\\\leq\|\mfrak{f}^{(3),V}_{m_1,a,b}\|_{L^\infty([-M_V,M_V]^c)}^2\|\mfrak{g}(.,x_1)\|_{H^{n_1+1}(\R)}^2
\\\leq2\|\mfrak{f}^{(3),V}_{m_1,a,b}\|_{L^\infty([-M_V,M_V]^c)}^2 \left(1+4P^2\pi^2\|\rho_V\|_{L^\infty(\R)}C_3(V,n)\right )
\\\times\|\partial_2^{m_2}\dots\partial_{a-1}^{m_{a-1}}\partial_{a+1}^{m_a}\dots\partial_{p+1}^{m_p}f(.,x_2,\dots,x_{a-1},x_1,x_a,\dots,x_p)\|_{H^{m_1+1}(\R)}^2.
\end{multline*}

We proceed in the exact same way for the second term and do the same thing on $]-\infty,-M_V]$, for every term we dealt with. Finally, by integrating with respect to $x_1$, collecting all the terms and then integrating over $x_1\in[-M_V,M_V]^c$ and over $x_2,\dots,x_p\in\R$, we get

$$\left\|\Theta^{(a)}\Big[\widetilde{\Xi^{-1}}[\partial_2^{m_2}\dots\partial_{a-1}^{m_{a-1}}\partial_{a+1}^{m_a}\dots\partial_{p+1}^{m_p}f]^{(m_1)}\Big]\right \|_{L^2([-M_V,M_V]^c\times\R^{p-1})}\leq C_{10}(V,n)\|f\|_{H^{n+1}(\R^{p+1})}$$

with $C_{10}(V,n)$ defined by

\begin{multline}\label{eq:C_10(V,m_1)}
C_{10}(V,n)^2:=C(n)\max_{m_1\leq n }\Bigg\{\big( \|\rho_V\|_{L^\infty(\R)}+4P^2\pi^2\|\rho_V\|_{L^\infty(\R)}^3C_\mcal{L}^2)\max_{\substack{0\leq a<m_1\\j\in\{1,2\}}}\|\mfrak{f}_{m_1,a,m_1-a}^{(j),V}\|_{L^\infty([-M_V,M_V]^c)}^2
\\+\left[1+4P^2\pi^2\|\rho_V\|_{L^\infty(\R)}C_3(V,n)\right]\max_{\substack{0\leq a <m_1\\0\leq b\leq m_1-a\\j\in\{1,3\}}}\|\mfrak{f}^{(j),V}_{m_1,a,b}\|_{L^\infty([-M_V,M_V]^c)}^2 \Bigg\}.
\end{multline}

It just remains to bound
$\displaystyle\int_{-M_V}^{M_V}\widetilde{\Xi^{-1}}[h(.,x_1)]^{(m_1)}(x_1)^2dx_1$. For that, we use (<ref>)

\begin{multline*}
\displaystyle\int\limits_{-M_V}^{M_V}\widetilde{\Xi^{-1}}[h(.,x_1)]^{(m_1)}(x_1)^2dx_1
\\\leq C(m_1)\displaystyle\int\limits_{-M_V}^{M_V}dx_1\Bigg\{\dfrac{S^{m_1}(\theta,\dots,\theta^{(m_1)})(x_1)^2}{\rho_V(x_1)^2}\left(\int\limits_{x_1}^{\mrm{sgn}(x_1)\infty}g(t,x_1)\rho_V(t)dt\right)^2
\\+\sum_{a=0}^{m_1-2}R_{m_1-a}^{m_1}\left(\theta,\dots,\theta^{(m_1-1-a)}\right)(x_1)^2[\partial_1^ag(x_1,x_1)]^2+[\partial_1^{m_1-1}g(x_1,x_1)]^2\Bigg\}.
\end{multline*}

By the same procedure as before, we first deal with $\mfrak{c}$ defined at the beginning of the proof, this yields, with $\mcal{I}_a$ defined in Theorem <ref>

\begin{multline*}
\int\limits_{-M_V}^{M_V}\mfrak{c}(x_1)^2dx_1\Bigg[\dfrac{S^{m_1}(\theta,\dots,\theta^{(m_1)})(x_1)^2}{\rho_V(x_1)^2}\left(\int\limits_{x_1}^{\mrm{sgn}(x_1)\infty}\rho_V(t)dt\right)^2+R_{m_1}^{m_1}\left(\theta,\dots,\theta^{(m_1-1-a)}\right)(x_1)^2 \Bigg]
\\\leq C(m_1)\|h\|_{H^1(\R^2)}^2\|\rho_V\|_{L^\infty(\R)}\left(1+4P^2\pi^2\|\rho_V\|_{L^\infty(\R)}^2C_\mcal{L} ^2\right)\Big\{ \|\mcal{I}_1^{V}\|_{\infty}^2\Big\|S^{m_1}(\theta,\dots,\theta^{(m_1)})\Big\|_{L^\infty([-M_V,M_V])}^2
\\+\Big\|R_{m_1}^{m_1}\left(\theta,\dots,\theta^{(m_1-1)}\right)\Big\|_{L^\infty([-M_V,M_V])}^2\Big\}.
\end{multline*}

Hence as before, we can replace $g$ by $\mfrak{g}$ and conclude with the last bounds:

\begin{multline*}\displaystyle\int\limits_{-M_V}^{M_V}\dfrac{S^{m_1}(\theta,\dots,\theta^{(m_1)})(x_1)^2}{\rho_V(x_1)^2}\left(\int\limits_{x_1}^{\mrm{sgn}(x_1)\infty}\mfrak{g}(t,x_1)\rho_V(t)dt\right)^2dx_1
\\\leq \|\mfrak{g}\|_{L^2(\R^2)}^2\Big\|\mcal{I}_2^{V}S^{m_1}(\theta,\dots,\theta^{(m_1)})\Big\|_{L^\infty([-M_V,M_V])}^2
\end{multline*}

and by Cauchy-Schwarz inequality,

\begin{multline*}
\int\limits_{-M_V}^{M_V}dx_1\Big(\sum_{a=0}^{m_1-2}R_{m_1-a}^{m_1}\left(\theta,\dots,\theta^{(m_1-1-a)}\right)(x_1)^2\partial_1^a\mfrak{g}(x_1,x_1)^2+\partial_1^{m_1-1}\mfrak{g}(x_1,x_1)^2\Big)
\\\leq\Bigg\{1+\max_{a\in\llbracket0,m_1-2\rrbracket}\Big\| R_{m_1-a}^{m_1}\left(\theta,\dots,\theta^{(m_1-1-a)}\right)\Big\|_{L^\infty([-M_V,M_V])}^2\Bigg\}
\\\times\displaystyle\sum_{a=0}^{m_1-1}\int\limits_{-M_V}^{M_V}dx_1\int\limits_{+\infty}^{x_1}dt\partial_1^{a+1}\mfrak{g}(t,x_1)\partial_1^{a}\mfrak{g}(t,x_1)
\\\leq C(m_1)\Bigg\{1+\max_{a\in\llbracket0,m_1-2\rrbracket}\Big\| R_{m_1-a}^{m_1}\left(\theta,\dots,\theta^{(m_1-1-a)}\right)\Big\|_{L^\infty([-M_V,M_V])}^2\Bigg\}\|\mfrak{g}\|_{H^{m_1}(\R^2)}^2.
\end{multline*}

Moreover, by Lemma <ref>

$$\|\mfrak{g}\|_{H^{m_1}(\R^2)}\leq\|h\|_{H^{m_1}(\R^2)}+2P\pi\left\|\rho_V\widetilde{\Xi^{-1}_1}[h]\right\|_{H^{m_1}(\R^2)}\leq \left[1+2P\pi C_3(V,n)\right]\|h\|_{H^{m_1}(\R^2)}.$$

We can then conclude that

\begin{multline*}
\int\limits_{-M_V}^{M_V}dx_1\Big(\sum_{a=0}^{m_1-2}R_{m_1-a}^{m_1}\left(\theta,\dots,\theta^{(m_1-1-a)}\right)(x_1)^2\partial_1^a\mfrak{g}(x_1,x_1)^2+\partial_1^{m_1-1}\mfrak{g}(x_1,x_1)^2\Big)
\\\leq C(m_1)\left[1+2P\pi C_3(V,n)\right]\Bigg\{1+\max_{a\in\llbracket0,m_1-2\rrbracket}\Big\| R_{m_1-a}^{m_1}\left(\theta,\dots,\theta^{(m_1-1-a)}\right)\Big\|_{L^\infty([-M_V,M_V])}^2\Bigg\}
\|h\|_{H^{m_1}(\R^2)}^2.
\end{multline*}

Thus by integrating with respect to $z\in\R$, we get,

$$\left \|\Theta^{(a)}\Big[\widetilde{\Xi^{-1}}[\partial^{m_2}_2\dots\partial_{a-1}^{m_{a-1}}\partial_{a+1}^{m_a}\dots\partial_{p+1}^{m_p}f]^{(m_1)}\Big]\right \|_{L^2([-M_V,M_V]\times\R^{p-1})}\leq C_{11}(V,n)\|\partial_3^{m_3}f\|_{H^{m_1+1}(\R^{p+1})}$$

with $C_{11}(V,n)>0$ defined by

\begin{multline}\label{eq:C_11(V,n)}
C_{11}(V,n)^2:=\max_{m_1\leq n}C(m_1)\Bigg\{\|\rho_V\|_{L^\infty(\R)}\left(1+4P^2\pi^2\|\rho_V\|_{L^\infty(\R)}^2C_\mcal{L} ^2\right)
\\\times\Big\{ \|\mcal{I}_1^{V}\|_{\infty}^2\Big\|S^{m_1}(\theta,\dots,\theta^{(m_1)})\Big\|_{L^\infty([-M_V,M_V])}^2+\Big\|R_{m_1}^{m_1}\left(\theta,\dots,\theta^{(m_1-1)}\right)\Big\|_{L^\infty([-M_V,M_V])}^2
\\+\left[1+2P\pi C_3(V,n)\right]^2\Bigg(\|\mcal{I}_2^{V}\|_{\infty}^2\Big\|S^{m_1}(\theta,\dots,\theta^{(m_1)})\Big\|_{L^\infty([-M_V,M_V])}^2
\\+1+\max_{a\in\llbracket0,m_1-2\rrbracket}\Big\| R_{m_1-a}^{m_1}\left(\theta,\dots,\theta^{(m_1-1-a)}\right)\Big\|_{L^\infty([-M_V,M_V])}^2\Bigg)  \Bigg\}.
\end{multline}
Collecting the bounds on the $L^2$-norms of $\Theta^{(a)}\Big[\widetilde{\Xi^{-1}}[\partial^{m_2}_2\dots\partial_{a-1}^{m_{a-1}}\partial_{a+1}^{m_a}\dots\partial_{p+1}^{m_p}f]^{(m_1)}\Big]$ on $[-M_V,M_V]^c\times\R$ and $[-M_V,M_V]\times\R$, we obtain:

$$\left \|\Theta^{(a)}\Big[\widetilde{\Xi^{-1}}[\partial^{m_2}_2\dots\partial_{a-1}^{m_{a-1}}\partial_{a+1}^{m_a}\dots\partial_{p+1}^{m_p}f]^{(m_1)}\Big]\right \|_{L^2(\R^p)}\leq 2\max_{i\in\{10,11\}}C_{i}(V,n)\|h\|_{H^{m_1+1}(\R^{p+1})}.$$

By combining the above equation together with (<ref>) and taking the supremum over $m\leq n$, we get

$$\Big\|\Theta^{(a)}\circ\widetilde{\Xi^{-1}_1}[f]\Big\|_{H^n(\R^{p})}\leq \mathtt{C}(\Theta^{(a)}\circ\widetilde{\Xi_1^{-1}},H^n)\|f\|_{H^{n+1}(\R^{p+1})}$$

with, $C_{10}$ and $C_{11}$ being given in (<ref>) and (<ref>), $\mcal{I}_a^{V}$ being given in (<ref>);

\begin{multline}\label{eq:C12}
\mathtt{C}(\Theta^{(a)}\circ\widetilde{\Xi_1^{-1}},H^n):=2\max_{i\in\{10,11\}}C_{i}(V,n)
\\+2\sqrt{(1+4P^2\pi^2\|\rho_V\|_{L^\infty(\R)}^2C_{\mcal{L}}^2)\Bigg(\|\mcal{I}_2^{V} \|_{L^\infty(\R)}^2+\|\rho_V\|_{L^\infty(\R)}\Big \|\mcal{I}_1^{V}\Big \|_{L^\infty(\R)}^2\Bigg)}.
\end{multline}

This yields the conclusion. The fact that, upon choosing the potential $V_{\phi,t}$ with $\phi\in\cap_{k\geq0}H^k(\R)$, $t\mapsto\mathtt{C}(\Theta^{(a)}\circ\widetilde{\Xi_1^{-1}},H^n)$ is continuous is shown in Proposition <ref>.

§ ASYMPTOTIC EXPANSION OF THE LINEAR STATISTICS

§.§ Loop equations for general functions

In the derivation of the loop equations, we will first prove the relations for $\psi\in\Xi\left(\mcal{C}_c^\infty(\R)\right)$ and extend them by density to more general functions. This is why we need the following lemma:

[Extension lemma]
$\Xi\left(\mcal{C}_c^\infty(\R)\right)$ is dense in $\bigcap_{k\geq0}H^k(\R)\cap\msf{H}$ for $\|.\|_{L^2(\mu_V)}$.

Furthermore $\Xi_1\left(\mcal{C}_c^\infty(\R^n)\right)$ is dense in for $\|.\|_{L^2(\mu_V\otimes dx\otimes\dots\otimes dx)}$
$$\msf{H}^{\infty}_n:=\bigcap_{k\geq0}\displaystyle H^k(\R^n)\cap\left\{u\in H^\infty(\R^n),\,\int_\R u(t,\xi_2,\dots,x_n)d\mu_V(t)=0\right\}.$$

This is the case because if $g\in H^\infty(\R)\cap\msf{H}$,  there exists a unique $f\in H^\infty(\R)\cap\mfrak{D}(\Xi)$ ($f$ is regular because of Theorem <ref>), such that $\Xi [f] =g$.  Since $\mcal{C}^\infty_c(\R)$ is dense in $H^\infty(\R)\cap\mfrak{D}(\Xi)\subset H^1(\R)$ for $\|.\|_{H^1(\R)}$, there exists $(f_q)_{q\geq0}\in\mcal{C}^\infty_c(\R)^\N$ such that $\|f_q-f\|_{H^1(\R)}$ goes to zero as $q$ goes to infinity. Hence:

\begin{multline*}
\|g-\Xi [f_q]\|_{L^2(\mu_V)}\leq \|\rho_V\|_{\infty}^{1/2}\|f'-f_q'\|_{L^2(\R)}+\left \|\rho_V\left(\dfrac{{\rho_V}\,'}{\rho_V}\right) \right \|_{\infty}^{1/2}\|f-f_q\|_{L^2(\R)}+2P\pi\|\rho_V\|_{\infty}\|f-f_q\|_{L^2(\R)}
\\\leq C\|f-f_q\|_{H^1(\R)}.
\end{multline*}

Since $f_q$ approximates $f$ then $\Xi [f_q]$ approximates $g$ hence $\Xi\left(\mcal{C}_c^\infty(\R)\right)$ is dense in $H^\infty(\R)\cap\msf{H}$ for $\|.\|_{L^2(\mu_V)}$. The second part of the lemma is a direct generalization.

We are now ready to state the loop equations, we recall the definition of a linear statistic was defined in (<ref>).

[Dyson-Schwinger equations]
The level $1$ Schwinger-Dyson equation holds for all $\psi_1\in \cap_{k\geq0}H^k(\R)$ and takes the form:

\begin{equation}\label{DSlvl1}
\Braket{\psi_1}_{\mcal{L}_N}=
\dfrac{P}{N}\Braket{\partial_1\widetilde{\Xi^{-1}}[\psi_1]}_{\mu_V}+\dfrac{P}{N}\Braket{\partial_1\widetilde{\Xi^{-1}}[\psi_1]}_{\mcal{L}_N}-P\Braket{\mcal{D}\circ\widetilde{\Xi^{-1}}[\psi_1]}_{\mcal{L}_N\otimes \mcal{L}_N}.
\end{equation}

For all $\psi_n\in\cap_{k\geq0}H^k(\R^n)$, the level $n>1$ Schwinger-Dyson equations reads:

\begin{multline}\label{DSlvlnthm}
\Braket{\psi_{n}}_{\overset{n}{\bigotimes}\mcal{L}_N}=\dfrac{P}{N}\Braket{\partial_1\widetilde{\Xi_1^{-1}}[\psi_{n}]}_{\mu_V\overset{n-1}{\bigotimes}\mcal{L}_N}+\dfrac{P}{N}\Braket{\partial_1\widetilde{\Xi_1^{-1}}[\psi_{n}]}_{\overset{n}{\bigotimes}\mcal{L}_N}-P\Braket{\mcal{D}_1\circ\widetilde{\Xi_1^{-1}}[\psi_{n}]}_{\overset{n+1}{\bigotimes}\mcal{L}_N}
\\-\dfrac{1}{N}\sum_{a=2}^{n}\Bigg(\Braket{\Theta^{(a)} \circ\widetilde{\Xi_1^{-1}}\big[\partial_a\psi_{n}\big] }_{\overset{n-1}{\bigotimes}\mcal{L}_N}+\Braket{\Theta^{(a)} \circ\widetilde{\Xi_1^{-1}}\big[\partial_a\psi_{n}\big] }_{\mu_V\overset{n-2}{\bigotimes}\mcal{L}_N}\Bigg).
\end{multline}

See for example [12][Prop 3.2.3] done in a similar context.

We take $\big(\phi^{(a)}\big)_{a=1}^{n+1}\in\mcal{C}_c^{\infty}(\R)^{n+1}$, set $V_{\{\varepsilon_a\}}(\lambda):=V(\lambda)+\displaystyle\sum_{a=2}^{n+1}\epsilon_a\phi^{(a)}(\lambda_a)$ and then define

\begin{equation}\label{key}
\end{equation}

The new normalisation constant $Z_N\big(\{\varepsilon_a\}\big)$ is such that $p_N^{(\{\varepsilon_a\})}$ is still a density probabilty on $\R^N$. We then define $G_t(\mu)=\mu+t\phi^{(1)}(\mu)$. $\partial_1\phi^{(1)}$ being bounded, for $t$ small enough $G_t$ can be considered a diffeomorphism over $\R$. By a change of variable $\lambda_a=G_t(\mu_a)$, we obtain :

\begin{equation}\label{key}
\end{equation}

Making an asymptotic expansion to the first order in $t$ of the right-hand side of this equality yields:

\begin{multline}\label{key}
1=\int_{\R^N}d^N\lambda.p_N^{(\{\varepsilon_a\})}(\bm{\lambda}).\Bigg\{1+t\sum_{a=1}^{N}\partial_1\phi^{(1)}(\lambda_a)\Bigg\}.\Bigg\{1+t\dfrac{P}{N}\sum_{a\neq b}\dfrac{\phi^{(1)}(\lambda_a)-\phi^{(1)}(\lambda_b)}{\lambda_a-\lambda_b}\Bigg\}\\.\Bigg\{1-t\sum_{a=1}^NV_{\{\varepsilon_a\}}'(\lambda_a)\phi^{(1)}(\lambda_a)\Bigg\}+O(t^2).
\end{multline}

Identifying the terms linear in $t$ leads to:

\begin{equation}
0=-N\Braket{\partial_1V_{\{\varepsilon_a\}}\phi^{(1)}}_{L_N}^{(\{\varepsilon_a\})}+PN\Braket{\mcal{D}[\phi^{(1)}]}_{L_N\otimes L_N}^{(\{\varepsilon_a\})}+(N-P)\Braket{\partial_1\phi^{(1)}}_{L_N}^{(\{\varepsilon_a\})}.
\end{equation}

By definition of $V_{\{\varepsilon_a\}}$, we get:

\begin{equation}
\Braket{\partial_1V\phi^{(1)}}_{L_N}^{(\{\varepsilon_a\})}+\sum_{a=2}^{n+1}\varepsilon_a\Braket{\phi^{(1)}\partial_1\phi^{(a)}}_{L_N}^{(\{\varepsilon_a\})}-P\Braket{\mcal{D}[\phi^{(1)}]}_{L_N\otimes L_N}^{(\{\varepsilon_a\})}-\Big(1-\dfrac{P}{N}\Big)\Braket{\partial_1\phi^{(1)}}_{L_N}^{(\{\varepsilon_a\})}=0.
\end{equation}

It becomes, after recentring the empirical measures against $\mu_V$ we obtain:

\begin{multline}\label{key}
\Braket{\partial_1V\phi^{(1)}}_{\mcal{L}_N}^{(\{\varepsilon_a\})}+\Braket{\partial_1V\phi^{(1)}}_{\mu_V}-P\Braket{\mcal{D}\left [\phi^{(1)}\right ]}_{\mcal{L}_N\otimes \mcal{L}_N}^{(\{\varepsilon_a\})}-2P\Braket{\mcal{D}\left [\phi^{(1)}\right ]}_{\mcal{L}_N\otimes \mu_V}^{(\{\varepsilon_a\})}-P\Braket{\mcal{D}\left [\phi^{(1)}\right ]}_{\mu_V\otimes \mu_V}
\\-\Big(1-\dfrac{P}{N}\Big)\left (\Braket{\partial_1\phi^{(1)}}_{\mcal{L}_N}^{(\{\varepsilon_a\})}+\Braket{\partial_1\phi^{(1)}}_{\mu_V}\right)+\sum_{a=2}^{n+1}\varepsilon_a\Big(\Braket{\phi^{(1)}\partial_1\phi^{(a)}}_{\mcal{L}_N}^{(\{\varepsilon_a\})}+\Braket{\phi^{(1)}\partial_1\phi^{(a)}}_{\mu_V}\Big)=0.
\end{multline}

Using the fact that for all signed measure $\nu$, $\Braket{\mcal{D}\left [\phi^{(1)}\right ]}_{\nu\otimes \mu_V}^{(\{\varepsilon_a\})}=-\Braket{\phi^{(1)}\mcal{H}[\rho_V]}_{\nu}^{(\{\varepsilon_a\})}+\Braket{\mcal{H}[\phi^{(1)}\rho_V]}_{\nu}^{(\{\varepsilon_a\})}$, that $V'(x)+2P\mcal{H}[\rho_V](x)=-\dfrac{\partial_1\rho_V}{\rho_V}(x)$ a.e, that $\Braket{\mcal{H}[\phi^{(1)}\rho_V]}_{\mu_V}=-\Braket{\phi^{(1)}\mcal{H}[\rho_V]}_{\mu_V}$ and
$\Braket{\dfrac{\partial_1\rho_V}{\rho_V}\phi^{(1)}}_{\mu_V}=-\Braket{\partial_1\phi^{(1)}}_{\mu_V}$ we get:

\begin{multline}
-\Braket{\dfrac{\partial_1\rho_V}{\rho_V}\phi^{(1)}}_{\mcal{L}_N}^{(\{\varepsilon_a\})}-\Braket{\partial_1\phi^{(1)}}_{\mcal{L}_N}^{(\{\varepsilon_a\})}-2P\Braket{\mcal{H}[\phi^{(1)}\rho_V]}_{\mcal{L}_N}^{(\{\varepsilon_a\})}+\dfrac{P}{N}\Braket{\partial_1\phi^{(1)}}_{\mcal{L}_N}^{(\{\varepsilon_a\})}-P\Braket{\mcal{D}[\phi^{(1)}]}_{\mcal{L}_N\otimes \mcal{L}_N}^{(\{\varepsilon_a\})}
\\+\dfrac{P}{N} \Braket{\partial_1\phi^{(1)}}_{\mu_V}+\sum_{a=2}^{n+1}\varepsilon_a\Big(\Braket{\phi^{(1)}\partial_1\phi^{(a)}}_{\mcal{L}_N}^{(\{\varepsilon_a\})}+\Braket{\phi^{(1)}\partial_1\phi^{(a)}}_{\mu_V}\Big)=0.
\end{multline}

which leads:

\begin{multline}\label{DSbuilding}
\Braket{\Xi\phi^{(1)}}_{\mcal{L}_N}^{(\{\varepsilon_a\})}=\dfrac{P}{N} \Braket{\partial_1\phi^{(1)}}_{\mu_V}+\dfrac{P}{N}\Braket{\partial_1\phi^{(1)}}_{\mcal{L}_N}^{(\{\varepsilon_a\})}-P\Braket{\mcal{D}\left [\phi^{(1)}\right ]}_{\mcal{L}_N\otimes \mcal{L}_N}^{(\{\varepsilon_a\})}
\\+\sum_{a=2}^{n+1}\varepsilon_a\Big(\Braket{\phi^{(1)}\partial_1\phi^{(a)}}_{\mcal{L}_N}^{(\{\varepsilon_a\})}+\Braket{\phi^{(1)}\partial_1\phi^{(a)}}_{\mu_V}\Big)
\end{multline}

