(2) The slope identity eq. 4.28 comes from eq. 4.26, (4.6), (4.7), and (4.8).

(3) We use the expression in eq. 4.26 with $w:=u^{\prime}-v(\cdot,U,p)$, then
we use the quadratic response for $\nu^{*}(U,q)$ eq. 4.10 that

$\displaystyle\mathbb{E}_{\rho}\left[\frac{1}{\rho|U|}\int_{U}\left(\frac{1}{2}\nabla(w-v(\cdot,U,p,q))\cdot\mathbf{a}\nabla(w-v(\cdot,U,p,q))\right)\,{\mathrm{d}}\mu\right]$
$\displaystyle\qquad=\mathbb{E}_{\rho}\left[\frac{1}{\rho|U|}\int_{U}\left(\frac{1}{2}\nabla(u^{\prime}-u(\cdot,U,q))\cdot\mathbf{a}\nabla(u^{\prime}-u(\cdot,U,q))\right)\,{\mathrm{d}}\mu\right]$
$\displaystyle\qquad=\nu^{*}(U,q)-\mathbb{E}_{\rho}\left[\frac{1}{\rho|U|}\int_{U}\left(-\frac{1}{2}\nabla
u^{\prime}\cdot\mathbf{a}\nabla u^{\prime}+q\cdot\nabla
u^{\prime}\right)\,{\mathrm{d}}\mu\right].$

Then we add back the term $\nu(U,p)$ and it gives the desired result

$\displaystyle\mathbb{E}_{\rho}\left[\frac{1}{\rho|U|}\int_{U}\left(\frac{1}{2}\nabla(w-v(\cdot,U,p,q))\cdot\mathbf{a}\nabla(w-v(\cdot,U,p,q))\right)\,{\mathrm{d}}\mu\right]$
$\displaystyle\qquad=J(U,p,q)-\left(\nu(U,p)+\mathbb{E}_{\rho}\left[\frac{1}{\rho|U|}\int_{U}\left(-\frac{1}{2}\nabla
u^{\prime}\cdot\mathbf{a}\nabla u^{\prime}+q\cdot\nabla
u^{\prime}\right)\,{\mathrm{d}}\mu\right]-p\cdot q\right)$
$\displaystyle\qquad=J(U,p,q)-\mathbb{E}_{\rho}\left[\frac{1}{\rho|U|}\int_{U}\left(-\frac{1}{2}\nabla
w\cdot\mathbf{a}\nabla w-p\cdot\mathbf{a}\nabla w+q\cdot\nabla
w\right)\,{\mathrm{d}}\mu\right].$

(4) Equation 4.30 is a consequence of eq. 4.11 and eq. 4.19. ∎

We conclude this section with the following lemma.

######  Lemma 4.4 (Comparison between two scales).

For every $n,k\in\mathbb{N}$ with $k\leqslant n$, and
$p,q\in{\mathbb{R}^{d}}$, writing $v(U)$ as shorthand for $v(\cdot,U,p,q)$, we
have

(4.31)
$\frac{1}{|\mathcal{Z}_{n,k}|}\sum_{z\in\mathcal{Z}_{n,k}}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{k}|}\int_{z+{\scaleobj{1.2}{\square}}_{k}}\frac{1}{2}|\nabla
v({\scaleobj{1.2}{\square}}_{n})-\nabla
v(z+{\scaleobj{1.2}{\square}}_{k})|^{2}\,{\mathrm{d}}\mu\right]\\\ \leqslant
J({\scaleobj{1.2}{\square}}_{k},p,q)-J({\scaleobj{1.2}{\square}}_{n},p,q).$

###### Proof.

For any $z\in\mathcal{Z}_{n,k}$, since
$v({\scaleobj{1.2}{\square}}_{n})\in\mathcal{A}(z+{\scaleobj{1.2}{\square}}_{k})$,
we use the quadratic response eq. 4.29 for
$J(z+{\scaleobj{1.2}{\square}}_{k},p,q)$ that

$\displaystyle\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{k}|}\int_{z+{\scaleobj{1.2}{\square}}_{k}}\frac{1}{2}|\nabla
v({\scaleobj{1.2}{\square}}_{n})-\nabla
v(z+{\scaleobj{1.2}{\square}}_{k})|^{2}\,{\mathrm{d}}\mu\right]$
$\displaystyle\qquad\leqslant\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{k}|}\int_{z+{\scaleobj{1.2}{\square}}_{k}}\frac{1}{2}(\nabla
v({\scaleobj{1.2}{\square}}_{n})-\nabla
v(z+{\scaleobj{1.2}{\square}}_{k}))\cdot\mathbf{a}(\nabla
v({\scaleobj{1.2}{\square}}_{n})-\nabla
v(z+{\scaleobj{1.2}{\square}}_{k}))\,{\mathrm{d}}\mu\right]$
$\displaystyle\qquad=J(z+{\scaleobj{1.2}{\square}}_{k},p,q)$
$\displaystyle\qquad\qquad-\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{k}|}\int_{z+{\scaleobj{1.2}{\square}}_{k}}\left(-\frac{1}{2}\nabla
v({\scaleobj{1.2}{\square}}_{n})\cdot\mathbf{a}\nabla
v({\scaleobj{1.2}{\square}}_{n})-p\cdot\mathbf{a}\nabla
v({\scaleobj{1.2}{\square}}_{n})+q\cdot\nabla
v({\scaleobj{1.2}{\square}}_{n})\right)\,{\mathrm{d}}\mu\right].$

We sum this expression over all $z\in\mathcal{Z}_{n,k}$ to obtain that

$\displaystyle\frac{1}{|\mathcal{Z}_{n,k}|}\sum_{z\in\mathcal{Z}_{n,k}}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{k}|}\int_{z+{\scaleobj{1.2}{\square}}_{k}}\frac{1}{2}|\nabla
v({\scaleobj{1.2}{\square}}_{n})-\nabla
v(z+{\scaleobj{1.2}{\square}}_{k})|^{2}\,{\mathrm{d}}\mu\right]$
$\displaystyle\quad\leqslant\frac{1}{|\mathcal{Z}_{n,k}|}\sum_{z\in\mathcal{Z}_{n,k}}\bigg{(}J(z+{\scaleobj{1.2}{\square}}_{k},p,q)$
$\displaystyle\quad\qquad-\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{k}|}\int_{z+{\scaleobj{1.2}{\square}}_{k}}\left(-\frac{1}{2}\nabla
v({\scaleobj{1.2}{\square}}_{n})\cdot\mathbf{a}\nabla
v({\scaleobj{1.2}{\square}}_{n})-p\cdot\mathbf{a}\nabla
v({\scaleobj{1.2}{\square}}_{n})+q\cdot\nabla
v({\scaleobj{1.2}{\square}}_{n})\right)\,{\mathrm{d}}\mu\right]\bigg{)}$
$\displaystyle\quad=J({\scaleobj{1.2}{\square}}_{k},p,q)-J({\scaleobj{1.2}{\square}}_{n},p,q).$

In the last step, we use the stationarity of $J$ and also eq. 4.20 for
$v({\scaleobj{1.2}{\square}}_{n})$. ∎

##  5\. Quantitative rate of convergence

We are now ready to prove Theorem 2.1. We decompose the argument into a series
of four steps.

