\tau_\infty(T,w,f) = \;& \int_{[0,1]} F(w, f)(\xi, \rd z_1,\dots, \rd z_{|T|}) \;\rd \xi.
\end{aligned}
\end{equation*}

For any $\varphi \in C_c(\R^{|T|})$, by Lemma <ref>,

\begin{equation*}
\begin{aligned}
\;& \lim_{N \to \infty} \int_{z \in \R^{|T|}} \tau_\infty(T,\tilde w_{N;\#},\tilde f_{N;\#}) (\rd z_1,\dots, \rd z_{|T|})
\\
= \;& \lim_{N \to \infty} \int_{[0,1]} \int_{z \in \R^{|T|}} \varphi(z_1,\dots,z_{|T|}) F(\tilde w_N, \tilde f_N)(\Phi_N(\xi), \rd z_1,\dots, \rd z_{|T|}) \;\rd \xi
\\
= \;& \int_{[0,1]} \int_{z \in \R^{|T|}} \varphi(z_1,\dots,z_{|T|}) F(w, f)(\xi, \rd z_1,\dots, \rd z_{|T|}) \;\rd \xi
\\
= \;& \int_{z \in \R^{|T|}} \tau_\infty(T,w,f) (\rd z_1,\dots, \rd z_{|T|}).
\end{aligned}
\end{equation*}

Since $\varphi \in C_c(\R^{|T|})$ is arbitrary we conclude (<ref>), restated here:

\begin{equation*}
\begin{aligned}
\tau_\infty(T,\tilde w_N, \tilde f_N) \overset{\ast}{\rightharpoonup} \tau_\infty(T,w,f) \in \mathcal{M}(\R^{|T|}), \quad \forall T \in \mathcal{T}.
\end{aligned}
\end{equation*}

§ PROOFS OF THE QUANTITATIVE RESULTS

§.§ The hierarchy of equations

The subsection provides the main proofs of Proposition <ref>, <ref> and <ref>, which derive the hierarchy of equations from the Liouville equation (<ref>) and the Vlasov equation (<ref>)-(<ref>).

We begin with the proof of Proposition <ref>, showing that the observables corresponding to the laws of $(X^{1;N}_0,\dots, X^{N;N}_0)$ solving (<ref>) satisfy the extended BBGKY hierarchy (<ref>)-(<ref>).

Since the coefficients are bounded Lipschitz, the well-posedness of the SDE system (<ref>) and the Liouville-type equation (<ref>) are classical results.
For simplicity of the presentation, we avoid using weak formulations but only present a formal calculation.

Consider any distinct indexes $i_1,\dots,i_k \in \{1,\dots,N\}$. It is easy to verify the following identity deriving the marginal laws from the full joint law,

\begin{equation*}
\begin{aligned}
f_{N}^{i_1,\dots, i_k}(t,z_1,\dots,z_{k}) \defeq \;& \law (X^{i_1;N}_t,\dots, X^{i_{k};N}_t)
\\
=\;&  \bigg(\int_{\R^{N - k}} f_{N}(t,x_1,\dots,x_N) \textstyle{\prod_{i \neq i_1,\dots, i_k} \rd x_i}\bigg)\bigg|_{\forall l = 1,\dots, k, \; x_{i_l} = z_l}.
\end{aligned}
\end{equation*}

By integrating Liouville equation (<ref>) along spatial directions $i \notin \{i_1,\dots,i_{k}\}$ and calculate the summation $i \in \{i_1,\dots,i_{k}\}$ and $i \notin \{i_1,\dots,i_{k}\}$ separately, we obtain equations for the marginals,

\begin{equation} \label{eqn:hierarchy_equation_marginal_integrate}
\begin{aligned}
& \partial_t f_{N}^{i_1,\dots, i_{k}}(t,z_1,\dots,z_{k})
\\
&\quad =  \sum_{m = 1}^{k} \Bigg\{ \bigg[ - \partial_{z_m}(\mu(z_m) f_{N}^{i_1,\dots, i_{k}}(t,z)) + \frac{\sigma^2}{2} \partial_{z_m}^2 f_{N}^{i_1,\dots, i_{k}}(t,z)
\\
&\qquad - \nu(z_m) f_{N}^{i_1,\dots, i_{k}}(t,z) + \delta_0(z_m) \bigg( \int_{\R} \nu(u_m) f_{N}^{i_1,\dots, i_{k}}(t,u - {w_{N;i_m}^{i_1,\dots, i_{k}}}) \Big) \;\rd u_m \bigg)\bigg|_{\forall n \neq m,\, u_n = z_n} \bigg] \Bigg\}
\\
&\qquad\  + \sum_{i \neq i_1,\dots,i_{k}} \int_{\R} \nu(z_{k+1}) \bigg( f_{N}^{i_1,\dots, i_{k}, i}(t,z - {w_{N;i}^{i_1,\dots, i_{k}, i}}) - f_{N}^{i_1,\dots, i_{k}, i}(t,z) \bigg) \;\rd z_{k+1}.
\end{aligned}
\end{equation}

We can reformulate the last line as

\begin{equation*}
\begin{aligned}
& \sum_{i \neq i_1,\dots,i_{k}} \int_{\R} \nu(z_{k+1}) \bigg( f_{N}^{i_1,\dots, i_{k}, i}(t,z - {w_{N;i}^{i_1,\dots, i_{k}, i}}) - f_{N}^{i_1,\dots, i_{k}, i}(t,z) \bigg) \;\rd z_{k+1}
\\
&\quad= \sum_{i \neq i_1,\dots,i_{k}} \int_{\R} \nu(z_{k+1}) \bigg( \int_0^1 \sum_{m=1}^{k} - w_{i_m,i} \partial_{z_m} f_{N}^{i_1,\dots, i_{k}, i}(t,z - r {w_{N;i}^{i_1,\dots, i_{k}, i}}) \;\rd r \bigg) \;\rd z_{k+1}
\\
&\quad = \sum_{m=1}^{k} - \partial_{z_m} \bigg[ \sum_{i \neq i_1,\dots,i_{k}} w_{i_m,i;N} \int_{\R} \nu(z_{k+1}) \bigg( \int_0^1 f_{N}^{i_1,\dots, i_{k}, i}(t,z - r {w_{N;i}^{i_1,\dots, i_{k}, i}}) \;\rd r \bigg) \;\rd z_{k+1} \bigg],
\end{aligned}
\end{equation*}

changing it into an additional advection term $\partial_{z_m}[\dots]$ to the equation.

Introduce the simple identity

\begin{equation*}
\begin{aligned}
f_{N}^{i_1,\dots, i_{k}}(u - {w_{N;i_m}^{i_1,\dots, i_{k}}}) = f_{N}^{i_1,\dots, i_{k}}(u) - \big\{f_{N}^{i_1,\dots, i_{k}}(u) - f_{N}^{i_1,\dots, i_{k}}(u - {w_{N;i_m}^{i_1,\dots, i_{k}}})\big\},
\end{aligned}
\end{equation*}

and proceed to do the same for $f_{N}^{i_1,\dots, i_{k}, i}(z - r {w_{N;i}^{i_1,\dots, i_{k}, i}})$, so that
the marginal equations (<ref>) now read

\begin{equation} \label{eqn:hierarchy_equation_marginal}
\begin{aligned}
& \partial_t f_{N}^{i_1,\dots, i_{k}}(z_1,\dots,z_{k})
\\
&\quad = \sum_{m = 1}^{k} \Bigg\{ \bigg[ - \partial_{z_m}(\mu(z_m) f_{N}^{i_1,\dots, i_{k}}(z)) + \frac{\sigma^2}{2} \partial_{z_m}^2 f_{N}^{i_1,\dots, i_{k}}(z) - \nu(z_m) f_{N}^{i_1,\dots, i_{k}}(z)
\\
&\qquad + \delta_0(z_m) \bigg( \int_{\R} \nu(u_m) \Big( f_{N}^{i_1,\dots, i_{k}}(u) - \big\{f_{N}^{i_1,\dots, i_{k}}(u) - f_{N}^{i_1,\dots, i_{k}}(u - {w_{N;i_m}^{i_1,\dots, i_{k}}})\big\} \Big) \;\rd u_m \bigg)\bigg|_{\forall n \neq m,\, u_n = z_n} \bigg]
\\
& \qquad- \partial_{z_m} \bigg[ \sum_{i \neq i_1,\dots,i_{k}} w_{i_m,i;N} \int_{\R} \nu(z_{k+1}) \bigg( \int_0^1 f_{N}^{i_1,\dots, i_{k}, i}(z) \\
&\qquad\qquad\qquad\qquad\qquad- \big\{f_{N}^{i_1,\dots, i_{k}, i}(z) - f_{N}^{i_1,\dots, i_{k}, i}(z - r {w_{N;i}^{i_1,\dots, i_{k}, i}})\big\} \;\rd r \bigg) \;\rd z_{k+1} \bigg] \Bigg\},
\end{aligned}
\end{equation}

where we omit variable $t$ for simplicity.

By taking the time derivative to the definition of observables (<ref>), restated here

\begin{equation*}
\begin{aligned}
\tau_N (T,w_N,f_N)(t,z) \defeq \;& \frac{1}{N} \sum_{i_1,\dots, i_{|T|} = 1}^N w_{N,T}(i_1,\dots, i_{|T|}) f_{N}^{i_1,\dots, i_{|T|}}(t,z_1,\dots,z_{|T|})
\end{aligned}
\end{equation*}

and substituting the right hand side $\partial_t f_{N}^{i_1,\dots, i_{|T|}}$ by the marginal equation (<ref>) with $k = |T|$, we obtain that

\begin{equation*}
\begin{aligned}
& \partial_t \bigg( \frac{1}{N} \sum_{i_1,\dots, i_{|T|} = 1}^N w_{N,T}(i_1,\dots, i_{|T|}) f_{N}^{i_1,\dots, i_{|T|}}(z_1,\dots,z_{|T|}) \bigg)= \frac{1}{N} \sum_{i_1,\dots, i_{|T|} = 1}^N w_{N,T}(i_1,\dots, i_{|T|}) \sum_{m = 1}^{|T|} \Bigg\{ \\
&\ \bigg[ - \partial_{z_m}(\mu(z_m) f_{N}^{i_1,\dots, i_{|T|}}(z)) + \frac{\sigma^2}{2} \partial_{z_m}^2 f_{N}^{i_1,\dots, i_{|T|}}(z) - \nu(z_m) f_{N}^{i_1,\dots, i_{|T|}}(z)\\
&\ + \delta_0(z_m) \bigg( \int_{\R} \nu(u_m) \Big( f_{N}^{i_1,\dots, i_{|T|}}(u) - \big\{f_{N}^{i_1,\dots, i_{|T|}}(u)- f_{N}^{i_1,\dots, i_{|T|}}(u - {w_{N;i_m}^{i_1,\dots, i_{|T|}}})\big\} \Big) \;\rd u_m \bigg)\bigg|_{\forall n \neq m,\, u_n = z_n} \bigg]
\\
& - \partial_{z_m} \bigg[ \sum_{i \neq i_1,\dots,i_{k}} \!\!\!\! w_{i_m,i;N} \!\! \int_{\R} \! \nu(z_{|T|+1}) \! \bigg( \int_0^1 f_{N}^{i_1,\dots, i_{|T|}, i}(z) - \big\{f_{N}^{i_1,\dots, i_{|T|}, i}(z) \\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad - f_{N}^{i_1,\dots, i_{|T|}, i}(z - r {w_{N;i}^{i_1,\dots, i_{|T|}, i}})\big\} \rd r \bigg) \rd z_{|T|+1} \bigg] \Bigg\}.
\end{aligned}
\end{equation*}

Noticing the identity $w_{N,T + j}(i_1,\dots, i_{|T| + 1}) = w_{N,T}(i_1,\dots, i_{|T|}) w_{i_j, i_{|T|+1}}$,
we see that all the marginals, except the two terms of form
$\{f_N^{\dots}(\cdot) - f_N^{\dots}(\cdot - w)\}$, are expressed in the right way so they can be rewritten as observables, obtaining (<ref>) as the approximate hierarchy and (<ref>) as the explicit form of the remainders.

