\Ss_1(k,t) = \int_{\R^{2d}} \nu^{-1} h_0(\nu^{-1} k_1\cdot k_3,y) \, \varphi(k_1,k_3;t)\, dk_1 dk_3
\]

\[
\begin{split}
y& = -\gamma_k+\gamma_{\frac{k_1}{\sqrt{2}} +k} + \gamma_{\frac{k_1-k_3}{\sqrt{2}} +k} + \gamma_{k-\frac{k_3}{\sqrt{2}}}\\
\varphi(k_1,k_3)& =\varphi(k_1,k_3;t)= f \left(\frac{k_1}{\sqrt{2}}+k, t\right) \, f\left(\frac{k_1-k_3}{\sqrt{2}}+k, t\right)\, f\left(k-\frac{k_3}{\sqrt{2}},t\right).
\end{split}
\]

Note that $\varphi$ is a Schwartz function of $k_1,k_3$ uniformly in $t$, and that $\Omega=k_1\cdot k_3$ in these new variables. Similarly, note that

\begin{equation}\label{eq:approx_id_Ss2}
4\nu^{-1}\,\mbox{Re}\left( \frac{e^{i\tau T_{\mathrm{kin}} \, \Omega}}{y-i\nu^{-1}\Omega}\right)  = \nu^{-1} h_1(\nu^{-1}\Omega,y) - \nu^{-1} h_2(\nu^{-1}\Omega,y),
\end{equation}

\begin{equation}\label{eq:h12}
h_1(x,y) = \frac{4y\,\cos(\varrho \tau x)}{y^2+x^2}\qquad \qquad h_2(x,y) = \frac{4x\,\sin(\varrho \tau x)}{y^2+x^2}.
\end{equation}

\[
\Ss_2(k,t,\tau) = \int_{\R^{2d}} [\nu^{-1} h_1(\nu^{-1} \Omega,y)-\nu^{-1} h_2(\nu^{-1} \Omega,y)]  \, \varphi(k_1,k_3;t,\tau)\, dk_1 dk_3,
\]

Note that $\varphi(k_1,k_3;t,\tau)\in \Sc (\R^d_{k_1}\times \R^d_{k_3})$ uniformly in $t,\tau$ and $\varrho$. Indeed, we claim that $\pa_{k_{ij}}^n e^{-\varrho (y+\gamma_k) \tau} = \O_n( 1)$ for any $n\in\N_0$, $i=1,3$, $j=1,\ldots ,d$. This follows from the following estimates:

\[
\pa_{k_{ij}}^n e^{-\varrho \gamma_{k_{\ell}} \, \tau} = \O_n( \sum_{\alpha=0}^n (\varrho\tau)^{\alpha} \, \gamma_{k_{\ell}} \, e^{-\varrho \gamma_{k_{\ell}} \, \tau}) = \O_n( 1) ,
\]

which in turn follow from the fact that $z^n \, e^{-z}=\O_n(1)$ for $n\geq 0$, from $\pa_{k_{ij}}^n \gamma_{k_\ell} \lesssim_n \gamma_{k_\ell}$ for any $n\in\N_0$, $i=1,3$, $j=1,\ldots ,d$, and from $1\leq \gamma_{k_\ell}$. As a result we may treat $\varphi(k_1,k_3;t,\tau)$ as $\varphi(k_1,k_3)$, which we do from now on.

\[
\varphi(k_1,k_3;t,\tau)=e^{-\varrho\, (y+2\gamma_k) \, \tau}\,f \left(\frac{k_1}{\sqrt{2}}+k, t\right) \, f\left(\frac{k_1-k_3}{\sqrt{2}}+k, t\right)\, f\left(k-\frac{k_3}{\sqrt{2}},t\right).
\]

Using the inverse Fourier transform[Since the focus from here on is to estimate error terms which vanish in the limit, we shall remove the factor $2\pi$ from the Fourier transform in order to shorten formulas.], we may rewrite these expressions as:

\begin{equation}\label{eq:new_S}
\begin{split}
\Ss_1(k,t) & =\int_{\R^{2d}} \int_{\R} \widehat{h_0}(\nu \xi,y)\, e^{i\xi\Omega}\, \varphi(k_1,k_3)\,  d\xi\, dk_1 dk_3\\
\Ss_2(k,t,\tau) & = \int_{\R^{2d}} \int_{\R} [\widehat{h_1}(\nu \xi,y)-\widehat{h_2}(\nu \xi,y)] \, e^{i\xi\Omega}\, \varphi(k_1,k_3;t,\tau)\,  d\xi\, dk_1 dk_3.
\end{split}
\end{equation}

We note that

\begin{equation}\label{eq:h}
\begin{split}
\widehat{h_0}(\xi,y) & =  4\pi \, e^{-|\xi y|}\\
\widehat{h_1}(\xi,y) & = 2\pi\, e^{-|y|\,|\xi -\varrho \tau|} +2\pi\, e^{-|y|\, |\xi +\varrho \tau|}\\
\widehat{h_2}(\xi,y) & = -2\pi\, e^{-|y| \, |\xi -\varrho \tau|}  \, \mbox{sign}(y \xi -y\varrho \tau ) + 2\pi\, e^{-|y|\, |\xi +\varrho \tau|}\, \mbox{sign}(y\xi+y\varrho \tau ) \qquad y\neq 0,\\
\widehat{h_2}(\xi,0) & = 4\pi\,\mathbbm{1}_{[-1,1]}\left(\frac{\xi}{\varrho \tau}\right).
\end{split}
\end{equation}

We introduce a cut-off $\chi_0\in C_c^{\infty}(\R)$ such that $\mbox{supp}(\chi_0)\subset [-2,2]$, $\chi_0\geq 0$, $\chi_0=1$ in $[-1,1]$, and we let $\chi_1= 1-\chi_0$. We also add and subtract the top order behavior (a delta function).

\begin{align}
\Ss_1(k,t) =&\ \int_{\R^{2d}} \delta (\Omega)\, \widehat{h_0}(0,y)\, \varphi(k_1,k_3)\, dk_1\, dk_3 \label{eq:S1_delta}\\
& +\int_{\R^{2d}} \int_{\R} [\widehat{h_0}(\nu \xi,y)-\widehat{h_0}(0,y)]\, e^{i\xi\Omega}\, \varphi(k_1,k_3)\,d\xi\, dk_1 dk_3. \label{eq:S1_error}
\end{align}

A similar decomposition can be done for $\Ss_2$. Note, however, that for any $y\geq 0$

\begin{equation}
\widehat{h_1}(0,y) = \widehat{h_2}(0,y)= 4\pi \, e^{-\varrho \tau\, y},
\end{equation}

and therefore the delta function vanishes for positive values of $y$ in the case of $\Ss_2$. Therefore:

\begin{align}
\Ss_2(k,t,\tau) =&\  \int_{\R^{2d}} \delta (\Omega)\, [\widehat{h_1}(0,y)-\widehat{h_2}(0,y)] \, \mathbbm{1}_{y<0} \,\varphi(k_1,k_3)\, dk_1\, dk_3 \label{eq:S2_delta}\\
& +  \int_{\R^{2d}} \int_{\R} \left[ \widehat{h_1}(\nu \xi,y)-\widehat{h_1}(0,y) \right] \, e^{i\xi\Omega}\, \varphi(k_1,k_3)\,d\xi\, dk_1 dk_3 \label{eq:S2_error1}\\
& -  \int_{\R^{2d}} \int_{\R} \left[ \widehat{h_2}(\nu \xi,y)-\widehat{h_2}(0,y) \right] \, e^{i\xi\Omega}\, \varphi(k_1,k_3)\,d\xi\, dk_1 dk_3 \label{eq:S2_error2}.
\end{align}

Moreover, we proved in <Ref> that $y\geq 1$ whenever $|\Omega|\leq 1$. As a result, $\{\Omega=0\}\cap \{y<0\} =\emptyset$ and thus the right-hand side of (<ref>) vanishes.

Over the next steps, we will estimate the terms  (<ref>), (<ref>) and (<ref>), which will finish the proof of <Ref>. In order to do so, we will work with a general kernel $h$ which satisfies some common properties to all $h_0,h_1,h_2$. We introduce the following:

A kernel $h(x,y)$ is an acceptable kernel if its Fourier transform in $x$ satisfies the following properties for some $\theta\in [0,1)$ as small as desired.

*  $\widehat{h}(\xi,y)$ is twice differentiable in both variables except on the lines $(\xi,0)$ and $\{(\xi_j, y)\}_{j=1,\ldots, n}$ for some finite number $n\in\N_0$. Let $\Ac(h) = \cup_{j=1}^{n} [\xi_j-\nu^{\theta} ,\xi_j+\nu^{\theta}]$.

*  $\sup_{\xi} | \widehat{h}(\xi, y)|\lesssim 1$ for all $(\xi, y)$.

*  $\Big | \widehat{h}(\nu \xi, y) - \widehat{h}(0, y) \Big | \lesssim
\begin{cases}
\nu\,|y| \, |\xi| & \mbox{if}\ |\xi|\leq \nu^{-\theta},\\
\nu^{1-\theta} \langle y\rangle \, |\xi|& \mbox{if}\  |\xi|>\nu^{-\theta}.
\end{cases}$

*  $\Big | \frac{\pa^j}{\pa \xi^j} \widehat{h}(\nu \xi, y ) \Big | \lesssim (\nu\, |y|)^j $ for $j=1,2$, $\xi\in \R-\nu^{-1} \Ac(h)$.

*  $\int_{\R-\nu^{-1} \Ac (h)} \Big | (\pa_{\xi}^j \widehat{h}) (\nu \xi, y) \Big | \, d\xi \lesssim (\nu |y|)^{j-1}$ for $j=1,2$.

*   $\Big | \frac{\pa}{\pa y} [\widehat{h}(\nu \xi, y) - \widehat{h}(0, y) ]\Big | \lesssim \nu\, |\xi|$, whenever $y\neq 0$ and $\xi\in \R- \nu^{-1} \Ac(h)$.