###  5.1. Step 1: setup

We use the shorthand
${\overbracket{\mathbf{a}}}_{n}:={\overbracket{\mathbf{a}}}_{*}({\scaleobj{1.2}{\square}}_{n})$,
so that by eq. 4.28, the average slope of the function
$v(\cdot,{\scaleobj{1.2}{\square}}_{n},p,q)$ is
${\overbracket{\mathbf{a}}}_{n}^{-1}q-p$, in the sense that

(5.1)
$\displaystyle\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\int_{U}\nabla
v(\cdot,{\scaleobj{1.2}{\square}}_{n},p,q)\,{\mathrm{d}}\mu\right]$
$\displaystyle={\overbracket{\mathbf{a}}}_{n}^{-1}q-p.$

We let $\tau_{n}$ denote a measure of the defect in the subadditivity of $J$,
precisely,

(5.2) $\begin{split}\tau_{n}:=&\sup_{p,q\in
B_{1}}(J({\scaleobj{1.2}{\square}}_{n},p,q)-J({\scaleobj{1.2}{\square}}_{n+1},p,q))\\\
=&\sup_{p\in
B_{1}}(\nu({\scaleobj{1.2}{\square}}_{n},p)-\nu({\scaleobj{1.2}{\square}}_{n+1},p))+\sup_{q\in
B_{1}}(\nu^{*}({\scaleobj{1.2}{\square}}_{n},q)-\nu^{*}({\scaleobj{1.2}{\square}}_{n+1},q)).\end{split}$

A direct corollary from eq. 5.2 is that for any integers $n<m$,

(5.3)
$\displaystyle|{\overbracket{\mathbf{a}}}_{n}^{-1}-{\overbracket{\mathbf{a}}}_{m}^{-1}|=\sup_{q\in
B_{1}}q\cdot({\overbracket{\mathbf{a}}}_{n}^{-1}-{\overbracket{\mathbf{a}}}_{m}^{-1})q=\sup_{q\in
B_{1}}\left(\nu^{*}({\scaleobj{1.2}{\square}}_{n},q)-\nu^{*}({\scaleobj{1.2}{\square}}_{m},q)\right)\leqslant
C\sum_{k=n}^{m-1}\tau_{k}.$

We recall that
$\\{{\overbracket{\mathbf{a}}}({\scaleobj{1.2}{\square}}_{m})\\}_{m\geqslant
0}$ is decreasing and
$\\{{\overbracket{\mathbf{a}}}_{*}({\scaleobj{1.2}{\square}}_{m})\\}_{m\geqslant
0}$ is increasing, with the comparison
${\overbracket{\mathbf{a}}}_{*}({\scaleobj{1.2}{\square}}_{m})\leqslant{\overbracket{\mathbf{a}}}({\scaleobj{1.2}{\square}}_{m})$.
From eq. 4.21, we know that

$\displaystyle|{\overbracket{\mathbf{a}}}({\scaleobj{1.2}{\square}}_{m})-{\overbracket{\mathbf{a}}}|\leqslant|{\overbracket{\mathbf{a}}}({\scaleobj{1.2}{\square}}_{m})-{\overbracket{\mathbf{a}}}_{*}({\scaleobj{1.2}{\square}}_{m})|\leqslant
C\sup_{p\in
B_{1}}(J({\scaleobj{1.2}{\square}}_{m},p,{\overbracket{\mathbf{a}}}_{m}p))^{\frac{1}{2}}.$

From now on, we thus fix $p\in B_{1}$, and focus on estimating
$J({\scaleobj{1.2}{\square}}_{m},p,{\overbracket{\mathbf{a}}}_{m}p)$. We also
assume without further notification that $m$ is sufficiently large that
$3^{m}\geqslant R_{0}$, for the constant $R_{0}$ appearing in Proposition 3.9.
We use
${{\mathsf{A}}_{3^{m}+2}v(\cdot,{\scaleobj{1.2}{\square}}_{m+1},p,{\overbracket{\mathbf{a}}}_{m}p)}$
to compare with eq. 4.27 and apply the quadratic response eq. 4.29. In the
rest of Step 1, we write $v(U)$ as a shorthand for
$v(\cdot,U,p,{\overbracket{\mathbf{a}}}_{m}p)$, and decompose

(5.4)
$\begin{split}(J({\scaleobj{1.2}{\square}}_{m},p,{\overbracket{\mathbf{a}}}_{m}p))^{\frac{1}{2}}&=\left(\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\frac{1}{2}\nabla
v({\scaleobj{1.2}{\square}}_{m})\cdot\mathbf{a}\nabla
v({\scaleobj{1.2}{\square}}_{m})\,{\mathrm{d}}\mu\right]\right)^{\frac{1}{2}}\\\
&\leqslant\text{\lx@cref{creftype~refnum}{eq:JamDecom}-a}+\text{\lx@cref{creftype~refnum}{eq:JamDecom}-b},\end{split}$

with

$\text{\lx@cref{creftype~refnum}{eq:JamDecom}-a}\\\
=\left(\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\frac{1}{2}(\nabla
v({\scaleobj{1.2}{\square}}_{m})-\nabla{\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1}))\cdot\mathbf{a}(\nabla
v({\scaleobj{1.2}{\square}}_{m})-\nabla{\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1}))\,{\mathrm{d}}\mu\right]\right)^{\frac{1}{2}},$

and

$\text{\lx@cref{creftype~refnum}{eq:JamDecom}-b}=\left(\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\frac{1}{2}\nabla{\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1})\cdot\mathbf{a}\nabla{\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1})\,{\mathrm{d}}\mu\right]\right)^{\frac{1}{2}}.$

We treat the two terms separately. For eq. 5.4-a, since
${{\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1})\in\mathcal{A}({\scaleobj{1.2}{\square}}_{m})}$
(see Proposition A.2 for details), we use eq. 4.29 to get

$\displaystyle|\text{\lx@cref{creftype~refnum}{eq:JamDecom}-a}|^{2}$
$\displaystyle\qquad=J({\scaleobj{1.2}{\square}}_{m},p,{\overbracket{\mathbf{a}}}_{m}p)-\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\left(-\frac{1}{2}\nabla{\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1})\cdot\mathbf{a}\nabla{\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1})\right)\,{\mathrm{d}}\mu\right]$
$\displaystyle\qquad\qquad-\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\left(-p\cdot\mathbf{a}\nabla{\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1})+{\overbracket{\mathbf{a}}}_{m}p\cdot\nabla{\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1})\right)\,{\mathrm{d}}\mu\right].$

Using Jensen’s inequality, we have

$\mathbb{E}_{\rho}\left[\int_{{\scaleobj{1.2}{\square}}_{m}}\left(\frac{1}{2}\nabla{\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1})\cdot\mathbf{a}\nabla{\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1})\right)\,{\mathrm{d}}\mu\right]\\\
\leqslant\mathbb{E}_{\rho}\left[\int_{{\scaleobj{1.2}{\square}}_{m}}\left(\frac{1}{2}\nabla
v({\scaleobj{1.2}{\square}}_{m+1})\cdot\mathbf{a}\nabla
v({\scaleobj{1.2}{\square}}_{m+1})\right)\,{\mathrm{d}}\mu\right],$

and the conditional expectation also implies that

$\mathbb{E}_{\rho}\left[\int_{{\scaleobj{1.2}{\square}}_{m}}\left(-p\cdot\mathbf{a}\nabla{\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1})+{\overbracket{\mathbf{a}}}_{m}p\cdot\nabla{\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1})\right)\,{\mathrm{d}}\mu\right]\\\
=\mathbb{E}_{\rho}\left[\int_{{\scaleobj{1.2}{\square}}_{m}}\left(-p\cdot\mathbf{a}\nabla
v({\scaleobj{1.2}{\square}}_{m+1})+{\overbracket{\mathbf{a}}}_{m}p\cdot\nabla
v({\scaleobj{1.2}{\square}}_{m+1})\right)\,{\mathrm{d}}\mu\right].$

