$\displaystyle\qquad\cdot\Big{(}G(Y_{t}^{\varepsilon})-\frac{\nabla\omega(Y_{t}^{\varepsilon})\cdot
F(Y_{t}^{\varepsilon},\tau_{t}^{\varepsilon})}{\mu(|p|^{2\beta}+|p^{\prime}|^{2\beta})-2i\pi(n+m)\omega(Y^{\varepsilon}_{t})}\Big{)}$
$\displaystyle\qquad\cdot\Big{(}V\Big{(}\frac{t}{\varepsilon^{2}},dp\Big{)}V\Big{(}\frac{t}{\varepsilon^{2}},dp^{\prime}\Big{)}-r(p)\delta(p-p^{\prime})\,dp\,dp^{\prime}\Big{)}W_{n,m}(p,Y^{\varepsilon}_{t})\,$
$\displaystyle+\varepsilon
v\Big{(}\frac{t}{\varepsilon^{2}}\Big{)}\sum_{\begin{subarray}{c}n,m\in\mathbb{Z}^{*}\\\
n+m\neq
0\end{subarray}}\iint\frac{e^{2i\pi(n+m)\tau^{\varepsilon}_{t}}}{\mu(|p|^{2\beta}+|p^{\prime}|^{2\beta})-2i\pi(n+m)\omega(Y^{\varepsilon}_{t})}$
$\displaystyle\qquad\cdot\Big{(}V\Big{(}\frac{t}{\varepsilon^{2}},dp\Big{)}V\Big{(}\frac{t}{\varepsilon^{2}},dp^{\prime}\Big{)}-r(p)\delta(p-p^{\prime})\,dp\,dp^{\prime}\Big{)}$
$\displaystyle\qquad\cdot\nabla_{y}W_{n,m}(p,y)_{|y=Y_{t}^{\varepsilon}}\cdot
F(Y_{t}^{\varepsilon},\tau_{t}^{\varepsilon})\,.$

From here, we can see that $\mathcal{R}_{1}^{\varepsilon}(t)$ can be treated
in the same manner as we treated $h^{\varepsilon}_{2}(t)$ in Section 7.4. The
completes the proof.

###  7.6. Proof of Lemma 7.9

We devote this section to proving Lemma 7.9. The bounds for
$h_{3}^{\varepsilon}(t)$ can be obtained in the same manner as the bounds for
$h_{22}^{\varepsilon}$ in equation (7.4). As a result, we obtain

$\lim_{\varepsilon\to 0}\sup_{t\in[0,T]}\bm{E}|h^{\varepsilon}_{3}(t)|=0\,.$

Treating $R_{2}^{\varepsilon}(t)$ in the same manner as
$R_{1}^{\varepsilon}(t)$ we obtain

$\displaystyle\mathcal{R}_{2}^{\varepsilon}(t)=\varepsilon
v\Big{(}\frac{t}{\varepsilon^{2}}\Big{)}\sum_{\begin{subarray}{c}n,m\in\mathbb{Z}^{*}\\\
n+m\neq
0\end{subarray}}\int\,dp\,\frac{2i\pi(n+m)r(p)e^{2i\pi(n+m)\tau^{\varepsilon}_{t}}}{\varepsilon^{3}-2i\pi(n+m)\omega(Y^{\varepsilon}_{t})}$
$\displaystyle\qquad\cdot\Big{(}G(Y_{t}^{\varepsilon})-\frac{\nabla\omega(Y_{t}^{\varepsilon})\cdot
F(Y_{t}^{\varepsilon},\tau_{t}^{\varepsilon})}{\varepsilon^{3}-2i\pi(n+m)\omega(Y^{\varepsilon}_{t})}\Big{)}W_{n,m}(p,Y^{\varepsilon}_{t})\,$
$\displaystyle+\varepsilon^{3}\sum_{\begin{subarray}{c}n,m\in\mathbb{Z}^{*}\\\
n+m\neq
0\end{subarray}}\int\,dp\,\frac{r(p)e^{2i\pi(n+m)\tau^{\varepsilon}_{t}}}{\varepsilon^{3}-2i\pi(n+m)\omega(Y^{\varepsilon}_{t})}W_{n,m}(p,Y^{\varepsilon}_{t})$
$\displaystyle+\varepsilon
v\Big{(}\frac{t}{\varepsilon^{2}}\Big{)}\sum_{\begin{subarray}{c}n,m\in\mathbb{Z}^{*}\\\
n+m\neq
0\end{subarray}}\int\,dp\,\frac{r(p)e^{2i\pi(n+m)\tau^{\varepsilon}_{t}}}{\varepsilon^{3}-2i\pi(n+m)\omega(Y^{\varepsilon}_{t})}\nabla_{y}W_{n,m}(p,y)_{|y=Y_{t}^{\varepsilon}}\cdot
F(Y_{t}^{\varepsilon},\tau_{t}^{\varepsilon})\,.$

From here, we can see that each of the terms in
$\mathcal{R}_{2}^{\varepsilon}(t)$ can be treated in the same manner as
$h^{\varepsilon}_{2}(t)$ in Section 7.4. This concludes the proof.

###  7.7. Proof of Lemma 7.6

We devote this section to proving Lemma 7.6. The proof analyzes how
$V(t/\varepsilon^{2},dp)V(t/\varepsilon^{2},dp^{\prime})$ in
$\mathcal{A}_{2}^{\varepsilon}$ (defined through Formula (7.10)) averages in
the limit $\varepsilon\to 0$. This is essentially similar to the analysis of
the covariance in the proof of Proposition 1.1 in Section 2 in the case when
the Hamiltonian was quadratic.

Let $\eta>0$ and write

$\begin{split}\int_{0}^{t}(\mathcal{A}^{\varepsilon}_{2}(u)-\mathcal{L}^{M}h(Y^{\varepsilon}_{u}))\,du&=\Big{(}\int_{0}^{[t/(\varepsilon\eta)]\varepsilon\eta}+\int_{[t/(\varepsilon\eta)]\varepsilon\eta}^{t}\Big{)}(\mathcal{A}^{\varepsilon}_{2}(u)-\mathcal{L}h(Y^{\varepsilon}_{u}))\,du\\\
&=\sum_{q=0}^{[t/(\varepsilon\eta)]-1}\int_{q\varepsilon\eta}^{(q+1)\varepsilon\eta}(\mathcal{A}^{\varepsilon}_{2}(u)-\mathcal{L}h(Y^{\varepsilon}_{u}))\,du\\\
&\qquad+\int_{[t/(\varepsilon\eta)]\varepsilon\eta}^{t}(\mathcal{A}^{\varepsilon}_{2}(u)-\mathcal{L}h(Y^{\varepsilon}_{u}))\,du\\\
&\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}R^{\varepsilon}_{1}(t)+R^{\varepsilon}_{2}(t)\,.\end{split}$

Following the computations of Lemma 7.7, we have

$\bm{E}|R^{\varepsilon}_{2}(t)|\leqslant\varepsilon\tilde{C}\,.$

so that we only have to take care of $R^{\varepsilon}_{1}(t)$. For this term
we consider the decomposition

$\begin{split}R^{\varepsilon}_{1}(t)&=\sum_{q=0}^{[t/(\varepsilon\eta)]-1}\int_{q\varepsilon\eta}^{(q+1)\varepsilon\eta}(\mathcal{A}^{\varepsilon}_{2}(u)-\mathcal{L}h(Y^{\varepsilon}_{u}))\,du\\\
&=\sum_{q=0}^{[t/(\varepsilon\eta)]-1}\int_{q\varepsilon\eta}^{(q+1)\varepsilon\eta}\,du\\\
&\qquad\qquad\qquad\cdot\sum_{n\in\mathbb{Z}^{*}}\iint\Big{(}V\Big{(}\frac{u}{\varepsilon^{2}},dp\Big{)}V\Big{(}\frac{u}{\varepsilon^{2}},dp^{\prime}\Big{)}-r(p)\delta(p-p^{\prime})\,dp\,dp^{\prime}\Big{)}\\\
&\qquad\qquad\cdot W_{n}(p,Y^{\varepsilon}_{q\varepsilon\eta})\\\
&\qquad+\sum_{q=0}^{[t/(\varepsilon\eta)]-1}\int_{q\varepsilon\eta}^{(q+1)\varepsilon\eta}\,du\\\
&\qquad\qquad\qquad\cdot\sum_{n\in\mathbb{Z}^{*}}\iint\Big{(}V\Big{(}\frac{u}{\varepsilon^{2}},dp\Big{)}V\Big{(}\frac{u}{\varepsilon^{2}},dp^{\prime}\Big{)}-r(p)\delta(p-p^{\prime})\,dp\,dp^{\prime}\Big{)}\\\
&\qquad\qquad\cdot\Big{(}W_{n}(p,Y^{\varepsilon}_{u})-W_{n}(p,Y^{\varepsilon}_{q\varepsilon\eta})\Big{)}\\\
&\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}R^{\varepsilon}_{11}(t)+R^{\varepsilon}_{12}(t)\,.\end{split}$

For $R^{\varepsilon}_{12}(t)$, we have

$\begin{split}\bm{E}|R^{\varepsilon}_{12}(t)|&\leqslant\frac{C_{h,\omega_{0},\nabla\omega,F}}{\varepsilon}\sum_{q=0}^{[t/(\varepsilon\eta)]-1}\int_{q\varepsilon\eta}^{(q+1)\varepsilon\eta}du\int_{q\varepsilon\eta}^{u}du^{\prime}\int
dp\,r(p)\leqslant\eta\tilde{C}\,,\end{split}$

following the computations in the proof of Lemma 7.7. It only remains to treat
$R^{\varepsilon}_{11}(t)$, and this term requires more care. We have

$\begin{split}\bm{E}|R^{\varepsilon}_{11}(t)|&\leqslant
C_{h,\omega_{0},\nabla\omega,F^{M}}\sum_{q=0}^{[t/(\varepsilon\eta)]-1}\sqrt{I^{\varepsilon}_{q,\eta}}\,,\end{split}$

with

$I^{\varepsilon}_{q,\eta}\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\bm{E}\Bigl{[}\Bigl{|}\int_{q\varepsilon\eta}^{(q+1)\varepsilon\eta}du\iint\Big{[}V\Big{(}\frac{u}{\varepsilon^{2}},dp\Big{)}V\Big{(}\frac{u}{\varepsilon^{2}},dp^{\prime}\Big{)}-r(p)\delta(p-p^{\prime})\,dp\,dp^{\prime}\Big{]}\Bigr{|}^{2}\Bigr{]}.$

Using Gaussian property of $V$, we have

$\begin{split}I^{\varepsilon}_{q,\eta}&=\int_{q\varepsilon\eta}^{(q+1)\varepsilon\eta}du_{1}\int_{q\varepsilon\eta}^{(q+1)\varepsilon\eta}du_{2}\\\
&\qquad\cdot\int\Big{(}\bm{E}\Big{[}V\Big{(}\frac{u_{1}}{\varepsilon^{2}},dp_{1}\Big{)}V\Big{(}\frac{u_{2}}{\varepsilon^{2}},dp_{2}\Big{)}\Big{]}\bm{E}\Big{[}V\Big{(}\frac{u_{1}}{\varepsilon^{2}},dp^{\prime}_{1}\Big{)}V\Big{(}\frac{u_{2}}{\varepsilon^{2}},dp^{\prime}_{2}\Big{)}\Big{]}\\\
&\qquad\qquad\qquad+\bm{E}\Big{[}V\Big{(}\frac{u_{1}}{\varepsilon^{2}},dp_{1}\Big{)}V\Big{(}\frac{u_{2}}{\varepsilon^{2}},dp^{\prime}_{2}\Big{)}\Big{]}\bm{E}\Big{[}V\Big{(}\frac{u_{1}}{\varepsilon^{2}},dp^{\prime}_{1}\Big{)}V\Big{(}\frac{u_{2}}{\varepsilon^{2}},dp_{2}\Big{)}\Big{]}\Big{)},\end{split}$

and by the symmetry between $u_{1}$ and $u_{2}$, we can write

$\begin{split}I^{\varepsilon}_{q,\eta}&\leqslant
C_{h,\omega_{0},\nabla\omega,F}\iint\,dp\,dp^{\prime}\,r(p)r(p^{\prime})\\\
&\qquad\cdot\int_{q\varepsilon\eta}^{(q+1)\varepsilon\eta}du_{1}\int_{q\varepsilon\eta}^{u_{1}}du_{2}\,e^{-\mu(|p|^{2\beta}+|p^{\prime}|^{2\beta})(u_{1}-u_{2})/\varepsilon^{2}}.\end{split}$