We now turn to the proof of Proposition <ref>. It is worth noting that the main Gronwall estimate could also be written in the probabilistic language of Itô calculus. However, we prefer to keep an approach and notation similar to the rest of the proofs presented.

To simplify the argument, we only present a formal calculation where the tensorized weight $\eta^{\otimes |T|}$ is directly used as the test function, while, strictly speaking, the valid test functions for distributional solutions should have compact support. Given that the remaining coefficients are bounded Lipschitz and all terms in the subsequent calculation are non-negative, passing the limit to justify the use of unbounded weight on the dual side poses no problems.

The weighted total variation $\||\tau_N|(T) \eta^{\otimes |T|}\|_{\mathcal{M}(\R^{|T|})}$ can be decomposed as

\begin{equation*}
\begin{aligned}
\||\tau|(T)(t,\cdot) \eta^{\otimes |T|}\|_{\mathcal{M}(\R^{|T|})} = \;& \int_{\R^{|T|}} \frac{1}{N} \sum_{i_1,\dots, i_{|T|} = 1}^N \big| w_{N,T}(i_1,\dots, i_{|T|}) \big| f_{N}^{i_1,\dots, i_{|T|}}(t,z) \eta^{\otimes |T|}(z) \;\rd z
\\
= \;& \frac{1}{N} \sum_{i_1,\dots, i_{|T|} = 1}^N \big| w_{N,T}(i_1,\dots, i_{|T|}) \big| \int_{\R^{|T|}} f_{N}^{i_1,\dots, i_{|T|}}(t,z) \eta^{\otimes |T|}(z) \;\rd z.
\end{aligned}
\end{equation*}

For any distinct indexes $i_1,\dots,i_k$, we have

\begin{equation*}
\begin{aligned}
\int_{\R^{|T|}} f_{N}^{i_1,\dots, i_{|T|}}(t,z) \eta^{\otimes |T|}(z) \;\rd z = \int_{\R^N} f_{N}(t,x) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \;\rd x.
\end{aligned}
\end{equation*}

The forthcoming estimate is not exclusive to our specific choice $\eta = \eta_\alpha$, but for any weight function adhering to the form

\begin{equation*}
\begin{aligned}
\eta(x) = \exp( h(x)), \quad \forall x \in \R
\end{aligned}
\end{equation*}

such that $\|h'\|_{L^\infty}$, $\|h''\|_{L^\infty}$ are bounded and $h(0) \leq h(x)$.
Our choice of $\eta = \eta_\alpha$ is clearly included by choosing
$h(x) = \sqrt{1 + \alpha^2 x^2}$, resulting in $\|h'\|_{L^\infty} \leq \alpha$ and $\|h''\|_{L^\infty} \leq \alpha^2$.
The following inequalities are immediate results by chain rule and fundamental theorem of calculus.

For any weight function of form $\eta(x) = \exp( h(x))$ such that $\|h'\|_{L^\infty}$, $\|h''\|_{L^\infty}$ are bounded and $h(0) \leq h(x)$, one has that

\begin{equation*}
\begin{aligned}
|\eta'/\eta|(x) \leq \|h'\|_{L^\infty}, \quad |\eta''/\eta|(x) \leq \|h''\|_{L^\infty} + \|h'\|_{L^\infty}^2,
\end{aligned}
\end{equation*}

\begin{equation*}
\begin{aligned}
\eta(x + y) - \eta(x) %\leq [ \exp(\|h'\|_{L^\infty} |y|) - 1 ] \eta(x)
\leq \|h'\|_{L^\infty} |y| \exp(\|h'\|_{L^\infty} |y|) \eta(x).
\end{aligned}
\end{equation*}

The last inequality can be extended to the tensorized case $\eta^{\otimes k}(x) = \prod_{l = 1}^{k} \eta(x_{i_l})$ as

\begin{equation*}
\begin{aligned}
\eta^{\otimes k}(x + y) - \eta^{\otimes k}(x) %\leq [ \exp(\|h'\|_{L^\infty} \|y\|_{\ell^1}) - 1 ] \eta^{\otimes k}(x)
\leq \|h'\|_{L^\infty} \|y\|_{\ell^1} \exp(\|h'\|_{L^\infty} \|y\|_{\ell^1}) \eta^{\otimes k}(x).
\end{aligned}
\end{equation*}

We are now ready to prove Proposition <ref> under the more general assumption that $\eta(x) = \exp( h(x))$.
Since $f_N$ solves (<ref>) in the distributional sense, it is easy to verify that

\begin{equation*}
\begin{aligned}
& \int_{\R^N} f_{N}(t,x) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \;\rd x
= \int_{\R^N} f_{N}(0,x) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \;\rd x
\\
&\quad + \int_0^t \int_{\R^N} f_{N}(s,x) \Bigg[ \sum_{m=1}^{|T|} \bigg( \mu(x_{i_m}) (\eta'/\eta) (x_{i_m}) + \frac{1}{2}\sigma^2 (\eta''/\eta) (x_{i_m}) \bigg) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})}
\\
&\quad + \sum_{j = i_1,\dots, i_{|T|}} \nu(x_j) \bigg( \frac{\eta(0)}{\eta(x_{j})} {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l} + w_{i_l,j;N})} - {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \bigg)
\\
&\quad + \sum_{j \neq i_1,\dots, i_{|T|}} \nu(x_j) \bigg( {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l} + w_{i_l,j;N})} - {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \bigg)
\Bigg] \;\rd x \rd s.
\end{aligned}
\end{equation*}

By Lemma <ref>, we have that

\begin{equation*}
\begin{aligned}
& \int_{\R^N} f_{N}(t,x) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \;\rd x
\leq \int_{\R^N} f_{N}(0,x) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \;\rd x
\\
&\qquad + \int_0^t \int_{\R^N} f_{N}(s,x) \Bigg[ \sum_{m=1}^{|T|} \bigg( \|\mu\|_{L^\infty} \|h'\|_{L^\infty} + \frac{1}{2}\sigma^2 (\|h''\|_{L^\infty} + \|h'\|_{L^\infty}^2) \bigg) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})}
\\
&\qquad + \sum_{j = 1}^{N} \|\nu\|_{L^\infty} \; \|h'\|_{L^\infty} {\textstyle \sum_{m=1}^{|T|} |w_{i_m,j;N}|} \exp\Big(\|h'\|_{L^\infty} \; {\textstyle \max_{j} \sum_{i} |w_{i,j;N}|}\Big) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \Bigg] \;\rd x \rd s
\\
&\quad= \int_{\R^N} f_{N}(0,x) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \;\rd x
+ \Bigg[ \sum_{m=1}^{|T|} \bigg( \|\mu\|_{L^\infty} \|h'\|_{L^\infty} + \frac{1}{2}\sigma^2 (\|h''\|_{L^\infty} + \|h'\|_{L^\infty}^2) \bigg)
\\
& \qquad+ \sum_{j = 1}^{N} \sum_{m=1}^{|T|} |w_{i_m,j;N}| \; \|\nu\|_{L^\infty} \|h'\|_{L^\infty} \exp\Big(\|h'\|_{L^\infty} \; {\textstyle \max_{j} \sum_{i} |w_{i,j;N}|}\Big) \Bigg] \int_0^t \int_{\R^N} f_{N}(s,x) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \;\rd x \rd s,
\end{aligned}
\end{equation*}

where the summations of $j = i_1,\dots, i_{|T|}$ and $j \neq i_1,\dots, i_{|T|}$ are combined together by the simple fact that $h(0) \leq h(x_j)$, hence ${\eta(0)}/{\eta(x_{j})} \leq 1$.

Furthermore, we have that

\begin{equation*}
\begin{aligned}
\sum_{j = 1}^{N} \sum_{m=1}^{|T|} |w_{i_m,j;N}| \leq |T| \; {\textstyle \max_{i} \sum_{j} |w_{i,j;N}|}.
\end{aligned}
\end{equation*}

Hence by choosing

\begin{equation*}
\begin{aligned}
C_\mathcal{W} = \;& \textstyle \max\left(\max_{i} \sum_{j} |w_{i,j;N}| ,\ \max_{j} \sum_{i} |w_{i,j;N}|\right),
\\
A_\eta = \;& \Big( \|\mu\|_{L^\infty} \|h'\|_{L^\infty} + \frac{1}{2}\sigma^2 (\|h''\|_{L^\infty} + \|h'\|_{L^\infty}^2) + \|\nu\|_{L^\infty} \|h'\|_{L^\infty} C_{\mathcal{W}} \exp(\|h'\|_{L^\infty} C_{\mathcal{W}}) \Big),
\end{aligned}
\end{equation*}

we conclude that

\begin{equation*}
\begin{aligned}
& \int_{\R^N} f_{N}(t,x) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \;\rd x \\
&\quad \leq \int_{\R^N} f_{N}(0,x) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \;\rd x + \int_0^t |T| A_\eta \int_{\R^N} f_{N}(s,x) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \;\rd x \;\rd s.
\end{aligned}
\end{equation*}

By Gronwall lemma, this implies that

\begin{equation*}
\begin{aligned}
\;& \int_{\R^N} f_{N}(t,x) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \;\rd x \leq \exp \big( |T| A_\eta t \big) \int_{\R^N} f_{N}(0,x) {\textstyle \prod_{l = 1}^{|T|} \eta(x_{i_l})} \;\rd x.
\end{aligned}
\end{equation*}

Taking the summation over $i_1,\dots, i_{|T|}$, we have that

\begin{equation*}
\begin{aligned}
\||\tau|(T) \eta^{\otimes |T|}(t,\cdot)\|_{\mathcal{M}(\R^{|T|})} \leq \;& \exp \big( |T| A_\eta t \big) \||\tau|(T) \eta^{\otimes |T|}(0,\cdot)\|_{\mathcal{M}(\R^{|T|})}
\\
\leq \;& C_\eta \big( M_\eta \exp(A_\eta t_*) \big)^{|T|}
\end{aligned}
\end{equation*}

Finally, by applying Lemma <ref> to the left hand side, we immediately obtain (<ref>), restated here,

\begin{equation*}
\begin{aligned}
\| |\tau_N| (T) (t,\cdot) \|_{H^{-1\otimes |T|}_\eta} \leq C_\eta(T) \big(\|K\|_{L^2(\R)} \exp(A_\eta t_*) \big)^{|T|}
\end{aligned}
\end{equation*}

for all $T \in \mathcal{T}, t \in [0,t_*]$.

Finally, we give the proof of Proposition <ref>.