Thus we combine these terms with the quadratic response eq. 4.29 to obtain

$\displaystyle|\text{\lx@cref{creftype~refnum}{eq:JamDecom}-a}|^{2}$
$\displaystyle\leqslant
J({\scaleobj{1.2}{\square}}_{m},p,{\overbracket{\mathbf{a}}}_{m}p)-\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\left(-\frac{1}{2}\nabla
v({\scaleobj{1.2}{\square}}_{m+1})\cdot\mathbf{a}\nabla
v({\scaleobj{1.2}{\square}}_{m+1})\right)\,{\mathrm{d}}\mu\right]$
$\displaystyle\qquad-\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\left(-p\cdot\mathbf{a}\nabla
v({\scaleobj{1.2}{\square}}_{m+1})+{\overbracket{\mathbf{a}}}_{m}p\cdot\nabla
v({\scaleobj{1.2}{\square}}_{m+1})\right)\,{\mathrm{d}}\mu\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\left(\frac{1}{2}\nabla(v({\scaleobj{1.2}{\square}}_{m+1})-v({\scaleobj{1.2}{\square}}_{m}))\cdot\mathbf{a}\nabla(v({\scaleobj{1.2}{\square}}_{m+1})-v({\scaleobj{1.2}{\square}}_{m}))\right)\,{\mathrm{d}}\mu\right],$

and we use Lemma 4.4 between ${\scaleobj{1.2}{\square}}_{m}$ and
${\scaleobj{1.2}{\square}}_{m+1}$ to get

(5.5) $\displaystyle|\text{\lx@cref{creftype~refnum}{eq:JamDecom}-a}|^{2}$
$\displaystyle\leqslant
3^{d}(J({\scaleobj{1.2}{\square}}_{m},p,{\overbracket{\mathbf{a}}}_{m}p)-J({\scaleobj{1.2}{\square}}_{m+1},p,{\overbracket{\mathbf{a}}}_{m}p))\leqslant
C(d,\Lambda)\tau_{m},$

where the quantity $\tau_{m}$ is defined in eq. 5.2.

For the term eq. 5.4-b, we can apply the modified Caccioppoli inequality eq.
3.37: there exist two finite positive constants $C(d,\Lambda)$ and
$\theta(d,\Lambda)\in(0,1)$ such that

(5.6)
$\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\nabla({\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1}))\cdot\mathbf{a}\nabla({\mathsf{A}}_{3^{m}+2}v({\scaleobj{1.2}{\square}}_{m+1}))\,{\mathrm{d}}\mu\right]\\\
\leqslant\frac{C}{3^{2m}\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\mathbb{E}_{\rho}[(v({\scaleobj{1.2}{\square}}_{m+1}))^{2}]+\theta\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\int_{{\scaleobj{1.2}{\square}}_{m+1}}\nabla
v({\scaleobj{1.2}{\square}}_{m+1})\cdot\mathbf{a}\nabla
v({\scaleobj{1.2}{\square}}_{m+1})\,{\mathrm{d}}\mu\right].$

Using eq. 4.27, we see that the averaged gradient term on the right side of
eq. 5.6 is
$J({\scaleobj{1.2}{\square}}_{m+1},p,{\overbracket{\mathbf{a}}}_{m}p)$, and
eq. 4.30 asserts that
${J({\scaleobj{1.2}{\square}}_{m+1},p,{\overbracket{\mathbf{a}}}_{m}p)\leqslant
J({\scaleobj{1.2}{\square}}_{m},p,{\overbracket{\mathbf{a}}}_{m}p)}$.
Therefore, we get the bound for eq. 5.4-b

(5.7)
$\displaystyle|\text{\lx@cref{creftype~refnum}{eq:JamDecom}-b}|^{2}\leqslant\frac{C}{3^{2m}\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\mathbb{E}_{\rho}[(v({\scaleobj{1.2}{\square}}_{m+1}))^{2}]+\theta
J({\scaleobj{1.2}{\square}}_{m},p,{\overbracket{\mathbf{a}}}_{m}p).$

We put eq. 5.5 and eq. 5.7 back to eq. 5.4, obtaining

$\displaystyle(J({\scaleobj{1.2}{\square}}_{m},p,{\overbracket{\mathbf{a}}}_{m}p))^{\frac{1}{2}}$
$\displaystyle\leqslant
C\tau_{m}^{\frac{1}{2}}+\left(\frac{C}{3^{2m}\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\left\|{v({\scaleobj{1.2}{\square}}_{m+1})}\right\|^{2}_{\mathscr{L}^{2}}+\theta
J({\scaleobj{1.2}{\square}}_{m},p,{\overbracket{\mathbf{a}}}_{m}p)\right)^{\frac{1}{2}}$
$\displaystyle\leqslant
C\tau_{m}^{\frac{1}{2}}+\frac{C}{3^{m}(\rho|{\scaleobj{1.2}{\square}}_{m+1}|)^{\frac{1}{2}}}\left\|{v({\scaleobj{1.2}{\square}}_{m+1})}\right\|_{\mathscr{L}^{2}}+\theta^{\frac{1}{2}}(J({\scaleobj{1.2}{\square}}_{m},p,{\overbracket{\mathbf{a}}}_{m}p))^{\frac{1}{2}}.$

Since $\theta<1$, this gives

(5.8) $\displaystyle
J({\scaleobj{1.2}{\square}}_{m},p,{\overbracket{\mathbf{a}}}_{m}p)\leqslant
C\left(\tau_{m}+\frac{1}{3^{2m}\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\left\|{v(\mu,{\scaleobj{1.2}{\square}}_{m+1},p,{\overbracket{\mathbf{a}}}_{m}p)}\right\|^{2}_{\mathscr{L}^{2}}\right).$

###  5.2. Step 2: flatness estimate

In this step, we estimate the $\mathscr{L}^{2}$-flatness of optimizers of $J$.
Notice that, using the result of Lemma 5.10 with
$v(\cdot,{\scaleobj{1.2}{\square}}_{m+1},p,{\overbracket{\mathbf{a}}}_{m}p)$,
the corresponding affine function is $0$ and we obtain from eq. 5.8 that

(5.9) $\displaystyle
J({\scaleobj{1.2}{\square}}_{m},p,{\overbracket{\mathbf{a}}}_{m}p)\leqslant
C\left(3^{-\beta m}+\sum_{n=0}^{m}3^{-\beta(m-n)}\tau_{n}\right).$

######  Lemma 5.1 ($\mathscr{L}^{2}$-flatness estimate).

There exist $\beta(d)>0$ and $C(d,\Lambda,\rho)<\infty$ such that for every
$p,q\in B_{1}$ and $m\in\mathbb{N}$,

(5.10)
$\displaystyle\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\left\|{v(\cdot,{\scaleobj{1.2}{\square}}_{m+1},p,q)-\ell_{\overline{\mathbf{a}}_{m}^{-1}q-p,{\scaleobj{1.2}{\square}}_{m+1}}}\right\|^{2}_{\mathscr{L}^{2}}\leqslant
C3^{2m}\left(3^{-\beta m}+\sum_{n=0}^{m}3^{-\beta(m-n)}\tau_{n}\right).$

###### Proof.

In the rest of the proof, we write $v(U):=v(\cdot,U,p,q)$ as we will not
change $p,q$ in the proof. Since
$\mathbb{E}_{\rho}\left[v({\scaleobj{1.2}{\square}}_{m+1})-\ell_{\overline{\mathbf{a}}_{m}^{-1}q-p,{\scaleobj{1.2}{\square}}_{m+1}}|\mathcal{G}_{m+1}\right]=0$,
we can use the multiscale Poincaré inequality eq. 3.23