Now, we split the integration domain of $p$ into the region where
$|p|\leqslant\nu$, and the region where $|p|>\nu$. This leads to the
decomposition

$I^{\varepsilon}_{q,\eta}\leqslant
J^{\varepsilon}_{1,q,\nu}+J^{\varepsilon}_{2,q,\nu}.$

For the second term (when $|p|>\nu$), we have

$\begin{split}\int_{q\varepsilon\eta}^{(q+1)\varepsilon\eta}\,du_{1}\int_{q\varepsilon\eta}^{u_{1}}\,du_{2}\,e^{-\mu(|p|^{2\beta}+|p^{\prime}|^{2\beta})(u_{1}-u_{2})/\varepsilon^{2}}&\leqslant\frac{\varepsilon^{2}}{\mu|p|^{2\beta}}\int_{0}^{\varepsilon\eta}(1-e^{-\mu|p|^{2\beta}u_{1}/\varepsilon^{2}})\,du\\\
&\leqslant C_{\eta,\nu}\varepsilon^{3}\,,\end{split}$

so that

$\sum_{q=0}^{[t/(\varepsilon\eta)]-1}\sqrt{J^{\varepsilon}_{2,q,\eta}}\leqslant\varepsilon^{1/2}C_{\eta,\nu,T}\,.$

For the first term (when $\lvert p\rvert\leqslant\nu$), we have

$\begin{split}\sum_{q=0}^{[t/(\varepsilon\eta)]-1}\sqrt{J^{\varepsilon}_{1,q,\nu}}&\leqslant
C_{T,\eta}\Big{(}\int_{\\{|p|\leqslant\nu\\}\times
S}\,dp\,dp^{\prime}\,r(p)r(p^{\prime})\Big{)}^{1/2}\\\ &\leqslant
C_{T,\eta}\Big{(}\int_{\\{|p|\leqslant\nu\\}}\frac{dp}{|p|^{2\alpha}}\Big{)}^{1/2}\,.\end{split}$

Since $\alpha<1/2$, the dominated convergence theorem implies that this
converges to $0$. This concludes the proof of Lemma 7.6.

###  7.8. Well-posedness of the martingale problem

We devote this section to showing that the martingale problem with generator
$\mathcal{L}$ (see equation (7.9)) is well-posed. First, we rewrite
$\mathcal{L}h(y)$ as

$\mathcal{L}h(y)=\frac{1}{2}\sum_{j,l=1}^{2}\mathfrak{a}_{jl}(y)\partial^{2}_{x_{l}x_{j}}h(y)+\sum_{j=1}^{2}\partial_{x_{j}}h(y)\mathfrak{b}_{j}(y),$

where

(7.16)
$\begin{split}\mathfrak{a}_{jl}(y)\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\sum_{n\in\mathbb{Z}^{*}}R_{n}(y)F_{j,n}(y)F_{-l,n}(y)=\sum_{n\in\mathbb{Z}^{*}}R_{n}(y)F_{j,n}(y)\overline{F_{l,n}(y)}\end{split}$

with (equivalently as in formula (5.2))

(7.17)
$\begin{split}R_{n}(y)&\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\int
dp\,r(p)\Big{(}\frac{1}{\mu|p|^{2\beta}-2i\pi
n\omega(y)}+\frac{1}{\mu|p|^{2\beta}+2i\pi n\omega(y)}\Big{)}\\\ &=\int
dp\,r(p)\frac{2\mu|p|^{2\beta}}{\mu|p|^{4\beta}+4\pi^{2}n^{2}\omega^{2}(y)}\\\
&=2\int_{0}^{\infty}du\,\bm{E}[v(u)v(0)]\cos(2\pi n\omega(y)u)\,,\end{split}$

and

$\displaystyle\mathfrak{b}_{j}(y)\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\sum_{n\in\mathbb{Z}^{*}}\int
dp\,\frac{r(p)}{\mu|p|^{2\beta}-2i\pi n\omega(y)}$ (7.18)
$\displaystyle\cdot\Bigl{(}\sum_{l=1}^{2}\Bigl{(}\partial_{x_{l}}F_{j,n}(y)F_{l,-n}(y)-\frac{2i\pi
n}{\mu|p|^{2\beta}-2i\pi
n\omega(y)}F_{j,n}(y)F_{l,-n}(y)\partial_{x_{l}}\omega(y)\Bigr{)}\Bigr{)}.$

The uniqueness of the martingale problem is guaranteed by [KS91, Corollary 4.9
p. 317, Proposition 3.20 p. 309, and Theorem 2.5 p. 287] together with [Fri06,
Theorem 1.2 p. 129], where the latter provides the locally Lipschitz behavior
of a nonnegative square root. It is not hard to see that $\mathfrak{b}$ is
locally Lipschitz with respect to $y$, as well as $\mathfrak{a}(y)$ is a
symmetric nonnegative matrix, which is $\mathcal{C}^{2}(\mathbb{R}^{2})$ with
respect to $y$. The smoothness property is direct from (7.16) and (7.17).
Regarding the nonnegativity, we have for any $x\in\mathbb{R}^{2}$

(7.19)
$x^{T}\mathfrak{a}(y)x=\big{<}{\sigma}^{*}(y)(x),{\sigma}^{*}(y)(x)\big{>}_{\mathscr{H}}\geqslant
0$

with the Hilbert space

$\mathscr{H}=\Big{\\{}u:\mathbb{Z}^{*}\to\mathbb{C}:\overline{u_{-n}}=u_{n}\quad\text{and}\quad\sum_{n\in\mathbb{Z}^{*}}|u_{n}|^{2}<\infty\Big{\\}}\,,$

the adjoint ${\sigma}_{M}^{*}$ is an operator from $\mathbb{R}^{2}$ to
$\mathscr{H}$ defined by

${\sigma}_{M}^{*}(y)(n,x)\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}R^{1/2}_{n}(y)F_{1,n}(y)x_{1}+R^{1/2}_{n}(y)F_{2,n}(y)x_{2}\,,$

and $\sigma$, from $\mathscr{H}$ to $\mathbb{R}^{2}$, is defined by

(7.20)
${\sigma}(y)(u,j)\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\sum_{n\in\mathbb{Z}^{*}}R^{1/2}_{n}(y)F_{j,n}(y)u_{-n}\qquad
j=1,2\,.$

Hence, we have an explicit self-adjoint Hilbert-Schmidt square root for
$\mathfrak{a}$ that we use in Section 8 to derive a SDE for the process
$(Y_{t})_{t\geqslant 0}$. It is also clear that $\sigma$ is locally Lipschitz
with respect to $y$, so that we can conclude the uniqueness of the martingale
problem.

Let us finish by rewriting the martingale problem in a more convenient way and
let us start with $\mathfrak{b}$. Using the fact that

$\begin{split}\frac{-2i\pi n\partial_{x_{l}}\omega(y)}{(\mu|p|^{2\beta}-2i\pi
n\omega(y))^{2}}&=\partial_{y_{l}}\int_{0}^{\infty}e^{-(\mu|p|^{2\beta}-2i\pi
n\omega(y))u}\,du\\\ &=2i\pi
n\partial_{x_{l}}\omega(y)\int_{0}^{\infty}du\,u\,e^{-(\mu|p|^{2\beta}-2i\pi
n\omega(y))u}\,,\end{split}$

we sum Fourier series in (7.8) to obtain

$\mathfrak{b}(y)=\int_{0}^{1}d\tau\int_{0}^{\infty}du\,R(u)\textrm{Jac}_{y}[\tilde{F}(u,y,\tau)]\tilde{F}(0,y,\tau)\,,$

where

$\tilde{F}(u,y,\tau)=F(y,\tau+\omega(y)u)\,.$

For $\mathfrak{a}$ we have the following formulation

$\begin{split}\mathfrak{a}_{jl}(y)&=\int_{0}^{1}d\tau\int_{0}^{\infty}du\,R(u)\Big{(}\tilde{F}_{j}(u,y,\tau)\tilde{F}_{l}(0,y,\tau)+\tilde{F}_{j}(0,y,\tau)\tilde{F}_{l}(u,y,\tau)\Big{)}\\\
&=2\int_{0}^{1}d\tau\int_{0}^{\infty}du\,R(u)\tilde{F}_{j}(u,y,\tau)\tilde{F}_{l}(0,y,\tau),\quad
j,l\in\\{1,2\\}\,.\end{split}$

Finally, to conclude, we just have to remark that according to (6.4), as well
as the definition of $\mathfrak{a}$ involving the integration of the periodic
variable over a whole period, we obtain the formulation (7.9).

##  8\. Proof of Proposition 6.2

The proof of this proposition follows the treatment in [KS91, Proposition 4.6
p. 315] and consists of deriving the SDE corresponding to the martingale
problem with generator (6.7). To prove this correspondence, we consider first
the function $h_{j}(y)=y_{j}$ for any $j\in\\{1,2\\}$, so that

(8.1) $M^{0}_{j}(t)=Y_{j,t}-Y_{j,0}-\int_{0}^{t}\mathfrak{b}_{j}(Y_{s})\,ds$

is a martingale, where $\mathfrak{b}$ is defined as $\mathfrak{b}^{M}$ in
(7.8). Proceeding the same way choosing $h_{jl}(y)=y_{j}y_{l}$ one can show
that for any $j,l\in\\{1,2\\}$

$M^{0}_{j}(t)M^{0}_{l}(t)-\int_{0}^{t}\mathfrak{a}_{jl}(Y_{s})\,ds$

is a martingale. (Recall, $\mathfrak{a}$ is defined by (7.16).) Therefore
$M^{0}$ is a martingale with quadratic variation given by

$t\mapsto\int_{0}^{t}\mathfrak{a}_{jl}(Y_{s})\,ds\,.$

Using (7.19), we note $\mathfrak{a}=\sigma{\sigma}^{*}$, where $\sigma$ is
defined by (7.20). We can now apply a standard martingale representation
theorem involving Hilbert spaces (see [DPZ96, Theorem 8.2] for instance). This
guarantees the existence of a complex valued cylindrical Brownian motion $B$
($B=(B^{1}-iB^{2})/\sqrt{2}$, with $B^{1}$ and $B^{2}$ being two independent
real valued cylindrical Brownian motions) on the Hilbert space $\mathcal{T}$,
possibly defined on an extension of the probability space under consideration,
and such that with probability one

$M^{0}(t)=\int_{0}^{t}\sigma(Y_{s})\,dB_{s}\\\ $

for all $t\geqslant 0$. As a result, from (8.1) we obtain the following SDE
for $(Y_{t})_{t\geqslant 0}$,

$dY_{t}=\sigma(Y_{t})dB_{t}+\mathfrak{b}(Y_{t})dt\,.$

Note that, because of symmetries, and that for the cylindrical Brownian motion
on $\mathcal{T}$ we necessarily have $B^{1}_{-n}=B^{1}_{n}$ and
$B^{2}_{-n}=-B^{2}_{n}$, this defines a real valued equation.