We show the well-posedness of Vlasov equation (<ref>)-(<ref>) by a classical fixed point argument. Let us first define the mapping $f \mapsto \mathcal{L} f$ as the solution of

\begin{equation*}
\begin{aligned}
\partial_t \mathcal{L} f(t,\xi,x) + \partial_x \Big(\mu^*_{f}(t,\xi,x) \mathcal{L} f(t,\xi,x) \Big) - \frac{\sigma^2}{2} \partial_{xx} \Big( \mathcal{L} f(t,\xi,x) \Big)
\\
+ \nu(x) \mathcal{L} f(t,\xi,x) - \delta_0(x) J_f(t,\xi) = 0
\end{aligned}
\end{equation*}

If $f$ is given, then $J_f$ and $\mu^*_{f}$ are determined, making the above identity a linear equation with respect to $\mathcal{L} f$.
We are going to see that if $f \in L^\infty([0,t_*] \times [0,1] ; H^{-1}_\eta \cap \mathcal{M}_+(\R) )$, then $\mathcal{L} f$ belongs to the same space.

By multiplying  the equation by the weight function $\eta$ and applying Leibniz formula, we obtain that

\begin{equation*}
\begin{aligned}
& \partial_t \mathcal{L} f(t,\xi,x) \eta(x)
\\
& \quad=- \partial_x \Big(\mu^*_{f}(t,\xi,x) \mathcal{L} f(t,\xi,x) \eta(x) \Big) + \frac{\sigma^2}{2} \partial_{xx} \Big( \mathcal{L} f(t,\xi,x) \eta(x) \Big) - \nu(x) \mathcal{L} f(t,\xi,x) \eta(x)\\
&\qquad + \delta_0(x) \eta(0) J_f(t,\xi) + \mu^*_{f}(t,\xi,x) (\eta'/\eta)(x) \mathcal{L} f(t,\xi,x) \eta(x)\\
&\qquad + \frac{\sigma^2}{2} \bigg[ - \partial_x \Big( 2(\eta'/\eta)(x) \mathcal{L} f(t,\xi,x) \eta(x) \Big) + (\eta''/\eta)(x) \mathcal{L} f(t,\xi,x) \eta(x) \bigg].
\end{aligned}
\end{equation*}

We start the a priori estimate of the linear mapping $\mathcal{L}$ by the total mass.
It is straightforward to verify that

\begin{equation} \label{eqn:a_priori_invariance_1}
\begin{aligned}
\| \mathcal{L}f(t,\cdot,\xi)\|_{\mathcal{M}(\R)} \leq \;& \| f(0,\cdot,\xi)\|_{\mathcal{M}(\R)} + \int_0^t J_f(s,\xi) \;\rd s
\\
\leq \;& \| f(0,\cdot,\xi)\|_{\mathcal{M}(\R)} + \int_0^t \|\nu\|_{L^\infty} \| f(s,\cdot,\xi) \|_{\mathcal{M}(\R)} \;\rd s.
\end{aligned}
\end{equation}

Note that by choosing $t_1 = 1/(2\|\nu\|_{L^\infty})$, we have that

\begin{equation*}
\begin{aligned}
\sup_{t \in [0,t_1]} \| f(t,\cdot,\xi)\|_{\mathcal{M}(\R)} \leq 2 \| f(0,\cdot,\xi)\|_{\mathcal{M}(\R)} \implies \sup_{t \in [0,t_1]} \| \mathcal{L}f(t,\cdot,\xi)\|_{\mathcal{M}(\R)} \leq 2 \| f(0,\cdot,\xi)\|_{\mathcal{M}(\R)}.
\end{aligned}
\end{equation*}

Next, consider the $\eta$-weighted total moment,

\begin{equation*}
\begin{aligned}
& \| \mathcal{L}f(t,\cdot,\xi) \eta\|_{\mathcal{M}(\R)}
\\
&\quad \leq \| f(0,\cdot,\xi) \eta\|_{\mathcal{M}(\R)} + \int_0^t \Big\{\eta(0)J_f(s,\xi) + \Big[ \|\mu_f^*(s,\cdot,\xi)\|_{L^\infty} \|\eta'/\eta\|_{L^\infty} + \frac{\sigma^2}{2} \|\eta''/\eta\|_{L^\infty} \Big] \\
&\hspace{330pt} \|\mathcal{L}f(s,\cdot,\xi) \eta\|_{\mathcal{M}(\R)}\Big\} \;\rd s
\\
&\quad \leq  \| f(0,\cdot,\xi) \eta\|_{\mathcal{M}(\R)} + \int_0^t \Big\{\eta(0)\|\nu\|_{L^\infty} \| f(s,\cdot,\xi) \|_{\mathcal{M}(\R)}
\\
& \qquad+ \Big[ \big( \|\mu\|_{L^\infty} + \|w\|_{\mathcal{W}} \|\nu\|_{L^\infty} \| f(s,\cdot,\cdot) \|_{L^\infty_\xi \mathcal{M}_x} \big) \|\eta'/\eta\|_{L^\infty} + \frac{\sigma^2}{2} \|\eta''/\eta\|_{L^\infty} \Big] \| \mathcal{L}f(s,\cdot,\xi) \eta\|_{\mathcal{M}(\R)}\Big\} \;\rd s.
\end{aligned}
\end{equation*}

By taking the supremum over $\xi \in [0,1]$, we have, for $t \in [0,t_1]$,

\begin{equation} \label{eqn:a_priori_invariance_2}
\begin{aligned}
& \| \mathcal{L}f(t,\cdot,\cdot) \eta\|_{L^\infty_\xi \mathcal{M}_x}
\leq \| f(0,\cdot,\cdot) \eta\|_{L^\infty_\xi \mathcal{M}_x} +  \int_0^t \eta(0)\|\nu\|_{L^\infty} \| f \|_{L^\infty_{t,\xi} \mathcal{M}_x}
\\
& \qquad + \Big[ \big( \|\mu\|_{L^\infty} + \|w\|_{\mathcal{W}} \|\nu\|_{L^\infty} \| f \|_{L^\infty_{t,\xi} \mathcal{M}_x} \big) \|\eta'/\eta\|_{L^\infty} + \frac{\sigma^2}{2} \|\eta''/\eta\|_{L^\infty} \Big] \| \mathcal{L}f(s,\cdot,\cdot) \eta\|_{L^\infty_\xi \mathcal{M}_x} \;\rd s
\\
& \quad \leq \bigg( \| f(0,\cdot,\cdot) \eta\|_{L^\infty_\xi \mathcal{M}_x} +  \int_0^t \eta(0)\|\nu\|_{L^\infty} \| f \|_{L^\infty_{t,\xi} \mathcal{M}_x} \;\rd s\bigg)
\\
& \qquad\qquad\exp \bigg( \Big[ \big( \|\mu\|_{L^\infty} + \|w\|_{\mathcal{W}} \|\nu\|_{L^\infty} \| f \|_{L^\infty_{t,\xi} \mathcal{M}_x} \big) \|\eta'/\eta\|_{L^\infty} + \frac{\sigma^2}{2} \|\eta''/\eta\|_{L^\infty} \Big] t \bigg),
\end{aligned}
\end{equation}

where the $L^\infty_t$ should be understood as the supremum over $t \in [0,t_1]$.

We construct the invariance set and show $\mathcal{L}$-contractivity on the set by the following procedure:
For any $ R > R_0 \defeq \| f(0,\cdot,\cdot) \eta\|_{L^\infty_\xi \mathcal{M}_x}$, and any $t_* > 0$, denote

\begin{equation*}
\begin{aligned}
E_{R;t} \defeq \{ f \in \mathcal{M}_+ :  \sup_{s \in [0,t]} \|f (s,\cdot,\cdot) \eta\|_{L^\infty_{\xi} \mathcal{M}_x} < R\}.
\end{aligned}
\end{equation*}

By taking sufficiently small $t_2$, for example

\begin{equation*}
\begin{aligned}
t_2 \leq \min\bigg( \frac{1}{2 \|\nu\|_{L^\infty}} , \frac{R - R_0}{2 \eta(0) \|\nu\|_{L^\infty} R} , \frac{\log \frac{2R}{R+R_0}}{\big( \|\mu\|_{L^\infty} + \|w\|_{\mathcal{W}} \|\nu\|_{L^\infty} R \big) \|\eta'/\eta\|_{L^\infty} + \frac{\sigma^2}{2} \|\eta''/\eta\|_{L^\infty}} \bigg),
\end{aligned}
\end{equation*}

we can make $E_{R;t_2}$ an invariance set, i.e. $\mathcal{L}(E_{R;t_2}) \subset E_{R;t_2}$.

To show that $f \mapsto \mathcal{L} f$ is contracting in the $H^{-1}_\eta$-sense, we consider the following energy estimate: Along each fiber $\xi \in [0,1]$,

\begin{equation*}
\begin{aligned}
& \frac{\rd}{\rd t} \bigg( \frac{1}{2} \int_{\R} \Big[ \Lambda \star \big((\mathcal{L} f - \mathcal{L} g) \eta\big) \Big] (\mathcal{L} f - \mathcal{L} g) \eta \;\rd x \bigg) = \int_{\R} \Big[ \Lambda \star \big( (\mathcal{L} f - \mathcal{L} g) \eta \big) \Big] \partial_t (\mathcal{L} f - \mathcal{L} g) \eta \;\rd x
\\
&\quad= \int_{\R} - \frac{\sigma^2}{2} \Big[ \Lambda \star \partial_x \big( (\mathcal{L} f - \mathcal{L} g) \eta\big) \Big] \Big[ \partial_x \big( (\mathcal{L} f - \mathcal{L} g) \eta\big) \Big]
\\
&\qquad + \Big[ \Lambda \star \partial_x \big( (\mathcal{L} f - \mathcal{L} g) \eta\big) \Big] \bigg[ \mu^*_{f} (\mathcal{L} f - \mathcal{L} g) \eta + (\mu^*_{f} - \mu^*_{g}) (\mathcal{L} g) \eta + \sigma^2 (\eta'/\eta) (\mathcal{L} f - \mathcal{L} g) \eta \bigg]
\\
&\qquad + \Big[ \Lambda \star \big( (\mathcal{L} f - \mathcal{L} g) \eta\big) \Big] \bigg[ - \nu (\mathcal{L} f - \mathcal{L} g) \eta + \delta_0 \eta(0) (J_f - J_g)
\\
& \qquad + \mu^*_{f} (\eta'/\eta) (\mathcal{L} f - \mathcal{L} g) \eta + (\mu^*_{f} - \mu^*_{g}) (\eta'/\eta) (\mathcal{L} g) \eta + \frac{\sigma^2}{2} (\eta''/\eta) (\mathcal{L} f - \mathcal{L} g) \eta \bigg] \;\rd x.
\end{aligned}
\end{equation*}

Apply Cauchy-Schwartz inequality, we obtain that

\begin{equation*}
\begin{aligned}
& \frac{\rd}{\rd t} \bigg( \int_{\R} \Big[ \Lambda \star \big((\mathcal{L} f - \mathcal{L} g) \eta\big) \Big] (\mathcal{L} f - \mathcal{L} g) \eta \;\rd x \bigg)
\\
&\quad \leq \frac{4}{\sigma^2} \|\mu^*_{f} (\mathcal{L} f - \mathcal{L} g)\|_{H^{-1}_\eta}^2 + \frac{4}{\sigma^2} \|(\mu^*_{f} - \mu^*_{g}) (\mathcal{L} g)\|_{H^{-1}_\eta}^2 + 4 \| (\eta'/\eta) (\mathcal{L} f - \mathcal{L} g)\|_{H^{-1}_\eta}^2
\\
& \qquad+ \Big( 4 + \frac{\sigma^2}{2} \Big) \|(\mathcal{L} f - \mathcal{L} g)\|_{H^{-1}_\eta}^2 + \|\nu (\mathcal{L} f - \mathcal{L} g)\|_{H^{-1}_\eta}^2 + \|\delta_0 (J_f - J_g)\|_{H^{-1}_\eta}^2
\\
& \qquad+ \|\mu^*_{f} (\eta'/\eta) (\mathcal{L} f - \mathcal{L} g)\|_{H^{-1}_\eta}^2 + \|(\mu^*_{f} - \mu^*_{g}) (\eta'/\eta) (\mathcal{L} g)\|_{H^{-1}_\eta}^2 + \frac{\sigma^2}{2} \|(\eta''/\eta) (\mathcal{L} f - \mathcal{L} g)\|_{H^{-1}_\eta}^2.
\end{aligned}
\end{equation*}