(5.11)
$\begin{split}&\frac{1}{(\rho|{\scaleobj{1.2}{\square}}_{m+1}|)^{\frac{1}{2}}}\left\|{v({\scaleobj{1.2}{\square}}_{m+1})-\ell_{\overline{\mathbf{a}}_{m}^{-1}q-p,{\scaleobj{1.2}{\square}}_{m+1}}}\right\|_{\mathscr{L}^{2}}\\\
&\leqslant
C\left(\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\int_{{\scaleobj{1.2}{\square}}_{m+1}}|\nabla
v({\scaleobj{1.2}{\square}}_{m+1})-({\overbracket{\mathbf{a}}}_{m}^{-1}q-p)|^{2}\,{\mathrm{d}}\mu\right]\right)^{\frac{1}{2}}\\\
&\qquad+C\sum_{n=0}^{m+1}3^{n}\left(\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\int_{{\scaleobj{1.2}{\square}}_{m+1}}|\mathsf{S}_{m+1,n}\nabla
v({\scaleobj{1.2}{\square}}_{m+1})-({\overbracket{\mathbf{a}}}_{m}^{-1}q-p)|^{2}\,{\mathrm{d}}\mu\right]\right)^{\frac{1}{2}}.\end{split}$

The first term on the right side above is of constant order, by eq. 4.27. For
the second term, we use a two-scale comparison for every $0\leqslant
n\leqslant m+1$ that

(5.12)
$\begin{split}&\left(\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\int_{{\scaleobj{1.2}{\square}}_{m+1}}|\mathsf{S}_{m+1,n}\nabla
v({\scaleobj{1.2}{\square}}_{m+1})-({\overbracket{\mathbf{a}}}_{m}^{-1}q-p)|^{2}\,{\mathrm{d}}\mu\right]\right)^{\frac{1}{2}}\\\
&\leqslant|{\overbracket{\mathbf{a}}}_{m}^{-1}-{\overbracket{\mathbf{a}}}_{n}^{-1}|+\left(\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\sum_{z\in\mathcal{Z}_{m+1,n}}\int_{z+{\scaleobj{1.2}{\square}}_{n}}|\mathsf{S}_{m+1,n}\nabla
v({\scaleobj{1.2}{\square}}_{m+1})-\mathsf{S}_{m+1,n}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n})|^{2}\,{\mathrm{d}}\mu\right]\right)^{\frac{1}{2}}\\\
&\qquad+\left(\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\sum_{z\in\mathcal{Z}_{m+1,n}}\int_{z+{\scaleobj{1.2}{\square}}_{n}}|\mathsf{S}_{m+1,n}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n})-({\overbracket{\mathbf{a}}}_{n}^{-1}q-p)|^{2}\,{\mathrm{d}}\mu\right]\right)^{\frac{1}{2}}.\end{split}$

For the first term
$|{\overbracket{\mathbf{a}}}_{m}^{-1}-{\overbracket{\mathbf{a}}}_{n}^{-1}|$ we
have

$\displaystyle|{\overbracket{\mathbf{a}}}_{m}^{-1}-{\overbracket{\mathbf{a}}}_{n}^{-1}|^{2}\leqslant
C(d,\Lambda)|{\overbracket{\mathbf{a}}}_{m}^{-1}-{\overbracket{\mathbf{a}}}_{n}^{-1}|\leqslant\sum_{k=n}^{m-1}\tau_{k}.$

For the second term in eq. 5.12, recalling eq. 3.15, we use Jensen’s
inequality and eq. 4.31 to get

$\displaystyle\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\sum_{z\in\mathcal{Z}_{m+1,n}}\int_{z+{\scaleobj{1.2}{\square}}_{n}}|\mathsf{S}_{m+1,n}\nabla
v({\scaleobj{1.2}{\square}}_{m+1})-\mathsf{S}_{m+1,n}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n})|^{2}\,{\mathrm{d}}\mu\right]$
$\displaystyle\quad\leqslant\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\sum_{z\in\mathcal{Z}_{m+1,n}}\int_{z+{\scaleobj{1.2}{\square}}_{n}}|\nabla
v({\scaleobj{1.2}{\square}}_{m+1})-\nabla
v(z+{\scaleobj{1.2}{\square}}_{n})|^{2}\,{\mathrm{d}}\mu\right]$
$\displaystyle\quad\leqslant\sum_{k=n}^{m}\tau_{k}.$

For the third term eq. 5.12, we use eq. 3.17, Jensen’s inequality, and
stationarity. Here we remark that the operator $\mathsf{S}_{n,n}^{z}$ is a
conditional expectation with more information than $\mathsf{S}_{m+1,n}$.

$\displaystyle\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\sum_{z\in\mathcal{Z}_{m+1,n}}\int_{z+{\scaleobj{1.2}{\square}}_{n}}|\mathsf{S}_{m+1,n}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n})-({\overbracket{\mathbf{a}}}_{n}^{-1}q-p)|^{2}\,{\mathrm{d}}\mu\right]$
$\displaystyle\quad\leqslant\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\sum_{z\in\mathcal{Z}_{m+1,n}}\int_{z+{\scaleobj{1.2}{\square}}_{n}}|\mathsf{S}_{n,n}^{z}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n})-({\overbracket{\mathbf{a}}}_{n}^{-1}q-p)|^{2}\,{\mathrm{d}}\mu\right]$
$\displaystyle\quad=\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\int_{{\scaleobj{1.2}{\square}}_{n}}|\mathsf{S}_{n}\nabla
v({\scaleobj{1.2}{\square}}_{n})-({\overbracket{\mathbf{a}}}_{n}^{-1}q-p)|^{2}\,{\mathrm{d}}\mu\right].$

The estimation of this term is postponed to the next step. We will prove in
Lemma 5.2 below that

$\displaystyle\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\int_{{\scaleobj{1.2}{\square}}_{n}}|\mathsf{S}_{n}\nabla
v({\scaleobj{1.2}{\square}}_{n})-({\overbracket{\mathbf{a}}}_{n}^{-1}q-p)|^{2}\,{\mathrm{d}}\mu\right]\leqslant
C3^{-\beta n}+\sum_{k=0}^{n-1}3^{-\beta(n-k)}\tau_{k}.$

We put these estimates back to eq. 5.11 and obtain that

$\displaystyle\frac{1}{(\rho|{\scaleobj{1.2}{\square}}_{m+1}|)^{\frac{1}{2}}}\left\|{v({\scaleobj{1.2}{\square}}_{m+1})-\ell_{{\overbracket{\mathbf{a}}}_{m}^{-1}q-p,{\scaleobj{1.2}{\square}}_{m+1}}}\right\|_{\mathscr{L}^{2}}\leqslant
C\sum_{n=0}^{m}3^{n}\left(3^{-\beta
n}+\sum_{k=0}^{n-1}3^{-\beta(n-k)}\tau_{k}+\sum_{k=n}^{m}\tau_{k}\right)^{\frac{1}{2}}.$

We square the two sides and use the Cauchy-Schwarz inequality to obtain

$\displaystyle\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m+1}|}\left\|{v({\scaleobj{1.2}{\square}}_{m+1})-\ell_{{\overbracket{\mathbf{a}}}_{m}^{-1}q-p,{\scaleobj{1.2}{\square}}_{m+1}}}\right\|_{\mathscr{L}^{2}}^{2}$
$\displaystyle\qquad\leqslant
C\left(\sum_{n=0}^{m}3^{n}\right)\left(\sum_{n=0}^{m}3^{n}\left(3^{-\beta
n}+\sum_{k=0}^{n-1}3^{-\beta(n-k)}\tau_{k}+\sum_{k=n}^{m}\tau_{k}\right)\right)$
$\displaystyle\qquad\leqslant C3^{2m}\left(3^{-\beta
m}+\sum_{n=0}^{m}3^{-\beta(m-n)}\tau_{n}\right),$

as announced. ∎

###  5.3. Step 3: variance estimate

In this part, we prove the following variance estimate, which was used in Step
2.