Regarding $\sigma$, one can write

(8.2)
$\begin{split}\sigma_{j}(Y_{t})dB_{t}&=\frac{1}{\sqrt{2}}\sum_{n\in\mathbb{Z}^{*}}\tilde{F}_{j,n}(Y_{t})R^{1/2}_{n}(Y_{t})(dB^{1}_{t,n}-idB^{2}_{t,n})\\\
&=\int_{0}^{1}d\tau\tilde{F}_{j}(Y_{t},\tau)\,dW_{t}(Y_{t},\tau)\end{split}$

where

$dW_{t}(y,\tau)=\frac{1}{\sqrt{2}}\sum_{n\in\mathbb{Z}^{*}}e^{2i\pi
n\tau}R^{1/2}_{n}(y)(dB^{1}_{t,n}-idB^{2}_{t,n})$

is a real valued Brownian field with covariance function

$\begin{split}\bm{E}W_{t}(y,\tau)W_{s}(y,\tau^{\prime})&=t\wedge
s\sum_{n\in\mathbb{Z}^{*}}e^{2i\pi n(\tau-\tau^{\prime})}R_{n}(y)\\\ &=t\wedge
s\sum_{n\in\mathbb{Z}^{*}}e^{2i\pi n(\tau-\tau^{\prime})}\int dp\,r(p)\\\
&\qquad\cdot\int_{0}^{\infty}du\,\Big{[}e^{-(\mu|p|^{2\beta}+2i\pi
n\omega(y))u}+e^{-(\mu|p^{\prime}|^{2\beta}-2i\pi n\omega(y))u}\Big{]}\\\
&=t\wedge s\int_{0}^{\infty}du\,R(u)\\\
&\qquad\cdot\sum_{n\in\mathbb{Z}^{*}}\Big{[}e^{2i\pi
n(\tau-\tau^{\prime}+\omega(y)u)}+e^{2i\pi
n(\tau-\tau^{\prime}-\omega(y)u)}\Big{]}.\end{split}$

so that for any test functions $\phi_{1}$, $\phi_{2}$ in the Hilbert space

$L^{2}_{0}(\mathbb{T})=\Big{\\{}\phi\in
L^{2}(\mathbb{T}):\int_{0}^{1}\phi(\tau)d\tau=0\Big{\\}}\,,$

the covariance function reads

$\begin{split}\bm{E}W_{t}(y,\phi_{1})W_{s}(y,\phi_{2})&=t\wedge
s\int_{0}^{\infty}\,du\,R(u)\\\
&\cdot\int_{0}^{1}d\tau\Big{[}\phi_{1}(\tau+\omega(y)u)\phi_{2}(\tau)+\phi_{1}(\tau)\phi_{2}(\tau+\omega(y)u)\Big{]}\,.\end{split}$

To determine the SDE satisfied by $(I^{M}_{t})_{t\geqslant 0}$ we just have to
consider the first coordinate of $(Y_{t})_{t\geqslant 0}$. Then, we have

$dI^{M}_{t}=\int_{0}^{1}d\tau\,a^{M}(I^{M}_{t},\tau)\,dW_{t}(I^{M}_{t},\tau)+\int_{0}^{1}d\tau\int_{0}^{\infty}du\,R(u)\tilde{A}(u,I^{M}_{t},\tau)\,dt$

with

$\begin{split}\tilde{A}(u,I,\tau)&=\partial_{I}\big{(}a^{M}(I,\tau+\omega(I)u)\big{)}a^{M}(I,\tau)\\\
&+\partial_{\tau}a^{M}(I,\tau+\omega(I)u)\big{(}b^{M}(I,\tau)-\langle
b^{M}(I)\rangle\big{)}\,.\end{split}$

Using the fact that $a^{M}$ is mean-zero in $\tau$, and integrating by parts,
we have

$\begin{split}\int_{0}^{1}d\tau\,\partial_{\tau}&a^{M}(I,\tau+\omega(I)u)\big{(}b^{M}(I,\tau)-\langle
b^{M}(I)\rangle\big{)}\\\
=&\int_{0}^{1}d\tau\partial_{\tau}a^{M}(I,\tau+\omega(I)u)b^{M}(I,\tau)-\int_{0}^{1}d\tau
a^{M}(I,\tau+\omega(I)u)\partial_{\tau}b^{M}(I,\tau)\,.\end{split}$

Now, we restrict our study for $t\leqslant\eta_{M}$ so that, according to
(4.10), we have for $I\in(1/M,M)$

$\partial_{\tau}b^{M}(I,\tau)=\partial_{\tau}b(I,\tau)=\partial_{\tau}\partial_{I}\varphi_{1}^{-1}(I,\tau)=\partial_{I}\partial_{\tau}\varphi_{1}^{-1}(I,\tau)=-\partial_{I}a(I,\tau)\,,$

so that

$\int_{0}^{1}d\tau\tilde{A}(u,I,\tau)=\int_{0}^{1}d\tau\,\partial_{I}\big{(}a(I,\tau+\omega(I)u)\big{)}a(I,\tau)+a(I,\tau+\omega(I)u)\partial_{I}a(I,\tau).$

We then obtain the desired result for the stopped process
$\tilde{I}^{M}_{t}=I^{M}_{t\wedge\eta_{M}}$. Now for the process
$\psi^{M}_{t\wedge\eta_{M}}$, we look at the second coordinate of
$(Y_{t})_{t\geqslant 0}$, and we have

$d\psi^{M}_{t}=\int_{0}^{1}d\tau\,b^{M}(I^{M}_{t},\tau)\,dW_{t}(I^{M}_{t},\tau)+\int_{0}^{1}d\tau\int_{0}^{\infty}du\,R(u)\tilde{B}(u,I^{M}_{t},\tau)\,dt$

with

$\begin{split}\tilde{B}(u,I,\tau)&=\partial_{I}\big{(}b^{M}(I,\tau+\omega(I)u)-\langle
b^{M}(I)\rangle\big{)}a^{M}(I,\tau)\\\
&\qquad+\partial_{\tau}(b^{M}(I,\tau+\omega(I)u)-\langle
b^{M}(I)\rangle)\big{(}b^{M}(I,\tau)-\langle
b^{M}(I)\rangle\big{)}\,.\end{split}$

Using again that $a^{M}$ and $\partial_{\tau}b^{M}$ are mean-zero in $\tau$,
together with an integration by parts, we obtain

$\begin{split}\int_{0}^{1}d\tau\,\tilde{B}(u,I,\tau)&=\int_{0}^{1}d\tau\,\partial_{I}\big{(}b^{M}(I,\tau+\omega(I)u)\big{)}a^{M}(I,\tau)\\\
&\qquad-\int_{0}^{1}d\tau\,b^{M}(I,\tau+\omega(I)u)\partial_{\tau}b^{M}(I,\tau)\,.\end{split}$

Since for $I\in(1/M,M)$ we know

$\partial_{\tau}b^{M}(I,\tau)=-\partial_{I}a(I,\tau)\,,$

the proof of Proposition 6.2 is complete.

##  Appendix A Proofs of Lemmas 2.1 and 2.2

In this appendix we prove Lemmas 2.1 and 2.2. While the proofs are a bit
lengthy, they are a direct computation.

###### Proof of Lemma 2.1.

From (2.1) we note

(A.1)
$w_{1}^{\varepsilon}(t)=\varepsilon\int_{0}^{t/\varepsilon^{2}}\sin(\tau)v(\tau)\,d\tau\,,$

and hence

$\displaystyle\bm{E}(w_{1}^{\varepsilon}(t)-w_{1}^{\varepsilon}(s))^{2}$
$\displaystyle=\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\sin(\theta)\sin(\tau)R(\theta-\tau)\,d\theta\,d\tau$
$\displaystyle=\frac{\varepsilon^{2}}{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\left[\cos(\theta-\tau)-\cos(\theta+\tau)\right]R(\theta-\tau)\,d\theta\,d\tau$
(A.2)
$\displaystyle=\frac{1}{2}\big{(}I_{1}^{\varepsilon}(s,t)-I_{2}^{\varepsilon}(s,t)\big{)}\,,$

where

$\displaystyle I_{1}^{\varepsilon}(s,t)$
$\displaystyle\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\cos(\theta-\tau)R(\theta-\tau)\,d\theta\,d\tau\,,$
$\displaystyle\text{and}\qquad I_{2}^{\varepsilon}(s,t)$
$\displaystyle\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\cos(\theta+\tau)R(\theta-\tau)\,d\theta\,d\tau\,.$

We first analyze $I_{1}^{\varepsilon}(s,t)$. Making a change of variables
$z=\theta-\tau$ and integrating by parts with respect to variable $\tau$, we
have

$\displaystyle I_{1}^{\varepsilon}(s,t)$
$\displaystyle=\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}-\tau}^{t/\varepsilon^{2}-\tau}\cos(z)R(z)\,dz\,d\tau$
$\displaystyle={\varepsilon^{2}}\Big{[}\tau\int_{s/{\varepsilon^{2}}-\tau}^{t/{\varepsilon^{2}}-\tau}\cos(z)R(z)\,dz\Big{]}_{\tau=s/{\varepsilon^{2}}}^{\tau=t/{\varepsilon^{2}}}$
$\displaystyle\qquad-\varepsilon^{2}\int_{s/{\varepsilon^{2}}}^{t/{\varepsilon^{2}}}\tau\Bigl{(}\cos\Big{(}\frac{s}{\varepsilon^{2}}-\tau\Big{)}R\Big{(}\frac{s}{\varepsilon^{2}}-\tau\Big{)}-\cos\Big{(}\frac{t}{\varepsilon^{2}}-\tau\Big{)}R\Big{(}\frac{t}{\varepsilon^{2}}-\tau\Big{)}\Bigr{)}\,d\tau$

The first term on the right reduces to

${\varepsilon^{2}}\Big{[}\tau\int_{s/{\varepsilon^{2}}-\tau}^{t/{\varepsilon^{2}}-\tau}\cos(z)R(z)\,dz\Big{]}_{\tau=s/{\varepsilon^{2}}}^{\tau=t/{\varepsilon^{2}}}=(t-s)\int_{0}^{(t-s)/{\varepsilon^{2}}}\cos(z)R(z)\,dz\,.$

For the second term on the right, we note

${\varepsilon^{2}}\int_{s/{\varepsilon^{2}}}^{t/{\varepsilon^{2}}}\tau\cos\left(\frac{s}{\varepsilon^{2}}-\tau\right)R\left(\frac{s}{\varepsilon^{2}}-\tau\right)\,d\tau\\\
={\varepsilon^{2}}\int_{0}^{(t-s)/{\varepsilon^{2}}}\cos(z)R(z)zdz+s\int_{0}^{(t-s)/{\varepsilon^{2}}}\cos(z)R(z)dz\,,$

and

${\varepsilon^{2}}\int_{s/{\varepsilon^{2}}}^{t/{\varepsilon^{2}}}\cos\left(\frac{t}{\varepsilon^{2}}-\tau\right)R\left(\frac{t}{\varepsilon^{2}}-\tau\right)\tau
d\tau\\\
=-{\varepsilon^{2}}\int_{0}^{(t-s)/{\varepsilon^{2}}}\cos(z)R(z)zdz+t\int_{0}^{(t-s)/{\varepsilon^{2}}}\cos(z)R(z)dz\,.$

Hence,

$\displaystyle I_{1}^{\varepsilon}(s,t)$
$\displaystyle=2(t-s)\int_{0}^{(t-s)/\varepsilon^{2}}\cos(z)R(z)\,dz-2\varepsilon^{2}\int_{0}^{(t-s)/\varepsilon^{2}}z\cos(z)R(z)\,dz$
(A.3) $\displaystyle=I_{11}^{\varepsilon}-2I_{12}^{\varepsilon}\,.$

Clearly

(A.4) $\lvert I_{11}^{\varepsilon}\rvert\leqslant
2(t-s)\sup_{x\in\mathbb{R}}\int_{0}^{x}\cos(z)R(z)\,dz\,.$

Note that the supremum on the right-hand side is finite for all
$\gamma\in(0,2)$, as by Remark 1.2 we know
$\lim_{x\to\infty}\int_{0}^{x}\cos(z)R(z)\,dz$ exists and is finite.