Applying Lemma <ref>, we further have that

\begin{equation*}
\begin{aligned}
& \frac{\rd}{\rd t} \| (\mathcal{L} f - \mathcal{L} g)\|_{H^{-1}_\eta}^2 \leq  \bigg( \frac{16}{\sigma^2} \|\mu^*_{f}\|_{W^{1,\infty}}^2 + 16 \|\eta'/\eta\|_{W^{1,\infty}}^2 + \Big( 4 + \frac{\sigma^2}{2} \Big)
\\
&\qquad + 4 \|\nu\|_{W^{1,\infty}}^2 + 4 \|\mu^*_{f}\|_{W^{1,\infty}}^2 \|\eta'/\eta\|_{W^{1,\infty}}^2 + 2 \sigma^2 \|\eta''/\eta\|_{W^{1,\infty}}^2 \bigg) \| (\mathcal{L} f - \mathcal{L} g)\|_{H^{-1}_\eta}^2
\\
&\qquad + \bigg( \frac{4}{\sigma^2} \| (\mathcal{L} g)\|_{H^{-1}_\eta}^2 + 4\|\eta'/\eta\|_{W^{1,\infty}}^2 \| (\mathcal{L} g)\|_{H^{-1}_\eta}^2 \bigg) |\mu^*_{f} - \mu^*_{g}|^2 + \|\delta_0\|_{H^{-1}_\eta}^2 |J_f - J_g|^2.
\end{aligned}
\end{equation*}

Now let us consider the integration over $\xi \in [0,1]$. Firstly, using that $w \in \mathcal{W}$ combined with classical interpolation,

\begin{equation*}
\begin{aligned}
& \int_{[0,1]} |\mu^*_{f}(t,\xi,x) - \mu^*_{g}(t,\xi,x)|^2 \;\rd \xi = \int_{[0,1]} \bigg( \int_{[0,1]} w(\xi, \zeta) \big( J_f(t,\zeta) - J_g(t,\zeta) \big) \;\rd \zeta \bigg)^2 \;\rd \xi
\\
&\qquad \leq  \|w\|_{\mathcal{W}}^2 \|J_f(t,\cdot) - J_g(t,\cdot)\|_{L^2_\xi}^2.
\end{aligned}
\end{equation*}

Secondly, by Lemma <ref>,

\begin{equation*}
\begin{aligned}
\big| J_f(t,\xi) - J_g(t,\xi) \big| \leq \bigg| \int_{\R} \nu(x) \big( f(t,\xi,x) - g(t,\xi,x) \big) \;\rd x \bigg| \leq C (\alpha) \|\nu\|_{W^{1,\infty}} \|f(t,\cdot,\xi) - g(t,\cdot,\xi)\|_{H^{-1}_\eta}.
\end{aligned}
\end{equation*}

Hence, we have that

\begin{equation*}
\begin{aligned}
& \|J_f(t,\cdot) - J_g(t,\cdot)\|_{L^2_\xi}^2 = \int_{[0,1]} \big| J_f(t,\xi) - J_g(t,\xi) \big|^2 \;\rd \xi\\
&\quad \leq  C(\alpha)^2 \|\nu\|_{W^{1,\infty}}^2 \int_{[0,1]} \|f(t,\cdot,\xi) - g(t,\cdot,\xi)\|_{H^{-1}_\eta}^2 \;\rd \xi
=  C(\alpha)^2 \|\nu\|_{W^{1,\infty}}^2 \|f - g\|_{L^2_\xi (H^{-1}_\eta)_x}^2.
\end{aligned}
\end{equation*}

Therefore, by integrating over $\xi \in [0,1]$,

\begin{equation} \label{eqn:Vlasov_Gronwall}
\begin{aligned}
\| (\mathcal{L} f - \mathcal{L} g)(t,\cdot,\cdot)\|_{L^2_\xi (H^{-1}_\eta)_x}^2
\leq \;& \int_0^t M_0 \| (\mathcal{L} f - \mathcal{L} g) (s,\cdot,\cdot)\|_{L^2_\xi (H^{-1}_\eta)_x}^2 +  M_1 \|(f - g) (s,\cdot,\cdot)\|_{L^2_\xi (H^{-1}_\eta)_x}^2 \;\rd s
\\
\leq \;& \exp(M_0 t) \int_0^t M_1 \|(f - g) (s,\cdot,\cdot)\|_{L^2_\xi (H^{-1}_\eta)_x}^2 \;\rd s
\end{aligned}
\end{equation}

where $M_0,M_1$ are required to satisfy that

\begin{equation*}
\begin{aligned}
M_0 \geq \;&\sup_{t \in [0,t_2]} \bigg( \frac{16}{\sigma^2} \|\mu^*_{f}\|_{L^\infty_\xi W^{1,\infty}_x}^2 + 16 \|\eta'/\eta\|_{W^{1,\infty}}^2 + \Big( 4 + \frac{\sigma^2}{2} \Big)
\\
\;&\quad + 4 \|\nu\|_{W^{1,\infty}}^2 + 4 \|\mu^*_{f}\|_{L^\infty_\xi W^{1,\infty}_x}^2 \|\eta'/\eta\|_{W^{1,\infty}}^2 + 2 \sigma^2 \|\eta''/\eta\|_{W^{1,\infty}}^2 \bigg)
\\
M_1 \geq \;& \sup_{t \in [0,t_2]} \bigg[ \bigg( \frac{4}{\sigma^2} \| (\mathcal{L} g)\|_{L^\infty_\xi (H^{-1}_\eta)_x}^2 + 4\|\eta'/\eta\|_{W^{1,\infty}}^2 \| (\mathcal{L} g)\|_{L^\infty_\xi (H^{-1}_\eta)_x}^2 \bigg) \|w\|_{\mathcal{W}}^2 + \|\delta_0\|_{H^{-1}_\eta}^2 \bigg] C(\alpha)^2 \|\nu\|_{W^{1,\infty}}^2.
\end{aligned}
\end{equation*}

In addition, by $w \in \mathcal{W}$ and Lemma <ref>, we can derive

\begin{equation*}
\begin{aligned}
\|\mu^*_{f}\|_{L^\infty_\xi W^{1,\infty}_x} \leq \;& \|\mu\|_{W^{1,\infty}_x} + \sup_{\xi \in [0,1]} \bigg| \int_0^1 w(\xi, \zeta) J_f(t,\zeta) \;\rd \zeta \bigg|
\\
\leq \;& \|\mu\|_{W^{1,\infty}_x} + \|w\|_{\mathcal{W}} \; \|J_f(t,\cdot)\|_{L^\infty}
\\
\leq \;& \|\mu\|_{W^{1,\infty}_x} + \|w\|_{\mathcal{W}} \; C (\alpha) \|\nu\|_{W^{1,\infty}} \|f\|_{L^\infty_\xi (H^{-1}_\eta)_x}.
\end{aligned}
\end{equation*}

When $f,g,\mathcal{L}f,\mathcal{L}g \in E_{R;t_2}$, by Lemma <ref>, we have that

\begin{equation*}
\begin{aligned}
\|f\|_{L^\infty_\xi (H^{-1}_\eta)_x} \leq \frac{R}{2}, \quad \|(\mathcal{L}g)\|_{L^\infty_\xi (H^{-1}_\eta)_x} \leq \frac{R}{2},
\end{aligned}
\end{equation*}

for $t \in [0,t_2]$.

Hence $M_0,\; M_1$ in (<ref>) can be chosen such that they only depend on $R$ and the regularity of the various fixed coefficients in the system. By choosing sufficiently small $t_* > 0$, for example,

\begin{equation*}
\begin{aligned}
t_* \leq \max\left(t_2,\ \frac{1}{3M_1},\ \frac{\log 2}{M_0}\right),
\end{aligned}
\end{equation*}

by (<ref>) we conclude that $\mathcal{L}$ is contracting on the set $\mathcal{L}(E_{R;t_*})$ for the $L^2_\xi (H^{-1}_\eta)_x$ norm.
Repeating the argument allows extending the weak solution to any finite time interval as usual, since the a priori estimates (<ref>) and (<ref>) do not blow up in finite time.

We now turn to the derivation of the limiting hierarchy.
Taking the derivative of $\tau_\infty(T) = \tau_\infty(T,w,f)$ in Definition <ref>, we first obtain

\begin{equation*}
\begin{aligned}
&\partial_t \tau_\infty(T,w,f)(t,z) =  \sum_{m=1}^{|T|} \bigg[ -\partial_{z_m}\Big( \mu(z_m) \tau_\infty(T)(t,z)\Big) + \frac{\sigma^2}{2} \partial_{z_m}^2 \tau_\infty(T)(t,z)
\\
&\qquad -\nu(z_m) \tau_\infty(T)(t,z) + \delta_0(z_m) \bigg( \int_{\R} \nu(u_m) \tau_\infty(T)(t,u) \bigg) \bigg|_{\forall n \neq m, u_n=z_n}
\\
& \qquad - \partial_{z_m} \bigg( \int_{[0,1]^{|T|}} w_T(\xi_1,\dots,\xi_{|T|}) f^{\otimes |T|}(t,z_1,\xi_1,\dots,z_{|T|},\xi_{|T|})
\\
& \qquad \bigg( \int_0^1 w(\xi_m,\xi_{|T|+1}) \int_{\R} \nu(z_{|T|+1}) f(t,z_{|T|+1},\xi_{|T|+1}) \;\rd z_{|T|+1} \rd \xi_{|T|+1} \bigg) \;\rd \xi_1,\dots,\xi_{|T|} \bigg) \bigg].
\end{aligned}
\end{equation*}

The last term can be rewritten by using the observables with one more leaf, resulting the limiting hierarchy (<ref>), restated here:

\begin{equation*}
\begin{aligned}
& \partial_t \tau_\infty (T)(t,z)
\\
&\quad=  \sum_{m = 1}^{|T|} \Bigg\{ \bigg[ - \partial_{z_m}(\mu(z_m) \tau_\infty (T)(t,z)) + \frac{\sigma^2}{2} \partial_{z_m}^2 \tau_\infty (T)(t,z)
\\
&\qquad - \nu(z_m) \tau_\infty (T)(t,z) + \delta_0(z_m) \bigg( \int_{\R} \nu(u_m) \tau_\infty (T)(t,u) \;\rd u_m \bigg)\bigg|_{\forall n \neq m,\, u_n = z_n} \bigg]
\\
&\qquad - \partial_{z_m} \bigg[ \int_{\R} \nu(z_{|T|+1}) \tau_\infty (T+m)(t,z) \;\rd z_{|T|+1} \bigg] \Bigg\}.
\end{aligned}
\end{equation*}

§.§ Quantitative stability

This subsection focuses on the proof of the main quantitative estimate of the article. The technical Lemma <ref> about recursive differential inequalities is given separately in the next subsection.

