_2 +\beta \partial_\alpha f(\alpha - \beta)}{\beta^2 + (f(\alpha - \beta) + h_2)^2} \Omega_1(\alpha - \beta)\; d\beta. \label{Omega2}
\end{align}
The equations \eqref{eq: interface}, \eqref{omega1equation}, and \eqref{omega2equation} give a coupled system for the evolution of fluid-fluid graph interface $f(\alpha,t)$.
\begin{comment}
Iterated convolutions will be denoted as
\begin{equation*}
(\ast^{n} g)(x) = (\underbrace{g \ast \cdots *g}_{n\text{ times}})(x)
\end{equation*}
with the convention $(*^0 g)(x) = \delta(x)$.

The following product rule type inequalities will be useful throughout this paper when applied to the norms \eqref{Fspnu}.
\begin{lemma}
Given functions $f_{k} : \mathbb{R}^{d}\rightarrow \mathbb{R}$ for $1 \leq k \leq n$ we have
\begin{equation}\label{convolutionexp}
e^{\nu t |\xi|} (|f_{1}|\ast |f_{2}| \ast \cdots \ast |f_{n}|)  \leq (e^{\nu t |\xi|}|f_{1}|)\ast (e^{\nu t |\xi|}|f_{2}|) \ast \cdots \ast (e^{\nu t |\xi|}|f_{n}|)
\end{equation}
and for $0 < s \leq 1$
\begin{equation}\label{productrule}
|\xi|^{s} (|f_{1}|\ast |f_{2}| \ast \cdots \ast |f_{n}|)  \leq \sum_{k=1}^{n} (|\xi|^{s} |f_{k}|)\ast (\ast_{j\neq k}|f_{j}|)
\end{equation}
where
$$\ast_{j\neq k}|f_{j}|$$ indicates a convolution over the absolute values of all functions $f_{j}$ except $f_{k}$.
\end{lemma}
\begin{proof}
By the triangle inequality, we have for $0< s \leq 1$
$$ |\xi_{0}|^{s} \leq |\xi_{0}-\xi_{1}|^{s} + |\xi_{1}-\xi_{2}|^{s} + \ldots + |\xi_{n-2}-\xi_{n-1}|^{s} + |\xi_{n-1}|^{s}.$$
Applying this triangle inequality to the convolution gives us \eqref{productrule}.
Moreover,
$$e^{\nu t |\xi_{0}|} \leq \prod_{j=0}^{n} e^{\nu t |\xi_{j}-\xi_{j+1}|}.$$
Plugging this inequality into the function convolution, we obtain \eqref{convolutionexp}.
\end{proof}
Finally, we also have the interpolation inequality by a direct application of Holder's inequality:
\begin{proposition}
If $g\in \mathcal{F}^{s_{1},1}_{\nu} \cap \mathcal{F}^{s_{2},1}_{\nu}$ for $s_{1} < s_{2}$, then  $g\in \mathcal{F}^{s,1}_{\nu}$ for each $s\in [s_{1},s_{2}]$ and satisfies
\begin{equation}\label{interpolation}
\|g\|_{\mathcal{F}^{s,1}_{\nu}} \leq \|g\|_{\mathcal{F}^{s_{1},1}_{\nu}}^{\theta}\|g\|_{\mathcal{F}^{s_{2},1}_{\nu}}^{1-\theta}
\end{equation}
for $\theta \in [0,1]$ such that $s = \theta s_{1} + (1-\theta) s_{2}$.
\end{proposition}

The following Fourier transforms can be computed by a semicircle contour argument in the complex plane and will be used throughout the paper.

\begin{proposition}
For $a \in \mathbb{R}$, we have
\begin{equation}\label{fouriercalc1}
\calF\Big[ \frac{a}{x^{2}+a^{2}} \Big](\xi) = \pi e^{-a|\xi|}
\end{equation}
and
\begin{equation}\label{fouriercalc2}
\calF\Big[ \frac{x}{x^{2}+a^{2}} \Big](\xi) = -i\pi \sgn(\xi) e^{-a|\xi|}.
\end{equation}
\end{proposition}
\end{comment}
\section{Main Results}
To study the evolution of the interface, we adopt the weighted Fourier norms defined as follows. For a function $g:\mathbb{R}^{d} \times \R_{\geq 0} \rightarrow \mathbb{R}$, and for $s > -d$ define the norm
\begin{equation}\label{Fspnu}
\|g\|_{\mathcal{F}^{s,p}_{\nu}} (t) = \norm{ e^{\nu t \abs{\xi}} \, |\xi|^{s}\, \hat{g}(\xi,t) }_{L^p} = \Big(\int_{\mathbb{R}^{d}} e^{\nu t  p|\xi|} |\xi|^{sp} |\hat{g}(\xi, t)|^{p} d\xi\Big)^{\frac{1}{p}}
\end{equation}
where $\hat g$ is the Fourier transform of $g$ in the spatial variable
\[
    \hat g(\xi, t) = \calF(g(\,\cdot\, , t))(\xi) = \int_{\R^d} g(x, t) e^{-ix\cdot \xi}\; dx.
\]
Let $\calF^{s,p}_{\nu}$ be the space of all functions with finite $\norm{\, \cdot\, }_{\calF^{s,p}_{\nu}}$ norm.

Let $\theta$ be the constant defined in \eqref{infconst} and observe $\theta > 0$ because $\abs{A_\kappa} < 1$ in the setting of \eqref{darcylaw}. Let $\sigma_{s} = \sigma_{s}(\|f_{0}\|_{\mathcal{F}^{0,1}_{0}}, \|f_{0}\|_{\mathcal{F}^{1,1}_{0}})$, $s=0,1,2$ be continuous functions defined in \eqref{sigma0}, \eqref{sigma1}, \eqref{sigma2}. Note that $\sigma_{s}(0,0) = 0$. Since the $\sigma_{s}$ are continuous, we define the constants $ k_{0}(|A_{\mu}|,|A_{\kappa}|)  < h_{2}$ and $ k_{1}(|A_{\mu}|,|A_{\kappa}|)  < 1$ such that $$\theta  -  \sigma_{s}(k_{0}(|A_{\mu}|,|A_{\kappa}|),k_{1}(|A_{\mu}|,|A_{\kappa}|)) > 0$$ for $s= 0,1,2$.
\begin{theorem}\label{MainTheorem}
    Suppose $f_0 \in L^{2}\cap \mathcal{F}^{1,1}_{0}$ such that $\|f_{0}\|_{\mathcal{F}^{0,1}_{0}} < k_{0}(|A_{\mu}|,|A_{\kappa}|)$  and $\|f_{0}\|_{\mathcal{F}^{1,1}_{0}} < k_{1}(|A_{\mu}|,|A_{\kappa}|)$ hold. Then there exists a unique solution $f \in L^{\infty}([0, T]; L^{2}\cap\mathcal{F}^{1,1}_{0} ) \cap L^{1}([0, T]; \mathcal{F}^{2,1}_{0})$ for all $T>0$ and a constant $\nu > 0$ satisfying for $s=0,1$
\begin{equation}\label{maininequality}
\norm{f}_{\mathcal{F}^{s,1}_{\nu}}(t) +(A_\rho\theta  - A_{\rho} \sigma_{s} -\nu)\int_{0}^{t} \norm{f}_{\mathcal{F}^{s+1,1}_{\nu}}(s)\; ds \leq \|f_{0}\|_{\mathcal{F}^{s,1}_{0}}
\end{equation}
and
\begin{equation}\label{L2inequality}
  \norm{f}_{L^2_\nu}^2 (t) \leq \norm{f_0}_{L^2}^2 \cdot \exp(R(\norm{f_0}_{\mathcal{F}^{0,1}_{0}}, \norm{f_0}_{\mathcal{F}^{1,1}_{0}},t))
\end{equation}
for any $t \geq 0$ for a positive function $R$ that is bounded for $\|f_{0}\|_{\mathcal{F}^{0,1}_{0}} < k_{0}(|A_{\mu}|,|A_{\kappa}|)$  and $\|f_{0}\|_{\mathcal{F}^{1,1}_{0}} < k_{1}(|A_{\mu}|,|A_{\kappa}|)$ and $t\geq 0$.
\end{theorem}
\begin{remark}
By \eqref{maininequality} and \eqref{L2inequality}, it can be seen from the exponential weight, that the solution gains analytic regularity instantly in time for all $t>0$.
\end{remark}
Next, to show decay to the flat solution, we will  need the Decay Lemma proved in \cite{GGPS}, which we have restated for our setting below.
\begin{lemma}[Decay Lemma]
\label{lem: decay}
    Suppose $\norm{g}_{\mathcal{F}^{s_{1},1}}(t) \leq C_0$ and
    \[
        \frac{d}{dt} \norm{g}_{\mathcal{F}^{s_{2},1}}(t) \leq -C \norm{g}_{\mathcal{F}^{s_{2}+1,1}}(t)
    \]
    where $s_1 < s_2$. Then
    \[
        \norm{g}_{\mathcal{F}^{s_{2},1}}(t) \lesssim (1+t)^{s_1 - s_2}.
    \]
\end{lemma}

This Decay Lemma along with \eqref{maininequality} implies the large time decay of solutions to the inhomogeneous Muskat problem. Specifically, \eqref{maininequality} implies uniform in time bounds of $\norm{f}_{\mathcal{F}^{s,1}}$ for $s=0,1$ and then we use the Decay Lemma to obtain the following result.
\begin{theorem}
    Suppose $f_0(\alpha)$ is initial data satisfying the conditions of Theorem \ref{MainTheorem}. Then the solution $f(\alpha,t)$ to the \eqref{eq: interface} decays with the rate
    \[
        \norm{f}_{\mathcal{F}^{1,1}}(t) \lesssim (1+t)^{-1}.
    \]
\end{theorem}

To prove Theorem \ref{MainTheorem}, we employ the following collection of useful facts. First, letting iterated convolutions be denoted as
\begin{equation*}
(\ast^{n} g)(x) = (\underbrace{g \ast \cdots *g}_{n\text{ times}})(x)
\end{equation*}
we have the product rule inequalities
\begin{lemma}
Given functions $f_{k} : \mathbb{R}^{d}\rightarrow \mathbb{R}$ for $1 \leq k \leq n$ we have
\begin{equation}\label{convolutionexp}
e^{\nu t |\xi|} (|f_{1}|\ast |f_{2}| \ast \cdots \ast |f_{n}|)  \leq (e^{\nu t |\xi|}|f_{1}|)\ast (e^{\nu t |\xi|}|f_{2}|) \ast \cdots \ast (e^{\nu t |\xi|}|f_{n}|)
\end{equation}
and for $0 < s \leq 1$
\begin{equation}\label{productrule}
|\xi|^{s} (|f_{1}|\ast |f_{2}| \ast \cdots \ast |f_{n}|)  \leq \sum_{k=1}^{n} (|\xi|^{s} |f_{k}|)\ast (\ast_{j\neq k}|f_{j}|)
\end{equation}
where
$$\ast_{j\neq k}|f_{j}|$$ indicates a convolution over the absolute values of all functions $f_{j}$ except $f_{k}$.
\end{lemma}
\begin{proof}
By the triangle inequality, we have for $0< s \leq 1$
$$ |\xi_{0}|^{s} \leq |\xi_{0}-\xi_{1}|^{s} + |\xi_{1}-\xi_{2}|^{s} + \ldots + |\xi_{n-2}-\xi_{n-1}|^{s} + |\xi_{n-1}|^{s}.$$
Applying this triangle inequality to the convolution gives us \eqref{productrule}.
Moreover,
$$e^{\nu t |\xi_{0}|} \leq \prod_{j=0}^{n} e^{\nu t |\xi_{j}-\xi_{j+1}|}.$$
Plugging this inequality into the function convolution, we obtain \eqref{convolutionexp}.
\end{proof}
We also have the interpolation inequality:
\begin{proposition}
If $g\in \mathcal{F}^{s_{1},1}_{\nu} \cap \mathcal{F}^{s_{2},1}_{\nu}$ for $s_{1} < s_{2}$, then  $g\in \mathcal{F}^{s,1}_{\nu}$ for each $s\in [s_{1},s_{2}]$ and satisfies
\begin{equation}\label{interpolation}
\|g\|_{\mathcal{F}^{s,1}_{\nu}} \leq \|g\|_{\mathcal{F}^{s_{1},1}_{\nu}}^{\theta}\|g\|_{\mathcal{F}^{s_{2},1}_{\nu}}^{1-\theta}
\end{equation}
for $\theta \in [0,1]$ such that $s = \theta s_{1} + (1-\theta) s_{2}$.
\end{proposition}
Finally, to compute the linearization of \eqref{eq: interface} and of the vorticity terms, the following Fourier transforms will be needed.

\begin{proposition}
For $a \in \mathbb{R}$, we have
\begin{equation}\label{fouriercalc1}
\calF\Big[ \frac{a}{x^{2}+a^{2}} \Big](\xi) = \pi e^{-a|\xi|}
\end{equation}
and
\begin{equation}\label{fouriercalc2}
\calF\Big[ \frac{x}{x^{2}+a^{2}} \Big](\xi) = -i\pi \sgn(\xi) e^{-a|\xi|}.
\end{equation}
\end{proposition}
\begin{proof}
    Suppose $\xi < 0$ and let $\gamma$ be the upper semi-circle contour with base $[-R, R] \subset \R \subset \C$. Using the residue theorem we can compute the contour integral
    \begin{align*}

        \int_\gamma \frac{a}{z^2 + a^2} e^{-i\xi z} \; dz = 2\pi i \cdot \text{Res}\Big(\frac{a}{z^2 + a^2} e^{-i\xi z}, ia\Big)  = 2\pi i \cdot \lim_{z \to ia} (z-ia) \frac{a}{z^2 + a^2} e^{-i\xi z} = \pi e^{-a \abs{\xi}}.
    \end{align*}
    Now, taking the limit as $R \to \infty$ the left hand side becomes
    \begin{align*}
        \lim_{R \to \infty}
        \int_\gamma \frac{a}{z^2 + a^2} e^{-i\xi z} \; dz &= \calF\Big[ \frac{a}{x^{2}+a^{2}} \Big](\xi) + \lim_{R \to \infty} \int_{_{\substack{

        0 < \theta < \pi \\ z = Re^{i\theta}}}} \frac{a}{z^2 + a^2} e^{-i\xi z} \; dz
    \end{align*}
    where
    \[
        \Big | \int_{_{\substack{

        0 < \theta < \pi \\ z = Re^{i\theta}}}} \frac{a}{z^2 + a^2} e^{-i\xi z} \; dz \Big| \leq  \int_{_{\substack{

        0 < \theta < \pi \\ z = Re^{i\theta}}}} \Big|\frac{a}{z^2 + a^2}\Big|  e^{\xi R\sin\theta} \; Rd\theta \to 0
    \]
    since $\xi < 0$. In the case that $\xi > 0$ we instead set $\gamma$ to be the lower semi-circle contour and derive the same result. The proof of \eqref{fouriercalc2} follows this same outline.

\end{proof}

\subsubsection*{Outline of the Paper}
In Section \ref{potentialjuimpestimates}, we compute bounds on the vorticity terms by Taylor expanding the expressions of $\omega_{i}$ and $\Omega_{i}$ and then computing the Fourier transforms. Next, in Section \ref{sec:interfacedecay}, we decompose \eqref{eq: interface} into its linear and nonlinear parts and prove \eqref{maininequality} for $s=0,1$. In Section \ref{sec:l2est}, we prove \eqref{L2inequality} and then a higher order Sobolev estimate that is used for the existence argument. We conclude in Section \ref{sec:existence} by proving Theorem \ref{MainTheorem}.

\section{Potential Jump and Vorticity}
\label{potentialjuimpestimates}
In this section, we will decompose the potential jumps $\Omega_{1}$ and $\Omega_{2}$ and the vorticity terms $\omega_{1}$ and $\omega_{2}$ into linear and nonlinear parts. We can then use this decomposition to bound $\Omega_{i}$ and $\omega_i$ in terms of the interface function $f$.

First, we compute the Fourier transform of $\Omega_{2}$. Write $\Omega_{2} = \Omega_{21} + \Omega_{22}$ where
$$\Omega_{21}(\alpha) = -\frac{A_\kappa}{\pi} \int_\R \frac{f(\alpha - \beta) + h_2 }{\beta^2 + (f(\alpha - \beta) + h_2)^2} \Omega_1(\alpha - \beta)\; d\beta,$$
and
$$\Omega_{22}(\alpha) = -\frac{A_\kappa}{\pi} \int_\R \frac{\beta \partial_\alpha f(\alpha - \beta)}{\beta^2 + (f(\alpha - \beta) + h_2)^2} \Omega_1(\alpha - \beta)\; d\beta.$$
Taking the Fourier transform of $\Omega_{21}$, we obtain using \eqref{fouriercalc1}
\begin{align}
   \hat\Omega_{21}(\xi) &= -\frac{A_\kappa}{\pi} \int_\R \int_\R  \frac{f(\alpha - \beta) + h_2}{\beta^2 + (f(\alpha - \beta) + h_2)^2}\ \Omega_1(\alpha - \beta) e^{-i\xi \alpha}\; d\beta d\alpha \nonumber \\
    &= -\frac{A_\kappa}{\pi}\int_\R \int_\R  \frac{f(y) + h_2}{\beta^2 + (f(y) + h_2)^2}\  e^{-i\xi \beta}\; d\beta\; \Omega_1(y) e^{-i\xi y}\; dy \nonumber \\
    &= -\frac{A_\kappa}{\pi}\int_\R \pi e^{-(f(y) + h_2) \abs \xi }\cdot \Omega_1(y)  e^{-i\xi y}\; dy \nonumber \\
    &= -A_\kappa e^{-h_2 \abs \xi}  \int_\R e^{-f(y) \abs \xi} \cdot \Omega_1(y) e^{-i\xi y}\; dy \nonumber \\
    &= -A_\kappa \sum_{n=0}^{\infty}e^{-h_2 \abs \xi}  \int_\R \frac{(-f(y)|\xi|)^{n}}{n!} \cdot \Omega_1(y) e^{-i\xi y}\; dy \nonumber \\
    &= -A_\kappa\sum_{n=0}^{\infty}e^{-h_2 \abs \xi} \frac{(-1)^{n}|\xi|^{n}}{n!} \Big((\ast^{n}\hat{f}) \ast \hat \Omega_1 \Big)(\xi). \label{Omega21fourierexpansion}
\end{align}
For $\Omega_{22}$, we have
\begin{align}
    \hat{ \Omega}_{22} (\xi) &= -\frac{A_\kappa}{\pi} \int_\R d\beta \int_\R d\alpha\;  e^{-i \xi \alpha}  \frac{(\alpha - \beta) \partial_\alpha f(\beta)}{(\alpha - \beta)^2 + (f(\beta) + h_2)^2} \Omega_1(\beta)  \nonumber \\
    &= -\frac{A_\kappa}{\pi} \int_\R d\beta\;  \partial_\alpha f(\beta) \Omega_1(\beta) \calF\left[\tau_\beta\left(\frac{\alpha }{\alpha^2 + (f(\beta) + h_2)^2} \right) \right](\xi) \nonumber \\
    &= -\frac{A_\kappa}{\pi} \int_\R d\beta \; e^{-i\xi \beta} \partial_\alpha f(\beta) \Omega_1(\beta) \cdot -i\pi \sgn(\xi) e^{-(f(\beta) + h_2)\abs{\xi}} \nonumber\\
    &= A_\kappa \sum_{n= 0}^\infty i \sgn(\xi) e^{-h_2\abs{\xi} } \frac{(-1)^n \abs{\xi}^n}{n!} \Big((\ast^n \hat f) * \widehat{\partial_\alpha f} * \hat \Omega_1 \Big)(\xi). \label{Omega22fourierexpansion}
\end{align}
Next, we compute similarly for $\Omega_{1}$. By \eqref{Omega1}, we write the term $\Omega_{1}=\Omega_{11}+ \Omega_{12} - 2A_{\rho}f(\alpha)$ where
\begin{equation}\label{Omega11}
    \Omega_{11} (\alpha) = -\frac{A_\mu}{\pi} \int_\R \frac{\Delta_\beta f(\alpha) - \partial_\alpha f(\alpha - \beta)}{1 + \Delta_\beta f(\alpha)^2} \frac{\Omega_1(\alpha - \beta)}{\beta}\; d\beta,
\end{equation}
and
$$\Omega_{12}(\alpha) = - \frac{A_\mu}{\pi} \int_\R \frac{f(\alpha) + h_2}{\beta^2 + (f(\alpha) + h_2)^2}\Omega_2(\alpha - \beta) \; d\beta.$$
It can be seen that $\Omega_{11}$ has no part that is linear in $\Omega_{i}$. For $\Omega_{12}$, taking the Fourier transform, we obtain
\begin{align}
    \hat{\Omega}_{12}(\xi)
    &= -\frac{A_{\mu}}{\pi}\int_\R e^{-i\xi \alpha} \int_\R \int_\R \hat \Omega_2(\xi_1) e^{i\xi_{1}(\alpha-\beta)}d\xi_{1}\frac{f(\alpha) + h_2}{\beta^2 + (f(\alpha) + h_2)^2}\; d\beta d\alpha \nonumber\\
    &= -\frac{A_{\mu}}{\pi}\int_\R e^{-i\xi \alpha} \int_\R  \hat \Omega_2(\xi_1) e^{i\xi_{1}\alpha} \left(\int_\R e^{-i\xi_{1}\beta}\frac{f(\alpha) + h_2}{\beta^2 + (f(\alpha) + h_2)^2}\; d\beta \right) d\xi_{1} d\alpha \nonumber\\
     &= -\frac{A_{\mu}}{\pi}\int_\R e^{-i\xi\alpha} \int_\R e^{i \xi_1 \alpha} \; \hat\Omega_2(\xi_1) \ \pi e^{-(f(\alpha) + h_2)\abs{\xi_1}}\; d\xi_1 d\alpha \nonumber\\
    &= -A_{\mu}\sum_{n = 0}^{\infty} \frac{(-1)^n}{n!} \int_\R  e^{-i\xi\alpha} f(\alpha)^n \int_\R \hat \Omega_2(\xi_1) \  e^{-h_2\abs{\xi_1} } \ \abs{\xi_1}^n  e^{i \xi_1 \alpha}  \;d\xi_1 d\alpha  \nonumber\\
    &=-A_{\mu} \sum_{n = 0}^{\infty} \frac{(-1)^n}{n!}\Big((\hat \Omega_2 \  e^{-h_2\abs{\xi } }\ \abs{\xi}^n ) * (\ast^n \hat f) \Big)(\xi). \label{omega12fourierlinearpartsum}
\end{align}
The first term $\Omega_{11}$ does not have an explicit computation as the other terms, but it satisfies the following bound. The proof technique is used from \cite{CCGS}.
\begin{lemma}
For $\norm{f}_{\mathcal{F}^{1,1}_{0}} < 1$, we have the bound
\begin{equation}\label{Omega11abs}
    |\hat \Omega_{11}(\xi)| \leq  2 |A_\mu| \sum_{n=0}^{\infty}(\ast^{2n+1} |\widehat{\partial_\alpha f}| \ast \abs{\hat \Omega_1})(\xi).
\end{equation}
\end{lemma}
\begin{proof}
We first consider the term:
\begin{align*} \Omega_{11}(\alpha) &= -\frac{A_\mu}{\pi} \int_\R \frac{\Delta_\beta f(\alpha) - \partial_\alpha f(\alpha - \beta)}{1 + \Delta_\beta f(\alpha)^2} \frac{\Omega_1(\alpha - \beta)}{\beta}\; d\beta\\
&\eqdef \Omega_{111}(\alpha) + \Omega_{112} (\alpha).
\end{align*}
Taking the Fourier transform of the first term $\Omega_{111}$ and Taylor expanding the denominator for $|\Delta_\beta f(\alpha)| < 1$, we obtain
\begin{align*}
   \hat \Omega_{111} (\xi) &= -\frac{A_{\mu}}{\pi}\int_\R \calF \left[ \frac{\Delta_\beta f(\alpha)}{1 + \Delta_\beta f(\alpha)^2} \frac{\Omega_1(\alpha - \beta)}{\beta} \right](\xi)\; d\beta\\
    &= -\frac{A_{\mu}}{\pi}\sum_{n = 0}^\infty (-1)^n \int_\R \calF\left[ \Delta_\beta f(\alpha)^{2n+1} \frac{\Omega_1(\alpha - \beta)}{\beta}\right](\xi) \; d\beta \\
    &= -\frac{A_{\mu}}{\pi}\sum_{n = 0}^\infty  (-1)^n \int_\R (\ast^{2n+1} (\hat f m_\beta) * \widehat{\tau_\beta \Omega_1})(\xi)\; \frac{d\beta}{\beta}
\end{align*}
in which
\[
    m_\beta(\xi) = \frac{1 - e^{-i\xi\beta}}{\beta}.
\]
Expanding the convolution of the $n$-th term in the sum, we have
\begin{align*}
    \int_\R (\ast^{2n+1} (\hat f m_\beta) * \widehat{\tau_\beta \Omega_1})(\xi) \frac{d\beta}{\beta} = \int_\R d\xi_1 \ldots \int_\R d\xi_{2n+1} \hat f(\xi - \xi_1) \cdot \ldots \cdot \hat f(\xi_{2n} - \xi_{2n+1}) \cdot \hat \Omega_1(\xi_{2n+1}) \cdot M_n
\end{align*}
where
\begin{align*}
    M_n = M_n(\xi, \xi_1, \ldots, \xi_{2n+1}) = \int_\R d\beta \frac{e^{-i\xi_{2n+1} \beta}}{\beta} m_\beta(\xi - \xi_1) \cdot \ldots \cdot m_{\beta}(\xi_{2n} - \xi_{2n+1} ).
\end{align*}
Since
\[
    m_\beta(\xi) = i\xi \int_0^1 ds\; e^{i\beta(s-1)\xi}
\]
we have
\begin{align*}
    |M_n| &= \Big|i^{2n}(\xi - \xi_1) \cdot \ldots \cdot (\xi_{2n} - \xi_{2n+1}) \int_0^1 ds_1 \ldots \int_0^1 ds_{2n+1} \int_\R d\beta \frac{e^{-i\xi_{2n+1} \beta}}{\beta} e^{i \beta A}\Big|\\
    &= \Big| (\xi - \xi_1) \cdot \ldots \cdot (\xi_{2n} - \xi_{2n+1}) \int_0^1 ds_1 \ldots \int_0^1 ds_{2n+1}\  i\pi \sgn(A- \xi_{2n+1})\Big|\\
    &\leq \pi \cdot \abs{\xi - \xi_1} \cdot \ldots \cdot \abs{\xi_{2n} - \xi_{2n+1}}
\end{align*}
where
\[
    A = (s_1 - 1) (\xi - \xi_1) + \ldots + (s_{2n+1} - 1) (\xi_{2n} - \xi_{2n+1}).
\]
Therefore,
\begin{align*}
    \dabs{\int_\R (\ast^{2n+1} (\hat f m_\beta) * \widehat{\tau_\beta \omega_1})(\xi) \frac{d\beta}{\beta}}
    &\leq \int_\R d\xi_1 \ldots \int_\R d\xi_{2n+1} \abs{ \hat f(\xi - \xi_1)} \cdot \ldots \cdot \abs{\hat f(\xi_{2n} - \xi_{2n+1})} \cdot \abs{\hat \Omega_1(\xi_{2n+1})} \cdot \abs{M_n} \\
    &\leq \pi (\ast^{2n+1} |\widehat{\partial_\alpha f}| \ast \abs{\hat \Omega_1})(\xi).
\end{align*}
The term $\Omega_{112}$ is bounded by the same quantity using a similar computation. This concludes the proof.
\end{proof}