To estimate $I_{12}$, we divide the analysis into cases.

_Case I: $\gamma\in(0,1)$._ Integrating by parts we note

$\displaystyle I_{12}^{\varepsilon}$
$\displaystyle=\varepsilon^{2}\Big{[}\sin(z)zR(z)\Big{]}_{z=0}^{z=(t-s)/{\varepsilon^{2}}}-\varepsilon^{2}\int_{0}^{(t-s)/{\varepsilon^{2}}}\sin(z)(R^{\prime}(z)z+R(z))\,dz\,$
(A.5)
$\displaystyle=\varepsilon^{2\gamma}(t-s)^{1-\gamma}L\Big{(}\frac{t-s}{\varepsilon^{2}}\Big{)}-\varepsilon^{2}\int_{0}^{(t-s)/{\varepsilon^{2}}}\sin(z)(R^{\prime}(z)z+R(z))\,dz\,.$

For the first term on the right we note that

$\varepsilon^{2\gamma}(t-s)^{1-\gamma}L\Big{(}\frac{t-s}{\varepsilon^{2}}\Big{)}\leqslant
2\varepsilon^{2\gamma}L\Big{(}\frac{1}{\varepsilon^{2}}\Big{)}(t-s)^{1-\gamma}\,,$

for all sufficiently small $\varepsilon$.

For the second term, we note that (1.8) implies

$|R^{\prime}(z)z|=|L^{\prime}(z)z^{1-\gamma}-\gamma L(z)z^{-\gamma}|\leqslant
2|R(z)|\,,$

for all sufficiently large $z$. Thus

(A.6)
$\displaystyle\varepsilon^{2}\int_{0}^{(t-s)/{\varepsilon^{2}}}\sin(z)(R^{\prime}(z)z+R(z))\,dz\leqslant
3\varepsilon^{2}\int_{0}^{(t-s)/\varepsilon^{2}}\lvert R(z)\rvert\,dz$
$\displaystyle\leqslant
4\varepsilon^{2\gamma}L\Big{(}\frac{1}{\varepsilon^{2}}\Big{)}(t-s)^{1-\gamma}\,,$

where the last inequality followed from Karamata’s theorem [BIKS18, Th 6.2.1],
and the fact that $L$ is slowly varying. Combining these, we see

(A.7) $\lvert I_{12}^{\varepsilon}\rvert\leqslant
10\varepsilon^{2\gamma}L\Big{(}\frac{1}{\varepsilon^{2}}\Big{)}(t-s)^{1-\gamma}\,,$

for all sufficiently small $\varepsilon$.

_Case II: $\gamma=1$._ We follow the proof in the case $\gamma<1$, with a few
minor changes. For the first term on the right of (A), we note that
$L(z)\leqslant 2R(0)z$ for all sufficiently small $z$. Since $L$ is slowly
varying at infinity we must certainly have $L(z)\leqslant z$ for all
sufficiently large $z$. Hence we can find a finite constant $C$ such that
$L(z)\leqslant Cz$ for all $z\geqslant 0$. As a result, we have

$\varepsilon^{2}L\Big{(}\frac{t-s}{\varepsilon^{2}}\Big{)}\leqslant C(t-s)\,.$

For the remaining terms, note that the function
$g(x)\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\int_{0}^{x}R(z)\,dz$
is slowly varying (see for instance [BGT89] Proposition 1.5.9a, p. 26). Thus
using (A.6) we see that

(A.8)
$\displaystyle\varepsilon^{2}\int_{0}^{(t-s)/{\varepsilon^{2}}}\sin(z)(R^{\prime}(z)z+R(z))\,dz$
$\displaystyle\leqslant 3\varepsilon^{2}\int_{0}^{(t-s)/\varepsilon^{2}}\lvert
R(z)\rvert\,dz$ (A.9)
$\displaystyle=3\varepsilon^{2}g\Big{(}\frac{t-s}{\varepsilon^{2}}\Big{)}\,.$

Since $g(0)=0$, $g^{\prime}(0)=R(0)<\infty$, and $g$ is slowly varying at
infinity, we must have $g(z)\leqslant Cz$ for all $z\geqslant 0$ and a finite
constant $C$. (As before, we allow the constant $C$ to change from line to
line, provided it does not depend on $\varepsilon$, $t$ and $s$.)
Consequently,

$\lvert I^{\varepsilon}_{21}\rvert\leqslant C(t-s)\,,$

as desired.

_Case III: $\gamma\in(1,2)$._ Directly integrating $I^{\varepsilon}_{12}$ and
using Karamata’s theorem we see

$\lvert I_{12}^{\varepsilon}\rvert\leqslant
4\varepsilon^{2(\gamma-1)}L\Big{(}\frac{1}{\varepsilon^{2}}\Big{)}(t-s)^{2-\gamma}\,.$

We now turn our attention to the term $I_{2}^{\varepsilon}$. Substituting
$z=\theta-\tau$, we note

$\displaystyle I_{2}^{\varepsilon}$
$\displaystyle=\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}-\tau}^{t/\varepsilon^{2}-\tau}\cos(z+2\tau)R(z)\,dz\,d\tau$
(A.10)
$\displaystyle=\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}-\tau}^{t/\varepsilon^{2}-\tau}[\cos(z)\cos(2\tau)-\sin(z)\sin(2\tau)]R(z)\,dz\,d\tau$
(A.11) $\displaystyle=I_{21}^{\varepsilon}-I_{22}^{\varepsilon}\,,$

where

(A.12) $\displaystyle I_{21}^{\varepsilon}$
$\displaystyle\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}-\tau}^{t/\varepsilon^{2}-\tau}\cos(z)\cos(2\tau)R(z)\,dz\,d\tau\,,$
$\displaystyle\text{and}\qquad I_{22}^{\varepsilon}$
$\displaystyle\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}-\tau}^{t/\varepsilon^{2}-\tau}\sin(z)\sin(2\tau)R(z)\,dz\,d\tau\,.$

We again divide the analysis into cases.

_Case I: $\gamma\in(0,1)$._ Integrating by parts with respect to $\tau$, we
observe

(A.13) $\displaystyle I_{21}^{\varepsilon}$
$\displaystyle=\frac{\varepsilon^{2}}{2}\Big{[}\sin(2\tau)\int_{s/{\varepsilon^{2}}-\tau}^{t/{\varepsilon^{2}}-\tau}\cos(z)R(z)\,dz\Big{]}_{\tau=s/{\varepsilon^{2}}}^{\tau=t/{\varepsilon^{2}}}$
(A.16)
$\displaystyle\qquad-\varepsilon^{2}\int_{s/{\varepsilon^{2}}}^{t/{\varepsilon^{2}}}\Bigl{(}\begin{multlined}\cos\Big{(}\frac{s}{\varepsilon^{2}}-\tau\Big{)}R\Big{(}\frac{s}{\varepsilon^{2}}-\tau\Big{)}\\\
-\cos\Big{(}\frac{t}{\varepsilon^{2}}-\tau\Big{)}R\Big{(}\frac{t}{\varepsilon^{2}}-\tau\Big{)}\Bigr{)}\sin(2\tau)\,d\tau\,.\end{multlined}\cos\Big{(}\frac{s}{\varepsilon^{2}}-\tau\Big{)}R\Big{(}\frac{s}{\varepsilon^{2}}-\tau\Big{)}\\\
-\cos\Big{(}\frac{t}{\varepsilon^{2}}-\tau\Big{)}R\Big{(}\frac{t}{\varepsilon^{2}}-\tau\Big{)}\Bigr{)}\sin(2\tau)\,d\tau\,.$

and hence by Karamata’s theorem,

(A.17) $\lvert I_{21}^{\varepsilon}\rvert\leqslant
C\varepsilon^{2}\int_{0}^{(t-s)/\varepsilon^{2}}\lvert R(z)\rvert\,dz\leqslant
C\varepsilon^{2\gamma}L\Big{(}\frac{1}{\varepsilon^{2}}\Big{)}(t-s)^{1-\gamma}\,,$

for some finite constant $C$.

_Case II: $\gamma=1$._ As in the previous case, we know from (A.17) that

$\lvert I^{\varepsilon}_{21}\rvert\leqslant
C\varepsilon^{2}\int_{0}^{(t-s)/\varepsilon^{2}}\lvert R(z)\rvert\,dz\,.$

Using the same argument as that used to bound (A.9) we obtain

$\lvert I^{\varepsilon}_{21}\rvert\leqslant C(t-s)$

as desired.

_Case III: $\gamma\in(1,2)$._ As before, set
$g(x)\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\int_{0}^{x}\lvert
R(z)\rvert\,dz$, and note

$\lvert
I_{21}^{\varepsilon}\rvert\leqslant\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\Big{(}g\Big{(}\frac{t}{\varepsilon^{2}}-\tau\Big{)}-g\Big{(}\frac{s}{\varepsilon^{2}}-\tau\Big{)}\Big{)}\,d\tau=2\varepsilon^{2}\int_{0}^{(t-s)/\varepsilon^{2}}g(z)\,dz\,.$

Let $G(x)=\int_{0}^{x}g(z)\,dz$ and note that $G$ is regularly varying with
index $2-\gamma$. Thus

$\lvert I^{\varepsilon}_{21}\rvert\leqslant
2\varepsilon^{2}G\Big{(}\frac{t-s}{\varepsilon^{2}}\Big{)}=3\varepsilon^{2\gamma-2}\tilde{L}\Big{(}\frac{1}{\varepsilon^{2}}\Big{)}(t-s)^{2-\gamma}$

The estimates for $I_{22}^{\varepsilon}$ are identical to those for
$I_{21}^{\varepsilon}$. Combining the above estimates we obtain (2.3) for
$i=1$. The proof when $i=2$ is identical, and this finishes the proof of Lemma
2.1. ∎

###### Proof of Lemma 2.2.

We first prove equation (2.4) for $i=1$. To do this, we follow the same
computation as in the proof of Lemma 2.1 up to (A.3). Now note

$\lim_{\varepsilon\to
0}I_{11}^{\varepsilon}=2\int_{0}^{\infty}\cos(z)R(z)\,dz\,,$

where the above integral converges absolutely for $\gamma\in(1,2)$, and
conditionally for $\gamma\in(0,1]$ (see Remark 1.2). When $\gamma\neq 1$, the
proof of Lemma 2.1 already shows that $I_{12}^{\varepsilon}$,
$I_{21}^{\varepsilon}$ and $I_{22}^{\varepsilon}$ all vanish as
$\varepsilon\to 0$. When $\gamma=1$, the proof of Lemma 2.1 shows

$\lvert I^{\varepsilon}_{12}\rvert+\lvert I^{\varepsilon}_{21}\rvert\leqslant
C\varepsilon^{2}\int_{0}^{(t-s)/\varepsilon^{2}}\lvert
R(z)\rvert\,dz=C\varepsilon^{2}g\Big{(}\frac{t-s}{\varepsilon^{2}}\Big{)}\,,$

where $g(x)=\int_{0}^{x}\lvert R(z)\rvert\,dz$. Since $g$ is slowly varying
this vanishes as $\varepsilon\to 0$. This proves (2.4) as claimed.