For simplicity, let us recall the notation

\begin{equation*}
\begin{aligned}
\nu_m = 1 \otimes \dots \otimes \nu \otimes \dots \otimes,
\end{aligned}
\end{equation*}

where $\nu$ appears in the $m$-th coordinate, i.e. $\nu_m(z) = \nu(z_m)$. The same convention applies to $\mu$ and $\eta$.

Define the difference $\Delta_N (T) (t,z) \defeq \tau_N (T) (t,z) - \tau_\infty (T) (t,z)$.
By subtracting (<ref>) from (<ref>), one has that

\begin{equation*}
\begin{aligned}
& \partial_t \Delta_N (T)(t,z) \\
&\quad = \sum_{m = 1}^{|T|} \Bigg\{ \bigg[ - \partial_{z_m}(\mu(z_m) \Delta_N (T)(t,z)) + \frac{\sigma^2}{2} \partial_{z_m}^2 \Delta_N (T)(t,z) \\
&\qquad - \nu(z_m) \Delta_N (T)(t,z) + \delta_0(z_m) \bigg( \int_{\R} \nu(u_m) \Big( \Delta_N (T)(t,u) + \mathscr{R}_{N,T,m} (t,u) \Big) \;\rd u_m \bigg)\bigg|_{\forall n \neq m,\, u_n = z_n} \bigg] \\
&\qquad - \partial_{z_m} \bigg[ \int_{\R} \nu(z_{|T|+1}) \Big( \Delta_N (T+m)(t,z) + \mathscr{\tilde R}_{N,T+m,|T|+1} (t,z) \Big) \;\rd z_{|T|+1} \bigg] \Bigg\}, \quad \forall T \in \mathcal{T}.
\end{aligned}
\end{equation*}

We highlight that, for any fixed $N < \infty$, the above equalities and later inequalities involving $\Delta_N (T)$ can be understood as recursive relations that holds on all $T \in \mathcal{T}$.
At a first glance, one may think that the approximate hierarchy (<ref>) is only defined for observables $\tau_N(T)$ with $|T| \leq N$.
Nevertheless, by our formal definition that $f_{N}^{i_1,\dots, i_k} \equiv 0$ if there are duplicated indices among $i_1,\dots,i_k$, it is easy to verify that for any tree $T$ such that $|T| > N$,

\begin{equation*}
\begin{aligned}
\tau_N (T,w_N,f_N)(t,z) \defeq \frac{1}{N} \sum_{i_1,\dots, i_{|T|} = 1}^N w_{N,T}(i_1,\dots, i_{|T|}) f_{N}^{i_1,\dots, i_{|T|}}(t, z_1,\dots,z_{|T|})
\equiv 0
\end{aligned}
\end{equation*}

as in each marginal there must be duplicated indices. By a similar discussion, we see that $\mathscr{R}_{N,T,m} \equiv 0$ and $\mathscr{\tilde R}_{N,T+m,|T|+1} \equiv 0$ when $|T| > N$. With these formal definition, it is then straightforward to show that approximate hierarchy (<ref>) holds for all $T \in \mathcal{T}$.

By multiplying by the weight function $\eta^{\otimes |T|}$ and integrating, we obtain that

\begin{equation*}
\begin{aligned}
& \Big( \partial_t \Delta_N (T)(t,z) \Big) \eta^{\otimes |T|}(z) =  \sum_{m = 1}^{|T|} \Bigg\{ - \partial_{z_m}\Big(\mu_m \Delta_N (T) \eta^{\otimes |T|}\Big)(t,z) + \frac{\sigma^2}{2} \partial_{z_m}^2 \Big( \Delta_N (T) \eta^{\otimes |T|}\Big)(t,z) \\
&\qquad - \Big(\nu_m \Delta_N (T) \eta^{\otimes |T|}\Big)(t,z) + \Big( \mu_m (\eta_m'/\eta_m) \Delta_N (T) \eta^{\otimes |T|} \Big)(t,z)\\
&\qquad + \delta_0(z_m) \eta(z_m) \bigg( \int_{\R} \Big( (\nu_m/\eta_m) ( \Delta_N (T) + \mathscr{R}_{N,T,m} ) \eta^{\otimes |T|} \Big)(t,u) \;\rd u_m \bigg)\bigg|_{\forall n \neq m,\, u_n = z_n}
\\
&\qquad - \partial_{z_m} \bigg[ \int_{\R} \Big( (\nu_{|T|+1}/\eta_{|T|+1}) ( \Delta_N (T+m) + \mathscr{\tilde R}_{N,T+m,|T|+1} ) \eta^{\otimes |T|+1} \Big)(t,z) \;\rd z_{|T|+1} \bigg] \\
&\qquad + \frac{\sigma^2}{2} \bigg[ \partial_{z_m} \Big( -2(\eta_m'/\eta_m) \Delta_N (T) \eta^{\otimes |T|}\Big) +  (\eta_m''/\eta_m) \Delta_N (T) \eta^{\otimes |T|}
\bigg](t,z) \Bigg\}.
\end{aligned}
\end{equation*}

Substituting $\big(\partial_t \Delta_N (T)\big) \eta^{\otimes |T|}$ in the right hand side of

\begin{equation*}
\begin{aligned}
\;& \frac{\rd}{\rd t} \bigg( \frac{1}{2} \int_{\R^{|T|}} \Big( K^{\otimes |T|} \star \big(\Delta_N (T)\eta^{\otimes |T|}\big) (t,z)\Big)^2 \; \rd z \bigg)
\\
= \;& \int_{\R^{|T|}} \bigg(K^{\otimes |T|} \star \big(\Delta_N (T)\eta^{\otimes |T|}\big) (t,z)\bigg) \bigg(K^{\otimes |T|} \star \big( \partial_t \Delta_N (T)\eta^{\otimes |T|} \big) (t,z)\bigg) \; \rd z,
\end{aligned}
\end{equation*}

yields the extensive expression

\begin{equation*}
\begin{aligned}
& \frac{\rd}{\rd t} \bigg( \frac{1}{2} \int_{\R^{|T|}} \Big( K^{\otimes |T|} \star \big(\Delta_N (T)\eta^{\otimes |T|}\big) (t,z)\Big)^2 \; \rd z \bigg)
\\
&\quad = \int_{\R^{|T|}} \sum_{m = 1}^{|T|} \Bigg\{ - \frac{\sigma^2}{2} \bigg[ \partial_{z_m} K^{\otimes |T|} \star \big( \Delta_N (T)\eta^{\otimes |T|} \big) (t,z) \bigg]^2
\\
&\qquad + \bigg[ K^{\otimes |T|} \star \big( \Delta_N (T)\eta^{\otimes |T|} \big) (t,z) \bigg] \bigg[- K^{\otimes |T|} \star \big( \nu_m \Delta_N (T)\eta^{\otimes |T|} \big)(t,z)
\\
&\qquad + K(z_m)\eta(0) \bigg( \int_{\R} K^{\otimes |T|} \star \big( (\nu_m/\eta_m) \Delta_N (T) \eta^{\otimes |T|} \big)(t,u) \;\rd u_m \bigg)\bigg|_{\forall n \neq m,\, u_n = z_n}
\\
&\qquad + K(z_m)\eta(0) \bigg( \int_{\R} K^{\otimes |T|} \star \big( (\nu_m/\eta_m) \mathscr{R}_{N,T,m} \eta^{\otimes |T|} \big) (t,u) \;\rd u_m \bigg)\bigg|_{\forall n \neq m,\, u_n = z_n}
\\
&\qquad + K^{\otimes |T|} \star \big( \mu_m (\eta_m'/\eta_m) \Delta_N (T) \eta^{\otimes |T|} \big)(t,z) + \frac{\sigma^2}{2} K^{\otimes |T|} \star \big( (\eta_m''/\eta_m) \Delta_N (T) \eta^{\otimes |T|} \big)(t,z)
\bigg]
\\
&\qquad + \bigg[ \partial_{z_m} K^{\otimes |T|} \star \big( \Delta_N (T) \eta^{\otimes |T|} \big) (t,z) \bigg] \bigg[ K^{\otimes |T|} \star \big( \mu_m \Delta_N (T) \eta^{\otimes |T|} \big)(t,z)
\\
&\qquad + \int_{\R} K^{\otimes |T|+1} \star \big( (\nu_{|T|+1}/\eta_{|T|+1}) \Delta_N (T+m) \eta^{\otimes |T|+1} \big) (t,z) \;\rd z_{|T|+1}
\\
&\qquad + \int_{\R} K^{\otimes |T|+1} \star \big( (\nu_{|T|+1}/\eta_{|T|+1}) \mathscr{\tilde R}_{N,T+m,|T|+1} \eta^{\otimes |T|+1} \big) (t,z) \;\rd z_{|T|+1}
\\
&\qquad + \frac{\sigma^2}{2} K^{\otimes |T|} \star \big( 2(\eta_m'/\eta_m) \Delta_N (T) \eta^{\otimes |T|} \big)(t,z)
\bigg] \Bigg\}
\; \rd z.
\end{aligned}
\end{equation*}

We then apply Cauchy-Schwartz inequality to obtain,

\begin{equation} \label{eqn:energy_Cauchy_Schwartz}
\begin{aligned}
& \frac{\rd}{\rd t} \bigg( \frac{1}{2} \|\Delta_N (T)\|_{H^{-1 \otimes |T|}_\eta}^2 \bigg)
\leq \sum_{m = 1}^{|T|} \Bigg\{ \Big(2 + \frac{\sigma^2}{4}\Big) \|\Delta_N (T)\|_{H^{-1 \otimes |T|}_\eta}^2 + \frac{1}{2} \|\nu_m \Delta_N (T)\|_{H^{-1 \otimes |T|}_\eta}^2
\\
&\quad + \frac{1}{2}\|\mu_m (\eta_m'/\eta_m) \Delta_N (T)\|_{H^{-1 \otimes |T|}_\eta}^2 + \frac{\sigma^2}{4} \|(\eta_m''/\eta_m)\Delta_N (T)\|_{H^{-1 \otimes |T|}_\eta}^2
\\
&\quad + \frac{1}{2} \|K\|_{L^2}^2 \eta(0)^2 \int_{\R^{|T|-1}} \bigg( \int_{\R} K^{\otimes |T|} \star \big( (\nu_m/\eta_m) \Delta_N (T) \eta^{\otimes |T|} \big)(t,z) \;\rd z_m \bigg)^2 \prod_{n \neq m} \;\rd z_n
\\
&\quad + \frac{1}{2} \|K\|_{L^2}^2 \eta(0)^2 \int_{\R^{|T|-1}} \bigg( \int_{\R} K^{\otimes |T|} \star \big( (\nu_m/\eta_m) \mathscr{R}_{N,T,m} \eta^{\otimes |T|} \big)(t,z) \;\rd z_m \bigg)^2 \prod_{n \neq m} \;\rd z_n
\\
&\quad + \frac{2}{\sigma^2} \|\mu_m \Delta_N (T)\|_{H^{-1 \otimes |T|}_\eta}^2 + \frac{\sigma^2}{2} \|2(\eta_m'/\eta_m) \Delta_N (T)\|_{H^{-1 \otimes |T|}_\eta}^2
\\
&\quad + \frac{2}{\sigma^2} \int_{\R^{|T|}} \bigg( \int_{\R} K^{\otimes |T|+1} \star \big( (\nu_{|T|+1}/\eta_{|T|+1}) \Delta_N (T+m) \eta^{\otimes |T|+1} \big) (t,z) \;\rd z_{|T|+1} \bigg)^2 \prod_{n=1}^{|T|} \;\rd z_n
\\
&\quad + \frac{2}{\sigma^2} \int_{\R^{|T|}} \bigg( \int_{\R} K^{\otimes |T|+1} \star \big( (\nu_{|T|+1}/\eta_{|T|+1}) \mathscr{\tilde R}_{N,T+m,|T|+1} \eta^{\otimes |T|+1} \big) (t,z) \;\rd z_{|T|+1} \bigg)^2 \prod_{n=1}^{|T|} \;\rd z_n \Bigg\}.
\end{aligned}
\end{equation}