######  Lemma 5.2 (Variance estimate).

There exist $\beta(d)>0$ and $C(d,\Lambda,\rho)<\infty$ such that for every
$p,q\in B_{1}$ and $n\in\mathbb{N}$,

(5.13)
$\displaystyle\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\int_{{\scaleobj{1.2}{\square}}_{n}}|\mathsf{S}_{n}\nabla
v(\mu,{\scaleobj{1.2}{\square}}_{n},p,q)-({\overbracket{\mathbf{a}}}_{n}^{-1}q-p)|^{2}\,{\mathrm{d}}\mu\right]\leqslant
C3^{-\beta n}+\sum_{k=0}^{n-1}3^{-\beta(n-k)}\tau_{k}.$

###### Proof.

In the rest of the proof, we write $v(U):=v(\cdot,U,p,q)$, as we will not
change $p,q$ in the proof. From eq. 5.1, we know that the average slope of
$v({\scaleobj{1.2}{\square}}_{n})$ is
$({\overbracket{\mathbf{a}}}_{n}^{-1}q-p)$, and notice that
$v({\scaleobj{1.2}{\square}}_{n})$ is
$\mathcal{F}_{B_{1}({\scaleobj{1.2}{\square}}_{n})}$-measurable. Thus the idea
is to use $\\{v(z+{\scaleobj{1.2}{\square}}_{k})\\}_{z\in\mathcal{Z}_{n,k}}$
to approximate eq. 5.13 in scale $3^{k}$ with some error, and then apply the
independence for $v(z+{\scaleobj{1.2}{\square}}_{k})$ and
$v(z^{\prime}+{\scaleobj{1.2}{\square}}_{k})$ for
$\operatorname{dist}(z,z^{\prime})$ large. However, different from the
standard elliptic setting, here we will see a renormalization with random
weights.

We start by relaxing eq. 5.13 to $\mathcal{G}_{n,n-2}$. We observe that in
fact $\mathsf{S}_{n}\nabla v({\scaleobj{1.2}{\square}}_{n})$ is constant in
${\scaleobj{1.2}{\square}}_{n}$, so

$\displaystyle\int_{{\scaleobj{1.2}{\square}}_{n}}|\mathsf{S}_{n}\nabla
v({\scaleobj{1.2}{\square}}_{n})-({\overbracket{\mathbf{a}}}_{n}^{-1}q-p)|^{2}\,{\mathrm{d}}\mu=\frac{1}{\mu({\scaleobj{1.2}{\square}}_{n})}\left|\int_{{\scaleobj{1.2}{\square}}_{n}}\left(\mathsf{S}_{n}\nabla
v({\scaleobj{1.2}{\square}}_{n})-({\overbracket{\mathbf{a}}}_{n}^{-1}q-p)\right)\,{\mathrm{d}}\mu\right|^{2}.$

We denote by $V_{n}$ the left hand side of eq. 5.13. By triangle inequality,
we have

(5.14)
$\displaystyle(V_{n})^{\frac{1}{2}}\leqslant\text{\lx@cref{creftype~refnum}{eq:mainVarianceDecom}-a}+\text{\lx@cref{creftype~refnum}{eq:mainVarianceDecom}-b}+\text{\lx@cref{creftype~refnum}{eq:mainVarianceDecom}-c},$

with

eq. 5.14-a
$\displaystyle=|{\overbracket{\mathbf{a}}}_{n}^{-1}-{\overbracket{\mathbf{a}}}_{n-2}^{-1}|,$
eq. 5.14-b
$\displaystyle=\left(\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\frac{1}{\mu({\scaleobj{1.2}{\square}}_{n})}\left|\sum_{z\in\mathcal{Z}_{n,n-2}}\int_{z+{\scaleobj{1.2}{\square}}_{n-2}}\left(\mathsf{S}_{n,n}\nabla
v({\scaleobj{1.2}{\square}}_{n})-\mathsf{S}_{n,n-2}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n-2})\right)\,{\mathrm{d}}\mu\right|^{2}\right]\right)^{\frac{1}{2}},$
eq. 5.14-c
$\displaystyle=\left(\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\frac{1}{\mu({\scaleobj{1.2}{\square}}_{n})}\left|\sum_{z\in\mathcal{Z}_{n,n-2}}\int_{z+{\scaleobj{1.2}{\square}}_{n-2}}\left(\mathsf{S}_{n,n-2}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n-2})-({\overbracket{\mathbf{a}}}_{n-2}^{-1}q-p)\right)\,{\mathrm{d}}\mu\right|^{2}\right]\right)^{\frac{1}{2}}.$

The term eq. 5.14-a can be controlled by eq. 5.3:

(5.15)
$\displaystyle\text{\lx@cref{creftype~refnum}{eq:mainVarianceDecom}-a}\leqslant
C(\tau_{n-2}+\tau_{n-1})^{\frac{1}{2}}.$

For the term eq. 5.14-b, recalling eq. 3.17 and eq. 3.15, we use Jensen’s
inequality and the two-scale comparison eq. 4.31 to get

(5.16)
$\begin{split}\text{\lx@cref{creftype~refnum}{eq:mainVarianceDecom}-b}&\leqslant\left(\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\sum_{z\in\mathcal{Z}_{n,n-2}}\int_{z+{\scaleobj{1.2}{\square}}_{n-2}}|\mathsf{S}_{n,n}\nabla
v({\scaleobj{1.2}{\square}}_{n})-\mathsf{S}_{n,n-2}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n-2})|^{2}\,{\mathrm{d}}\mu\right]\right)^{\frac{1}{2}}\\\
&\leqslant\left(\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\sum_{z\in\mathcal{Z}_{n,n-2}}\int_{z+{\scaleobj{1.2}{\square}}_{n-2}}|\nabla
v({\scaleobj{1.2}{\square}}_{n})-\nabla
v(z+{\scaleobj{1.2}{\square}}_{n-2})|^{2}\,{\mathrm{d}}\mu\right]\right)^{\frac{1}{2}}\\\
&\leqslant(\tau_{n-2}+\tau_{n-1})^{\frac{1}{2}}.\end{split}$

The term eq. 5.14-c is the key for our result. To simplify a little more the
notation, we write

(5.17) $\left\\{\begin{array}[]{ll}X_{z}:=\mathsf{S}_{n,n-2}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n-2})(\mu,z)-({\overbracket{\mathbf{a}}}_{n-2}^{-1}q-p),\\\
m_{z}:=\mu(z+{\scaleobj{1.2}{\square}}_{n-2}).\end{array}\right.$

Notice that $X_{z},m_{z}$ are
$\mathcal{F}_{z+{\scaleobj{1.2}{\square}}_{n-1}}$-measurable. With this
notation in place, we have

$\displaystyle\int_{z+{\scaleobj{1.2}{\square}}_{n-2}}\left(\mathsf{S}_{n,n-2}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n-2})-({\overbracket{\mathbf{a}}}_{n-2}^{-1}q-p)\right)\,{\mathrm{d}}\mu$
$\displaystyle=m_{z}X_{z},$

and by eq. 5.1,

$\mathbb{E}_{\rho}[m_{z}X_{z}]=0.$

The term eq. 5.14-c we want to estimate can be rewritten as

eq. 5.14-c
$\displaystyle=\left(\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\frac{\left(\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}X_{z}\right)^{2}}{\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}}\right]\right)^{\frac{1}{2}}.$