Next, we consider the vorticity terms. The Fourier transform of $\omega_{2}$ can be computed similarly to $\Omega_{21}$

\begin{equation}\label{omega2fourierexpansion}
   \hat\omega_2(\xi) = -A_\kappa\sum_{n=0}^{\infty}e^{-h_2 \abs \xi} \frac{(-1)^{n}|\xi|^{n}}{n!} \Big((\ast^{n}\hat{f}) \ast \hat \omega_1 \Big)(\xi).
\end{equation}
For $ \omega_{1}$, similarly to $\Omega_{1}$, we decompose $\omega_{1} = \omega_{11} + \omega_{12} -2A_{\rho}\partial_{\alpha}f$ where we have the analogous bound on $\hat \omega_{11}$
\begin{equation}\label{omega11est}
|\hat\omega_{11}(\xi)| \leq 2 |A_\mu|  \sum_{n=0}^{\infty}(\ast^{2n+1} |\widehat{\partial_\alpha f}| \ast \abs{\hat \omega_1})(\xi)
\end{equation}
and, $\hat \omega_{12}(\xi)$ is explicitly computed as $-i\xi \hat \Omega_{12}(\xi)$ using \eqref{productrule}
\begin{align}
\hat \omega_{12}(\xi) &=A_{\mu}\sum_{n = 0}^\infty \frac{(-1)^{n+1}}{n!} \Big((\ i \sgn(\cdot) e^{-h_2 \abs{\cdot }} \ \abs{ \cdot }^n \hat{\omega}_2) \ast \widehat{\partial_\alpha f} * (\ast^n \hat f) \Big)(\xi) \label{omega12calc} \\ &\hspace{1in} - A_{\mu}\sum_{n = 0}^{\infty} \frac{(-1)^n}{n!}\Big((\hat \omega_2 \ e^{-h_2\abs{\cdot } }\ \abs{\cdot}^n ) * (\ast^n \hat f) \Big)(\xi) \nonumber
\end{align}
Using the computations above, we can bound the vorticity terms in the frequency space norms $\mathcal{F}^{s,1}_{\nu}$ for $s=0,1$.
\begin{proposition}
The term $\omega_2$ satisfies

\begin{align}\label{lem:omega2fzeroneinitial}
     \|\omega_{2}\|_{\mathcal{F}^{0,1}_{\nu}} &\leq   \abs{A_\kappa}C_{0} \|\omega_1\|_{\mathcal{F}^{0,1}_{\nu}}
\end{align}
and
\begin{align}\label{lem:omega2foneoneinitial}
     \|\omega_{2}\|_{\mathcal{F}^{1,1}_{\nu}} &\leq   \abs{A_\kappa} \frac{C_{2}}{\norm{f}_{\mathcal{F}^{0,1}_{\nu}}} \|\omega_1\|_{\mathcal{F}^{0,1}_{\nu}}\|f\|_{\mathcal{F}^{1,1}_{\nu}} + \abs{A_\kappa}C_{0} \|\omega_1\|_{\mathcal{F}^{1,1}_{\nu}}
\end{align}
where
\begin{equation}\label{C0}C_{0} = C_{0}(\|f\|_{\mathcal{F}^{0,1}_{\nu}}) = \sum_{n = 0}^\infty \dnorm{ e^{-h_2\abs\xi} \frac{\abs{\xi}^n}{n!}}_{L^\infty} \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n = \sum_{n = 0}^\infty \frac{n^{n}}{e^nn!} \left(\frac{\norm{f}_{\mathcal{F}^{0,1}_{\nu}}}{h_2}\right)^n
\end{equation}
and
\begin{equation}\label{C2}
C_2 = \sum_{n=1}^{\infty} n \dnorm{ e^{-h_2\abs\xi} \frac{\abs{\xi}^n}{n!}}_{L^\infty} \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n = \sum_{n = 1}^\infty  \frac{n^{n+1}}{e^nn!} \left(\frac{\norm{f}_{\mathcal{F}^{0,1}_{\nu}}}{h_2}\right)^n
\end{equation}
which converge for $\norm{f}_{\mathcal{F}^{0,1}_{\nu}} < h_{2}$. Note that $C_0 \to 1$ and $C_2 \to 0$ in the limit $\norm{f}_{\mathcal{F}^{0,1}_{\nu}} \to 0$.
\end{proposition}
\begin{proof}
Using Young's inequalities for convolutions, \eqref{convolutionexp}, and \eqref{omega2fourierexpansion}:
\begin{align*}
\norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}} &\leq \abs{A_\kappa}   \sum_{n=0}^{\infty} \Big\| e^{\nu t \abs{\xi}} e^{-h_2 \abs \xi} \frac{(-1)^{n}|\xi|^{n}}{n!} \Big((\ast^{n}\hat{f}) \ast \hat \omega_1 \Big)(\xi)\Big\|_{L^1}\\
&\leq  \abs{A_\kappa} \sum_{n=0}^{\infty} \dnorm{ e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \Big\| e^{\nu t \abs{\xi}} (\ast^{n}\hat{f}) \ast \hat \omega_1 \Big\|_{L^1}\\
&\leq \abs{A_\kappa}\Big( \sum_{n = 0}^\infty \dnorm{ e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \|f\|_{\mathcal{F}^{0,1}_{\nu}}^n \Big) \|\omega_1\|_{\mathcal{F}^{0,1}_{\nu}}.
\end{align*}
This computation yields \eqref{lem:omega2fzeroneinitial}. Applying \eqref{productrule}, we obtain \eqref{lem:omega2foneoneinitial}.
\end{proof}
We will use \eqref{lem:omega2fzeroneinitial} and \eqref{lem:omega2foneoneinitial} implicitly to bound $\omega_{1}$ in the same norms.
\begin{lemma}
The vorticity term $\omega_{1}$ satisfies the bounds
\begin{equation}\label{omega1fzeronenubound}
\|\omega_1\|_{\mathcal{F}^{0,1}_{\nu}} \leq 2A_\rho C_1\norm{f}_{\mathcal{F}^{1,1}_{\nu}}
\end{equation}
and
\begin{equation}\label{omega1foneonenubound}
\|\omega_1\|_{\mathcal{F}^{1,1}_{\nu}} \leq 2A_\rho C_1 C_3  \norm{f}_{\mathcal{F}^{2,1}_{\nu}}
\end{equation}
where
\begin{equation}\label{C1}
     C_1 = \left(1- \abs{A_\mu} \left[\frac{2\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} +   \abs{A_\kappa} C_0^2  (1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}})\right]\right)\inv
     \end{equation}
     and
\begin{equation}\label{C3}
C_3 =  1 + 2 \abs{A_\mu} C_1 \bigg(\frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}} (1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2)}{(1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2)^2} + \frac{1}{2} \abs{A_\kappa} C_0 \Big[(C_0 + 2C_2) \norm{f}_{\mathcal{F}^{1,1}_{\nu}} + C_2(1 +  \norm{f}_{\mathcal{F}^{1,1}_{\nu}}  )\Big] \bigg)
\end{equation}
 are defined for $\|f\|_{\mathcal{F}^{0,1}_{\nu}}< k_0(|A_{\mu}|,|A_{\kappa}|)$ and $\|f\|_{\mathcal{F}^{1,1}_{\nu}} <  k_1(|A_{\mu}|,|A_{\kappa}|)$. As $\norm{f}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \to 0$, we have $C_1 \to (1- \abs{A_\kappa} \abs{A_\mu} )\inv $ and $C_3 \to 1$.
\end{lemma}
\begin{proof}
Similarly to the proof of \eqref{lem:omega2fzeroneinitial}, we use \eqref{convolutionexp} and Young's inequality to obtain from \eqref{omega11est} that
\begin{align}
\|\omega_{11}\|_{\mathcal{F}^{0,1}_{\nu}} &\leq 2|A_{\mu}| \sum_{n=0}^{\infty}\|f\|_{\mathcal{F}^{1,1}_{\nu}}^{2n+1}\|\omega_{1}\|_{\mathcal{F}^{0,1}_{\nu}}\nonumber\\
&\leq |A_{\mu}|\frac{2\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2}    \|\omega_{1}\|_{\mathcal{F}^{0,1}_{\nu}}.\label{omega11fzerone}
\end{align}
We can also bound $\omega_{12}$ from \eqref{omega12calc} by
\begin{align}
\|\omega_{12}\|_{\mathcal{F}^{0,1}_{\nu}} &\leq |A_{\mu}|C_{0}(1+\|f\|_{\mathcal{F}^{1,1}_{\nu}})\|\omega_{2}\|_{\mathcal{F}^{0,1}_{\nu}} \nonumber\\
&\leq |A_\kappa| |A_{\mu}|C_{0}^{2}(1+\|f\|_{\mathcal{F}^{1,1}_{\nu}})\|\omega_1\|_{\mathcal{F}^{0,1}_{\nu}} \label{omega12fzerone}
\end{align}
where we used \eqref{lem:omega2fzeroneinitial} in the second inequality. Hence, we now have that
$$\|\omega_{1}\|_{\mathcal{F}^{0,1}_{\nu}} \leq 2 |A_{\mu}|\frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} \|\omega_{1}\|_{\mathcal{F}^{0,1}_{\nu}} + |A_\kappa |  |A_{\mu}|C_{0}^{2}(1+\|f\|_{\mathcal{F}^{1,1}_{\nu}})\|\omega_{1}\|_{\mathcal{F}^{0,1}_{\nu}} + 2A_\rho\|f\|_{\mathcal{F}^{1,1}_{\nu}}.$$
Using \eqref{lem:omega2fzeroneinitial} solving for $\|\omega_{1}\|_{\mathcal{F}^{0,1}_{\nu}}$ in the inequality implies \eqref{omega1fzeronenubound}. The estimate \eqref{omega1foneonenubound} follows similarly by applying \eqref{productrule}. Next, via \eqref{interpolation} we compute
\begin{align*}
\norm{\omega_{11}}_{\mathcal{F}^{1,1}_{\nu}} &\leq 2|A_{\mu}|\frac{1+\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{2}}{(1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2)^{2}}\|f\|_{\mathcal{F}^{2,1}_{\nu}}    \|\omega_{1}\|_{\mathcal{F}^{0,1}_{\nu}} + |A_{\mu}|\frac{2\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2}    \|\omega_{1}\|_{\mathcal{F}^{1,1}_{\nu}},\\
\norm{\omega_{12}}_{\mathcal{F}^{1,1}_{\nu}} &\leq \abs{A_\mu} (C_0 + C_2)\norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}}\norm{f}_{\mathcal{F}^{2,1}_{\nu}} + \abs{A_\mu} C_0   \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{\omega_2}_{\mathcal{F}^{1,1}_{\nu}}\\
&\hspace{5em} + \abs{A_\mu} \Big( \frac{C_2}{\norm{f}_{\mathcal{F}^{0,1}_{\nu}}} \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}}  + C_0 \norm{\omega_2}_{\mathcal{F}^{1,1}_{\nu}} \Big).
\end{align*}
Now, using \eqref{omega1fzeronenubound} and \eqref{lem:omega2fzeroneinitial} gives \eqref{eq:omega2fzerone} from the next proposition. Using \eqref{eq:omega2fzerone}, \eqref{lem:omega2foneoneinitial}, \eqref{omega1fzeronenubound} and applying the interpolation \eqref{interpolation}, we obtain
\begin{align}
\norm{\omega_{11}}_{\mathcal{F}^{1,1}_{\nu}} &\leq 4 A_\rho \abs{A_\mu} C_1 \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}} (1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2)}{(1- \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2)^2} \norm{f}_{\mathcal{F}^{2,1}_{\nu}} + 2\abs{A_\mu} \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}} \label{omega11foneone}\\
\norm{\omega_{12}}_{\mathcal{F}^{1,1}_{\nu}} &\leq 2 A_\rho \abs{A_\kappa} \abs{A_\mu} C_0 C_1 (C_0 + 2C_2) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{f}_{\mathcal{F}^{2,1}_{\nu}} +  \abs{A_\kappa} \abs{A_\mu} C_0^2(1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}}) \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}}\nonumber\\
&\hspace{5em} + 2 A_\rho \abs{A_\kappa}\abs{A_\mu} C_0 C_1 C_2 (1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}}) \norm{f}_{\mathcal{F}^{2,1}_{\nu}}  \label{omega12foneone}.
\end{align}
Computing implicitly as before yields the bound.
\end{proof}

Plugging the estimates \eqref{omega1fzeronenubound} and \eqref{omega1foneonenubound} into \eqref{lem:omega2fzeroneinitial} and \eqref{lem:omega2foneoneinitial}, and then using \eqref{interpolation} we obtain
\begin{proposition}
The term $\omega_{2}$ is bounded as
   \begin{align}
\|\omega_{2}\|_{\mathcal{F}^{0,1}_{\nu}} &\leq   2  A_\rho \abs{A_\kappa} C_0 C_1 \|f\|_{\mathcal{F}^{1,1}_{\nu}}  \label{eq:omega2fzerone}\\
\|\omega_2\|_{\mathcal{F}^{1,1}_{\nu}} &\leq 2 A_\rho \abs{A_\kappa} C_1 C_4 \norm{f}_{\mathcal{F}^{2,1}_{\nu}}\label{omega2foneone}
\end{align}
where
\begin{equation}\label{C4}
C_{4} = C_{2} + C_{0}C_{3} \to 1 \quad \text{as} \quad \norm{f}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \to 0.
\end{equation}
\end{proposition}

\begin{comment}
The linear expression \eqref{Omega2linearpart} allows us to prove the following estimate.
\begin{lemma} The term $\Omega_2$ satisfies the bound
\label{lem:Omega2fzerone}
\begin{align}\label{Omega2zeroest}
     \|\Omega_{2}\|_{\mathcal{F}^{0,1}_{\nu}} &\leq  |A_{\kappa}|(1+ \|f\|_{\mathcal{F}^{1,1}_{\nu}})C_{1}(\norm{f}_{\mathcal{F}^{0,1}_{\nu}})\|\Omega_1\|_{\mathcal{F}^{0,1}_{\nu}}
\end{align}
and
\begin{multline}\label{Omega2oneest}
     \|\Omega_{2}\|_{\mathcal{F}^{1,1}_{\nu}} \leq  |A_{\kappa}|(1+ \|f\|_{\mathcal{F}^{1,1}_{\nu}})C_{1}'(\norm{f}_{\mathcal{F}^{0,1}_{\nu}})\|f\|_{\mathcal{F}^{1,1}_{\nu}}|\Omega_1\|_{\mathcal{F}^{0,1}_{\nu}} +  |A_{\kappa}|C_{1}(\norm{f}_{\mathcal{F}^{0,1}_{\nu}})\|\Omega_1\|_{\mathcal{F}^{0,1}_{\nu}}\|f\|_{\mathcal{F}^{2,1}_{\nu}} \\ +  |A_{\kappa}|(1+ \|f\|_{\mathcal{F}^{1,1}_{\nu}})C_{1}(\norm{f}_{\mathcal{F}^{0,1}_{\nu}})\|\Omega_1\|_{\mathcal{F}^{1,1}_{\nu}}
\end{multline}
for
\begin{align}
    C_{1}(\norm{f}_{\mathcal{F}^{0,1}_{\nu}}) &= \sum_{n = 0}^\infty \dnorm{ e^{-h_2\abs\xi} \frac{\abs{\xi}^n}{n!}}_{L^\infty} \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n = \sum_{n = 0}^\infty \frac{n^{n}}{e^nn!} \left(\frac{\norm{f}_{\mathcal{F}^{0,1}_{\nu}}}{h_2}\right)^n.
\end{align}
\end{lemma}
\begin{proof}
Using the Fourier transform calculations \eqref{Omega21fourierexpansion} and \eqref{Omega22fourierexpansion} and applying Young's inequality with \eqref{convolutionexp}, we obtain
\begin{align*}
\norm{\Omega_2}_{\mathcal{F}^{0,1}_{\nu}} &\leq \abs{A_\kappa}   \sum_{n=0}^{\infty} \Big\| e^{\nu t |\xi|} e^{-h_2 \abs \xi} \frac{(-1)^{n}|\xi|^{n}}{n!} \Big(((\ast^{n}\hat{f})  \ast \hat \Omega_1 \Big)(\xi)\Big\|_{L^1} \\
&\hspace{0.25in}+ \abs{A_\kappa}   \sum_{n=0}^{\infty} \Big\| e^{\nu t \abs{\xi}} e^{-h_2 \abs \xi} \frac{(-1)^{n}|\xi|^{n}}{n!} \Big((\ast^{n}\hat{f})\ast \widehat{\partial_{\alpha}f} \ast \hat \Omega_1\Big)(\xi)\Big\|_{L^1}\\
&\leq  \abs{A_\kappa} \sum_{n=0}^{\infty} \dnorm{ e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \Big\| e^{\nu t \abs{\xi}} (\ast^{n}\hat{f}) \ast \hat \Omega_1 \Big\|_{L^1} \\
&\hspace{0.75in} +\dnorm{e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \Big\| e^{\nu t \abs{\xi}} (\ast^{n}\hat{f}) \ast \widehat{\partial_{\alpha}f} \ast \hat \Omega_1 \Big\|_{L^1}\\
&\leq \abs{A_\kappa} \sum_{n = 0}^\infty \dnorm{e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \|f\|_{\mathcal{F}^{0,1}_{\nu}}^{n} \|\Omega_1\|_{\mathcal{F}^{0,1}_{\nu}} \\
&\hspace{0.75in}+ \abs{A_\kappa} \sum_{n = 0}^\infty \dnorm{ e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \|f\|_{\mathcal{F}^{0,1}_{\nu}}^{n} \|f\|_{\mathcal{F}^{1,1}_{\nu}}\|\Omega_1\|_{\mathcal{F}^{0,1}_{\nu}}\\
&\eqdef |A_{\kappa}|c_{1}(1+ \|f\|_{\mathcal{F}^{1,1}_{\nu}})\|\Omega_1\|_{\mathcal{F}^{0,1}_{\nu}}.
\end{align*}
Performing a similar estimate in $\mathcal{F}^{1,1}_{\nu}$, we can apply the product rule for convolutions \eqref{productrule} to obtain \eqref{Omega2oneest}.

Applying Young's inequalities for convolutions and  \eqref{convolutionexp} to the nonlinear part of \eqref{Omega2linearnonlinear}, we obtain:
\begin{align*}
\norm{N(\Omega_2)\frac{A_\kappa A_\mu}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}}_{\mathcal{F}^{0,1}_{\nu}} &\leq \abs{A_\kappa}   \sum_{n=0}^{\infty} \Big\|\frac{A_\kappa A_\mu}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu} e^{\nu t \abs{\xi}} e^{-h_2 \abs \xi} \frac{(-1)^{n}|\xi|^{n+1}}{(n+1)!} \Big(((\ast^{n+1}\hat{f})  \ast \hat \Omega_1 \Big)(\xi)\Big\|_{L^1} \\
&\hspace{0.25in}+ \abs{A_\kappa}   \sum_{n=0}^{\infty} \Big\|\frac{A_\kappa A_\mu}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu} e^{\nu t \abs{\xi}} e^{-h_2 \abs \xi} \frac{(-1)^{n}|\xi|^{n}}{n!} \Big((\ast^{n}\hat{f})\ast \widehat{\partial_{\alpha}f} \ast \hat \Omega_1\Big)(\xi)\Big\|_{L^1}\\
&\leq  \abs{A_\kappa} \sum_{n=0}^{\infty} \dnorm{\frac{A_\kappa A_\mu}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu} e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n+1}}{(n+1)!}}_{L^\infty} \Big\| e^{\nu t \abs{\xi}} (\ast^{n+1}\hat{f}) \ast \hat \Omega_1 \Big\|_{L^1} \\
&\hspace{0.75in} +\dnorm{\frac{A_\kappa A_\mu}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu} e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \Big\| e^{\nu t \abs{\xi}} (\ast^{n}\hat{f}) \ast \widehat{\partial_{\alpha}f} \ast \hat \Omega_1 \Big\|_{L^1}\\
&\leq \abs{A_\kappa} \sum_{n = 0}^\infty \dnorm{ \frac{A_\kappa A_\mu}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n+1}}{(n+1)!}}_{L^\infty} \|f\|_{\mathcal{F}^{0,1}_{\nu}}^{n+1} \|\Omega_1\|_{\mathcal{F}^{0,1}_{\nu}} \\
&\hspace{0.75in}+ \abs{A_\kappa} \sum_{n = 0}^\infty \dnorm{ \frac{A_\kappa A_\mu}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{(n)!}}_{L^\infty} \|f\|_{\mathcal{F}^{0,1}_{\nu}}^{n} \|f\|_{\mathcal{F}^{1,1}_{\nu}}\|\Omega_1\|_{\mathcal{F}^{0,1}_{\nu}}\\
&\eqdef (c_{1} \|f\|_{\mathcal{F}^{1,1}_{\nu}}+c_{2}\|f\|_{\mathcal{F}^{0,1}_{\nu}})\|\Omega_1\|_{\mathcal{F}^{0,1}_{\nu}}.
\end{align*}
Next, for the nonlinear part of \eqref{Omega1linearnonlinear}, we use the estimate from \eqref{Omega11abs} to obtain
\begin{equation}\label{NOmega1abs}
|N(\Omega_{1})(\xi)| \leq  2 A_\mu \sum_{n=0}^{\infty} (\ast^{2n+1} |\cdot||\hat{f}| \ast \abs{\hat \Omega_1})(\xi) + A_{\mu} \sum_{n = 1}^{\infty} \Big((\hat \Omega_2  \frac{e^{-h_2\abs{\cdot }}|\cdot|^{n}}{n!} ) \ast (\ast^n \hat f) \Big)(\xi).
\end{equation}
Applying Young's inequality and \eqref{convolutionexp}, we obtain
\begin{align*}
\|N(\Omega_{1})\|_{\mathcal{F}^{0,1}_{\nu}}  & \leq 2A_\mu \sum_{n = 0}^\infty \norm{f}^{2n+1}_{\mathcal{F}^{1,1}_{\nu}} \norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}} + A_{\mu}\sum_{n=1}^{\infty} \norm{f}^{n}_{\mathcal{F}^{0,1}_{\nu}} \norm{\Omega_2}_{\mathcal{F}^{0,1}_{\nu}} \Big\| \frac{e^{-h_2\abs{\cdot }}|\cdot|^{n}}{n!}\Big\|_{L^{\infty}}\\
& = A_\mu \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} \norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}}+ A_{\mu}\sum_{n=1}^{\infty} \norm{f}^{n}_{\mathcal{F}^{0,1}_{\nu}} \norm{\Omega_2}_{\mathcal{F}^{0,1}_{\nu}} \Big\| \frac{e^{-h_2\abs{\cdot }}|\cdot|^{n}}{n!}\Big\|_{L^{\infty}}\\
&= c_{3} \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}} + c_{4}\norm{f}_{\mathcal{F}^{0,1}_{\nu}} \norm{\Omega_2}_{\mathcal{F}^{0,1}_{\nu}}.
\end{align*}
Hence, we obtain from \eqref{Omega2linearnonlinear} that

Combining the above estimates on $N(\Omega_{1})$ and $N(\Omega_{2})$, we conclude the proof.