Taking $\varepsilon_a$ go to 0 for all $a$, $\phi^{(1)}=\Xi^{-1}\psi$ leads to:

\begin{equation*}
\Braket{\Xi\phi^{(1)}}_{\mcal{L}_N}=\dfrac{P}{N} \Braket{\partial_1\phi^{(1)}}_{\mu_V}+\dfrac{P}{N}\Braket{\partial_1\phi^{(1)}}_{\mcal{L}_N}-P\Braket{\mcal{D}\left [\phi^{(1)}\right ]}_{\mcal{L}_N\otimes \mcal{L}_N}
\end{equation*}

By Lemma <ref>, we can take $\phi^{(1)}$ close to a general function $\phi\in H^\infty(\R)\cap\mfrak{D}(\Xi)$, which implies that $\Xi\phi^{(1)}\in\msf{H}$ will be close to $\Xi\phi$ in $L^2(\mu_V)$-norm. Since $\phi\mapsto	\Braket{\phi}_{\mcal{L}_N}$ is continuous with respect to $L^2(\mu_V)$ norm, the first term will converge to $\Braket{\Xi\phi}_{\mcal{L}_N}$. The two first terms are clearly continuous  with respect to $H^1(\R)$-norm and the last one is also continuous because of Theorem <ref>. We can then conclude that this equation holds for any $\phi\in H^\infty(\R)\cap\mfrak{D}(\Xi)$, ie taking $\psi=\Xi^{-1}\phi \in H^\infty(\R)\cap\msf{H}$ leads to:

\begin{equation*}
\Braket{\psi}_{\mcal{L}_N}=
\dfrac{P}{N}\Braket{\partial_1\Xi^{-1}\psi}_{\mu_V}+\dfrac{P}{N}\Braket{\partial_1\Xi^{-1}\psi}_{\mcal{L}_N}-P\Braket{\mcal{D}\circ\Xi^{-1}\psi}_{\mcal{L}_N\otimes \mcal{L}_N}
\end{equation*}

Furthermore, multiplying by $\dfrac{Z_N\big(\{\varepsilon_a\}\big)}{Z_N[V]}$ and applying $\partial_{\varepsilon_2}\dots\partial_{\varepsilon_{n+1}}$ to (<ref>) and evaluate at $\varepsilon_a=0$ leads to

\begin{multline}\label{DSbuildinglvl}
(-N)^{n}\Braket{\Xi\phi^{(1)} (\xi_1).\prod_{a=2}^{n+1}\phi^{(a)}(\xi_a)}_{\overset{n+1}{\bigotimes}\mcal{L}_N}=(-N)^{n}\dfrac{P}{N}\Braket{\partial_1\phi^{(1)}(\xi_1)\prod_{a=2}^{n+1}\phi^{(a)}(\xi_a)}_{\mu_V\overset{n}{\bigotimes}\mcal{L}_N}
\\(-N)^{n}\dfrac{P}{N}\Braket{\partial_1\phi^{(1)}(\xi_1)\prod_{a=2}^{n+1}\phi^{(a)}(\xi_a)}_{\overset{n+1}{\bigotimes}\mcal{L}_N}-P(-N)^{n}\Braket{\mcal{D}[\phi^{(1)}](\xi_1,\xi_2)\prod_{a=2}^{n+1}\phi^{(a)}(\xi_{a+1})}_{\overset{n+2}{\bigotimes}\mcal{L}_N}
\\+(-N)^{n-1}\sum_{a=2}^{n+1}\Bigg(\Braket{\phi^{(1)}(\xi_1)\partial_1\phi^{(a)}(\xi_1)\prod_{\substack{b=2\\b\neq a}}^{n+1}\phi^{(b)}(\xi_b)}_{\overset{n}{\bigotimes}\mcal{L}_N}+\Braket{\phi^{(1)}(\xi_1)\partial_1\phi^{(a)}(\xi_1)\prod_{\substack{b=2\\b\neq a}}^{n+1}\phi^{(b)}(\xi_b)}_{\mu_V\overset{n-1}{\bigotimes}\mcal{L}_N}\Bigg)
\end{multline}

Dividing by $(-N)^n$ and defining $\phi_{n+1}(\xi_1,\dots,\xi_{n+1}):=\prod_{a=1}^{n+1}\phi^{(a)}(\xi_a)\in\mcal{C}_c^{\infty}(\R)\otimes\dots\mcal{C}_c^{\infty}(\R)$ and using the operators defined earlier leads to

\begin{multline}\label{DSbuiltl}
\Braket{\Xi_1\phi_{n+1}}_{\overset{n+1}{\bigotimes}\mcal{L}_N}=\dfrac{P}{N}\Braket{\partial_1\phi_{n+1}}_{\mu_V\overset{n}{\bigotimes}\mcal{L}_N}+\dfrac{P}{N}\Braket{\partial_1\phi_{n+1}}_{\overset{n+1}{\bigotimes}\mcal{L}_N}-P\Braket{\mcal{D}_1[\phi_{n+1}]}_{\overset{n+2}{\bigotimes}\mcal{L}_N}
\\-\dfrac{1}{N}\sum_{a=2}^{n+1}\Bigg(\Braket{\Theta^{(a)}\left [\partial_a\phi_{n+1}\right ]}_{\overset{n}{\bigotimes}\mcal{L}_N}+\Braket{\Theta^{(a)}\left [\partial_a\phi_{n+1}\right ]}_{\mu_V\overset{n-1}{\bigotimes}\mcal{L}_N}\Bigg)
\end{multline}

It then not hard to see that, since $\mcal{C}_c^{\infty}(\R)\otimes\dots\mcal{C}_c^{\infty}(\R)$ is dense in $\mcal{C}_c^\infty(\R^{n+1})$ for $\|.\|_{W_1^\infty(\R^{n+1})}$, (<ref>) remains valid for all $\phi_{n+1}\in\mcal{C}_c^\infty(\R^{n+1})$.
Defining $\psi_{n+1}:=\Xi_1\phi_{n+1}$ this yields:

\begin{multline}\label{DSlvln}
\Braket{\psi_{n+1}}_{\overset{n+1}{\bigotimes}\mcal{L}_N}=\dfrac{P}{N}\Braket{\partial_1\Xi_1^{-1}\psi_{n+1}}_{\mu_V\overset{n}{\bigotimes}\mcal{L}_N}+\dfrac{P}{N}\Braket{\partial_1\Xi_1^{-1}\psi_{n+1}}_{\overset{n+1}{\bigotimes}\mcal{L}_N}-P\Braket{\mcal{D}_1\circ\Xi_1^{-1}\psi_{n+1}}_{\overset{n+2}{\bigotimes}\mcal{L}_N}
\\-\dfrac{1}{N}\sum_{a=2}^{n+1}\Bigg(\Braket{\Theta^{(a)} \circ\Xi_1^{-1}\partial_a\psi_{n+1} }_{\overset{n}{\bigotimes}\mcal{L}_N}+\Braket{\Theta^{(a)} \circ\Xi_1^{-1}\partial_a\psi_{n+1} }_{\mu_V\overset{n-1}{\bigotimes}\mcal{L}_N}\Bigg)
\end{multline}

Again by Lemma <ref>, we can take $\psi_{n+1}\in \msf{H}_{n+1}^{\infty}$ which concludes the proof.

[Regularity of the master operator]
Let $\psi\in\msf{H}$ such that $\mcal{K}_\kappa[\psi]\in W_n^\infty(\R)$ for all $\kappa>0$, $n>0$, then $\mcal{K}_{\kappa}\left [\Xi^{-1}\psi\right ]\in W_n^\infty(\R)$ for all $\kappa>0$. The same results holds for $\Xi^{-1}_1$.

$$\psi=\dfrac{\left(\rho_V\Xi^{-1}\psi\right)' }{\rho_V}+2P\mcal{H}\left [\Xi^{-1}\psi\rho_V\right ]=\left (\Xi^{-1}\psi\right )'+\dfrac{\rho_V'}{\rho_V}\Xi^{-1}\psi+2P\mcal{H}\left [\Xi^{-1}\psi\rho_V\right ]$$

By integrating, we get that

$$\Xi^{-1}\psi(x)=\dfrac{1}{\rho_V(x)}\int_x^{+\infty}dt\rho_V(t)\left[\psi(t)-2P\mcal{H}\left [\Xi^{-1}\psi\rho_V\right ](t)\right] $$

\begin{multline*}
|\Xi^{-1}\psi(x)|\leq\dfrac{1}{\rho_V(x)}\int_x^{+\infty}dt\rho_V(t)e^{\kappa V(t)}\left[\mcal{K}_\kappa[\psi](t)-2P\mcal{K}_\kappa\left[\mcal{H}\left [\Xi^{-1}\psi\rho_V\right ]\right ](t)\right]
\\\leq\left (\|[\mcal{K}_\kappa[\psi]\|_{W^\infty_0(\R)}+2P\|\mcal{K}_\kappa\left[\mcal{H}\left [\Xi^{-1}\psi\rho_V\right ]\right ]\|_{W^\infty_0(\R)}\right )\dfrac{1}{\rho_V(x)}\int_x^{+\infty}dt\rho_V(t)e^{\kappa V(t)}
\\\leq\left (\|[\mcal{K}_\kappa[\psi]\|_{W^\infty_0(\R)}+2P\|\mcal{K}_\kappa\left[\mcal{H}\left [\Xi^{-1}\psi\rho_V\right ]\right ]\|_{W^\infty_0(\R)}\right )\dfrac{1}{\rho_V(x)}\int_x^{+\infty}dt\rho_V(t)e^{\kappa V(t)}
\end{multline*}

Hence by integration by parts, one gets:

\begin{multline*}
|\mcal{K}_{\kappa}[\Xi^{-1}\psi](x)|\leq\left (\|[\mcal{K}_\kappa[\psi]\|_{W^\infty_0(\R)}+2P\|\mcal{K}_\kappa\left[\mcal{H}\left [\Xi^{-1}\psi\rho_V\right ]\right ]\|_{W^\infty_0(\R)}\right )\\\times\left\{\dfrac{1}{-(1-\kappa)V'(x)+2P\mcal{H}[\rho_V](x)}-\int_{x}^{+\infty}dt\underset{t\rightarrow+\infty}{O}\left (\rho_V(t)e^{\kappa V(t)}\right )\right\}
\end{multline*}

Since the last term is bounded, we conclude that $\mcal{K}_{\kappa}[\Xi^{-1}\psi]\in W_0^\infty(\R)$ for all $\kappa>0$. Hence, since

$$\left (\Xi^{-1}\psi\right )'=\psi-\dfrac{\rho_V'}{\rho_V}\Xi^{-1}\psi-2P\mcal{H}\left [\Xi^{-1}\psi\rho_V\right ]$$

we can conclude by Assumption <ref> that for all $\kappa'$, $\mcal{K}_{\kappa'}\left [\left(\Xi^{-1}\psi\right)'\right ]\in W_n^\infty(\R)$ if $\mcal{K}_{\kappa}[\psi]\in W_n^\infty(\R)$ for all $\kappa>0$. For that one needs to show that for any $\phi$ such that $\phi\in W_n^\infty(\R)$, $\mcal{K}_{\kappa}\left[\mcal{H}\left [\phi'\rho_V\right ]\right]\in W_n^\infty(\R)$, which is easy checkable by the fact that all its derivative (see <ref>), will be bounded.

It is clear that from Theorem <ref>,

$$|\Braket{f_{n}}_{\overset{n}{\bigotimes}\mcal{L}_N}|\leq\dfrac{C_n}{N^{\frac{n}{2}-\delta}}\left\|\mcal{K}_\kappa[f_n]\right\|_{W_n^{\infty}(\R^n)} $$

\begin{multline*}
\\\leq\dfrac{C_n'}{N^{\frac{n}{2}-\delta}}\left \|\mcal{K}_{\kappa'}[f_n]\right\|_{W_{n+1}^{\infty}(\R^n)}
\end{multline*}

for $0<\kappa'<\kappa$ using that $V'e^{-(\kappa-\kappa')V}$ is bounded.

§.§ Asymptotic expansion of linear statistics

The a priori bound on the linear statistics of Theorem <ref>, provides a starting point for obtaining the existence of their large-$N$ asymptotic expansion in powers of $N^{-1}$ up to any order through an analysis of the loop equations.

Let $\psi_k\in \displaystyle\bigcap_{m\geq0}H^m(\R^k)$, then for all integer $K$, there exists a sequence $(d_{a}^{(k),V})_{a\geq\lceil k/2\rceil}\in\R^N$ such that

$$\Braket{\psi_k}_{\overset{k}{\bigotimes}\mcal{L}_N}=\sum_{a=\lceil k/2\rceil}^{K}\dfrac{d_{a}^{(k),V}(\psi_k)}{N^a}+O\left (N^{-(K+1)}\right )$$

$$d_{1}^{(1),V}[\psi_1]=P\Braket{\partial_1\widetilde{\Xi^{-1}}[\psi_1]}_{\mu_V}+P\Braket{\Theta^{(2)} \circ\widetilde{\Xi_1^{-1}}\left[\partial_2\mcal{D}\circ\widetilde{\Xi^{-1}}[\psi_1]\right]  }_{\mu_V}.$$

Furthermore, there exists a sequence $\left(m_{K,k}\right) >0$, increasing in $K$, such that for all $k\geq1$ and $K\geq0$, all $\psi_k\in \displaystyle\bigcap_{m\geq0}H^m(\R^k)$,

\begin{equation}\label{eq:controleraminedercorrelator}
\Bigg|\Braket{\psi_k}_{\overset{k}{\bigotimes}\mcal{L}_N}-\sum_{a=\lceil k/2\rceil}^{K}\dfrac{d_{a}^{(k),V}(\psi_k)}{N^a}\Bigg|\leq \dfrac{\mathtt{C}_{\mrm{rem}}(V,K,k)}{N^{K+1}}\mcal{N}^{(k)}_{m_{K,k}}(\psi_k).
\end{equation}

Above $\mcal{N}_{m}^{(n)}(\psi_m):=\max\left(\|\psi_n\|_{W_n^{\infty}(\R^m)},\|\psi_n\|_{H^{n}(\R^m)}\right)$, while $\mathtt{C}_{\mrm{rem}}(V,K,k)>0$ is a constant depending on $V$, $K$ and $k$. Finally, under the choice of potential $V_{\phi,t}$ defined in Theorem <ref>, for $\phi\in\cap_{k\geq0}H^k(\R)$ the map $t\in[0,1]\mapsto\mathtt{C}_{\mrm{rem}}(V_{\phi,t},K,k)$ is continuous.
Using the first loop equation given in Theorem <ref>, we get:

\begin{equation}\label{premiere expansion}
\Braket{\psi_1}_{\mcal{L}_N}=
\dfrac{P}{N}\Braket{\partial_1\widetilde{\Xi^{-1}}[\psi_1]}_{\mu_V}+\dfrac{P}{N}\Braket{\partial_1\widetilde{\Xi^{-1}}[\psi_1]}_{\mcal{L}_N}-P\Braket{\mcal{D}\circ\widetilde{\Xi^{-1}}[\psi_1]}_{\mcal{L}_N\otimes \mcal{L}_N}.
\end{equation}

where we recall that $\widetilde{\Xi^{-1}}=\Xi^{-1}\circ\mcal{X}$ defined in <ref>.
The idea is to verify the hypotheses of Theorems <ref> for each function involved in the Dyson-Schwinger equations. By Proposition <ref> and Theorem <ref>, <ref>,  <ref>, <ref> and <ref> and the fact that $\psi_k\in\displaystyle\cap_{m\geq0}H^m(\R^k)$, we're ensured that all the norms are finite and that a $n$-linear statistic will be a $O\left (N^{-\frac{n}{2}(1-\varepsilon)}\right )$ where $\varepsilon>0$ is fixed but can be chosen arbitrarly small.

We show by induction on $K$ that there exists an asymptotic expansion up to $o(N^{-K})$ for any function $\psi_k\in \displaystyle\cap_{m\geq0}H^m(\R^k)$ for all $k\leq 2K$.

For $K=1$, since the first term in (<ref>) clearly contributes to the asymptotic expansion of $	\Braket{\psi_1}_{\mcal{L}_N}$ up to $o(N^{-1})$ so we focus on the two other terms. In (<ref>), the second term is clearly a $o(N^{-1})$ since by Theorem <ref>, Theorem <ref> and Theorem <ref>

\begin{multline*}
|\Braket{\partial_1\widetilde{\Xi^{-1}}[\psi_1]}_{\mcal{L}_N}|\leq C_{1,\varepsilon}e^{K_V}N^{-(1-\varepsilon)/2}\left(\|\partial_1\widetilde{\Xi^{-1}}[\psi_1]\|_{H^{1/2}(\R)}+\|\partial_1\widetilde{\Xi^{-1}}[\psi_1]\|_{W^{\infty}_1(\R)} \right)
\\\leq C_{1,\varepsilon}e^{K_V}N^{-(1-\varepsilon)/2}\left(\mathtt{C}(\widetilde{\Xi_1^{-1}},H^2)+\mathtt{C}(\widetilde{\Xi_1^{-1}},W_2^\infty)\right) \mcal{N}_3(\psi_1).
\end{multline*}

To obtain the expansion of the $2$-linear statistic up to $o(N^{-1})$, we will need to use the loop equation at level 2 with $\psi_2:=\mcal{D}\circ\widetilde{\Xi^{-1}}[\psi]$. Let $\psi_2\in \cap_{k\geq0}H^k(\R^2)$ be arbitrary for now. The level $2$ equation reads:

\begin{multline}\label{deuxieme expansion}
\Braket{\psi_{2}}_{\overset{2}{\bigotimes}\mcal{L}_N}=\dfrac{P}{N}\Braket{\partial_1\widetilde{\Xi_1^{-1}}[\psi_{2}]}_{\mu_V\bigotimes\mcal{L}_N}+\dfrac{P}{N}\Braket{\partial_1\widetilde{\Xi_1^{-1}}[\psi_{2}]}_{\overset{2}{\bigotimes}\mcal{L}_N}-\dfrac{1}{N}\Braket{\Theta^{(2)} \circ\widetilde{\Xi_1^{-1}}\big[\partial_2\psi_{2}\big] }_{\mcal{L}_N}
\\-P\Braket{\mcal{D}_1\circ\widetilde{\Xi_1^{-1}}[\psi_{2}]}_{\overset{3}{\bigotimes}\mcal{L}_N}-\dfrac{1}{N}\Braket{\Theta^{(2)} \circ\widetilde{\Xi_1^{-1}}\big[\partial_2\psi_{2}\big]}_{\mu_V}.
\end{multline}

The first term is a $o(N^{-1})$ as a 1-linear statistic $\Braket{\psi}_{\mcal{L}_N}$ where $\psi(x):=\displaystyle\int_\R\partial_1\widetilde{\Xi_1^{-1}}[\psi_{2}](x,y)d\mu_V(y)$. This function is indeed in $H^1(\R)$ because of Theorem  <ref>

$$\|\psi\|_{H^1(\R)}^2\leq\|\rho_V\|_{\infty}\|\partial_1\widetilde{\Xi_1^{-1}[}\psi_{2}]\|_{H^1(\R^2)}^2\leq\|\rho_V\|_{\infty}\mathtt{C}(\widetilde{\Xi_1^{-1}},H^2) ^2\|\psi_2\|_{H^3(\R^2)}^2$$

and in $W_1^{\infty}(\R)$ by Theorem <ref>

Thus by the a priori bound Theorem <ref>, we get

\begin{equation}\label{eq:controle sur les correlateurs}
\Big|\dfrac{P}{N}\Braket{\partial_1\widetilde{\Xi_1^{-1}}[\psi_{2}]}_{\mu_V\bigotimes\mcal{L}_N}\Big|\leq\dfrac{PC_{1,\varepsilon}e^{K_V}}{N^{1+(1-\varepsilon)/2}}\left[\mathtt{C}(\widetilde{\Xi_1^{-1}},W_2^\infty)+\|\rho_V\|_{\infty}^{1/2}\mathtt{C}(\widetilde{\Xi_1^{-1}},H^2)\right] \mcal{N}^{(2)}_3(\psi_2).
\end{equation}

The following two terms in (<ref>) are also a $o(N^{-1})$ by the same reasons as before. By Theorem <ref>, the $3$-linear statistics is a $o(N^{-1})$ for $\varepsilon>0$ small enough. Hence, we obtain the expansion:

$$\Braket{\psi}_{\mcal{L}_N}= \dfrac{d_{1}^{(1)}(\psi_1)}{N}+o(N^{-1}) \text{ and } \Braket{\psi_2}_{\mcal{L}_N\otimes\mcal{L}_N}=\dfrac{d_{1}^{(2)}(\psi_2)}{N}+o(N^{-1})$$

$$d_{1}^{(1)}(\psi_1):=P\Braket{\partial_1\widetilde{\Xi^{-1}}[\psi_1]}_{\mu_V}+P\Braket{\Theta^{(2)} \circ\widetilde{\Xi_1^{-1}}\left[\partial_2\mcal{D}\circ\widetilde{\Xi^{-1}}[\psi_1]\right]  }_{\mu_V}$$

$$d_{1}^{(2)}(\psi_2):=-\Braket{\Theta^{(2)} \circ\widetilde{\Xi_1^{-1}}\big[\partial_2\psi_{2}\big] }_{\mu_V}.$$

If $K=2$, having the expansion till $o(N^{-1})$ of the $1$-linear statistics, we know the expansion till $o(N^{-2})$ of the second term in (<ref>), so we just have to obtain the expansion till this order for the last term namely the $2$-linear statistics. Since in (<ref>), the loop equation involves only $1$ and $2$-linear statistics with a prefactor $N^{-1}$ and a $3$-linear statistic, we just need to obtain the expansion for the latter till $o(N^{-2})$. This equation at lvl $3$ reads:

\begin{multline}\label{troisieme expansion}
\Braket{\psi_{3}}_{\overset{3}{\bigotimes}\mcal{L}_N}=\dfrac{P}{N}\Braket{\partial_1\widetilde{\Xi^{-1}_1}[\psi_{3}]}_{\mu_V\overset{2}{\bigotimes}\mcal{L}_N}-\dfrac{1}{N}\sum_{a=2}^{3}\Bigg(\Braket{\Theta^{(a)} \circ\widetilde{\Xi^{-1}_1}\big[\partial_a\psi_{3}\big] }_{\overset{2}{\bigotimes}\mcal{L}_N}+\Braket{\Theta^{(a)} \circ\widetilde{\Xi^{-1}_1}\big[\partial_a\psi_{3}\big] }_{\mu_V\overset{}{\bigotimes}\mcal{L}_N}\Bigg)
\\+\dfrac{P}{N}\Braket{\partial_1\widetilde{\Xi^{-1}_1}[\psi_{3}]}_{\overset{3}{\bigotimes}\mcal{L}_N}-P\Braket{\mcal{D}_1\circ\widetilde{\Xi^{-1}_1}[\psi_{3}]}_{\overset{4}{\bigotimes}\mcal{L}_N}.
\end{multline}

The first line involves only $1$ and $2$-linear statistics with prefactors $N^{-1}$ so we know its asymptotic expansion till $o(N^{-2})$. The $3$-linear statistic together with its prefactor will yield a $O\left (N^{-5(1-\varepsilon)/2}\right )$ error, hence we just need to obtain the asymptotic expansion of the $4$-linear statistic till $o(N^{-2})$. The loop equation at level four gives:

\begin{multline}\label{quatrieme expansion}
\Braket{\psi_{4}}_{\overset{4}{\bigotimes}\mcal{L}_N}=\dfrac{P}{N}\Braket{\partial_1\widetilde{\Xi^{-1}_1}[\psi_{4}]}_{\mu_V\overset{3}{\bigotimes}\mcal{L}_N}+\dfrac{P}{N}\Braket{\partial_1\widetilde{\Xi^{-1}_1}[\psi_{4}]}_{\overset{4}{\bigotimes}\mcal{L}_N}-P\Braket{\mcal{D}_1\circ\widetilde{\Xi^{-1}_1}[\psi_{4}]}_{\overset{5}{\bigotimes}\mcal{L}_N}
\\-\dfrac{1}{N}\sum_{a=2}^{4}\Bigg(\Braket{\Theta^{(a)} \circ\widetilde{\Xi^{-1}_1}\big[\partial_a\psi_{4}\big] }_{\overset{3}{\bigotimes}\mcal{L}_N}\Bigg)-\dfrac{1}{N}\sum_{a=2}^{4}\Bigg(\Braket{\Theta^{(a)} \circ\widetilde{\Xi^{-1}_1}\big[\partial_a\psi_{4}\big] }_{\mu_V\overset{2}{\bigotimes}\mcal{L}_N}\Bigg).
\end{multline}

Because of the a priori estimates, every term in this equation will be $o(N^{-2})$ except the last sum. Indeed, its summands are $2$-linear statistic for which we know the asymptotic expansion till $o(N^{-1})$. Therefore we obtain that:

\begin{multline}
\Braket{\psi_{4}}_{\overset{4}{\bigotimes}\mcal{L}_N}=-\dfrac{1}{N}\sum_{a=2}^{4}\Bigg(\Braket{\Theta^{(a)} \circ\widetilde{\Xi^{-1}_1}\big[\partial_a\psi_{4}\big] }_{\mu_V\overset{2}{\bigotimes}\mcal{L}_N}\Bigg)+o(N^{-2})
\\=-\dfrac{1}{N^2}\underbrace{\sum_{a=2}^{4}d_{-1}^{(2)}(\phi_2^{(a)})}_{=:-d_{-2}^{(4)}(\psi_4)}+o(N^{-2})
\end{multline}

where $\phi_2^{(a)}(x,y)=\displaystyle\int_\R d\mu_V(t)\left[\Theta^{(a)} \circ\Xi_1^{-1}\big[\partial_a\psi_{4}\big]\right](t,x,y)$. We thus obtain the expansion of $\Braket{\psi_{3}}_{\overset{3}{\bigotimes}\mcal{L}_N}$ till $o(N^{-2})$ and so the one for $\Braket{\psi_{2}}_{\overset{2}{\bigotimes}\mcal{L}_N}$ at the same order which allows to obtained the expansion of $\Braket{\psi_{k}}_{\overset{k}{\bigotimes}\mcal{L}_N}$ till $o(N^{-2})$ for all $k=1,\dots,4$.