* For all $j=1,2$,

\[
|\pa_{y}^{j} \widehat{h}(0, y) | \lesssim |y|^{-j}
\]

*  There exists some $R_0\geq 0$ (which may depend on $\nu$, $T_{\mathrm{kin}}$ and $\tau$) such that $\Ac(h)\subset [-\nu R_0,\nu R_0]$ and for any $R\geq R_0$ any $j=0,1,2$ and any $y\neq 0$, we have that

\begin{equation}\label{k:R_large}
\int_{|\xi|>R}  |(\pa_{y}^{j} \widehat{h})(\nu \xi, y) | \, \frac{d\xi}{|\xi|^j} \lesssim  |y|^{-1} \,  \nu^{j-1} \, e^{- \nu |y| (R-R_0)}\ .
\end{equation}

Moreover, for $j=1,2$ and $y\neq 0$,

\begin{equation}\label{k:R_small}
\int_{\{\nu^{\theta} <|\xi|<R_0\}-\nu^{-1}\Ac(h)}  |\pa_{y}^{j}[\widehat{h}(\nu \xi, y)- \widehat{h}(0, y) ]|\, \frac{d\xi}{|\xi|^2} \lesssim \max\{ 1, |y|^{1-j}\} \, \nu \,\log (\nu^{-\theta} \langle R_0\rangle) \ .
\end{equation}

The kernels $h_0,h_1$, defined in (<ref>) and (<ref>), satisfy <ref>-<ref>. The kernel $h_2$ in (<ref>) also satisfies <ref>-<ref> provided $\varrho\tau \geq \nu^{\theta}$. When $\varrho\tau < \nu^{\theta}$, $h_2$ satisfies <ref>-<ref> except for <ref>.
In the case of $\widehat{h_0}$, the only singular lines from <ref> are $\xi=0$ and $y=0$, and we may choose $R_0=\nu^{-1+\theta}$ (note that (<ref>) is holds with $R_0=0$ but we must make $R_0$ large enough so that $\Ac(h_0)=[-\nu^{\theta},\nu^{\theta}]\subset [-\nu R_0,\nu R_0]$). In the case of $\widehat{h_1}$ and $\widehat{h_2}$, we remove the singular lines $y=0$, and $\xi=0,\pm \varrho \tau$, with $R_0 = \max\{ T_{\mathrm{kin}}\,\tau,\nu^{-1}\}$.

We shall first show that for any acceptable kernel $h$ with $R_0\leq \exp(\nu^{-1+})$, there holds that

\begin{equation}\label{sec4aim}
\left|\int_{\R^{2d}} \int_{\R} [\widehat{h}(\nu \xi,y)-\widehat{h}(0,y)]\, e^{i\xi\Omega}\, \varphi(k_1,k_3)\,d\xi\, dk_1 dk_3\right|\lesssim \nu^\theta.
\end{equation}

This will cover the kernels $h_0$ and $h_1$, as well as $h_2$ in the case when $\varrho \tau\geq \nu^\theta$. The remaining case of $h_2$ in the regime $\varrho \tau\leq \nu^\theta$ will be easily treated afterwards. Finally, we shall give the proof of <Ref>.

To prove (<ref>) for any admissible kernel $h$, it suffices to consider the case when $n=1$ in <ref> and $\xi_1=0$. The cases when $n >1$ are treated in exactly the same way.
As such, we decompose:

\begin{equation}\label{eq:isolated_points}
\widehat{h}(\nu\xi,y) : = \widehat{h}(\nu\xi,y)  \chi_0 (\nu^{-\theta} \xi)  + \widehat{h}(\nu\xi,y)  \chi_1 (\nu^{-\theta} \xi) \chi_0 (\nu^{-\theta} y) + \widehat{h}(\nu\xi,y)  \chi_1 (\nu^{-\theta} \xi)  \chi_1 (\nu^{-\theta} y)
\end{equation}

for a fixed $0<\theta\ll 1$ as small as desired. These decomposition may be adapted to deal with any finite number of singular points $\xi_j$ of $\widehat{h}$ by adding adequate cutoff functions in addition to $ \chi_0 (\nu^{-\theta}  \xi)$ such as $ \chi_0 (\nu^{-\theta}  (\xi-\nu^{-1}\xi_j))$.

We will show that the contribution of the first two summands is negligible, and thus we may focus on the kernel $\widehat{h}(\xi,y) \chi_1 (\nu^{-\theta}  \xi) \chi_1 (\nu^{-\theta}  y) $.

Step 0. Integration around singularities.

Consider the contribution of the first summand in (<ref>). Using <ref> and the fast decay of $\varphi$, we have that:

\[
\Big | \int_{\R^{2d}} \int_{\R} [\widehat{h}(\nu\xi,y)-\widehat{h}(0,y)]\, \chi_0 (\nu^{-\theta} \xi) \, e^{i\xi\Omega}\, \varphi(k_1,k_3)\, d\xi\, dk_1 dk_3 \Big | \lesssim \nu^{\theta}.
\]

Consider the second summand in (<ref>). We integrate by parts $e^{i\Omega\xi}$ twice in $\xi$, which yields:

\begin{align}
\int_{\R^{2d}} \int_{\R} & [\widehat{h}(\nu\xi,y)-\widehat{h}(0,y)]\, \chi_1 (\nu^{-\theta} \xi)\, \chi_0 (\nu^{-\theta} y)\, e^{i\xi\Omega}\, \varphi(k_1,k_3)\, d\xi\, dk_1 dk_3 \nonumber \\
= &\ \int_{\R^{2d}} \int_{\R} \frac{1}{(i\Omega)^2} \, e^{i\xi\Omega}   \frac{\pa^2}{\pa \xi^2}\left[  \left( \widehat{h}(\nu\xi,y)-\widehat{h}(0,y) \right)\, \chi_1 (\nu^{-\theta} \xi)  \right] \, \varphi(k_1,k_3)\, \chi_0 (\nu^{-\theta}\, y)\, d\xi\, dk_1 dk_3. \label{eq:y_small_1}
\end{align}

* Suppose first that no derivative hits $\chi_1 (\nu^{-\theta} \xi)$. Using <ref>, we may bound this contribution by

\begin{equation}\label{eq:y_small_11}
\int_{\R^{2d}}\frac{\nu |y|}{\Omega^2} \, |\varphi(k_1,k_3)|\, |\chi_0 (\nu^{-\theta}\, y)|\, dk_1 dk_3 \lesssim \nu^{\theta} \, \int_{\R^{2d}}\frac{\nu}{\Omega^2} \, |\varphi(k_1,k_3)|\, |\chi_0 (\nu^{-\theta}\, y)|\, dk_1 dk_3 .
\end{equation}

By <Ref>, i.e. the fact[As a matter of fact, <Ref> yields the stronger bound $\Omega\gtrsim 1$, but we do not need such a strong result here.] that $y^2 + (\nu^{-1}\Omega)^2\geq 1$, we must have that $\Omega\gtrsim \nu$ in this region, and thus we change variables in order to integrate in $\Omega$. Given that $\Omega=k_1\cdot k_3$, we divide $\R^{2d}$ into regions $\Rc_{ij}$ where each variable $|k_{ij}|$, $i=1,3$, $j=1,\ldots,d$, is maximal (up to a multiple) than the remaining $2d-1$ variables. We do so via a smooth partition of unity, i.e.

\[
1 =\sum_{i=1,3} \sum_{j=1,\ldots,d} \chi_{\Rc_{ij}}.
\]

Without loss of generality let us assume that $|k_{31}|$ is maximal. We single out $k_{11}$, define $k_1'=(k_{12},\ldots,k_{1d})$, and make the change of variables $k_{11} \mapsto \Omega$. The Jacobian is proportional to $k_{31}^{-1}$, which is locally integrable in $\R^{2d-1}$. As a result, we have that

\[
|\eqref{eq:y_small_11}|\lesssim \nu^{1+\theta} \, \int_{\R^{2d-1}} |k_{31}|^{-1}\, \int_{|\Omega|>\nu}  |\Omega|^{-2} \, |\varphi ( k_1 (\Omega),k_1',k_3)|\, d\Omega dk_1' dk_3  = \O ( \nu^{\theta}).
\]

* Back to (<ref>), if a single derivative hits $\chi_1$, we note that $\pa_{\xi}  \chi_1 (\nu^{-\theta} \xi) = - \nu^{-\theta} \chi_0' (\nu^{-\theta} \xi)$. Then <ref> yields:

\[
\Big | \int_{\R^{2d}} \int_{\R}  \frac{1}{|\Omega|^2} \, \nu^{1-\theta} |y|\, |\chi_0' (\nu^{-\theta} \xi)| \, |\varphi(k_1,k_3)|\, \chi_0 (\nu^{-\theta}\, y)\, d\xi\, dk_1 dk_3\Big |  \leq \int_{\R^{2d}} \frac{1}{|\Omega|^2} \, \nu^{1+\theta} \, |\varphi(k_1,k_3)|\, \chi_0 (\nu^{-\theta}\, y)\, dk_1 dk_3
\]

and the same integration in $\Omega$ as for (<ref>) yields a contribution of size $\O( \nu^{\theta})$.

* A similar strategy based on <ref> works if two derivatives hit $\chi_1$ in (<ref>). Indeed,

\begin{align*}
\Big | \int_{\R^{2d}} \int_{\R} & \frac{1}{(i\Omega)^2} \, e^{i\xi\Omega}    \left( \widehat{h}(\nu\xi,y)-\widehat{h}(0,y) \right)\, \nu^{-2\theta} \, \chi_0'' (\nu^{-\theta} \xi)  \, \varphi(k_1,k_3)\, \chi_0 (\nu^{-\theta}\, y)\, d\xi\, dk_1 dk_3\Big | \\
& \lesssim \int_{\R^{2d}} \frac{1}{\Omega^2} \, \int_{|\xi|\leq \nu^{\theta}} \nu^{1-2\theta}\, |y|\,  |\xi| \, |\varphi(k_1,k_3)|\, \chi_0 (\nu^{-\theta}\, y)\, d\xi\, dk_1 dk_3\\
& \lesssim \int_{\R^{2d}} \frac{1}{\Omega^2} \, \nu^{1+\theta}\, |\varphi(k_1,k_3)|\,\chi_0 (\nu^{-\theta}\, y)\, d\xi\, dk_1 dk_3\lesssim \nu^{\theta},
\end{align*}

the last step coming from integration over $|\Omega|>\nu$.