\end{proof}
We will now use the bounds \eqref{Omega2zeroest} and \eqref{Omega2oneest}  in the a similar estimate for $\Omega_{1}$ to obtain and bound it purely in the interface $f(\alpha,t)$.
\begin{lemma}
\label{lem:omega1fzerone}
The term $\Omega_1$ satisfies
\begin{equation}\label{Omega1zeroest}
\|\Omega_{1}\|_{\mathcal{F}^{0,1}_{\nu}}\leq 2 A_\rho(1- C_{2})^{-1} \norm{f}_{\mathcal{F}^{1,1}_{\nu}}
\end{equation}
and
\begin{equation}\label{Omega1oneest}
\|\Omega_{1}\|_{\mathcal{F}^{1,1}_{\nu}}\leq  \norm{f}_{\mathcal{F}^{2,1}_{\nu}}
\end{equation}
where
\begin{equation}\label{c2}
C_{2} =  |A_{\mu}|\Big( 2 \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} + |A_{\kappa}| C_{1}(\|f\|_{\mathcal{F}^{0,1}_{\nu}})^{2}(1+\|f\|_{\mathcal{F}^{1,1}_{\nu}}) \Big)
\end{equation}
\end{lemma}

\begin{proof}
In the decomposition $\Omega_{1}=\Omega_{11}+ \Omega_{12} - 2A_{\rho}\partial_{\alpha}f(\alpha)$, we first use \eqref{Omega11abs} and apply \eqref{convolutionexp} and Young's convolution inequality, to obtain the bound
\begin{align}
    \norm{\Omega_{11}}_{\mathcal{F}^{0,1}_{\nu}} &\leq 2|A_\mu|  \sum_{n=0}^\infty \int_{\R} e^{\nu t\abs{\xi}} (\ast^{2n+1} |\cdot||\hat{f}|\ast \abs{\hat \Omega_1})(\xi) \; d\xi \nonumber \\
    & \leq 2|A_\mu| \sum_{n = 0}^\infty \norm{f}^{2n+1}_{\mathcal{F}^{1,1}_{\nu}} \norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}} \nonumber \\
    & = 2|A_\mu| \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} \norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}}. \label{Omega11zeroest}
\end{align}
Next, we can similarly compute from \eqref{omega12fourierlinearpartsum} that
\begin{equation}\label{Omega12zeroest}
\norm{\Omega_{12}}_{\mathcal{F}^{0,1}_{\nu}} \leq |A_{\mu}| \sum_{n=0}^\infty \dnorm{\frac{ e^{-h_2 \abs \xi}\abs{\xi}^{n}}{n!}}_{L^\infty} \norm{f}^n_{\mathcal{F}^{0,1}_{\nu}} \norm{\Omega_2}_{\mathcal{F}^{0,1}_{\nu}} =  |A_{\mu}| C_{1}(\|f\|_{\mathcal{F}^{0,1}_{\nu}}) \norm{\Omega_2}_{\mathcal{F}^{0,1}_{\nu}}.
\end{equation}
Next, by the bound \eqref{Omega2zeroest}, we obtain
\begin{equation}
\norm{\Omega_{12}}_{\mathcal{F}^{0,1}_{\nu}} \leq |A_{\mu}A_{\kappa}| C_{1}(\|f\|_{\mathcal{F}^{0,1}_{\nu}})^{2}(1+\|f\|_{\mathcal{F}^{1,1}_{\nu}})   \|\Omega_1\|_{\mathcal{F}^{0,1}_{\nu}}
\end{equation}
Combining \eqref{Omega11zeroest} and \eqref{Omega12zeroest}, we obtain that
$$\norm{\Omega_{1}}_{\mathcal{F}^{0,1}_{\nu}} \leq |A_{\mu}|\Big( 2 \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} + |A_{\kappa}| C_{1}(\|f\|_{\mathcal{F}^{0,1}_{\nu}})^{2}(1+\|f\|_{\mathcal{F}^{1,1}_{\nu}}) \Big)\norm{\Omega_{1}}_{\mathcal{F}^{0,1}_{\nu}}  + 2A_\rho \|f\|_{\mathcal{F}^{1,1}_{\nu}}$$
which implies the final estimate \eqref{Omega1zeroest}.
Next, for $\Omega_{1}$ in $\mathcal{F}^{1,1}_{\nu}$, we have by \eqref{productrule} that
\begin{align*}
\|\Omega_{1}\|_{\mathcal{F}^{1,1}_{\nu}} &\leq  |A_{\mu}|\Big( 2 \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{2}+ 1}{(1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2)^{2}}\|f\|_{\mathcal{F}^{2,1}_{\nu}} + |A_{\kappa}| 2C_{1}(\|f\|_{\mathcal{F}^{0,1}_{\nu}})C_{1}'(\|f\|_{\mathcal{F}^{0,1}_{\nu}})\|f\|_{\mathcal{F}^{1,1}_{\nu}}(1+\|f\|_{\mathcal{F}^{1,1}_{\nu}}) \\ &\hspace{0.5in}+  |A_{\kappa}| C_{1}(\|f\|_{\mathcal{F}^{0,1}_{\nu}})^{2}\|f\|_{\mathcal{F}^{2,1}_{\nu}}\Big)\norm{\Omega_{1}}_{\mathcal{F}^{0,1}_{\nu}} + C_{2}\norm{\Omega_{1}}_{\mathcal{F}^{1,1}_{\nu}}  + 2A_\rho \|f\|_{\mathcal{F}^{2,1}_{\nu}}.
\end{align*}
Now, using \eqref{Omega1zeroest}, we obtain \eqref{Omega1oneest}.

By \eqref{Omega1}, we write the term $\Omega_{1}=\Omega_{11}+ \Omega_{12} - 2A_{\rho}\partial_{\alpha}f(\alpha)$ where
$$ \Omega_{11} = -\frac{A_\mu}{\pi} \int_\R \frac{\Delta_\beta f(\alpha) - \partial_\alpha f(\alpha - \beta)}{1 + \Delta_\beta f(\alpha)^2} \frac{\Omega_1(\alpha - \beta)}{\beta}\; d\beta = \Omega_{111} + \Omega_{112}$$
and
$$\Omega_{12} = - \frac{A_\mu}{\pi} \int_\R \frac{f(\alpha) + h_2}{\beta^2 + (f(\alpha) + h_2)^2}\Omega_2(\alpha - \beta) \; d\beta$$
Taking the Fourier transform of the first term $\Omega_{111}$ and Taylor expanding the denominator for $|\Delta_\beta f(\alpha)| < 1$, we obtain
\begin{align*}
   \widehat{\Omega_{111}}(\xi) &= -\frac{A_{\mu}}{\pi}\int_\R \calF \left[ \frac{\Delta_\beta f(\alpha)}{1 + \Delta_\beta f(\alpha)^2} \frac{\Omega_1(\alpha - \beta)}{\beta} \right](\xi)\; d\beta\\
    &= -\frac{A_{\mu}}{\pi}\sum_{n = 0}^\infty (-1)^n \int_\R \calF\left[ \Delta_\beta f(\alpha)^{2n+1} \frac{\Omega_1(\alpha - \beta)}{\beta}\right](\xi) \; d\beta \\
    &= -\frac{A_{\mu}}{\pi}\sum_{n = 0}^\infty  (-1)^n \int_\R (\ast^{2n+1} (\hat f m_\beta) * \widehat{\tau_\beta \Omega_1})(\xi)\; \frac{d\beta}{\beta}
\end{align*}
in which
\[
    m_\beta(\xi) = \frac{1 - e^{-i\xi\beta}}{\beta}.
\]
Expanding the convolution of the $n$-th term in the sum, we have
\begin{align*}
    \int_\R (\ast^{2n+1} (\hat f m_\beta) * \widehat{\tau_\beta \omega_1})(\xi) \frac{d\beta}{\beta} = \int_\R d\xi_1 \ldots \int_\R d\xi_{2n} \hat f(\xi - \xi_1) \cdot \ldots \cdot \hat f(\xi_{2n-1} - \xi_{2n}) \cdot \hat \Omega_1(\xi_{2n}) \cdot M_n
\end{align*}
where
\begin{align*}
    M_n = M_n(\xi, \xi_1, \ldots, \xi_{2n}) = \int_\R d\beta \frac{e^{-i\xi_{2n} \beta}}{\beta} m_\beta(\xi - \xi_1) \cdot \ldots \cdot m_{\beta}(\xi_{2n-1} - \xi_{2n} ).
\end{align*}
Since
\[
    m_\beta(\xi) = i\xi \int_0^1 ds\; e^{i\beta(s-1)\xi}
\]
we have
\begin{align*}
    |M_n| &= \Big|i^{2n}(\xi - \xi_1) \cdot \ldots \cdot (\xi_{2n-1} - \xi_{2n}) \int_0^1 ds_1 \ldots \int_0^1 ds_{2n} \int_\R d\beta \frac{e^{-i\xi_{2n} \beta}}{\beta} e^{i \beta A}\Big|\\
    &= \Big| (\xi - \xi_1) \cdot \ldots \cdot (\xi_{2n-1} - \xi_{2n}) \int_0^1 ds_1 \ldots \int_0^1 ds_{2n}\  i\pi \sgn(A- \xi_{2n})\Big|\\
    &\leq \pi \cdot \abs{\xi - \xi_1} \cdot \ldots \cdot \abs{\xi_{2n-1} - \xi_{2n}}
\end{align*}
where
\[
    A = (s_1 - 1) (\xi - \xi_1) + \ldots + (s_{2n} - 1) (\xi_{2n-1} - \xi_{2n}).
\]
Therefore,
\begin{align*}
    \dabs{\int_\R (\ast^{2n+1} (\hat f m_\beta) * \widehat{\tau_\beta \omega_1})(\xi) \frac{d\beta}{\beta}}
    &\leq \int_\R d\xi_1 \ldots \int_\R d\xi_{2n} \abs{ \hat f(\xi - \xi_1)} \cdot \ldots \cdot \abs{\hat f(\xi_{2n-1} - \xi_{2n})} \cdot \abs{\hat \Omega_1(\xi_{2n})} \cdot \abs{M_n} \\
    &\leq \pi (\ast^{2n+1} \abs{\widehat{\partial_\alpha f}} * \abs{\hat \Omega_1})(\xi).
\end{align*}
Applying \eqref{convolutionexp} and Young's convolution inequality, we obtain the bound
\begin{align}
    \norm{\Omega_{111}}_{\mathcal{F}^{0,1}_{\nu}} &\leq A_\mu  \sum_{n=0}^\infty \int_{\R} e^{\nu t\abs{\xi}} (\ast^{2n+1} \abs{\widehat{\partial_\alpha f}} * \abs{\hat \Omega_1})(\xi) \; d\xi \nonumber \\
    & \leq A_\mu \sum_{n = 0}^\infty \norm{f}^{2n+1}_{\mathcal{F}^{1,1}_{\nu}} \norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}} \nonumber \\
    & = A_\mu \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} \norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}}. \label{Omega111zeroest}
\end{align}
By a similar calculation, we also have
\begin{equation}\label{Omega112zeroest}
    \norm{\Omega_{112}}_{\mathcal{F}^{0,1}_{\nu}} \leq A_\mu \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} \norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}}.
\end{equation}
Next, we consider the term $\Omega_{12}$.
Apply Young's inequality and \eqref{convolutionexp}, we can compute

\end{proof}
\end{comment}

\begin{comment}

We can perform a similar iteration for $\Omega_{1}$:
\begin{align*}
    L(\hat \Omega_1) &=-A_\mu e^{-h_2 \abs \xi}  (  L(\hat \Omega_2)+ N(\hat \Omega_2))- 2A_{\rho}\widehat{\partial_{\alpha}f}(\xi)\\
    &= A_{\kappa} A_{\mu} e^{-2 h_2 \abs \xi} (  L(\hat \Omega_1)+ N(\hat \Omega_1)) - A_\mu e^{-h_2 \abs \xi}  N(\hat \Omega_2) - 2A_{\rho}\widehat{\partial_{\alpha}f}(\xi)\\
    &= A_{\kappa} A_{\mu} e^{-2 h_2 \abs \xi} ( -A_\mu e^{-h_2 \abs \xi}  \widehat{\Omega_{2}}(\xi) - 2A_{\rho}\widehat{\partial_{\alpha}f}(\xi)) + A_{\kappa} A_{\mu} e^{-2 h_2 \abs \xi} N(\hat \Omega_1) - A_\mu e^{-h_2 \abs \xi}  N(\hat \Omega_2) - 2A_{\rho}\widehat{\partial_{\alpha}f}(\xi)\\
    &= -A_{\kappa} A_{\mu}^{2} e^{-3 h_2 \abs \xi}  ( L(\hat \Omega_2)+ N(\hat \Omega_2) ) - A_\mu e^{-h_2 \abs \xi} ()  N(\hat \Omega_2) - 2A_{\rho}\widehat{\partial_{\alpha}f}(\xi)(1-A_{\kappa}A_{\mu}).
\end{align*}
\end{comment}

\begin{comment}
\section{Vorticity Estimates}
\label{sec:vorticity_ests}
In this section, we will decompose the vorticity terms into the linear and nonlinear components and then prove the relevant bounds on the vorticity terms $\omega_{1}$ and $\omega_{2}$.
\begin{proposition}
We have the estimates
\begin{align}
\label{eq:omega1fzerone}
 \|\omega_1\|_{\mathcal{F}^{0,1}_{\nu}} &\leq 2A_\rho C_1\norm{f}_{\mathcal{F}^{1,1}_{\nu}}\\
 \label{eq:omega2fzerone}
\|\omega_{2}\|_{\mathcal{F}^{0,1}_{\nu}} &\leq   2  A_\rho \abs{A_\kappa} C_0 C_1 \|f\|_{\mathcal{F}^{1,1}_{\nu}}\\
\label{eq:omega1foneone}
\|\omega_1\|_{\mathcal{F}^{1,1}_{\nu}} &\leq 2A_\rho C_1 C_3 \norm{f}_{\mathcal{F}^{2,1}_{\nu}}\\
\label{omega2foneone}
\|\omega_2\|_{\mathcal{F}^{1,1}_{\nu}} &\leq 2 A_\rho \abs{A_\kappa} C_1 C_4 \norm{f}_{\mathcal{F}^{2,1}_{\nu}}
\end{align}

for small $f$ (size to be specified) where
\begin{align}\label{C0}
C_0 &= \sum_{n = 0}^\infty \dnorm{ e^{-h_2\abs\xi} \frac{\abs{\xi}^n}{n!}}_{L^\infty} \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n = \sum_{n = 0}^\infty \frac{e^{-n}}{n!} \Big( \frac{n}{h_2}\Big)^n \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n\\
\label{C1}
     C_1 &= \left(1- 2\abs{A_\mu} \left[\frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} + \frac{1}{2}  \abs{A_\kappa} C_0^2  (1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}})\right]\right)\inv\\
\label{C2}
C_2 &= \sum_{n=1}^{\infty} n \dnorm{ e^{-h_2\abs\xi} \frac{\abs{\xi}^n}{n!}}_{L^\infty} \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n = \sum_{n = 1}^\infty n\frac{e^{-n}}{n!} \Big( \frac{n}{h_2}\Big)^n \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n.\\
\label{C3}
C_3 &=  1 + 2 \abs{A_\mu} C_1 \bigg(\frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2 (\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2 + 1)}{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2 - 1} + \frac{1}{2} \abs{A_\kappa} C_0 \Big[(C_2 + (C_0 + 2C_2) \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \Big] \bigg) \\
C_4 &= (C_2 + C_0 C_3)
\end{align}
\end{proposition}
\begin{proof}
    Idea: combine the following four lemmas to get the estimates. Roughly speaking the first two (resp. last two) lemmas give the first two (resp. last two) estimates. We need $f$ to be small enough for the constants to make sense.
\end{proof}

\begin{lemma} The term $\omega_2$ satisfies
\label{lem:omega2fzerone}
\begin{align}
     \|\omega_{2}\|_{\mathcal{F}^{0,1}_{\nu}} &\leq   \abs{A_\kappa}C_0 \|\omega_1\|_{\mathcal{F}^{0,1}_{\nu}}
\end{align}
\end{lemma}
\begin{proof}
Taking the Fourier transform of $\omega_{2}$, we obtain
\begin{align}
   \hat\omega_2(\xi) &= -\frac{A_\kappa}{\pi} \int_\R \int_\R  \frac{f(\alpha - \beta) + h_2}{\beta^2 + (f(\alpha - \beta) + h_2)^2}\ \omega_1(\alpha - \beta) e^{-i\xi \alpha}\; d\beta d\alpha \nonumber \\
    &= -\frac{A_\kappa}{\pi}\int_\R \int_\R  \frac{f(y) + h_2}{\beta^2 + (f(y) + h_2)^2}\  e^{-i\xi \beta}\; d\beta\; \omega_1(y) e^{-i\xi y}\; dy \nonumber \\
    &= -\frac{A_\kappa}{\pi}\int_\R \pi e^{-(f(y) + h_2) \abs \xi }\cdot \omega_1(y)  e^{-i\xi y}\; dy \nonumber \\
    &= -A_\kappa e^{-h_2 \abs \xi}  \int_\R e^{-f(y) \abs \xi} \cdot \omega_1(y) e^{-i\xi y}\; dy \nonumber \\
    &= -A_\kappa \sum_{n=0}^{\infty}e^{-h_2 \abs \xi}  \int_\R \frac{(-f(y)|\xi|)^{n}}{n!} \cdot \omega_1(y) e^{-i\xi y}\; dy \nonumber \\
    &= -A_\kappa\sum_{n=0}^{\infty}e^{-h_2 \abs \xi} \frac{(-1)^{n}|\xi|^{n}}{n!} \Big((\ast^{n}\hat{f}) \ast \hat \omega_1 \Big)(\xi). \label{omega2fourierexpansion}
\end{align}
Above, the Fourier integral in the third equality can be computed via a semicircle contour in the complex plane. Next, using Young's inequalities for convolutions and \eqref{convolutionexp}:
\begin{align*}
\norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}} &\leq \abs{A_\kappa}   \sum_{n=0}^{\infty} \Big\| e^{\nu t \abs{\xi}} e^{-h_2 \abs \xi} \frac{(-1)^{n}|\xi|^{n}}{n!} \Big((\ast^{n}\hat{f}) \ast \hat \omega_1 \Big)(\xi)\Big\|_{L^1}\\
&\leq  \abs{A_\kappa} \sum_{n=0}^{\infty} \dnorm{ e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \Big\| e^{\nu t \abs{\xi}} (\ast^{n}\hat{f}) \ast \hat \omega_1 \Big\|_{L^1}\\
&\leq \abs{A_\kappa}\Big( \sum_{n = 0}^\infty \dnorm{ e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \|f\|_{\mathcal{F}^{0,1}_{\nu}}^n \Big) \|\omega_1\|_{\mathcal{F}^{0,1}_{\nu}}.
\end{align*}

\end{proof}

Next,

\begin{align}
   \hat\Omega_{22}(\xi) &= -\frac{A_\kappa}{\pi} \int_\R d\beta\;  \widehat{\tau_\beta \Omega_1} * \widehat{\tau_\beta\partial_\alpha f} * \Big(\calF \left[\frac{\beta}{\beta^2 + (f(\alpha-\beta) + h_2)^2}\right]\Big)(\xi) \nonumber \\
     &=-\frac{A_\kappa}{\pi} \int_\R d\beta \int_\R d \xi_1 \int_\R d\xi_2\;  e^{-i\xi_2 \beta} \hat \Omega_1(\xi_2) e^{-i(\xi_{1}-\xi_{2})\beta} \widehat{\partial_\alpha f}(\xi_1 - \xi_2) \int_\R d\alpha\; e^{-i(\xi - \xi_1)\alpha} \frac{\beta}{\beta^2 + (f(\alpha-\beta) + h_2)^2} \nonumber\\
      &= -\frac{A_\kappa}{\pi} \int_\R d \xi_1 \int_\R d\xi_2\;  \hat \Omega_1(\xi_2)  \widehat{\partial_\alpha f}(\xi_1 - \xi_2) \int_\R d\alpha \  e^{-i(\xi-\xi_{1})\alpha}\int_\R d\beta\; e^{-i\xi_1\beta} \frac{\beta}{\beta^2 + (f(\alpha-\beta) + h_2)^2} \nonumber\\
       &= -\frac{A_\kappa}{\pi} \int_\R d \xi_1 \int_\R d\xi_2\;  \hat \Omega_1(\xi_2)  \widehat{\partial_\alpha f}(\xi_1 - \xi_2) \int_\R d\gamma \  e^{-i(\xi-\xi_{1})\gamma}\int_\R d\beta\; e^{-i\xi \beta} \frac{\beta}{\beta^2 + (f(\gamma) + h_2)^2} \nonumber
      \end{align}
      by a change of variables $\alpha = \gamma + \beta$.

      \begin{align}
 \hat\Omega_{22}(\xi)   &= -\frac{A_\kappa}{\pi} \int_\R d\xi_1 \int_\R d\xi_2\; \hat \Omega_1(\xi_2) \widehat{\partial_\alpha f}(\xi_1 - \xi_2) \int_\R d\gamma\;  e^{-i(\xi - \xi_1) \gamma}  \Big(-i \pi \sgn(\xi) e^{-(f(\gamma) + h_2)|\xi|}\,\Big) \nonumber\\
    & = A_{\kappa} \sum_{n = 0}^\infty \frac{(-1)^{n}}{n!} \int_\R d\xi_1 \int_\R d\xi_2 \;  \hat \Omega_1 (\xi_2) \widehat{\partial_\alpha f}(\xi_1 - \xi_2) \ i \sgn(\xi) e^{-h_2 \abs{\xi}}\  \abs{\xi}^n \int_\R d\gamma\; e^{-i(\xi - \xi_1)\gamma} f(\gamma)^n \nonumber\\
        & = A_{\kappa}\sum_{n = 0}^\infty \frac{(-1)^{n}}{n!} \int_\R d\xi_1 \int_\R d\xi_2 \;  \hat \Omega_1 (\xi_2) \widehat{\partial_\alpha f}(\xi_1 - \xi_2) \ i \sgn(\xi) e^{-h_2 \abs{\xi}}\  \abs{\xi}^n (\ast^{n}\hat{f})(\xi-\xi_{1}) \nonumber\\
    &=A_{\kappa} \sum_{n = 0}^\infty \frac{(-1)^{n}}{n!} \Big((\ i \sgn(\xi) e^{-h_2 \abs{\xi }} \ \abs{ \xi }^n \widehat{\Omega_1}(\xi)) \ast(\widehat{\partial_\alpha f(\xi)}) * (\ast^n \hat f(\xi)) \Big)(\xi).
\end{align}
Alternative derivation of $\Omega_{22}$

\begin{lemma}
\label{lem:omega1fzerone}
The term $\omega_1$ satisfies
\begin{equation}\label{omega2fzerone}
\|\omega_{1}\|_{\mathcal{F}^{0,1}_{\nu}}\leq 2 A_\rho C_1 \norm{f}_{\mathcal{F}^{1,1}_{\nu}}
\end{equation}
\end{lemma}

\begin{proof}
By \eqref{omega1equation}, the term $\omega_{1}$ is given by
\begin{align*}
    \omega_1(\alpha) = 2 A_\mu (BR(\omega_1, z)(z(\alpha)) + BR(\omega_2, h)(z(\alpha)) )\cdot \partial_\alpha z(\alpha) - 2 A_\rho \partial_\alpha f(\alpha).
\end{align*}
We first estimate the Birkoff-Rott integrals. We decompose the first term as
\begin{align*}
    BR(\omega_1, z)(z(\alpha)) \cdot \partial_\alpha z(\alpha) &= \frac{1}{2\pi} (-B_1(\alpha) + B_2(\alpha)) \\
    &= \frac{1}{2\pi} \left(-\int_\R \frac{\Delta_\beta f(\alpha)}{1 + \Delta_\beta f(\alpha)^2} \frac{\omega_1(\alpha - \beta)}{\beta} \; d\beta + \int_\R \frac{\partial_\alpha f(\alpha)}{1 + \Delta_\beta f(\alpha)^2} \frac{\omega_1(\alpha - \beta)}{\beta} \; d\beta
    \right).
\end{align*}
Taking the Fourier transform and Taylor expanding the denominator for $|\Delta_\beta f(\alpha)| < 1$, we obtain
\begin{align*}
    \hat B_1(\xi) &= \int_\R \calF \left[ \frac{\Delta_\beta f(\alpha)}{1 + \Delta_\beta f(\alpha)^2} \frac{\omega_1(\alpha - \beta)}{\beta} \right] \; d\beta\\
    &= \sum_{n = 0}^\infty (-1)^n \int_\R \calF\left[ \Delta_\beta f(\alpha)^{2n+1} \frac{\omega_1(\alpha - \beta)}{\beta}\right] \; d\beta \\
    &= \sum_{n = 0}^\infty  (-1)^n \int_\R (\ast^{2n+1} (\hat f m_\beta) * \widehat{\tau_\beta \omega_1})(\xi)\; \frac{d\beta}{\beta}
\end{align*}
in which
\[
    m_\beta(\xi) = \frac{1 - e^{-i\xi\beta}}{\beta}.
\]
Expanding the convolution of the $n$-th term in the sum, we have
\begin{align*}
    \int_\R (\ast^{2n+1} (\hat f m_\beta) * \widehat{\tau_\beta \omega_1})(\xi) \frac{d\beta}{\beta} = \int_\R d\xi_1 \ldots \int_\R d\xi_{2n} \hat f(\xi - \xi_1) \cdot \ldots \cdot \hat f(\xi_{2n-1} - \xi_{2n}) \cdot \hat \omega_1(\xi_{2n}) \cdot M_n
\end{align*}
where
\begin{align*}
    M_n = M_n(\xi, \xi_1, \ldots, \xi_{2n}) = \int_\R d\beta \frac{e^{-i\xi_{2n} \beta}}{\beta} m_\beta(\xi - \xi_1) \cdot \ldots \cdot m_{\beta}(\xi_{2n-1} - \xi_{2n} ).
\end{align*}
Since
\[
    m_\beta(\xi) = i\xi \int_0^1 ds\; e^{i\beta(s-1)\xi}
\]
we have
\begin{align*}
    |M_n| &= \Big|i^{2n}(\xi - \xi_1) \cdot \ldots \cdot (\xi_{2n-1} - \xi_{2n}) \int_0^1 ds_1 \ldots \int_0^1 ds_{2n} \int_\R d\beta \frac{e^{-i\xi_{2n} \beta}}{\beta} e^{i \beta A}\Big|\\
    &= \Big| (\xi - \xi_1) \cdot \ldots \cdot (\xi_{2n-1} - \xi_{2n}) \int_0^1 ds_1 \ldots \int_0^1 ds_{2n}\  i\pi \sgn(A- \xi_{2n})\Big|\\
    &\leq \pi \cdot \abs{\xi - \xi_1} \cdot \ldots \cdot \abs{\xi_{2n-1} - \xi_{2n}}
\end{align*}
where
\[
    A = (s_1 - 1) (\xi - \xi_1) + \ldots + (s_{2n} - 1) (\xi_{2n-1} - \xi_{2n}).
\]
Therefore,
\begin{align*}
    \dabs{\int_\R (\ast^{2n+1} (\hat f m_\beta) * \widehat{\tau_\beta \omega_1})(\xi) \frac{d\beta}{\beta}}
    &\leq \int_\R d\xi_1 \ldots \int_\R d\xi_{2n} \abs{ \hat f(\xi - \xi_1)} \cdot \ldots \cdot \abs{\hat f(\xi_{2n-1} - \xi_{2n})} \cdot \abs{\hat \omega_1(\xi_{2n})} \cdot \abs{M_n} \\
    &\leq \pi (\ast^{2n+1} \abs{\widehat{\partial_\alpha f}} * \abs{\hat \omega_1})(\xi).
\end{align*}
Applying \eqref{convolutionexp} and Young's convolution inequality, we obtain the bound
\begin{align*}
    \norm{B_1}_{\mathcal{F}^{0,1}_{\nu}} &\leq \pi  \sum_{n=0}^\infty \int_{\R} e^{\nu t\abs{\xi}} (\ast^{2n+1} \abs{\widehat{\partial_\alpha f}} * \abs{\hat \omega_1})(\xi) \; d\xi\\
    & \leq \pi \sum_{n = 0}^\infty \norm{f}^{2n+1}_{\mathcal{F}^{1,1}_{\nu}} \norm{\omega_1}_{\mathcal{F}^{0,1}_{\nu}}\\
    & = \pi \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} \norm{\omega_1}_{\mathcal{F}^{0,1}_{\nu}}.
\end{align*}
By a similar calculation, we also have
\[
    \norm{B_2}_{\mathcal{F}^{0,1}_{\nu}} \leq \pi \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} \norm{\omega_1}_{\mathcal{F}^{0,1}_{\nu}}.
\]
Now let
\begin{align*}
    BR(\omega_2, h)(z(\alpha)) \cdot \partial_\alpha z(\alpha) &= \frac{1}{2\pi} (B_3 - B_4) \\
    &= \frac{1}{2\pi} \left(\int_{\R} \frac{ \beta \partial_\alpha f(\alpha)}{\beta^2 + (f(\alpha)+h_2)^2} \omega_2(\alpha - \beta)\; d\beta - \int_{\R} \frac{f(\alpha) +h_2}{\beta^2 + (f(\alpha)+h_2)^2} \omega_2(\alpha - \beta)\; d\beta \right).
\end{align*}
The Fourier transform of $B_{3}$ is given by
\begin{align*}
    \hat{B_3}(\xi) &= \int_\R d\beta\;  \widehat{\tau_\beta \omega_2} * \widehat{\partial_\alpha f} * \Big(\calF \left[\frac{\beta}{\beta^2 + (f(\alpha) + h_2)^2}\right]\Big)(\xi)\\