To prove (2.5), we note

$\displaystyle\bm{E}(w_{1}^{\varepsilon}(t)-w_{1}^{\varepsilon}(s))(w_{2}^{\varepsilon}(t)-w_{2}^{\varepsilon}(s))$
$\displaystyle=\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}R(\theta-\tau)\sin(\theta)\cos(\tau)\,d\theta\,d\tau$
$\displaystyle=\frac{\varepsilon^{2}}{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}R(\theta-\tau)[\sin(\theta+\tau)+\sin(\theta-\tau)]\,d\theta\,d\tau\,,$
$\displaystyle=\tilde{I}_{1}^{\varepsilon}+\tilde{I}_{2}^{\varepsilon}\,,$

where

$\displaystyle\tilde{I}_{1}^{\varepsilon}$
$\displaystyle\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\sin(\theta-\tau)R(\theta-\tau)\,d\theta\,d\tau,$
$\displaystyle\tilde{I}_{2}^{\varepsilon}$
$\displaystyle\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\sin(\theta+\tau)R(\theta-\tau)\,d\theta\,d\tau.$

We claim $\tilde{I}_{1}^{\varepsilon}=0$. To see this, make the changes of
variables

(A.18)
$\theta^{\prime}\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\frac{(t+s)}{\varepsilon^{2}}-\theta\,,\quad\tau^{\prime}\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\frac{(t+s)}{\varepsilon^{2}}-\tau\,,$

and rewrite $\tilde{I}_{1}^{\varepsilon}$ as

$\displaystyle\tilde{I}_{1}^{\varepsilon}$
$\displaystyle=\varepsilon^{2}\int_{t/\varepsilon^{2}}^{s/\varepsilon^{2}}\int_{t/\varepsilon^{2}}^{s/\varepsilon^{2}}\sin(\tau^{\prime}-\theta^{\prime})R(\tau^{\prime}-\theta^{\prime})\,d\theta^{\prime}\,d\tau^{\prime}$
$\displaystyle=-\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\sin(\theta^{\prime}-\tau^{\prime})R(\theta^{\prime}-\tau^{\prime})\,d\theta^{\prime}\,d\tau^{\prime}$
$\displaystyle=-\tilde{I}_{1}^{\varepsilon}\,.$

This implies $\tilde{I}_{1}^{\varepsilon}=0$ as claimed.

We now turn our attention to $\tilde{I}_{2}^{\varepsilon}$. Making the change
of variable $z=\theta-\tau$ we note

$\displaystyle\tilde{I}_{2}^{\varepsilon}$
$\displaystyle=\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}-\tau}^{t/\varepsilon^{2}-\tau}\sin(z+2\tau)R(z)\,dz\,d\tau$
$\displaystyle=\varepsilon^{2}\int_{s/\varepsilon^{2}}^{t/\varepsilon^{2}}\int_{s/\varepsilon^{2}-\tau}^{t/\varepsilon^{2}-\tau}[\sin(z)\cos(2\tau)+\cos(z)\sin(2\tau)]R(z)\,dz\,d\tau\,.$

This is similar to the expression for $I^{\varepsilon}_{2}$ (equation (A.10))
and following the proof of Lemma 2.1 we see $\tilde{I}^{\varepsilon}_{2}\to 0$
as $\varepsilon\to 0$. Since $\tilde{I}_{1}^{\varepsilon}$ and
$\tilde{I}_{2}^{\varepsilon}$ both vanish as $\varepsilon\to 0$ we obtain
(2.5) as claimed. ∎

##  Appendix B Proof of Proposition 1.3

In this section we show convergence of the rescaled processes
$u^{\varepsilon}$ (equation (1.11)) to fBm. The result is similar to well
known results [Taq75, Taq77, Mar05], and the proof is presented here for
completeness.

###### Proof of Proposition 1.3 when $\gamma\in(0,1)$.

Using (1.5) and (1.11) we note

$\displaystyle\bm{E}(u^{\varepsilon}(t)-u^{\varepsilon}(s))^{2}$
$\displaystyle=\frac{2}{\sigma(\varepsilon)^{2}}\int_{r_{1}=s}^{t}\int_{r_{2}=s}^{r_{1}}R\Big{(}\frac{r_{1}-r_{2}}{\varepsilon^{2}}\Big{)}\,dr_{2}\,dr_{1}\,.$

By (1.7) and the uniform convergence theorem [BGT89, Th. 1.2.1], we know

(B.1)
$\frac{1}{\sigma(\varepsilon)^{2}}\Big{\lvert}R\Big{(}\frac{r_{1}-r_{2}}{\varepsilon^{2}}\Big{)}\Big{\rvert}=\frac{1}{L(\varepsilon^{-2})\lvert
r_{1}-r_{2}\rvert^{\gamma}}L\Big{(}\frac{r_{1}-r_{2}}{\varepsilon^{2}}\Big{)}\leqslant\frac{2}{(r_{1}-r_{2})^{\gamma}}\,,$

for all $r_{1},r_{2}\in[s,t]$ and all sufficiently small $\varepsilon$. Hence

(B.2) $\bm{E}(u^{\varepsilon}(t)-u^{\varepsilon}(s))^{2}\leqslant
4\int_{r_{1}=s}^{t}\int_{r_{2}=s}^{r_{1}}\frac{dr_{2}\,dr_{1}}{(r_{1}-r_{2})^{\gamma}}=\frac{4(t-s)^{2-\gamma}}{(1-\gamma)(2-\gamma)}\,.$

Moreover (1.7), (B.1) and the dominated convergence theorem imply

$\displaystyle\lim_{\varepsilon\to
0}\bm{E}(u^{\varepsilon}(t)-u^{\varepsilon}(s))^{2}$
$\displaystyle=2\int_{r_{1}=s}^{t}\int_{r_{2}=s}^{r_{1}}\frac{dr_{2}\,dr_{1}}{(r_{1}-r_{2})^{\gamma}}=\frac{(t-s)^{2\mathcal{H}}}{(2\mathcal{H}-1)\mathcal{H}}\,.$

Since $u^{\varepsilon}$ is Gaussian this implies that the finite-dimensional
distributions of $u^{\varepsilon}$ converge to that of
$\sigma_{\mathcal{H}}B^{\mathcal{H}}$. For convergence in law, note (B.2)
implies

$\displaystyle\bm{E}[(u^{\varepsilon}(s)-u^{\varepsilon}(r))^{2}(u^{\varepsilon}(t)-u^{\varepsilon}(s))^{2}]$
$\displaystyle\leqslant\bm{E}[(u^{\varepsilon}(s)-u^{\varepsilon}(r))^{4}]^{1/2}\bm{E}[(u^{\varepsilon}(t)-u^{\varepsilon}(s))^{4}]^{1/2}$
$\displaystyle\leqslant
3\bm{E}[(u^{\varepsilon}(s)-u^{\varepsilon}(r))^{2}]\bm{E}[(u^{\varepsilon}(t)-u^{\varepsilon}(s))^{2}]$
$\displaystyle\leqslant C|t-r|^{4{\mathcal{H}}}\,.$

Since $4{\mathcal{H}}>1$, Theorems 13.4 and 13.5 in [Bil99] imply that
$u^{\varepsilon}$ converges to $\sigma_{\mathcal{H}}B^{\mathcal{H}}$ in law. ∎

###### Proof of Proposition 1.3 when $\gamma\in(1,2)$.

Using (1.5), (1.11) and a change of variable we note

$\displaystyle\bm{E}(u^{\varepsilon}(t)-u^{\varepsilon}(s))^{2}$
$\displaystyle=\frac{2}{\sigma(\varepsilon)^{2}}\int_{r_{1}=s}^{t}\int_{r_{2}=s}^{r_{1}}R\Big{(}\frac{r_{1}-r_{2}}{\varepsilon^{2}}\Big{)}\,dr_{2}\,dr_{1}$
$\displaystyle=\frac{2}{\sigma(\varepsilon)^{2}}\int_{r_{1}=s}^{t}\int_{z=0}^{r_{1}-s}R\Big{(}\frac{z}{\varepsilon^{2}}\Big{)}\,dz\,dr_{1}$
(B.3)
$\displaystyle=\frac{-2}{\sigma(\varepsilon)^{2}}\int_{r_{1}=s}^{t}\int_{z=r_{1}-s}^{\infty}R\Big{(}\frac{z}{\varepsilon^{2}}\Big{)}\,dz\,dr_{1}\,,$

where the last equality followed from (1.10). Since $L$ is slowly varying,
(1.7) implies

(B.4)
$\frac{1}{\sigma(\varepsilon)^{2}}R\Big{(}\frac{z}{\varepsilon^{2}}\Big{)}\leqslant
R(0)\wedge\Big{(}\frac{C}{z^{\gamma^{\prime}}}\Big{)}\,,$

where $\gamma^{\prime}=(1+\gamma)/2$ and $C$ is a finite constant that is
independent of $\varepsilon$. We will subsequently allow $C$ to change from
line to line, as long as it doesn’t depend on $\varepsilon$.

Note (B.3) and (B.4) immediately imply

(B.5) $\bm{E}(u^{\varepsilon}(t)-u^{\varepsilon}(s))^{2}\leqslant
2C\int_{r_{1}=s}^{t}\int_{z=r_{1}-s}^{\infty}\frac{dz\,dr_{1}}{z^{\gamma^{\prime}}}=C(t-s)^{2-\gamma^{\prime}}\,.$

Moreover, by (1.7), (B.4) and the dominated convergence theorem, we see

$\displaystyle\lim_{\varepsilon\to
0}\bm{E}(u^{\varepsilon}(t)-u^{\varepsilon}(s))^{2}$
$\displaystyle=2\int_{r_{1}=s}^{t}\int_{z=r_{1}-s}^{\infty}\frac{dz\,dr_{2}}{z^{\gamma}}=\frac{(t-s)^{2\mathcal{H}}}{(1-2\mathcal{H})\mathcal{H}}\,.$

Now the remainder of the proof is identical to the case when $\gamma\in(0,1)$.
∎

###### Proof of Proposition 1.3 when $\gamma=1$.

In this case we write

$\displaystyle\bm{E}(u^{\varepsilon}(t)-u^{\varepsilon}(s))^{2}=\frac{2}{\sigma(\varepsilon)^{2}}\int_{r=s}^{t}\int_{z=0}^{r-s}R\Big{(}\frac{z}{\varepsilon^{2}}\Big{)}\,dz\,dr$
$\displaystyle=\frac{2}{L(\varepsilon^{-2})\lvert\ln\varepsilon\rvert}\int_{r=s}^{t}\int_{z=0}^{1}R(z)\,dz\,dr+\frac{2}{\sigma(\varepsilon)^{2}}\int_{r=s}^{t}\int_{z=\varepsilon^{2}}^{r-s}R\Big{(}\frac{z}{\varepsilon^{2}}\Big{)}\,dz\,dr$

The first term on the right vanishes as $\varepsilon\to 0$. Using the uniform
convergence theorem [BGT89, Th 1.2.1] on the second term we see