If the coefficients $m_{z}$ were deterministic, then we would be able to
leverage on the finite range of dependence of $X_{z}$ in this variance term.
However, since the number of particles $m_{z}$ is random, we introduce the
event

(5.18)
$\displaystyle\mathcal{C}_{n,\rho,\delta}:=\left\\{\mu\in\mathcal{M}_{\delta}({\mathbb{R}^{d}}):\forall
z\in\mathcal{Z}_{n,n-2},\left|\frac{\mu(z+{\scaleobj{1.2}{\square}}_{n-2})}{\rho|{\scaleobj{1.2}{\square}}_{n-2}|}-1\right|\leqslant\delta,\text{
and
}\left|\frac{\mu({\scaleobj{1.2}{\square}}_{n})}{\rho|{\scaleobj{1.2}{\square}}_{n}|}-1\right|\leqslant\delta\right\\},$

thus we can divide eq. 5.14-c into two terms

eq. 5.14-c
$\displaystyle\leqslant\text{\lx@cref{creftype~refnum}{eq:mainVarianceDecom}-c1}+\text{\lx@cref{creftype~refnum}{eq:mainVarianceDecom}-c2},$
eq. 5.14-c1
$\displaystyle=\left(\mathbb{E}_{\rho}\left[\frac{\mathbf{1}_{\left\\{(\mathcal{C}_{n,\rho,\delta})^{c}\right\\}}}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\frac{\left(\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}X_{z}\right)^{2}}{\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}}\right]\right)^{\frac{1}{2}},$
eq. 5.14-c2
$\displaystyle=\left(\mathbb{E}_{\rho}\left[\frac{\mathbf{1}_{\left\\{\mathcal{C}_{n,\rho,\delta}\right\\}}}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\frac{\left(\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}X_{z}\right)^{2}}{\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}}\right]\right)^{\frac{1}{2}}.$

For the term eq. 5.14-c1, we know that $(\mathcal{C}_{n,\rho,\delta})^{c}$ is
not typical in large scales, and we have the Chernoff bound

$\displaystyle\mathbb{P}_{\rho}[\mu\notin\mathcal{C}_{n,\rho,\delta}]\leqslant
3^{2d+1}\exp\left(-\frac{\rho|{\scaleobj{1.2}{\square}}_{n-2}|\delta^{2}}{4}\right).$

Moreover, by the Cauchy-Schwarz inequality,

$\displaystyle\frac{\left(\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}X_{z}\right)^{2}}{\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}}\leqslant\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}|X_{z}|^{2}.$

We need a bound for the term $|X_{z}|^{2}$: recalling the definition in eq.
3.15 and eq. 4.28,

$\displaystyle\mathsf{S}_{n-2,n-2}^{z}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n-2})(\mu,z)$
$\displaystyle=\mathbb{E}_{\rho}\left[\strokedint_{z+{\scaleobj{1.2}{\square}}_{n-2}}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n-2})\,{\mathrm{d}}\mu\ \Bigg{|}\
\mathcal{G}_{n-2,n-2}^{z}\right]$
$\displaystyle=\mathbf{a}(z+{\scaleobj{1.2}{\square}}_{n-2};\mathcal{G}_{z+{\scaleobj{1.2}{\square}}_{n-2}})^{-1}q-p.$

Using the martingale structure of eq. 3.17, we have

$\displaystyle X_{z}$ $\displaystyle=\mathsf{S}_{n,n-2}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n-2})(\mu,z)-({\overbracket{\mathbf{a}}}_{n-2}^{-1}q-p)$
$\displaystyle=\mathbb{E}_{\rho}\left[\strokedint_{z+{\scaleobj{1.2}{\square}}_{n-2}}\mathsf{S}_{n-2,n-2}^{z}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n-2})\,{\mathrm{d}}\mu\ \Bigg{|}\
\mathcal{G}_{n,n-2}\right]-({\overbracket{\mathbf{a}}}_{n-2}^{-1}q-p)$
$\displaystyle=\mathbb{E}_{\rho}\left[\mathbf{a}(z+{\scaleobj{1.2}{\square}}_{n-2};\mathcal{G}_{z+{\scaleobj{1.2}{\square}}_{n-2}})^{-1}-{\overbracket{\mathbf{a}}}_{n-2}^{-1}\
\Bigg{|}\ \mathcal{G}_{n,n-2}\right]q.$

Then we use Jensen’s inequality and the bound of
${\mathsf{Id}\leqslant\mathbf{a}(z+{\scaleobj{1.2}{\square}}_{n-2};\mathcal{G}_{z+{\scaleobj{1.2}{\square}}_{n-2}})\leqslant\Lambda\mathsf{Id}}$

(5.19) $\begin{split}|X_{z}|^{2}&=|\mathsf{S}_{n,n-2}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n-2})-({\overbracket{\mathbf{a}}}_{n-2}^{-1}q-p)|^{2}\\\
&=\mathbb{E}_{\rho}\left[|\mathbf{a}(z+{\scaleobj{1.2}{\square}}_{n-2};\mathcal{G}_{z+{\scaleobj{1.2}{\square}}_{n-2}})-{\overbracket{\mathbf{a}}}_{n-2}^{-1}|^{2}\,|\,\mathcal{G}_{n,n-2}\right]\leqslant\Lambda^{2}.\end{split}$

This concludes that

(5.20)
$\begin{split}\text{\lx@cref{creftype~refnum}{eq:mainVarianceDecom}-c1}\leqslant\Lambda^{2}\mathbb{E}_{\rho}\left[\frac{\mathbf{1}_{\left\\{(\mathcal{C}_{n,\rho,\delta})^{c}\right\\}}}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\mu({\scaleobj{1.2}{\square}}_{n})\right]&\leqslant
C(d,\Lambda)\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n-2}|}\exp\left(-\frac{\rho|{\scaleobj{1.2}{\square}}_{n-2}|\delta^{2}}{4}\right)\\\
&\leqslant C(d,\Lambda,\rho)3^{-dn}.\end{split}$

Finally, we treat eq. 5.14-c2. We calculate eq. 5.14-c2 at first with the
conditional expectation with respect to $\mathcal{G}_{n,n-2}$. Clearly,
$\mathcal{C}_{n,\rho,\delta}$ is $\mathcal{G}_{n,n-2}$-measurable, and under
this condition
${\mu({\scaleobj{1.2}{\square}}_{n})\geqslant(1-\delta)\rho|{\scaleobj{1.2}{\square}}_{n}|}$,
so we have

(5.21)
$\begin{split}|\text{\lx@cref{creftype~refnum}{eq:mainVarianceDecom}-c2}|^{2}&=\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\mathbb{E}_{\rho}\left[\frac{\mathbf{1}_{\left\\{\mathcal{C}_{n,\rho,\delta}\right\\}}}{\mu({\scaleobj{1.2}{\square}}_{n})}\mathbb{E}_{\rho}\left[\left(\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}X_{z}\right)^{2}\
\Bigg{|}\ \mathcal{G}_{n,n-2}\right]\right]\\\
&\leqslant\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\mathbb{E}_{\rho}\left[\frac{1}{(1-\delta)\rho|{\scaleobj{1.2}{\square}}_{n}|}\mathbb{E}_{\rho}\left[\mathbf{1}_{\left\\{\mathcal{C}_{n,\rho,\delta}\right\\}}\left(\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}X_{z}\right)^{2}\
\Bigg{|}\ \mathcal{G}_{n,n-2}\right]\right].\end{split}$

We would like to develop the term
$|\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}X_{z}|^{2}$ and also drop out the
indicator term. The argument here is deterministic