More generally, suppose the desired expansion for $\Braket{\psi_{k}}_{\overset{k}{\bigotimes}\mcal{L}_N}$ holds up to $o(N^{-n})$ for all $k\in\llbracket0,2n\rrbracket$ and for any function $\psi_k\in \displaystyle\cap_{m\geq0}H^m(\R^k)$. Additionnaly, suppose that (<ref>) is true for all $k\in\llbracket1,K-2\rrbracket$. Then, taking a general function $\psi_{2n+2}\in \displaystyle\cap_{m\geq0}H^m(\R^{2n+2})$, the $(2n+2)$-th equation involves the $2n+3$ linear statistic $\Braket{\mcal{D}_1\circ\Xi_1^{-1}[\psi_{2n+2}]}_{\overset{2n+3}{\bigotimes}\mcal{L}_N}$ (see (<ref>)). By Theorem <ref>, it will be a $o\left (N^{-(n+1)}\right )$ for $\varepsilon$ small enough. The other terms will be either, $2n+1$-linear statistics with a $N^{-1}$ prefactor and therefore behave like $O(N^{-(n+1)})$ for $\varepsilon$ small enough, or either be $2n$-linear statistics with a prefactor $N^{-1}$. For the latter, by hypothesis, we know the asymptotic expansion up to $O(N^{-n})$, thus with the prefactor $N^{-1}$, we deduce the following expansion for $\braket{\psi_{2n+2}}_{\overset{2n+2}{\bigotimes}\mcal{L}_N}$

$$\braket{\psi_{2n+2}}_{\overset{2n+2}{\bigotimes}\mcal{L}_N}=\dfrac{d_{n+1}^{(2n+2)}(\psi_{2n+2})}{N^{n+1}}+o\left(N^{-(n+1)}\right) $$

We will deduce from that, the asymptotic expansion of $\braket{\psi_{2n+1}}_{\overset{2n+1}{\bigotimes}\mcal{L}_N}$ for a general $\psi_{2n+1}$ belonging  to $\cap_{k\geq0}H^k(\R^{2n+1})$. In the $(2n+1)$-th loop equation (<ref>), the $2n+2$ linear statistic will yield a non-trivial term of order $N^{-(n+1)}$ ie

$$\Braket{\mcal{D}_1\circ\Xi_1^{-1}[\psi_{2n+1}]}_{\overset{2n+2}{\bigotimes}\mcal{L}_N}=\dfrac{d_{n+1}^{(2n+2)}\left(\mcal{D}_1\circ\widetilde{\Xi_1^{-1}}[\psi_{2n+1}]\right) }{N^{n+1}}+o\left(N^{-(n+1)}\right).$$

Again, the $(2n+1)$-linear statistics with a prefactor $N^{-1}$ will be $o(N^{-(n+1)})$. Finally for the $2n$  and $2n-1$ linear statistics with the prefactor $N^{-1}$ appearing in the $(2n+1)$-th loop equation, we know the asymptotic expansion up to $o(N^{-(n+1)})$ by hypothesis.

To conclude on the asymptotic expansion up to $o(N^{-(n+1)})$ for the $2n$-linear statistics, just notice that for each term appearing in the $2n$ equation each term will either be a $2n+1$-linear statistics for which we know the asymptotic expansion up to $o(N^{-(n+1)})$, or a linear statistic for which we know, by hypothesis, the asymptotic expansion up to $o(N^{-n})$ (of order $2n$,$2n-1$ or $2n-2$), preceded by a factor $N^{-1}$. We can therefore conclude on the existence of the asymptotic expansion up to $o(N^{-(n+1)})$for the $2n$-linear statistics. Then applying the same arguments for $2n-1$,…$1$-linear statistics allows us to conclude that the induction step is established.

Finally, to conclude on (<ref>), one just has to notice that for all $n\geq 1$ and $\psi_n\in\cap_{k\geq0}H^{k}(\R^n)$, for all $K\geq\lceil n/2\rceil$,

$$\Braket{\psi_n}_{\overset{n}{\bigotimes}\mcal{L}_N}-\sum_{a=\lceil n/2\rceil}^{K}\dfrac{d_{a}^{(n)}(\psi_n)}{N^a}=\dfrac{d_{K+1}^{(n)}(\psi_n)}{N^{K+1}}+\mfrak{R}_{K+1}^{(n)}(\psi_n)\hspace{0,5cm}\text{ and }\hspace{0,5cm}\mfrak{R}_{K+1}^{(n)}(\psi_n)=o(N^{-(K+1)}).$$

Above, the remainder $\mfrak{R}_{K+1}^{(n)}(\psi_n)$ contains all the negligible (by the a priori bound) statistics involving the operators $\widetilde{\Xi_1^{-1}}$, $\mcal{D}_1\circ\widetilde{\Xi_1^{-1}}$ and $\Theta^{(a)} \circ\widetilde{\Xi_1^{-1}}$, thus just as in (<ref>), by using continuity of the different operators involved in each of the statistics, there exists $m^{(1)}_{K,n}>0$, a polynomial $\msf{Q}^{(1)}_{K,n}$ in $e^{K_V}$, $\|\rho_V\|_{\infty}^{1/2}$, $\left(\mathtt{C}(\widetilde{\Xi_1^{-1}},H^i)\right)_{0\leq i\leq m^{(1)}_{K,n}}$ $\left(\mathtt{C}(\widetilde{\Xi_1^{-1}},W_i^{\infty})\right)_{0\leq i\leq m^{(1)}_{K,n}}$ and $\Big( \mathtt{C}(\Theta^{(a)}\circ\widetilde{\Xi_{1}^{-1}},H^n)\Big)_{0\leq i\leq m^{(1)}_{K,n}}$ with coefficients independent of $V$ and a constant $C(K,n)>0$ such that

$$\bigg|\mfrak{R}_{K+1}^{(n)}[\psi_n]\bigg|\leq \dfrac{C^{(1)}(K,n)}{N^{K+1}}\msf{Q}^{(1)}_{K,n}\mcal{N}_{m_{K,n}^{(1)}}^{(n)}(\psi_n).$$

To bound, $d_{K+1}^{(n)}(\psi_n)$ and extract the $V$-dependance, one just notices that it is a sum of linear statistics, involving as before the previous operators. By contintuity of the operators, there exists a polynomial $\msf{Q}_{K,n}^{(2)}$ in the previous operator norms and $\|\rho_V\|_{\infty}^{1/2}$ with coefficients independent of $V$ such that

$$\bigg|d_{K+1}^{(n)}(\psi_n)\bigg|\leq C^{(2)}(K,n)\msf{Q}^{(2)}_{K,n}\mcal{N}^{(n)}_{m^{(2)}_{K,n}}(\psi_n).$$

Thus setting $\mathtt{C}_{\mrm{rem}}(V,K,n)=C^{(1)}(K,n)\msf{Q}^{(1)}_{K,n}+C^{(2)}(K,n)\msf{Q}^{(2)}_{K,n}$ and $m_{K,n}:=\max\left(m^{(1)}_{K,n},m^{(2)}_{K,n}\right)$ allows us to conclude about <ref>. The fact that $t\in[0,1]\mapsto\mathtt{C}_{\mrm{rem}}(V_{\phi,t},K,n)$ is continuous  follows from the fact it is a polynomial in building blocks which are continuous as it is shown in Appendix <ref>, Lemma <ref>, Proposition <ref>.

§ PARAMETER CONTINUITY OF THE EQUILIBRIUM MEASURE

We want to conclude about the asymptotic expansion of $\log\mcal{Z}_N\left[V_{G,\phi}\right]$ for a smooth $\phi,$ by inserting the asymptotic expansion of the linear statistics of Theorem <ref> in Lemma <ref>. In order to make that step rigorous, it is necessary to prove that all the linear statistics integrated with respect to the probability measure $\mbb{P}_N^{V_{G,\phi,t}}$ with $t\in[0,1]$, when integrating them with respect to $t$, yield a definite and finite integral. Since all the quantities depend on $t$ through $\mu_{V_{G,\phi,t}}$, we first prove a continuity result for $t\mapsto\rho_{V_{G,\phi,t}}$. The result that we are going to prove does not depend on the specificity of the Gaussian potential so in the following, we set $V_{\phi,t}:x\mapsto V(x)+t\phi(x)$ with $V$ satisfying the assumptions <ref>.

While from the measure point of view, it is easy to show that $t\mapsto\mu_{V_{\phi,t}}$ is continuous for the weak topology of measures, it is not sufficient to deduce the continuity of the quantities involved in our problem. Indeed, in the controls we showed in Section <ref> quantities like $L^\infty$-norm of derivatives of $\rho_{V_t}$ and $C_{\mcal{L}}$, it will be necessary to show that $\|\rho_{V_{\phi,t}}-\rho_{V_{\phi,t_0}}\|_{W_i^{\infty}(\R)}\tend{t\rightarrow t_0}0$ for all $t_0\in[0,1]$.

§.§ Setting for Banach fixed-point theorem

Let $\phi\in\mcal{C}^{\infty}(\R)$, such that for all $k\in\N$, $\phi^{(k)}\in L^2(\R)$, let $t_0\in[0,1]$, we define the function $u_t$ by

\begin{equation}\label{def:u_t}
\rho_{V_{\phi,t}}=(1+\delta t u_t)\rho_{V_{\phi,t_0}}\;\;\;\; \textit{ ie    }\;\;\;\; u_t=\dfrac{\rho_{V_{\phi,t}}-\rho_{V_{\phi,t_0}}}{\delta t}\dfrac{1}{\rho_{V_{\phi,t_0}}}
\end{equation}

where $t\neq t_0$ and $\delta t:=t-t_0$. We will show, by Banach fixed-point theorem, that $x\mapsto u_t(x)\in\mcal{C}^\infty(\R)$, by Lemma <ref>, is the unique fixed-point of a $t$ continuous operator. This will allow us to deduce that $t\mapsto u_t$ is continuous for the $\bigcap_{k\in\N}W^\infty_{k}$-norm. The continuity of $t\mapsto\rho_{V_t}$ will then follow.

In order to construct the operator of interest, we start with the following lemma

Let $t,t_0\in[0,1]$,

\begin{multline}
\label{eq:lambda_t}
\lambda_{V_{\phi,t}}=\lambda_{V_{\phi,t_0}}+\delta t\int_\R\phi(x) d\mu_{V_{\phi,t_0}}(x)-2P\delta t\iint_{\R^2}\log|x-y|u_t(x)d\mu_{V_{\phi,t_0}}(x)d\mu_{V_{\phi,t_0}}(y)
\\+\int_\R\left[\log\big(1+\delta t u_t(x)\big)-\delta t u_t(x)\right]  d\mu_{V_{\phi,t_0}}(x).
\end{multline}

Here $\lambda_{V_{\phi,t}}$ denotes the constant appearing in (<ref>) with potential $V_{\phi,t}$.

We integrate with respect to $\mu_{V_{\phi,t_0}}$ (<ref>) to get

\begin{multline*}
\lambda_{V_{\phi,t}}=\int_\R V_{\phi,t_0}(x)d\mu_{V_{\phi,t_0}}(x)+\delta t\int_\R\phi(y) d\mu_{V_{\phi,t_0}}(y)-2P\iint_{\R^2}\log|x-y|d\mu_{V_{\phi,t}}(x)d\mu_{V_{\phi,t_0}}(y)
\\+\int_\R \log\rho_{V_{\phi,t}}(x)d\mu_{V_{\phi,t_0}}(x).
\end{multline*}

After using the fact that $\displaystyle\int_\R u_td\mu_{V_{\phi,t_0}}=0$, that $\rho_{V_{\phi,,t}}=(1+\delta t u_t)\rho_{V_{\phi,t_0}}$ and the characterization (<ref>) of $\mu_{V_{\phi,t_0}}$, this yields the result.

To show that $u_t$ is a fixed point of a $t$-continuous operator, we need to invert and control the operator $\mcal{T}:=\mcal{L}\circ\mcal{A}^{-1}$ (these operators were inroduced in Definition <ref>)which will appear naturally when comparing $\rho_{V_{\phi,t}}$ to $\rho_{V_{\phi,t_0}}$.

via Fredholm theory of determinants (see [27]). We now define the 2-determinant which is a powerful tool to show that kernel operators that are invertible.

We define the 2-determinant of a kernel operator $\mcal{K}$ of kernel $k$ on $L^2(\mu_{V_{t_0}})$ by

\\k(t_2,t_1)&0&\dots&k(t_2,t_n)
\\\vdots& & &\vdots
\\k(t_n,t_1)&k(t_n,t_2)&\hdots&0
\end{vmatrix}\prod_{i=1}^{n}d\mu_{V_{t_0}}(t_i)$$

With this definition, it can be shown that $I+\mcal{K}$ is invertible if and only if $\underset{2}{\det}\left(I+\mcal{K}\right)\neq0$ (see [27]).

We define the operator $\mcal{T}$  by
$\mcal{T}[v]:=v-\mcal{K}[v]$ for all $v\in L^2(\mu_{V_{\phi,t_0}})$,

$$\mcal{K}[v](x):=2P\int_\R k(x,y)v(y)\rho_{V_{\phi,t_0}}(y)dy$$

$$k(x,y):=\left (\log\dfrac{|x-y|}{1+|x|}-\displaystyle\int_\R\log\dfrac{|z-y|}{1+|z|}\rho_{V_{\phi,t_0}}(z)dz\right).$$

$\mcal{T}:L^2(\mu_{V_{\phi,t_0}})\rightarrow L^2(\mu_{V_{\phi,t_0}})$ is bijective and for all $n\geq0$, $\mcal{T}\left[W_n^\infty(\R)\right]= W_n^\infty(\R)$.

of inverse given by:

\begin{multline}\label{eq:inverse fredholm}
\mcal{T}^{-1}[v](x):=v(x)\\+\dfrac{1}{\underset{2}{\det}(I-\mcal{K})}\sum_{n\geq1}\dfrac{(-1)^n}{n!}\int_{\R^{n+1}}\begin{vmatrix}
\\k(t_1,s)&0&\dots&k(t_1,t_n)
\\\vdots& & &\vdots
\\k(t_n,s)&k(t_n,t_1)&\hdots&0
\end{vmatrix}v(s)d\mu_{V_{t_0}}(s)\prod_{i=1}^{n}d\mu_{V_{t_0}}(t_i).
\end{multline}

Finally, for all $n\in\N$, there exists $C_{\mcal{T},n}>0$ such that for any $v\in W_n^\infty(\R)$,

\begin{equation}\label{eq:controleTLinfini}
\left \|\mcal{T}^{-1}[v]\right \|_{W_n^\infty(\R)}\leq C_{\mcal{T},n}\|v\|_{W_n^\infty(\R)}.
\end{equation}

It was shown in [24] that $\mcal{T}[v]=-\mcal{L}\circ\mcal{A}^{-1}[v].$
$\mcal{A}$ and $\mcal{L}$ are unbounded operators on $\msf{H}$, it was also shown there that
$$\mcal{A}:\mcal{D}(\mcal{A}):=\left\{\phi\in\msf{H},\,\mcal{A}[\phi]\in\msf{H}\right\}\rightarrow\msf{H} \hspace{0,5cm}\text{and}\hspace{0,5cm}\mcal{L}:\mcal{D}(\mcal{A})\rightarrow\msf{H}$$
are bijective, thus so is $\mcal{T}:\msf{H}\rightarrow\msf{H}$.

The fact that $k$ verifies $\|k(x,y)\|_{L^2(\mu_{V_{\phi,t_0}}\otimes\mu_{V_{\phi,t_0}})}<+\infty$ implies that $\mcal{K}:L^2(\mu_{V_{\phi,t_0}})\rightarrow L^2(\mu_{V_{\phi,t_0}})$ is an Hilbert-Schmidt operator thus compact and so $\mcal{T}$ is a Fredholm operator. We now show that the kernel of $\mcal{T}$ is trivial. Let $v\in L^2(\mu_{V_{\phi,t_0}})$ such that $\mcal{T}[v]=0$ so $v=\mcal{K}[v]$. The RHS is in $H^1(\mu_{V_{\phi,t_0}})$, indeed we have
$$\mcal{K}[v]'(x)=-\mcal{H}[v\rho_{V_{\phi,t_0}}](x)-\dfrac{\mrm{sgn}(x)}{1+|x|}\int_{\R}v(y)d\mu_{V_{\phi,t_0}}(y)\in L^2(\mu_{V_{\phi,t_0}}).$$

Moreover since $\int_\R\mcal{K}[v]d\mu_{V_{\phi,t_0}}=0$, we conclude that $v\in\msf{H}$. We can now conclude that $v=0$ by the bijectivity of $\mcal{T}$ on $\msf{H}$. Finally, by Fredhom alternative, $\mcal{T}$ is invertible on $L^2(\mu_{V_{\phi,t_0}})$ since it is injective.

Since $\mcal{T}$ is invertible on $L^2(\mu_{V_{t_0}})$, $\underset{2}{\det}(\mcal{T})\neq0$. This implies, by [27][Section XII.2], that $\mcal{T}^{-1}$ can be expressed explicitly as a Fredholm's serie, this yields exactly (<ref>).

For the bound with $L^\infty$-norm, we can use the fact that for any $v\in L^2_0(\mu_{V_{t_0}})$, we can replace every kernel $k$ appearing in the determinant in the integral by the kernel $k$ defined by:

$$k(x,y):=\left (\log\dfrac{|x-y|}{1+|x|}-\displaystyle\int_\R\log|y-z|\rho_{V_{t_0}}(z)dz\right).$$

Indeed this is ensured because $\int_\R v(s)\rho_{V_{t_0}}(s)ds=0$ so
$$\mcal{K}[v](x)=\mcal{K}[v](x)-2P\log\left(1+|x|\right)\int_\R v(s)\rho_{V_{t_0}}(s)ds=.$$

$$k(x,y):=\left (\log\dfrac{|x-y|}{1+|x|}-\displaystyle\int_\R\log|y-z|\rho_{V_{t_0}}(z)dz\right).$$

Let $v\in L^\infty(\R)\subset L^2_0(\mu_{V_{t_0}})$, let $x\in\R$, we use direct bounds to show that
$$|\mcal{T}^{-1}[v](x)|\leq (f(x)+1)\|v\|_{\infty}$$
where $f$ is defined by

\\k(t_1,s)&0&\dots&k(t_1,t_n)
\\\vdots& & &\vdots
\\k(t_n,s)&k(t_n,t_1)&\hdots&0
\end{vmatrix}^2}\dfrac{d\mu_{V_{t_0}}(s)}{n!}\prod_{i=1}^{n}d\mu_{V_{t_0}}(t_i).$$

It just remains to show that $f$ is bounded. First we use Hadamard's and Jensen's inequality to show that

\begin{multline*}
\int_{\R^{n+1}}\sqrt{\begin{vmatrix}
\\k(t_1,s)&0&\dots&k(t_1,t_n)
\\\vdots& & &\vdots
\\k(t_n,s)&k(t_n,t_1)&\hdots&0
\end{vmatrix}^2}d\mu_{V_{t_0}}(s)\prod_{i=1}^{n}d\mu_{V_{t_0}}(t_i)
\\\leq\left(\int_{\R^{n+1}}\left(k(x,s)^2+\sum_{i=1}^{n}k(t_i,s)^2 \right).\prod_{j=1}^{n}\left(k(x,t_j)^2+\sum_{i\neq j}k
\\\leq\Bigg\{\int_{\R^{n+1}}\left[k(x,s)^2\sqrt{\rho_{V_{t_0}}(s)}+\sum_{i=1}^{n}k(t_i,s)^2\sqrt{\rho_{V_{t_0}}(s)}\right].\prod_{j=1}^{n}\left(k(x,t_j)^2+\sum_{i\neq j}k
\end{multline*}

Moreover, $M_2:=\sup_{x\in\R}\int_{\R}k(x,y)^2d\mu_{V_{t_0}}(y)$ is finite so in the previous product by selecting only the terms having $x$ as the first coordinate, we get:

$$\int_{\R^{n+1}}k(x,s)^2\prod_{j=1}^{n}k(x,t_j)^2d\mu_{V_{t_0}}(s)\prod_{i=1}^{n}d\mu_{V_{t_0}}(t_i)=\left(\int_{\R}k(x,y)^2d\mu_{V_{t_0}}(y)\right)^{n+1}\leq M_2^{n+1}.$$

Furthermore, by setting $\msf{A}_n:=\left\{g:\llbracket1,n\rrbracket\rightarrow\llbracket1,n\rrbracket,\,g\text{ has no fixed point}\right\}$, we get

$$\int_{\R^{n+1}}k(x,s)^2\prod_{j=1}^{n}\sum_{i\neq j}k
(t_i,t_j)^2d\mu_{V_{t_0}}(s)\prod_{i=1}^{n}d\mu_{V_{t_0}}(t_i)\leq M_2\sum_{g\in\msf{A}_n}\int_{\R^n}\prod_{j=1}^{n}k\left(t_{f(j)},t_j\right)^2d\mu_{V_{t_0}}(t_j) .$$

\begin{multline*}
\left(\int_{\R^{n+1}}\left(k(x,s)^2+\sum_{i=1}^{n}k(t_i,s)^2 \right).\prod_{j=1}^{n}\left(k(x,t_j)^2+\sum_{i\neq j}k(t_i,t_j)^2\right)d\mu_{V_{t_0}}(s)\prod_{i=1}^{n}d\mu_{V_{t_0}}(t_i)\right)^{1/2}
\\\leq\Bigg(M_2^{n+1}+M_2\int_{\R^{n}}\prod_{j=1}^{n}\sum_{i\neq j}k(t_i,t_j)^2\prod_{i=1}^{n}d\mu_{V_{t_0}}(t_i)
\\+\sum_{i=1}^{n}\sum_{p=1}^{n}\sum_{j_1<\dots<j_p}\sum_{g\in A_p}\int_{\R^{n+1}}k(t_i,s)^2\prod_{m=1}^pk(x,t_{j_m})^2\prod_{\substack{j=1\\j\neq j_1,\dots,j_p}}^nk(t_{g(j)},t_j)^2d\mu_{V_{t_0}}(s)\prod_{i=1}^{n}d\mu_{V_{t_0}}(t_i)
\\+\sum_{i=1}^{n}\sum_{g\in A_n}\int_{\R^{n+1}}k(t_i,s)^2.\prod_{j=1}^{n}k(t_{g(j)},t_j)^2d\mu_{V_{t_0}}(s)\prod_{i=1}^{n}d\mu_{V_{t_0}}(t_i)\Bigg)^{1/2}
\\\leq \left(M_4^{n+1}n^nP(n)\right)^{1/2}
\end{multline*}

where $M_2:=\sup_{x\in\R}\int_{\R}k(x,y)^2d\mu_{V_{t_0}}(y)$, $M_4:=\left(\sup_{x\in\R}\int_{\R}k(x,y)^4d\mu_{V_{t_0}}(y)\right)^{1/2} \vee\sup_{x\in\R}\int_{\R}k(x,y)^2d\mu_{V_{t_0}}(y)$ for all $p\in\llbracket1,n\rrbracket$,  $A_p:=\left\{g\in\llbracket1,p\rrbracket^{\llbracket1,p\rrbracket}, g \text{ has no fixed point}\right\}$. In the last step, we consecutively apply Jensen and Cauchy Schwarz inequality to separate product like $k\left(t_j,t_k\right)k(t_l,t_j)$ which led to the $4$-th power in the definition of $M_4$.

Dividing by $n!$, we obtain a serie of general term $\dfrac{n^{n/2}}{n!}M^{n/2}P(n)(n)^{1/2}$, which converges by D'Alembert criterion. Hence setting, $C_\mcal{T}:=\|f\|_{\infty}$ allow us to conclude.