$\lim_{\varepsilon\to
0}\bm{E}(u^{\varepsilon}(t)-u^{\varepsilon}(s))^{2}=\lim_{\varepsilon\to
0}\frac{2}{\lvert\ln\varepsilon\rvert}\int_{r=s}^{t}\int_{z=\varepsilon^{2}}^{r-s}\frac{1}{z}\,dz\,dr=t-s\,,$

and the convergence is uniform when $s,t$ belong to any bounded interval.
Hence by the same argument as in the previous two cases we see that
$(u^{\varepsilon})$ converges in law to $\sigma_{\mathcal{H}}B^{\mathcal{H}}$
as claimed. ∎

##  Appendix C Proof of Lemma 3.4

First, to prove (3.5) it suffices to show, for all $n\geqslant 1$ and
$0\leqslant t_{1}\leqslant\dots\leqslant t_{n+1}$, that

$\bm{E}\big{[}V(t_{n+1},dp)|V(t_{1},\cdot),\dots,V(t_{n},\cdot)\big{]}=e^{-\mu|p|^{2\beta}(t_{n+1}-t_{n})}V(t_{n},dp).$

For $n=1$ and $0\leqslant t_{1}\leqslant t_{2}$, we write

$V(t_{2},dp)=e^{-\mu|p|^{2\beta}(t_{2}-t_{1})}V(t_{1},dp)+Y,$

where $Y$ and $V(t_{1},dp)$ are independent. In fact, since they are mean-zero
Gaussian variables, we have

$\begin{split}\bm{E}[Y(\varphi)V(t_{1},\psi)]&=\bm{E}[V(t_{2},\varphi)V(t_{1},\psi)]-\bm{E}[V(t_{1},\varphi_{t_{2}-t_{1}})V(t_{1},\psi)]\\\
&=\int
dp\,r(p)\varphi(p)\psi(p)(e^{-\mu|p|^{2\beta}(t_{2}-t_{1})}-e^{-\mu|p|^{2\beta}(t_{2}-t_{1})})\\\
&=0,\end{split}$

for all $\varphi,\psi$ bounded continuous functions where
$\varphi_{s}(p)=e^{-\mu|p|^{2\beta}s}\varphi(p)$. As a result, we have

$\bm{E}\big{[}V(t_{2},dp)|V(t_{1},\cdot)\big{]}=e^{-\mu|p|^{2\beta}(t_{2}-t_{1})}V(t_{1},dp).$

Now, let us fix $n\geqslant 2$ and assume that for all family
$(s_{j})_{j\in\\{1,\dots,n\\}}$ such that $0\leqslant
s_{1}\leqslant\dots\leqslant s_{n}$

$\bm{E}\big{[}V(s_{n},dp)|V(s_{1},\cdot),\dots,V(s_{n-1},\cdot)\big{]}=e^{-\mu|p|^{2\beta}(s_{n}-s_{n-1})}V(s_{n-1},dp),$

Then, we write

$V(t_{n+1},dp)=e^{-\mu|p|^{2\beta}(t_{n+1}-t_{n})}V(t_{n},dp)+Y,$

where $Y$ and $V(t_{n},dp)$ are independent as explained above, so that

$\begin{split}&\bm{E}\big{[}V(t_{n+1},dp)|V(t_{1},\cdot),\dots,V(t_{n},\cdot)\big{]}\\\
=&\;e^{-\mu|p|^{2\beta}(t_{n+1}-t_{n})}V(t_{n},dp)+\bm{E}\big{[}Y|V(t_{1},\cdot),\dots,V(t_{n-1},\cdot)\big{]}\\\
=&\;e^{-\mu|p|^{2\beta}(t_{n+1}-t_{n})}V(t_{n},dp)+\bm{E}\big{[}V(t_{n+1},dp)|V(t_{1},\cdot),\dots,V(t_{n-1},\cdot)\big{]}\\\
&-e^{-\mu|p|^{2\beta}(t_{n+1}-t_{n})}\bm{E}\big{[}V(t_{n},dp)|V(t_{1},\cdot),\dots,V(t_{n-1},\cdot)\big{]}\\\
=&\;e^{-\mu|p|^{2\beta}(t_{n+1}-t_{n})}V(t_{n},dp)\\\
&+(e^{-\mu|p|^{2\beta}(t_{n+1}-t_{n-1})}-e^{-\mu|p|^{2\beta}(t_{n+1}-t_{n})}e^{-\mu|p|^{2\beta}(t_{n}-t_{n-1})})V(t_{n-1},dp)\\\
=&\;e^{-\mu|p|^{2\beta}(t_{n+1}-t_{n})}V(t_{n},dp),\end{split}$

which concludes the proof of (3.5) by induction.

Second, to prove (3.6) it suffices to show that for all $n\geqslant 1$,
$0\leqslant t_{1}\leqslant\dots\leqslant t_{n+1}\leqslant\tilde{t}_{n+1}$ and
$\varphi,\psi$ bounded continuous functions, that

$\displaystyle\bm{E}\big{[}V(\tilde{t}_{n+1},\varphi)V(t_{n+1},\psi)|V(t_{1},\cdot),\dots,V(t_{n},\cdot)\big{]}$
$\displaystyle=\bm{E}\big{[}V(\tilde{t}_{n+1},\varphi)|V(t_{1},\cdot),\dots,V(t_{n},\cdot)\big{]}\bm{E}\big{[}V(t_{n+1},\psi)|V(t_{1},\cdot),\dots,V(t_{n},\cdot)\big{]}$
$\displaystyle\qquad+\int
dp\,r(p)\varphi(p)\psi(p)\begin{multlined}\Bigl{(}e^{-\mu|p|^{2\beta}(\tilde{t}_{n+1}-t_{n+1})}\\\
-e^{-\mu|p|^{2\beta}(\tilde{t}_{n+1}-t_{n})}e^{-\mu|p|^{2\beta}(t_{n+1}-t_{n})}\Bigr{)}\,.\end{multlined}\Bigl{(}e^{-\mu|p|^{2\beta}(\tilde{t}_{n+1}-t_{n+1})}\\\
-e^{-\mu|p|^{2\beta}(\tilde{t}_{n+1}-t_{n})}e^{-\mu|p|^{2\beta}(t_{n+1}-t_{n})}\Bigr{)}\,.$

This last relation is a consequence of the following lemma, which is a
consequence of [Roz87, Theorem 10.1 and Theorem 10.2].

######  Lemma C.1.

Let $(X,Y,Z_{1},\dots,Z_{n})$ be a Gaussian vector on a probability space
$(\Omega,\mathcal{F},\bm{P})$, and $\mathcal{G}=\sigma(Z_{1},\dots,Z_{n})$ be
the $\sigma$-field generated by $Z_{1},\dots,Z_{n}$. Then, the couple
$(X-\bm{E}[X|\mathcal{G}],Y-\bm{E}[Y|\mathcal{G}])$ is independent of
$\mathcal{G}$, and

(C.1)
$\bm{E}[XY|\mathcal{G}]=\bm{E}[X|\mathcal{G}]\bm{E}[Y|\mathcal{G}]+\bm{E}[(X-\bm{E}[X|\mathcal{G}])(Y-\bm{E}[Y|\mathcal{G}])].$

###### Proof of Lemma C.1.

The proof of the independence of
$(X-\bm{E}[X|\mathcal{G}],Y-\bm{E}[Y|\mathcal{G}])$ with respect to
$\mathcal{G}$ is a consequence of Theorem 10.1 and Theorem 10.2 of [Roz87].
Consequently,

$\bm{E}\Big{[}\Big{(}X-\bm{E}[X|\mathcal{G}]\Big{)}\Big{(}Y-\bm{E}[Y|\mathcal{G}]\Big{)}\Big{|}\mathcal{G}\Big{]}=\bm{E}\Big{[}\Big{(}X-\bm{E}[X|\mathcal{G}]\Big{)}\Big{(}Y-\bm{E}[Y|\mathcal{G}]\Big{)}\Big{]},$

and then

$\begin{split}\bm{E}[XY|\mathcal{G}]&=\bm{E}\Big{[}\Big{(}X-\bm{E}[X|\mathcal{G}]+\bm{E}[X|\mathcal{G}]\Big{)}\Big{(}Y-\bm{E}[Y|\mathcal{G}]+\bm{E}[Y|\mathcal{G}]\Big{)}\Big{|}\mathcal{G}\Big{]}\\\
&=\bm{E}[X|\mathcal{G}]\bm{E}[Y|\mathcal{G}]+\bm{E}\Big{[}\Big{(}X-\bm{E}[X|\mathcal{G}]\Big{)}\Big{(}Y-\bm{E}[Y|\mathcal{G}]\Big{)}\Big{]}\\\
&+\bm{E}[Y|\mathcal{G}]\underbrace{\bm{E}\big{[}\big{(}X-\bm{E}[X|\mathcal{G}]\big{)}\big{|}\mathcal{G}\big{]}}_{=0}+\bm{E}[X|\mathcal{G}]\underbrace{\bm{E}\big{[}\big{(}Y-\bm{E}[Y|\mathcal{G}]\big{)}\big{|}\mathcal{G}\big{]}}_{=0}.\end{split}$

∎

As a result, for all $\varphi,\psi$ bounded continuous functions, with

$X=V(\tilde{t}_{n+1},\varphi)\qquad\text{and}\qquad Y=V(t_{n+1},\psi),$

we have

$\begin{split}&\bm{E}\big{[}V(\tilde{t}_{n+1},\varphi)V(t_{n+1},\psi)|V(t_{1},\cdot),\dots,V(t_{n},\cdot)\big{]}\\\
&=\bm{E}\big{[}V(\tilde{t}_{n+1},\varphi)|V(t_{1},\cdot),\dots,V(t_{n},\cdot)\big{]}\\\
&\qquad\cdot\bm{E}\big{[}V(t_{n+1},\psi)|V(t_{1},\cdot),\dots,V(t_{n},\cdot)\big{]}\\\
&\quad+P,\end{split}$

where, using (3.5),

$\begin{split}P&=\bm{E}\Big{[}\Big{(}X-\bm{E}\big{[}X|V(t_{1},\cdot),\dots,V(t_{n},\cdot)\big{]}\Big{)}\Big{(}Y-\bm{E}\big{[}Y|V(t_{1},\cdot),\dots,V(t_{n},\cdot)\big{]}\Big{)}\Big{]}\\\
&=\bm{E}\Big{[}\Big{(}V(\tilde{t}_{n+1},\varphi)-V(t_{n},\varphi_{\tilde{t}_{n+1}-t_{n}})\Big{)}\Big{(}V(t_{n+1},\psi)-V(t_{n},\psi_{t_{n+1}-t_{n}})\Big{)}\Big{]}\\\
&=\int
dp\,r(p)\varphi(p)\psi(p)(e^{-\mu|p|^{2\beta}(\tilde{t}_{n+1}-t_{n+1})}-e^{-\mu|p|^{2\beta}(\tilde{t}_{n+1}-t_{n})}e^{-\mu|p|^{2\beta}(t_{n+1}-t_{n})}),\end{split}$

which concludes the proof of (3.6).