$\displaystyle\left|\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}X_{z}\right|^{2}$
$\displaystyle=\sum_{\begin{subarray}{c}z,z^{\prime}\in\mathcal{Z}_{n,n-2}\\\
|z-z^{\prime}|_{\infty}<3^{n-1}\end{subarray}}m_{z}m_{z^{\prime}}X_{z}\cdot
X_{z^{\prime}}+\sum_{\begin{subarray}{c}z,z^{\prime}\in\mathcal{Z}_{n,n-2}\\\
|z-z^{\prime}|_{\infty}\geqslant
3^{n-1}\end{subarray}}m_{z}m_{z^{\prime}}X_{z}\cdot X_{z^{\prime}}$
$\displaystyle\leqslant\frac{1}{2}\sum_{\begin{subarray}{c}z,z^{\prime}\in\mathcal{Z}_{n,n-2}\\\
|z-z^{\prime}|_{\infty}<3^{n-1}\end{subarray}}\left((m_{z})^{2}|X_{z}|^{2}+(m_{z^{\prime}})^{2}|X_{z^{\prime}}|^{2}\right)+\sum_{\begin{subarray}{c}z,z^{\prime}\in\mathcal{Z}_{n,n-2}\\\
|z-z^{\prime}|_{\infty}\geqslant
3^{n-1}\end{subarray}}m_{z}m_{z^{\prime}}X_{z}\cdot X_{z^{\prime}},$

where $|z-z^{\prime}|_{\infty}:=\max_{1\leqslant i\leqslant
d}|z_{i}-z^{\prime}_{i}|$. We now add back the indicator
$\mathbf{1}_{\left\\{\mathcal{C}_{n,\rho,\delta}\right\\}}$ and develop it

(5.22)
$\begin{split}&\mathbf{1}_{\left\\{\mathcal{C}_{n,\rho,\delta}\right\\}}\left|\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}X_{z}\right|^{2}\\\
&\leqslant\mathbf{1}_{\left\\{\mathcal{C}_{n,\rho,\delta}\right\\}}\left(\frac{(1+\delta)\rho|{\scaleobj{1.2}{\square}}_{n-2}|}{2}\sum_{\begin{subarray}{c}z,z^{\prime}\in\mathcal{Z}_{n,n-2}\\\
|z-z^{\prime}|<3^{n-1}\end{subarray}}\left(m_{z}|X_{z}|^{2}+m_{z^{\prime}}|X_{z^{\prime}}|^{2}\right)+\sum_{\begin{subarray}{c}z,z^{\prime}\in\mathcal{Z}_{n,n-2}\\\
|z-z^{\prime}|\geqslant 3^{n-1}\end{subarray}}m_{z}m_{z^{\prime}}X_{z}\cdot
X_{z^{\prime}}\right)\\\
&\leqslant\frac{(1+\delta)\rho|{\scaleobj{1.2}{\square}}_{n-2}|}{2}\sum_{\begin{subarray}{c}z,z^{\prime}\in\mathcal{Z}_{n,n-2}\\\
|z-z^{\prime}|<3^{n-1}\end{subarray}}\left(m_{z}|X_{z}|^{2}+m_{z^{\prime}}|X_{z^{\prime}}|^{2}\right)+\sum_{\begin{subarray}{c}z,z^{\prime}\in\mathcal{Z}_{n,n-2}\\\
|z-z^{\prime}|\geqslant 3^{n-1}\end{subarray}}m_{z}m_{z^{\prime}}X_{z}\cdot
X_{z^{\prime}}.\end{split}$

From the first line to the second line above, we use that
$m_{z}\leqslant(1+\delta)\rho|{\scaleobj{1.2}{\square}}_{n-2}|$ under the
event $\mathcal{C}_{n,\rho,\delta}$. We then keep in mind that the quantity in
$\Big{(}\cdots\Big{)}$ on the second line of eq. 5.22 is always larger than
$\left|\sum_{z\in\mathcal{Z}_{n,n-2}}m_{z}X_{z}\right|^{2}$, so it is
nonnegative. Therefore, from the second line to the third line, we can drop
the indicator function in front. Inserting this estimate into eq. 5.21, we
obtain that

$\displaystyle|\text{\lx@cref{creftype~refnum}{eq:mainVarianceDecom}-c2}|^{2}\leqslant\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\frac{(1+\delta)|{\scaleobj{1.2}{\square}}_{n-2}|}{(1-\delta)|{\scaleobj{1.2}{\square}}_{n}|}\sum_{\begin{subarray}{c}z,z^{\prime}\in\mathcal{Z}_{n,n-2}\\\
|z-z^{\prime}|_{\infty}<3^{n-1}\end{subarray}}\mathbb{E}_{\rho}\left[\frac{1}{2}\left(m_{z}|X_{z}|^{2}+m_{z^{\prime}}|X_{z^{\prime}}|^{2}\right)\right]$
$\displaystyle+\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\frac{1}{(1-\delta)|{\scaleobj{1.2}{\square}}_{n}|}\sum_{\begin{subarray}{c}z,z^{\prime}\in\mathcal{Z}_{n,n-2}\\\
|z-z^{\prime}|_{\infty}\geqslant
3^{n-1}\end{subarray}}\mathbb{E}_{\rho}\left[m_{z}m_{z^{\prime}}X_{z}\cdot
X_{z^{\prime}}\right].$

The sum in the second line is $0$, because for
$|z-z^{\prime}|_{\infty}\geqslant 3^{n-1}$, $m_{z}X_{z}$ and
$m_{z^{\prime}}X_{z^{\prime}}$ are independent,

$\displaystyle\mathbb{E}_{\rho}\left[m_{z}m_{z^{\prime}}X_{z}\cdot
X_{z^{\prime}}\right]=\mathbb{E}_{\rho}\left[m_{z}X_{z}\right]\cdot\mathbb{E}_{\rho}\left[m_{z^{\prime}}X_{z^{\prime}}\right]=0.$

For the sum in the first line, $\mathbb{E}_{\rho}[m_{z}|X_{z}|^{2}]$ is
nothing but

$\displaystyle\mathbb{E}_{\rho}\left[{\int_{z+{\scaleobj{1.2}{\square}}_{n-2}}|\mathsf{S}_{n,n-2}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n-2})-({\overbracket{\mathbf{a}}}_{n-2}^{-1}q-p)|^{2}\,d\mu}\right].$

We use Jensen’s inequality to shrink the operator to
$\mathsf{S}_{n-2,n-2}^{z}$ that

$\displaystyle\mathbb{E}_{\rho}\left[{\int_{z+{\scaleobj{1.2}{\square}}_{n-2}}|\mathsf{S}_{n,n-2}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n-2})-({\overbracket{\mathbf{a}}}_{n-2}^{-1}q-p)|^{2}\,d\mu}\right]$
$\displaystyle\quad\leqslant\mathbb{E}_{\rho}\left[{\int_{z+{\scaleobj{1.2}{\square}}_{n-2}}|\mathsf{S}_{n-2,n-2}^{z}\nabla
v(z+{\scaleobj{1.2}{\square}}_{n-2})-({\overbracket{\mathbf{a}}}_{n-2}^{-1}q-p)|^{2}\,d\mu}\right]$
$\displaystyle\quad=\mathbb{E}_{\rho}\left[{\int_{{\scaleobj{1.2}{\square}}_{n-2}}|\mathsf{S}_{n-2}\nabla
v({\scaleobj{1.2}{\square}}_{n-2})-({\overbracket{\mathbf{a}}}_{n-2}^{-1}q-p)|^{2}\,d\mu}\right].$