We now prove that for all $n\in\N$, $\mcal{T}\left[W_n^\infty(\R)\right]= W_n^\infty(\R)$. We proceed by induction. For $n=0$, let $f\in L^\infty(\R)\subset L^2(\mu_{V_{\phi,t_0}})$. There exists a unique $v\in L^2(\mu_{V_{\phi,t_0}})$ such that $\mcal{T}[v]=f$ so $v=f+\mcal{K}[v]$ but since $f$ and $\mcal{K}[v]$ are bounded, so is $v\in L^\infty(\R)$. Reciprocally, if $v\in L^\infty(\R)$ so is $\mcal{T}[v]$, hence $\mcal{T}\left[L^\infty(\R)\right]= L^\infty(\R)$. Finally let $v\in L^\infty(\R)$,

\begin{equation}\label{eq:t linfini}
\Big\|\mcal{T}[v]\Big\|_{L^\infty(\R)}\leq\left(1+2P\max_{x\in\R}\int_{\R}|k(x,y)|d\mu_{V_{\phi,t_0}}(y)\right)\|v\|_{L^\infty(\R)}.
\end{equation}

Now suppose $\mcal{T}\left[W_n^\infty(\R)\right]= W_n^\infty(\R)$ is true and let's show it for $n+1$. Let $f \in W_{n+1}^\infty(\R)\subset W_n^\infty(\R)$, so by hypothesis, there exists $v\in W_n^{\infty}(\R)$ such that

\begin{multline}\label{eq:t(v)^n}
f^{(n)}(x)=v^{(n)}(x)+2P\left(\log(1+|.|)\right)^{(n)}(x)\int_\R v(y)d\mu_{V_{\phi,t_0}}(y)-2P\int_\R\log|x-y| (v\rho_{V_{\phi,t_0}})^{(n)}(y)dy
\\+2P\delta_{n,0}\iint_{\R^2}\dfrac{\log|z-y|}{1+|z|}v(y)d\mu_{V_{\phi,t_0}}(y)d\mu_{V_{\phi,t_0}}(z).
\end{multline}

We deduce that $v^{(n)}$ is differentiable of derivative

\begin{equation}\label{eq:v n+1}
v^{(n+1)}(x)=f^{(n+1)}(x)-2P\left(\log(1+|.|)\right)^{(n+1)}(x)\int_\R v(y)d\mu_{V_{\phi,t_0}}(y)-2P\mcal{H}\left[(v\rho_{V_{\phi,t_0}})^{(n)}\right](x),
\end{equation}

where $\left(\log(1+|.|)\right)^{(n)}(x)=\dfrac{n!\mrm{sgn}(x)}{(1+|x|)^n}$ if $n$ is odd and $\dfrac{n!}{(1+|x|)^n}$ if $n$ is even. Since the two first terms in the RHS of (<ref>) are clearly bounded, we just have to show that  $\mcal{H}\big[(v\rho_{V_{\phi,t_0}})^{(n)}\big]\in H^1(\R)$. By boundedness of $f^{(n+1)}$ and $v^{(i)}$ for all $i\leq n$, we have

\begin{multline*}
\\=\rho_{V_{\phi,t_0}}\Bigg(f^{(n+1)}-2P\left(\log(1+|.|)\right)^{(n+1)}\int_\R v(y)d\mu_{V_{\phi,t_0}}(y)-2P\mcal{H}\left[(v\rho_{V_{\phi,t_0}})^{(n)}\right]
\\+\sum_{k=0}^{n}\binom{n+1}{k}\dfrac{\rho_{V_{\phi,t_0}}^{(n+1-k)}}{\rho_{V_{\phi,t_0}}}v^{(k)}\Bigg)\in L^2(\R).
\end{multline*}

Thus, it holds that $\mcal{H}\left[(v\rho_{V_{\phi,t_0}})^{(n)}\right]\in H^1(\R)$ and that it is bounded. Hence it proves that $v^{(n+1)}\in L^\infty(\R)$ and hence $W_{n+1}^\infty(\R)\subset \mcal{T}\left[W_{n+1}^\infty(\R)\right]$. Conversely, if $v\in W_{n+1}^\infty(\R)$, then $f\in W_{n}^\infty(\R)$ by hypothesis and just as before, we show that (<ref>) holds. We conclude that $f^{(n+1)}\in L^\infty(\R)$ again by showing that $\mcal{H}\left[(v\rho_{V_{\phi,t_0}})^{(n)}\right]$ is bounded by the fact that $v\in W_{n+1}^\infty(\R)$. This establishes that $W_{n+1}^\infty(\R)=\mcal{T}\left[W_{n+1}^\infty(\R)\right]$.

Thus for all $n\in\N$, $\mcal{T}:W_n^\infty(\R)\rightarrow W_n^\infty(\R)$ is a bijective operator. Furthermore, it is a bounded operator by the fact that for all $1\leq i\leq n$, there exists $C>0$ such that for all $v\in W_n^{\infty}(\R)$, by Leibniz formula

\begin{multline*}
\big|\mcal{T}[v]^{(i)}(x)\big|\leq \|v^{(i)}\|_{L^\infty(\R)}+2P\big\|\left(\log(1+|.|)\right)^{(i)}\big\|_{L^\infty(\R)}\|v\|_{L^\infty(\R)}\\+2P\|v\|_{W_i^{\infty}(\R)}\sum_{k=0}^{i}\binom{i}{k}\sup_{z\in\R}\int_{\R}\Big|\log\dfrac{|z-y|}{1+|z|}\Big|.\Bigg|\dfrac{\rho_{V_{\phi,t_0}}^{(k)}}{\rho_{V_{\phi,t_0}}}(y)\Bigg|d\mu_{V_{\phi,t_0}}(y)\leq C\|v\|_{W_i^{\infty}(\R)}.
\end{multline*}

Above we used (<ref>) and the fact that $\int_{\R}(v\rho_{V_{\phi,t_0}})^{(i)}(y)dy=0$ so

\begin{multline*}
\int_\R\log|x-y|(v\rho_{V_{\phi,t_0}})^{(i)}(y)dy=\int_\R\log|x-y|(v\rho_{V_{\phi,t_0}})^{(i)}(y)dy-\log(1+|x|)\int_\R(v\rho_{V_{\phi,t_0}})^{(i)}(y)dy
\\=\int_\R\log\dfrac{|x-y|}{1+|x|}(v\rho_{V_{\phi,t_0}})^{(i)}(y)dy.
\end{multline*}

Thus we conclude that $\sup_{i\in\llbracket1,n\rrbracket}\big\|\mcal{T}[v]^{(i)}\big\|_{L^\infty(\R)}\leq C\|v\|_{W_n^{\infty}(\R)}$. The bound on $\big\|\mcal{T}[v]\big\|_{L^\infty(\R)}$ was shown in (<ref>). We finally conclude that $\mcal{T}:W_n^\infty(\R)\rightarrow W_n^\infty(\R)$ is bounded bijective between Banach spaces and by Banach isomorphism theorem so is $\mcal{T}^{-1}$, this establishes (<ref>) and completes the proof.

An explicit expression for $\mcal{T}^{-1}$ is available Fredholm determinant theory for invertible Hilbert-Schmidt operators, see [27][Section XII]. For all $v\in L^2_0(\mu_{V_{\phi,t_0}})$,

\begin{multline}\label{eq:inverse fredholm}
\mcal{T}^{-1}[v](x):=v(x)\\+\dfrac{1}{\underset{2}{\det}(I-\mcal{K})}\sum_{n\geq1}\dfrac{(-1)^n}{n!}\int_{\R^{n+1}}\begin{vmatrix}
\\k(t_1,s)&0&\dots&k(t_1,t_n)
\\\vdots& & &\vdots
\\k(t_n,s)&k(t_n,t_1)&\hdots&0
\end{vmatrix}v(s)d\mu_{V_{\phi,t_0}}(s)\prod_{i=1}^{n}d\mu_{V_{\phi,t_0}}(t_i).
\end{multline}

Above $\underset{2}{\det}$ stands for the $2$-determinant. This formula was established in [24][Theorem 6.11]

In the previous proof, if $(x,y)\mapsto k(x,y)\rho_{V_{t}}(y)$ was bounded by $M$ on $\R^2$, it would be easy to bounded the determinant in the integrals of $f$ by $M^{n}n^{n/2}$ but because since $k$ has a singularity on the diagonal, one has to integrate this term to take care of that divergence. This is the reason why we have to developp the product obtained by Hadamard's inequality.

[Differentiability of the Fredholm expression]
Let $v\in\mcal{C}^{\infty}(\R)$, such that for all $k\in\N$ $v^{(k)}$ is bounded, then $\mcal{T}^{-1}[v]\in\mcal{C}^{\infty}(\R)$. Moreover, for all $k\in\N$ there exists $C_k>0$ such that for all $f\in\mcal{C}^{\infty}(\R)$ with $f^{(k)}$ bounded for all $k\in\N$ and $\int_\R fd\mu_{V_{t_0}}=0$:

\begin{equation}\label{eq:controleoperateurinverse7}
\left \|\mcal{T}^{-1}[f]^{(k)}\right\|_\infty\leq C_{\mcal{T},k}\sup_{i\in\llbracket0,k\rrbracket}\|f^{(i)}\| _{\infty}
\end{equation}

The statement is proven for $k=0$. Let's show it for $k=1$, we use the fact that for all $f\in\mcal{C}^{1}(\R)$, $\left(\displaystyle\int_\R \log|.-y|f(s)ds\right)'(x)=\displaystyle\int_\R \log|x-y|f'(s)ds$ for all $x\in\R$ which can be shown by an integration by part. Let $n\geq 1$,

\begin{multline}\label{dem:deriveeu}
\left(\int_{\R^{n+1}}\begin{vmatrix}
\\k(t_1,s)&0&\dots&k(t_1,t_n)
\\\vdots& & &\vdots
\\k(t_n,s)&k(t_n,t_1)&\hdots&0
\end{vmatrix}f(s)d\mu_{V_{t_0}}(s)\prod_{i=1}^{n}d\mu_{V_{t_0}}(t_i)\right)'(x)
\\=\left(\int_\R l(x,s)(f\rho_{V_{t_0}})'(s)\right)\int_{\R^{n}}\begin{vmatrix}
\\\vdots& & &\vdots
\\k(t_n,t_1)&k(t_n,t_2)&\hdots&0
\end{vmatrix}\prod_{i=1}^{n}d\mu_{V_{t_0}}(t_i)
\\+\sum_{j=1}^{n}\int_{\R^{n+1}}d\mu_{V_{t_0}}(s)\prod_{i=1}^{n}d\mu_{V_{t_0}}(t_i)l(x,t_j)\sum_{\sigma\in\mfrak{S}_n}\varepsilon(\sigma)l\left (\sigma^{-1}(s),s\right )
\\\times\prod_{i=1}^{n}l\left (t_i,\sigma(t_i)\right )\left(f(s)\left (\dfrac{\rho_{V_{t_0}}'(s)}{\rho_{V_{t_0}}(s)}+\sum_{p=1}^{n}\dfrac{\rho_{V_{t_0}}'(t_i)}{\rho_{V_{t_0}}(t_i)}\right )+f'(s)\right).
\end{multline}

Here $l(x,y):=-\log|x-y|$ with the convention $l(z,z)=0$, in the sum over $\sigma$, $\sigma$ has to be understood mapping $\{t_1,t_2,\dots,t_n\}$ to $\{s,t_1,\dots,t_{j-1},t_{j+1},\dots,t_n\}$ and we redefined $\sigma\left(\sigma^{-1}(t_j)\right):=s$. Since $\displaystyle\int_\R \left(f(s)\left (\dfrac{\rho_{V_{t_0}}'(s)}{\rho_{V_{t_0}}(s)}+\sum_{p=1}^{n}\dfrac{\rho_{V_{t_0}}'(t_i)}{\rho_{V_{t_0}}(t_i)}\right )+f'(s)\right)d\mu_{V_{t_0}}(s)\prod_{i=1}^{n}d\mu_{V_{t_0}}(t_i)=0$, it is possible to replace the kernel $l(x,t_j)$ by $-\log\dfrac{|x-t_j|}{1+|x|}$ which is a bounded function of $x$, allowing one to take the infinite norm and get that each of these integrals are bounded by a summable $\dfrac{C_{n,k}}{n!}\|v\|_{W^\infty_1(\R)}$.
Since the convergence of the serie of general term given by $\ref{dem:deriveeu}$ is normal, we can conclude that the sum of the serie is indeed $\mcal{T}^{-1}[v]$. Indeed, proceeding in the same way as in the proof of Proposition <ref> holds with $C_k=1+\sum_{k=1}^{\infty}C_{n,k}$
The proof for $k>1$ is similar.

We are now able to show that $u_t$ is a fixed point of a certain operator. We recall that $\phi$ was introduced in the beginning of subsection <ref>.

For all $t\in[0,1]$, $u_t$ defined in (<ref>) is the unique measurable function such that $\int_\R u_td\mu_{V_{\phi,t_0}}=0$ and which satisfies:

where $\mcal{V}_t[u]:=-\phi+\int_\R\phi(y) d\mu_{V_{\phi,t_0}}(y)+\delta t\,\mcal{U}_t[u]$,   and

\begin{multline*}
\mcal{U}_t[v](x):=\left(-\phi(x)+\int_\R\phi(y) d\mu_{V_{\phi,t_0}}(y)+\mcal{K}[v](x)+\int_\R\dfrac{\log\Big(1+\delta t v(y)\Big)-\delta t v(y)}{\delta t} d\mu_{V_{\phi,t_0}}(y)\right)^2
\\\times\int_{0}^{1}(1-s)ds\exp\Bigg\{s\delta t\left(-\phi(x)+\int_\R\phi(y) d\mu_{V_{\phi,t_0}}(y)+\mcal{K}[v](x)\right)
\\+\int_\R\frac{\log\Big(1+\delta t v(y)\Big)-\delta t v(y)}{\delta t} d\mu_{V_{\phi,t_0}}(y)\Bigg\} +\int_\R\dfrac{\log\Big(1+\delta t v(y)\Big)-\delta t v(y)}{(\delta t)^2} d\mu_{V_{\phi,t_0}}(y).
\end{multline*}

Lemma <ref> allows one to substitute $\lambda_{V_{\phi,t}}$ in the representation for $\rho_{V_{\phi,t}}$ by (<ref>) hence leading to

\begin{multline*}
\rho_{V_{\phi,t}}=(1+\delta t u_t)\rho_{V_{\phi,t_0}}=\exp\Bigg(-V_{\phi,t_0}-2PU^{\rho_{V_{\phi,t_0}}}+\lambda_{V_{\phi,t_0}}-\delta t\phi+\delta t\int_\R\phi(y) d\mu_{V_{\phi,t_0}}(y)-2P\delta t U^{u_t\rho_{V_{\phi,t_0}}}\\-2P\delta t\iint_{\R^2}\log|
y-z|u_t(z)d\mu_{V_{\phi,t_0}}(z) d\mu_{V_{\phi,t_0}}(y)+\int_\R\left[\log\Big(1+\delta t u_t(y)\Big)-\delta t u_t(y)\right]  d\mu_{V_{\phi,t_0}}(y)\Bigg).
\end{multline*}

Recognizing $\rho_{V_{\phi,t_0}}$ via the first three terms in the exponential, $u_t$ has to satisfy the following relation for all $x\in\R$,

\begin{multline*}
1+\delta t u_t(x)=\exp\Bigg\{\delta t\Big(-\phi(x)+\int_\R\phi(y) d\mu_{V_{\phi,t_0}}(y)+2P \int_\R\log\dfrac{|x-y|}{1+|x|}u_t(y)d\mu_{V_{\phi,t_0}}(y)
\\-2P\iint_{\R^2}\log\dfrac{|y-z|}{1+|y|}u_t(z)d\mu_{V_{\phi,t_0}}(z)d\mu_{V_{\phi,t_0}}(y)+\dfrac{1}{\delta t}\int_\R\left[\log\left(1+\delta t u_t\right)-\delta t u_t\right]  d\mu_{V_{\phi,t_0}}\Big) \Bigg\}.
\end{multline*}

Above, we have used that $-U^{u_t\rho_{V_{\phi,t_0}}}(x)=\displaystyle\int_\R\log\dfrac{|x-y|}{1+|x|} u_t(y)d\mu_{V_{\phi,t_0}}(y)$ which is justified by the fact that $\displaystyle\int_\R u(y)d\mu_{V_{\phi,t_0}}(y)=0$.
Conversely, any $u$ such that $\int_\R u(y)d\mu_{V_{\phi,t_0}}(y)=0$ and satisfying the previous relation, verifies for all $x\in\R$,

$$V_t(x)+2PU^{w}(x)+\log w(x)=\int_\R\Big\{V_t(y)+2PU^{w}(y)+\log w(y)\Big\}d\mu_{V_{\phi,t_0}}(y)$$

where we have set $w:=(1+\delta tu)\rho_{V_{\phi,t_0}}$. Because of this equation, $w$ can be written in exponential form as in (<ref>), it is thus positive and of mass $1$ which makes $d\mu(x):=w(x)dx$ a probability measure which satisfies the equation characterizing $\mu_{V_{\phi,t}}$, hence, by unicity of the solution of (<ref>), $ \mu_{V_{\phi,t}}=\mu$ and thus $u=u_t$.

We now expand $\exp$ into its Taylor-integral series of order 2, ie $e^x=1+x+x^2\int_{0}^{1}(1-s)e^{sx}ds$. By using that
$$\mcal{K}[v]=2P\displaystyle\int_\R\log\dfrac{|x-y|}{1+|x|} u(y)d\mu_{V_{\phi,t_0}}(y)-2P\iint_{\R^2}\log\dfrac{|y-z|}{1+|y|}v(z)d\mu_{V_{\phi,t_0}}(z)d\mu_{V_{\phi,t_0}}(y)$$

and $\mcal{T}[v]=v-\mcal{K}[v]$, we get:

\begin{multline*}
\mcal{T}[u_t](x)=-\phi(x)+\int_\R\phi d\mu_{V_{\phi,t_0}}
\\+\delta t\Bigg[\left(-\phi(x)+\int_\R\phi d\mu_{V_{\phi,t_0}}+\mcal{K}[u_t](x)+\dfrac{1}{\delta t}\int_\R\left[\log\left(1+\delta t u_t\right)-\delta t u_t\right]  d\mu_{V_{\phi,t_0}}\right)^2
\\\times\int_{0}^{1}\exp\left(s\delta t\left(-\phi(x)+\int_\R\phi d\mu_{V_{\phi,t_0}}+\mcal{K}[u_t](x) +\int_\R\frac{\log\left(1+\delta t u_t\right)-\delta t u_t}{\delta t} d\mu_{V_{\phi,t_0}}\right)\right) (1-s)ds
\\+\int_\R\dfrac{\log\left(1+\delta t u_t\right)-\delta t u_t}{\left(\delta t\right)^2 } d\mu_{V_{\phi,t_0}}\Bigg].
\end{multline*}

We next use the invertibility of $\mcal{T}$ to conclude.

The next theorem shows that for each $t$ sufficiently close to $t_0$, $\mcal{T}^{-1}\circ\mcal{V}_t$ is contractive on a ball of fixed radius. Let $n\geq0$, denote for all $R>0$, $\overline{\mscr{B}_n(0,R)}$ the closed ball of radius $R_n$ for the $W^{\infty}_n(\R)$-norm.

For all $\varepsilon_n>0$ small enough, there exists $R_n>0,$ such that for all $t\in]t_0-\varepsilon_n,t_0+\varepsilon_n[$, the operator
$$\mcal{T}^{-1}\circ\mcal{V}_t:\left(\overline{\mscr{B}_n(0,R_n)},\|.\|_{W^{\infty}_n(\R)}\right) \longrightarrow \left(\overline{\mscr{B}_n(0,R_n)},\|.\|_{W^{\infty}_n(\R)}\right)$$
is well-defined and continuous. Furthermore it is contractive, ie there exists $k_n\in]0,1[$, such that for all $v,w\in \overline{\mscr{B}_n(0,R_n)}$,
$$\left \|\mcal{T}^{-1}\circ\mcal{V}_t[v]-\mcal{T}^{-1}\circ\mcal{V}_t[w]\right \|_{W^{\infty}_n(\R)}\leq k_n\|v-w\|_{W^{\infty}_n(\R)}.$$
Moreover $k_n$ is independent of $t$ on $]t_0-\varepsilon_n,t_0+\varepsilon_n[$ for $\varepsilon_n>0$ small enough.

Let $\varepsilon>0$, $t\in]t_0-\varepsilon,t_0+\varepsilon[$, $v\in\overline{\mscr{B}_0(0,R)}$ for an arbitrary $0<R<\dfrac{1}{\delta t}$ (because otherwise the term $\int_\R\log(1+\delta tv)\rho_{V_{\phi,t_0}}$ in $\mcal{V}_t[v]$ might be ill-defined). We first show that $\|\mcal{V}_t[v]\|_{L^\infty(\R)}<+\infty$.

Furthermore, since for all $x\in\R, \displaystyle\left| \int_\R\log|x-y|v(y)\rho_{V_{t_0}}(y)dy\right|\leq\|v\|_{\infty}\int_\R\left|\log\dfrac{|x-y|}{1+|x|}\right|\rho_{V_{t_0}}(y)dy$ (because $\int_\R v\rho_{V_{t_0}}=0$) and b

First, by Taylor-Lagrange inequality, we deduce that

$$\Big|\int_\R\dfrac{\log\left[1+\delta t v(y)\right]-\delta t v(y)}{\delta t} d\mu_{V_{\phi,t_0}}(y)\Big|\leq \dfrac{\|v\|_{{L^\infty(\R)}}^2\delta t}{2}\sup_{x\in[-\|v\|_{\infty},\|v\|_{\infty}]}\dfrac{1}{(1+\delta t x)^2}\leq \dfrac{R^2\delta t}{2(1-\delta t R)^2}.$$

By recalling the definition of $\mcal{V}_t$ in Proposition <ref> and using the convexity of $x\mapsto x^2$, we get:

\begin{multline*}
\left \|\mcal{V}_t[v]\right \|_{L^\infty(\R)}\leq \delta t\Bigg[3\Bigg(4\|\phi\|_{L^\infty(\R)}^2+R^2\left \|\int_\R\left|\log\dfrac{|.-y|}{1+|.|}\right|d\mu_{V_{\phi,t_0}}(y)dy \right\|_{L^\infty(\R)}^2+\delta t^2\dfrac{R^4}{4(1-\delta t R)^4} \Bigg)
\\\times\exp\left\{\delta t\left( 2\|\phi\|_{L^\infty(\R)}+R\left \|\int_\R\left|\log\dfrac{|.-y|}{1+|.|}\right|d\mu_{V_{\phi,t_0}}(y)dy \right\|_{L^\infty(\R)} +\delta t\dfrac{R^2}{2(1-\delta t R)^2}\right)\right\}+\dfrac{R^2}{2(1-\delta t R)^2} \Bigg]
\\+2\|\phi\|_{L^\infty(\R)}.
\end{multline*}

The RHS is of the form $2\|\phi\|_{\infty}+\delta t g_0(\delta t)$ where $g_0$ is a positive function and with these notations

$$\big\|\mcal{T}^{-1}\circ\mcal{V}_t[v]\big\|_{L^\infty(\R)}\leq\mcal{C}_{\mcal{T},0}\left(2\|\phi\|_{L^\infty(\R)}+\delta t g_0(\delta t)\right).$$

Therefore, by choosing $t$ such that  $|t-t_0|<\varepsilon_0$ for $\varepsilon_0>0$ small enough, there exists $R_0>0$ such that $2\|\phi\|_{\infty}C_{\mcal{T},0}<R_0<\dfrac{1}{\varepsilon_0}$ and
This makes the operator  $\mcal{T}^{-1}\circ\mcal{V}_t:\left(\overline{\mscr{B}_n(0,R_n)},\|.\|_{W^{\infty}_n(\R)}\right) \longrightarrow \left(\overline{\mscr{B}_n(0,R_n)},\|.\|_{W^{\infty}_n(\R)}\right)$ well-defined for all $|t-t_0|<\varepsilon_0$. For the contractivity, let $u,v\in\overline{\mscr{B}(0,R_0)}$, we get by Proposition <ref>,

$$\left \|\mcal{T}^{-1}\Big[\mcal{V}_t[u]-\mcal{V}_t[v]\Big]\right \|_{L^\infty(\R)}\leq C_{\mcal{T},0}\left \|\mcal{U}_t[u]-\mcal{U}_t[v]\right \|_{L^\infty(\R)}.$$

where $\mcal{U}_t$ was defined in Proposition <ref>. We now want to control $\big|\mcal{U}_t[u](x)-\mcal{U}_t[v](x)\big|$. We have, by decomposing the sum

$$\big|\mcal{U}_t[u](x)-\mcal{U}_t[v](x)\big|\leq\delta t \left(\Delta_1+\Delta_2(x)\msf{E}[u](x) +\Delta_3(x)\mfrak{h}[v](x)^2\right) $$

\begin{align*}
\mfrak{h}[w](x)&:=-\phi(x)+\int_\R\phi d\mu_{V_{\phi,t_0}}+\mcal{K}[w](x) +\int_\R\frac{\log\left(1+\delta t w\right)-\delta t w}{\delta t} d\mu_{V_{\phi,t_0}}
\\\msf{E}[w](x)&:=\int_{0}^{1}\exp\Big[s\delta t\mfrak{h}[w](x)\Big] (1-s)ds
\\\Delta_1&:=\int_\R\left|\dfrac{\log(1+\delta t u)-\delta tu-\log(1+\delta t v)+\delta t v}{(\delta t)^2}\right|d\mu_{V_{\phi,t_0}}
\\\Delta_2(x)&:=\big|\mfrak{h}[u](x)^2-\mfrak{h}[v](x)^2\big|
\\\Delta_3(x)&:=\big|\msf{E}[u](x)-\msf{E}[v](x)\big|.
\end{align*}

First, Taylor Lagrange inequality leads to:

$$\Delta_1\leq\sup_{y\in[-R_0,R_0]}\dfrac{|y|}{1+\delta t y}\|u-v\|_{L^\infty(\R)}\leq\dfrac{R_0}{1-\delta t R_0}\|u-v\|_{L^\infty(\R)}.$$

Furthermore, by using $a^2-b^2=(a+b)(a-b)$, that $\mfrak{h}[u]$ and $\mfrak{h}[v]$ are bounded

\begin{multline*}
\Delta_2\leq 2 \max\Big(\|\mfrak{h}[u]\|_{L^\infty(\R)},\|\mfrak{h}[v]\|_{L^\infty(\R)}\Big)\Bigg[\Big|\mcal{K}[u-v](x)\Big|+\delta t \Delta_1\Bigg]
\\\leq\left(4\|\phi\|_{L^\infty(\R)}+8PR_0\left \|\int_\R\left|\log\dfrac{|.-y|}{1+|.|}\right|d\mu_{V_{\phi,t_0}}(y) \right\|_{L^\infty(\R)}+\dfrac{\delta tR_0^2}{(1-\delta t R_0)^2}\right) \|u-v\|_{L^\infty(\R)}
\\\times\left(\dfrac{R_0\delta t}{1-\delta t R_0}+4P\left \|\int_\R\left|\log\dfrac{|.-y|}{1+|.|}\right|.\rho_{V_{\phi,t_0}}(y)dy \right\|_{L^\infty(\R)}\right)\leq C(R_0)\|u-v\|_{L^\infty(\R)}
\end{multline*}

Similarly, there exists $C(R_0)>0$ such that:

\begin{equation*}
\Delta_3\|\mfrak{h}[v]\|_{L^\infty(\R)}\leq C(R_0)\delta t\|u-v\|_{L^\infty(\R)}
\end{equation*}

which finally leads to the existence of $C_0>0$ such that:

\begin{equation*}
\big\|\mcal{U}_t[u]-\mcal{U}_t[v]\big\|_{L^\infty(\R)}\leq C_0\delta t\|u-v\|_{L^\infty(\R)}.
\end{equation*}

We now choose $\varepsilon_0$ again small enough such that $k_0:=\delta tC_0C_{\mcal{T},0}<1$, this concludes the proof that $\mcal{T}^{-1}\circ\mcal{V}_t$ is contractive on $\overline{\mscr{B}_0(0,R_0)}$ with contractivity constant $k_0$.