##  Appendix D Proof of Lemma 3.5

Let us start with the following remark. For any $(t,\varphi)\in D_{k,M}$, it
is straightforward that $\varphi(t,\cdot)\in W_{k,M}$, so that

$\bm{E}\sup_{(t,\varphi)\in
D_{k,M}}\Big{|}V\Big{(}\frac{t}{\varepsilon^{2}},\varphi(t,\cdot)\Big{)}\Big{|}\leqslant\bm{E}\sup_{(t,\varphi)\in\tilde{D}_{k,M}}\Big{|}V\Big{(}\frac{t}{\varepsilon^{2}},\varphi\Big{)}\Big{|}\,,$

with

$\tilde{D}_{k,M}\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}[0,T]\times
W_{k,M}.$

As a result, to prove inequality (3.7), we only need to prove that

$\bm{E}\sup_{(t,\varphi)\in\tilde{D}_{k,M}}\Big{|}V\Big{(}\frac{t}{\varepsilon^{2}},\varphi\Big{)}\Big{|}\leqslant
C+\frac{C(\varepsilon)}{\varepsilon}\,,$

where $C>0$ and $C(\varepsilon)>0$ goes to $0$ as $\varepsilon\to 0$. To this
end we first remark that there is a compact embedding of $\tilde{D}_{k,M}$
into $[0,T]\times C^{0}(S)$ equipped with the metric

(D.1)
$\|(t,\varphi)\|_{\varepsilon}:=\sqrt{|t|}/\varepsilon+\|\varphi\|_{\infty}\,.$

Let us also recall [Bre11, Theorem 8.8] that there exists $C_{k}>0$ such that
for any $\varphi\in W^{1,k}(S)$, we have

$\|\varphi\|_{\infty}\leqslant C_{k}\|\varphi\|_{W^{1,k}}\,.$

We also remark that for the pseudometric

$d_{V,\varepsilon}((t,\varphi),(s,\psi))=\bm{E}[|V(t/\varepsilon^{2},\varphi)-V(s/\varepsilon^{2},\psi)|^{2}]^{1/2}\,,$

for which

(D.2)
$\begin{split}d^{2}_{V,\varepsilon}((t,\varphi),(s,\psi))&=\int_{S}dp\,r(p)(\varphi^{2}(p)+\psi^{2}(p)-2\varphi(p)\psi(p)e^{-\mu|p|^{2\beta}|t-s|/\varepsilon^{2}})\\\
&=\int_{S}dp\,r(p)(\varphi^{2}(p)+\psi^{2}(p))(1-e^{-\mu|p|^{2\beta}|t-s|/\varepsilon^{2}})\\\
&+\int_{S}dp\,r(p)(\varphi(p)-\psi(p))^{2}e^{-\mu|p|^{2\beta}|t-s|/\varepsilon^{2}}\\\
&\leqslant C_{k,M,r}\|(t,\varphi)-(s,\psi)\|^{2}_{\varepsilon}\,,\end{split}$

we have

$diam_{d_{V,\varepsilon}}(\tilde{D}_{k,M})\leqslant\sqrt{2C_{k,M,r}(T+M)}\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}c_{k,M,r}\,.$

Note now that it is not easy to see if $\tilde{D}_{k,M}$ is relatively compact
equipped with $d_{V,\varepsilon}$, while it is required to apply [AT07,
Theorem 1.5.1 p. 41]. The problem is that $d_{V,\varepsilon}$ is only a
pseudometric. For instance, $d_{V,\varepsilon}$ does not separate well points,
since we have for any $s$ and $t$

$d_{V,\varepsilon}((t,0),(s,0))=0\,.$

However, according to (D.2), the application $d_{V,\varepsilon}$ provides a
true metric on $\tilde{D}_{k,M}-[0,T]\times\\{0\\}$. Then, instead of working
directly with $\tilde{D}_{k,M}$ let us consider now the increasing sequence of
$\|\cdot\|_{\varepsilon}$-relatively compact subset of $\tilde{D}_{k,M}$

$D^{n}_{k,M}\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\tilde{D}_{k,M}\cap[0,T]\times\big{\\{}\varphi\in
W^{1,k}(S):\,\|\varphi\|_{\infty}\geqslant 1/n\big{\\}}\,.$

Now, since $d_{V,\varepsilon}$ is a metric on every $D^{n}_{k,M}$, it is
straightforward to see from (D.2) that all the $D^{n}_{k,M}$ are also
$d_{V,\varepsilon}$-relatively compact from the sequential characterization of
compactness.

We can now apply [AT07, Theorem 1.5.1 p. 41 and Lemma 1.5.2 p. 44] over all
the $\overline{D^{n}_{k,M}}^{d_{V,\varepsilon}}$, we have

$\begin{split}\bm{E}\sup_{(t,\varphi)\in
D^{n}_{k,M}}V\Big{(}\frac{t}{\varepsilon^{2}},\varphi\Big{)}&\leqslant
K\Big{(}c_{k,M,r}\ln(c_{k,M,r})+\int_{0}^{c_{k,M,r}}\sqrt{\ln(N_{\varepsilon}(r))}dr\Big{)}\,,\end{split}$

where $N_{\varepsilon}(r)$ is the smallest number of balls covering
$D^{n}_{k,M}$ with radius $r$, which are defined by

$B(X,r)=\\{Y\in D^{n}_{k,M}:\,d_{V,\varepsilon}(X,Y)<r\\}\,.$

Because of (D.2), it is clear that

$N_{\varepsilon}(r)\leqslant\mathcal{N}_{\varepsilon}\big{(}r/C^{1/2}_{k,M,r}\big{)}\,,$

where $\mathcal{N}_{\varepsilon}(u)$ stands for the smallest number of balls
with radius $u$ associated to the metric defined by the norm
$\|\cdot\|_{\varepsilon}$. Since $D^{n}_{k,M}$ is defined as a product space,
one can determine the smallest number of balls with radius $r$ that cover each
of its components. For each parts, the metric is defined by the corresponding
parts of the r.h.s of (D.1). Therefore, the smallest number of balls with
radius $r$ that cover $[0,T]$ is of order $1/(r^{2}\varepsilon^{2})$, and for
$\\{\varphi\in W^{1,k}(S):\quad\|\varphi\|_{W^{1,k}}\leqslant M\\}$ it is of
order $\exp(1/r)$ [BS67, Theorem 5.2 p. 311]. As a result, for any
$n\in\mathbb{N}^{*}$

$\begin{split}\bm{E}\sup_{(t,\varphi)\in
D^{n}_{k,M}}V\Big{(}\frac{t}{\varepsilon^{2}},\varphi\Big{)}&\leqslant
C_{1}+C_{2}\int_{0}^{c_{k,M,r}}\sqrt{\ln\Big{(}\frac{C\exp(1/r)}{r^{2}\varepsilon^{2}}\Big{)}}\,dr\,,\\\
&\leqslant
C^{\prime}_{1}+C^{\prime}_{2}\int_{0}^{c_{k,M,r}}\frac{dr}{\sqrt{r}}+\frac{C^{\prime}_{3}}{\varepsilon}\int_{0}^{\varepsilon
c_{k,M,r}}\sqrt{\ln\Big{(}\frac{1}{u}\Big{)}}\,du\,.\end{split}$

Finally, setting

$C(\varepsilon)\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}C^{\prime}_{3}\int_{0}^{\varepsilon
c_{k,M,r}}\sqrt{\ln\Big{(}\frac{1}{u}\Big{)}}\,du\,,$

and using the monotone convergence theorem, we have

$E_{1}\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\bm{E}\sup_{(t,\varphi)\in\tilde{D}_{k,M}}V\Big{(}\frac{t}{\varepsilon^{2}},\varphi\Big{)}=\lim_{n\to\infty}\bm{E}\sup_{(t,\varphi)\in
D^{n}_{k,M}}V\Big{(}\frac{t}{\varepsilon^{2}},\varphi\Big{)}\leqslant
C^{\prime\prime}_{1}+\frac{C(\varepsilon)}{\varepsilon}\,.$

We conclude the proof of the bound (3.7) using that

$\bm{E}\sup_{(t,\varphi)\in\tilde{D}_{k,M}}\Big{|}V\Big{(}\frac{t}{\varepsilon^{2}},\varphi\Big{)}\Big{|}\leqslant
2E_{1}$

by symmetry.

For proving (3.8), we first remark that by stationarity in $t$ we have

$\bm{E}\sup_{\varphi\in
W_{k,M}}\Big{|}V\Big{(}\frac{t}{\varepsilon^{2}},\varphi\Big{)}\Big{|}^{n}=\bm{E}\sup_{\varphi\in
W_{k,M}}|V(0,\varphi)|^{n}\,,$

so that following the same lines as above by removing the $t$-dependence we
have for the first order moment,

$\bm{E}\sup_{\varphi\in W_{k,M}}|V(0,\varphi)|\leqslant
2\bm{E}\sup_{\varphi\in W_{k,M}}V(0,\varphi)\leqslant C_{1}\,.$

Finally, for the arbitrary order moments we write

$\begin{split}E^{\prime}_{n}\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\bm{E}\sup_{\varphi\in
W_{k,M}}|V(0,\varphi)|^{n}&=\int_{0}^{\infty}du\,\bm{P}\Big{(}\sup_{\varphi\in
W_{k,M}}|V(0,\varphi)|^{n}>u\Big{)}\\\
&=\int_{0}^{\infty}du\,\bm{P}\Big{(}\sup_{\varphi\in
W_{k,M}}|V(0,\varphi)|>u^{1/n}\Big{)}\\\ &\leqslant
2\int_{0}^{\infty}du\,\bm{P}\Big{(}\sup_{\varphi\in
W_{k,M}}V(0,\varphi)>u^{1/n}\Big{)}\,.\end{split}$

Denoting

$E^{\prime\prime}_{1}=\bm{E}\sup_{\varphi\in W_{k,M}}V(0,\varphi)\,,$

we have

$\begin{split}E^{\prime}_{n}&\leqslant
2{E^{\prime\prime}_{1}}^{n}+2\int_{{E^{\prime\prime}_{1}}^{n}}^{\infty}du\,\bm{P}\Big{(}\sup_{\varphi\in
W_{k,M}}V(0,\varphi)-E^{\prime\prime}_{1}>u^{1/n}-E^{\prime\prime}_{1}\Big{)}\\\
&\leqslant
2{E^{\prime\prime}_{1}}^{n}+2\int_{{E^{\prime\prime}_{1}}^{n}}^{\infty}du\,e^{-(u^{1/n}-E^{\prime\prime}_{1})^{2}/(2\sigma_{D}^{2})}\,,\end{split}$

where the last line is given by [AT07, Theorem 2.1.1 p. 50] and

$\sigma_{D}^{2}\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}\sup_{\varphi\in
W_{k,M}}\bm{E}V(0,\varphi)^{2}=\sup_{\varphi\in W_{k,M}}\int
dp\,r(p)\varphi(p)^{2}\leqslant C_{k}^{2}M^{2}\int dp\,r(p)\,.$

As a result,

$\begin{split}E^{\prime}_{n}&\leqslant
2{E^{\prime\prime}_{1}}^{n}+2(n-1)\int_{E^{\prime\prime}_{1}}^{\infty}dv\,v^{n-1}\,e^{-(v-E^{\prime\prime}_{1})^{2}/(2\sigma_{D}^{2})}\\\
&\leqslant
2{E^{\prime\prime}_{1}}^{n}+2(n-1)\int_{0}^{\infty}dv\,v^{n-1}\,e^{-v^{2}/(2\sigma_{D}^{2})}\,,\end{split}$

which concludes the proof of the bound (3.8). $\hfill\square$

##  Appendix E Proof of Proposition 4.2

The estimates for $a$, $\partial_{\theta}a$, and $\partial_{I}a$ we prove in
this section are based on the following estimates on the inverse map
$\varphi^{-1}$.

######  Lemma E.1.

There exist $r>0$ small enough, and a constant $C_{r}>0$ such that for any
$I\in(0,r)$

$\sup_{\theta\in\mathbb{T}}\Big{(}\|\varphi^{-1}(I,\theta)\|+\|\partial_{\theta}\varphi^{-1}(I,\theta)\|\Big{)}\leqslant
C_{r}\sqrt{I}\,,$

and

$\sup_{\theta\in\mathbb{T}}\|\partial_{I}\varphi^{-1}(I,\theta)\|\leqslant\frac{C_{r}}{\sqrt{I}}\,.$

The proof of this lemma is postponed to the end of this appendix.