This is where the proper choice of weak distance becomes critical as we need to bound the various terms in the right-hand side by the norm $ \|\Delta_N (T)\|_{H^{-1 \otimes |T|}_\eta}^2$. The commutator estimate in Lemma <ref> can directly bound all the terms with an explicit $H^{-1 \otimes |T|}_\eta$-norms as the coefficients $\mu,\nu$ are $W^{1,\infty}$ and $\eta$ is smooth. For example

\[
\|\nu_m \Delta_N (T)\|_{H^{-1 \otimes |T|}_\eta}^2\leq 4\,\|\nu\|_{W^{1,\infty}(\R)}^2\,\|\Delta_N (T)\|_{H^{-1 \otimes |T|}_\eta}^2.
\]

This leads to the simplified expression for some constant $\tilde C_0$,

\begin{equation} \label{eqn:energy_Cauchy_Schwartz2}
\begin{aligned}
& \frac{\rd}{\rd t} \bigg( \frac{1}{2} \|\Delta_N (T)\|_{H^{-1 \otimes |T|}_\eta}^2 \bigg)
\leq \sum_{m = 1}^{|T|} \Bigg\{ \tilde C_0\, \|\Delta_N (T)\|_{H^{-1 \otimes |T|}_\eta}^2
\\
&\quad + \frac{1}{2} \|K\|_{L^2}^2 \eta(0)^2 \int_{\R^{|T|-1}} \bigg( \int_{\R} K^{\otimes |T|} \star \big( (\nu_m/\eta_m) \Delta_N (T) \eta^{\otimes |T|} \big)(t,z) \;\rd z_m \bigg)^2 \prod_{n \neq m} \;\rd z_n
\\
&\quad + \frac{1}{2} \|K\|_{L^2}^2 \eta(0)^2 \int_{\R^{|T|-1}} \bigg( \int_{\R} K^{\otimes |T|} \star \big( (\nu_m/\eta_m) \mathscr{R}_{N,T,m} \eta^{\otimes |T|} \big)(t,z) \;\rd z_m \bigg)^2 \prod_{n \neq m} \;\rd z_n
\\
&\quad + \frac{2}{\sigma^2} \int_{\R^{|T|}} \bigg( \int_{\R} K^{\otimes |T|+1} \star \big( (\nu_{|T|+1}/\eta_{|T|+1}) \Delta_N (T+m) \eta^{\otimes |T|+1} \big) (t,z) \;\rd z_{|T|+1} \bigg)^2 \prod_{n=1}^{|T|} \;\rd z_n
\\
&\quad + \frac{2}{\sigma^2} \int_{\R^{|T|}} \bigg( \int_{\R} K^{\otimes |T|+1} \star \big( (\nu_{|T|+1}/\eta_{|T|+1}) \mathscr{\tilde R}_{N,T+m,|T|+1} \eta^{\otimes |T|+1} \big) (t,z) \;\rd z_{|T|+1} \bigg)^2 \prod_{n=1}^{|T|} \;\rd z_n \Bigg\}.
\end{aligned}
\end{equation}

The remaining integrals terms in (<ref>) can be bounded by first applying Lemma <ref> followed by Proposition <ref>.
For example, consider the first remainder term and write by Lemma <ref>,

\begin{equation*}
\begin{aligned}
& \int_{\R^{|T|-1}} \bigg( \int_{\R} K^{\otimes |T|} \star \big( (\nu_m/\eta_m) \mathscr{R}_{N,T,m} \eta^{\otimes |T|} \big)(t,z) \;\rd z_m \bigg)^2 \prod_{n \neq m} \;\rd z_n
\\
%= \;& \int_{\R^{|T|-1}} \Bigg( K^{\otimes |T|-1} \star \Big(\int_{\R} K \star_m \big( (\nu_m/\eta_m) \mathscr{R}_{N,T,m} \eta_m \big) \;\rd z_m \; \eta^{\otimes |T|-1} \Big) \Bigg)^2 \prod_{n \neq m} \;\rd z_n
%\leq \;& {\color{red} \int_{\R^{|T|-1}} \Bigg( K^{\otimes |T|-1} \star \Big( C (\alpha) \|\nu\|_{W^{1,\infty}} \Big(\int_{\R} K \star_m \mathscr{R}_{N,T,m} \eta_m \; \rd z_m\Big)^{1/2} \; \eta^{\otimes |T|-1} \Big) \Bigg)^2 \prod_{n \neq m} \;\rd z_n}
%\leq \;& C (\alpha)^2 \|\nu\|_{W^{1,\infty}}^2 \int_{\R^{|T|-1}} \int_{\R} \bigg( K^{\otimes |T|} \star \big( \mathscr{R}_{N,T,m} \eta^{\otimes |T|} \big)(t,z) \bigg)^2 \;\rd z_m  \prod_{n \neq m} \;\rd z_n
&\qquad \leq  C (\alpha)^2 \|\nu\|_{W^{1,\infty}}^2 \|\mathscr{R}_{N,T,m}\|_{H^{-1 \otimes |T|}_\eta}^2.
\end{aligned}
\end{equation*}

Next, apply Proposition <ref> to the right hand side to conclude that

\begin{equation*}
\begin{aligned}
& \int_{\R^{|T|-1}} \bigg( \int_{\R} K^{\otimes |T|} \star \big( (\nu_m/\eta_m) \mathscr{R}_{N,T,m} \eta^{\otimes |T|} \big)(t,z) \;\rd z_m \bigg)^2 \prod_{n \neq m} \;\rd z_n
\\
&\qquad\leq  C (\alpha)^2 \|\nu\|_{W^{1,\infty}}^2 \big[ \exp \big( (2 + 2\alpha) c(w,|T|) \big) - 1 \big] \||\tau_N|(T)\|_{H^{-1 \otimes |T|}_\eta}^2.
\end{aligned}
\end{equation*}

The method applies for the other integrals terms in (<ref>), which yields

\begin{equation*}
\begin{aligned}
& \int_{\R^{|T|-1}} \bigg( \int_{\R} K^{\otimes |T|} \star \big( (\nu_m/\eta_m) \Delta_N (T) \eta^{\otimes |T|} \big)(t,z) \;\rd z_m \bigg)^2 \prod_{n \neq m} \;\rd z_n
\\
&\qquad \leq  C (\alpha)^2 \|\nu\|_{W^{1,\infty}}^2 \|\Delta_N (T)\|_{H^{-1 \otimes |T|}_\eta}^2,
\\
& \int_{\R^{|T|}} \bigg( \int_{\R} K^{\otimes |T|+1} \star \big( (\nu_{|T|+1}/\eta_{|T|+1}) \Delta_N (T+m) \eta^{\otimes |T|+1} \big) (t,z) \;\rd z_{|T|+1} \bigg)^2 \prod_{n=1}^{|T|} \;\rd z_n
\\
&\qquad \leq C (\alpha)^2 \|\nu\|_{W^{1,\infty}}^2 \|\Delta_N (T+m)\|_{H^{-1 \otimes (|T|+1)}_\eta}^2,
\end{aligned}
\end{equation*}

together with

\begin{equation*}
\begin{aligned}
& \int_{\R^{|T|}} \bigg( \int_{\R} K^{\otimes |T|+1} \star \big( (\nu_{|T|+1}/\eta_{|T|+1}) \mathscr{\tilde R}_{N,T+m,|T|+1} \eta^{\otimes |T|+1} \big) (t,z) \;\rd z_{|T|+1} \bigg)^2 \prod_{n=1}^{|T|} \;\rd z_n
\\
&\qquad \leq  C (\alpha)^2 \|\nu\|_{W^{1,\infty}}^2 \big[ \exp \big( (2 + 2\alpha) c(w,|T|) \big) - 1 \big] \||\tau_N|(T)\|_{H^{-1 \otimes |T|}_\eta}^2.
\end{aligned}
\end{equation*}

Inserting those bounds into the energy estimate (<ref>), we obtain a recursive differential inequality: for all $T \in \mathcal{T}$,

\begin{equation} \label{eqn:differential_inequality_hierarchy}
\begin{aligned}
& \frac{\rd}{\rd t} \|\Delta_N (T) (t,\cdot) \|_{H^{-1 \otimes |T|}_\eta}^2
\leq  \sum_{m = 1}^{|T|} \Bigg\{ \tilde C_0 \| \Delta_N (T) (t,\cdot) \|_{H^{-1 \otimes |T|}_\eta}^2 + \tilde C_1 \| \Delta_N (T+m) (t,\cdot) \|_{H^{-1 \otimes (|T|+1)}_\eta}^2
\\
& \qquad + \varepsilon_0(T) \| |\tau_N| (T) (t,\cdot) \|_{H^{-1 \otimes |T|}_\eta}^2 + \varepsilon_1(T) \| |\tau_N| (T+m) (t,\cdot) \|_{H^{-1 \otimes (|T|+1)}_\eta}^2 \Bigg\},
\end{aligned}
\end{equation}

where we can even provide the explicit expressions for the constants

\begin{equation*}
\begin{aligned}
\;& \tilde C_0 = 4 + \frac{\sigma^2}{2} + 4\bigg(\|\nu\|_{W^{1,\infty}}^2 + \|\mu(\eta'/\eta)\|_{W^{1,\infty}}^2 + \frac{\sigma^2}{2} \|\eta''/\eta\|_{W^{1,\infty}}^2 + \frac{4}{\sigma^2} \|\mu\|_{W^{1,\infty}}^2 + 2\sigma^2 \|(\eta'/\eta)\|_{W^{1,\infty}}^2 \bigg)
\\
\;& \quad\quad + \|K\|_{L^2}^2 \eta(0)^2 C (\alpha)^2 \|\nu\|_{W^{1,\infty}}^2,
\\
\;& \tilde C_1 = \frac{4C(\alpha)^2}{\sigma^2} \|\nu\|_{W^{1,\infty}}^2,
\\
\;& \varepsilon_0(T) = \|K\|_{L^2}^2 \eta(0)^2 C (\alpha)^2 \|\nu\|_{W^{1,\infty}}^2 \big[ \exp \big( (2 + 2\alpha) c(w,|T|) \big) - 1 \big],
\\
\;& \varepsilon_1(T) = \frac{4 C (\alpha)^2}{\sigma^2} \|\nu\|_{W^{1,\infty}}^2 \big[ \exp \big( (2 + 2\alpha) c(w,|T|) \big) - 1 \big].
\end{aligned}
\end{equation*}

We can now restrict the recursion relations by truncating them at any given depth $n \geq 1$, meaning that we only consider the inequalities (<ref>) for all $T \in \mathcal{T}$ such that $|T| \leq n - 1$.
In such a case, since

\begin{equation*}
\begin{aligned}
c(w,|T|) \leq |T| \big(\max_{i,j}|w_{i,j;N}| \big) \leq n \bar w_N,
\end{aligned}
\end{equation*}

the coefficients $\varepsilon_0$, $\varepsilon_1$ can take the vanishing expression

\begin{equation*}
\begin{aligned}
\;& \varepsilon_0(n) = \|K\|_{L^2}^2 \eta(0)^2 C (\alpha)^2 \|\nu\|_{W^{1,\infty}}^2 \big[ \exp \big( (2 + 2\alpha) n \bar w_N \big) - 1 \big],
\\
\;& \varepsilon_1(n) = \frac{4 C (\alpha)^2}{\sigma^2} \|\nu\|_{W^{1,\infty}}^2 \big[ \exp \big( (2 + 2\alpha) n \bar w_N \big) - 1 \big].
\end{aligned}
\end{equation*}

For a fixed depth $n \geq 1$, $\varepsilon_0(n)$ and $\varepsilon_1(n)$ now vanish as $\bar w_N \to 0$.