To get the contractivity property for $\mcal{T}^{-1}\circ\mcal{V}_t$ on $W_n^{\infty}(\R)$, we adapt a similar strategy.
Let $u\in\overline{\mscr{B}_n(0,R)}$ with $\delta tR<1$, by Proposition <ref>,

$$\|\mcal{T}^{-1}[\mcal{V}_t[u]]\|_{W_n^\infty(\R)}\leq C_{\mcal{T},n}\|\mcal{V}_t[u]\|_{W_n^\infty(\R)}.$$

Furthermore, it is clear that every term appearing in the definition of $\mcal{V}_t[u]$ belongs to $W_n(\R)$, thus by the same argument as before there exists a positive function $g_n$ such that,
$$\|\mcal{T}^{-1}\circ\mcal{V}_t[u]\|_{W_n(\R)}\leq C_{\mcal{T},n}\Big(2\|\phi\|_{W_n^{\infty}(\R)}+\delta t g_n(R)\Big).$$
We conclude just as before that by taking $\delta t$ small enough, $\mcal{T}^{-1}\circ\mcal{V}_t:\overline{\mscr{B}_n(0,R)}\rightarrow \overline{\mscr{B}_n(0,R)}$ is well-defined.
Finally, just as before since for all $u,v\in W_n^{\infty}(\R)$

\begin{align*}
\mcal{U}_t[u]^{(n)}(x)&=\delta t\sum_{k=0}^{n}\binom{n}{k}\left(\mfrak{h}[u]^2\right)^{(k)}(x)\msf{E}[u]^{(n-k)}(x)
\\&=\delta t\sum_{0\leq i\leq k\leq n}\binom{n}{k}\binom{k}{i}\mfrak{h}[u]^{(i)}(x)\mfrak{h}[u]^{(k-i)}(x)\msf{E}[u]^{(n-k)}(x).
\end{align*}

Moreover, by the same controls as before it is easy to derive that for all $0\leq i\leq k\leq n$, for all $u,v\in \overline{\mscr{B}_n(0,R_n)}$,
$$\|\mfrak{h}[u]^{(i)}\mfrak{h}[u]^{(k-i)}\msf{E}[u]^{(n-k)}-\mfrak{h}[v]^{(i)}\mfrak{h}[v]^{(k-i)}\msf{E}[v]^{(n-k)}\|_{L^\infty(\R)}\leq C(R_n,n,i,k)\|u-v\|_{W_n^{\infty}(\R)}.$$
This is enough to conclude that
$$\left \|\mcal{T}^{-1}\circ\mcal{V}_t[u]-\mcal{T}^{-1}\circ\mcal{V}_t[v]\right \|_{W_n^{\infty}(\R)}\leq \delta tC_{\mcal{T},n}C(R_n,n,i,k)\|u-v\|_{W_n^{\infty}(\R)}.$$

Finally, by taking $\delta t$ small enough, we conclude that $\mcal{T}^{-1}\circ\mcal{V}_t$ is contractive on $\overline{\mscr{B}_n(0,R_n)}$ with contractivity constant  $k_n:=\delta tC_{\mcal{T},n}C(R_n,n,i,k)<1$.

The form of the derivatives of $\mcal{V}_t[u]$ is similar to the form of $\mcal{V}_t[u]$, more explicitly for all
$u\in W_{j}^{\infty}(\R)\cap \{u\in L^\infty(\R),\int u\rho_{V_{t_0}}=0\}$, $\mcal{V}_t[u]'(x)=-\phi'(x)+\delta t\mcal{U}_t[u]'(x)$ and

\begin{multline*}
\mcal{U}_t'[u](x)=\int_{0}^{1}\exp\left(s\delta t\left(-\phi(x)+\int_\R\phi d\mu_{V_{t_0}}+\mcal{K}[u](x) +\int_\R\frac{\log\left(1+\delta t u\right)-\delta t u}{\delta t} d\mu_{V_{t_0}}\right)\right) (1-s)ds
\\\times	\left(-\phi(x)+\int_\R\phi d\mu_{V_{t_0}}+\mcal{K}[u](x)+\dfrac{1}{\delta t}\int_\R\left[\log\left(1+\delta t u\right)-\delta t u\right]  \rho_{V_{t_0}}\right)
\left(-\phi'(x)+\int_{\R}\log|x-y|(u\rho_{V_{t_0}})'(y)dy\right)
\\\times\left (2+s\delta t\left(-\phi(x)+\int_\R\phi d\mu_{V_{t_0}}+\mcal{K}[u](x)+\dfrac{1}{\delta t}\int_\R\left[\log\left(1+\delta t u\right)-\delta t u\right]  \rho_{V_{t_0}}\right) \right ).
\end{multline*}

It is then obvious that with the same estimates, by lowering $\delta t$ enough, using that $\int_\R \left(u\rho_{V_{t_0}}\right)'=0$ so we can replace all the $\log|x-y|$ by $\log\dfrac{|x-y|}{1+|x|}$, there exists $R_1>R_0$ such that if $u\in\overline{\mscr{B}(0,R_1)}$ (for $\|.\|_{W^\infty_1(\R)}$), $\left \|\mcal{T}^{-1}\circ \mcal{V}_t[u]\right \|_{W^{\infty}_1(\R)}\leq R_1$ even if it means lowering $\delta t$. Moreover, we can show that there exists $\widetilde{\widetilde{C}}(R_1)>0$ such that:

$$\left \|\mcal{V}_t[u]-\mcal{V}_t[v]\right \|_{W^\infty_1(\R)}\leq \widetilde{\widetilde{C}}(R_1)\delta t\|u-v\|_{W^\infty_1(\R)}\leq \widetilde{\widetilde{C}}(R_1)\delta t\|u-v\|_{W^{\infty}_i(\R)}.$$

Again lowering $\delta t$ enough, we can suppose $\widetilde{\widetilde{C}}(R_1)\delta t<1$. Itering this procedure leads to the existence of a constant $k_i>0$ such that:

$$\left \|\mcal{V}_t[u]-\mcal{V}_t[v]\right \|_{W^\infty_i(\R)}\leq k_i\|u-v\|_{W^{\infty}_i(\R)}.$$

This allow us to conclude that $\mcal{T}^{-1}\circ\mcal{V}_t$ is contractive on $X_i$ for the $W_i^{\infty}(\R)$-norm which makes this space complete. By Banach fixed point theorem, there exists a unique fixed point $v_t$ of $\mcal{T}^{-1}\circ\mcal{V}_t$ belonging to $\overline{\mscr{B}(0,R_i)}\cap\{w\in L^\infty(\R),\int_\R w\rho_{V_{t_0}}=0\}$. By construction $v_t=u_t$.

since $\mcal{T}^{-1}$ is continuous we have:

$$\left \|\mcal{T}^{-1}[\mcal{V}_t[v]]\right \|_\infty\leq C_{\mcal{T}}\left \|\mcal{V}_t[v]\right \|_\infty.$$

Note that the definition of $u_t$ as an element of $W^{\infty}_n(\R)$ depends on $t_0$ and that we only proved the characterization of $u_t$ as a fixed point for $t\in]t_0-\varepsilon_n,t_0+\varepsilon_n[\setminus\{t_0\}$ with $\varepsilon_n>0$ small enough (we stress that we successively lowered $\delta t$ when increasing $i$.) Furthermore, since  for all $v\in L^\infty(\R)$

$$\mcal{T}^{-1}\circ\mcal{V}_{t_0}[v]=-\phi+\displaystyle\int_\R\phi d\mu_{V_{\phi,t_0}},$$

then we can set $u_{t_0}:=-\phi+\displaystyle\int_\R\phi d\mu_{V_{\phi,t_0}}$ which is obviously the unique fixed point in $ \displaystyle\bigcap_{n\in\N}W_n^\infty{(\R)}$ of $\mcal{T}^{-1}\circ\mcal{V}_{t_0}$.

§.§ Regularity of the equilibrium measure

We now prove the continuity of $t\mapsto u_t\in W_n^{\infty}(\R)$.

Let $n\in\N$, $ \varepsilon_n>0$ and $R_n>0$ be as in Theorem <ref>. For all $v\in\overline{\mscr{B}_n(0,R_n)}$, for all $(t_p)_p\in\left(]t_0-\varepsilon_n,t_0+\varepsilon_n[\right)^{\N}$ such that $t_p\tend{p\rightarrow\infty}t\in]t_0-\varepsilon_n,t_0+\varepsilon_n[$ then
$$\|\mcal{T}^{-1}\circ\mcal{V}_{t_p}[v]-\mcal{T}^{-1}\circ\mcal{V}_{t}[v]\|_{W_n^{\infty}(\R)}\underset{p\rightarrow \infty}{\longrightarrow}0.$$

The proof is based on the fact that there exists a neighboorhood $U_n$ of $t_0$ such that for all $v\in \overline{\mscr{B}_n(0,R_n)}$, $t\in U_n\mapsto\mcal{V}_t[v]\in W_n^{\infty}(\R)$ is continuous. Since $\mcal{T}^{-1}$ is also continuous in $W_n^{\infty}(\R)$-norm, we can conclude.

[Continuity of the fixed-point]Let $n\in\N$, for all $t,t'\in]t_0-\varepsilon_n,t_0+\varepsilon_n[$,
$$\|u_t-u_{t'}\|_{W_n^{\infty}(\R)}\underset{t\rightarrow t'}{\longrightarrow}0.$$

Let $(t_n)_n\in\left(]t_0-\varepsilon,t_0+\varepsilon[\right)^\N $ such that $t_n\underset{n\rightarrow\infty}{\longrightarrow}t\in]t_0-\varepsilon_n,t_0+\varepsilon_n[$. First
$$\left \|u_{t_n}-\mcal{T}^{-1}\circ\mcal{V}_{t_n}[u_{t}]\right \|_{W_n^{\infty}(\R)}=\left \|\mcal{T}^{-1}\circ\mcal{V}_{t_n}[u_{t_n}]-\mcal{T}^{-1}\circ\mcal{V}_{t_n}[u_{t}]\right \|_{W_n^{\infty}(\R)}\leq k_i\left \|u_{t_n}-u_t\right \|_{W_n^{\infty}(\R)}$$

and by the triangle inequality, we obtain:

\begin{multline*}
\left \|u_{t}-\mcal{T}^{-1}\circ\mcal{V}_{t_n}[u_{t}]\right \|_{W_n^{\infty}(\R)}\geq \left \|u_{t_n}-u_t\right \|_{\infty}-\left \|u_{t_n}-\mcal{T}^{-1}\circ\mcal{V}_{t_n}[u_{t}]\right \|_{W_n^{\infty}(\R)}
\geq (1-k_n)\left \|u_{t_n}-u_t\right \|_{W_n^{\infty}(\R)}.
\end{multline*}

Nevertheless by Lemma <ref>, $\left \|u_{t}-\mcal{T}^{-1}\circ\mcal{V}_{t_n}[u_{t}]\right \|_{W_n^\infty(\R)}\underset{n\rightarrow\infty}{\longrightarrow}0$ because $u_t$ is the fixed point of $\mcal{T}^{-1}\circ\mcal{V}_t$ which is a continuous operator with respect to $t$ hence $\left \|u_{t_n}-u_t\right \|_{W_n^\infty(\R)}\underset{n\rightarrow\infty}{\longrightarrow}0$.

Let $t,t_0\in[0,1]$,

$$\left \|\rho_{V_{t}}-\rho_{V_{\phi,t_0}}\right \|_{W_n^{\infty}(\R)}\underset{t\rightarrow t_0}{\longrightarrow}0.$$

Furthermore, for all $x\in\R$, $k\in\N$,  $t\mapsto\partial_x^{k}\rho_{V_{t}}(x)\in\mcal{C}^\infty(\R)$ and satisfies the following partial differential equation:

$$\partial_t\partial_{x}^{k}\rho_{V_{t}}(x)=\partial_x^{k}\left[ \left(-\phi+\int_\R\phi(s)\rho_{V_{t}}(s)ds\right) \rho_{V_{t}}\right] (x).$$

By setting $u_t:=\dfrac{\rho_{V_t}-\rho_{V_{t'}}}{t-t'}\dfrac{1}{\rho_{V_{t'}}}$,

$$\|\rho_{V_{t}}-\rho_{V_{t'}}\|_{W_n^{\infty}(\R)}=|t-t'|\left \|\rho_{V_{t'}}u_t \right \|_{W_n^{\infty}(\R)}\leq 2^n|t-t'| \left \|\rho_{V_{t'}}\right \|_{W_n^{\infty}(\R)}.\left \|u_t \right \|_{W_n^{\infty}(\R)}.$$

By Corollary <ref>, $\left \|u_t \right \|_{W_n^{\infty}(\R)}\tend{t\rightarrow t'}\left \|u_{t'}\right \|_{W_n^{\infty}(\R)}$, thus the right-hand side goes to zero proving the claim.

For the second point, we notice that

$$\left \|\dfrac{\rho_{V_{t}}-\rho_{V_{\phi,t_0}}}{\delta t}-\rho_{V_{\phi,t_0}}u_{t_0}\right \|_{W_n^{\infty}(\R)}=\left \|(u_t-u_{t_0})\rho_{V_{\phi,t_0}}\right \|_{W_n^{\infty}(\R)}\leq2^n\|u_t-u_{t_0} \|_{W_n^{\infty}(\R)}\|\rho_{V_{\phi,t_0}} \|_{W_n^{\infty}(\R)}.$$

Since the RHS goes to zero as $t\rightarrow t_0$, $u_{t_0}=-\phi+\displaystyle\int_\R\phi d\mu_{V_{\phi,t_0}}$ and $n$ is arbitrary, we conclude that, $x\in\R$, $t\mapsto\rho_{V_{t}}(x)$ is differentiable at every $t\in[0,1]$ of derivative
$$\partial_t\rho_{V_{t}}^{(k)}(x)=-\left(\rho_{V_{t}}\phi\right)^{(k)} (x)+\rho_{V_{t}}^{(k)}(x)\int_\R\phi d\mu_{V_{t}}(x).$$

Since the above expression is again differentiable in $t$ (one deals with the integral by dominated convergence theorem with the domination $|\phi(x)\partial_t\rho_{V_{t}}(x)|\leq2\|\phi\|_\infty^2(1+\max_{s\in[0,1]}\|u_s\|_{\infty})\rho_{V_{\phi,t_0}}(x)$ for an arbitrary $t_0$), we conclude that for all $x\in\R$, $t\mapsto\rho_{V_{t}}(x)\in\mcal{C}^\infty(\R)$.

[Convergence of moments]
Let $h\in\N$, $t\in[0,1]$ by denoting $m(t,h):=\displaystyle\int_\R x^{h}d\mu_{V_t}$, we have for all $t_0\in[0,1]$

$$m(t,h)\underset{t\rightarrow t_0}{\rightarrow}m(t_0,h).$$

For all $t\in[0,1]$, for all $x\in\R$, $\rho_{V_t}(x)\leq (1+\max_{s\in[0,1]}\|u_s\|_{\infty})\rho_{V_{\phi,t_0}}(x)$, hence by dominated convergence theorem $|m(t,h)-m(t_0,h)|\underset{t\rightarrow t'}{\rightarrow}0$.

§ PROOF OF THEOREM <REF>

§.§ Asymptotic expansion of the partition function for the Gaussian potential

The asymptotic behaviour of $\mcal{Z}_N\left [V_G\right ]$ can be deduced from Mehta's formula <cit.>
\begin{equation}\label{partition,function gaussian}
\mcal{Z}_N\left [V_G\right ]=(2\pi)^{N/2}\prod_{a=1}^N\dfrac{\Gamma\left (1+\dfrac{aP}{N}\right )}{\Gamma\left (1+\dfrac{P}{N}\right )}.
\end{equation}

This will allow us to use this formula in our interpolating integration formula to deduce the asymptotic expansion of $\log\mcal{Z}_N[V_\phi]$. From the previous equation, we can deduce the asymptotic behaviour of $\log	\mcal{Z}_N\left [V_G\right ]$. It is given by the following theorem:

There exists a sequence $(g_k)_{k\geq0}\in\R^N$, such that for all $K\geq0$,

\begin{equation}\label{thm:asymptfctpartgauss}
\dfrac{1}{N}\log\mcal{Z}_N\left [V_G\right ]=\sum_{k=0}^{K}\dfrac{g_k}{N^{k}}+O\left(N^{-(K+1)}\right)
\end{equation}

$$g_1:=\gamma\dfrac{P}{2}+\dfrac{\log(1+P)}{2}+\dfrac{1}{2}\sum_{j\geq1}\left(\log\left(1+\dfrac{P+1}{j}\right)-\log\left(1+\dfrac{1}{j}\right)-\dfrac{P}{j}    \right).$$

Above $\gamma$ denotes the Euler-Mascheroni constant.

We first use (<ref>) to deduce

\begin{equation}\label{eq:logpartifuncgaussian}
\log\mcal{Z}_N\left [V_G\right ]=\dfrac{N\log(2\pi)}{2}+\sum_{a=1}^N\Gamma\left (1+\dfrac{aP}{N}\right )-N\log\Gamma\left (1+\dfrac{P}{N}\right ).
\end{equation}

Let $K>0$, using the Taylor series expansion of $\log\Gamma$ around 1 (see <cit.>), one has

\begin{equation}\label{taylor:loggamma}
-N\log\Gamma\left (1+\dfrac{P}{N}\right )=\gamma P-\sum_{k=1}^{+\infty}\dfrac{\zeta(k+1)}{k+1}\dfrac{(-P)^{k+1}}{N^{k}}.
\end{equation}

where $\zeta$ denotes the Riemann $\zeta$ function. The second term in (<ref>) can be estimated by using the Weierstrass product formula for $\dfrac{1}{\Gamma}$:

\begin{equation}\label{key}
\dfrac{1}{\Gamma(z)}=e^{\gamma z}z\prod_{j=1}^{+\infty}(1+\dfrac{z}{j})e^{-z/j}
\end{equation}

which is valid for any $z\in\C$. Hence we deduce that:

\begin{multline}\label{otherterm}
\sum_{k=1}^N\log\Gamma\left (1+\dfrac{kP}{N}\right )=-\sum_{k=1}^N\left (\gamma\Big(1+\dfrac{k P}{N}\Big)+\log\left (1+\dfrac{kP}{N}\right )-S_N(k) \right )
\\=-\gamma N-\gamma\dfrac{(N+1)P}{2}-\sum_{k=1}^N\log\left (1+\dfrac{kP}{N}\right )+\sum_{k=1}^{N}S_N(k)
\end{multline}
where $S_N(k):=-\displaystyle\sum_{j=1}^{+\infty}\left[\log\left (1+\dfrac{1}{j}+\dfrac{kP}{Nj}\right )-\dfrac{1}{j}-\dfrac{kP}{Nj}\right] $.
By the Euler-Maclaurin formula, we have the following identity for any $K>0$:

\begin{equation}\label{eulermaclaurin1}
\sum_{k=1}^N\log\left (1+\dfrac{kP}{N}\right )=\int_{0}^{N}f_N(t)dt+\sum_{k=1}^{K+2}\dfrac{B_k}{k!}\left (f_N^{(k-1)}(N)-f_N^{(k-1)}(0)\right )+R_{K+2}^{(N)}
\end{equation}

where $f_N(x):=\log \left (1+\dfrac{xP}{N}\right )$ and $B_k$ is the $k$-th Bernoulli number. The remainder $R_{K+2}^{(N)}$ is defined by $R_{K+2}^{(N)}:=(-1)^{K+1}\displaystyle\int_{0}^{N}f_N^{(K+2)}(t)\dfrac{\widetilde{B}_{K+2}(t-\lfloor t \rfloor)}{(K+2)!}dt$, where $\widetilde{B}_{K+2}$ is the $(K+2)$-th Bernoulli polynomial. By using the following bound on Bernoulli polynomials,

$$\forall x\in[0,1], \forall k>0,\hspace{1,5cm} |\widetilde{B}_k(x)|\leq2\dfrac{k!}{(2\pi)^k}\zeta(k)$$

where $\zeta$ is the Riemann zeta function, $R_{K+2}^{(N)}$ can be controlled by the following inequalities:

\begin{align*}
|R_{K+2}^{(N)}|\leq\dfrac{2\zeta(K+2)}{(2\pi)^{K+2}}\int_{0}^{N}|f_N^{(K+2)}(t)|dt&=\dfrac{2\zeta(K+2)}{(2\pi)^{K+2}}\int_{0}^{N}\dfrac{P^{K+2}}{N^{K+2}}\dfrac{(K+1)!}{\left (1+\dfrac{P}{N}t\right )^{K+2}}dt
\\&=\dfrac{2\zeta(K+2)(K+1)!}{(2\pi)^{K+2}}\dfrac{P^{K+1}}{N^{K+1}}\int_{0}^{P}\dfrac{du}{\left (1+u\right )^{K+2}}=O\Big(N^{-{(K+1)}}\Big).
\end{align*}
Extracting the large $N$-behaviour in (<ref>) leads to

\begin{multline}\label{euleurmaclaurinNextracted}
\sum_{k=1}^N\log\left (1+\dfrac{kP}{N}\right )=\dfrac{N}{P}\int_{0}^{P}\log(1+t)dt+B_1\log(1+P)\\+\sum_{k=2}^{K+2}\dfrac{B_k}{k!}(-1)^{k}(k-2)!\dfrac{P^{k-1}}{N^{k-1}}\left (\dfrac{1}{(1+P)^{k-1}}-1\right )+O\Big(N^{-(K+1)}\Big)
\\=Nc_{-1}+\sum_{k=0}^{K}\dfrac{c_{k}}{N^{k}}+O\Big(N^{-(K+1)}\Big)
\end{multline}

where $c_{-1}:=(1+P^{-1})\log(1+P)-1$, $c_0:=\dfrac{\log(1+P)}{2}$ and for all $k\in\llbracket1,K\rrbracket$,

$$c_{k}=\dfrac{-B_{k+1}(-P)^{k}}{k(k+1)}\left (\dfrac{1}{(1+P)^{k}}-1\right ).$$
Also by Fubini's theorem, we get,

\begin{align*}\label{Fubini}
\sum_{k=1}^NS_N(k)&=-\sum_{j=1}^{+\infty}	\left\{\sum_{k=1}^N\log\left (1+\dfrac{1}{j}+\dfrac{kP}{Nj}\right)-\dfrac{1}{j}-\dfrac{kP}{Nj}\right\}
\\&=\sum_{j=1}^{+\infty}\left\{\dfrac{N}{j}+\dfrac{(N+1)P}{2j}+\sum_{k=1}^Ng_{N,j}(k)\right\}
\end{align*}

where $g_{N,j}(x)=\log\left (1+\dfrac{1}{j}+\dfrac{P}{jN}x\right )$. The first equality clearly shows that the RHS is a serie of general term bounded by $O\left (j^{-2}\right )$, so it converges and justifies the application of Fubini's theorem. Again by Euler-Maclaurin formula, we get:

\begin{multline}\label{key}
\sum_{k=1}^Ng_{N,j}(k)=-\int_{0}^{N}\log\left (1+\dfrac{1}{j}+t\dfrac{P}{Nj}\right )dt-\dfrac{1}{2}\left [\log\left (1+\dfrac{P+1}{j}\right )-\log\left (1+\dfrac{1}{j}\right )\right ]\\-\sum_{k=2}^{K+2}\dfrac{(-1)^{k}B_k}{k(k-1)}\left\{\left (1+\dfrac{P+1}{j}\right )^{1-k}-\left (1+\dfrac{1}{j}\right )^{1-k}\right\}\left (\dfrac{P}{Nj}\right )^{k-1}+R_{K+2}^{(N)}(j)
\end{multline}

where again the new remainder $R_{K+2}^{(N)}(j)$ can be controlled via
\begin{multline*}
|R_{K+2}^{(N)}(j)|\leq\dfrac{2\zeta(K+2)}{(2\pi)^{K+2}}\int_{0}^{N}\dfrac{(K+1)!\left(\dfrac{P}{Nj}\right)^{K+2} }{\left(1+\dfrac{1}{j}+t\dfrac{P}{jN}\right)^{K+2}}dt\\=\dfrac{2\zeta(K+2)}{(2\pi)^{K+2}}\left(\dfrac{P}{Nj}\right)^{K+1}\left\{\left (1+\dfrac{P+1}{j}\right )^{-(K+1)}-\left (1+\dfrac{1}{j}\right )^{-(K+1)}\right\}=O\Bigg (\dfrac{1}{(Nj)^{K+1}}\Bigg )
\end{multline*}

where $O\left (\dfrac{1}{j^{K+1}}\right )$ depends on $K$ and $P$ but not on $N$. Hence we deduce that

\begin{multline*}
\sum_{k=1}^Ng_{N,j}(k)=-N\int_{0}^{1}\log\left (1+\dfrac{1+sP}{j}\right )ds-\dfrac{1}{2}\left [\log\left (1+\dfrac{P+1}{j}\right )-\log\left (1+\dfrac{1}{j}\right )\right ]\\+\sum_{k=1}^{K}\dfrac{(-P)^{k}B_{k+1}}{k(k+1)}\left\{\left (1+\dfrac{P+1}{j}\right )^{-k}-\left (1+\dfrac{1}{j}\right )^{-k}\right\}\dfrac{1}{(Nj)^k}+O\left ((jN)^{-(K+1)}\right ).
%+u_j^{(\text{rem})}O\left (N^{-(K+1)}\right )
\end{multline*}