By (<ref>) and the above simplifications,

\[
\eqref{eq:S1_error}  = \int_{\R^{2d}} \int_{\R} [\widehat{h}(\nu \xi,y)-\widehat{h}(0,y)]\, e^{i\xi\Omega}\, \varphi(k_1,k_3)\,  \chi_1 (\nu^{-\theta} \xi) \, \chi_1 (\nu^{-\theta} y)\, d\xi\, dk_1 dk_3 + \O ( \nu^{\theta}).
\]

Step 1. Localization around $\Omega=0$. We write,

\begin{align}
\eqref{eq:S1_error}  = &\  \int_{\R^{2d}} \int_{\R} [\widehat{h}(\nu \xi,y)-\widehat{h}(0,y)]\, e^{i\xi\Omega}\, \varphi(k_1,k_3)\,  \chi_1 (\nu^{-\theta} \xi) \, \chi_1 (\nu^{-\theta} y)\, \chi_0 (\nu^{-\beta} \Omega) \, d\xi\, dk_1 dk_3 \nonumber \\
& + \int_{\R^{2d}} \int_{\R} [\widehat{h}(\nu \xi,y)-\widehat{h}(0,y)]\, e^{i\xi\Omega}\, \varphi(k_1,k_3)\,  \chi_1 (\nu^{-\theta} \xi) \, \chi_1 (\nu^{-\theta} y)\, \chi_1 (\nu^{-\beta} \Omega) \, d\xi\, dk_1 dk_3 + \O ( \nu^{\theta} ) \label{eq:large_omega}
\end{align}

for $\beta\in (0,1)$ to be fixed later. As in Step 0, we integrate $e^{i\xi\Omega}$ by parts twice in the $\xi$ variable

\[
\begin{split}
\eqref{eq:large_omega}
%=&\  \int_{\R^{2d}} \int_{\R} \frac{e^{i\xi\Omega} }{(i\Omega)^2} \frac{\pa^2}{\pa \xi^2} \left[ ( \widehat{h}(\nu\xi,y)-\widehat{h}(0,y)) \, \chi_1 (\nu^{-\theta} \xi) \right] \, \varphi(k_1,k_3)\, \chi_1 (\nu^{-\theta}\, y)\, \chi_1 (\nu^{-\beta} \Omega) \, d\xi\, dk_1 dk_3\\
=&\ \int_{\R^{2d}} \int_{\R} \frac{e^{i\xi\Omega}}{(i\Omega)^2}  \frac{\pa^2}{\pa \xi^2} \left[ ( \widehat{h}(\nu\xi,y)-\widehat{h}(0,y)) \, [1- \chi_0 (\nu^{-\theta} \xi) ]\right] \, \varphi(k_1,k_3)\, \chi_1 (\nu^{-\theta}\, y)\, \chi_1 (\nu^{-\beta} \Omega) \, d\xi\, dk_1 dk_3.
\end{split}
\]

If both derivatives hit  $\widehat{h}(\nu\xi,y)-\widehat{h}(0,y)$, then we use <ref> and proceed as in Step 0 to obtain a contribution of size $\O (\nu^{1-\beta})$. If both derivatives hit $\chi_0 (\nu^{-\theta} \xi)$, we use the rapid decay of $\varphi$ to reduce the integral to $|(k_1,k_3)|\leq \nu^{-\theta}$, which implies that $|y|\leq \nu^{-2\theta}$. Then <ref> yields

\begin{align*}
& \nu^{-2\theta}\, \Big | \int_{\R^{2d}} \int_{\R} \frac{1}{(i\Omega)^2} \, e^{i\xi\Omega} \, [ \widehat{h}(\nu\xi,y)-\widehat{h}(0,y)] \, \chi_0'' (\nu^{-\theta} \xi) \, \varphi(k_1,k_3)\, \chi_1 (\nu^{-\theta}\, y)\, \chi_1(\nu^{-\beta} \Omega) \, d\xi\, dk_1 dk_3 \Big | \\
&\ \lesssim  \int_{\R^{2d}} \frac{1}{\Omega^2}\, \nu^{1-3\theta}\, |\chi_1 (\nu^{-\theta}\, y)|\, |\chi_1 (\nu^{-\beta} \Omega)| \, dk_1\, dk_3 = \O(\nu^{1-\beta-3\theta} ),
\end{align*}

which is small by imposing $4\theta <1-\beta$.

The case where one derivative hits each term follows from an analogous argument and <ref>. All in all, we have shown that $\eqref{eq:large_omega} = \O ( \nu^{\theta})$.

Step 2. Localization in $\xi$.  Our goal is to show that

\begin{equation}\label{eq:S1_error_new}
\int_{\R^{2d}} \int_{\R} [\widehat{h}(\nu \xi,y)-\widehat{h}(0,y)]\, e^{i\xi\Omega}\, \varphi(k_1,k_3)\,  \chi_1 (\nu^{-\theta} \xi) \, \chi_1 (\nu^{-\theta} y)\, \chi_0 (\nu^{-\beta} \Omega) \, d\xi\, dk_1 dk_3
= \O (\nu^{\theta}).
\end{equation}

In order to prove (<ref>), we add an additional cut-off in the $\xi$-variable. More precisely, we write:

\begin{align}
\eqref{eq:S1_error_new}=&\ \int_{\R^{2d}} \int_{\R} (\widehat{h}(\nu \xi, y) - \widehat{h}(0, y) )\, e^{i\xi\Omega}\, \varphi(k_1,k_3)\, \chi_0 (\nu^{-\beta}\, \Omega)\,\chi_1 (\nu^{-\theta}\xi)\, \chi_0 (\xi/R)\, \chi_1 (\nu^{-\theta} y)\,d\xi\, dk_1 dk_3\nonumber \\
& + \int_{\R^{2d}} \int_{\R}  \widehat{h}(\nu \xi, y) \, e^{i\xi\Omega}\, \varphi(k_1,k_3)\, \chi_0 (\nu^{-\beta}\, \Omega)\, \chi_1 (\xi/R) \, \chi_1 (\nu^{-\theta} y)\, d\xi\, dk_1 dk_3\label{eq:xi_large1}\\
& - \int_{\R^{2d}} \int_{\R}  \widehat{h}(0, y) \, e^{i\xi\Omega}\, \varphi(k_1,k_3)\, \chi_0 (\nu^{-\beta}\, \Omega)\, \chi_1 (\xi/R) \, \chi_1 (\nu^{-\theta}y)\, d\xi\, dk_1 dk_3\label{eq:xi_large2}
\end{align}

for some $R>\nu^{\theta}$ to be fixed later.

Firstly, we show that the contribution of (<ref>) is negligible. Once again, let us assume that $|k_{31}|$  is maximal among the components of $(k_1,k_3)$ as an example. Let us integrate $e^{i\xi\Omega}$ by parts in the variable $k_{11}$:

\[
\begin{split}
\eqref{eq:xi_large2} =&\ \int_{\R^{2d}} \int_{\R}  \frac{1}{(ik_{31}\xi)^2} e^{i\xi\Omega}  \frac{\pa^2}{\pa k_{11}^2} \left[ \widehat{h}(0, y) \, \varphi(k_1,k_3) \chi_0 (\nu^{-\beta}\, \Omega) \chi_1 (\nu^{-\theta}y) \right]  \chi_1 (\frac{\xi}{R}) \, \chi_{\Rc_{31}} \, d\xi \, dk_1 \, dk_3 .
\end{split}
\]

Note that $\frac{\pa y}{\pa k_{11}}$ and $\frac{\pa^2 y}{\pa k_{11}^2}$ are non-singular and easily bounded by

\begin{equation}\label{eq:deriv_y}
\Big | \frac{\pa^j y}{\pa k_{11}^j} \Big | \lesssim \langle k\rangle^{j} + \sum_{i=1}^3 \langle k_i\rangle^{j}\qquad j=1,2.
\end{equation}

Therefore the worst terms come from $\chi_0 (\nu^{-\beta}\,\Omega)$ and $\widehat{h}(0, y)$
which we bound as examples.

* If both derivatives hit $\chi_0 (\nu^{-\beta}\,\Omega)$, <ref> yields:

\begin{align*}
& \nu^{-2\beta}\, \Big | \int_{\R^{2d}} \int_{\R}  \frac{1}{(ik_{31}\xi)^2} e^{i\xi\Omega}\, \widehat{h}(0, y) \, \varphi(k_1,k_3) \chi_0'' (\nu^{-\beta}\, \Omega) \chi_1 (\nu^{-\theta}y)\,  \chi_1 (\frac{\xi}{R}) \, \chi_{\Rc_{31}} \, d\xi \, dk_1 \, dk_3\Big |  \\
&\ \lesssim \nu^{-2\beta-\theta}\, \int_{\R} \xi^{-2}\, \chi_1 (\xi/R) \,d\xi \int_{\R^{2d}} |\varphi(k_1,k_3)|\, \chi_0'' (\nu^{-\beta} \Omega) \,dk_1 dk_3 \lesssim  R^{-1} \,\nu^{-\beta-\theta} \lesssim \nu^{\theta}
\end{align*}

after imposing

\begin{equation}\label{eq:choose_R}
R \geq  \nu^{-\beta-2\theta} \,\gg 1.
\end{equation}

* If both derivatives hit $\widehat{h}(0, y)$, we use the fast decay of $\varphi$ to reduce integration to $|(k_1,k_3)|\leq \nu^{-\theta}$. Then <ref> and (<ref>) yield

\begin{multline*}
\Big | \int_{\R^{2d}} \int_{\R}  \frac{1}{(ik_{31}\xi)^2} e^{i\xi\Omega}\, \pa_{k_{11}}^2 \widehat{h}(0, y) \, \varphi(k_1,k_3) \chi_0 (\nu^{-\beta}\, \Omega) \chi_1 (\nu^{-\theta}y)\,  \chi_1 (\frac{\xi}{R}) \, \chi_{\Rc_{31}} \, d\xi \, dk_1 \, dk_3\Big |  \\
\lesssim \nu^{-4\theta}\,  \int_{\R} \xi^{-2}\, \chi_1 (\xi/R) \,d\xi \int_{\R^{2d}} |k_{31}|^{-2}\, |\varphi(k_1,k_3)|\, \chi_0(\nu^{-\beta} \Omega) \,dk_1 dk_3  \lesssim R^{-1} \,\nu^{-4\theta+\beta} \lesssim \nu^{\theta}
\end{multline*}

after imposing

\begin{equation}\label{eq:choose_R_1}
R \geq  \nu^{\beta-5\theta} .
\end{equation}

Next we show that the contribution from (<ref>) is negligibile. Assuming that $k_{31}$  is maximal among the components of $(k_1,k_3)$, we integrate $e^{i\xi\Omega}$ by parts in the variable $k_{11}$. As in the case of (<ref>), the worst terms come from the derivative hitting $\widehat{h}$ and $\chi_0 (\nu^{-\beta}\, \Omega)$. We deal with these using <ref> together with $|y|>\nu^{\theta}$, i.e.