Let us now rescale the energy inequality through some $\lambda^{|T|}$ factor: For all $T \in \mathcal{T}$ such that $|T| \leq n - 1$,

\begin{equation} \label{eqn:recursive_energy_estimate}
\begin{aligned}
& \frac{\rd}{\rd t} \lambda^{|T|} \|\Delta_N (T) (t,\cdot) \|_{H^{-1 \otimes |T|}_\eta}^2
\\
&\quad \leq  \sum_{m = 1}^{|T|} \Bigg\{ \tilde C_0 \lambda^{|T|} \| \Delta_N (T) (t,\cdot) \|_{H^{-1 \otimes |T|}_\eta}^2 + (\tilde C_1/\lambda) \lambda^{|T| + 1} \| \Delta_N (T+m) (t,\cdot) \|_{H^{-1 \otimes (|T|+1)}_\eta}^2
\\
&\qquad + \varepsilon_0(n) \lambda^{|T|} \| |\tau_N| (T) (t,\cdot) \|_{H^{-1 \otimes |T|}_\eta}^2 + (\varepsilon_1(n)/\lambda) \lambda^{|T| + 1} \| |\tau_N| (T+m) (t,\cdot) \|_{H^{-1 \otimes (|T|+1)}_\eta}^2 \Bigg\}.
\end{aligned}
\end{equation}

We also recall the a priori bound (<ref>) for $\tau_N,\tau_\infty$ assumed in Theorem <ref>:

\begin{equation*} %\label{eqn:hierarchy_boundedness_1}
% \||\tau_N|(\cdot,w_N,f_N)(t,\cdot)\|_{\lambda;K;\eta} \leq C_{\lambda;\eta}, \quad \forall t \in [0,t_*]
\sup_{t\leq t_*} \ \max_{|T| \leq \max(n,\ |T_*|)} \lambda^{\frac{|T|}{2}}\, \left(\||\tau_N|(T,w_N,f_N)(t,\cdot)\|_{H^{-1\otimes |T|}_\eta} %\leq ,  \quad \forall t \in [0,t_*],
% \|\tau_\infty(\cdot)(t,\cdot)\|_{\lambda;K;\eta} \leq C_{\lambda;\eta}, \quad \forall t \in [0,t_*].
+\|\tau_\infty(T)(t,\cdot)\|_{H^{-1\otimes |T|}_\eta}\right) \leq C_{\lambda;\eta}, %\leq C_{\lambda;\eta}, \quad \forall t \in [0,t_*],
\end{equation*}

where $T_* \in \mathcal{T}$ is the tree index in the final estimate (<ref>).

By a triangle inequality, this implies the following uniform bound of $\Delta_N$,

\begin{equation} \label{eqn:uniform_energy_bound}
\begin{aligned}
\sup_{t\leq t_*} \ \max_{|T| \leq \max(n,\ |T_*|)} \lambda^{|T|} \|\Delta_N (T) (t,\cdot) \|_{H^{-1\otimes |T|}_\eta}^2 \leq C_{\lambda;\eta}^2.
\end{aligned}
\end{equation}

\begin{equation*}
\begin{aligned}
M_k(t) = \;& \max_{|T| \leq k} \lambda^{|T|} \|\Delta_N (T) (t,\cdot) \|_{H^{-1\otimes |T|}_\eta}^2,
\\
C = \;& \tilde C_0 + \tilde C_1/\lambda,
\\
\varepsilon = \;& \big[ \varepsilon_0(n) + \varepsilon_1(n)/\lambda \big] C_{\lambda;\eta}^2,
\\
L = \;& C_{\lambda;\eta}^2,
\\
n = \;& n, \quad n' = |T_*|,
\end{aligned}
\end{equation*}

so that (<ref>) and (<ref>) can be summarized as follows,

\begin{align} \label{eqn:differential_inequality_recursion}
\frac{\rd}{\rd t} M_k(t)
\leq \;& k \Big( C M_{k+1}(t) + \varepsilon \Big), && \forall 1 \leq k \leq n-1,
\\ \label{eqn:bound_inequality_recursion}
M_k(t) \leq \;& L, && \forall 1 \leq k \leq \max(n,\ n'), \; t \in [0,t_*].
\end{align}

We now invoke the following result.

Consider a sequence of non-negative functions $(M_k(t))_{k=1}^\infty$ on $t \in [0,t_*]$ that satisfies the inequalities
(<ref>)-(<ref>) with $\big[ \varepsilon/CL + (2\theta)^n \big] \leq 1$.

\begin{equation} \label{eqn:convergence_inequality_hierarchy}
\begin{aligned}
\max_{1 \leq k \leq \max(n , \ n')} \big[ \theta^k M_k(t) \big]
\leq \;& L (Ct/\theta + 2) \,\max\left( \big[\varepsilon/CL + (2\theta)^n \big],\ \max_{1 \leq k \leq n - 1} \big[ \theta^k M_k(0) \big] / L \right)^{\frac{1}{p^{(Ct/\theta + 1)}}},
\end{aligned}
\end{equation}

holds for any $1 < p < \infty$, $0 < \theta < 2^{-p'}$ where $1/p + 1/p' = 1$, and any $t \in [0,t_*]$.

Assume for the time being that Lemma <ref> holds and apply it to (<ref>) and (<ref>).
Choose $p = 2$, $\theta = 1/8$ and substitute $\varepsilon, C, L$ by its explicit expression to find that

\begin{equation*}
\begin{aligned}
\varepsilon/CL = \frac{\varepsilon_0(n) + \varepsilon_1(n)/\lambda}{\tilde C_0 + \tilde C_1/\lambda} = C_1 \big[ \exp \big( (2 + 2\alpha) \bar w n \big) - 1 \big],
\end{aligned}
\end{equation*}

where $C_1$ depends only on $\lambda$, the $W^{1,\infty}$-regularity of coefficients $\mu$, $\nu$ and constant $\sigma > 0$ in (<ref>), but neither on $\bar w_N$ nor on $n$. Choosing $C_0 = C/\theta$, and as $\bar w_N\to 0$ as $N\to\infty$, we deduce that for $N$ large enough

\begin{equation*}
\begin{aligned}
\bar{\varepsilon} = \varepsilon/CL + (2\theta)^n = C_1 \big[ \exp \big( (2 + 2\alpha) n \bar w_N \big) - 1 \big] + (1/4)^n \leq 1.
\end{aligned}
\end{equation*}

The conclusion of Lemma <ref> hence holds, showing that

\begin{equation*}
\begin{aligned}
& \max_{|T| \leq \max(n , \ |T_*|)} (\lambda / 8)^{|T|} \| \tau_N(T,w_N,f_N)(t,\cdot) - \tau_\infty(T)(t,\cdot) \|_{H^{-1\otimes |T|}_\eta}^2
\\
&\quad \leq  C_{\lambda;\eta}^2\, \Big( C_0 t + 2 \Big)\,\max \left( \bar{\varepsilon},\  \max_{|T| \leq n-1} (\lambda / 8)^{|T|} \| \tau_N(T,w_N,f_N)(0,\cdot) - \tau_\infty(T)(0,\cdot) \|_{H^{-1\otimes |T|}_\eta}^2 / C_{\lambda;\eta}^2 \right)^{\frac{1}{2^{(C_0 t + 1)}}}.
\end{aligned}
\end{equation*}

This can be further simplified to (<ref>) by relaxing the maximum on the left hand side as $T = T_*$, taking the maximum on the right hand side over $|T| \leq \max(n,\ |T_*|)$, and choosing $C_2$ in $\eqref{eqn:stable_power_norm}$ as $C_2 = \max\big(C_0 t + 2, \ 2^{(C_0 t + 1)} \big)$.

§.§ Proof of Lemma <ref>

Let us restate here the recursive differential inequality (<ref>),

\begin{equation*}
\begin{aligned}
\frac{\rd}{\rd t} M_k(t)
\leq \;& k \Big( C M_{k+1}(t) + \varepsilon \Big), && \forall 1 \leq k \leq n-1,
\end{aligned}
\end{equation*}

which directly yields

\begin{equation*}
\begin{aligned}
\frac{\rd}{\rd t} \Big( M_k(t) + (\varepsilon/C) \Big)
\leq k C \Big( M_{k+1}(t) + (\varepsilon/C) \Big), && \forall 1 \leq k \leq n-1.
\end{aligned}
\end{equation*}

For any $1 \leq k \leq n - 1$ and $t \in [0,t_*]$, by inductively integrating the inequalities in time, we obtain that

\begin{equation*}
\begin{aligned}
\Big( M_k(t) + (\varepsilon/C) \Big) \leq \;&
C^{n - k} \int_s^t \begin{pmatrix} n-1 \\ k-1 \end{pmatrix} \frac{(t - r)^{n-k-1}}{n-k-1} \Big( M_n(r) + (\varepsilon/C) \Big) \;\rd r
\\
\;& + \sum_{l = k}^{n-1} C^{l - k} \begin{pmatrix} l-1 \\ k-1 \end{pmatrix} (t - s)^{l-k} \Big( M_l(s) + (\varepsilon/C) \Big),
\end{aligned}
\end{equation*}

We estimate the increase on $M_k$ within time steps of size

\begin{equation*}
\begin{aligned}
t-s = \theta / C.
\end{aligned}
\end{equation*}

First, we bound the constant terms,

\begin{equation*}
\begin{aligned}
& C^{n - k} \int_s^t \begin{pmatrix} n-1 \\ k-1 \end{pmatrix} \frac{(t - r)^{n-k-1}}{n-k-1} (\varepsilon/C) \;\rd r
+ \sum_{l = k}^{n-1} C^{l - k} \begin{pmatrix} l-1 \\ k-1 \end{pmatrix} (t - s)^{l-k} (\varepsilon/C)
\\
&\quad = (\varepsilon/C) \bigg\{ C^{n - k} \begin{pmatrix} n-1 \\ k-1 \end{pmatrix} (t - s)^{n-k-1}
+ \sum_{l = k}^{n-1} C^{l - k} \begin{pmatrix} l-1 \\ k-1 \end{pmatrix} (t - s)^{l-k} \bigg\}
\\
&\quad = (\varepsilon/C) \sum_{l = k}^n C^{l - k} \begin{pmatrix} l-1 \\ k-1 \end{pmatrix} (t - s)^{l-k}
\\
&\quad \leq \theta^{-k} (\varepsilon/C) \sum_{l = k}^n \begin{pmatrix} l-1 \\ k-1 \end{pmatrix} \theta^l,
\end{aligned}
\end{equation*}

where the last inequality uses our choice of time step $(t-s) \leq \theta / C$.