     &=\int_\R d\beta \int_\R d \xi_1 \int_\R d\xi_2\;  e^{-i\xi_2 \beta} \hat \omega_2(\xi_2) \widehat{\partial_\alpha f}(\xi_1 - \xi_2) \int_\R d\alpha\; e^{-i(\xi - \xi_1)\alpha} \frac{\beta}{\beta^2 + (f(\alpha) + h_2)^2}\\
    &= \int_\R d\xi_1  \int_\R d \xi_2 \; \hat \omega_2(\xi_2) \widehat{\partial_\alpha f}(\xi_1 - \xi_2) \int_\R d\alpha \; e^{-i(\xi - \xi_1) \alpha} \int_\R d\beta\; e^{-i\xi_2 \beta}  \frac{\beta}{\beta^2 + (f(\alpha) + h_2)^2}\\
    &= \int_\R d\xi_1 \int_\R d\xi_2\; \hat \omega_2(\xi_2) \widehat{\partial_\alpha f}(\xi_1 - \xi_2) \int_\R d\alpha\;  e^{-i(\xi - \xi_1) \alpha} \cdot \Big(-i \pi \sgn(\xi_2) e^{-(f(\alpha) + h_2)|\xi_2|}\,\Big)\\
    & = \sum_{n = 0}^\infty \frac{(-1)^{n+1}}{n!} \int_\R d\xi_1 \int_\R d\xi_2 \;  \hat \omega_2 (\xi_2) \widehat{\partial_\alpha f}(\xi_1 - \xi_2) \ i\pi \sgn(\xi_2) e^{-h_2 \abs{\xi_2}}\  \abs{\xi_2}^n \int_\R d\alpha\; e^{-i(\xi - \xi_1)\alpha} f(\alpha)^n\\

    &= \sum_{n = 0}^\infty \frac{(-1)^{n+1}}{n!} \Big((\ i \pi\sgn(\xi) e^{-h_2 \abs{\xi }} \ \abs{ \xi }^n \hat{\omega_2}(\xi)) \ast(\widehat{\partial_\alpha f(\xi)}) * (\ast^n \hat f(\xi)) \Big)(\xi)
\end{align*}
and similarly
\begin{align}
    \hat B_4 &= \int_\R \hat \omega_2(\xi_1) \int_\R e^{-i(\xi - \xi_1) \alpha} \int_\R e^{-i \xi_1 \beta}\frac{f(\alpha) + h_2}{\beta^2 + (f(\alpha) + h_2)^2}\; d\beta d\alpha d\xi_1\nonumber\\
    &= \int_\R \hat \omega_2(\xi_1) \int_\R e^{-i(\xi - \xi_1) \alpha} \ \pi e^{-(f(\alpha) + h_2)\abs{\xi_1}}\;  d\alpha d\xi_1\nonumber\\
    &= \sum_{n = 0}^{\infty} \frac{(-1)^n}{n!} \int_\R \hat \omega_2(\xi_1) \ \pi e^{-h_2\abs{\xi_1} } \ \abs{\xi_1}^n \int_\R e^{-i(\xi - \xi_1) \alpha}  f(\alpha)^n\; d\alpha d\xi_1 \nonumber\\
    &= \sum_{n = 0}^{\infty} \frac{(-1)^n}{n!}\Big((\hat \omega_2 \ \pi e^{-h_2\abs{\xi } }\ \abs{\xi}^n ) * (\ast^n \hat f) \Big)(\xi) \label{omega2fourierlinearpartsum}
\end{align}
From these, we can compute
\begin{align*}
    \norm{B_3}_{\mathcal{F}^{0,1}_{\nu}} &\leq \pi \sum_{n=0}^\infty \dnorm{\frac{ e^{-h_2 \abs \xi}\abs{\xi}^{n}}{n!}}_{L^\infty} \norm{f}^n_{\mathcal{F}^{0,1}_{\nu}} \norm{f}_{\mathcal{F}^{1,1}_{\nu}}  \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}} = \pi C_0 \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}}\\
    \norm{B_4}_{\mathcal{F}^{0,1}_{\nu}} &\leq \pi \sum_{n=0}^\infty \dnorm{\frac{ e^{-h_2 \abs \xi}\abs{\xi}^{n}}{n!}}_{L^\infty} \norm{f}^n_{\mathcal{F}^{0,1}_{\nu}} \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}} = \pi C_0 \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}}
\end{align*}
Combining all these estimates gives our desired result
\end{proof}

\begin{lemma}
\label{lem:omega2foneone}
The following estimate holds
\begin{equation}
    \norm{\omega_2}_{\mathcal{F}^{1,1}_{\nu}} \leq \abs{A_\kappa} (2A_\rho C_1 C_2 \norm{f}_{\mathcal{F}^{2,1}_{\nu}} + C_0 \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}})
\end{equation}
\end{lemma}
\begin{proof}
Following the proof of \ref{lem:omega2fzerone}, applying \eqref{productrule} and \ref{lem:omega1fzerone} yields
\begin{align*}
    \norm{\omega_2}_{\mathcal{F}^{1,1}_{\nu}} &\leq \abs{A_\kappa} \sum_{n=0}^{\infty} \dnorm{ e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \Big\| e^{\nu t \abs{\xi}}  \abs{\xi}  (\ast^{n}\hat{f}) \ast \hat \omega_1 \Big\|_{L^1}\\
    &\leq \abs{A_\kappa} \sum_{n=0}^\infty \dnorm{ e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} (n\norm{f}_{\mathcal{F}^{0,1}_{\nu}}^{n-1} \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\omega_1}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}})\\
    & \leq \abs{A_\kappa} \sum_{n=0}^\infty \dnorm{ e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} (2A_\rho nC_1\, \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^{n-1} \norm{f}^2_{\mathcal{F}^{1,1}_{\nu}} + \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}})\\
    & \leq \abs{A_\kappa} \sum_{n=0}^\infty \dnorm{ e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} (2 A_\rho nC_1\, \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^{n} \norm{f}_{\mathcal{F}^{2,1}_{\nu}} + \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}})
\end{align*}
where the last line follows from \eqref{interpolation}.
\end{proof}

\begin{lemma}
\label{lem:omega1foneone}
\begin{equation}
    \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}} \leq 2A_\rho C_1 C_3 \norm{f}_{\mathcal{F}^{2,1}_{\nu}}
\end{equation}
\end{lemma}

\begin{proof}
    Estimates similar to those in Lemma \ref{lem:omega1fzerone} lead to
\begin{align*}
        \frac{1}{2\pi}\norm{B_1 + B_2}_{\mathcal{F}^{1,1}_{\nu}} &\leq \sum_{n = 0}^\infty (2n+1) \norm{f}^{2n}_{\mathcal{F}^{1,1}_{\nu}} \norm{f}_{\mathcal{F}^{2,1}_{\nu}} \norm{\omega_1}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{2n+1} \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}} \\
\begin{split}
            \frac{1}{2\pi}\norm{ B_3}_{\mathcal{F}^{1,1}_{\nu}} &\leq \frac{1}{2} \sum_{n=0}^\infty \dnorm{ e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \Big[

            (n+1) \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n \norm{f}_{\mathcal{F}^{2,1}_{\nu}} \norm{\omega_2}_{\mathcal{F}^{0,1}_{0}}\\
            & \hspace{13.5em} + \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\omega_2}_{\mathcal{F}^{1,1}_{0}}\Big]
\end{split}\\
\begin{split}
    \frac{1}{2\pi} \norm{ B_4}_{\mathcal{F}^{1,1}_{\nu}} &\leq \frac{1}{2} \sum_{n=0}^\infty \dnorm{ e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \Big[n \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^{n-1} \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n \norm{\omega_2}_{\mathcal{F}^{1,1}_{\nu}} \Big]
\end{split}
\end{align*}
Simplifying and making use of lemmas \ref{lem:omega2fzerone}, \ref{lem:omega1fzerone}, and \ref{lem:omega2foneone} gives
\begin{align*}
        \frac{1}{2\pi} \norm{B_1 + B_2}_{\mathcal{F}^{1,1}_{\nu}} &\leq  \frac{ \norm{f}_{\mathcal{F}^{1,1}_{\nu}}(1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2)}{(1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2)^2 } \norm{f}_{\mathcal{F}^{2,1}_{\nu}} \norm{\omega_1}_{\mathcal{F}^{0,1}_{\nu}} + \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}} }{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}}\\
        & \leq 2A_\rho C_1 \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2 (1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2)}{(1- \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2)^2} \norm{f}_{\mathcal{F}^{2,1}_{\nu}} + \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}} \\
        \frac{1}{2\pi} \norm{B_3}_{\mathcal{F}^{1,1}_{\nu}} &\leq \frac{1}{2} (C_0 + C_2)\norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}}\norm{f}_{\mathcal{F}^{2,1}_{\nu}} + \frac{1}{2} C_0   \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{\omega_2}_{\mathcal{F}^{1,1}_{\nu}}\\
        & \leq A_\rho \abs{A_\kappa} C_0 C_1 (C_0 + 2C_2) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{f}_{\mathcal{F}^{2,1}_{\nu}} + \frac{1}{2} \abs{A_\kappa} C_0^2 \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}}\\
        \frac{1}{2\pi} \norm{ B_4}_{\mathcal{F}^{1,1}_{\nu}} &\leq \frac{1}{2} \sum_{n=0}^\infty \dnorm{ e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \Big[n \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^{n-1} \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n  \norm{\omega_2}_{\mathcal{F}^{1,1}_{\nu}} \Big]\\
        &\leq \frac{1}{2} \Big( \frac{C_2}{\norm{f}_{\mathcal{F}^{0,1}_{\nu}}} \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}}  + C_0 \norm{\omega_2}_{\mathcal{F}^{1,1}_{\nu}} \Big)\\
        & \leq  A_\rho \abs{A_\kappa} C_0 C_1 C_2 \norm{f}_{\mathcal{F}^{2,1}_{\nu}} + \frac{1}{2} \abs{A_\kappa} C_0^2 \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}}
\end{align*}
    Plugging these into the equation for $\omega_1$ tells us
    \begin{align*}
        \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}} &\leq 2\abs{A_\mu} \bigg( 2A_\rho C_1 \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2 (\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2 + 1)}{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2 - 1} \norm{f}_{\mathcal{F}^{2,1}_{\nu}} + \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}} \\
        & \hspace{5em} +  A_\rho \abs{A_\kappa} C_0 C_1 (C_0 + 2C_2) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{f}_{\mathcal{F}^{2,1}_{\nu}} + \frac{1}{2} \abs{A_\kappa} C_0^2 \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}}\\
        & \hspace{5em}  + A_\rho \abs{A_\kappa} C_0 C_1 C_2 \norm{f}_{\mathcal{F}^{2,1}_{\nu}} + \frac{1}{2} \abs{A_\kappa} C_0^2 \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}} \bigg) + 2A_\rho \norm{f}_{\mathcal{F}^{2,1}_{\nu}}
    \end{align*}
    Reorganizing terms gives
    \begin{align*}
         C_1\inv  \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}} &\leq 2A_\rho C_3 \norm{f}_{\mathcal{F}^{2,1}_{\nu}}.
    \end{align*}

\end{proof}
The naive bound on $L(\hat \omega_2)$, linear term of $\omega_2$, is
\[
   \norm{L(\omega_2)}_{\mathcal{F}^{0,1}_{\nu}} \leq 2 A_\rho \abs{A_\kappa} C_1 \norm{f}_{\mathcal{F}^{1,1}_{\nu}} = 2 A_\rho \abs{A_\kappa} \norm{f}_{\mathcal{F}^{1,1}_{\nu}} + \mathcal{O}(\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2)
\]
We can bound this term in a different way using the relationship
Next, we look at the part of \eqref{omega1equation} and \eqref{omega2equation} that are linear in $\omega_{i}$. From the Fourier transform computation of \eqref{omega2fourierexpansion} and \eqref{omega2fourierlinearpartsum}, the terms that are linear in $\omega_{i}$ are
\begin{align*}
    L(\hat \omega_2) &= -A_\kappa e^{-h_2 \abs{\xi} } \hat \omega_1,\\
    L(\hat \omega_1) &= -A_\mu e^{-h_2 \abs{\xi}} \hat \omega_2 - 2A_\rho \widehat{\partial_\alpha f}(\xi).
\end{align*}
Writing the decomposition into linear and nonlinear terms as $\hat \omega_j = L(\hat \omega_j) + N(\hat \omega_j)$ for $j=1,2$ and iterating this relationship gives
\begin{align*}
    L(\hat \omega_2) &= -A_\kappa e^{-h_2 \abs{\xi} } ( L(\hat \omega_1) + N(\hat \omega_1) )\\
    &= A_\kappa A_\mu e^{-2h_2\abs{\xi} }(L(\hat \omega_2) + N(\hat \omega_2)) + 2A_\kappa A_\rho e^{-h_2 \abs{\xi} }\widehat{\partial_\alpha f} - A_\kappa e^{-h_2 \abs{\xi}} N(\hat \omega_1)\\
    &= -A_\kappa^2 A_\mu e^{-3h_2 \abs{\xi}} (L(\hat \omega_1) + N(\hat \omega_1))\\
    & \qquad +2A_\kappa A_\rho e^{-h_2 \abs{\xi} }\widehat{\partial_\alpha f}  - A_\kappa e^{-h_2 \abs{\xi}} N(\hat \omega_1) + A_\kappa A_\mu e^{-2h_2\abs{\xi} } N(\hat \omega_2)\\
    &= A_\kappa^2 A_\mu^2 e^{-4h_2\abs{\xi}} (L(\hat \omega_2) + N(\hat \omega_2))\\
    & \qquad  + 2 A_\kappa A_\rho \widehat{\partial_\alpha f} (e^{-h_2\abs{\xi}} + A_\kappa A_\mu e^{-3 h_2\abs{\xi}}) \\
    &\qquad - A_\kappa N(\hat \omega_1) (e^{-h_2 \abs{\xi}} + A_\kappa A_\mu e^{-3h_2 \abs{\xi}})\\
    & \qquad + A_\kappa A_\mu e^{-2h_2\abs{\xi} } N(\hat \omega_2)\\
    & = \ldots\\
    & = 2A_\kappa A_\rho \widehat{\partial_\alpha f} \sum_{n = 0}^\infty (A_\kappa A_\mu)^n e^{-(2n+1) h_2\abs{\xi}}\\
    &\qquad - A_\kappa N(\hat \omega_1)\sum_{n=0}^\infty (A_\kappa A_\mu)^n e^{-(2n+1)h_2 \abs{\xi}}\\
    &\qquad + N(\hat \omega_2) \sum_{n=1}^\infty (A_\kappa A_\mu)^n e^{-2nh_2\abs{\xi}}
\end{align*}
The sums converge as geometric series, so
\begin{align}
    \begin{split}
    \label{eq:recursion}
    L(\hat \omega_2) &= 2A_\kappa A_\rho \widehat{\partial_\alpha f} \frac{e^{h_2 \abs{\xi}}}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}\\
    & \qquad  - A_\kappa N(\hat \omega_1) \frac{e^{h_2 \abs{\xi}}}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}\\
    & \qquad - N(\hat \omega_2) \frac{A_\kappa A_\mu}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}.
    \end{split}
\end{align}
\end{comment}

\section{Instant Analyticity and Decay Inequality for the Interface}
\label{sec:interfacedecay}
In this section, we will prove the inequality \eqref{maininequality} which will imply the instantaneous gain of analytic regularity and the decay to the flat solution of the density jump interface for initial data of an explicitly calculable size. We perform these estimates in the spaces defined in \eqref{Fspnu} for $p=1$, $\nu > 0$ and $0\leq s \leq 1$. We first need to linearize the contour equation for the fluid-fluid interface $f(\alpha,t)$. To do so, we need to extract the linear part of $\omega_{i}$. First, by \eqref{omega2fourierexpansion}, we can write the decomposition in of $\omega_{2}$ in frequency space:
\begin{equation}\label{omega2linearnonlinear}
\hat{\omega}_{2}(\xi) = L(\hat\omega_{2})(\xi) + N(\hat\omega_{2})(\xi)
\end{equation}
where
$$L(\hat\omega_{2})(\xi) = -A_\kappa e^{-h_2 \abs \xi}  \hat \omega_{1}(\xi)$$
and
\begin{align*}
N(\hat\omega_{2})(\xi) &= -A_\kappa\sum_{n=1}^{\infty}e^{-h_2 \abs \xi} \frac{(-1)^{n}|\xi|^{n}}{n!} \Big((\ast^{n}\hat{f}) \ast \hat \omega_1 \Big)(\xi)
\end{align*}
From \eqref{omega12calc}, we can write
\begin{equation}\label{Omega1linearnonlinear}
\hat \omega_{1}(\xi) = L(\hat\omega_{1})(\xi) + N(\hat\omega_{1})(\xi)
\end{equation}
where
$$L(\hat \omega_{1})(\xi) = -A_\mu e^{-h_2 \abs \xi}  \hat \omega_{2}(\xi) - 2A_{\rho}\widehat{\partial_{\alpha}f}(\xi)$$
and
\begin{align}
\label{NOmega1hat}
N(\hat \omega_{1})(\xi) &= \hat \omega_{11}(\xi) -A_{\mu} \sum_{n = 1}^{\infty} \frac{(-1)^n}{n!}\Big((\hat \omega_2 \  e^{-h_2\abs{\xi } }\ \abs{\xi}^n ) * (\ast^n \hat f) \Big)(\xi)\\
&\hspace{1in}+ A_{\mu}\sum_{n = 0}^\infty \frac{(-1)^{n+1}}{n!} \Big((\ i \sgn(\xi) e^{-h_2 \abs{\xi }} \ \abs{ \xi }^n \hat \omega_2) \ast\widehat{\partial_\alpha f} * (\ast^n \hat f) \Big)(\xi) \nonumber.
\end{align}
We can now use the relations \eqref{omega2linearnonlinear} and \eqref{Omega1linearnonlinear} to sa
\begin{align*}
    L(\hat \omega_2) &= -A_\kappa e^{-h_2 \abs{\xi} } ( L(\hat \omega_1)(\xi) + N(\hat \omega_1)(\xi) )\\
    &= A_\kappa A_\mu e^{-2h_2\abs{\xi} }(L(\hat \omega_2)(\xi)+ N(\hat \omega_2)(\xi)) + 2A_\kappa A_\rho e^{-h_2 \abs{\xi} } \widehat{\partial_{\alpha}f}(\xi) - A_\kappa e^{-h_2 \abs{\xi}} N(\hat \omega_1

\end{align*}
Combining terms and solving for $L(\hat \omega_2)$ we find
\begin{equation}\label{Omega2linearpart}
    L(\hat \omega_2)(\xi) = 2A_\kappa A_\rho  \widehat{\partial_{\alpha}f}(\xi) \frac{e^{h_2 \abs{\xi}}}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu} - A_\kappa N(\hat \omega_1)(\xi) \frac{e^{h_2 \abs{\xi}}}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu} - N(\hat \omega_2)(\xi) \frac{A_\kappa A_\mu}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}.
\end{equation}
With \eqref{Omega2linearpart} in hand, we are ready to begin analyzing the interface decay. Differentiating in time, we obtain
\begin{align*}
    \frac{d}{dt} \norm{f}_{\mathcal{F}^{s,1}_{\nu}} &= \frac{d}{dt} \int_\R e^{\nu t\abs{\xi} } \abs{\xi}^s|\hat{f}(\xi)|\; d\xi \\
    &= \nu \int_\R e^{\nu t\abs{\xi} } \abs{\xi}^{s+1}|\hat{f}(\xi)|\; d\xi + \frac{1}{2}\int_\R e^{\nu t\abs{\xi} } \abs{\xi}^s\frac{\hat{f}(\xi)\overline{\widehat{\partial_{t}f}(\xi)}
    +\widehat{\partial_{t}f}(\xi)\overline{\hat{f}(\xi)}}{|\hat{f}(\xi)|}\; d\xi.
\end{align*}
Next, to obtain the decay term in the expression above, we need to decompose the evolution equation for the interface into the linear and nonlinear terms
\begin{align*}
    \widehat{\partial_t f}(\xi) &= \frac{1}{2\pi} ( \hat I_1(\xi) + \hat I_2(\xi) + \hat I_3(\xi) + \hat I_4(\xi ) ).
\end{align*}
Defining $N_0$ as
\begin{align*}
    I_1(\alpha) &= \pv \int_\R \frac{\omega_1(\alpha - \beta)}{ \beta}\; d\beta - \int_\R \frac{\Delta_\beta f(\alpha)^2}{1 + \Delta_\beta f(\alpha)^2} \frac{\omega_1(\alpha - \beta)}{\beta}\; d\beta\\
    &\eqdef \pi H(\omega_1)(\alpha) + 2\pi N_0(\alpha),
\end{align*}
where $H$ is the Hilbert transform, we use equation \eqref{Omega1linearnonlinear} to find
\begin{align}
    \begin{split}
    \label{eq:Ionehat_first}
        \frac{1}{2\pi} \hat I_1(\xi) &= \frac{1}{2} \widehat{H(\omega_1)}(\xi) + \hat N_0(\xi) \\
        &= -A_\rho |\xi| \hat f(\xi) + \frac{i}{2}\sgn(\xi) A_\mu e^{-h_2|\xi|} \hat \omega_2(\xi) -  \frac{i}{2}\sgn(\xi) N( \hat \omega_1)(\xi) +  \hat N_0(\xi).
    \end{split}
\end{align}
Expanding $\hat \omega_2$ via \eqref{omega2linearnonlinear} and \eqref{Omega2linearpart} gives
\begin{align}
    \label{Ionehat_second}
    \frac{i}{2}\sgn(\xi) A_\mu e^{-h_2|\xi|} \hat \omega_2(\xi)
    &= -A_\rho |\xi| \hat f(\xi) \frac{A_\kappa A_\mu}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu} - A_\mu(\hat N_1 + \hat N_2 + \hat N_3)(\xi)

\end{align}
in which
\begin{align}
\label{N1}
    \hat N_1(\xi) &= -\frac{i}{2}\sgn(\xi) e^{-h_2\abs{\xi}} N(\hat \omega_2)(\xi),\\
\label{N2}
    \hat N_2(\xi) &= \frac{i}{2} \sgn(\xi) \frac{A_\kappa}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu} N(\hat \omega_1)(\xi), \\
\label{N3}
    \hat N_3(\xi) &= \frac{i}{2} \sgn(\xi) \frac{A_\kappa A_\mu e^{-h_2\abs{\xi}}}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}N(\hat \omega_2)(\xi).
\end{align}
Combining \eqref{eq:Ionehat_first} and \eqref{Ionehat_second} leads to
\begin{align*}
    \frac{1}{2\pi}\hat I_1(\xi) = -A_\rho |\xi| \hat f(\xi) \left (1 + \frac{A_\kappa A_\mu}{e^{2h_2 \abs{\xi}} - A_\kappa A_\mu} \right) +  (\hat N_0 + \hat N_1 + \hat N_2 + \hat N_3)(\xi) - \frac{i}{2} \sgn(\xi) N(\hat\omega_1)(\xi) .
\end{align*}

Next,
\begin{align}
\label{I3linearnonlinear}
     I_3(\alpha) &= \int_\R \frac{\beta}{\beta^2 + h_2^2}\, \omega_2(\alpha - \beta) \; d\beta - \int_\R \frac{f(\alpha) + 2h_2}{\beta^2 + (f(\alpha) + h_2)^2} \frac{\beta f(\alpha) }{\beta^2 + h_2^2}\, \omega_2(\alpha - \beta)\; d\beta,
\end{align}
denoting the nonlinear part as
\[
     N_4 = -\frac{1}{2\pi} \int_\R \frac{f(\alpha) + 2h_2}{\beta^2 + (f(\alpha) + h_2)^2} \frac{\beta f(\alpha) }{\beta^2 + h_2^2}\, \omega_2(\alpha - \beta)\; d\beta,
\]
we apply \eqref{fouriercalc2} and \eqref{Omega2linearpart} once again
\begin{align}
    \frac{1}{2\pi} \hat I_3(\xi) &= -\frac{i}{2} \sgn(\xi) e^{-h_2\abs\xi} \hat \omega_2(\xi) + \hat N_4(\xi) \nonumber\\
     \label{eq:I3hat}
    &= A_\rho |\xi| \hat f(\xi) \frac{A_\kappa }{e^{2h_2\abs{\xi}} - A_\kappa A_\mu} +  (\hat N_1 + \hat N_2 + \hat N_3 + \hat N_4)(\xi).
\end{align}
Collecting terms we have
\begin{align}
\label{interface_neat}
\begin{split}
    \widehat{\partial_t f}(\xi) &= - A_\rho |\xi| \hat f(\xi) \left(1 - \frac{A_\kappa(1-A_\mu) }{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}\right) + \frac{1}{2\pi}(\hat I_2 + \hat I_4)(\xi)\\
    & \hspace{2.5em} +\hat N_0 (\xi) +  (1 + A_\mu)(\hat N_1 + \hat N_2 + \hat N_3)(\xi)  + \hat N_4(\xi) - \frac{i}{2} \sgn(\xi) N(\omega_1)(\xi).
    %