\[
\begin{split}
|\eqref{eq:xi_large1}| & = \Big | \int_{\R^{2d}} \int_{\R} \frac{e^{i\xi\Omega}}{i\xi k_{31}}\, \frac{\pa}{\pa k_{11}} \left[ \widehat{h}(\nu\xi, y)\, \varphi(k_1,k_3)\, \chi_0 (\nu^{-\beta}\, \Omega) \,\chi_1 (\nu^{-\theta}y)\, \right]\, \chi_1 (\xi/R) \, d\xi\, dk_1 dk_3\Big | \\
& \lesssim  \nu^{-\theta}\, \int_{\R^{2d}\cap \{ |\Omega|\leq \nu^{\beta}\}\cap \{|y|>\nu^{\theta}\}\cap\{ |k_j|\leq \nu^{-\theta}\}} \int_{|\xi|>R} \frac{1}{|\xi| |k_{31}|} \left(  |(\pa_{y} \widehat{h})(\nu \xi, y) |  + \nu^{-\beta}  |\widehat{h}(\nu \xi, y) |\right) d\xi\, dk_1 dk_3\\
& \lesssim  \nu^{-\theta}\,  \int_{\R^{2d}\cap \{ |\Omega|\leq \nu^{\beta}\}\cap \{|y|>\nu^{\theta}\}\cap\{ |k_j|\leq \nu^{-\theta}\}} |y|^{-1}\, e^{- \nu |y| (R-R_0)} \, \left(1+ \nu^{-\beta-1} \, R^{-1}   \right)\, |k_{31}|^{-1} dk_1 dk_3 \\
& \lesssim \nu^{-2\theta}\, e^{- \nu^{1+\theta} \,(R-R_0)} \left(  \nu^{\beta} + \nu^{-1} R^{-1} \right) \lesssim_N \nu^{\theta N},
\end{split}
\]

by fixing $R$ sufficiently large, i.e.

\begin{equation}\label{eq:choose_R_2}
R > R_0 + \nu^{-1-2\theta}.
\end{equation}

All in all, we have proved that

\[
\eqref{eq:S1_error_new}  =  \int_{\R^{2d}} \int_{\R} (\widehat{h}(\nu \xi, y) - \widehat{h}(0, y) )\, e^{i\xi\Omega}\, \varphi(k_1,k_3)\, \chi_0 (\nu^{-\beta}\, \Omega)\,\chi_1 (\nu^{-\theta}\xi)\, \chi_0 (\xi/R)\, \chi_1 (\nu^{-\theta} y)\,d\xi\, dk_1 dk_3  + \O (\nu^{\theta}).
\]

Our final goal is to show that this last term also has size $\O (\nu^{\theta})$.

Step 3. Main terms. Without loss of generality, let us assume that we are in the region $\Rc_{31}$ where $|k_{31}|$ is maximal, and let us integrate by parts the term $e^{i\xi\Omega}$  in the variable $k_{11}$. Thanks to the fast decay of $\varphi$ we may assume that $|(k_1,k_3)|\leq \nu^{-\theta}$ and thus $|y|\leq \nu^{-2\theta}$. We find that

\begin{multline}\label{eq:final_IBP}
\int_{\R^{2d}} \int_{\R}  e^{i\xi\Omega}\, (\widehat{h}(\nu \xi, y) - \widehat{h}(0, y))\, \varphi(k_1,k_3)\, \chi_0 (\nu^{-\beta}\Omega)\,\chi_0 (\frac{\xi}{R})\, \chi_1 (\nu^{-\theta}\xi)\, \chi_1 (\nu^{-\theta} y )\, d\xi \, dk_1\, dk_3\\
= \int_{\R^{2d}} \int_{\R} \frac{e^{i\xi\Omega}}{(i\xi \, k_{31})^2}  \frac{\pa^2}{\pa k_{11}^2} \left[ (\widehat{h}(\nu \xi, y) - \widehat{h}(0, y))\, \varphi(k_1,k_3)\, \chi_0 (\nu^{-\beta}\Omega)\,\chi_1 (\nu^{-\theta} y ) \right]\, \chi_0 (\frac{\xi}{R})\, \chi_1 (\nu^{-\theta}\xi)\,d\xi \, dk_1\, dk_3.
\end{multline}

We describe the estimates depending on which term is hit by $\pa_{k_{11}}$.

* If both derivatives hit any of the terms $\varphi(k_1,k_3)$, $\chi_0 (\nu^{-\beta}\Omega)$ or $\chi_1 (\nu^{-\theta} y )$ we proceed in the same manner. Let us show how to bound the worst case scenario, which is both derivatives hitting $\chi_0 (\nu^{-\beta}\Omega)$. Using <ref> and <ref> to estimate the kernel, we may bound the contribution of these terms by

\[
\nu^{-2\beta} \int_{\{ |\Omega|\leq \nu^{\beta}\} \cap \{ |k_j|\leq \nu^{-\theta}\}} \int_{\nu^{\theta}<|\xi|< R} \frac{\min \{ \nu^{1-3\theta} \, |\xi| ,1 \} }{|\xi|^2} \,  d\xi \, dk_1\, dk_3 \lesssim \nu^{1-\beta-6\theta} .
\]

The argument for the terms $\varphi(k_1,k_3)$ and $\chi_1 (\nu^{-\theta} y )$ is analogous, only one does not have a $\nu^{-2\beta}$ loss, and there is the additional term $|k_{31}|^{-2}$ which is integrable. As a result these terms are $\O (\nu^{1+\beta-6\theta} )$.

* Consider the term where both derivatives in (<ref>) hit $\widehat{h}(\nu \xi, y) - \widehat{h}(0, y)$. These derivatives give rise to

\[
(\pa_{k_{11}} y)^2\, \pa_y^2 \left[ \widehat{h}(\nu \xi, y) - \widehat{h}(0, y)\right] + (\pa_{k_{11}}^2 y)\, \pa_y \left[ \widehat{h}(\nu \xi, y) - \widehat{h}(0, y)\right] .
\]

The derivatives of $y$ may be controlled with $\nu^{-2\theta}$ in view of (<ref>). Then we use <ref> and <ref> with $R=R_0$ to finish the proof.

\begin{align*}
& \int_{\{ |\Omega|\leq \nu^{\beta}\} \cap \{ |k_j|\leq \nu^{-\theta}\}}\frac{ \chi_1 (\nu^{-\theta} y )}{|k_{31}|^2}  \int_{\nu^{\theta}<|\xi|< R}  \frac{\nu^{-2\theta}}{|\xi|^2}\, \sum_{j=1}^2 \Big | \pa_y^j  \left[ \widehat{h}(\nu \xi, y) - \widehat{h}(0, y)\right]\Big | \, d\xi \, dk_1\, dk_3\\
&\ \lesssim \int_{\{ |\Omega|\leq \nu^{\beta}\} \cap \{ |k_j|\leq \nu^{-\theta}\}}\frac{ \chi_1 (\nu^{-\theta} y )}{|k_{31}|^2}  \int_{\nu^{\theta}<|\xi|< R_0}  \frac{\nu^{-2\theta}}{|\xi|^2}\, \sum_{j=1}^2\Big | \pa^j_y \left[ \widehat{h}(\nu \xi, y) - \widehat{h}(0, y)\right]\Big | \, d\xi \, dk_1\, dk_3
\\
&\ \ + \int_{\{ |\Omega|\leq \nu^{\beta}\} \cap \{ |k_j|\leq \nu^{-\theta}\}}\frac{ \chi_1 (\nu^{-\theta} y )}{|k_{31}|^2}  \int_{|\xi|>\max\{R_0,\nu^{\theta}\}}   \frac{\nu^{-2\theta}}{|\xi|^2}\, \sum_{j=1}^2 \Big | \pa^j_y \widehat{h}(\nu \xi, y) \Big | \, d\xi \, dk_1\, dk_3\\
&\ \ + \int_{\{ |\Omega|\leq \nu^{\beta}\} \cap \{ |k_j|\leq \nu^{-\theta}\}}\frac{ \chi_1 (\nu^{-\theta} y )}{|k_{31}|^2}  \int_{\max\{R_0,\nu^{\theta}\}<|\xi|< R}   \frac{\nu^{-2\theta}}{|\xi|^2}\, \sum_{j=1}^2 \Big | \pa^j_y \widehat{h}(0, y)\Big | \, d\xi \, dk_1\, dk_3\\
&\ \lesssim  \nu^{1+\beta-3\theta} \,  \log( \nu^{-\theta} \langle R_0\rangle)+ \nu^{\beta-\theta}\lesssim \nu^{\theta}
\end{align*}

after imposing that $\beta>2\theta$ and

\begin{equation}\label{eq:choose_R_3}
R_0 \leq \exp (\nu^{-1-\beta+4\theta} ).
\end{equation}

* Let us show one example on how to handle cross-terms. Suppose that one derivative hits $\widehat{h}(\nu\xi,y)$ and the other one hits, say, $\chi_0 (\nu^{-\beta} \Omega)$. One may bound such a contribution using <ref>

\begin{align*}
& \nu^{-\beta-2\theta}\, \int_{\{ |\Omega|\leq \nu^{\beta}\} \cap \{ |k_j|\leq \nu^{-\theta}\}} \int_{\nu^{-\theta}<|\xi|< R} \frac{1}{|\xi|^2} \, \Big | \pa_y \left[ \widehat{h}(\nu \xi, y) - \widehat{h}(0, y)\right]\Big | \, d\xi \, dk_1\, dk_3\\
&\ \lesssim \nu^{-\beta-2\theta}\, \int_{\{ |\Omega|\leq \nu^{\beta}\} \cap \{ |k_j|\leq \nu^{-\theta}\}} \int_{\nu^{\theta} <|\xi|< R} \frac{1}{|\xi|^2} \, \nu |\xi| \, d\xi \, dk_1\, dk_3
\lesssim \nu^{1-2\theta} \, \log (R\, \nu^{-\theta}) \lesssim \nu^{\theta}
\end{align*}

after imposing

\begin{equation}\label{eq:choose_R_4}
R \leq \exp (\nu^{-1+3\theta}).
\end{equation}

Step 4. Choice of $R$. In order to finish the proof, we show that a choice of $(R_0,R)$ satisfying (<ref>), (<ref>), (<ref>), (<ref>) and (<ref>) is possible.

In the case of $h_0$, we fix $R_0=\nu^{-1+\theta}$ and $R=\nu^{-1-2\theta}$ which satisfy all requirements. In the case of $h_1$ and $h_2$, we have that $R_0 = \max\{ T_{\mathrm{kin}}\tau,\nu^{-1}\}$ and $R= T_{\mathrm{kin}}\tau + \nu^{-2\theta-1}$ satisfy all requirements. In particular, the upper bounds (<ref>)-(<ref>) hold since $T_{\mathrm{kin}}$ and $\nu$ depend on $L$ polynomially.

Step 5. Small $\tau$. In the case of $\widehat{h}_2$, property <ref> only holds when for $\varrho \tau \geq \nu^{\theta}$. However, the rest of the properties <ref>, <ref> and <ref>-<ref> hold even when $\varrho \tau < \nu^{\theta}$.