This leads to:

\begin{multline}\label{}
\sum_{k=1}^NS_N(k)=\sum_{j=1}^{+\infty}\left[\dfrac{N}{j}+\dfrac{(N+1)P}{2j}+\sum_{k=1}^Ng_{N,j}(k)\right] =\sum_{j=1}^{+\infty}\left(u_j^{(1)}N+\sum_{k=0}^{K}u_j^{(k)}N^{-k}\right) +O(N^{-(K+1)})\\=d_1N+\sum_{k=0}^{K}d_{-k}N^{-k} +O\Big(N^{-(K+1)}\Big)
\end{multline}

where for all $k=-1,0\dots K$, $\left( u_j^{(k)}\right )_{j>0}\in\ell^1(\N^*)$ and $d_{-k}\in\R$. This establishes the existence of the asymptotic expansion of $	\log	\mcal{Z}_N\left [V_G\right]$ up to $O\Big(N^{-(K+1)}\Big)$. Collecting the different terms, leads to the formula for $g_1$.

Note that in, a lot of coefficients simplifie due to the fact that $B_{2k+1}=0$ for $k\geq1$.

§.§ Free energy of the model

Only, in this subsection, since the parameter $P$ varies, we include the $P$-dependance of $\mcal{Z}_N[V]$ in the notation and write $\mcal{Z}_N^P[V]$ instead.
[Free energy formula for Gaussian Potential]
Let $P>0$, the free energy associated with the Gaussian potential is

\begin{equation}
F(P):=\lim_{N\rightarrow\infty}N^{-1}\log\mcal{Z}_N^P\left [2PV_G\right ]=-\dfrac{1+P}{2}\log(2P)+\dfrac{\log(2\pi)}{2}+\int_{0}^{1}\log\Gamma\left(1+Px\right)dx
\end{equation}

As $P$ goes to $+\infty$, we have:

\begin{equation}\label{free:energy formula asympto}
F(P)=-P\left(\dfrac{3+\log 2}{2}\right) -\dfrac{1+\log 2}{2}+\log(2\pi)+\dfrac{\log P}{12 P}+O\left(P^{-1}\right)
\end{equation}

By a change of variable, it holds that $\mcal{Z}_N^P\left [2PV_G\right ]=\left(\sqrt{2P}\right)^{-N-P(N-1)} \mcal{Z}_N^P\left[V_G\right]. $ Hence by Mehta's formula (<ref>),

$$F(P):=\lim_{N\rightarrow\infty}N^{-1}\log\mcal{Z}_N^P\left [2PV_G\right ]=-\dfrac{1+P}{2}\log(2P)+\dfrac{\log(2\pi)}{2}+\int_{0}^{1}\log\Gamma\left(1+Px\right)dx.$$

We can replace the last term by its asymptotic expansion so that

$$\int_{0}^{1}\log\Gamma\left(1+Px\right)dx=\dfrac{(P+1)}{2}\log P-\dfrac{3P}{2}+\dfrac{\log(2\pi)-1}{2}+\dfrac{1}{12P}\log P+O\left(P^{-1}\right).$$

We used the classic formula to conclude

$$\log \Gamma(1+Px)=(1+Px)\log (1+Px)-1-Px-\dfrac{\log(1+Px)-\log(2\pi)}{2}+\dfrac{1}{12(1+Px)}+O\left(\dfrac{1}{(1+Px)^3}\right).$$

§.§ Interpolation with general potential

We first establish the link between the 1-linear statistics and the partition function with general potential and the one with Gaussian potential.

Let $V_t(x)=tV(x)+(1-t)V_G(x)$ with $t\in[0,1]$. We have

\begin{equation}\label{eq:interpolation}
\log\dfrac{\mcal{Z}_N[V]}{	\mcal{Z}_N\left [V_G\right ]}=-N\int_{0}^{1}\Braket{V-V_G}_{L_N}^{V_t}dt
\end{equation}

By the fundamental theorem of calculus:

$$\log\dfrac{\mcal{Z}_N[V]}{\mcal{Z}_N\left [V_G\right ]}=\int_{0}^{1}\partial_t\log \mcal{Z}_N[V_t]dt=-\int_{0}^{1}dt\int_{\R^N}p_N^{V_t}(\underline{x})\sum_{i=1}^{N}\partial_t V_t(x_i)d^N\underline{x}.$$

Since $\displaystyle\sum_{i=1}^{N}\partial_t V_t(x_i)=N\displaystyle\int_\R \left[V(x)-V_G(x)\right] dL_N(x)$, where $L_N$ is the empricial measure associated to the external potential $V_t$, it concludes the proof.

For all $\phi\in\cap_{k\geq0}H^k(\R)$, there exists a sequence $(c_i)_{i\geq0}\in\R^{\N}$ depending on $\phi$ and $P$ such that for all $K\geq0$

$$\dfrac{1}{N}\log \mcal{Z}_N\left[V_{G,\phi}\right]=\sum_{i=0}^{K}\dfrac{c_i}{N^i}+O\left(N^{-(K+1)}\right) .$$

The leading term $c_{0}$ is equal to the following expression:

$$\int_\R V_{G,\phi}(x)d\mu_{V_{G,\phi}}(x)-P\iint_{\R^2}\log|x-y|d\mu_{V_{G,\phi}}(x)d\mu_{V_{G,\phi}}(y)+\int_{\R}\log\left(\dfrac{d\mu_{V_{G,\phi}}(x)}{dx}\right) d\mu_{V_{G,\phi}}(x).$$

The subleading term $c_1$ can be written as

\begin{multline}
\\-P\int_{0}^{1}\left[\Braket{\partial_1\widetilde{\Xi^{-1}}\phi}_{\mu_{V_{G,\phi,t}}}+\Braket{\Theta^{(2)} \circ\widetilde{\Xi_1^{-1}}\left[\partial_2\mcal{D}\circ\widetilde{\Xi^{-1}}\phi\right]  }_{\mu_{V_{G,\phi,t}}}\right] dt.
\end{multline}

By Lemma <ref> and Theorem <ref>, to establish the asymptotic expansion of $\mcal{Z}_N\left[V_{G,\phi}\right]$, it suffices to obtain the one for $\displaystyle\int_{0}^{1}\Braket{\phi}_{L_N}^{V_{G,\phi,t}}dt$. By Theorem <ref>, we get

Finally, we conclude that the last integral is a $O\left(N^{-(K+1)}\right)$ by (<ref>) and the continuity on  $t\mapsto\msf{P}_{K,k}^{V_t}$ obtained in Proposition <ref>. Furthermore, by collecting order $1$ for $\log\mcal{Z}_N\left[V_G\right]$ and $\displaystyle\int_0^1d_{a}^{(1),V_{G,\phi,t}}(\phi)dt$ for $a=1$ in Theorem <ref>, we infer on the value of$c_{1}$.

§ CONCLUSION

This work adapted the analysis of the loop equations method to prove the existence of a $N^{-1}$ asymptotic expansion for a general class of potential. This class include all potentials given by $x^2+\phi$ where $\phi$ is a smooth bounded function. An immediate continuation of this result would be to extend it to more general confining potentials like $x^4$ for example. Our method relied on new controls on the equilibrium measures resulting from an energy minimization and entropy maximization. A natural question would be to extend these ideas to more general interactions.

§ APPENDIX: LEMMAS AND TECHNICAL RESULTS

[Properties of the Hilbert transform]

i) As a consequence, $\pi^{-1}\mathcal{H}$ is an isometry of $L^2(\R)$, and $\mathcal{H}$ satisfies on $L^2(\R)$ the identity $\mathcal{H}^2=-\pi^2I$.

ii) Derivative: For any $f\in H^1(\R)$, $\mathcal{H}[f]$ is also $H^1(\R)$ and $\mcal{H}[f]'=\mcal{H}[f']$.

iii) For all $p>1$, the Hilbert transform can be extended as a bounded operator $\mathcal{H}:L^p(\R)\to L^p(\R)$.

iv) Skew-self adjointness: For any $f,g\in L^2(\R)$, $\Braket{\mcal{H}[f],g}_{L^2(\R)}=-\langle f,\mcal{H}[g] \rangle_{L^2(\R)}$.

v) For all $\delta>0$, for all $f\in L^1(\R)$ such that $f'\in L^\infty(\R)$, $\|\mcal{H}[f]\|_{\infty}\leq(\delta^{-1}\|f\|_{1}+2\delta\|f'\|_\infty)$

We refer to [34] for the proofs of properties i)-iv). To prove v), let $f$ be such a function,

\leq \lim_{\varepsilon\rightarrow0}\Big|\int_{\varepsilon\leq|x-y|
\leq\delta}\dfrac{f(y)dy}{y-x}\Big|+\lim_{\varepsilon\rightarrow0}\Big|\int_{\delta\leq|x-y|\leq\varepsilon^{-1}}\dfrac{f(y)dy}{y-x}\Big|.$$

The second term in the RHS can be bounded by $\delta^{-1}\|f\|_{L^1(\R)}$ while the first term verifies

The first term in the RHS can be bounded by $2\delta\|f'\|_{\infty}$ while the second is equal to $0$. This allows to conclude.

We recall some results obtained in [24].

Let $u\in L^2(\R)$ be such that $\int_\R u(t)dt$ exists and let $f:t\mapsto tu(t)\in H^1(\R)$ then
$$\mcal{H}[u](x)\underset{|x|\rightarrow\infty}{\sim}\displaystyle\dfrac{-\int_\R u(t)dt}{x}.$$

Moreover if  $\displaystyle\int_\R u(t)dt=0$, $\displaystyle\int_\R f(t)dt$ exists and $g:t\mapsto t^2u(t)\in H^1(\R)$, then
$$\mcal{H}[u](x)\underset{|x|\rightarrow\infty}{\sim}\displaystyle\dfrac{-\displaystyle\int_\R tu(t)dt}{x^2}.$$

As a consequence, we obtain that $\mcal{H}[\rho_V](x)\underset{|x|\rightarrow\infty}{\sim}-x^{-1}$ and the logarithmic potential $U^{\rho_V}$ is Lipschitz bounded, with bounded derivative $\mathcal{H}[\rho_V]$.

There exists a complete orthonormal set $(\varphi_n)_{n\geq 1}$ of $\mathsf{H}$ of eigenfunctions of $-\mathcal{L}$, associated to the eigenvalues $0<\lambda_1(-\mathcal{L})<\lambda_2(-\mathcal{L})<\ldots$ going to infinity. Furthermore, denoting by $0<\lambda_1(\mathcal{A})<\lambda_2(\mathcal{A})<\ldots$ the eigenvalues of $\mathcal{A}$, we have for all $n\geq 1$

$$ \lambda_n(-\mathcal{L}) \geq \lambda_n(\mathcal{A})\,.$$

Furthermore, for all $u\in \mathsf{H}$,

$$ \mathcal{L}^{-1}u = \sum_{n\geq 1} \lambda_n(\mathcal{L})^{-1}\Braket{u,\varphi_n}_{\mathsf{H}}\varphi_n\,,$$
and $\mathcal{L}:\mathcal{D}(\mathcal{L})\to \mathsf{H}$ is a bijection.

By the second point of Lemma <ref>, $\mathcal{L}$ is self-adjoint on $\mathsf{H}$. It suffices to establish that $\mathcal{L}$ has a compact inverse, and then apply the spectral theorem. Using <cit.>[Theorem XIII.64] we show that $-\mathcal{L}=A+2P\mathcal{W}$ has compact resolvent, as a sum of the positive operator with compact resolvent $\mathcal{A}$ and the positive operator $\mathcal{W}$: Indeed, by <cit.>[Theorem XIII.64], it suffices to show that the set

$$Q^\mathsf{H}_{-\mathcal{L}}:=\left\{ u\in \mathsf{H}\ |\ \Braket{-\mathcal{L}u,u}_{\mathsf{H}}<+\infty\right\} =\left\{u\in \mathsf{H}\ |\ \Braket{(\mathcal{A}+2P\mathcal{W})u,u}_{\mathsf{H}}<+\infty\right\}$$

is relatively compact in $\mathsf{H}$. By positivity of $\mathcal{W}$,

$$ Q^\mathsf{H}_{-\mathcal{L}} = Q^\mathsf{H}_{\mathcal{A}} \cap Q^\mathsf{H}_{\mathcal{W}}\subset Q^\mathsf{H}_{\mathcal{A}}\,.$$

Now, because of Proposition <ref>, $\mathcal{A}^{-1}$ is a compact operator of $\mathsf{H}$, and using again <cit.>[Theorem XIII.64] we deduce that $Q^\mathsf{H}_{\mathcal{A}}$ is relatively compact in $\mathsf{H}$. We conclude that $\mathcal{L}$ has compact resolvent.

We now show that $\lambda_n(-\mathcal{L})\geq \lambda_n(\mathcal{A})$. This is a consequence of the positivity of $\mathcal{W}$, and it is done with the same argument as in [31][Proposition 6.3], using the min-max theorem, <cit.>[Theorem XIII.2]. We thus deduce the diagonalization of $\mathcal{L}^{-1}$. We emphasize the fact that for $f\in \mathsf{H}$, $(-\mathcal{L})^{-1}f$ is the weak inverse of $f$, meaning that it is the unique element of $Q^\mathsf{H}(-\mathcal{L})$Pour moi, on est dans D(L) directement et du coup il manque un paragraphe où on définit D(L) dans le sens usuel that satisfies: for any $v\in Q^\mathsf{H}(-\mathcal{L})$, we have

$$ q_{-\mathcal{L}}((-\mathcal{L})^{-1}f,v)=\braket{f,v}_{\mathsf{H}}\,.$$

Là ce que tu fais pour moi c'est que tu montres que $\mcal{D}(\mcal{L})=\mcal{D}(\mcal{A})\cap\mcal{D}(\mcal{W})$It remains to show that $\mathcal{L}^{-1}f$ is in $\mathcal{D}(\mathcal{L})$ for $f\in \mathsf{H}$, from which we will deduce that $\mathcal{L}^{-1}u\blue{f*}$ is the strong inverse of $f$, and the bijection result.

Let then $f\in \mathsf{H}$, and $u=(-\mathcal{L})^{-1}f\in Q^{\mathsf{H}}_{-\mathcal{L}}$trop compliqué pour moi ici, faut juste montrer que $Lu\in\msf{H}$ donc $u\in D(A)\cap D(W)$. In particular, $u$ is in $\mathsf{H}$ and thus $u'\rho_P$ is $L^2(dx)$. Now we have $\mathcal{W}u=-\mathcal{H}[u'\rho_P]$, thus $\mathcal{W}u$ is $L^2(dx)$, and so it is $L^2(\rho_P)$. By assumption we have

\begin{equation}
\label{eq:sensfaible}
\mathcal{A}u = f-2P\mathcal{W}u
\end{equation}

in the weak sense. The right hand side being $L^2(\rho_P)$, we deduce that $\mathcal{A}u$ is $L^2(\rho_P)$. Additionaly, because $\mathcal{A}$ is self adjoint in $L^2(\rho_P)$

$$ \braket{\mathcal{A}u,\Tilde{1}}_{L^2(\rho_P)}=\braket{u,\mathcal{A}\Tilde{1}}_{L^2(\rho_P}=0\,,$$

which means that $\mathcal{A}u$ has mean zero against $\rho_P$. Therefore $\mathcal{A}u\in L^2_0(\rho_P)$. By Theorem <ref>, $u$ is $H^2_{V',0}$ici il faut utiliser que l'intégrale de u par rapport à $\rho$ est nulle sinon t'as à priori $u+Ker\mcal{A}$ solutions. We deduce that both $\mathcal{A}u$ and $\mathcal{W}u$ make sense pointwise.

Now $(\mathcal{W}u)'=\mathcal{H}[\rho_P \mathcal{A}u]$. Because $\mathcal{A}u\in L^2(\rho_P)$, $\rho_P \mathcal{A}u$ is $L^2(dx)$, and so is $(\mathcal{W}u)'$. We deduce that $\mathcal{W}u$ is in $H^1(\rho_P)$, and also is $\mathcal{A}u$ by equation <ref>. Since we already showed that $\mathcal{A}u$ has zero mean against $\rho_P$, we deduce the same for $\mathcal{W}u$. Consequently, $\mathcal{A}u$ and $\mathcal{W}u$ are elements of $\mathsf{H}$, which is the definition of $\mathcal{D}(\mathcal{L})$.

[Inversion of the operator]
$\mcal{L}:\mcal{D}(\mcal{L})\longrightarrow\msf{H}$ is a positive (w.r.t $\Braket{.,.}_{\msf{H}}$) bijective operator where
$\mcal{D}(\mcal{L}):=\left\{u\in\mcal{D}_{L^2(\R),0}(\mcal{A}), \mcal{L}u\in\msf{H}\right\}$. Moreover there exists $C_{\mcal{L}}>0$, such that
$$\vertiii{\mcal{L}^{-1}}_{\mscr{L}(\msf{H})}\leq C_{\mcal{L}}.$$

The smallest $C_{\mcal{L}}>0$ which verifies the inequality is the inverse of the smallest eigenvalue of $\mcal{L}$, denoted $\lambda_1\left (\mcal{L}\right)^{-1}$.

Indeed it was proven in $[24]$, that $-\mcal{L}$ was positive and diagonalizable with countable spectrum ie for all $f\in\msf{H}$,

$$\mcal{L}^{-1}[f]=\sum_{i\geq0}\dfrac{\Braket{f,\phi_i}_{\msf{H}}}{\lambda_i\left (\mcal{L}\right )}\phi_i$$

where $\left(\lambda_i(\mcal{L})\right)_i$ is the sequence of eigenvalues of $\mcal{L}$ in increasing order and $(\phi_i)_i$ the corresponding hilbertian basis of corresponding eigenvectors.

We define
$$\mcal{T}:=\Big\{f\in\mcal{C}^1(\R), f(x)=\underset{|x|\rightarrow\infty}{O}(x^{-1}),\;f'(x)=\underset{|x|\rightarrow\infty}{O}(x^{-2}), \int_\R f\rho_V=0\Big\}$$

[Regularity of the solutions]
For all $f\in \mcal{T}$, there exists a unique $u\in \mcal{C}^3(\R)$ such that $u'\in H^2(\R)$ with $u^{(3)}$ bounded wich verifies:

* $u'(x)=\underset{|x|\rightarrow\infty}{O}\Big(\dfrac{1}{xV'(x)}\Big)$

* $u''(x)=\underset{|x|\rightarrow\infty}{O}\Big(\dfrac{1}{xV'(x)}\Big)$

* $u^{(3)}(x)=\underset{|x|\rightarrow\infty}{O}\Big(\dfrac{1}{x}\Big)$

such that $f=\mcal{L}[u]$.

Let $n\geq1$, and $h\in H^n(\R)$,

$$\|\rho_V\widetilde{\Xi^{-1}}[h] \|_{H^{n}(\R)}\leq C_3(V,n)\|h\|_{H^{n}(\R)}.$$

with a constant $C_3(V,n),$ only depending on $V$ and $n$. For the choice of potentiel $V=V_{\phi,t}$, for $\phi\in\mcal{C}^{\infty}(\R)$ with $\phi^{(k)}\in L^2(\R)$ for all $k\in\N$ and $t\in[0,1]$, $t\mapsto C_3(V_{\phi,t},n)$ is a continuous function.

Moreover, for all $h\in H^n(\R)\cap W_n^{\infty}(\R)$,

$$\|\rho_V\widetilde{\Xi^{-1}}[h] \|_{H^{n}(\R)}\leq C_5(V,n)\|h\|_{W_{n}^\infty(\R)}$$

with a constant $C_5(V,n),$ only depending on $V$ and $n$. The function $t\mapsto C_5(V_{\phi,t},n)$ is also continuous.