\end{split}
\end{align}

\begin{comment}
\todo{Nikhil: ignore the rest of this section for now}
\begin{align*}
     & \frac{1}{2} \widehat{H(\omega_1)}(\xi) - \frac{i}{2} \sgn(\xi) e^{-h_2 \abs{\xi}} \hat \omega_2(\xi) + \frac{1}{2\pi}(\hat I_2(\xi) + \hat I_4(\xi))\\
    & \hspace{2.5em} \frac{1}{2\pi} \calF \left[\int_\R \frac{\Delta_\beta f(\alpha)^2}{1 + \Delta_\beta f(\alpha)^2} \frac{\omega_1(\alpha - \beta)}{\beta}\; d\beta\right] +\frac{i \pi}{2} \sum_{n=1}^\infty \left(\hat \omega_2 \frac{\xi^n e^{-h_2 \abs{\xi} }}{n!} *(\ast^n \hat f)\right)(\xi)
\end{align*}
\begin{align*}
     \widehat{\partial_t f}(\xi) &= -A_\rho \abs{\xi} \hat f(\xi) + A_\rho A_\kappa \abs{\xi} \hat f(\xi) \frac{1}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu} + \frac{1}{2\pi}(\hat I_2(\xi) + \hat I_4(\xi))\\
    &\hspace{2.5em} + A_\mu \calF\big (H\big([BR(\omega_1, z)(z(\alpha)) + BR(\omega_2, h)(h(\alpha))]\cdot \partial_\alpha z\big)\big)(\xi) \\
    & \hspace{2.5em}  -\frac{i}{2} \sgn(\xi) \frac{A_\kappa}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu} N(\hat \omega_1)(\xi) + \frac{i}{2}\sgn(\xi) \frac{e^{-h_2\abs{\xi}}}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu} N(\hat \omega_2)(\xi) \\
    & \hspace{2.5em}  + \frac{1}{2\pi} \calF \left[\int_\R \frac{\Delta_\beta f(\alpha)^2}{1 + \Delta_\beta f(\alpha)^2} \frac{\omega_1(\alpha - \beta)}{\beta}\; d\beta \right] + \frac{i \pi}{2} \sum_{n=1}^\infty \left(\hat \omega_2 \frac{\xi^n e^{-h_2 \abs{\xi} }}{n!} *(\ast^n \hat f)\right)(\xi).
\end{align*}
Additionally,
\begin{align*}
    \calF\big (H\big( BR(\omega_2, h)(z(\alpha)) \cdot \partial_\alpha z(\alpha) \big) \big)(\xi) &= \frac{i}{2} \sgn(\xi) e^{-h_2\abs{\xi}} \hat \omega_2(\xi) - \frac{i}{2\pi}\sgn(\xi) \hat B_3(\xi)\\
    &\hspace{2.5em} + \frac{i}{2} \sgn(\xi) \sum_{n = 1}^{\infty} \frac{(-1)^n}{n!}\Big((\hat \omega_2 \ e^{-h_2\abs{\xi } }\ \abs{\xi}^n ) * (\ast^n \hat f) \Big)(\xi)
\end{align*}
so
\begin{align*}
    \widehat{\partial_t f}(\xi) &= -A_\rho \abs{\xi} \hat f(\xi) \left(1 - \frac{A_\kappa(1+A_\mu)}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}\right) + \frac{1}{2\pi}(\hat I_2(\xi) + \hat I_4(\xi))\\
    &\hspace{2.5em} - A_\mu \calF\big (H\big(BR(\omega_1, z)(z(\alpha))\cdot \partial_\alpha z\big)\big)(\xi) + \frac{i}{2} \sgn(\xi) e^{-h_2\abs{\xi}}N(\hat \omega_2)(\xi)\\
    & \hspace{2.5em} + \frac{i}{2\pi} A_\mu \sgn(\xi) \hat B_3(\xi) - \frac{i}{2\pi} A_\mu \sgn(\xi) \sum_{n = 1}^{\infty} \frac{(-1)^n}{n!}\Big((\hat \omega_2 \ \pi e^{-h_2\abs{\xi } }\ \abs{\xi}^n ) * (\ast^n \hat f) \Big)(\xi)\\
    & \hspace{2.5em}  -\frac{i}{2} \sgn(\xi) \frac{A_\kappa(1+A_\mu)}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu} N(\hat \omega_1)(\xi) + \frac{i}{2}\sgn(\xi) \frac{(1+A_\mu)e^{-h_2\abs{\xi}}}{e^{2h_2\abs{\xi}} + A_\kappa A_\mu} N(\hat \omega_2)(\xi) \\
    & \hspace{2.5em}  \frac{1}{2\pi} \calF \left[\int_\R \frac{\Delta_\beta f(\alpha)^2}{1 + \Delta_\beta f(\alpha)^2} \frac{\omega_1(\alpha - \beta)}{\beta}\; d\beta \right] + \frac{i \pi}{2} \sum_{n=1}^\infty \left(\hat \omega_2 \frac{\xi^n e^{-h_2 \abs{\xi} }}{n!} *(\ast^n \hat f)\right)(\xi)\\
    &=  -A_\rho \abs{\xi} \hat f(\xi) \left(1 - \frac{A_\kappa(1+A_\mu)}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}\right) + \mathcal{O}(\norm{f}^2)
\end{align*}

The linear term is interesting
\[
    \left(1 - \frac{A_\kappa(1-A_\mu)}{e^{2h_2\abs{\xi}} + A_\kappa A_\mu}\right) > 0
\]
whenever $A_\kappa(1-2A_\mu) < 1$. The troublesome case is when $A_\mu < 0$ and $A_\kappa \gg 0$. The most stable case is when $A_\kappa < 0$, then
\[
    \left(1 - \frac{A_\kappa(1-A_\mu)}{e^{2h_2\abs{\xi}} + A_\kappa A_\mu}\right) > 1
\]
and the permeability interface aids in the decay of the fluid interface.
\end{comment}
In \eqref{interface_neat}, the linear terms will give the decay of the interface as long as the nonlinear terms are sufficiently bounded. So let us bound the nonlinear terms by following analogous computations to those in Section \ref{potentialjuimpestimates}. In all the nonlinear bounds in this section, the constants grow arbitrarily small as $\norm{f}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \to 0$.

Defining
\begin{align}
\label{C5}
C_5 &= \bnorm{\frac{A_\kappa A_\mu e^{-h_2\abs{\xi}}}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}}_{L^\infty}\\
\label{C6}
C_{6} &= \frac{\|f\|_{\mathcal{F}^{1,1}_{\nu}}}{1-\|f\|_{\mathcal{F}^{1,1}_{\nu}}^{2}} \to 0 \quad \text{as} \quad \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \to 0
\end{align}
we have the estimates
\begin{align}\label{I2fzeroone}
    \frac{1}{2\pi}\norm{I_2}_{\mathcal{F}^{0,1}_{\nu}} &\leq \frac{1}{2}  \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} {1- \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} \norm{\omega_1}_{\mathcal{F}^{0,1}_{\nu}}   \leq
     A_\rho C_1 C_6 \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2, \\
     \frac{1}{2\pi}\norm{I_2}_{\mathcal{F}^{1,1}_{\nu}} &\leq \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{( 1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2)^2}( \norm{\omega_1}_{\mathcal{F}^{0,1}_{\nu}} \norm{f}_{\mathcal{F}^{2,1}_{\nu}} + \frac{1}{2}(1- \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2) \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}}) \nonumber\\
     &\leq   2A_\rho C_1(1 + \frac{1}{2}C_3(1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2))C_6^2 \norm{f}_{\mathcal{F}^{2,1}_{\nu}}.    \label{I2foneone}
\end{align}
Next, similar to \eqref{omega12calc},
\begin{align*}
    \hat I_4(\xi) = \pi \sum_{n = 0}^{\infty} \frac{(-1)^n}{n!}\Big((\hat \omega_2 \ e^{-h_2\abs{\cdot } }\ \abs{\cdot}^n ) * (\ast^n \hat f)  * \widehat{\partial_\alpha f} \Big)(\xi)
\end{align*}
the bounds on $I_{4}$ are
\begin{align}
    \frac{1}{2\pi }\norm{I_4}_{\mathcal{F}^{0,1}_{\nu}} & \leq \frac{1}{2} C_0 \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}} \nonumber \\
    &\leq   A_\rho \abs{A_\kappa} C_0^2 C_1 \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2    \label{I4fzeroone}\\

    \frac{1}{2\pi}\norm{I_4}_{\mathcal{F}^{1,1}_{\nu}} & \leq A_\rho \abs{A_\kappa} C_0 C_1 (C_0 + C_2) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{f}_{\mathcal{F}^{2,1}_{\nu}} + C_0 \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\omega_2}_{\mathcal{F}^{1,1}_{\nu}} \nonumber \\
    &\leq  A_\rho \abs{A_\kappa} C_0C_1(C_0 + C_2 +C_4) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{f}_{\mathcal{F}^{2,1}_{\nu}} \nonumber\\
    &\eqdef A_\rho \abs{A_\kappa} C_1\lambda_0   \norm{f}_{\mathcal{F}^{2,1}_{\nu}}  \label{I4foneone}
\end{align}
where
\begin{equation}\label{lambda0}
\lambda_{0} =  C_0(C_0 + C_2 +C_4)\norm{f}_{\mathcal{F}^{1,1}_{0}} \to 0 \quad \text{as} \quad \norm{f}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \to 0.
\end{equation}
Next, like the term $\Omega_{111}$,
\[
    |\hat N_0| \leq \frac{1}{2} \sum_{n =0}^\infty ((\ast^{2n+2} |\widehat{\partial_\alpha f}|)*|\hat \omega_1|)(\xi)
\]
which leads to
\begin{align*}
    \norm{N_0}_{\mathcal{F}^{0,1}_{\nu}} &\leq A_\rho C_1 C_6 \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2\\
    \norm{N_0}_{\mathcal{F}^{1,1}_{\nu}} &\leq A_\rho C_1 C_{6}\Big(C_3 \norm{f}_{\mathcal{F}^{1,1}_{\nu}} + \frac{2C_6}{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}} \Big)\norm{f}_{\mathcal{F}^{2,1}_{\nu}}.
\end{align*}
Now, reusing techniques from the Fourier transforms in section \ref{potentialjuimpestimates}, we have
\begin{align*}
    -2\pi \hat N_{4}(\xi) &= \calF \left[\int_\R d\beta \; \frac{f(\alpha) + 2h_2}{\beta^2 + (f(\alpha) + h_2)^2} \frac{\beta f(\alpha) }{\beta^2 + h_2^2} \omega_2(\alpha - \beta) \right](\xi)\\
    &= \int_\R d\beta \;  \left( \calF\left(\frac{f(\alpha) + 2h_2}{\beta^2 + (f + h_2)^2} \frac{\beta f(\alpha)}{\beta^2 + h_2^2} \right) * \widehat{\tau_\beta \omega_2}\right)(\xi) \\
    &= \int_\R d\beta \int_\R d\xi_1  \; \calF\left( \frac{(f(\alpha)  + 2h_2)}{\beta^2 + (f(\alpha) + h_2)^2} \frac{\beta f(\alpha)}{\beta^2 + h_2^2} \right)(\xi - \xi_1) \   \ e^{-i\beta \xi_1}\hat \omega_2(\xi_1)\\
    &= \int_\R d\xi_1 \;  \hat \omega_2 (\xi_1) \int_\R d\alpha \; e^{-i(\xi - \xi_1)\alpha}f(\alpha)  \int_\R d\beta \;e^{-i\beta \xi_1} \frac{f(\alpha) + 2h_2}{\beta^2 + (f(\alpha) + h_2)^2} \frac{\beta }{\beta^2 + h_2^2}.
\end{align*}
By \eqref{fouriercalc1} and \eqref{fouriercalc2} the integral in $\beta$ is the convolution
\begin{align*}
   T(\xi_1) &\eqdef \int_\R d\beta \;e^{-i\beta \xi_1} \frac{f(\alpha) + 2h_{2}}{\beta^2 + (f(\alpha) + h_2)^2} \frac{\beta}{\beta^2 + h_2^2}\\
   &= \Bigg(\frac{\pi (f(\alpha)+2h_2)}{f(\alpha) + h_2} e^{-(f(\alpha) + h_2)\abs{\cdot} } \ast -i\pi \sgn(\cdot) e^{-h_2 \abs{\cdot}}\Bigg)(\xi_1)
\end{align*}
which can be calculated via the identity
\begin{align*}
    \Big( e^{-a |\cdot| } * \sgn(\cdot) e^{-b |\cdot| } \Big) (x)= - \sgn(x)
    \frac{2a(e^{-a|x|} - e^{-b|x|})}{a^2-b^2}
\end{align*}
as
\begin{align*}
    T(\xi_1) &= i\pi^2 \frac{ f(\alpha) + 2h_2}{f(\alpha) + h_2}\cdot \sgn(\xi_1) \frac{2(f(\alpha) + h_2)(e^{-(f(\alpha) + h_2)\abs{\xi_1}} - e^{-h_2|\xi_1|})}{(f(\alpha) + h_2)^2 - h_2^2}\\
    &= 2i\pi^2 \sgn(\xi_1) e^{-h_2|\xi_1|}  \frac{(e^{-f(\alpha)\abs{\xi_1}} - 1)}{f(\alpha)}\\
    &= 2i\pi^2 \sgn(\xi_1) e^{-h_2|\xi_1|} \sum_{n=1}^\infty \frac{(-f(\alpha))^{n-1} |\xi_1|^n}{n!}
\end{align*}
and so
\begin{align*}
     \hat N_{4}(\xi) &= -i\pi\sum_{n=1}^\infty  \frac{(-1)^{n-1}}{n!} \int_\R d\xi_1 \; \hat \omega_2(\xi_1) \sgn(\xi_1) \abs{\xi_1}^n e^{-h_2|\xi_1|} \int_\R d\alpha \; e^{-i(\xi - \xi_1)\alpha}f(\alpha)^{n}.
\end{align*}
Hence,
\begin{align}
\label{N4hat}
    \hat N_4(\xi) = i\pi \sum_{n=1}^\infty \frac{(-1)^n}{n!} \Big( (\hat \omega_2\ \sgn(\xi) \ \abs{\xi}^n\ e^{-h_2 \abs{\xi}} ) * (\ast^n \hat f) \Big) (\xi).
\end{align}
Using \eqref{interpolation} gives us the bounds
\begin{align*}
    \norm{N_4}_{\mathcal{F}^{0,1}_{\nu}} &\leq \pi (C_0 - 1)
    \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}} \nonumber\\
    &\leq 2\pi A_\rho \abs{A_\kappa} C_0(C_0-1)C_1 \norm{f}_{\mathcal{F}^{1,1}_{\nu}},\\
    \norm{N_4}_{\mathcal{F}^{1,1}_{\nu}} &\leq 2\pi A_\rho \abs{A_\kappa} C_0 C_1 C_2 \norm{f}_{\mathcal{F}^{2,1}_{\nu}} + \pi (C_0 - 1) \norm{\omega_2}_{\mathcal{F}^{1,1}_{\nu}} \nonumber\\
    &\leq 2\pi A_\rho \abs{A_\kappa} C_1(C_0 C_2 + (C_0 - 1)C_4)\norm{f}_{\mathcal{F}^{2,1}_{\nu}}.
\end{align*}
At this point, bounds on $N(\omega_1)$ and $N(\omega_2)$ will lead to quick estimates of all the remaining terms. In $\calF^{0,1}$
\begin{align*}
    \norm{N(\omega_2)}_{\mathcal{F}^{0,1}_{\nu}} &\leq \abs{A_\kappa} (C_0 - 1) \norm{\omega_1}_{\mathcal{F}^{0,1}_{\nu}} \nonumber\\
    &\leq 2A_\rho | A_\kappa| (C_0 - 1) C_1\norm{f}_{\mathcal{F}^{1,1}_{\nu}}
\end{align*}
and
\begin{align*}
    \norm{N(\omega_1)}_{\mathcal{F}^{0,1}_{\nu}}
    &\leq 2 |A_\mu| C_6 \norm{\omega_1}_{\mathcal{F}^{0,1}_{\nu}} + |A_\mu| (C_0 \norm{f}_{\mathcal{F}^{1,1}_{\nu}} + (C_0-1) ) \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}} \nonumber\\
    &\leq A_\rho |A_\mu|C_1 \lambda_1  \norm{f}_{\mathcal{F}^{1,1}_{\nu}}

\end{align*}
where the second inequality follows from  \eqref{omega1fzeronenubound} and \eqref{eq:omega2fzerone}. Above,
\begin{align*}
    \lambda_1 = 4C_6 + 2\abs{A_\kappa} C_0 (C_0 \norm{f}_{\mathcal{F}^{1,1}_{\nu}} + (C_0 - 1)) \to 0 \quad \text{as} \quad \norm{f}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \to 0.
\end{align*}
Next in $\calF^{1,1}$
\begin{align*}
    \norm{N(\omega_2)}_{\mathcal{F}^{1,1}_{\nu}} &\leq \abs{A_\kappa} \Big((C_0 - 1) \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}} + \frac{C_2}{\norm{f}_{\mathcal{F}^{0,1}_{\nu}}} \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\omega_1}_{\mathcal{F}^{0,1}_{\nu}} \Big)\\
    &\leq A_\rho \abs{A_\kappa}  C_1 \lambda_2 \norm{f}_{\mathcal{F}^{2,1}_{\nu}}
\end{align*}
where $\lambda_2$ is defined by applying \eqref{omega1fzeronenubound}, \eqref{omega1foneonenubound}, and then \eqref{interpolation}
\begin{align*}
    \lambda_2 = 2(C_0 - 1)C_3 + 2C_2 \to 0 \quad \text{as} \quad \norm{f}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \to 0.
\end{align*}
Finally, calculating as in \eqref{omega12foneone},
\begin{align*}
\begin{split}
      \norm{N(\omega_1)}_{\mathcal{F}^{1,1}_{\nu}}  &\leq  \norm{\omega_{11}}_{\mathcal{F}^{1,1}_{\nu}} + |A_\mu| \Big( (C_0 + C_2) \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}} \norm{f}_{\mathcal{F}^{2,1}_{\nu}} + C_0 \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\omega_2}_{\mathcal{F}^{1,1}_{\nu}} +\\
      & \hspace{12em} \frac{C_2}{\norm{f}_{\mathcal{F}^{0,1}_{\nu}} } \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}} + (C_0 - 1)\norm{\omega_2}_{\mathcal{F}^{1,1}_{\nu}} \Big)

\end{split}\\
      &\leq A_\rho \abs{A_\mu} C_1 \lambda_3 \norm{f}_{\mathcal{F}^{2,1}_{\nu}}
\end{align*}
in which $\lambda_3$ is defined by using \eqref{omega1fzeronenubound}, \eqref{omega1foneonenubound}, \eqref{omega11foneone}, \eqref{eq:omega2fzerone}, \eqref{omega2foneone},
\begin{align*}
\lambda_3 &= 4 \frac{1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} {(1-\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2)^2} \norm{f}_{\mathcal{F}^{1,1}_{\nu}}  + 4C_3 C_6 + 2\abs{A_\kappa} (C_0(C_0+C_2 + C_4) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\\
&\hspace{5em} + 2\abs{A_\kappa}( C_0C_2 + (C_0-1) C_4),
\end{align*}
and $\lambda_3 \to 0$ as $\norm{f}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \to 0$.

\begin{comment}
\begin{align}
    \norm{N(\Omega_1)}_{\mathcal{F}^{1,1}_{\nu}} &\leq \norm{\Omega_{11}}_{\mathcal{F}^{1,1}_{\nu}} + \abs{A_\mu} \sum_{n=1}^\infty \dnorm{e^{-h_2|\xi|} \frac{|\xi|^n}{n!}}_{L^\infty} (\norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n \norm{\Omega_2}_{\mathcal{F}^{1,1}_{\nu}} + n\norm{f}_{\mathcal{F}^{0,1}_{\nu}}^{n-1} \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\Omega_2}_{\mathcal{F}^{0,1}_{\nu}})\\
    &\leq 2|A_\mu| \sum_{n=0}^\infty \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{2n+1} \norm{\Omega_1}_{\mathcal{F}^{1,1}_{\nu}} + (2n+1) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{2n}\norm{f}_{\mathcal{F}^{2,1}_{\nu}} \norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}}
\end{align}
and
\begin{align}
\begin{split}
     \norm{N(\Omega_2)}_{\mathcal{F}^{1,1}_{\nu}} &\leq \abs{A_\kappa} ((C_0 - 1) \norm{\Omega_1}_{\mathcal{F}^{1,1}_{\nu}}\\
     &\hspace{2.5em}+ C_2\norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\Omega}_{\mathcal{F}^{0,1}_{\nu}}) + \abs{A_\kappa}(C_0(\norm{f}_{\mathcal{F}^{2,1}_{\nu}}\norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}} )+ C_2\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2 \norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}}).
\end{split}
\end{align}
\end{comment}
\begin{comment}

from \eqref{Omega111zeroest}, \eqref{Omega112zeroest}

Next, for the nonlinear part of \eqref{Omega1linearnonlinear}, we use the estimate from \eqref{Omega11abs} to obtain
\begin{equation}\label{NOmega1abs}
|N(\Omega_{1})(\xi)| \leq  2 A_\mu \sum_{n=0}^{\infty} (\ast^{2n+1} |\cdot||\hat{f}| \ast \abs{\hat \Omega_1})(\xi) + A_{\mu} \sum_{n = 1}^{\infty} \Big((\hat \Omega_2  \frac{e^{-h_2\abs{\cdot }}|\cdot|^{n}}{n!} ) \ast (\ast^n \hat f) \Big)(\xi).
\end{equation}
Applying Young's inequality and \eqref{convolutionexp}, we obtain
\begin{align*}
\|N(\Omega_{1})\|_{\mathcal{F}^{0,1}_{\nu}}  & \leq 2A_\mu \sum_{n = 0}^\infty \norm{f}^{2n+1}_{\mathcal{F}^{1,1}_{\nu}} \norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}} + A_{\mu}\sum_{n=1}^{\infty} \norm{f}^{n}_{\mathcal{F}^{0,1}_{\nu}} \norm{\Omega_2}_{\mathcal{F}^{0,1}_{\nu}} \Big\| \frac{e^{-h_2\abs{\cdot }}|\cdot|^{n}}{n!}\Big\|_{L^{\infty}}\\
& = A_\mu \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2} \norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}}+ A_{\mu}\sum_{n=1}^{\infty} \norm{f}^{n}_{\mathcal{F}^{0,1}_{\nu}} \norm{\Omega_2}_{\mathcal{F}^{0,1}_{\nu}} \Big\| \frac{e^{-h_2\abs{\cdot }}|\cdot|^{n}}{n!}\Big\|_{L^{\infty}}\\
&= c_{3} \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}} + c_{4}\norm{f}_{\mathcal{F}^{0,1}_{\nu}} \norm{\Omega_2}_{\mathcal{F}^{0,1}_{\nu}}.
\end{align*}
Hence, we obtain from \eqref{Omega2linearnonlinear} that

Combining the above estimates on $N(\Omega_{1})$ and $N(\Omega_{2})$, we conclude the proof.

\begin{align*}
    \norm{N_1}_{\mathcal{F}^{0,1}_{\nu}} &\leq \abs{A_\mu} \sum_{n=0}^\infty \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{2n+1} \norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}} + \frac{\abs{A_\mu}}{2}\sum_{n=1}^\infty \dnorm{e^{-h_2 |\xi|} \frac{|\xi|^n}{n!}}_{L^\infty} \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{\Omega_{1}}_{\mathcal{F}^{0,1}_{\nu}}\\
    &\leq \abs{A_\mu} \big(C_6 + \frac{1}{2}\norm{f}_{\mathcal{F}^{1,1}_{\nu}} (C_0-1) \big) \norm{\Omega_1}_{\mathcal{F}^{0,1}_{\nu}}
\end{align*}

We can compute

\begin{align*}
    \hat N_1(\xi) &= -\frac{1}{2\pi} \calF\left[\int_\R \frac{\Delta_\beta f(\alpha)^2}{1 + \Delta_\beta f(\alpha)^2} \frac{\omega_1(\alpha - \beta)}{\beta}\; d\beta\right](\xi),\\
    \hat N_2(\xi) &= \frac{i}{2} \sgn(\xi) \sum_{n = 1}^{\infty} \frac{(-1)^n}{n!}\Big((\hat \omega_2 \ e^{-h_2\abs{\xi } }\ \abs{\xi}^n ) * (\ast^n \hat f) \Big)(\xi),\\
        \hat N_3(\xi) &= -\frac{i}{2}\sgn(\xi) e^{-h_2\abs{\xi}} N(\hat \omega_2)(\xi),\\
    \hat N_4(\xi) &= \frac{i}{2} \sgn(\xi) \frac{A_\kappa}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu} N(\hat \omega_1)(\xi), \\
    \hat N_5(\xi) &= -\frac{i}{2} \sgn(\xi) \frac{A_\kappa A_\mu e^{-h_2\abs{\xi}}}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}N(\hat \omega_2)(\xi),\\
    \hat N_6(\xi) &=  - \frac{1}{2\pi}\calF \left[ \int_\R \frac{f(\alpha) + h_2}{\beta^2 + (f(\alpha) + h_2)^2} \frac{\beta f(\alpha) }{\beta^2 + h_2^2}\, \omega_2(\alpha - \beta)\; d\beta \right]