In the latter regime, we want to prove that $\Ss_2(k,t,\tau)=\O(|\log \nu|)$. In order to do so, it suffices to show that (<ref>) is $\O(|\log \nu|)$. This bound follows from Steps 0-4 with the choice $\theta=\beta=0$ with minor modifications, which we detail next.

In Step 0, the integration around the singularities in $\xi$ admits the same bound. The integration around $y=0$ in (<ref>) requires the following modification:

\begin{align}
\int_{\R^{2d}} \int_{\R} & e^{i\xi\Omega}\, [\widehat{h}(\nu\xi,y)-\widehat{h}(0,y)]\, \chi_0 (2y)\,  \varphi(k_1,k_3) \widetilde{\chi}_1 (\xi)\, d\xi\, dk_1 dk_3 \nonumber \\
= &\ \int_{\R^{2d}} \int_{\R} \frac{1}{i\Omega} \, e^{i\xi\Omega}   \frac{\pa}{\pa \xi}\left[  \left( \widehat{h}(\nu\xi,y)-\widehat{h}(0,y) \right)\, \widetilde{\chi}_1 (\xi)  \right] \, \varphi(k_1,k_3)\, \chi_0 ( 2y)\, d\xi\, dk_1 dk_3, \label{eq:y_small_1_alt}
\end{align}

where $\widetilde{\chi}_1 (\xi)= \chi_1 (\xi) \chi_1 (\xi-T\varrho)\, \chi_1 (\xi+T\varrho)$.

* If the derivative hits $\widetilde{\chi}_1 (\xi)$, we use <ref> and the localization in $\xi$ allows us to bound the $\xi$-integral.  Since $|y|<1/2$, <Ref> guarantees[Once again, <Ref>  actually yields $|\Omega|\gtrsim 1$, and thus the contribution of such a term is only $\O(1)$. However, the argument we present allows us to carry out the proof using only with the weaker bound $y^2 + (\nu^{-1}\Omega)^2\geq 1$, which is satisfied by many more dissipation choices $\gamma_k$.]
that $|\Omega|\gtrsim \nu$. Integrating $|\Omega|^{-1}$ then yields a bound $\O(|\log \nu |)$ for this contribution.

* If the derivative hits $\widehat{h}(\nu\xi,y)-\widehat{h}(0,y)$, we use <ref> to integrate in $\xi$. Then  <Ref> allows us to integrate in $\Omega$ and obtain a contribution of size $\O(|\log \nu|)$.

In Step 1, it suffices to replace the usage of <ref> by <ref>. All in all, this yields a contribution of size $\O(1)$ since $\theta=\beta=0$.
Step 2 holds unchanged, yielding $\O(1)$ contributions.
In Step 3, all arguments hold with $\theta=\beta=0$ except one uses  <ref> instead of <ref> in the first case. As a result, such terms are $\O (1)$ without gain in $\nu$.

Kernel $h_0$. First consider $\widehat{h_0}(\xi,y)=2\pi\, e^{-|y\xi|}$. Properties <ref>-<ref> are trivial, and so is <ref>. Note that

\[
|\pa_{\xi}^j \widehat{h_0}(\nu \xi,y)| = 4\pi \, (\nu |y|)^j \, e^{-\nu |y| \, |\xi|}, \qquad j=1,2,
\]

and thus

\[
\int_{\R} |\pa_{\xi}^j \widehat{h_0}(\nu \xi,y)|  \, d\xi = 4\pi\, (\nu \, |y|)^{j-1}, \qquad j=1,2,
\]

which yields <ref>.

Since $R_0=\nu^{-1}$ and $\Ac (h_0)=[-\nu^{\theta},\nu^{\theta}]$, (<ref>) is trivially true. For any $R>0$ we have that

\[
\int_{|\xi|>R}\frac{1}{|\xi|^j}\,  (\nu |\xi|)^j e^{-\nu |\xi|\, |y|}\, d\xi \lesssim \nu^{j-1} |y|^{-1} \, e^{-\nu |y| \, R},
\]

which in particular implies (<ref>) and thus <ref>.

Kernels $h_1$ and $h_2$. By inspection of (<ref>), one verifies that <ref> holds with the singular lines being $y=0$, $\xi=0, \pm \varrho \tau$. <ref> also follows easily from the expressions in (<ref>).

Consider next <ref> for $h_1$. We write

\[
\frac{1}{2\pi}\, \left[ \widehat{h_1} (\nu \xi, y) -\widehat{h_1} (0, y)\right] = \left( e^{-\nu |y|\,|\xi -T_{\mathrm{kin}} \tau |} - e^{-|y\,\varrho \tau |} \right) + \left( e^{-\nu |y|\,|\xi + T_{\mathrm{kin}} \tau |}- e^{-|y\,\varrho \tau |} \right).
\]

We explain how to handle the first summand, the second being analogous. It suffices to note that

\[
|e^{-\nu |y|\,|\xi -T_{\mathrm{kin}} \tau|} - e^{-|y\,\varrho \tau|}| \leq  \Big | \int_0^{\xi} \pa_{\eta} e^{-\nu |y|\,|\eta -T_{\mathrm{kin}} \tau|} \, d\eta\Big |  =  \int_0^{|\xi|} \nu |y|\, d\eta= \nu |y| |\xi|.
\]

Since $h_1$ is a linear combination of such exponential terms, <ref> follows.

Let us consider <ref> for $h_2$ when $\varrho \tau>\nu^{\theta}$. Assume that $y\neq 0$. Given that $\widehat{h}_2 (\nu\xi,y)$ has a jump at $\nu\xi = \varrho \tau$, one cannot expect the Lipschitz property to hold everywhere. If $\nu\xi \in (-\varrho \tau,\varrho \tau)$, then

\[
|\widehat{h}_2 (\nu\xi,y)- \widehat{h}_2 (0,y)| \lesssim |\nu \xi| \, \sup_{\eta  \in (-\varrho \tau,\varrho \tau)} |\pa_{\eta} e^{-|y|\, |\eta -\varrho\tau|}| \lesssim \nu \, |\xi| |y|.
\]

In particular, the above bound holds whenever $|\xi|\leq \nu^{-\theta}$, since $\nu |\xi| \leq \nu^{1-\theta}\leq \nu^{\theta}<\varrho\tau$.

Suppose next that $|\nu\xi|> \varrho\tau$. Then we use <ref> to obtain

\[
|\widehat{h}_2 (\nu \xi,y)- \widehat{h}_2 (0,y)| \lesssim 1\lesssim \frac{\nu |\xi|}{\varrho\tau} \leq \nu^{1-\theta} |\xi|
\]

since $\varrho\tau\geq \nu^{\theta}$. Finally, if $y=0$, $\widehat{h}_2 (\nu \xi,y)- \widehat{h}_2 (0,y)=0$ whenever $\nu|\xi|<\varrho\tau$ (in particular, this is the case if $|\xi|\leq \nu^{-\theta}$). When $\nu|\xi|>\varrho\tau$,

\[
|\widehat{h}_2 (\nu \xi,0)- \widehat{h}_2 (0,0)|=4\pi \lesssim \frac{\nu |\xi|}{\varrho\tau} \leq \nu^{1-\theta} \, |\xi|,
\]

which concludes the proof of <ref>.

Next we study <ref> for $h_1$ and $h_2$. If $y=0$, away from singularities in $\xi$, we notice that $\widehat{h}_1$ and $\widehat{h}_2$ are constant in $\xi$, and thus <ref> is trivial. Let us therefore assume that $y\neq 0$.
Away from $\eta = T_{\mathrm{kin}}\tau$, we have that

\begin{equation}\label{eq:block}
|\pa_{\eta}^j ( e^{-\nu |y|\,|\eta -T_{\mathrm{kin}} \tau|}) |= (\nu |y|)^j e^{-\nu |y|\,|\eta -T_{\mathrm{kin}} \tau|}
\end{equation}

Since $h_1$ and $h_2$ are a linear combination of similar exponential terms, <ref> follows.

Consider next <ref>. As before, we may assume $y\neq 0$ since the derivative vanishes when $y=0$.
Using (<ref>) we have that

\[
\int_{\eta>T_{\mathrm{kin}}\tau +\nu^{\theta}} (\nu |y|)^j \, e^{-\nu |y|\,(\eta -T_{\mathrm{kin}} \tau )} \, d\eta + \int_{\eta<T_{\mathrm{kin}}\tau-\nu^{\theta}} (\nu |y|)^j \, e^{-\nu |y|\,(T_{\mathrm{kin}}\tau-\eta)} \, d\eta =  2 \,(\nu |y|)^{j-1} \, e^{-\nu^{1+\theta} |y|} \lesssim (\nu\, |y|)^{j-1},
\]

which implies <ref>.

We now turn to <ref>. Consider $h_1$ first:

\begin{multline}\label{eq:der_h1}
\frac{\pa}{\pa y} \left[ e^{-\nu |y|\,|\xi -T_{\mathrm{kin}} \tau|}+  e^{-\nu |y|\,|\xi +T_{\mathrm{kin}} \tau |} - 2e^{-|y\,\varrho \tau |}\right] \\
= \mbox{sign}(y)\, \left( - |\nu \xi -\varrho \tau | e^{-\nu |y|\,|\xi -T_{\mathrm{kin}} \tau |} - |\nu \xi + \varrho \tau | e^{-\nu |y|\,|\xi +T_{\mathrm{kin}} \tau |}+ 2\varrho \tau e^{-|y\,\varrho \tau |} \right).
\end{multline}

We separate this expression into two summands and exploit the following cancellation between each of them:

\begin{equation}\label{eq:block2}
| \varrho\tau e^{-|y\,\varrho \tau |} - |\nu \xi -\varrho \tau | e^{-\nu |y|\,|\xi -T_{\mathrm{kin}} \tau |}| \lesssim \nu |\xi| .
\end{equation}

The above bound follows immediately if $|\xi|>T_{\mathrm{kin}}\tau$ (since it implies $\nu |\xi| > \varrho \tau$). If $\xi \in (-T_{\mathrm{kin}}\tau,T_{\mathrm{kin}}\tau)$ we write

\[
\varrho \tau e^{-|y\,\varrho \tau |} - (\varrho \tau  - \nu \xi ) \,  e^{-\nu |y|\,(T_{\mathrm{kin}} \tau -\xi)}  = \frac{1}{|y|}\, [F(\eta_1) -F(\eta_1-\eta_2)]
\]

where $F(z) = z e^{-z}$, $\eta_1= |y\,\varrho \tau |>\eta_2= \nu |y \xi|$. Note that $F$ is bounded for all $z\geq 0$, and thus

\begin{equation}\label{eq:block2_trick}
\frac{1}{|y|} \, | F(\eta_1) -F(\eta_1-\eta_2)| \leq \frac{1}{|y|} \,\left| \int_{\eta_1-\eta_2}^{\eta_1} (e^{-z} -F(z) )\, dz \right|\lesssim \frac{|\eta_2|}{|y|} = \nu |\xi|.
\end{equation}

It follows that (<ref>) holds for all $\xi$. Since $\widehat{h_1}(\nu\xi ,y) - \widehat{h_1}(0 ,y)$ is a sum of two such terms, <ref> holds for the kernel $h_1$.