We first prove that for all $k\geq0$, $h\in\dfrac{1}{\rho_V}H^{k}(\R)$, there exists finite sets of indices $\mfrak{I}_{l,a}^k$, $\mfrak{J}^{k}_{l}$ and $\mfrak{K}_{l}^{k}$ independent of $V$ and polynomials
$p_{a,1,1}^k$ ,$p_{a,b,c,d}^{k}$, $q_{a,b,c}^{k}$ in $\theta,\dots,\theta^{(k-1)}$, with coefficients independent of $V$ and of degree at most $k$, such that for all $x\in\R$,

\begin{multline}\label{eq:l-1enfonctiondehilbert}
\widetilde{\Xi^{-1}}[h]^{(k)}=\sum_{a=0}^{k-1}p_{a,1,1}^{k}h^{(a)}+\sum_{a=0}^{k-1}\sum_{b\in\mfrak{I}^k_{2,a}}p_{a,b,2,1}^{k}\mcal{H}\left[\rho_Vp_{a,b,2,2}^kh^{(a)}\right]+\hdots
\\+\sum_{a=0}^{k-1}\sum_{b\in\mfrak{I}^k_{k,a}}p_{a,b,k,1}^{k}\mcal{H}\Bigg[\rho_Vp_{a,b,k,2}^{k}\mcal{H}\Big[\rho_Vp_{a,b,k,3}^{k}\mcal{H}\Big[ \hdots\mcal{H}\Big[\rho_V p_{a,b,k,k}^{k}h^{(a)} \underbrace{\Big]\Big]\hdots\Bigg ]}_{k-1}+q^k_{1,1}\widetilde{\Xi^{-1}}[h] \\+\sum_{b\in\mfrak{J}_{2}^k}q^k_{b,2,1}\mcal{H}\left[\rho_V q^k_{b,2,2}\widetilde{\Xi^{-1}}[h] \right]+\hdots+\sum_{b\in\mfrak{J}_{k+1}^k}q^k_{b,k+1,1}\mcal{H}\Bigg[\rho_Vq^k_{b,k+1,2}\mcal{H}\Big[\hdots\mcal{H}\Big[\rho_Vq^k_{b,k+1,k+1}\widetilde{\Xi^{-1}}[h]\underbrace{\Big]\hdots\Bigg]}_{k}
\\+\left(r^k_{1,1}+\sum_{b\in\mfrak{K}^k_{2}}r_{b,2,1}^{k}\mcal{H}\left[\rho_Vr_{b,2,2}^k\right]+\hdots+\sum_{b\in\mfrak{K}^k_{k}}r^k_{b,k,1}\mcal{H}\Bigg[\rho_Vr^k_{b,k,2}\mcal{H}\Big[\hdots\mcal{H}\Big[\rho_Vr^k_{b,k,k}\underbrace{\Big]\hdots\Bigg]}_{k-1}\right) \\\times\left(2P\int_\R\mcal{H}\left[\rho_V \widetilde{\Xi^{-1}}[h]\right]d\mu_V-\int_\R hd\mu_V\right).
\end{multline}

We prove it by induction, where for $n=1$ one just uses the definition of $\Xi$ for the initial case ie

$$		\left(\widetilde{\Xi^{-1}}[h]\right)'=h-\int_\R hd\mu_V-\dfrac{\rho_V'}{\rho_V}\left(\widetilde{\Xi^{-1}}[h]\right)'-2P\mcal{H}\left[\rho_V\widetilde{\Xi^{-1}}[h]\right]+2P\int_\R\mcal{H}\left[\rho_V\widetilde{\Xi^{-1}}[h]\right]d\mu_V.$$
For the induction step, use a bootstrap argument. Suppose (<ref>) holds at rank $k$, then differentiate and replace $\left(\widetilde{\Xi^{-1}}[h]\right)'$ by the RHS of the above relation to show that (<ref>) holds at rank $k+1$.

Now, by the Leibniz formula, for all $k\in\llbracket0,n\rrbracket$, it holds that

$$\left(\rho_V\Xi^{-1}[h]\right) ^{(k)}=\sum_{i=0}^{k}\binom{k}{i}\rho_{V}^{(k-i)}\Xi^{-1}[h]^{(i)}.$$

Furthermore by (<ref>), by using successively that $\pi^{-1}\mcal{H}$ is an isometry of $L^2(\R)$, inequality (<ref>) and Jensen's inequality, we obtain:

$$\max_{0\leq k\leq n}\|\left(\rho_V\Xi^{-1}[h]\right) ^{(k)}(x)\|_{L^2(\R)}\leq C_3(V,n)\|h\|_{H^k(\R)}$$

with $C_3(V,n)$ given by

\begin{multline}\label{eq:def C_3(V,n)}
\\\times\max_{i\leq k\leq n}\Bigg\{i\max_{0\leq a<i}\left\|\rho_{V}^{(k-i)} p_{a,b,1,1}^{i}\right\|_{\infty}+i\pi\max_{0\leq a<i}|\mfrak{I}^i_{2,a}|\max_{b\in\mfrak{I}^i_{2,a}}\left (\left\|\rho_{V}^{(k-i)}p_{a,b,2,1}^{i}\right\|_{\infty}\|\rho_{V} p_{a,b,2,2}^i\|_{\infty}	\right)+\dots
\\+i\pi^{i-1}\max_{0\leq a<i}|\mfrak{I}^i_{i,a}|\max_{b\in\mfrak{I}^i_{i,a}}\left(\left\|\rho_{V}^{(k-i)} p_{a,b,i,1}^{i}\right\|_{\infty}.\prod_{l=2}^{i}\|\rho_{V} p_{a,b,i,l}^i\|_{\infty}\right) +C_{\mcal{L}}\left \|\rho_{V}^{(k-i)} q_{1,1}^{i}\right\|_{\infty}\|\rho_V^{1/2}\|_\infty
\\+C_{\mcal{L}}\pi|\mfrak{J}^i_{2}|\max_{b\in\mfrak{J}^i_{2}}\left \|\rho_{V}^{(k-i)} q_{b,2,1}^{i}\right\|_{\infty}\|\rho_V^{1/2}q_{b,2,2}^i\|_\infty\|\rho_V^{1/2}\|_\infty+\dots
\\+C_{\mcal{L}}\pi^{i}|\mfrak{J}^i_{i+1}|\max_{b\in\mfrak{J}^i_{i+1}}\left \|\rho_{V}^{(k-i)} q_{b,i+1,1}^{i}\right\|_{\infty}\prod_{l=2}^{i}\|\rho_V q_{b,i+1,l}^{i}\|_{\infty}\|\rho_V^{1/2} q_{b,i+1,i+1}^{i}\|_{\infty}\|\rho_V^{1/2}\|_{\infty}
\\+2\|\rho_V\|_{\infty}\left(1+2P\pi C_{\mcal{L}}\right) \Big[\|r_{1,1}^{i}\dfrac{\rho_{V}^{(k-i)}}{\rho_V}\sqrt{\rho_V}\|_{\infty}+\pi|\mfrak{K}^i_{2}|\max_{b\in\mfrak{K}^i_{2}}\|\rho_V^{(k-i)}r_{b,2,1}^{i}\|_{\infty}\|\sqrt{\rho_V}r_{b,2,1}^{i}\|_{\infty}+\dots
\\+\pi^{i-1}|\mfrak{K}^i_{i}|\max_{b\in\mfrak{K}^i_{i}}\|\rho_V^{(k-i)}r_{b,i,1}^{i}\|_{\infty}\prod_{l=2}^{i-1}\|\rho_V r_{b,i,l}^{i}\|_{\infty}\|\sqrt{\rho_V}r_{b,i,i}^{i}\|_{\infty}\Big] \Bigg\}
\end{multline}

For the second inequality, if $h\in H^n(\R)\cap W_n^{\infty}(\R)$, we use the fact the same inequalities but we use the following integrals at the end $\displaystyle\int_\R \left(p_{a,b,i,l}^{i} h^{(a)} \rho_V(t)\right)^2 dt\leq \|h\|_{W_a^\infty(\R)}^2\int_\R (p_{a,b,i,l}^{i}\rho_{V_{t}})^2dt$. This leads to

$$\|\rho_V\widetilde{\Xi^{-1}}[h] \|_{H^{n}(\R)}\leq C_5(V,n)\|h\|_{W_{n}^\infty(\R)}$$

with $C_5(V,n)$ given by

\begin{multline}\label{eq:def C_5(V,n)}
\\\times\max_{i\leq k\leq n}\Bigg\{i\max_{0\leq a<i}\left\|\dfrac{\rho_{V}^{(k-i)}}{\rho_{V}} p_{a,b,1,1}^{i}\sqrt{\rho_V}\right\|_{\infty}+i\pi\max_{0\leq a<i}|\mfrak{I}^i_{2,a}|\max_{b\in\mfrak{I}^i_{2,a}}\left (\left\|\rho_{V}^{(k-i)}p_{a,b,2,1}^{i}\right\|_{\infty}\|\sqrt{\rho_{V}} p_{a,b,2,2}^i\|_{\infty}	\right)+\dots
\\+i\pi^{i-1}\max_{0\leq a<i}|\mfrak{I}^i_{i,a}|\max_{b\in\mfrak{I}^i_{i,a}}\left(\left\|\rho_{V}^{(k-i)} p_{a,i,1}^{i}\right\|_{\infty}.\prod_{l=2}^{i-1}\|\rho_{V} p_{a,i,l}^i\|_{\infty}\|\sqrt{\rho_{V} }p_{a,i,i}^i\|_{\infty}\right)
\\ +C_{\mcal{L}}\left \|\rho_{V}^{(k-i)} q_{1,1}^{i}\right\|_{\infty}\|\rho_V\|_\infty^{1/2}+C_{\mcal{L}}\pi|\mfrak{J}^i_{2}|\max_{b\in\mfrak{J}^i_{2}}\left \|\rho_{V}^{(k-i)} q_{b,2,1}^{i}\right\|_{\infty}\|\rho_V^{1/2}q_{b,2,2}^i\|_\infty+\dots
\\+C_{\mcal{L}}\pi^{i}|\mfrak{J}^i_{i+1}|\max_{b\in\mfrak{J}^i_{i+1}}\left \|\rho_{V}^{(k-i)} q_{b,i+1,1}^{i}\right\|_{\infty}\prod_{l=2}^{i}\|\rho_V q_{b,i+1,l}^{i}\|_{\infty}\|\rho_V^{1/2} q_{b,i+1,i+1}^{i}\|_{\infty}
\\+2\left(1+2P\pi\|\rho_V\|_\infty^{1/2}C_{\mcal{L}}\right) \Big[\|r_{1,1}^{i}\dfrac{\rho_{V}^{(k-i)}}{\rho_{V}}\sqrt{\rho_V}\|_{L^2(\R)}+\pi|\mfrak{K}^i_{2}|\max_{b\in\mfrak{K}^i_{2}}\|\rho_V^{(k-i)}r_{b,2,1}^{i}\|_{\infty}\|\sqrt{\rho_V}r_{b,2,1}^{i}\|_{\infty}+\dots
\\+\pi^{i-1}|\mfrak{K}^i_{i}|\max_{b\in\mfrak{K}^i_{i}}\|\rho_V^{(k-i)}r_{b,i,1}^{i}\|_{\infty}\prod_{l=2}^{i-1}\|\rho_V r_{b,i,l}^{i}\|_{\infty}\|\sqrt{\rho_V}r_{b,i,i}^{i}\|_{\infty}\Big] \Bigg\}.
\end{multline}

The fact that $t\mapsto C_i(V_{\phi,t},n)$ is shown in Appendix <ref>.

With $\theta=\dfrac{\rho_V'}{\rho_V}$, $g:=\widetilde{\Xi^{-1}}[f]$ and $c:=\displaystyle\int_\R\left( 2P\mcal{H}\left[\rho_Vg\right]-f\right) d\mu_V$ we have

\begin{align*}
g'&=f-\theta g-2P\mcal{H}\left[\rho_Vg\right] +c
\\g''&=-\theta f+f'-2P\mcal{H}\left[\rho_Vf\right]+\left(\theta^2-\theta'\right)g+2P\theta\mcal{H}\left[\rho_Vg\right] +4P^2\mcal{H}\left[\rho_V\mcal{H}\left[\rho_V g\right]\right]
\\&+\left(-\theta-2P\mcal{H}[\rho_V]\right)c.
\\g^{(3)}&=\left(\theta^2-2\theta'\right)f-\theta f'+f''+2P\theta\mcal{H}\left[\rho_Vf\right]-2P\mcal{H}\left[\rho_V\theta f\right]-2P\mcal{H}\left[\rho_Vf'\right]+4P^2\mcal{H}\big[\rho_V\mcal{H}\left[\rho_V f\right]\Big]
\\&+\left[\left(\theta^2-\theta'\right)'-\left(\theta^3-\theta\theta'\right)\right]g +2P\left(\theta^2-2\theta'\right)\mcal{H}[\rho_Vg]-4P^2\theta\mcal{H}\left[\rho_V\mcal{H}\left[\rho_V g\right]\right ]+4P^2\mcal{H}\left[\rho_V\theta\mcal{H}\left[\rho_V g\right]\right]
\\&+4P^2\mcal{H}\left[\rho_V\mcal{H}\left[\rho_V\theta g\right]\right]-8P^3\mcal{H}\Big[\rho_V\mcal{H}\big[\rho_V\mcal{H}\left[\rho_V g \right]\big]\Big]
\\&+\left((\theta^2-2\theta')+2P\theta\mcal{H}[\rho_V]-2P\mcal{H}[\rho_V\theta]+4P^2\mcal{H}\Big [\rho_V\mcal{H}[\rho_V]\right] \Big)c
\end{align*}

§ INTEGRABILITY OF THE CONSTANTS

§.§ Parameter continuity of norms of certain functions

In this appendix, we work with $V=V_{G,\phi,t}$, $t\in[0,1]$ and $\phi\in\cap_{k\geq0}H^k(\R)$.
We will show, that the constant $C_i(V_{G,\phi,t},n)$ appearing in our problem, see Theorem <ref>, Theorem <ref> and <ref> will be continuous in $t$ hence integrable on $[0,1]$. In this section, we denote for all $t\in[0,1]$,

\begin{equation*}
\alpha_t:=\dfrac{\rho_{V_{G,\phi,t}}}{\rho_{V_{G,\phi,t}}'}\hspace{1cm}and\hspace{1cm}\theta_t:=\dfrac{\rho_{V_{G,\phi,t}}'}{\rho_{V_{G,\phi,t}}}.
\end{equation*}

First of all, by Theorem <ref>, the map $t\mapsto\|\rho_{V_{G,\phi,t}}\|_{W_n^{\infty}(\R)}$ is continuous for all $n\in\N$. This allows to conclude that

Let $t,t_0\in[0,1]$, for all $n\in\N$,

$$\left \|\mcal{H}[\rho_{V_{G,\phi,t}}-\rho_{V_{G,\phi,t_0}}]\right \|_{W_n^{\infty}(\R)}\underset{t\rightarrow t_0}{\longrightarrow}0.$$

We prove it by induction and use Lemma <ref> and Theorem <ref>. For $n=0$, we know that there exists $C>0$, such that:

$$\left \|\mcal{H}[\rho_{V_{G,\phi,t}}-\rho_{V_{G,\phi,t_0}}]\right \|_{\infty}\leq C\left(\|\rho_{V_{G,\phi,t}}-\rho_{V_{G,\phi,t_0}}\right\|_{L^1(\R)}+\|\rho_{V_{G,\phi,t}}'-\rho_{V_{G,\phi,t_0}}'\|_{\infty}).$$

By Scheffé's lemma, the $L^1$ norm goes to zero and by Theorem <ref> goes also to zero as $t$ goes to $t_0$. Now suppose that $\left \|\mcal{H}[\rho_{V_{G,\phi,t}}-\rho_{V_{G,\phi,t_0}}]\right \|_{W_n^{\infty}(\R)}\underset{t\rightarrow t_0}{\longrightarrow}0$ for some $n\geq0$. We have that

$$\left \|\mcal{H}[\rho_{V_{G,\phi,t}}^{(n+1)}-\rho_{V_{G,\phi,t_0}}^{(n+1)}]\right \|_{\infty}\leq C\left(\|\rho_{V_{G,\phi,t}}^{(n+1)}-\rho_{V_{G,\phi,t_0}}^{(n+1)}\right\|_{L^1(\R)}+\|\rho_{V_{G,\phi,t}}^{(n+2)}-\rho_{V_{G,\phi,t_0}}^{(n+2)}\|_{\infty}).$$

Since for all $x\in\R$, $\rho_{V_{G,\phi,t_0}}^{(n+1)}(x)$ goes to zero and we have the following domination by Leibniz formula

By dominated convergence theorem, the $L^1$-norm goes to zero and the last term also trivially goes to zero by theorem <ref>.

Secondly, we can set  $M_{V_{G,\phi,t}}$  (see Lemma <ref>) equal to any value $M$ such that
$$M>\max\left(1+\|\phi'\|_{\infty}+2P\max_{t\in[0,1]}\|\mcal{H}[\rho_{V_{G,\phi,t}}]\|_{\infty},2\left(\|\phi'\|_{\infty}+2P\max_{t\in[0,1]}\|\mcal{H}[\rho_{V_{G,\phi,t}}]\|_{\infty} \right)\right).$$
We choose such a $M$, it is well-defined because of Lemma <ref>.

For all $t\in[0,1]$, for all $i,j\in\N$, for all $|x|\geq  M$,

$$\bigg|\dfrac{\rho_{V_{G,\phi,t}}'}{\rho_{V_{G,\phi,t}}}(x)\bigg|\geq 1 \hspace{0,5cm}\text{ and }\hspace{0,5cm}\Big|\left(\dfrac{\rho_{V_{G,\phi,t}}}{\rho_{V_{G,\phi,t}}'}\right) ^{(i)}(x)^j\Big|\leq \delta_{i,0}\dfrac{C_{0,j}}{|x|^j}+\dfrac{C_{i,j}}{|x|^{2j}}$$

for constants $C_{i,j}>0$  independent of $t$.

Let $x\in\R$, $-\dfrac{\rho_{V_{G,\phi,t}}'}{\rho_{V_{G,\phi,t}}}(x)=x+t\phi'(x)+2P\mcal{H}[\rho_{V_{G,\phi,t}}](x).$ Thus if $|x|\geq M$,

$$\bigg|\dfrac{\rho_{V}'}{\rho_V}(x)\bigg|\geq1+\left (\|\phi'\|_{\infty}-t|\phi'(x)|\right )+2P\left (\max_{s\in[0,1]}\|\mcal{H}[\rho_{V_{G,\phi,s}}]\|_{\infty}-|\mcal{H}[\rho_{V_{G,\phi,t}}](x)|\right )\geq1.$$

For the second point, one notices by differentiation and (<ref>) that there exists polynomials $\msf{P}_k$ with coefficients independent of $t$ such that

$$\left( \dfrac{\rho_{V_{G,\phi,t}}}{\rho_{V_{G,\phi,t}}'}\right) ^{(i)}(x)=\sum_{k=1}^{i}\dfrac{\msf{P}_k\left(t\phi'(x),\dots,t\phi^{(i+1)}(x),\mcal{H}[\rho_{V_{G,\phi,t}}](x),\dots,\mcal{H}\left [\rho_{V_{G,\phi,t}}^{(i)}\right ](x)\right) }{\left(x+t\phi'(x)+2P\mcal{H}[\rho_{V_{G,\phi,t}}](x)\right)^{k+1} }.$$

Furthermore, since $|x|\geq2\left(\|\phi'\|_{\infty}+2P\max_{t\in[0,1]}\|\mcal{H}[\rho_{V_{G,\phi,t}}]\|_{\infty} \right)$ we have

\begin{equation*}
\Big|x+t\phi'(x)+2P\mcal{H}[\rho_{V_{G,\phi,t}}](x)\Big|\geq \dfrac{|x|}{2}+\left(\dfrac{|x|}{2}-\|\phi'\|_{\infty}-2P\max_{s\in[0,1]}\|\mcal{H}[\rho_{V_s}]\|_{\infty}\right)\geq\dfrac{|x|}{2}.
\end{equation*}

Finally, the whole dependence in $t$ and $x$ of the numerator are in the entries which are bounded uniformly in $t$ and $x$, we can conclude that each numerator in the sum is bounded by a constant $C_k>0$. We can conclude that $\Big|\left( \dfrac{\rho_{V_{G,\phi,t}}}{\rho_{V_{G,\phi,t}}'}\right) ^{(i)}(x)\Big|\leq4i\max_{k\leq i}C_k|x|^{-2}$, raising to the power $j$ leads to the conclusion.

The map $t\mapsto\|\rho_{V_{G,\phi,t}}^{-1}\|_{L^\infty([-M,M])}$ is continuous.

Let $x\in[-M,M]$,

$$|\rho_{V_{G,\phi,t}}(x)^{-1}-\rho_{V_{G,\phi,t_0}}(x)^{-1}|=\left|\dfrac{\delta t u_t(x)\rho_{V_{G,\phi,t_0}}(x)}{\rho_{V_{G,\phi,t_0}}(x)\rho_{V_{G,\phi,t}}(x)}\right| \leq|\delta t|\dfrac{\|u_t\|_{\infty}\|\rho_{V_{G,\phi,t_0}}^{-1}\|_{L^\infty([-M,M])}}{\left(1-|\delta t|\|u_t\|_{\infty}\right)}.$$

Taking the supremum over $x\in[-M,M]$ and let $t$ goes to $t_0$ establishes the result.

Now it remains to bound the $L^2$ or $L^\infty$ norms of the functions $\mfrak{f}^{(i),V_{G,\phi,t}}$ and $\mcal{I}_a^{V_{G,\phi,t}}$ appearing in Theorem <ref>, <ref> and <ref>.

$t\mapsto\|\mcal{I}_a^{V_{G,\phi,t}}\|_{\infty}$ for all $a\in\{1,2\}$ is continuous where $\mcal{I}_a^{V_{G,\phi,t}}$ is defined in (<ref>).

Let $x>0$, $t,t_0\in[0,1]$, by the mean-value theorem and with $u_t$ defined in Section <ref>, we get

\begin{multline}\label{eq:diff I_1}
\\\leq \dfrac{1}{\rho_{V_{G,\phi,t_0}}(x)}\int_x^{+\infty}\rho_{V_{G,\phi,t_0}}(s)ds\left| \dfrac{1+\delta tu_t(s)}{1+\delta tu_t(x)}-1\right|
\\=\dfrac{|\delta t|}{\rho_{V_{G,\phi,t_0}}(x)(1-|\delta t|\|u_t\|_{\infty})}\int_x^{+\infty}\rho_{V_{G,\phi,t_0}}(s)ds\left|u_t(s)-u_t(x)\right|
\\\leq\dfrac{|\delta t|\|u_t'\|_{\infty}}{(1-|\delta t|\|u_t\|_{\infty})}\dfrac{1}{\rho_{V_{G,\phi,t_0}}(x)}\int_x^{+\infty}(s-x)\rho_{V_{G,\phi,t_0}}(s)ds.
\end{multline}

One thus concludes by showing that $x\in]0,+\infty[\mapsto\dfrac{1}{\rho_{V_{G,\phi,t_0}}(x)}\displaystyle\int_x^{+\infty}(s-x)\rho_{V_{G,\phi,t_0}}(s)ds$ is bounded, since, in (<ref>), $\delta t$ goes to zero and $t\mapsto\|u_t\|_{W_1^{\infty}(\R)}$ is bounded.

This can be proven by integration by parts

\begin{multline*}
\dfrac{1}{\rho_{V_{G,\phi,t_0}}(x)}\int_x^{+\infty}(s-x)\rho_{V_{G,\phi,t_0}}(s)ds=\dfrac{1}{\rho_{V_{G,\phi,t_0}}(x)}\left [\dfrac{\rho_{V_{G,\phi,t_0}}}{\rho_{V_{G,\phi,t_0}}'}(s)\rho_{V_{G,\phi,t_0}}(s)s\right ]_{x}^{+\infty}
\\-\dfrac{1}{\rho_{V_{G,\phi,t_0}}(x)}\int_{x}^{+\infty}\left[\dfrac{\rho_{V_{G,\phi,t_0}}}{\rho_{V_{G,\phi,t_0}}'}(s)+\left(\dfrac{\rho_{V_{G,\phi,t_0}}}{\rho_{V_{G,\phi,t_0}}'}\right)'(s)s\right]\rho_{V_{G,\phi,t_0}}(s)ds.
\end{multline*}

The first term in the right hand side is bounded, while by assumption v), the last term is a
$$\dfrac{1}{\rho_{V_{G,\phi,t_0}}(x)}\int_{x}^{+\infty}\underset{x\rightarrow+\infty}{O}\left(\dfrac{\rho_{V_{G,\phi,t_0}}(s)}{s}\right)ds=\underset{x\rightarrow+\infty}{O}\left( \dfrac{1}{\rho_{V_{G,\phi,t_0}}(x)}\int_{x}^{+\infty}\dfrac{\rho_{V_{G,\phi,t_0}}(s)}{s}ds\right).$$

Again by an integration by parts, the last integral in the remainder is equal to
$$\dfrac{\rho_{V_{G,\phi,t_0}}(x)}{\rho_{V_{G,\phi,t_0}}'(x)x}+\underset{x\rightarrow+\infty}{O}\left( \dfrac{1}{\rho_{V_{G,\phi,t_0}}(x)}\int_{x}^{+\infty}\dfrac{\rho_{V_{G,\phi,t_0}}(s)}{s^2}ds\right).$$

Since $\rho_{V_{G,\phi,t_0}}$ is decreasing in a neighborhood at infinity the last remainder is a $\underset{x\rightarrow+\infty}{o}(1)$ while the first is behaves like $x^{-2}$ at infinity. Finally, $x\mapsto\dfrac{x}{\rho_{V_{G,\phi,t_0}}(x)}\displaystyle\int_{x}^{+\infty}\rho_{V_{G,\phi,t_0}}(s)ds$ is bounded by the same exact technique.

Doing the same thing over $]-\infty,0]$ establishes that $t\mapsto\|\mcal{I}_1^{V_{G,\phi,t}}\|_{\infty}$ is continuous.