\end{align*}

The term $N_1$ satisfies the same estimates as $I_2$ by a calculation analogous to that for $B_1$. Then, $N_2$ is the non-linear part of $B_4$ so
\begin{align}
\norm{N_2}_{\mathcal{F}^{0,1}_{\nu}} &\leq \frac{1}{2} (C_0 - 1) \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}}\\
&\leq A_\rho \abs{A_\kappa}  C_0(C_0 -1) C_1 \norm{f}_{\mathcal{F}^{1,1}_{\nu}},\\
\norm{N_2}_{\mathcal{F}^{1,1}_{\nu}} &\leq A_\rho \abs{A_\kappa} C_0 C_1 C_2 \norm{f}_{\mathcal{F}^{2,1}_{\nu}} + \frac{1}{2} \abs{A_\kappa} (C_0^2 - 1) \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}}\\
\label{N2foneone}
&\leq A_\rho \abs{A_\kappa} C_1(C_0 C_2 + (C_0^2-1)C_4)\norm{f}_{\mathcal{F}^{2,1}_{\nu}}.
\end{align}
To bound $N_4$ we estimate the pre-factor in $L^\infty$ and note that

\begin{align*}
    \norm{N(\omega_1)}_{\mathcal{F}^{s,1}_{\nu}} &\leq \frac{\abs{A_\mu}}{\pi} \Big( \norm{B_1}_{\mathcal{F}^{s,1}_{\nu}} + \norm{B_2}_{\mathcal{F}^{s,1}_{\nu}} +  \norm{B_3}_{\mathcal{F}^{s,1}_{\nu}} + \pi  \norm{N_2}_{\mathcal{F}^{s,1}_{\nu}} \Big),
\end{align*}
so
\begin{align}
    \norm{N(\omega_1)}_{\mathcal{F}^{0,1}_{\nu}} &\leq \abs{A_\mu} \Big( \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}} \norm{\omega_1}_{\mathcal{F}^{0,1}_{\nu}} + \frac{1}{2} C_0 \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}}  + \norm{N_2}_{\mathcal{F}^{0,1}_{\nu}}\Big)\\
    &\leq 2A_\rho \abs{A_\mu} C_1 \Big(  \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}} + \abs{A_\kappa} \Big( \frac{1}{2} C_0^2 \norm{f}_{\mathcal{F}^{1,1}_{\nu}} + C_0 (C_0 -1)  \Big) \Big) \norm{f}_{\mathcal{F}^{1,1}_{\nu}} ,\\
    \begin{split}
    \norm{N(\omega_1)}_{\mathcal{F}^{1,1}_{\nu}} &\leq  2A_\rho \abs{A_\kappa} \abs{A_\mu} C_1 \Bigg( 2 \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}} \left( \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}} (1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}})}{1 - \norm{f}_{\mathcal{F}^{1,1}_{\nu}}} + C_3 \right) \\
    &\hspace{10em} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}} (2C_0 + 2C_2 + C_3) + (C_0C_2 + (C_0^2 - 1)C_4) \Bigg) \norm{f}_{\mathcal{F}^{2,1}_{\nu}},
    \end{split}
\end{align}
For $N_3$ and $N_5$  we can transfer a factor of $\xi$ from $N(\hat \omega_2)$ to their pre-factors, and use
\begin{align}
    \norm{\xi \inv N(\omega_2)}_{\mathcal{F}^{0,1}_{\nu}} &\leq \abs{A_\kappa} h_2 C_5 \norm{\omega_1}\\
    &\leq 2 h_2 A_\rho \abs{A_\kappa} C_1 C_5 \norm{f}_{\mathcal{F}^{1,1}_{\nu}},\\
    \norm{\xi \inv N(\omega_2)}_{\mathcal{F}^{1,1}_{\nu}} &\leq \abs{A_\kappa} \sum_{n=1}^\infty \dnorm{e^{-h_2\abs\xi} \frac{\abs{\xi}^{n-1}}{n!}}_{L^\infty} (\norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n \norm{\omega}_{\mathcal{F}^{1,1}_{\nu}} + n\norm{f}_0^{n-1} \norm{\omega_1}_{\mathcal{F}^{0,1}_{\nu}})\\
    &\leq 2h_2 A_\rho \abs{A_\kappa} C_1(C_3C_5 + C_0 \norm{f}_{\mathcal{F}^{0,1}_{\nu}}) \norm{f}_{\mathcal{F}^{2,1}_{\nu}}
\end{align}
in which
\begin{align}
    \label{C5}
    C_5 = \frac{1}{h_2}\sum_{n = 1}^\infty \dnorm{e^{-h_2\abs\xi} \frac{\abs{\xi}^{n-1}}{n!}}_{L^\infty} \norm{f}_{\mathcal{F}^{0,1}_{0}}^n = \sum_{n=0}^\infty \frac{1}{n+1}\frac{e^{-n}}{n!} \left(\frac{n}{h_2}\right)^n \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^{n+1}.
\end{align}
Finally,
\begin{align*}
    \hat N_6 = -i\pi \sum_{n=1}^\infty \frac{(-1)^n}{n!} \Big( (\hat \omega_2\ \sgn(\xi) \ \abs{\xi}^n\ e^{-h_2 \abs{\xi}} ) * (\ast^n \hat f) \Big) (\xi)
\end{align*}
is bounded similarly to $B_4$ with
\begin{align}
    \norm{N_6}_{\mathcal{F}^{0,1}_{\nu}} &\leq \pi (C_0 - 1)
    \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}}\\
    &\leq 2\pi A_\rho \abs{A_\kappa} C_0(C_0-1)C_1 \norm{f}_{\mathcal{F}^{1,1}_{\nu}},\\
    \norm{N_6}_{\mathcal{F}^{1,1}_{\nu}} &\leq 2\pi A_\rho \abs{A_\kappa} C_0 C_1 C_2 \norm{f}_{\mathcal{F}^{2,1}_{\nu}} + \pi \abs{A_\kappa} (C_0^2 - 1) \norm{\omega_1}_{\mathcal{F}^{1,1}_{\nu}}\\
    &\leq 2\pi A_\rho \abs{A_\kappa} C_1(C_0 C_2 + C_3)\norm{f}_{\mathcal{F}^{2,1}_{\nu}} .
\end{align}
Examining \eqref{interface_neat} we'll group terms as follows. First combine the estimates for $I_2$, $B_1$, $B_2$ and $N_1$ to find
\begin{align}
    \frac{1}{2\pi} (\norm{I_2}_{\mathcal{F}^{0,1}_{\nu}} + \norm{N_1}_{\mathcal{F}^{0,1}_{\nu}}+  \abs{A_\mu} ( \norm{B_1}_{\mathcal{F}^{0,1}_{\nu}}  + \norm{B_2}_{\mathcal{F}^{0,1}_{\nu}})) &\leq   A_\rho C_1 \frac{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}} {1- \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2}( 2 \norm{f}_{\mathcal{F}^{1,1}_{\nu}} + \abs{A_\mu}) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\\
    &\leq A_\rho C_1 \lambda_0 \norm{f}_{\mathcal{F}^{1,1}_{\nu}}.
\end{align}
Doing similarly for $N_2$ and $N_6$ gives
\begin{align}
\abs{A_\mu} \norm{N_2}_{\mathcal{F}^{0,1}_{\nu}} + \norm{N_6}_{\mathcal{F}^{0,1}_{\nu}} & \leq A_\rho \abs{A_\kappa} C_1 (\abs{A_\mu} + 2\pi) C_0(C_0 - 1) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\\
&\leq A_\rho \abs{A_\kappa} C_1 \lambda_1 \norm{f}_{\mathcal{F}^{1,1}_{\nu}}.
\end{align}
Grouping the terms $N_3$ and $N_5$
\begin{align}
\begin{split}
	(1 + A_\mu) (\norm{N_3}_{\mathcal{F}^{0,1}_{\nu}} + \norm{N_5}_{\mathcal{F}^{0,1}_{\nu}}) &\leq A_\rho \abs{A_\kappa} C_1  C_5(1+ A_\mu) \\
	&\hspace{3em} \cdot h_2\left( \norm{e^{-h_2 \xi} \xi}_{L^\infty} + \dnorm{\frac{A_\kappa A_\mu \xi e^{-h_2 \abs\xi}}{e^{2h_2\abs\xi} - A_\kappa A_\mu}}_{L^\infty} \right) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}
\end{split}\\
	&\leq A_\rho \abs{A_\kappa} C_1 C_5 (1+A\mu)\Big(\frac{1}{e} +\frac{1}{2}\Big) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\\
	&\leq A_\rho \abs{A_\kappa} C_1 \lambda_2 \norm{f}_{\mathcal{F}^{1,1}_{\nu}}
\end{align}
Then,
\begin{align}
    (1+A_\mu) (\norm{N_4}_{\mathcal{F}^{0,1}_{\nu}}) &\leq A_\rho \abs{A_\mu} C_1 \bnorm{\frac{(1+A_\mu) A_\kappa}{e^{2h_2 \abs{\xi}} - A_\kappa A_\mu}}_{L^\infty} \lambda_3 \norm{f}_{\mathcal{F}^{1,1}_{\nu}}
\end{align}
and finally, we choose $\lambda_4$ so that
\begin{align}
\frac{1}{2\pi} (\norm{I_4}_{\mathcal{F}^{0,1}_{\nu}} + \norm{B_3}_{\mathcal{F}^{0,1}_{\nu}}) &\leq A_\rho \abs{A_\kappa} C_1 \lambda_4 \norm{f}_{\mathcal{F}^{1,1}_{\nu}}
\end{align}

\end{comment}
Now, fix
\begin{align}
\label{infconst}
    \theta = \inf_{\xi \in \R} \left(1 - \frac{A_\kappa(1-A_\mu) }{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}\right)
\end{align}
and note that $\theta > 0$ since $|A_\kappa| < 1$. Then by \eqref{interface_neat} and the above estimates
\begin{align*}
    \frac{d}{dt} \norm{f}_{\mathcal{F}^{0,1}_{\nu}} &\leq (-A_\rho\theta + \nu) \norm{f}_{\mathcal{F}^{1,1}_{\nu}} + A_\rho \sigma_0 \norm{f}_{\mathcal{F}^{1,1}_{\nu}}
\end{align*}
where
\begin{align}
\label{sigma0}
    \sigma_0 &\eqdef \sigma_{0}(\|f\|_{\mathcal{F}^{0,1}_{\nu}}, \|f\|_{\mathcal{F}^{1,1}_{\nu}}) \nonumber\\
    \begin{split}
    &= C_1 \Big[ (C_6 + \abs{A_\kappa} C_0^2) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\\
    &\hspace{2.5em} +
    \frac{1}{2} \abs{1+ A_\mu} \left( 2|A_\kappa| (C_0 - 1) + C_5 \Big(2 \abs{A_\kappa} (C_0 -1) + \lambda_1 \Big) \right)\\
    &\hspace{3.5em} + 2\pi \abs{A_\kappa} C_0(C_0-1) + \abs{A_\mu}\lambda_1 +C_6 \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \Big]
    \end{split}
\end{align}
and
\begin{align*}
    \frac{d}{dt} \norm{f}_{\mathcal{F}^{1,1}_{\nu}} &\leq (-A_\rho\theta + \nu) \norm{f}_{\mathcal{F}^{2,1}_{\nu}} + A_\rho \sigma_1 \norm{f}_{\mathcal{F}^{2,1}_{\nu}}
\end{align*}
where
\begin{align}
\label{sigma1}
    \sigma_1&\eqdef \sigma_{1}(\|f\|_{\mathcal{F}^{0,1}_{\nu}}, \|f\|_{\mathcal{F}^{1,1}_{\nu}}) \nonumber\\
    \begin{split}
    &= C_1 \Big[ 2(1+\frac{1}{2}C_3 (1-\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^2))C_6^2 + |A_\kappa| \lambda_0 \\
    &\hspace{2.5em} + \frac{1}{2} \abs{1+ A_\mu} \left( |A_\kappa| \lambda_2 + C_5 \Big(|A_\kappa| \lambda_2 + \lambda_3 \Big) \right)\\
    &\hspace{3.5em} + 2\pi \abs{A_\kappa} (C_0 C_2 + (C^2_0 - 1)C_4) + \frac{1}{2} \abs{A_\mu} \lambda_3 + C_3C_6 \norm{f}_{\mathcal{F}^{1,1}_{\nu}} + \frac{2C_6^2}{\norm{f}_{\mathcal{F}^{1,1}_{\nu}}}\Big]
    \end{split}
\end{align}
Note that $\sigma_{i}(\|f\|_{\mathcal{F}^{0,1}_{\nu}}, \|f\|_{\mathcal{F}^{1,1}_{\nu}})$, $i=1,2$, are continuous functions in $(\|f\|_{\mathcal{F}^{0,1}_{\nu}}, \|f\|_{\mathcal{F}^{1,1}_{\nu}})$ such that $\sigma_{i}(0,0) = 0$.

\begin{comment}
\section{Estimates for Decay/Analyticity}
We will make these estimates in $\calF^{s,1}$ to show decay. Similar calculations in $\calF^{s,1}_\nu$ also prove instant analyticity. First we will bound $\omega_2$ which simplifies future calculations.

\begin{proposition}
We have the estimate
\begin{equation}\label{omega2fzerone}
\|\omega_{2}\|_{\mathcal{F}^{0,1}_{\nu}} \leq \tfrac{2}{\pi}\calK\rho \,C_1 \|f\|_{\mathcal{F}^{1,1}_{\nu}}
\end{equation}
where
\begin{equation}\label{C1}
C_1 = \sum_{n = 0}^\infty \dnorm{\pi e^{-h_2\abs\xi} \frac{\abs{\xi}^n}{n!}}_{L^\infty}.
\end{equation}
Further, for $0< s \leq 1$

\begin{equation}\label{omega2foneone}
\|\omega_{2}\|_{\mathcal{F}^{s,1}_{\nu}} \leq \tfrac{2}{\pi}\calK\rho \Big( \sum_{n = 0}^\infty (n+1)\dnorm{ \frac{\pi e^{-h_2 \abs \xi}\abs{\xi}^{n}}{n!}}_{L^\infty} \|f\|_{\mathcal{F}^{0,1}_{\nu}}^{n} \Big) \|f\|_{\mathcal{F}^{2,1}_{\nu}}.
\end{equation}
\end{proposition}

\begin{proof}
Taking the Fourier transform of $\omega_{2}$, we obtain
\begin{align*}
    (\tfrac{2}{\pi}\calK\rho)\inv\ \hat\omega_2 &= \int_\R \int_\R  \frac{f(x - \gamma) + h_2}{\gamma^2 + (f(x - \gamma) + h_2)^2}\ \partial_\alpha f(x - \gamma)e^{-i\xi x}\; d\gamma dx\\
    &= \int_\R \int_\R  \frac{f(y) + h_2}{\gamma^2 + (f(y) + h_2)^2}\  e^{-i\xi \gamma}\; d\gamma\; \partial_\alpha f(y) e^{-i\xi y}\; dy\\
    &= \int_\R \pi e^{-(f(y) + h_2) \abs \xi }\cdot \partial_\alpha f(y)  e^{-i\xi y}\; dy\\
    &= \pi e^{-h_2 \abs \xi}  \int_\R e^{-f(y) \abs \xi} \cdot \partial_\alpha f(y) e^{-i\xi y}\; dy\\
    & =\sum_{n=0}^{\infty}\pi e^{-h_2 \abs \xi}  \int_\R \frac{(-f(y)|\xi|)^{n}}{n!} \cdot \partial_\alpha f(y) e^{-i\xi y}\; dy\\
    &= \sum_{n=0}^{\infty}\pi e^{-h_2 \abs \xi} \frac{(-1)^{n}|\xi|^{n}}{n!} \Big((\ast^{n}\hat{f}) \ast (i\xi\hat{f}) \Big)(\xi).
\end{align*}
Next, using Young's inequalities for convolutions and \eqref{convolutionexp}:
\begin{align*}
(\tfrac{2}{\pi}\calK\rho)\inv\ \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}} &\leq   \sum_{n=0}^{\infty} \Big\|\pi e^{-h_2 \abs \xi} \frac{(-1)^{n}|\xi|^{n}}{n!} \Big((\ast^{n}\hat{f}) \ast (i\xi\hat{f}) \Big)(\xi)\Big\|_{\mathcal{F}^{0,1}_{\nu}}\\
&\leq   \sum_{n=0}^{\infty} \dnorm{\pi e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \Big\| (\ast^{n}\hat{f}) \ast (i\xi\hat{f})\Big\|_{\mathcal{F}^{0,1}_{\nu}}\\
&\leq \Big( \sum_{n = 0}^\infty \dnorm{\pi e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \|f\|_{\mathcal{F}^{0,1}_{\nu}}^n \Big) \|f\|_{\mathcal{F}^{1,1}_{\nu}}.
\end{align*}
The estimate \eqref{omega2foneone} is derived similarly but additionally using \eqref{productrule}:
\begin{align*}
 (\tfrac{2}{\pi}\calK\rho)\inv\ \|\omega_{2}\|_{\mathcal{F}^{1,1}_{\nu}} &\leq \Big( \sum_{n = 0}^\infty \dnorm{ \frac{\pi e^{-h_2 \abs \xi}\abs{\xi}^{n}}{(n-1)!}}_{L^\infty} \|f\|_{\mathcal{F}^{0,1}_{\nu}}^{n-1}\|f\|_{\mathcal{F}^{1,1}_{\nu}} \Big) \|f\|_{\mathcal{F}^{1,1}_{\nu}}\\ &\hspace{0.5in}+ \Big( \sum_{n = 0}^\infty \dnorm{\pi e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \|f\|_{\mathcal{F}^{0,1}_{\nu}}^{n} \Big) \|f\|_{\mathcal{F}^{2,1}_{\nu}}\\
&\leq \Big( \sum_{n = 0}^\infty (n+1)\dnorm{ \frac{\pi e^{-h_2 \abs \xi}\abs{\xi}^{n}}{n!}}_{L^\infty} \|f\|_{\mathcal{F}^{0,1}_{\nu}}^{n} \Big) \|f\|_{\mathcal{F}^{2,1}_{\nu}}
\end{align*}
where in the last line we used the interpolation \eqref{interpolation}.
\end{proof}
\begin{remark}

By the ratio test and the following computation
\[
    \dnorm{\pi e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} = \pi \frac{e^{-n}}{n!} \Big( \frac{n}{h_2}\Big)^n,
\]
the power series in \eqref{omega2fzerone} and \eqref{omega2foneone} converge  for $\|f\|_{\mathcal{F}^{0,1}_{\nu}} < \frac{h_{2}}{e}$ \todo{Nikhil comment: I don't think the factor of $e$ is necessary}. An easier to manage, but looser version of this estimate is
\[
\pi \frac{e^{-n}}{n!} \Big( \frac{n}{h_2}\Big)^n \leq \frac{\pi}{\sqrt{2\pi n} \cdot h_2^n} \leq \frac{\pi}{h_2^n}, \qquad \forall n \geq 1.
\]
implying
\[
    \norm{\omega_2}_0 \leq \frac{2}{\pi}\calK \rho\, C_1 \norm{f}_1 \coloneqq  2\calK\rho \frac{\norm{f}_0}{h_2 - \norm{f}_0} \norm{f}_1.
\]
\end{remark}
Using \eqref{omega2fzerone} and \eqref{omega2foneone}, we can estimate $I_2$.

\begin{proposition}
\begin{equation}\label{I2fzerone}
\|I_{2}\|_{\mathcal{F}^{0,1}_{\nu}}\leq \frac{C_1 }{2\pi} \norm{\omega_2}_{\mathcal{F}^{0,1}_{\nu}}(1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}})
\end{equation}
and
\begin{equation}\label{I2foneone}
\|I_2\|_{\mathcal{F}^{1,1}_{\nu}}\leq \frac{C_1}{2\pi} \norm{\omega_2}_{\mathcal{F}^{1,1}_{\nu}}(1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}})
\end{equation}
where $C_1$ was defined in \eqref{C1}.
\end{proposition}
\begin{proof}
We can write $I_{2}= \tfrac{1}{2\pi}(A_1 + A_2)$ where
\begin{align*}
    A_1 &= \int_\R \frac{\beta}{\beta^2 + (f(\alpha) + h_2)^2}\ \omega_2(\alpha - \beta)\; d\beta,\\
    A_2 &= \int_\R \frac{\partial_\alpha f(\alpha) (f(\alpha) + h_2)}{\beta^2 + (f(\alpha) + h_2)^2}\ \omega_2(\alpha - \beta)\; d\beta.
\end{align*}
Well,
\end{proof}
\end{comment}

\section{$L^{2}$ Estimates}
\label{sec:l2est}
\begin{comment}
\begin{theorem}
Suppose $f_0 \in L^2 \cap \calF^{1,1}$ such that TBA hold. Then $f(t) \in L^2_\nu$ instantly for all $t > 0$. Moreover
\begin{align}
    \norm{f}_{L^2_\nu}^2 (t) \leq \norm{f_0}_{L^2}^2 \cdot \exp(R(\norm{f}_{\mathcal{F}^{0,1}_{\nu}}, \norm{f}_{\mathcal{F}^{1,1}_{\nu}})).
\end{align}
for a rational function $R$. In particular this implies $f(t)\in H^s$ instantly for all $t > 0$.
\end{theorem}
\end{comment}

\subsection{Analytic Estimates}
In this section, we will prove the $L^{2}$ estimate \eqref{L2inequality} of Theorem \ref{MainTheorem}. We will introduce the notation $\lesssim$ which indicates a constant depending on $\norm{f}_{\mathcal{F}^{0,1}_{\nu}}$ and $\norm{f}_{\mathcal{F}^{1,1}_{\nu}}$. Let us begin by differentiating
\begin{align}
\label{interfaceL2}
    \frac{1}{2} \frac{d}{dt}\norm{f}_{L^2_\nu}^2(t) = \nu \norm{f}_{\dot{H}^{1/2}_\nu}^2 - A_\rho \int_\R e^{2\nu t |\xi|} \abs{\xi} \abs{\hat f(\xi)}^2 \left(1- \frac{A_\kappa(1-A_\mu)}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}\right) \; d\xi + \langle \hat f, \text{h.o.t.} \rangle_{L^2_\nu}
\end{align}
in which the higher order terms are
\begin{align*}
\begin{split}
    \text{h.o.t.} &= \frac{1}{2\pi}(\hat I_2 + \hat I_4)
     + \hat N_0(\xi) + (1 + A_\mu)(\hat N_1 + \hat N_2 + \hat N_3)(\xi)  + \hat N_4(\xi) - \frac{i}{2} \sgn(\xi) N(\hat \omega_1)(\xi).
\end{split}
\end{align*}
In the following estimates, we will use the convention $\tilde{g}(x) = g(-x)$ and the convolution identities
\begin{align*}
   \int_{\mathbb{R}} g_{1}(x)(g_{2}\ast g_{3})(x) dx &= \int_{\mathbb{R}} g_{1}(x)\int_{\mathbb{R}}g_{2}(x-y)g_{3}(y) dy dx\\
   &= \int_{\mathbb{R}}g_{3}(y)  \int_{\mathbb{R}}g_{1}(x) g_{2}(x-y)dx dy\\
   &=  \int_{\mathbb{R}} g_{3}(y)(g_{1}\ast \tilde{g_{2}})(y) dy
\end{align*}
and
\begin{align*}
\widetilde{g_{1}\ast g_{2}} = \tilde{g_{1}}\ast \tilde{g_{2}}.
\end{align*}
Using the substitution $\abs{\hat \omega_1(\xi)} = \abs{\xi} \abs{\hat \Omega_1(\xi)}$, we obtain
\begin{align}
    \langle \hat f, \hat I_2 \rangle_{L^2_\nu} &\leq \int_\R e^{2\nu t\abs{\xi} } \abs{\hat f(\xi)} \cdot \pi \sum_{n=0}^\infty \Big((\ast^{2n+2} \abs{\widehat {\partial_\alpha f}}) * \abs{\cdot} \abs{\hat \Omega_1} \Big)(\xi)\; d\xi \nonumber\\
    &\leq \pi \sum_{n=0}^\infty \int_\R e^{\nu t\abs{\cdot} } \abs{\hat f(\xi)} \Big((\ast^{2n+2} e^{\nu t\abs{\cdot} }\abs{\widehat {\widetilde{\partial_\alpha f}}}) * e^{\nu t\abs{\cdot} }\abs{\cdot} \abs{\hat \Omega_1} \Big)(\xi)\; d\xi \nonumber\\
    &\leq \pi \sum_{n=0}^\infty \int_\R e^{\nu t\abs{\cdot} } \abs{\xi}\abs{\hat \Omega_1(\xi)} \Big((\ast^{2n+2} e^{\nu t\abs{\cdot} }   \abs{\widehat {\widetilde{\partial_\alpha f}}}) * e^{\nu t\abs{\cdot} }\abs{\hat f} \Big)(\xi)\; d\xi \nonumber\\
    &\leq \pi \sum_{n=0}^\infty \norm{\Omega_1}_{\dot{H}^{1/2}_\nu} \bnorm{\abs{\xi}^{1/2}\Big((\ast^{2n+2} e^{\nu t\abs{\cdot} }   \abs{\widehat {\widetilde{\partial_\alpha f}}}) * e^{\nu t\abs{\cdot} }\abs{\hat f} \Big)}_{L^2} \nonumber\\
    &\leq \pi  \sum_{n=0}^\infty  \norm{\Omega_1}_{\dot{H}^{1/2}_\nu}  \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{2n+1}\Big((2n+2) \norm{f}_{\calF^{3/2,1}_\nu} \norm{f}_{L^2_\nu} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{f}_{\dot{H}^{1/2}_\nu} \Big)\nonumber\\
    \begin{split}
    \label{youngsexample}
    &\leq \pi \sum_{n=0}^\infty \frac{\epsilon_n}{2} \norm{\Omega_1}_{\dot{H}^{1/2}_\nu}^2 + \frac{1}{2\epsilon_n} (2n+2)^2 \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{4n+2} \norm{f}_{\calF^{3/2,1}_\nu}^2 \norm{f}_{L^2_\nu}^2\\
    &\hspace{5em} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{2n+2} \norm{\Omega_1}_{\dot{H}^{1/2}_\nu}\norm{f}_{\dot{H}^{1/2}_\nu}
    \end{split}
\end{align}
where the last inequality is from Young's inequality for products where we choose $\epsilon_n = \epsilon/(1+n^2)$ for a small value $\epsilon> 0$.
The other non-linear terms can be estimated via similar methods
\begin{align*}
    \begin{split}
    \langle \hat f, \hat I_4 \rangle_{L^2_\nu} &\leq \pi \sum_{n=0}^\infty \dnorm{ e^{-h_2 \abs \xi} \frac{\abs{\xi}^{n}}{n!}}_{L^\infty} \norm{\Omega_2}_{\dot{H}^{1/2}_\nu} \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^{n}\Big( \norm{f}_{\mathcal{F}^{3/2,1}_{\nu}} \norm{f}_{L^2_\nu}