Let us now prove <ref> for $h_2$. Without loss of generality, suppose that $y>0$. Then

\begin{multline}\label{eq:der_h2}
\frac{1}{2\pi}\, \frac{\pa}{\pa y} \left[ \widehat{h_2}(\nu\xi ,y) - \widehat{h_2}(0 ,y)\right] =\frac{\pa}{\pa y} \left[  - e^{-\nu y\, |\xi -T_{\mathrm{kin}} \tau |}  \, \mbox{sign}(\xi -T_{\mathrm{kin}}\tau ) + e^{-\nu y\, |\xi +T_{\mathrm{kin}} \tau |}\, \mbox{sign}(\xi+T_{\mathrm{kin}} \tau  )  - 2 e^{-\varrho \tau y}\right] \\
=  \nu \, (\xi -T_{\mathrm{kin}} \tau )\, e^{-\nu y \, |\xi -T_{\mathrm{kin}} \tau |} - \nu\,  (\xi+T_{\mathrm{kin}} \tau )\, e^{-\nu y\, |\xi +T_{\mathrm{kin}} \tau |} + 2\varrho \tau \, e^{-\varrho \tau  y} .
\end{multline}

The result then follows from applying (<ref>) twice.

Let us now show <ref> for both $h_1$ and $h_2$. For $y\neq 0$,

\[
\widehat{h_1}(0 ,y)= 4\pi \, e^{- |y| \varrho \tau } = \mbox{sign}(y) \, \widehat{h_2}(0 ,y),
\]

thus it suffices to note that

\[
|\pa_y^{j} e^{- |y| \varrho \tau }| = (\varrho \tau )^j \, e^{- |y| \varrho \tau } = |y|^{-j}\, \left[ (\varrho \tau |y|)^j \, e^{- |y| \varrho \tau } \right]\lesssim |y|^{-j}
\]

given that $\eta^j e^{-\eta} = \O_j (1)$.

Finally, we set out to prove <ref>. In order to prove (<ref>), note that  for any $R> R_0 =\max\{T_{\mathrm{kin}}\tau,\nu^{-1}\}$,

\[
\int_{|\xi|>R} \frac{1}{|\xi|^j} \, (\nu \, |\xi - T_{\mathrm{kin}}\tau |)^j e^{-\nu |y| \, |\xi - T_{\mathrm{kin}}\tau |}\, d\xi \lesssim  \nu^j\, \int_{|\xi|>R} e^{-\nu |y| \, (\xi - T_{\mathrm{kin}}\tau)}\, d\xi = |y|^{-1} \, \nu^{j-1}\, e^{-\nu |y| \, (R- T_{\mathrm{kin}}\tau )}.
\]

Both $h_1$ and $h_2$ are a linear combination of such exponential terms, and thus (<ref>) follows.

Let us turn to (<ref>). Recall that $R_0 =\max\{T_{\mathrm{kin}}\tau,\nu^{-1}\}$. In view of the cancellation exploited in (<ref>) and (<ref>), it suffices to study the $y$-derivative of the expression in (<ref>). Assuming, without loss of generality, that $y>0$

\[
\pa_y \left( \varrho \tau e^{-|y\,\varrho \tau |} - |\nu \xi -\varrho \tau| e^{-\nu |y|\,|\xi -T_{\mathrm{kin}} \tau |}\right) = -(\varrho \tau)^2 e^{-|y\,\varrho \tau|} + |\nu \xi -\varrho \tau|^2 e^{-\nu |y|\,|\xi -T_{\mathrm{kin}} \tau|}.
\]

We want to integrate this quantity in the region where $|\xi|\geq \nu^\theta$, $|\nu \xi-\varrho \tau|\geq \nu^\theta$ and $|\nu \xi |\leq \max(\rho \tau, 1)$. We claim that

\begin{equation}\label{eq:block3}
\Big | |\nu \xi -\varrho \tau|^2 e^{-\nu |y|\,|\xi -T_{\mathrm{kin}} \tau|} -(\varrho \tau)^2 e^{-|y\,\varrho \tau|} \Big |\lesssim
\begin{cases}
\frac{\nu |\xi|}{|y|} \quad \mbox{when}\ |\nu \xi| <\varrho \tau,\\
\nu |\xi| \quad \mbox{when}\ 1>|\nu \xi| \geq \varrho \tau.
\end{cases}
\end{equation}

The first inequality follows from an argument such as (<ref>) with $F(z)=z^2 e^{-z}$, while the second inequality follows from a direct bound.

Using the fact that $h_1$ and $h_2$ are linear combinations of such terms, we find for $h\in \{h_1,h_2\}$ and $y\neq 0$ that

\[
\begin{split}
\int_{\{\nu^{\theta} <|\xi|<R_0\}-\nu^{-1}\Ac(h)}  \frac{1}{|\xi|^2} \, |\pa_{y}^{2}[\widehat{h}(\nu \xi, y)- \widehat{h}(0, y) ]| d\xi & \lesssim
\max\{ 1, |y|^{-1}\}\, \int_{\nu^{\theta}<|\xi|<R_0}  \frac{\nu}{|\xi|} \, d\xi \\
& \lesssim \nu\, \max\{ 1, |y|^{-1}\}\, \log (\nu^{-\theta}R_0).
\end{split}
\]

Using (<ref>) also yields:

\[
\int_{\{\nu^{\theta} <|\xi|<R_0\}-\nu^{-1}\Ac(h)}  \frac{1}{|\xi|^2} \, |\pa_{y}[\widehat{h}(\nu \xi, y)- \widehat{h}(0, y) ]| d\xi \lesssim
\int_{\nu^{\theta}<|\xi|<R_0}  \frac{\nu}{|\xi|} \, d\xi \lesssim \nu\, \log (\nu^{-\theta}R_0).
\]

This completes the proof <ref> and the claim.

We are ready to give the main convergence results in the case $\varrho\gtrsim 1$.

Let $T,T_{\mathrm{kin}}$ be as in <Ref> and let $0 \leq t T_{\mathrm{kin}} \leq T$. Then

* If $\varrho \in (0,\infty)$, then there exists some $0<\theta\ll 1$ such that

\begin{equation}\label{eq:towards_forcing}
\E |u_k (tT_{\mathrm{kin}})|^2 = f(t,k)+ \int_0^t  e^{-2\varrho \gamma_k(t-t')}\, \Kc (f(t',\cdot))(k)\, dt' +\O_{\ell^{\infty}_k} ( L^{-\theta}\, t ),
\end{equation}

where  $\Kc$ is as in (<ref>) and $f$ is defined in (<ref>). As a result, $\E |u(k,tT_{\mathrm{kin}})|^2$ is well-approximated by the solution $n(t,k)$ to

\begin{equation}\label{eq:damped_forced_WKE}
\pa_t n = \Kc (n) - 2\varrho \gamma_k \, n + 2\varrho \, b_k^2, \qquad n |_{t=0}=c_k^2.
\end{equation}

* If $\varrho \sim L^{\kappa}$ for some $\kappa> 0$, then there exists some $0<\theta\ll 1$ such that

\begin{equation}\label{eq:overforced_WKE}
\E |u(k,tT_{\mathrm{kin}})|^2 =
c_k^2 \, e^{-2\varrho\gamma_k t}+\frac{b_k^2}{\gamma_k} \left( 1- e^{-2\varrho \gamma_k t}\right) + \O_{\ell^{\infty}_k} ( L^{-\kappa}+L^{-\theta}\, t ),
\end{equation}

which is the exact solution to

\begin{equation}
\pa_t n = - 2\varrho \gamma_k \, n + 2\varrho \, b_k^2, \qquad n |_{t=0}=c_k^2.
\end{equation}

We recall that $f$ defined in (<ref>) is given by

\[
\E |w_k^{(0)}(t)|^2 = c_k^2 \, e^{-2\varrho\gamma_k t}+\frac{b_k^2}{\gamma_k} \left( 1- e^{-2\varrho \gamma_k t}\right).
\]

(i) By <Ref>, (<ref>), <Ref> and <Ref>, there exists some $0<\theta\ll 1$ such that

\[
\E|w_k^{(0)}(t) + w_k^{(1)} (t)+w_k^{(2)} (t)|^2 =
f(t,k) + \int_0^t  e^{-2\varrho \gamma_k(t-t')}\, \Kc (f(t',\cdot))(k)\, dt'  + \O_{\ell^{\infty}_k} \left(L^{-\theta} \,t\right).
\]

By <Ref> and (<ref>), we have that if $0\leq t T_{\mathrm{kin}}\leq T$,

\[
\begin{split}
\E|u_k(tT_{\mathrm{kin}})|^2 & = f(t,k) + \int_0^t  e^{-2\varrho \gamma_k(t-t')}\, \Kc (f(t',\cdot))(k)\, dt'  + \O_{\ell^{\infty}_k} \left(L^{-\theta} t\right)
\end{split}
\]

Notice that the right-hand side is the first Picard iterate of the kinetic equation (<ref>), and thus the result follows.

(ii) If $\varrho=L^{\kappa}\rightarrow\infty$, we have that

\begin{equation}
f(t,k)=c_k^2 \, e^{-2\varrho\gamma_k t}+\frac{b_k^2}{\gamma_k} \left( 1- e^{-2\varrho \gamma_k t}\right)=\begin{cases}
c_k^2 + \O_{\ell^{\infty}_k} (L^{-N})& \mbox{if}\ t\ll \varrho^{-1},\\
c_k^2 \, e^{-2\varrho\gamma_k t}+\frac{b_k^2}{\gamma_k} \left( 1- e^{-2\varrho \gamma_k t}\right)& \mbox{if}\ t\varrho \sim 1\\
\frac{b_k^2}{\gamma_k} + \O_{\ell^{\infty}_k} (L^{-N}) & \mbox{if}\ t\gg \varrho^{-1}.
\end{cases}
\end{equation}

for $N$ as large as desired. Since $\Kc (f)=\O_{\ell^{\infty}_k} (1)$, it follows that

\[
\int_0^t  e^{-2\varrho \gamma_k(t-t')}\, \Kc (f(t',\cdot))(k)\, dt' =\O_{\ell^{\infty}_k}  (L^{-\kappa}).
\]

Combining <Ref>, (<ref>), <Ref> and <Ref>, we find that there exists some $0<\theta\ll 1$ such that

\[
\E|w_k^{(0)}(t) + w_k^{(1)} (t)+  w_k^{(2)} (t)|^2  =  f(t,k) + \O_{\ell^{\infty}_k} \left(L^{-\kappa}+L^{-\theta} \, t\right).
\]

Finally, we use <Ref> and (<ref>) finish the proof of (<ref>).