In what follows we denote by

$B_{r}=\varphi^{-1}((0,r)\times\mathbb{T})$

with $r>0$ small enough so that $B_{r}$ describes a small neighborhood of
$(0,0)\in\mathbb{R}^{2}$.

###  E.1. Bound for $a$.

Let us remind the reader that

$a(I,\theta)=e_{2}\cdot\nabla\varphi_{1}(\varphi^{-1}(I,\theta))=\partial_{y}\varphi_{1}(\varphi^{-1}(I,\theta))\,,$

so that differentiating w.r.t. $y$ the relation $K(I(x,y))=H(x,y)$ we have

$\partial_{y}I(x,y)=\partial_{y}\varphi_{1}(x,y)=\partial_{y}H(x,y)/\omega(I)\,.$

Now, using that $\partial_{y}H(0,0)=0$, we have for any $I\in(0,r)$

$\begin{split}|a(I,\theta)|&\leqslant\omega^{-1}_{0}|\partial_{y}H(\varphi^{-1}(I,\theta))-\partial_{y}H(0,0)|\\\
&\leqslant\omega^{-1}_{0}\sup_{B_{r}}\|\nabla\partial_{y}H\|\cdot\|\varphi^{-1}(I,\theta)\|\leqslant
C_{r}\sqrt{I}.\end{split}$

###  E.2. Bound for $\partial_{\theta}a$

Using that

(E.1) $a(I,\theta)=\partial_{y}H(\varphi^{-1}(I,\theta))/\omega(I)\,,$

and now differentiating w.r.t. $\theta$, we obtain

$\partial_{\theta}a(I,\theta)=\partial_{\theta}\varphi^{-1}(I,\theta)\cdot\nabla\partial_{y}H(\varphi^{-1}(I,\theta))/\omega(I)\,.$

Then, by the Cauchy-Schwarz inequality

$\begin{split}|\partial_{\theta}a(I,\theta)|&\leqslant\omega^{-1}_{0}\|\partial_{\theta}\varphi^{-1}(I,\theta)\|\sup_{B_{r}}\|\nabla\partial_{y}H\|\leqslant
C_{r}\sqrt{I}\,.\end{split}$

###  E.3. Bound for $\partial_{I}a$

Differentiating (E.1) w.r.t. $I$, we obtain

$\begin{split}\partial_{I}a(I,\theta)&=-\frac{\omega^{\prime}(I)}{\omega^{2}(I)}\partial_{y}H(\varphi^{-1}(I,\theta))+\frac{1}{\omega(I)}\partial_{I}\varphi^{-1}(I,\theta)\cdot\nabla\partial_{y}H(\varphi^{-1}(I,\theta))\\\
&\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}T_{1}+T_{2}\,.\end{split}$

For $T_{1}$, using (4.12) and proceeding as for the function $a$ to deal with
the term $\partial_{y}H$ we have

$|T_{1}|\leqslant\frac{C_{r}}{\sqrt{I}}\,.$

For $T_{2}$, applying the Cauchy-Schwarz inequality together with Lemma E.1
yields

$|T_{2}|\leqslant\frac{C^{\prime}}{\sqrt{I}}\,,$

which concludes the proof of Proposition 4.2.

###  E.4. Proof of Lemma E.1

Before going into the proof of the lemma, let us remind the reader that $H$
admits a unique minimum at $(0,0)$ so that $\nabla^{2}H(0,0)$ is a positive
definite matrix

$\nabla^{2}H(0,0)=\begin{pmatrix}\lambda_{1}&\lambda_{12}\\\
\lambda_{12}&\lambda_{2}\end{pmatrix}$

with $\lambda_{1}$, $\lambda_{2}$ $>0$. In this case, the inner product
corresponding to this matrix defines a norm on $\mathbb{R}^{2}$ which is
equivalent to the Euclidean norm $|\cdot|$. In other words, there exist two
constants $\lambda$, $\bar{\lambda}>0$ such that for any
$X=(x,y)^{T}\in\mathbb{R}^{2}$

(E.2) $\lambda|X|^{2}\leqslant
X^{T}\nabla^{2}H(0,0)X\leqslant\bar{\lambda}|X|^{2}\,.$

####  E.4.1. Bound for $\varphi^{-1}$

Since $(0,0)$ is the unique minimum to $H$, a Taylor expansion at the second
order gives us

(E.3) $K(I)=H(X)=\frac{1}{2}X^{T}\nabla^{2}H(0,0)X+o(|X|^{2})\,,$

with $X=(x,y)^{T}$. For any $X\in B_{r}$ we can have using (E.2) that

(E.4) $K(I)\geqslant\tilde{\lambda}|X|^{2}\geqslant\tilde{\lambda}y^{2}$

with $\tilde{\lambda}=\lambda/2-\mu>0$ for some small $\mu$, yielding directly

$|\varphi^{-1}_{2}(I,\theta)|=|y|\leqslant\sqrt{K(I)/\tilde{\lambda}}=C\sqrt{K(I)-K(0)}\leqslant
C\sup_{J\in(0,r)}\sqrt{\omega(J)}\sqrt{I}=C^{\prime}\sqrt{I}.$

In the same way, we also have

$|\varphi^{-1}_{1}(I,\theta)|\leqslant C^{\prime}_{r}\sqrt{I}\,.$

####  E.4.2. Bound for $\partial_{\theta}\varphi^{-1}$

Differentiating w.r.t. $x$ and then $y$ the relation $K(I(X))=H(X)$ we have

$\partial_{x}\varphi_{1}(X)=\partial_{x}H(X)/\omega(I)\qquad\text{and}\qquad\partial_{y}\varphi_{1}(X)=\partial_{y}H(X)/\omega(I)\,,$

so that using (4.10) we obtain

(E.5)
$\partial_{\theta}\varphi^{-1}_{2}(I,\theta)=\frac{\partial_{x}H(\varphi^{-1}(I,\theta))}{\omega(I)}\quad\text{and}\quad\partial_{\theta}\varphi^{-1}_{1}(I,\theta)=\frac{-\partial_{y}H(\varphi^{-1}(I,\theta))}{\omega(I)}\,.$

Therefore, we have

$\|\partial_{\theta}\varphi^{-1}(I,\theta)\|\leqslant\|\nabla
H(\varphi^{-1}(I,\theta))\|/\omega_{0}\,,$

with

(E.6) $\begin{split}\|\nabla H(\varphi^{-1}(I,\theta))\|&=\|\nabla
H(\varphi^{-1}(I,\theta))-\nabla H(0,0)\|\\\
&\leqslant\sup_{B_{r}}\|\nabla^{2}H\|\cdot\|\varphi^{-1}(I,\theta)\|\leqslant
C_{r}\sqrt{I}\,.\end{split}$

####  E.4.3. Bound for $\partial_{I}\varphi^{-1}$

Differentiating (E.5) w.r.t. $I$ we obtain

$\begin{split}\partial^{2}_{I\theta}\varphi^{-1}_{2}(I,\theta)&=-\frac{\omega^{\prime}(I)}{\omega^{2}(I)}\partial_{x}H(\varphi^{-1}(I,\theta))+\frac{1}{\omega(I)}\partial_{I}\varphi^{-1}(I,\theta)\cdot\nabla\partial_{x}H(\varphi^{-1}(I,\theta))\\\
\partial^{2}_{I\theta}\varphi^{-1}_{1}(I,\theta)&=-\frac{\omega^{\prime}(I)}{\omega^{2}(I)}\partial_{y}H(\varphi^{-1}(I,\theta))+\frac{1}{\omega(I)}\partial_{I}\varphi^{-1}(I,\theta)\cdot\nabla\partial_{y}H(\varphi^{-1}(I,\theta))\,,\end{split}$

so that using (4.4) and (E.6)

$|\partial_{I}\varphi^{-1}_{1}(I,\theta)|\leqslant\frac{C_{1}}{\sqrt{I}}+\frac{1}{\omega_{0}}\sup_{B_{r}}\|\nabla^{2}H\|\int_{0}^{\theta}\|\partial_{I}\varphi^{-1}(I,\theta^{\prime})\|d\theta^{\prime}\,,$

and in the same way for $\varphi^{-1}_{2}$

$|\partial_{I}\varphi^{-1}_{2}(I,\theta)|\leqslant\frac{C_{2}}{\sqrt{I}}+|\partial_{I}\varphi^{-1}_{2}(I,0)|+\frac{1}{\omega_{0}}\sup_{B_{r}}\|\nabla^{2}H\|\int_{0}^{\theta}\|\partial_{I}\varphi^{-1}(I,\theta^{\prime})\|\,d\theta^{\prime}.$

Letting

$U(I,\theta)\stackrel{{\scriptstyle\scriptscriptstyle\textup{def}}}{{=}}|\partial_{I}\varphi^{-1}_{1}(I,\theta)|+|\partial_{I}\varphi^{-1}_{2}(I,\theta)|\,,$

and gathering the last two inequalities, we have

$U(I,\theta)\leqslant\frac{C}{\sqrt{I}}+|\partial_{I}\varphi^{-1}_{2}(I,0)|+C^{\prime}\int_{0}^{\theta}U(I,\theta^{\prime})\,d\theta^{\prime}\,.$

Then, applying the Gronwall lemma, we obtain

$\sup_{\theta\in(0,1)}U(I,\theta)\leqslant\Big{(}\frac{C}{\sqrt{I}}+|\partial_{I}\varphi^{-1}_{2}(I,0)|\Big{)}e^{C^{\prime}}\,.$

Moreover, differentiating w.r.t. $I$ the relation

$H(\varphi^{-1}(I,\theta))=K(I)$

we have

$\partial_{I}\varphi^{-1}_{1}(I,\theta)\partial_{x}H(\varphi^{-1}(I,\theta))+\partial_{I}\varphi^{-1}_{2}(I,\theta)\partial_{y}H(\varphi^{-1}(I,\theta))=\omega(I)$

and setting $\theta=0$ we have from (4.3) and (4.4)

$\partial_{I}\varphi^{-1}_{2}(I,0)\partial_{y}H(0,\varphi^{-1}_{2}(I,0))=\omega(I)\,.$

Also, using that

$\begin{split}\partial_{y}H(0,y)&=\partial_{y}H(0,0)+\partial^{2}_{yy}H(0,0)y+o(y)=\lambda_{2}y+o(y)\,,\end{split}$

so that for any $(0,y)\in B_{r}$ we have, for some small $\mu$,

$|\partial_{y}H(0,y)|\geqslant(\lambda_{2}-\mu)|y|\,.$

As a result,

$|\partial_{I}\varphi^{-1}_{2}(I,0)|\leqslant\frac{\omega(I)}{(\lambda_{2}-\mu)|\varphi^{-1}_{2}(I,0)|}\,,$

and from (4.3) together with (E.3), we have

$K(I)=H(0,\varphi^{-1}_{2}(I,0))\leqslant\bar{\lambda}_{2}\big{(}\varphi^{-1}_{2}(I,0)\big{)}^{2},$

with $\bar{\lambda}_{2}=\lambda_{2}/2+\mu$ so that

$|\varphi^{-1}_{2}(I,0)|\geqslant\sqrt{K(I)/\bar{\lambda}_{2}}\geqslant\sqrt{c_{1}I/\bar{\lambda}_{2}}\,.$

Finally, we obtain

$\sup_{\theta\in\mathbb{T}}|\partial_{I}\varphi^{-1}_{1}(I,\theta)|+|\partial_{I}\varphi^{-1}_{2}(I,\theta)|\leqslant\frac{C_{r}}{\sqrt{I}}\,,$

which gives the desired result and concludes the proof of the lemma.

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