    + (n+1) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}
    \norm{f}_{\dot{H}^{1/2}_\nu}\Big)
    \end{split}
\end{align*}
and
\begin{align*}
    \langle \hat f, \hat N_1 \rangle_{L^2_\nu} &\leq \frac{1}{2} \int_{\R} d\xi \; e^{2\nu t\abs{\xi}} e^{-h_2\abs{\xi}} \abs{\hat f(\xi)} \abs{N(\widehat{\partial_\alpha \Omega_2})(\xi)} \\
    &\leq \frac{\abs{A_\kappa}}{2} \sum_{n=1}^{\infty} \int_{\R} d\xi \; e^{2\nu t\abs{\xi}} e^{-2h_2\abs{\xi}} \frac{|\xi|^{n}}{n!} \cdot \abs{\hat f(\xi)}   \Big((\ast^{n}\abs{\hat{f}}) \ast \abs{\cdot} \abs{\hat \Omega_1} \Big)(\xi) \\

    &\leq  \frac{\abs{A_\kappa}}{2} \sum_{n=1}^{\infty} \dnorm{e^{-2h_2\abs{\xi}} \frac{|\xi|^{n}}{n!}}_{L^\infty} \int_{\R} d\xi \; e^{2\nu t\abs{\xi}} \abs{\xi} \abs{\hat \Omega_1(\xi)}   (\ast^{n}\abs{\hat{\tilde{f}}} * \abs{\hat f})(\xi) \\
    &\leq \frac{\abs{A_\kappa}}{2} \sum_{n=1}^{\infty} \dnorm{e^{-2h_2\abs{\xi}} \frac{|\xi|^{n}}{n!}}_{L^\infty} (n+1)\norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n \norm{\Omega_1}_{\dot{H}^{1/2}_\nu} \norm{f}_{\dot{H}^{1/2}_\nu}\\
    &\eqdef \abs{A_\kappa} \lambda_4 \norm{\Omega_1}_{\dot{H}^{1/2}_\nu} \norm{f}_{\dot{H}^{1/2}_\nu}
\end{align*}
where $\lambda_4 \to 0$ as $\norm{f}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \to 0$. Next

\begin{align}
\begin{split}
    \label{prodfN2}
    \langle \hat f, \hat N_2 \rangle_{L^2_\nu} &\leq \frac{1}{2} C_5 \Big( \norm{\Omega_1}_{\dot{H}^{1/2}_\nu}\sum_{n=0}^\infty 2 (\norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{2n+1} \norm{f}_{\dot{H}^{1/2}_\nu} + (2n+1) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{2n} \norm{f}_{\mathcal{F}^{3/2,1}_{\nu}}\norm{f}_{L^2_\nu}) \\
    &\hspace{3em} + \norm{\Omega_1}_{\dot{H}^{1/2}_\nu}\sum_{n= 0}^\infty \dnorm{ e^{-h_2|\xi|} \frac{|\xi|^n}{n!} }_{L^\infty} \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n( \norm{f}_{\mathcal{F}^{3/2,1}_{\nu}} \norm{f}_{L^2_\nu} + (n+1) \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{f}_{\dot{H}^{1/2}_\nu}) \\
    &\hspace{5em} + \norm{\Omega_2}_{\dot{H}^{1/2}_\nu}\sum_{n=1}^\infty  \dnorm{ e^{-h_2|\xi|} \frac{|\xi|^n}{n!} }_{L^\infty}   (n+1) \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n \norm{f}_{\dot{H}^{1/2}_\nu} \Big)
    \end{split}\\

    &\eqdef \frac{1}{2} C_5\left(\norm{\Omega_1}_{\dot{H}^{1/2}_\nu} (\lambda_5 \norm{f}_{\dot{H}^{1/2}_\nu} + \lambda_6 \norm{f}_{\mathcal{F}^{3/2,1}_{\nu}}\norm{f}_{L^2_\nu} ) + \lambda_7  \norm{\Omega_2}_{\dot{H}^{1/2}_\nu} \norm{f}_{\dot{H}^{1/2}_\nu} \right), \nonumber \\

    \langle \hat f, \hat N_3 \rangle_{L^2_\nu} &\leq \frac{1}{2} C_5 \int_{\R} d\xi \; e^{2\nu t\abs{\xi}} e^{-h_2\abs{\xi}} \abs{\hat f(\xi)} \abs{N(\hat \Omega_2)(\xi)} \nonumber \\
    &\leq \frac{1}{2} C_5 \abs{A_\kappa} \lambda_4 \norm{\Omega_1}_{\dot{H}^{1/2}_\nu} \norm{f}_{\dot{H}^{1/2}_\nu}, \nonumber \\
    \langle \hat f, \hat N_4 \rangle_{L^2_\nu} &\leq \pi \sum_{n=1}^\infty \dnorm{e^{-h_2\abs{\xi}} \frac{|\xi|^{n}}{n!}}_{L^\infty} (n+1) \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^n \norm{\Omega_2}_{\dot{H}^{1/2}_\nu} \norm{f}_{\dot{H}^{1/2}_\nu}, \nonumber
\end{align}
in which
\begin{align}
    \lambda_5, \lambda_7 \to 0 \quad \text{and} \quad \lambda_6 \lesssim 1 \quad \text{as} \quad \norm{f}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \to 0.
\end{align}

Related to the the computations in \eqref{prodfN2},
\begin{align*}
     \langle \hat f, \hat N_0 \rangle_{L^2_\nu} &\leq \frac{1}{2} \norm{\Omega_1}_{\dot{H}^{1/2}_\nu}\sum_{n=0}^\infty \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{2n+2} \norm{f}_{\dot{H}^{1/2}_\nu} + (2n+2) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{2n+1} \norm{f}_{\mathcal{F}^{3/2,1}_{\nu}}\norm{f}_{L^2_\nu},\\
    \langle \hat f, N(\widehat{\partial_\alpha \Omega_1 })\rangle_{L^2_\nu} &\leq \norm{\Omega_1}_{\dot{H}^{1/2}_\nu} (\lambda_5 \norm{f}_{\dot{H}^{1/2}_\nu} + \lambda_6 \norm{f}_{\mathcal{F}^{3/2,1}_{\nu}} \norm{f}_{L^2_\nu} ) + \lambda_7  \norm{\Omega_2}_{\dot{H}^{1/2}_\nu} \norm{f}_{\dot{H}^{1/2}_\nu}.
\end{align*}
So it is enough to control $\Omega_1$ and $\Omega_2$. We can write
\begin{align*}
    \hat \Omega_1 = \hat \Omega_{11} + \hat \Omega_{12} - 2A_\rho \hat f
\end{align*}
Using \eqref{omega12fourierlinearpartsum} and \eqref{Omega11abs} and  we have

\begin{align}
    \norm{\Omega_{11}}_{\dot{H}^{1/2}_\nu}  &\leq \abs{A_\mu} \sum_{n=0}^\infty \norm{f}^{2n+1}_{\mathcal{F}^{1,1}_{\nu}} \norm{\Omega_1}_{\dot{H}^{1/2}_\nu} + (2n + 1) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{2n} \norm{f}_{\mathcal{F}^{3/2,1}_{\nu}}\norm{\Omega_1}_{L^2_\nu} \nonumber\\
    \label{Omega11Hdotnu}
    &\defeq \abs{A_\mu} (C_6\norm{\Omega_1}_{\dot{H}^{1/2}_\nu} + C_7 \norm{\Omega_1}_{L^2_\nu})\\
    \label{Omega12Hdotnu}
    \norm{\Omega_{12}}_{\dot{H}^{1/2}_\nu} &\leq \abs{A_\mu} \Big(C_0 + C_8 \Big) \norm{\Omega_2}_{\dot{H}^{1/2}_\nu}
\end{align}
where
\begin{align*}
    C_7 &\to \norm{f}_{\mathcal{F}^{3/2,1}_{\nu}},\\
    C_8 &= \sum_{n = 1}^\infty n \bnorm{e^{-h_2\abs{\xi}} \frac{\abs{\xi}^{n-1/2}}{n!} }_{L^\infty} \norm{f}_{\mathcal{F}^{0,1}_{\nu}}^{n-1}\norm{f}_{\mathcal{F}^{1/2,1}_{\nu}} \to \frac{1}{\sqrt{2 e h_2} } \norm{f}_{\mathcal{F}^{1/2,1}_{\nu}} \to 0
\end{align*}
as $\norm{f}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \to 0$.
With \eqref{Omega21fourierexpansion} and \eqref{Omega22fourierexpansion} we have the bounds
\begin{align}
    \norm{\Omega_2}_{\dot{H}^{1/2}_\nu} &\leq \abs{A_\kappa} C_0(1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}})\norm{\Omega_1}_{\dot{H}^{1/2}_\nu} \nonumber \\
    &\hspace{2.5em} + \abs{A_\kappa} \Big(\frac{C_2}{\norm{f}_{\mathcal{F}^{0,1}_{\nu}}} \norm{f}_{\mathcal{F}^{1/2,1}_{\nu}} (1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}}) + C_0 \norm{f}_{\mathcal{F}^{3/2,1}_{\nu}}\Big) \norm{\Omega_1}_{L^2_\nu} \nonumber \\
    \label{Omega2Hdotnu}
    &\eqdef \abs{A_\kappa}( C_9 \norm{\Omega_1}_{\dot{H}^{1/2}_\nu} + C_{10}\norm{\Omega_1}_{L^2_\nu})
\end{align}
with $C_9 \to 1$ and $C_{10} \to \abs{A_\kappa} \norm{f}_{\mathcal{F}^{3/2,1}_{\nu}}$ as $\norm{f}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \to 0$.

Similarly to the above calculations, we derive
\begin{align*}
    \norm{\Omega_2}_{L^2_\nu} &\leq \abs{A_\kappa} C_0(1 + \norm{f}_{\mathcal{F}^{1,1}_{\nu}}) \norm{\Omega_1}_{L^2_\nu} = \abs{A_\kappa} C_9 \norm{\Omega_1}_{L^2_\nu}
\end{align*}
and therefore by \eqref{omega12fourierlinearpartsum} and \eqref{Omega11abs} we have
\begin{align*}
    (1-\abs{A_\mu} C_6)\norm{\Omega_1}_{L^2_\nu} &\leq  \abs{A_\mu} C_0 \norm{\Omega_2}_{L^2_\nu} + 2A_\rho \norm{f}_{L^2_\nu}\\
    &\leq |A_\kappa| \abs{A_\mu} C_0 C_9 \norm{\Omega_1}_{L^2_\nu} + 2A_\rho \norm{f}_{L^2_\nu}
\end{align*}
to give us
\begin{align}
    \label{Omega1elltwonu}
    \norm{\Omega_1}_{L^2_\nu} &\leq 2A_\rho C_{11} \norm{f}_{L^2_\nu},\\
    \label{Omega2elltwonu}
    \norm{\Omega_2}_{L^2_\nu} &\leq 2A_\rho \abs{A_\kappa} C_9 C_{11} \norm{f}_{L^2_\nu}
\end{align}
where
\begin{equation}\label{C11}
C_{11} = (1-\abs{A_{\mu}}C_{6}-\abs{A_{\kappa}}\abs{A_{\mu}}C_{0}C_{9})^{-1} \to (1- \abs{A_\kappa} \abs{A_\mu} )\inv \quad \text{as} \quad \norm{f}_{\mathcal{F}^{0,1}_{\nu}} + \norm{f}_{\mathcal{F}^{1,1}_{\nu}} \to 0.
\end{equation}
Collecting terms from \eqref{Omega11Hdotnu}, \eqref{Omega12Hdotnu}, \eqref{Omega2Hdotnu}, \eqref{Omega1elltwonu}, and \eqref{Omega2elltwonu} we find
\begin{align}
    \norm{\Omega_1}_{\dot{H}^{1/2}_\nu} &\leq 2A_\rho C_{12} ( |A_\mu| C_{11} (C_7 + \abs{A_\kappa} (C_0 + C_8) C_{10}) \norm{f}_{L^2_\nu} + \norm{f}_{\dot{H}^{1/2}_\nu})\nonumber\\
    \label{Omega1Hdot}
    &\eqdef 2A_\rho C_{12} (C_{13} \norm{f}_{L^2_\nu} + \norm{f}_{\dot{H}^{1/2}_\nu})\\
    \norm{\Omega_2}_{\dot{H}^{1/2}_\nu} &\leq 2A_\rho \abs{A_\kappa} \Big((C_9C_{12}C_{13} + C_{10}C_{11}) \norm{f}_{L^2_\nu} + C_9C_{12} \norm{f}_{\dot{H}^{1/2}_\nu} \Big) \nonumber\\
    \label{Omega2Hdot}
    &\eqdef 2A_\rho \abs{A_\kappa} (C_{14} \norm{f}_{L^2_\nu} + C_9C_{12} \norm{f}_{\dot{H}^{1/2}_\nu})
\end{align}
for
\begin{align*}
    C_{12} &= \Big( 1- \abs{A_\mu} (C_6 + \abs{A_\kappa} (C_0 + C_8) C_9 )\Big)\inv \to (1-\abs{A_\kappa} \abs{A_\mu} )\inv ,\\
    C_{13} &\lesssim 1+\norm{f}_{\mathcal{F}^{3/2,1}_{\nu}},\\
    C_{14} &\lesssim 1+ \norm{f}_{\mathcal{F}^{3/2,1}_{\nu}}
\end{align*}

We see two types of expressions. For the first type of expression, of the form $\norm{\Omega_i}_{\dot{H}^{1/2}_\nu} \norm{f}_{L^2_\nu}$, we can apply Young's inequality for products and control the $\dot{H}^{1/2}_\nu$ terms as in \eqref{youngsexample}, e.g. using \eqref{Omega1Hdot}:
\begin{align*}
    \norm{\Omega_1}_{\dot{H}^{1/2}_\nu} \norm{f}_{L^2_\nu} &\leq  2A_\rho C_{12} (C_{13} \norm{f}_{L^2_\nu}^{2}+ \norm{f}_{\dot{H}^{1/2}_\nu}\norm{f}_{L^2_\nu})\\
    &\leq \frac{C}{\varepsilon_{n}}\norm{f}_{L^2_\nu}^{2} + \varepsilon_{n}\norm{f}_{\dot{H}^{1/2}_\nu}^{2}.
\end{align*}
Here $\varepsilon_{n}$ as earlier in \eqref{youngsexample} can always be chosen arbitrarily small. For the second type of expression $\norm{\Omega_i}_{\dot{H}^{1/2}_\nu} \norm{f}_{\dot{H}^{1/2}_\nu}$, after applying \eqref{Omega1Hdot} or \eqref{Omega2Hdot} we can control the resulting $\norm{f}_{\dot{H}^{1/2}_\nu}^2$ term via the linear decay term in the interface equation, see $\sigma_{2}$ below. In terms with $\norm{f}_{L^{2}_{\nu}}$ that contain coefficient of $\norm{f}_{\mathcal{F}^{3/2,1}_{\nu}}^2$, such as the middle term in \eqref{youngsexample}, we use $\norm{f}_{\mathcal{F}^{3/2,1}_{\nu}}^2 \leq \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{f}_{\mathcal{F}^{2,1}_{\nu}}$ by \eqref{interpolation}. Collecting terms from above, \eqref{interfaceL2} becomes
\begin{align*}
    \frac{1}{2} \frac{d}{dt}\norm{f}_{L^2_\nu}^2(t) &\leq  (-A_\rho \theta + \nu + A_\rho \sigma_2 + \varepsilon) \norm{f}_{\dot{H}^{1/2}_\nu}^2  + (R_{1}+ R_{2}\cdot(\|f\|_{\mathcal{F}^{3/2,1}_{\nu}}+\|f\|_{\mathcal{F}^{2,1}_{\nu}})) \norm{f}_{L^2_\nu}^{2}
\end{align*}
where
\begin{align}
    \sigma_2 &\eqdef \sigma_2 (\norm{f}_{\mathcal{F}^{0,1}_{\nu}}, \norm{f}_{\mathcal{F}^{1,1}_{\nu}}) \nonumber\\
    \begin{split}
    \label{sigma2}
    &=  C_{12}(C_6 + C_0 + C_2) \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\\
    &\hspace{2.5em} + |1+A_\mu| C_{12} \left( 2\abs{A_\kappa}\lambda_4  +  C_5 (\lambda_5  + \abs{A_\kappa}  ( \lambda_7C_9 + \lambda_4 )) \right) \\
    &\hspace{3.5em} + 2\pi\abs{A_\kappa}(C_0-1 + C_2)  C_9C_{12} + C_6C_{12} \norm{f}_{\mathcal{F}^{1,1}_{\nu}} + 2\lambda_5 C_{12} + 2\abs{A_\kappa}\lambda_7 C_9C_{12},
    \end{split}
\end{align}
in which $\varepsilon = \varepsilon(\norm{f_{0}}_{\mathcal{F}^{0,1}_{0}}, \norm{f_{0}}_{\mathcal{F}^{1,1}_{0}})$ is arbitrarily small and $R_{i}=  R_{i}(\norm{f_0}_{\mathcal{F}^{0,1}_{\nu}}, \norm{f_0}_{\mathcal{F}^{1,1}_{\nu}})$ for $i=1,2$ is bounded for medium sized initial data. Again, recall that $\norm{f}_{\mathcal{F}^{3/2,1}_{\nu}}^2 \leq \norm{f}_{\mathcal{F}^{1,1}_{\nu}}\norm{f}_{\mathcal{F}^{2,1}_{\nu}}$ by \eqref{interpolation}. Hence, by \eqref{maininequality}, $\|f\|_{\mathcal{F}^{1,1}_{\nu}}(t)$ and $\|f\|_{\mathcal{F}^{2,1}_{\nu}}(t)$ are $L^{1}$ functions in time on $[0,T]$ for any $T>0$, and hence, so is $\|f\|_{\mathcal{F}^{3/2,1}_{\nu}}(t)$ with $L^{1}$ norm bounded by initial data. By Gronwall's inequality, we obtain \eqref{L2inequality}.

\subsection{Sobolev Space Estimates}
In this section, we will prove an evolution estimate for a subcritical Sobolev norm $H^{\frac{3}{2}+\epsilon}$ of the interface
\begin{align*}
        \frac{1}{2} \frac{d}{dt}\norm{f}_{\dot H^{\frac{3}{2} + \epsilon}}^2(t) =- A_\rho \int_\R  \abs{\xi}^{4+2\epsilon} \abs{\hat f(\xi)}^2 \left(1- \frac{A_\kappa(1-A_\mu)}{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}\right) \; d\xi + \langle \hat f, \text{h.o.t.} \rangle_{H^{\frac{3}{2} + \epsilon}}.
\end{align*}
The weighted triangle inequality
\begin{align*}
\abs{\sum_{j=1}^{n} a_{j}}^{1+\epsilon} \leq n^{\epsilon}\sum_{j=1}^{n} \abs{a_{j}}^{1+\epsilon}
\end{align*}
allows us to use the constants from prior sections with the convention that $C_{k,\epsilon}$ is $C_k$ adjusted for the small $\epsilon$ weight from the weighted triangle inequality. All of the $C_{k,\epsilon}$ are bounded and $C_{k,\epsilon} \rightarrow C_{k}$ as $\epsilon \rightarrow 0$.
By \eqref{omega2fourierexpansion} we have
\begin{align*}
        \norm{\Omega_2}_{\dot H^{2+\epsilon}} &= \norm{\omega_2}_{\dot H^{1 + \epsilon}} \\
        &\leq \abs{A_\kappa } \Big( \frac{C_{2,\epsilon}}{\norm{f}_{\mathcal{F}^{0,1}_{0}}} \norm{\omega_1}_{\mathcal{F}^{0,1}_{0}} \norm{f}_{\dot H^{1+\epsilon}} + C_{0,\epsilon} \norm{\Omega_1}_{\dot H^{2+\epsilon}} \Big).
\end{align*}
where
\begin{equation*}C_{0,\epsilon}= \sum_{n = 0}^\infty \frac{n^{n}(n+1)^{\epsilon}}{e^nn!} \left(\frac{\norm{f}_{\mathcal{F}^{0,1}_{\nu}}}{h_2}\right)^n, \qquad C_{2,\epsilon} = \sum_{n = 1}^\infty  \frac{n^{n+1}(n+1)^{\epsilon}}{e^nn!} \left(\frac{\norm{f}_{\mathcal{F}^{0,1}_{\nu}}}{h_2}\right)^n
\end{equation*}
Similarly, by \eqref{omega11est} and \eqref{omega12calc}
\begin{align*}
\norm{\Omega_1}_{\dot H^{2+\epsilon}} &\leq \abs{A_\mu} \Big( 2\frac{1 + \norm{f}_{\mathcal{F}^{1,1}_{0}}^2}{(1-\norm{f}_{\mathcal{F}^{1,1}_{0}}^2)^2} \norm{\omega_1}_{\mathcal{F}^{0,1}_{0}} \norm{f}_{\dot H^{2+\epsilon}} + 2C_{6,\epsilon} \norm{\Omega_1}_{\dot H^{2+\epsilon}} + C_{0,\epsilon} (1+\norm{f}_{\mathcal{F}^{1,1}_{0}}) \norm{\Omega_2}_{\dot H^{2+\epsilon}}\\
& \hspace{5em} + \frac{C_{2,\epsilon}}{\norm{f}_{\mathcal{F}^{0,1}_{0}}}(1+\norm{f}_{\mathcal{F}^{1,1}_{0}}) \norm{\omega_2}_{\mathcal{F}^{0,1}_{0}} \norm{f}_{\dot H^{1+\epsilon}} + C_{0,\epsilon} \norm{\omega_2}_{\mathcal{F}^{0,1}_{0}} \norm{f}_{\dot H^{2+\epsilon}}\Big) + 2A_\rho \norm{f}_{\dot H^{2+\epsilon}}
    \end{align*}
from these we derive
    \begin{align}
        \label{Omega1H2}
        \norm{\Omega_1}_{\dot H^{2+\epsilon}} \leq 2A_\rho C_{1,\epsilon} (C_{15,\epsilon} \norm{f}_{\dot H^{1+\epsilon}} + C_{16,\epsilon} \norm{f}_{\dot H^{2+\epsilon}})
    \end{align}
where for example, we recover the $\epsilon$ adjusted constant
    \[
        C_{1,\epsilon} = \Big(1 - 2\abs{A_\mu}C_{6,\epsilon} - \abs{A_\mu}\abs{A_\kappa} C_{0,\epsilon}^2 (1+\norm{f}_{\mathcal{F}^{1,1}_{0}}) \Big)\inv
    \]
and $\lim_{\epsilon\rightarrow 0}C_{16,\epsilon} < C_{3} $. This in turn implies
    \begin{align}
        \label{Omega2H2}
        \norm{\Omega_2}_{\dot H^{2+\epsilon}} \leq 2A_\rho (C_{17,\epsilon}\norm{f}_{\dot H^{1+\epsilon}} + C_{18,\epsilon} \norm{f}_{\dot H^{2+\epsilon}}).
    \end{align}
where $C_{18,\epsilon} = C_{0,\epsilon}C_{1,\epsilon} C_{16,\epsilon}$ which as $\epsilon \rightarrow 0$ approaches a value strictly less than $C_{1}C_{4}$.

Using \eqref{eq: I_2}, we have that
\begin{align*}
\langle \hat f, \hat I_2 \rangle_{\dot H^{\frac{3}{2}+\epsilon}} &\leq \pi \int_{\mathbb{R}} \abs{\xi}^{3+2\epsilon}\abs{\hat{f}(\xi)} \sum_{n=0}^{\infty}(\ast^{2n+2}\abs{\widehat{\partial_\alpha f}} \ast (\abs{\cdot}\abs{\Omega_{1}}))(\xi) d\xi\\
&\leq \pi \norm{f}_{\dot{H}^{2+\epsilon}} \sum_{n=0}^{\infty}\bnorm{\ast^{2n+2}\abs{\widehat{\partial_\alpha f}} \ast \abs{\cdot}\abs{\Omega_{1}}}_{\dot{H}^{1+\epsilon}}
\end{align*}
Applying the weighted triangle inequality and then using \eqref{Omega1H2} and \eqref{Omega2H2} we obtain
\begin{align*}
  \langle \hat f, \hat I_2 \rangle_{\dot H^{\frac{3}{2} + \epsilon}} &\leq \pi \norm{f}_{\dot H^{2+\epsilon}}\sum_{n=0}^\infty  (2n+3)^{\epsilon}\Big((2n+2) \norm{f}_{\mathcal{F}^{1,1}_{0}}^{2n+1} \norm{\Omega_1}_{\mathcal{F}^{1,1}_{0}} \norm{f}_{\dot H^{2+\epsilon}} + \norm{f}_{\mathcal{F}^{1,1}_{0}}^{2n+2} \norm{\Omega_1}_{\dot H^{2+\epsilon}}\Big)

\end{align*}
Similarly, making use of the fact that the $\norm{N(\Omega_i)}_{H^{2 + \epsilon}}$ can be bounded in the same way as \eqref{Omega1H2} or \eqref{Omega2H2}
\begin{align*}
     \langle \hat f, \hat I_4 \rangle_{\dot H^{\frac{3}{2} + \epsilon}} &\leq \pi \norm{f}_{\dot H^{2+\epsilon}} \Big( C_{0,\epsilon} (\norm{f}_{\mathcal{F}^{1,1}_{\nu}} \norm{\Omega_2}_{\dot H^{2+\epsilon}} + \norm{\omega_2}_{\mathcal{F}^{0,1}_{0}} \norm{f}_{\dot H^{2+\epsilon}})\\
     &\hspace{50mm}+ \frac{C_{2,\epsilon}}{\norm{f}_{\mathcal{F}^{0,1}_{0}}} \norm{f}_{\mathcal{F}^{1,1}_{0}} \norm{\omega_2}_{\mathcal{F}^{0,1}_{0}} \norm{ f}_{\dot H^{1+\epsilon}}\Big)\\

     \langle \hat f, \hat N_0 \rangle_{\dot H^{\frac{3}{2} + \epsilon}} &\leq \frac{1}{2} \norm{f}_{\dot H^{2+\epsilon}} (2n+3)^\epsilon \Big((2n+2) \norm{f}_{\mathcal{F}^{1,1}_{0}}^{2n+1} \norm{f}_{\mathcal{F}^{1,1}_{0}} \norm{\Omega_1}_{\dot H^{1+\epsilon}} + \norm{f}_{\mathcal{F}^{1,1}_{0}}^{2n+2} \norm{\Omega_1}_{\dot H^{2+\epsilon}} \Big)\\

     \langle \hat f, \hat N_1 \rangle_{\dot H^{\frac{3}{2} + \epsilon}} &\leq \frac{1}{2} \norm{f}_{\dot H^{2+\epsilon}} \norm{N( \Omega_2)}_{\dot H^{2+\epsilon}}\\

     \langle \hat f, \hat N_2 \rangle_{\dot H^{\frac{3}{2} + \epsilon}} &\leq \frac{1}{2} \bnorm{\frac{A_\kappa}{e^{2h_2 \abs \xi} - A_\kappa A_\mu} }_{L^\infty} \norm{f}_{\dot H^{2+\epsilon}} \norm{N( \Omega_1)}_{\dot H^{2+\epsilon}}\\

     \langle \hat f, \hat N_3 \rangle_{\dot H^{\frac{3}{2} + \epsilon}} &\leq \frac{1}{2} C_5 \norm{f}_{\dot H^{2+\epsilon}} \norm{N( \Omega_2)}_{\dot H^{2+\epsilon}}\\

     \langle \hat f, \hat N_4 \rangle_{\dot H^{\frac{3}{2} + \epsilon}} &\leq \pi \norm{f}_{\dot H^{2+\epsilon}} \Big((C_{0,\epsilon}-1) \norm{\Omega_2}_{\dot H^{2+\epsilon}} + \frac{C_{2,\epsilon}}{\norm{f}_{\mathcal{F}^{0,1}_{0}}} \norm{f}_{\mathcal{F}^{1,1}_{0}} \norm{\Omega_2}_{\dot H^{1+\epsilon}} \Big)\\

    \langle \hat f, N(\widehat{\partial_\alpha \Omega_1}) \rangle_{\dot H^{\frac{3}{2} + \epsilon}} &\leq \norm{f}_{\dot H^{2+\epsilon}} \norm{N( \Omega_1)}_{\dot H^{2+\epsilon}}

\end{align*}

    Combining terms via Young's inequality for products we have
    \begin{align}
    \label{sobolevestimate}
    \frac{1}{2} \frac{d}{dt}\norm{f}_{\dot H^{\frac{3}{2} +\epsilon} }^2(t) &\leq  -A_\rho( \theta - \tilde \sigma_{\epsilon}) \norm{f}_{\dot H^{2+\epsilon}}^2  + S(\norm{f}_{\mathcal{F}^{0,1}_{0}}, \norm{f}_{\mathcal{F}^{1,1}_{0}}) \norm{f}_{\dot H^{1+\epsilon}}^2
    \end{align}
    In which, as $\epsilon \to 0$
\begin{align}
    \tilde \sigma_{\epsilon} < \sigma_1
\end{align}
and $S$ is a rational function of $\norm{f}_{\mathcal{F}^{0,1}_{0}}$ and $\norm{f}_{\mathcal{F}^{1,1}_{0}}$ that vanishes at 0. So for every finite $T > 0$ we can bound $\frac{d}{dt}\norm{f}_{\dot H^{\frac{3}{2}+\epsilon}}(t)$ on $[0,T]$ giving the desired derivative in time bound.