Just as before, we get by the mean-value theorem,
\begin{multline}
|\mcal{I}_2^{V_{G,\phi,t}}(x)^2-\mcal{I}_2^{V_{G,\phi,t_0}}(x)^2|\leq \dfrac{1}{\rho_{V_{G,\phi,t_0}^2}(x)}\int_x^{+\infty}\rho_{V_{G,\phi,t_0}^2}(s)ds\left| \dfrac{(1+\delta tu_t(s))^2}{(1+\delta tu_t(x))^2}-1\right|
\\\leq\dfrac{|\delta t|\|u_t'\|_{\infty}(2+\|u_t\|_\infty)}{(1-|\delta t|\|u_t\|_{\infty})^2}\dfrac{1}{\rho_{V_{G,\phi,t_0}}(x)^2}\int_x^{+\infty}(s-x)\rho_{V_{G,\phi,t_0}}(s)^2ds.
\end{multline}

We conclude by showing that $x\in[0,+\infty[\mapsto\dfrac{1}{\rho_{V_{G,\phi,t_0}}(x)^2}\displaystyle\int_x^{+\infty}(s-x)\rho_{V_{G,\phi,t_0}}(s)^2ds$ is bounded which can again be proven by the same integration by parts and by doing the exact same thing on $]-\infty,0]$. Therefore by the fact that

$$0\leq\bigg|\|\mcal{I}_2^{V_{G,\phi,t}}\|_{\infty}^2-\|\mcal{I}_2^{V_{G,\phi,t_0}}\|^2_\infty\bigg|\leq\|\left(\mcal{I}_2^{V_{G,\phi,t}}\right) ^2-\left(\mcal{I}_2^{V_{G,\phi,t_0}}\right) ^2\|_{\infty}\tend{t\rightarrow t_0}0$$

we conclude that $t\mapsto\|\mcal{I}_2^{V_{G,\phi,t}}\|_{\infty}$ is continuous.

Next, we show that any polynomial in $\theta_t:=\dfrac{\rho_{V_{G,\phi,t}}'}{\rho_{V_{G,\phi,t}}}$ and its derivatives yield a continuous dependance in $t$.

Let $\msf{P}$ a polynomial in $\theta_t,\dots,\theta_t^{(k)}$ for some $k\geq0$ with coefficients independent of $t$, let $l\in\N$ then then the two following maps are continuous:

* $t\mapsto\|\msf{P}\big(\theta_t,\dots,\theta_t^{(k)}\big)\|_{L^\infty([-M,M])}$

* $t\mapsto\|\sqrt{\rho_{V_{G,\phi,t}}^{(l)}}\msf{P}\big(\theta_t,\dots,\theta_t^{(k)}\big)\|_{L^\infty(\R)}$

* $t\mapsto\|\rho_{V_{G,\phi,t}}^{(l)}\msf{P}\big(\theta_t,\dots,\theta_t^{(k)}\big)\|_{L^\infty(\R)}.$

Proving these continuity results for any monomial in those variables is enough. Furthermore, since by continuity $x\in[-M,M]\mapsto\theta_t^{(i)}$ is bounded for all $i\leq k$, thus this monomial in $(\theta_t^{(i)})_{0\leq i \leq k}$ converges uniformly to the monomial in $(\theta_{t_0}^{(i)})_{0\leq i \leq k}$ as $t$ goes to $t_0$. The arguments are that the product of two bounded, uniformly converging sequences of functions converges to the product of the limits and that for all $i>0$, $\theta_t^{(i)}(x)-\theta_{t_0}^{(i)}(x)=-\delta t\phi^{(i+1)}(x)-2P\mcal{H}\left[\rho_{V_{G,\phi,t}}^{(i)}-\rho_{V_{G,\phi,t_0}}^{(i)}\right]$. The latter, when taking the supremum over $x\in[-M,M]$, goes to zero by Lemma <ref>. This establishes (i).
Furthermore, notice that (iii) implies (ii) since $\|\sqrt{\rho_{V_{G,\phi,t}}^{(l)}}\msf{P}\left(\theta_t,\dots,\theta^{(k)}\right)\|_{L^\infty(\R)}=\|\rho_{V_{G,\phi,t}}^{(l)}\msf{P}\left(\theta_t,\dots,\theta_t^{(k)}\right)^2\|_{L^\infty(\R)}^{1/2}$ and $\msf{P}$ is arbitrary so we only prove (iii). Moreover since by Faà di Bruno's formula $\rho_{V_{G,\phi,t}}^{(l)}=\exp(\log\rho_{V_{G,\phi,t}})^{(l)}$ can be written as $\msf{Q}\left(\theta,\dots,\theta^{(l)}\right)\rho_V$  where $\msf{Q}$ is a polynomial with coefficients independent of $t$, it suffices to prove the result for $l=0$.

For all $i\in\N$, we have $\theta_t^{(i)}(x)=-\delta_{i,0}x- t\phi^{(i+1)}(x)-2P\mcal{H}\left[\rho_{V_{G,\phi,t}}^{(i)}\right]$. Noticing that by Leibniz formula and the mean value theorem that for all $j\in\N$, for all $0<\alpha<1$

\begin{multline*}
\bigg|x^{j}\rho_{V_{G,\phi,t}}(x)^\alpha-x^j\rho_{V_{G,\phi,t_0}}(x)^\alpha\bigg|\leq\bigg|x^j\rho_{V_{G,\phi,t_0}}(x)^\alpha\bigg|.\Big|\left (1+\delta tu_t(x)\right )^\alpha-1\Big|
\\\leq\dfrac{\alpha|\delta t| \max_{s\in[0,1]}\|u_s\|_{W_l^{\infty}(\R)}}{\left(1-{|\delta t| \max_{s\in[0,1]}\|u_s\|_{W_l^{\infty}(\R)}}\right)^{1-\alpha}}  \Big\|x\mapsto x^{j}\rho_{V_{G,\phi,t_0}}(x)^\alpha\Big\|_{\infty}
\end{multline*}

where the existence of the max is justified by Corollary <ref>. Taking the supremum over $x\in\R$ and let $t$ goes to $t_0$ shows that $t\mapsto\left(x\mapsto x^j\rho_{V_{G,\phi,t}}(x)^\alpha\right)\in L^\infty(\R)$ is continuous. By boundedness and continuity with respect to the $t$ parameter of $t\mapsto\phi^{(i+1)}(x)+2P\mcal{H}\left [\rho_{V_{G,\phi,t}}^{(i)}\right ]$, we deduce that for all $i\in\N$ and $\alpha>0$,

$$\|\theta_t^{(i)}\rho_{V_{G,\phi,t}}(x)^{\alpha}-\theta_{t_0}^{(i)}\rho_{V_{G,\phi,t_0}}(x)^{\alpha}\|_{\infty}\tend{t\rightarrow t_0}0.$$

From this last uniform convergence result, we show that by taking a monomial $\prod_{i=0}^{k}\left(\theta^{(i)}_t\right)^{l_i}$ such that $\sum_{i=0}^kl_i=m$, we deduce that $\rho_{V_{G,\phi,t}}\prod_{i=0}^{k}\left(\theta^{(i)}_t\right)^{l_i}=\prod_{i=0}^{k}\left(\theta^{(i)}_t\sqrt[m]{\rho_{V_{G,\phi,t}}}\right)^{l_i}$, as a product of bounded, uniformly converging $t$-sequences of functions, it converges uniformly. This concludes the proof.

[Continuity of uniform norms]
For all $j\in\{1,2,3,5,6\}$, $t\mapsto\|\mfrak{f}^{(j),V_{G,\phi,t}}_{n_1,a,b}\|_{L^\infty([-M,M]^c)}$ is continuous where we recall that

\begin{align*}
&\mfrak{f}^{(1),V_{G,\phi,t}}_{l,a,b}:x\mapsto Q_a^{l}\left(\theta_t,\dots,\theta_t^{(a)}\right)(x)\alpha_t(x) P_{b}^{l-a}(\alpha_t,\dots,\alpha_t^{(b)})(x),
\\&\mfrak{f}_{l,a,b}^{(2),V_{G,\phi,t}}:x\mapsto  \dfrac{Q_a^{l}\left(\theta_t,\dots,\theta_t^{(a)}\right)(x)}{\rho_{V_{G,\phi,t}}(x)}\int\limits_{x}^{\mrm{sgn}(x)\infty}\left[\alpha_t P_{b}^{l-a}\left (\alpha_t,\dots,\alpha_t^{(b)}\right)  \right] '(y)\rho_{V_{G,\phi,t}}(y)dy,
\\&\mfrak{f}^{(3),V_{G,\phi,t}}_{l,a,b}:x\mapsto \dfrac{Q_a^{l}\left(\theta_t,\dots,\theta_t^{(a)}\right)(x)}{\rho_{V_{G,\phi,t}}(x)}\Bigg|\int\limits_{x}^{\mrm{sgn}(x)\infty}\alpha_t(y)^2P_{b}^{l-a}\left (\alpha_t,\dots,\alpha_t^{(b)}\right )(y)^2\rho_{V_{G,\phi,t}}(y)^2dy\Bigg|^{1/2},
\\&\mfrak{f}^{(5),V_{G,\phi,t}}_{l,a,b}:x\mapsto \dfrac{Q_a^{l}\left(\theta_t,\dots,\theta_t^{(a)}\right)(x)}{\rho_{V_{G,\phi,t}}(x)}\int\limits_{x}^{\mrm{sgn}(x)\infty}\Big|P_{b}^{l-a}(\alpha_t,\dots,\alpha_t^{(b)})(y)\Big|\rho_{V_{G,\phi,t}}(y)dy,
\\&\mfrak{f}^{(6),V_{G,\phi,t}}_{l,a,b}:x\mapsto\dfrac{\Big|Q_a^{l}\left(\theta_t,\dots,\theta_t^{(a)}\right)(x)\Big|}{\rho_{V_{G,\phi,t}}(x)}\Bigg|\int\limits_{x}^{\mrm{sgn}(x)\infty}\Big|P_{b}^{l-a}(\alpha_t,\dots,\alpha_t^{(b)})(y)\Big|^2\rho_{V_{G,\phi,t}}(y)^2dy\Bigg|^{1/2}.
\end{align*}

First, one can check that, from Lemma <ref>, $\mfrak{f}^{(1),V_{G,\phi,t}}_{n_1,a,b}(x)=\underset{|x|\rightarrow\infty}{O}(x^{-1})$ for all $t\in[0,1]$. Noticing that since $\alpha(x)=(-x-t\phi'(x)-2P\mcal{H}[\rho_{V_{G,\phi,t}}](x))^{-1}=\theta(x)^{-1}$, there exists $n>0$ and a polynomial expression $\msf{P}$ with coefficients independent of $t$ such that

$$\mfrak{f}^{(1),V_{G,\phi,t}}_{n_1,a,b}(x)=\dfrac{\msf{P}\left(x^{-1},t\phi',\dots,t\phi^{(k)},P\mcal{H}[\rho_{V_{G,\phi,t}}],\dots,P\mcal{H}[\rho_{V_{G,\phi,t}}^{(k-1)}]\right) }{(1+t\phi'(x)x^{-1}+2P\mcal{H}[\rho_{V_{G,\phi,t}}](x)x^{-1})^{n}}.$$

In the above expression, the numerator must be a $\underset{|x|\rightarrow\infty}{O}(x^{-1})$. We conclude from this closed form, that $\mfrak{f}^{(1),V_{G,\phi,t}}_{n_1,a,b}$ converges uniformly to $\mfrak{f}^{(1),V_{G,\phi,t_0}}_{n_1,a,b}$ when $t$ goes to $t_0$ on $[-M,M]^c$. Indeed, $\mfrak{f}^{(1),V_{G,\phi,t}}_{n_1,a,b}$ is a bounded rational function such that the denominator is bounded from below uniformly in $t$ (see Lemma <ref>) and such that both the numerator and denominator converges uniformly. Thus, $t\mapsto\|\mfrak{f}^{(1),V_{G,\phi,t}}_{n_1,a,b}\|_{L^\infty([-M,M]^c)}$ is continuous.

We only prove the continuity of $t\mapsto\|\mfrak{f}^{(j),V_{G,\phi,t}}_{n_1,a,b}\|_{L^\infty([-M,M]^c)}$ in the case $j=5$, since the same arguments also prove the cases $j\in\{2,3,6\}$. Since $|\rho_{V_{G,\phi,t}}(x)-\rho_{V_{G,\phi,t_0}}(x)|\leq|\delta t |\max_{s\in[0,1]}\|u_s\|_{\infty}|\rho_{V_{G,\phi,t_0}}(x)$ and that the following map is uniformly bounded in $t\in[0,1]$ and $x>M$

$$g_t:x\mapsto \dfrac{Q_a^{l}\left(\theta_t,\dots,\theta_t^{(a)}\right)(x)}{\rho_{V_{G,\phi,t}}(x)}\int\limits_{x}^{+\infty}\Big|P_{b}^{l-a}(\alpha_t,\dots,\alpha_t^{(b)})(y)\Big|\rho_{V_{G,\phi,t_0}}(y)dy,$$

we can just show that $g_t$ converges uniformly to $t_0$ as $t$ goes to $t_0$. Moreover since
$$|\rho_{V_{G,\phi,t}}(x)^{-1}-\rho_{V_{G,\phi,t_0}}(x)^{-1}|\leq|\delta t |\dfrac{\max_{s\in[0,1]}\|u_s\|_{\infty}}{1-|\delta t|\max_{s\in[0,1]}\|u_s\|_{\infty}}\rho_{V_{G,\phi,t_0}}(x)^{-1},$$

it is enough to show the uniform convergence for $\dfrac{\rho_{V_{G,\phi,t}}}{\rho_{V_{G,\phi,t_0}}}g_t$. One can also notice that, for constants $C_{l,a}$ and $C_{l,a,b}$ independents of $t$,
$$|Q_a^l\left(\theta_t,\dots,\theta_t^{(a)}\right)(x)|\leq C_{l,a}|x|^{l-a}\hspace{0,5cm}\text{ and }\hspace{0,5cm}\Big|P_{b}^{l-a}(\alpha_t,\dots,\alpha_t^{(b)})(y)\Big|\leq C_{l,a,b}|x|^{-(l-a)}.$$
Finally, by writing $|x^{^{-(l-a)}}Q_a^l\left(\theta_t,\dots,\theta_t^{(a)}\right)(x)|$ as polynomial in $x^{-1}$, $t\phi^{(i+1)}(x)$ and $\mcal{H}[\rho_{V_{G,\phi,t}}^{(i)}]$ for $i\geq 0$ and $x^{l-a}P_{b}^{l-a}(\alpha_t,\dots,\alpha_t^{(b)})(x)$ as a rational function in those same variables, we conclude that these functions converge uniformly towards the same functions at $t_0$. Therefore

$$x\mapsto x^{-(l-a)}Q_a^{l}\left(\theta_t,\dots,\theta_t^{(a)}\right)(x)\dfrac{x^{l-a}}{\rho_{V_{G,\phi,t_0}}(x)}\int\limits_{x}^{+\infty}\dfrac{\rho_{V_{G,\phi,t_0}}(y)dy}{y^{l-a}}\Big|y^{l-a}P_{b}^{l-a}(\alpha_t,\dots,\alpha_t^{(b)})(y)\Big|$$

converges uniformly to the same functions at $t_0$. This establishes the proposition.

[Continuity $L^2$-norms]
For all $j\in\llbracket1,4\rrbracket$, the maps $t\mapsto\|\mfrak{f}^{(j),V_{G,\phi,t}}_{n_1,a,b}\|_{L^2([-M,M]^c)}$ are continuous where

\begin{equation}
\mfrak{f}^{(4),V_{G,\phi,t}}_{n_1,a,b}:x\mapsto\dfrac{Q_a^{n_1}\left(\theta_t,\dots,\theta_t^{(a)}\right)(x) }{\rho_{V_{G,\phi,t}}(x)}\Bigg|\int\limits_{x}^{\mrm{sgn}(x)\infty}\left[\alpha_t P_{b}^{n_1-a}\left (\alpha_t,\dots,\alpha_t^{(b)}\right)  \right] '(y)\rho_{V_{G,\phi,t}}(y)^2dy\Bigg|^{1/2}.
\end{equation}

For the continuity of $t\mapsto\|\mfrak{f}^{(1),V_{G,\phi,t}}_{n_1,a,b}\|_{L^2([-M,M]^c)}$, we use dominated convergence theorem. Since, we showed uniform convergence and that $\|\mfrak{f}^{(1),V_{G,\phi,t}}_{n_1,a,b}\|_{L^\infty([-M,M]^c)}<+\infty$, we conclude that for all $x\in[-M,M]^c$,

$$\mfrak{f}^{(1),V_{G,\phi,t}}_{n_1,a,b}(x)^2\tend{t\rightarrow t_0}\mfrak{f}^{(1),V_{G,\phi,t_0}}_{n_1,a,b}(x)^2.$$

The domination follows from the fact $\mfrak{f}^{(1),V_{G,\phi,t}}_{n_1,a,b}(x)=\underset{|x|\rightarrow\infty}{O}(x^{-2})$ and that all the dependance in $t$ is bounded, hence there exists a constant $C_{n_1,a,b}>0$ independent of $t$ such that, for all $x\in[-M,M]^c$ and all $t\in[0,1]$,

This establishes that $\|\mfrak{f}^{(1),V_{G,\phi,t}}_{n_1,a,b}\|_{L^2([-M,M]^c)}\tend{t\rightarrow t_0}\|\mfrak{f}^{(1),V_{G,\phi,t_0}}_{n_1,a,b}\|_{L^2([-M,M]^c)}$.

We now establish the continuity for $t\mapsto\|\mfrak{f}^{(3),V_{G,\phi,t}}_{n_1,a,b}\|_{L^2([-M,M]^c)}$, the case $j\in\{2,4\}$ is done with the exact same arguments. We want to use dominated convergence theorem, for $\left({\mfrak{f}^{(3),V_{G,\phi,t}}_{n_1,a,b}}\right) ^2$. The latter, when $t\rightarrow t_0\in[0,1]$, also converges uniformly since it is uniformly bounded and that we proved that $\mfrak{f}^{(3),V_{G,\phi,t}}_{n_1,a,b}$ converges uniformly. It just remains to verify the domination hypothesis. By Lemma <ref>, we know that  $\alpha(y)^2P_{b}^{n_1-a}\left (\alpha_t,\dots,\alpha_t^{(b)}\right )(y)^2=\underset{|y|\rightarrow\infty}{O}(y^{-2(n_1-a+1)}).$ We conclude by Lemma <ref> that there exists a constant $C_{n_1,a,b}>0$ such that for all $y>M$,

$$\alpha_t(y)^2P_{b}^{n_1-a}\left (\alpha_t,\dots,\alpha_t^{(b)}\right )(y)^2\leq \dfrac{C_{n_1,a,b}}{y^{2(n_1-a+1)}}.$$

Similarly $\Big|Q_a^{n_1}\left(\theta_t,\dots,\theta_t^{(a)}\right)(y)^2\Big|\leq C_{n_1,a}|x|^{2(n_1-a)}$ for all $y>M$ and for $C_{n_1,a}>0$ a constant independent of $t$ and $y$. Finally, we get the following domination for an arbitrary $t_0$ and all $y>M$

\begin{multline*}
\dfrac{Q_a^{n_1}\left(\theta_t,\dots,\theta_t^{(a)}\right)(x)^2}{\rho_{V_{G,\phi,t}}(x)^2}\int\limits_{x}^{+\infty}\alpha_t(y)^2P_{b}^{n_1-a}\left (\alpha_t,\dots,\alpha_t^{(b)}\right )(y)^2\rho_{V_{G,\phi,t}}(y)^2dy
\\\leq\dfrac{C_{n_1,a}|x|^{2(n_1-a)}}{\rho_{V_{G,\phi,t_0}}(x)(1-\max_{s\in[0,1]}\|u_s\|_\infty)}\int_x^{+\infty}\dfrac{C_{n_1,a,b}(1+\max_{s\in[0,1]}\|u_s\|_\infty)\rho_{V_{G,\phi,t_0}}(y)dy}{y^{2(n_1-a+1)}}.
\end{multline*}

The RHS is in $L^1([M,+\infty[)$ by integration by parts as it was done in the proof of Lemma <ref>. We conclude by doing the same on $]-\infty,-M]$.

With the choice of potential $V_{\phi,t}$, the following map is continuous

\begin{equation*}
\end{equation*}

By recalling the expression of those constants in (<ref>), (<ref>), (<ref>), since in this appendix, all the building blocks in these constants were shown to be continuous we can conclude.

§.§ Parameter-continuity of $C_{\mcal{L}_t}$ and $K_{V_{G,\phi,t}}$

In [24][App. A], the authors showed that for a general potential $V$, the operator $\mcal{A}$ considered as an unbounded operator on $\msf{H}$ has the same spectrum as the Schrödinger operator $\mcal{S}:\mcal{D}(\mcal{S})\rightarrow L^2(\R)$,defined by

\begin{equation*}
\mcal{S}:=-\Delta+w_V\hspace{0,5cm}with\hspace{0,5cm}\mcal{D}(\mcal{S}):=\left\{u\in H^1(\R),uV'\in L^2(\R),-u''+w_V u\in L^2(\R), \int_\R u(x)dx=0\right\}
\end{equation*}

\begin{equation}
\label{eq:w_p}
w_{V}:=\frac{1}{2}\left( \frac{1}{2}V'^2-V'' + 2PV'\mathcal{H}[\rho_V]-2P\mathcal{H}[\rho_V']+2P^2\mathcal{H}[\rho_V]^2 \right)=\dfrac{1}{2}\Big[(\ln\rho_V)''+\dfrac{1}{2}(\ln\rho_V)'^2\Big]\,.
\end{equation}

Since $C_{\mcal{L}}=\lambda_1(\mcal{A})^{-1/2}=\lambda_1(\mcal{S})^{-1/2}$ by Theorem <ref>, we just have to show that when choosing the potential $V=V_{G,\phi,t}$, the $t$-dependent Schrödinger operator $\mcal{S}_t$ with potential $w_{V_{G,\phi,t}}$ produces a continuous smallest eigenvalue $\lambda_1(\mcal{S}_t)$. Before that, we recall the essential material for manipulating $\mcal{S}_t$.

The map $t\mapsto C_{\mcal{L}_t}= \lambda_1(\mcal{S}_t)^{-1/2}$ is continuous.

First for all $t>0$, $\lambda_1(\mcal{S}_t)>0$.
Secondly, we have the following equalities:

\\ \|u\|_2=1}}\Braket{u,\mcal{S}_t [u]}_{L^2(\R)}=\inf_{\substack{u\in\mcal{C}^{\infty}_c(\R)
\\ \|u\|_2=1}}\int_{\R}(u')^2+\int_\R u^2w_{V_{G,\phi,t}}.$$

From the previous section $w_{V_{t'}}$ converges uniformly to $w_{V_{G,\phi,t}}$ when $t'$ goes to $t$. Hence for all $t,t'\in[0,1]$, $u\in\mcal{C}^{\infty}_c(\R)$ with $\|u\|_2=1$

$$\left|\int_{\R}(u')^2+\int_\R u^2w_{V_{G,\phi,t}}-\int_{\R}(u')^2+\int_\R u^2w_{V_{t'}}\right|\leq\|w_{V_{G,\phi,t}}-w_{V_{t'}}\|_{L^\infty(\R)}  $$

hence $\sup_{\substack{u\in\mcal{C}^{\infty}_c(\R)
\\ \|u\|_2=1}}\left|\int_{\R}(u')^2+\int_\R u^2w_P^t-\int_{\R}(u')^2+\int_\R u^2w_{V_{t'}}\right|$ goes to zero as $t'$ goes to $t$. Since uniform convergence is enough to ensure convergence of infinimums we get the result.

We know prove the continuity of the constant $K_{V_{G,\phi,t}}$ introduced in Theorem <ref>.

The following map is continuous
$$t\mapsto K_{V_{G,\phi,t}}=2P\|\mcal{H}[\rho_{V_{G,\phi,t}}]\|_{\infty}+C+P\displaystyle\Big|\iint_{\R^2}\log|x-y|d\mu_{V_{G,\phi,t}}(x)d\mu_{V_{G,\phi,t}}(y)\Big|$$

for $C$ some fixed constant.

We already proved the continuity of the map $t\mapsto\|\mcal{H}[\rho_{V_{G,\phi,t}}]\|_{L^\infty(\R)}$ in Lemma <ref> so it just remains to show that the double integral is continuous with respect to $t$. We prove this by dominated convergence theorem. The function $(x,y)\mapsto\log|x-y|\rho_{V_{G,\phi,t}}(x)\rho_{V_{G,\phi,t}}(y)$ converges almost everywhere to
as $t$ goes to $t_0$. Furthermore we have the following domination $(x,y)$-almost everywhere

This allows us to conclude on the continuity of $t\mapsto K_{V_{G,\phi,t}}$.

The map $t\mapsto\msf{P}_{K,k}^{V_{G,\phi,t}}$ where $\msf{P}_{K,k}^{V}$ appears in (<ref>) is integrable on $[0,1]$.

By the bounds on $\vertiii{\widetilde{\Xi_1}^{-1}}_{H^i}$, $\vertiii{\widetilde{\Xi_1}^{-1}}_{W_i^\infty}$ and $\vertiii{\Theta^{(a)}\circ\widetilde{\Xi_1}^{-1}}_{W_i^\infty}$ for $0\leq i\leq m_{K,k}$ in Theorem <ref>, <ref> and <ref> plus the continuity results of Appendix <ref>, we conclude on the finitness of $\int_{0}^{1}\msf{P}_{K,k}^{V_{G,\phi,t}}dt$.

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