§.§ Admissible dissipations

A key tool in the proof of <Ref> was the fact that $\Gamma_{-}^2 + (\nu^{-1} \Omega)^2\geq 1$ for our choice of dissipation $\gamma_k=(1+|k|_{\zeta}^2)^r$, $r\in (0,1]$. In particular, this means that there are no points in the intersection of a neighborhood of the manifolds $\Gamma_{-}=0$ and $\Omega=0$. This requirement may be weakened provided this intersection remains small:

Let $d\geq 2$, $T\leq L^2$, $L^{-2}<\alpha<1$ and fix a small $\theta \in (0,1)$. Consider the dissipation $\gamma_k=|k|_{\zeta}^2$. Let $k\in \Z_L^d$ satisfy $|k|\leq L^{\theta}$ and consider the set:

\[
\mathfrak{S}(k)= \{ (k_1,k_2,k_3)\in(\Z_L^d)^3 \ |\ k=k_1-k_2+k_3,\ k\neq k_1,k_3,\  |k_j|\leq L^{\theta},\ |\Omega_{13}^{2k}|<T^{-1},\ \Big |\sum_{j=1}^{3} \gamma_{k_j} - \gamma_k\Big |\leq \alpha\}.
\]

Then the number of points in $\mathfrak{S}(k)$ satisfies the following bound

\[
|\mathfrak{S}(k)|\lesssim L^{d\theta}\, L^{2d}T^{-1}\, (\alpha+T^{-1})^{\frac{d-1}{2}}.
\]

Given that $|\Omega_{13}^{2k}|<T^{-1}$ and $\gamma_k=|k|_{\zeta}^2$,

\[
2|k_2|_{\zeta}^2 - T^{-1} \leq  \sum_{j=1}^{3} \gamma_{k_j} - \gamma_k = \Omega_{13}^{2k} + 2 |k_2|_{\zeta}^2 .
\]

This implies that $\mathfrak{S}(k) \subseteq \widetilde{\mathfrak{S}}(k)$, where

\[
\widetilde{\mathfrak{S}}(k)=  \{ (k_1,k_2,k_3)\in(\Z_L^d)^3 \ |\ k=k_1-k_2+k_3,\ k\neq k_1,k_3,\  |k_j|\leq L^{\theta},\ |\Omega_{13}^{2k}|<T^{-1},\ |k_2|_{\zeta}^2 \leq \frac{T^{-1} + \alpha}{2} \} \ .
\]

It thus suffices to estimate $|\widetilde{\mathfrak{S}}(k)|$. In view of the equality $\Omega_{13}^{2k}= -2 (k_1-k)\cdot (k_3-k)$, we introduce the variables  $p=L\, (k_1-k)\in \Z^d$ and $q=L\, (k_3-k)\in \Z^d$. Suppose that $p_iq_i=0$ for $1\leq i\leq j$ and $p_iq_i\neq 0$ for $j+1\leq i\leq d$, where $j\in \{0,\ldots, d\}$. The condition $|\Omega_{13}^{2k}|<T^{-1}$ implies that

\begin{equation}\label{eq:last_component}
\zeta_d p_d q_d = -\sum_{i=j+1}^{d-1} \zeta_i p_i\, q_i + \O (L^2 T^{-1}).
\end{equation}

The conditions $|k_1|, |k_3|\leq L^{\theta}$ imply that each component of $p$ and $q$ belong to an interval of size $L^{1+\theta}$. Moreover, the condition on $|k_2|_{\zeta}$ implies that

\begin{equation}\label{eq:gamma_cond}
|p+q-k|_{\zeta} \lesssim L\, \sqrt{\alpha+T^{-1}}.
\end{equation}

In particular, there are $\O( \min\{L^{(1+\theta)\, j}, L^{j}\, (\alpha+T^{-1})^{j/2}\})$ choices for the first $j$ components of $p$ and $q$ (since $p_iq_i=0$). If $j=d$, we are done since $T\leq L^2$.

If $j=d-1$, (<ref>) yields $\O (L^2 T^{-1})$ choices for the product $p_d q_d$. It is well known that the number of integer pairs $(p_d,q_d)$, each smaller than $n$, such that an $p_dq_d$ is fixed is bounded by $n^{\theta}$ for $\theta\in (0,1)$ as small as desired (this is a consequence of the divisor bound in $\mathbb Z$). As a result, there are $\O (L^{2+\theta} T^{-1})$ choices for the pair $(p_d,q_d)$. All in all, $|\widetilde{\mathfrak{S}}(k)|=\O( L^{d-1} (\alpha+T^{-1})^{\frac{d-1}{2}} L^{2+\theta} T^{-1})$ which concludes the proof in the case $j=d-1$.

If $j\leq d-2$, let us consider the remaining $d-j$ components of $p$ and $q$. For $j+1\leq i\leq d-1$, we fix $p_i$ freely (there are $\O ( L^{(1+\theta)(d-j-1)})$ such choices). By (<ref>), $q_i$ lives in an interval of size $ L\, \sqrt{\alpha+T^{-1}}$ and thus there are $\O (L^{d-j-1}\, (\alpha+T^{-1})^{\frac{d-j-1}{2}})$ choices for the remaining components of $q$. Finally, there are $\O (L^{2+\theta} T^{-1})$ choices for $(p_d,q_d)$ in view of (<ref>). All in all,

\[
\begin{split}
|\widetilde{\mathfrak{S}}(k)| & \lesssim L^{j}\, (\alpha+T^{-1})^{j/2} \, L^{(1+\theta)(d-j-1)}\, L^{d-j-1}\, (\alpha+T^{-1})^{\frac{d-j-1}{2}}\, L^{2+\theta} T^{-1} \\
& \lesssim L^{d\theta}\,  L^{2d-j} \, T^{-1}\,  (\alpha+T^{-1})^{\frac{d-1}{2}}
\end{split}
\]

which is maximal when $j=0$, and yields the desired estimate.

Using this result, we may extend cases (i) and (ii) of <Ref> to the dissipation given by $\gamma_k=|k|^2_{\zeta}$ with forcing $b_k = \O (|k|_{\zeta})$ at $k=0$. Indeed, by choosing $\alpha=L^{-\mu}$ with $0<6\theta<\mu\ll 1$, one may show that the contribution of the sum over $\mathfrak{S}(k)$ towards, say, $\E |w^{(1)}(k,t)|^2$ in (<ref>) has size

\[
\O\left( \frac{T_{\mathrm{kin}}}{L^{2d}} \, |\mathfrak{S}(k)|\right)=\O ( L^{d\theta- \frac{d-1}{2}\, \mu})=\O(L^{-\theta}).
\]

In particular, the leading terms towards the kinetic equation are given by (<ref>) in the region $\mathfrak{S}(k)^c$. This is equivalent to adding a smooth cut-off $\chi(k_1,k_2,k_3)$ supported in $\mathfrak{S}(k)^c$ in all remaining calculations. Finally, one needs to verify that the term (<ref>) vanishes with this choice of $\gamma_k$. This easily follows from the fact that $\Gamma_{-} = \Omega + 2|k_2|_{\zeta}^2$, which implies that $\Gamma_{-}>0$ a.e. in the support of $\delta(\Omega)$. We remark that the condition $b_k = \O (|k|_{\zeta})$ at $k=0$ guarantees that the quantity $b_k^2/\gamma_k$ is well-defined at $k=0$. Similar conditions regarding the vanishing of the forcing around $k=0$ are not uncommon in the turbulence literature, see for instance [2, 3].

§ ANALYSIS OF THE REMAINDER TERMS

In this section we prove <Ref>. As derived in (<ref>), the remainder satisfies the following equation:

\begin{equation}\label{eq:erroreq2}
\bm{\mathcal{R}}_{N+1} = \mathcal{L}(\bm{\mathcal{R}}_{N+1}) + \mathcal{Q}(\bm{\mathcal{R}}_{N+1}) +\mathcal{C}(\bm{\mathcal{R}}_{N+1})+ \sum_{\substack{N\leq n_1+n_2+n_3 \\ n_1,n_2,n_3\leq N}}\mathcal{IW}(\bm{w}^{(n_1)}, \bm{w}^{(n_2)},\bm{w}^{(n_3)})
\end{equation}

The proof of <Ref> is an application of the contraction mapping theorem.
The worst terms on the right-hand side of (<ref>) are the linear ones, which are essentially given by the operators

\[ \mathcal{IW}(\bm{\mathcal{R}}_{N+1}, \bm{w}^{(n_1)},\bm{w}^{(n_2)}), \quad \mbox{and}\quad \mathcal{IW}(\bm{w}^{(n_1)},\bm{\mathcal{R}}_{N+1}, \bm{w}^{(n_2)}),\]

where $n_1,n_2\leq N$. In particular, we would like to estimate the operator norm of

\[
\bm{v}\in h^{s,b}\ \longmapsto \mathcal{IW}(\bm{v}, \bm{w}^{(n_1)},\bm{w}^{(n_2)})\in h^{s,b} \ .
\]

By <Ref>, it suffices to study the operators with $w^{(n_1)}$ and $w^{(n_2)}$ substituted by $\Jc_{\Tc_1}$ and $\Jc_{\Tc_2}$, where $\Tc_j$ is a ternary tree of order $n_j$, $j=1,2$. We have the following result:

Suppose that $T$ and $T_{\mathrm{kin}}$ are as in (<ref>)- (<ref>). Let $0\leq n_1,n_2\leq N$  and $\Tc_j$ be a ternary tree of order $n_j$ satisfying (<ref>)-(<ref>). Then the operators

\begin{equation}\label{eq:linear_error}
\Pc_{+}: \bm{v}\mapsto \mathcal{IW}(\Jc_{\Tc_1},\Jc_{\Tc_2},\bm{v}), \qquad \Pc_{-}: \bm{v}\mapsto \mathcal{IW}(\Jc_{\Tc_1},\bm{v}, \Jc_{\Tc_2})
\end{equation}

satisfy the following bounds $L$-certainly

\begin{equation}
\norm{\Pc_{\pm}}_{h^{s,b}\rightarrow h^{s,b}} \leq L^{\theta}\, L^{-\delta (n_1+n_2+\frac{1}{2})}
\end{equation}

where $0<\theta\ll 1$ is as small as desired.