\section{Proof of Theorem}
\label{sec:existence}
We argue similarly to \cite{GGPS} for the proof of existence and uniqueness of solutions to \eqref{eq: interface}. Uniqueness of solutions is proven at the level of $\mathcal{F}^{0,1}_{0}$ and follows exactly as in \cite{GGPS}. It yields an inequality of the type
\begin{align}\label{uniqueness}
\frac{d}{dt}\norm{f-g}_{\mathcal{F}^{0,1}_{0}} &\leq -c_{1}\norm{f-g}_{\mathcal{F}^{1,1}_{0}} + c_{2}\norm{f-g}_{\mathcal{F}^{0,1}_{0}}
\end{align}
 for two solutions $f,g$ of \eqref{eq: interface} and for some constants $c_{i}= c_{i}(\norm{f_{0}}_{\mathcal{F}^{0,1}_{0}},\norm{f_{0}}_{\mathcal{F}^{1,1}_{0}},\norm{g_{0}}_{\mathcal{F}^{0,1}_{0}},\norm{g_{0}}_{\mathcal{F}^{1,1}_{0}})>0$. Note that the key difference with \cite{GGPS} is that for some terms, the difference of solutions occurs at the level of $\mathcal{F}^{0,1}_{0}$ which, unlike the Muskat problem without a permeability jump, can not be absorbed into the decay term.

 Consider the mollified system with initial data $f_{0}^{\varepsilon} = \varphi^{\varepsilon} \ast f_{0}$ and the evolution equation
 \begin{equation}\label{existenceequation}
      \partial_t f^{\epsilon} =  \mathcal{L}(\varphi^{\varepsilon}\ast\varphi^{\varepsilon}\ast f^{\epsilon}) + \varphi^{\varepsilon}\ast\mathcal{N}(\varphi^{\varepsilon}\ast\varphi^{\varepsilon}\ast f^{\epsilon}, \omega_{1}^{\varepsilon},\omega_{2}^{\varepsilon})
 \end{equation}
 where
 \begin{align*}
    \widehat{\mathcal{L}g}(\xi) = - A_\rho |\xi| \hat g(\xi) \left(1 - \frac{A_\kappa(1-A_\mu) }{e^{2h_2\abs{\xi}} - A_\kappa A_\mu}\right)
 \end{align*}
 and $\mathcal{N}$ is the remaining nonlinear terms from \eqref{interface_neat}. Here $\omega_{i}^{\varepsilon} = \partial_{\alpha}\Omega_i^{\varepsilon}$ are given by the mollified \eqref{omega1equation}, \eqref{omega2equation}, \eqref{Omega1}, and \eqref{Omega2} where $f$ is replaced by the mollified $\varphi^{\varepsilon}\ast\varphi^{\varepsilon}\ast f^{\varepsilon}$. Given initial data of medium size from Theorem \ref{MainTheorem} and because $H^{\frac{3}{2} + \epsilon} \hookrightarrow \mathcal{F}^{0,1}_{0}\cap\mathcal{F}^{1,1}_{0}$, the mollified system satisfies the hypothesis to apply Picard's theorem. We get a local solution $f^{\varepsilon} \in C([0,T_{\varepsilon}); H^{\frac{3}{2} + \epsilon})$. Next, we can reproduce the analogous estimates for the medium size condition on $f_{0}$ for $0\leq s \leq 1$:
\begin{equation}\label{existenceinequality}
\norm{f^{\varepsilon}}_{\mathcal{F}^{s,1}_{\nu}}(t) +(A_\rho\theta  - A_{\rho} \sigma_{s} -\nu)\int_{0}^{t} \norm{\varphi^{\varepsilon}\ast\varphi^{\varepsilon}\ast f^{\varepsilon}}_{\mathcal{F}^{s+1,1}_{\nu}}(s)\; ds \leq \|f_{0}\|_{\mathcal{F}^{s,1}_{0}}
\end{equation}
and
\begin{equation}\label{eq:epsilonL2}
  \norm{f^{\varepsilon}}_{L^2_\nu}^2 (t) \leq \norm{f_0}_{L^2}^2 \cdot \exp(R(\norm{f_0}_{\mathcal{F}^{0,1}_{0}}, \norm{f_0}_{\mathcal{F}^{1,1}_{0}}))
\end{equation}

Due to the exponential weight in $L^{2}_{\nu}$, the estimate \eqref{eq:epsilonL2} implies that $\|f^{\varepsilon}\|_{H^{\frac{3}{2}+\epsilon}}(t) \leq  C_{\epsilon}(t)\norm{f_0}_{L^2}^2 \cdot \exp(R(\norm{f_0}_{\mathcal{F}^{0,1}_{0}}, \norm{f_0}_{\mathcal{F}^{1,1}_{0}}))$ for $t>0$ where $C_{\epsilon}(t)$ is a bounded decreasing constant in $t>0$. Moreover, it can be seen combining the proof of \eqref{L2inequality} and \eqref{sobolevestimate} that
$$\frac{d}{dt}\|f^{\varepsilon}\|_{H^{\frac{3}{2}+\epsilon}}^{2} \leq G(\|f^{\varepsilon}\|_{H^{\frac{3}{2}+\epsilon}}, \|f_{0}\|_{\mathcal{F}^{0,1}_{0}}, \|f_{0}\|_{\mathcal{F}^{1,1}_{0}})$$
for a continuous function $G$. Hence, the local solution can be extended to $C([0,T]; H^{\frac{3}{2}+\epsilon})$ for any $T>0$.

By \eqref{uniqueness} and following the argument in \cite{GGPS}, the sequence $f^{\varepsilon_{n}}$ is shown to be Cauchy in $L^{\infty}([0,T];\mathcal{F}^{0,1}_{0})$ for any $\varepsilon_{n} \rightarrow 0$. The main idea is that by the argument from uniqueness, we have
\begin{align*}
\frac{d}{dt}\norm{f^{\varepsilon} - f^{\varepsilon '}}_{\mathcal{F}^{0,1}_{0}} &\leq -c_{1}\norm{ \varphi^{\varepsilon} \ast \varphi^{\varepsilon} \ast f^{\varepsilon} -  \varphi^{\varepsilon'} \ast \varphi^{\varepsilon'} \ast f^{\varepsilon '}}_{\mathcal{F}^{1,1}_{0}} + c_{2}\norm{ \varphi^{\varepsilon} \ast \varphi^{\varepsilon} \ast f^{\varepsilon} -  \varphi^{\varepsilon'} \ast \varphi^{\varepsilon'} \ast f^{\varepsilon '}}_{\mathcal{F}^{0,1}_{0}}
\end{align*}
and hence
\begin{align*}
\norm{f^{\varepsilon} - f^{\varepsilon '}}_{\mathcal{F}^{0,1}_{0}}(t) \leq \norm{\varphi^{\varepsilon} \ast f_{0} - \varphi^{\varepsilon'} \ast f_{0}}_{\mathcal{F}^{0,1}_{0}} + c_{2} \int_{0}^{t}\norm{ \varphi^{\varepsilon} \ast \varphi^{\varepsilon} \ast f^{\varepsilon} -  \varphi^{\varepsilon'} \ast \varphi^{\varepsilon'} \ast f^{\varepsilon '}}_{\mathcal{F}^{0,1}_{0}}(s) ds.
\end{align*}
Using the Mean Value Theorem in the mollifiers in the Fourier variables and assuming $\varepsilon \geq \varepsilon'$
\begin{align*}
    \norm{\varphi^{\varepsilon} \ast f_{0} - \varphi^{\varepsilon'} \ast f_{0}}_{\mathcal{F}^{0,1}_{0}} \leq C\norm{f_{0}}_{\mathcal{F}^{1,1}_{0}}\varepsilon^{\frac{1}{2}}
\end{align*}
and
\begin{align*}
   \norm{ \varphi^{\varepsilon} \ast \varphi^{\varepsilon} \ast f^{\varepsilon} -  \varphi^{\varepsilon'} \ast \varphi^{\varepsilon'} \ast f^{\varepsilon '}}_{\mathcal{F}^{0,1}_{0}} &\leq  \norm{ \varphi^{\varepsilon} \ast \varphi^{\varepsilon} \ast f^{\varepsilon} -  \varphi^{\varepsilon} \ast \varphi^{\varepsilon} \ast f^{\varepsilon '}}_{\mathcal{F}^{0,1}_{0}} +  \norm{ \varphi^{\varepsilon} \ast \varphi^{\varepsilon} \ast f^{\varepsilon'} -  \varphi^{\varepsilon'} \ast \varphi^{\varepsilon'} \ast f^{\varepsilon '}}_{\mathcal{F}^{0,1}_{0}}\\
   &\leq \norm{ f^{\varepsilon} -  f^{\varepsilon '}}_{\mathcal{F}^{0,1}_{0}} +  C\norm{ f^{\varepsilon '}}_{\mathcal{F}^{1,1}_{0}}\varepsilon^{\frac{1}{2}}\\
     &\leq \norm{ f^{\varepsilon} -  f^{\varepsilon '}}_{\mathcal{F}^{0,1}_{0}} + C\norm{f_{0}}_{\mathcal{F}^{1,1}_{0}}\varepsilon^{\frac{1}{2}}.
\end{align*}
Thus,
\begin{align}\label{cauchy}
\norm{f^{\varepsilon} - f^{\varepsilon '}}_{\mathcal{F}^{0,1}_{0}}(t) \leq C(1+c_{2}t)\norm{f_{0}}_{\mathcal{F}^{1,1}_{0}}\varepsilon^{\frac{1}{2}} + c_{2} \int_{0}^{t}\norm{ f^{\varepsilon} -  f^{\varepsilon '}}_{\mathcal{F}^{0,1}_{0}}(s)  ds.
\end{align}
Gronwall's inequality finally yields
\begin{align*}
\norm{f^{\varepsilon} - f^{\varepsilon '}}_{\mathcal{F}^{0,1}_{0}}(t) \leq C(1+c_{2}t)e^{c_{2}t}\norm{f_{0}}_{\mathcal{F}^{1,1}_{0}}\varepsilon^{\frac{1}{2}}.
\end{align*}
Hence, there exists a limit $\ f^{\varepsilon}\rightarrow f$ in $L^{\infty}([0,T];\mathcal{F}^{0,1}_{0})$. Hence, we can obtain pointwise almost everywhere convergence of a subsequence $\hat f^{\varepsilon_{n}}(\xi,t)$ and $\hat{\varphi}^{\varepsilon_{n}}(\xi,t)^{2}\hat f^{\varepsilon_{n}}(\xi,t)$ to $\hat{f}(\xi,t)$. Thus, Fatou's lemma applied to \eqref{existenceinequality} allows us to conclude that the limit $f$ indeed satisfies the inequality \eqref{maininequality} for $s=0$ and $s=1$.

Interpolation \eqref{interpolation} with \eqref{cauchy} and \eqref{existenceinequality} yields strong convergence of $\varphi^{\varepsilon} \ast \varphi^{\varepsilon} \ast f^{\varepsilon}$ to $f$ in $L^{2}([0,T];\mathcal{F}^{1,1}_{0})$.
Finally, we can now take limits in \eqref{existenceequation}, we get the limiting function $f$ as the unique solution to \eqref{eq: interface}.

\begin{comment}
for $$R(t) =t R_{1} + R_{2} \int_{0}^{t} \|f\|_{\mathcal{F}^{2,1}_{\nu}}(s)\; ds \leq tR_{1} + \frac{1}{A_{\rho}(\theta-\sigma_{1})-\nu}R_{2}\|f_{0}\|_{\mathcal{F}^{1,1}_{0}}.$$
\end{comment}

\begin{comment}
\subsection{Old Estimates in $H^3$}
We would like to prove a decay-in-time estimate for the interface equation in $H^3$
\[
    \norm{f}_{H^3} = \norm{f}_{L^2} + \norm{\partial_x^3 f}_{L^2}.
\]
Adapting a proof from Berselli et al, the interface satisfies an $L^2$ maximum principle. So we need only estimate the latter term.
\[
    \partial_t \frac{1}{2}\norm{\partial_x^3 f}_{L^2}^2 = \partial_t \frac{1}{2}\norm{\xi^3 \hat f}_{L^2}^2 = \int_\R \abs{\xi}^6\, \text{Re}(\hat f \cdot\widehat{\partial_t f})\, d\xi.
\]
Recall we previously computed $\widehat{\partial_t f} = \hat I_1 + \hat I_2$ where $\hat I_2$ is the trickiest term. So for now, it is enough to show that
\[
    \int_\R \abs{\xi}^6 \abs{\hat f\cdot\hat I_2}\, d\xi
\]
doesn't grow too quickly in time.

\subsection*{Example Estimate}
It seems easiest to begin in real space and then move into frequency space. Roughly following Berselli's notation (note Berselli's $I_2$ is not the same as our $I_2$),
\[
    \int_\R \partial_x^3f \cdot \partial_x^3I_2 = \sum_{\ell = 3}^7 J_\ell + \text{l.o.t.}
\]
Before we estimate the $J_\ell$, let's collect the following useful fact
\begin{align*}
     \norm{\xi^2 \hat \omega_2}_{L^2} &\sim \dnorm{\xi^2 \pi e^{-h_2 \abs \xi}  \int_\R e^{-f(y) \abs \xi} \cdot \partial_\alpha f(y) e^{-i\xi y}\; dy}_{L^2}\\
     &\leq \sum_{n=0}^\infty \dnorm{\pi e^{-h_2\abs\xi} \frac{\abs{\xi}^n}{n!}}_{L^\infty}  \norm{\xi^2 \widehat{f^n} * \widehat{\partial_\alpha f}}_{L^2}\\
     &\leq \norm{\widehat{\partial_\alpha^3 f}}_{L^2} \sum_{n=0}^\infty \dnorm{\pi e^{-h_2\abs\xi} \frac{\abs{\xi}^n}{n!}}_{L^\infty} \norm{f}_0^n \\
     &\lesssim \norm{f}_{H^3}.
\end{align*}

Now let's estimate
\[
    J_3 = \frac{1}{2\pi} \int_\R \partial_x^3 f(x)\ \pv \int_\R \frac{\partial_x^3 \omega_2(x-\beta) \beta}{\beta^2 + (f(x) + h_2)^2}\, d\beta dx
\]
as an example. The other $J_\ell$ are fairly similar. Using the symmetry $\partial_x = -\partial_\beta$ we can integrate by parts
\begin{align*}
    J_3 = \frac{1}{2\pi} \int_\R \partial_x^3 f(x) \int_\R \partial_x^2 \omega_2(x-\beta) \partial_\beta\left(\frac{ \beta}{\beta^2 + (f(x) + h_2)^2}\right)\, d\beta dx.
\end{align*}
Then, by Plancherel's
\begin{align*}
    J_3 &\sim  \int_\R \xi^3 \hat f (\xi) \cdot \int_\R e^{-i\xi \beta} \left(\xi^2 \hat \omega_2 * \calF_x\Big( \partial_\beta \Big[ \frac{\beta}{\beta^2 + (f(x) + h_2)^2} \Big] \Big)\right)(\xi) \; d\beta dx\\
    &= \int_\R \xi^3 \hat f (\xi) \cdot \left(\xi^2 \hat \omega_2 * \calF_x\Big( \xi \calF_\beta \Big[ \frac{\beta}{\beta^2 + (f(x) + h_2)^2} \Big] \Big)\right)(\xi) dx\\
    &\leq \norm{\xi^3 \hat f}_{L^2} \norm{\xi^2 \hat \omega_2}_{L^2} \dnorm{\calF_x\Big( \xi \calF_\beta \Big[ \frac{\beta}{\beta^2 + (f(x) + h_2)^2} \Big] \Big)}_{L^1}\\
    &\lesssim \norm{f}_{H^3}^2.
\end{align*}

\section{Estimates for Uniqueness}
\begin{proposition}
    Suppose $f$ and $g$ are solutions to the interface equation TBD with initial conditions $f_0, g_0 \in L^2 \cap \mathcal{F}^{1,1}_{0}$ satisfying TBD. Then
    \[
    \partial_t \norm{f - g}_{\mathcal{F}^{0,1}_{\nu}} \leq -C \norm{f - g}_{\mathcal{F}^{1,1}_{\nu}}.
    \]
    Furthermore, if $f_0 = g_0$ the above inequality implies $\norm{f}_{L^\infty} = \norm{g}_{L^\infty}$ for all time.
\end{proposition}

\begin{proof}
Taking a derivative yields
\[
    \partial_t \norm{f - g}_{\calF^{0,1}_\nu}  \leq (\nu - \rho) \norm{f - g}_{\calF^{1,1}_\nu} + \norm{N(I_1^f) - N(I_1^g)}_{\calF^{0,1}_\nu} + \norm{I_2^f - I_2^g}_{\calF^{0,1}_\nu}
\]
In previous works CITATIONS it is shown that
\[
    \norm{N(I_1^f) - N(I_1^g)}_{\mathcal{F}^{0,1}_{\nu}} \leq C

    \norm{f - g}_{\mathcal{F}^{0,1}_{\nu}}.
\]
So it is enough to prove a similar inequality for  the $I_2$ term. Recall that $I_2 = \frac{1}{2\pi}(A_1 + A_2)$, so
\[
    \norm{I_2^f - I_2^g}_{\calF^{0,1}_\nu} \leq \frac{1}{2\pi}\norm{A_1^f - A_1^g}_{\calF^{0,1}_\nu} + \frac{1}{2\pi}\norm{A_2^f - A_2^g}_{\calF^{0,1}_\nu}.
\]
Let's analyze the $A_1$ term first,
\begin{align*}
    \norm{A_1^f - A_1^g}_{\calF^{0,1}_\nu} & = \int_\R e^{\nu t \abs \xi} \dabs{ \hat A_1^f(\xi) - \hat A_1^g(\xi) } \; d\xi\\
    &= \int_\R e^{\nu t \abs{\xi} } \dabs{ \sum_{n= 0}^\infty \frac{(-1)^n}{n!} \left[\Big(\hat \omega_2^f \pi e^{-h_2 \abs{\xi}} * (\ast^n \abs{\xi} \hat f) \Big)(\xi) - \Big(\hat \omega_2^g \pi e^{-h_2 \abs{\xi} } * (\ast^n \abs{\xi }\hat g) \Big)(\xi) \right]}\, d\xi \\

    &\leq \sum_{n = 0}^\infty \frac{\pi}{n!} \int_\R e^{\nu t \abs{\xi} } \cdot  \dabs{ \Big(\hat \omega_2^f  e^{-h_2 \abs{\xi}} * (\ast^n \abs{\xi}\hat f) \Big)(\xi) - \Big(\hat \omega_2^g  e^{-h_2 \abs{\xi} } * (\ast^n \abs{\xi}\hat g) \Big)(\xi)}\, d\xi.
\end{align*}
We can write the convolution terms as a telescoping sum
\begin{align*}
   \hat \omega_2^f e^{-h_2 \abs{\xi}} * (\ast^n \abs{\xi} \hat f ) - \hat \omega_2^g  e^{-h_2 \abs{\xi}} * (\ast^n \abs{\xi} \hat g) &= (\hat \omega_2^f - \hat \omega_2^g)  e^{-h_2 \abs{\xi}} * (\ast^n \abs{\xi} \hat f)  \\
   & \quad + \hat \omega_2^g  e^{-h_2 \abs{\xi}} * (\ast^{n-1} \abs{\xi} \hat f) * (\abs{\xi} \widehat{(f - g)})\\
   & \quad +  \ldots\\
   & \quad + \hat \omega_2^g  e^{-h_2 \abs{\xi}} * (\ast^{n-1}\abs{\xi} \hat g) * (\abs \xi \widehat{(f-g)}).
\end{align*}
Then, by applying \eqref{convolutionexp}
\begin{align*}
    \norm{A_1^f - A_1^g}_{\calF^{0,1}_\nu} &\leq \sum_{n = 0}^\infty \frac{\pi}{n!}\norm{\omega_2^f - \omega_2^g}_{\mathcal{F}^{0,1}_{\nu}} \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^n\\
    &\hspace{4em} + \norm{\omega_2^g}_{\mathcal{F}^{0,1}_{\nu}} \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{n-1} \norm{f-g}_{\mathcal{F}^{1,1}_{\nu}}\\
    &\hspace{4em} + \ldots \\
    &\hspace{4em} + \norm{\omega_2^g}_{\mathcal{F}^{0,1}_{\nu}} \norm{g}_{\mathcal{F}^{1,1}_{\nu}}^{n-1} \norm{f-g}_{\mathcal{F}^{1,1}_{\nu}}.
\end{align*}
Now, examining $\omega_2^f - \omega_2^g$
\begin{align*}
    (\tfrac{2}{\pi} \calK \rho)\inv \ \norm{\omega_2^f - \omega_2^g}_{\mathcal{F}^{0,1}_{\nu}} &= \int_\R e^{\nu t\abs{\xi} } \dabs{ \sum_{n = 0}^\infty \frac{(-1)^n}{n!} \pi e^{-h_2 \abs{\xi}} \left[ (\ast^{n+1} \abs{\xi} \hat f)(\xi) - (\ast^{n+1} \abs{\xi} \hat g)(\xi) \right]}\, d\xi \\

    &\leq  \sum_{n=0}^\infty \frac{\pi}{n!} \int_\R \dabs{(\ast^{n+1} \abs \xi \hat f) - (\ast^{n+1} \abs \xi \hat g)}\; d\xi \\
    &\leq \sum_{n=0}^\infty \frac{\pi}{n!} \norm{f-g}_{\mathcal{F}^{1,1}_{\nu}} \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^n  + \norm{f-g}_{\mathcal{F}^{1,1}_{\nu}}\norm{g}_{\mathcal{F}^{1,1}_{\nu}} \norm{f}_{\mathcal{F}^{1,1}_{\nu}}^{n-1} + \ldots + \norm{f-g}_{\mathcal{F}^{1,1}_{\nu}} \norm{g}_{\mathcal{F}^{1,1}_{\nu}}^n\\
    &\lesssim \norm{f-g}_{\mathcal{F}^{1,1}_{\nu}}
\end{align*}
which means
\[
    \norm{A_1^f - A_1^g}_{\calF^{0,1}_\nu} \lesssim \norm{f-g}_{\mathcal{F}^{1,1}_{\nu}}
\]
just like we wanted! A similar calculation completes the proof by showing $\norm{A_2^f - A_2^g}_{\calF^{0,1}_\nu} \lesssim \norm{f-g}_1$
\end{proof}
\end{comment}
\subsubsection*{Acknowledgments}
The authors would like to thank the University of Michigan Research Experience for Undergraduates (REU) program for their support of Nikhil Shankar during the early stages of this project. NP was partially supported by AMS-Simons Travel Grants, which are administered by the American Mathematical Society with support from the Simons Foundation.

\printbibliography

	\vspace{0.1in}
	\begin{tabular}{ll}
		\textbf{Neel Patel} \\
		{\small Instituto de Ciencias Matem\'aticas, ICMAT} \\
		{\small Madrid, Spain} \\
		{\small Email: neeljp@umich.edu}\\
		\small{\textbf{Former Affiliation:} }\\
		{\small Department of Mathematics}\\
		{\small University of Michigan} \\
		{\small Ann Arbor, Michigan, 48105}
	\end{tabular}\qquad
	\begin{tabular}{ll}
		\textbf{Nikhil Shankar} \\
		{\small Department of Mathematics}\\
		{\small Duke University} \\
		{\small Physics Building 274E}\\
		{\small 120 Science Dr, Durham, NC 27710}\\
		{\small Email: nikhil.shankar@duke.edu}\\[2.45em]
	\end{tabular}

\end{document}

redpajama_set_name: RedPajamaArXiv