On the other hand, for $\theta \leq 1/2$,

\begin{equation*}
\begin{aligned}
\sum_{l = k}^{\infty} \begin{pmatrix} l-1 \\ k-1 \end{pmatrix} \theta^l = \frac{1}{(\theta^{-1} - 1)^{k}} \leq 1.
\end{aligned}
\end{equation*}

\[
C^{n - k} \int_s^t \begin{pmatrix} n-1 \\ k-1 \end{pmatrix} \frac{(t - r)^{n-k-1}}{n-k-1} (\varepsilon/C) \;\rd r
+ \sum_{l = k}^{n-1} C^{l - k} \begin{pmatrix} l-1 \\ k-1 \end{pmatrix} (t - s)^{l-k} (\varepsilon/C)\leq \theta^{-k} \,\frac{\varepsilon}{C}.
\]

We now turn to the terms involving $M_l(s)$ and $M_n(r)$ (with $s \leq r \leq t$).
For $M_n(r)$ we have no choice but to take

\begin{equation*}
\begin{aligned}
M_n(r) \leq L
\end{aligned}
\end{equation*}

But for $M_l(s)$, $k \leq l \leq n-1$, we have

\begin{equation*}
\begin{aligned}
M_l(s) \leq \min\left(L,\ \max_{1 \leq m \leq n-1} \big[ \theta^m M_m(s) \big] \theta^{-l}\right),
\end{aligned}
\end{equation*}

together with any geometric average between the two terms. Choose $\frac{1}{p} + \frac{1}{p'} = 1$ so that

\begin{equation*}
\begin{aligned}
M_l(s) \leq \;& L^{\frac{1}{p'}} \Big( \max_{1 \leq m \leq n-1} \big[ \theta^m M_m(s) \big] \theta^{-l} \Big)^{\frac{1}{p}}
\\
= \;& L^{\frac{1}{p'}} \max_{1 \leq m \leq n-1} \big[ \theta^m M_m(s) \big]^{\frac{1}{p}} \big( \theta^{\frac{1}{p}} \big)^{-l}.
\end{aligned}
\end{equation*}

Then we may write

\begin{equation*}
\begin{aligned}
& C^{n - k} \int_s^t \begin{pmatrix} n-1 \\ k-1 \end{pmatrix} \frac{(t - r)^{n-k-1}}{n-k-1} M_n(r) \;\rd r
+ \sum_{l = k}^{n-1} C^{l - k} \begin{pmatrix} l-1 \\ k-1 \end{pmatrix} (t - s)^{l-k} M_l(s),
\\
&\quad\leq C^{n - k} \begin{pmatrix} n-1 \\ k-1 \end{pmatrix} (t - s)^{n-k} L
+ \sum_{l = k}^{n-1} C^{l - k} \begin{pmatrix} l-1 \\ k-1 \end{pmatrix} (t - s)^{l-k} L^{\frac{1}{p'}} \max_{1 \leq m \leq n-1} \big[ \theta^m M_m(s) \big]^{\frac{1}{p}} \big( \theta^{\frac{1}{p}} \big)^{-l}
\\
&\quad \leq  \theta^{-k} \bigg\{ L \begin{pmatrix} n-1 \\ k-1 \end{pmatrix} \theta^n + L^{\frac{1}{p'}} \max_{1 \leq m \leq n-1} \big[ \theta^m M_m(s) \big]^{\frac{1}{p}} \sum_{l = k}^{n-1} \begin{pmatrix} l-1 \\ k-1 \end{pmatrix} \big( \theta^{\frac{1}{p'}} \big)^{-l} \bigg\},
\end{aligned}
\end{equation*}

where again use our choice of time step $(t-s) \leq \theta / C$ in the last inequality.

Observe that

\begin{equation*}
\begin{aligned}
\begin{pmatrix} n-1 \\ k-1 \end{pmatrix} \theta^n \leq 2^{n-1} \theta^n, \quad \sum_{l = k}^{n-1} \begin{pmatrix} l-1 \\ k-1 \end{pmatrix} \big( \theta^{\frac{1}{p'}} \big)^{-l} \leq \frac{1}{(\theta^{-\frac{1}{p'}} - 1)^{k}} \leq 1
\end{aligned}
\end{equation*}

when choosing $\theta^{\frac{1}{p'}} \leq 1/2$, so that

\[\begin{split}
&C^{n - k} \int_s^t \begin{pmatrix} n-1 \\ k-1 \end{pmatrix} \frac{(t - r)^{n-k-1}}{n-k-1} M_n(r) \;\rd r
+ \sum_{l = k}^{n-1} C^{l - k} \begin{pmatrix} l-1 \\ k-1 \end{pmatrix} (t - s)^{l-k} M_l(s)\\
&\qquad\leq \theta^{-k}\,\left(L (2\theta)^n + L^{\frac{1}{p'}} \max_{1 \leq m \leq n - 1} \big[ \theta^m M_m(s) \big]^{\frac{1}{p}} \right).
\end{split}\]

Combining those bounds, provided that $\theta^{\frac{1}{p'}} \leq 1/2$, we have that, for all $1 \leq k \leq n-1$,

\begin{equation*}
\begin{aligned}
M_k(t) \leq \;& \theta^{-k} \bigg\{(\varepsilon/C) + L (2\theta)^n + L^{\frac{1}{p'}} \max_{1 \leq m \leq n - 1} \big[ \theta^m M_m(s) \big]^{\frac{1}{p}} \bigg\}.
\end{aligned}
\end{equation*}

On the other hand, for $n \leq k \leq \max(n,\ n')$, we simply have $M_k(t) \leq L$. As $\theta^{-k + n} \geq 1$,

\begin{equation*}
\begin{aligned}
M_k(t) \leq L \leq \theta^{-k} \bigg\{L (2\theta)^n \bigg\},
\end{aligned}
\end{equation*}

and we can combine the two cases to obtain that

\begin{equation*}
\begin{aligned}
\max_{1 \leq k \leq \max(n,\ n')} \big[ \theta^k M_k(t) \big] \leq \;& (\varepsilon/C) + L (2\theta)^n + L^{\frac{1}{p'}} \max_{1 \leq k \leq n-1} \big[ \theta^k M_k(s) \big]^{\frac{1}{p}}.
\end{aligned}
\end{equation*}

If $t\leq \theta/C$ we are done but otherwise we need to sum up the various bounds. Denote $t_j=j\,\theta/C$ and write that By the fact that , we have that

\begin{equation*}
\begin{aligned}
& \max_{1 \leq k \leq \max(n,\ n')} \big[ \theta^k M_k(t_j) \big]\\
&\quad \leq  (\varepsilon/C) + L (2\theta)^n + L^{\frac{1}{p'}} \max_{1 \leq k \leq n-1} \big[ \theta^k M_k(t_{j-1}) \big]^{\frac{1}{p}}\\
&\quad \leq (\varepsilon/C) + L (2\theta)^n + L^{\frac{1}{p'}} \bigg\{ (\varepsilon/C) + L (2\theta)^n + L^{\frac{1}{p'}} \max_{1 \leq k \leq n-1} \big[ \theta^k M_k(t_{j-2}) \big]^{\frac{1}{p}} \bigg\}^{\frac{1}{p}}\\
&\quad \leq (\varepsilon/C) + L (2\theta)^n + L^{\frac{1}{p'}} \bigg\{ (\varepsilon/C) + L (2\theta)^n \bigg\}^{\frac{1}{p}} + L^{1-\frac{1}{p^2}} \max_{1 \leq k \leq n-1} \big[ \theta^k M_k(t_{j-2}) \big]^{\frac{1}{p^2}}\\ &\qquad \dots \\
&\quad \leq  \sum_{i=0}^{j-1} L^{1-\frac{1}{p^i}} \bigg\{ (\varepsilon/C) + L (2\theta)^n \bigg\}^{\frac{1}{p^i}} \; + \; L^{1-\frac{1}{p^j}} \max_{1 \leq k \leq n-1} \big[ \theta^k M_k(0) \big]^{\frac{1}{p^j}},
\end{aligned}
\end{equation*}

where we use that $(a+b)^{1/p} \leq a^{1/p} + b^{1/p}$ by concavity.

For any $t\geq 0$, we hence have with $j(t)=\bigg\lfloor \frac{C t}{\theta} \bigg\rfloor + 1$,

\begin{equation} \label{eqn:convergence_inequality_hierarchy_tight}
\begin{aligned}
\max_{1 \leq k \leq \max(n, \ n')} \big[ \theta^k M_k(t) \big] \leq \;& \sum_{i=0}^{j(t)-1} L^{1-\frac{1}{p^i}} \bigg\{ \varepsilon/C + L (2\theta)^n \bigg\}^{\frac{1}{p^i}} \; + \; L^{1-\frac{1}{p^{j(t)}}} \max_{1 \leq k \leq n-1} \big[ \theta^k M_k(0) \big]^{\frac{1}{p^{j(t)}}}.
\end{aligned}
\end{equation}

Finally, by the assumption that $\big[ \varepsilon/CL + (2\theta)^n \big] \leq 1$,

\begin{equation*}
\begin{aligned}
\forall i \leq j, \quad L^{1-\frac{1}{p^i}} \bigg\{ \varepsilon/C + L (2\theta)^n \bigg\}^{\frac{1}{p^i}} = L \bigg\{ \varepsilon/CL + (2\theta)^n \bigg\}^{\frac{1}{p^i}} \leq L \bigg\{ \varepsilon/CL + (2\theta)^n \bigg\}^{\frac{1}{p^j}}.
\end{aligned}
\end{equation*}

Hence we can replace every $i$ and every $j(t)$ in (<ref>) by $(Ct/\theta + 1)$, which gives the looser bound (<ref>), restated here

\begin{equation*}
\begin{aligned}
\max_{1 \leq k \leq \max(n, \ n')} \big[ \theta^k M_k(t) \big]
\leq \;& L (Ct/\theta + 2) \max\bigg( \big[\varepsilon/CL + (2\theta)^n \big] , \sup_{1 \leq k \leq n-1} \big[ \theta^k M_k(0) \big] / L \bigg)^{\frac{1}{p^{(Ct/\theta + 1)}}}.
\end{aligned}
\end{equation*}

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The discrete indices are extended to $[0,1]$ in the following way:
For any $N$-dim vector $v_N = (v_{1;N},\dots,v_{N;N})^{\top}$ and $N \times N$ matrix $w_N \defeq (w_{i,j;N})_{i,j =1}^N$, consider the extension to piecewise constant function and kernel over $[0,1]$ as

\begin{equation*}
\begin{aligned}
\tilde v_N(\xi) \defeq \;& \sum_{i = 1}^N w_{i;N} \mathbbm{1}_{[\frac{i-1}{N}, \frac{i}{N})}(\xi), && \forall \xi \in [0,1],
\\
\tilde w_N(\xi,\zeta) \defeq \;& \sum_{i,j = 1}^N N w_{i,j;N} \mathbbm{1}_{[\frac{i-1}{N}, \frac{i}{N})}(\xi) \mathbbm{1}_{[\frac{j-1}{N}, \frac{j}{N})}(\zeta), && \forall \xi,\zeta \in [0,1].
\end{aligned}
\end{equation*}

Then for the vector $u_N = (u_{1;N},\dots,u_{N;N})^{\top}$ given by matrix multiplication $u_N = w_N v_N$, one has

\begin{equation*}
\begin{aligned}
\tilde u_N(\xi) = \int_{\zeta} \tilde w_N(\xi,\zeta) v_N(\zeta) \;\rd \zeta,
\end{aligned}
\end{equation*}

which translates matrix multiplication to kernel operation on $[0,1]$.