Let us now prove <Ref> assuming <Ref>.

Based on (<ref>), it suffices to prove that the mapping

\begin{equation}\label{eq:contraction}
\bm{v}\longmapsto \mathcal{L}(\bm{v}) + \mathcal{Q}(\bm{v}) +\mathcal{C}(\bm{v})+ \sum_{\substack{N\leq n_1+n_2+n_3 \\ n_1,n_2,n_3\leq N}}\mathcal{IW}(\bm{w}^{(n_1)}, \bm{w}^{(n_2)},\bm{w}^{(n_3)})
\end{equation}

is a contraction from the set

\[
\left\lbrace \bm{v}\in h^{s,b} : \norm{\bm{v}}_{h^{s,b}} < L^{-\delta N} \right\rbrace
\]

to itself. To handle the last term in (<ref>), we note that we can replace $\bm{w}^{(n_j)}$ by $\Jc_{\Tc_j}$ for trees $\Tc_j$ of order $n_j$. These trees uniquely determine a larger tree $\Tc$ of order $n_1+n_2+n_3+1$, thus <Ref> yields:

\[
\norm{\mathcal{IW}(\Jc_{\Tc_1},\Jc_{\Tc_2},\Jc_{\Tc_3})}_{h^{s,b}}=\norm{\Jc_{\Tc}}_{h^{s,b}}\lesssim L^{\theta+c(b-1/2)} L^{-\delta (n_1+n_2+n_3+1)}
\]

$L$-certainly. As a consequence, the $h^{s,b}$-norm of the rightmost term in (<ref>) is $L$-certainly bounded by

\[
\sum_{\substack{N\leq n_1+n_2+n_3 \\ n_1,n_2,n_3\leq N}} L^{\theta+c(b-1/2)} L^{-\delta (n_1+n_2+n_3+1)} \lesssim L^{\theta+c(b-1/2)} L^{-\delta (N+1)}\ll L^{-\delta N}.
\]

By <Ref>, the linear term admits the following bound $L$-certainly:

\[
\norm{\mathcal{L}(\bm{v})}_{h^{s,b}} \lesssim \sum_{0\leq n_1 , n_2 \leq N} L^{\theta}\, L^{-\delta (n_1+n_2+\frac{1}{2})} \norm{\bm{v}}_{h^{s,b}} \lesssim L^{\theta}\, L^{-\delta/2}\, L^{-\delta N} \ll  L^{-\delta N} .
\]

Finally, the quadratic and cubic terms are easy to estimate using the simple bound:

\[
\norm{\mathcal{IW}(\bm{f}_1,\bm{f}_2,\bm{f}_3)}_{h^{s,b}}\lesssim L^{\theta} \lambda T \, \prod_{j=1}^{3}\norm{\bm{f}_j}_{h^{s,b}}
\]

$L$-certainly for $\bm{f}_j\in \{ \bm{v},\, \bm{w}^{(n)}\, :\, n\in \N_0\}$, $j=1,2,3$. Using this bound together with the fact that $\lambda T\leq L^{d}$, we have that, $L$-certainly,

\[
\norm{\mathcal{Q}(\bm{v})}_{h^{s,b}} + \norm{\mathcal{C}(\bm{v})}_{h^{s,b}}\lesssim L^{\theta+c(b-1/2) + d} L^{-2\delta N} \ll L^{-\delta N}
\]

provided we choose $N>d/\delta$.

This shows that the operator (<ref>) $L$-certainly maps the set

\[
\left\lbrace \bm{v}\in h^{s,b}: \norm{\bm{v}}_{h^{s,b}} < L^{-\delta N} \right\rbrace
\]

to itself. The proof that it is a contraction is analogous.

The proof is an adaptation of <cit.> to our setting, so we only give a rough sketch of the main ideas. In order to show that $\Pc_{\pm}$ maps $h^{s,b}$ to $h^{s,b}$, one interpolates between a simple bound with a $L^{3d}$ loss

\begin{equation}\label{eq:rough_L2bound}
\norm{\Pc_{+}\bm{v}}_{h^{s,1}} \lesssim  L^{\theta + c (b-1/2)+3d} \, L^{-\delta (n_1+n_2)}\, \norm{\bm{v}}_{h^{s,0}},
\end{equation}

and a sharper bound

\begin{equation}\label{eq:sharp_L2bound}
\begin{split}
\norm{\Pc_{+}\bm{v}}_{h^{s,1-b}} & \lesssim L^{\theta} \, L^{-(n_1+n_2+1)\delta} \norm{\bm{v}}_{h^{s,b}} .
\end{split}
\end{equation}

The proof of (<ref>) requires a $TT^{\ast}$ argument applied $D\gg 1$ times, which leads to the analysis of the kernel of the operator $(\Pc_{+}\Pc_{+}^{\ast})^D$.
This kernel may be written in the form of $\Jc_{\Tc}$ in (<ref>) for some large auxiliary tree $\Tc=\Tc(D)$ which may be constructed by attaching $2D$ copies of the trees $\Tc_1,\Tc_2$ in (<ref>) successively.
In particular, giving sharp estimates of the kernel of the operator $(\Pc_{+}\Pc_{+}^{\ast})^D$ admits the same strategy as estimating the Picard iterates of our equation, see <Ref>  and <Ref>.
Choosing large enough $D\gg 1$ allows for $\theta$ in (<ref>) to be as small as desired. The full details may be found in <cit.>.

§ SOME USEFUL LEMMAS

Let $H$ be a Gaussian Hilbert space where our standard complex Wiener processeses, $\beta_k (t)$, and the Gaussian variables $\eta_k$ live. We define

\[
\mathcal{P}_n (H) = \{ p(X_1, \ldots , X_m) \mid \mbox{$p$ is a polynomial of degree} \leq n;\ X_1,\ldots,X_m\in H; m\in\N\}.
\]

We will often consider the closure $\overline{\mathcal{P}_n (H)}$ with respect to the topology given by the inner product in $L^2 (\Omega,\mathcal{F},\P)$.

We start with a corollary of Nelson's hypercontractivity estimate (see Theorem 5.10 and Remark 5.11 in [21]).

For any $1\leq p,q<\infty$ we have that

\begin{equation}\label{eq:hypercontractivity}
\left(\E |X|^q\right)^{1/q} \leq c(p,q)^{n} \, \left(\E |X|^p \right)^{1/p}
\end{equation}

for all $X\in \overline{\mathcal{P}_n (H)}$, $n\geq 0$.

It can be shown that $c(2,q)\leq (q-1)^{1/2}$ and $c(1,2)\leq e$, see [21]. We will often combine these and use the estimate

\begin{equation}\label{eq:hypercontractivity2}
\left(\E |X|^q\right)^{1/q} \leq (q-1)^{n/2} \, \left(\E |X|^2 \right)^{1/2} \leq  (q-1)^{n/2} \, e^{n}\,\E |X| .
\end{equation}

Another important result is Isserlis' theorem (also known as Wick's theorem). We record it below as stated in [21], Theorem 1.28.

Let $X_1,\ldots,X_n$ be centred jointly normal random variables. Then

\begin{equation}
\E \left[ X_1\cdots X_n\right] = \sum \prod_k \E [ X_{i_k} X_{j_k}],
\end{equation}

where we sum over all partitions of $\{ 1,\ldots, n\}$ into disjoint pairs $\{i_k,j_k\}$.

Beyond the combinatorial objects introduced in <Ref>, we need some additional definitions that will allow us to count the number of decorations.

First consider a tree $\Tc$ of order $n$ with a set of pairings $\Pc$ between the leaves. Suppose we fix some of the nodes of the tree in the counting process. These nodes are “colored” red and form the set $\Rc$. Then we have the following definition:

Suppose that we fix $n_{\sub{m}}\in\Z_L^d$ for each $\sub{m}\in\Rc\subset \Tc$ (the set of “red” nodes in the tree). A decoration $(k_{\sub{n}}:\sub{n}\in\Tc)$ is called strongly admissible with respect to the pairing $\Pc$, coloring $\Rc$, and $(n_{\sub{m}} :\sub{m}\in\Rc)$ if it satisfies (<ref>) and

\[
k_{\sub{m}}=n_{\sub{m}}\ \ \forall\sub{m}\in\Rc;\qquad |k_{\sub{l}}|\leq L^{\theta}\ \ \forall\sub{l}\in\Lc; \qquad k_{\sub{l}}=k_{\sub{l}'}\ \ \forall \ \{\sub{l},\sub{l}'\}\in \Pc .
\]

The main combinatorial result is the following.

Let $0\leq \theta\ll 1$. Suppose that $\Tc$ is a paired and colored ternary tree such that $\Rc\neq\emptyset$, and let $(n_{\sub{m}} : \sub{m}\in\Rc)$ be fixed. We also fix $\sigma_{\sub{n}}\in \R$ for each $\sub{n}\in\Nc$. Let $\Sc$ be the set of leaves which are not paired. Let $|\Lc|=l$ be the total number of leaves, $p=|\Pc|$ be the number of pairs, and $r=|\Rc|$ be the number of red nodes. Then the number of strongly admissible decorations $(k_{\sub{n}}:\sub{n}\in\Tc)$ that also satisfy

\[ |\Omega_{\sub{n}} - \sigma_{\sub{n}}|\leq T^{-1} \ \ \forall\sub{n}\in\Nc,\]

is bounded by

\begin{equation}\label{eq:main_counting}
\begin{cases}
L^{\theta}\, (L^d T^{-1} \rho)^{l-p-r}  & \quad \mbox{if}\ \Rc\neq\Sc \cup \{ \sub{r}\},\\
L^{\theta}\, (L^d T^{-1} \rho)^{l-p-r+1} & \quad \mbox{if}\ \Rc=\Sc \cup \{ \sub{r}\},
\end{cases}
\end{equation}

where $\rho$ is defined in (<ref>).

The proof is a description of a counting algorithm, and it may be found as Proposition 3.5 in [7].

The following result is Corollary 3.6 in [7]. The proof is based on a slight modification of the main counting algorithm.

Suppose that $\Rc=\{\sub{r}\}$  in the setting of <Ref>. Then (<ref>) can be improved to

\[
M\leq L^{\theta}\, (L^d T^{-1} \rho)^{l-p-3} \, L^{2d} T^{-1}.
\]

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