There are at most $9^{d}\times 5^{d}$ pairs
$z,z^{\prime}\in\mathcal{Z}_{n,n-2}$ such that
$|z-z^{\prime}|_{\infty}<3^{n-1}$; see Figure 3 for an illustration.
Therefore, we obtain

$\displaystyle|\text{\lx@cref{creftype~refnum}{eq:mainVarianceDecom}-c2}|^{2}$
$\displaystyle\leqslant\left(\frac{5}{9}\right)^{d}\left(\frac{1+\delta}{1-\delta}\right)\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n-2}|}\int_{{\scaleobj{1.2}{\square}}_{n-2}}|\mathsf{S}_{n-2}\nabla
v({\scaleobj{1.2}{\square}}_{n-2})-({\overbracket{\mathbf{a}}}_{n-2}^{-1}q-p)|^{2}\,d\mu\right]$
$\displaystyle=\left(\frac{5}{9}\right)^{d}\left(\frac{1+\delta}{1-\delta}\right)V_{n-2},$

where we recall that $V_{n}$ is the left hand side of eq. 5.13. We put this
estimate together with eq. 5.15, (5.16), (5.20) back to eq. 5.14 to obtain the
recurrence relation

$\displaystyle(V_{n})^{\frac{1}{2}}\leqslant\left(\frac{5}{9}\right)^{\frac{d}{2}}\left(\frac{1+\delta}{1-\delta}\right)^{\frac{1}{2}}(V_{n-2})^{\frac{1}{2}}+C(\tau_{n-2}+\tau_{n-1})^{\frac{1}{2}}+C3^{-dn}.$

By choosing $\delta(d)>0$ sufficiently small, we obtain the desired result eq.
5.13. ∎

Figure 3. In the cube $\square_{n}$ and all its sub-cubes
$\\{z+\square_{n-2}\\}_{z\in\mathcal{Z}_{n,n-2}}$, for a chosen sub-cube
$z_{0}+\square_{n-2}$ (the cube in dark red), the support of
${v(z_{0}+\square_{n-2})}$ is in $z_{0}+\square_{n-1}$ (the cube in light
red), so it has at most $5^{d}$ cubes of scale $3^{n-2}$ whose associated
function has a support intersecting with $z_{1}+\square_{n-1}$ (the cube in
blue). For example, $v(z_{2}+\square_{n-2})$ has correlation with
$v(z_{0}+\square_{n-2})$, while
${v(z_{1}+\square_{n-2})},{v(z_{3}+\square_{n-2})}$ do not. This gives us the
contraction factor $\left(\frac{5}{9}\right)^{d}$.

###  5.4. Step 4: iterations

Once we obtain the estimate eq. 5.9, it remains to do some numerical
iterations, similarly to [11, Page 59-60]. For the reader’s convenience, we
recall the main steps here. Let $\\{\mathbf{e}_{i}\\}_{1\leqslant i\leqslant
d}$ denote the canonical basis in ${\mathbb{R}^{d}}$, and define

$\displaystyle
F_{m}:=\sum_{i=1}^{d}J({\scaleobj{1.2}{\square}}_{m},\mathbf{e}_{i},{\overbracket{\mathbf{a}}}_{m}\mathbf{e}_{i}).$

In order to obtain an exponential decay for $(F_{m})_{m\geqslant 0}$, we first
introduce a weighted version of this quantity:

$\displaystyle\widetilde{F}_{m}:=\sum_{n=0}^{m}3^{-\frac{\beta}{2}(m-n)}F_{n}.$

Here the exponent $\beta$ is the same as in eq. 5.9. It is clear that
$F_{m}\leqslant\widetilde{F}_{m}$, so it suffices to prove an exponential
decay for $(\widetilde{F}_{m})_{m\geqslant 0}$. We will do so by proving a
recurrence equation of type $\widetilde{F}_{m+1}\leqslant
C(\widetilde{F}_{m}-\widetilde{F}_{m+1})$ for some constant
$C(d,\Lambda)<\infty$. Thus in the following, we calculate some bounds for
$(\widetilde{F}_{m}-\widetilde{F}_{m+1})$ and $\widetilde{F}_{m+1}$.

Starting with $(\widetilde{F}_{m}-\widetilde{F}_{m+1})$, we write

$\displaystyle\widetilde{F}_{m}-\widetilde{F}_{m+1}\geqslant\sum_{n=0}^{n}3^{-\frac{\beta}{2}(m-n)}(F_{n}-F_{n+1})-C3^{-\frac{\beta
m}{2}}.$

Noticing that ${\overbracket{\mathbf{a}}}_{n+1}p$ is the minimizer for the
mapping ${q\mapsto J({\scaleobj{1.2}{\square}}_{n+1},p,q)}$ in eq. 4.19, we
have

(5.23) $\displaystyle
F_{n+1}=\sum_{i=1}^{d}J({\scaleobj{1.2}{\square}}_{n+1},\mathbf{e}_{i},{\overbracket{\mathbf{a}}}_{n+1}\mathbf{e}_{i})\leqslant\sum_{i=1}^{d}J({\scaleobj{1.2}{\square}}_{n+1},\mathbf{e}_{i},{\overbracket{\mathbf{a}}}_{n}\mathbf{e}_{i}).$

Using also eq. 4.19, that
${\mathsf{Id}\leqslant{\overbracket{\mathbf{a}}}_{n}\leqslant\Lambda\mathsf{Id}}$,
and that
$p\mapsto\nu({\scaleobj{1.2}{\square}}_{n},p)-\nu({\scaleobj{1.2}{\square}}_{n+1},p)$
and
$q\mapsto\nu^{*}({\scaleobj{1.2}{\square}}_{n},q)-\nu^{*}({\scaleobj{1.2}{\square}}_{n+1},q)$
are positive semidefinite quadratic forms, we get

$\displaystyle F_{n}-F_{n+1}$
$\displaystyle\geqslant\sum_{i=1}^{d}(J({\scaleobj{1.2}{\square}}_{n},\mathbf{e}_{i},{\overbracket{\mathbf{a}}}_{n}\mathbf{e}_{i})-J({\scaleobj{1.2}{\square}}_{n+1},\mathbf{e}_{i},{\overbracket{\mathbf{a}}}_{n}\mathbf{e}_{i}))$
$\displaystyle=\sum_{i=1}^{d}(\nu({\scaleobj{1.2}{\square}}_{n},\mathbf{e}_{i})-\nu({\scaleobj{1.2}{\square}}_{n+1},\mathbf{e}_{i}))+\sum_{i=1}^{d}(\nu^{*}({\scaleobj{1.2}{\square}}_{n},{\overbracket{\mathbf{a}}}_{n}\mathbf{e}_{i})-\nu^{*}({\scaleobj{1.2}{\square}}_{n+1},{\overbracket{\mathbf{a}}}_{n}\mathbf{e}_{i}))$
$\displaystyle\geqslant C^{-1}\left(\sup_{p\in
B_{1}}\left(\nu({\scaleobj{1.2}{\square}}_{n},p)-\nu({\scaleobj{1.2}{\square}}_{n+1},p)\right)+\sup_{q\in
B_{1}}\left(\nu^{*}({\scaleobj{1.2}{\square}}_{n},q)-\nu^{*}({\scaleobj{1.2}{\square}}_{n+1},q)\right)\right)$
$\displaystyle\geqslant C^{-1}\tau_{n},$

and thus

(5.24) $\widetilde{F}_{m}-\widetilde{F}_{m+1}\geqslant
C^{-1}\sum_{n=0}^{m}3^{-\frac{\beta}{2}(m-n)}\tau_{n}-C3^{-\frac{\beta
m}{2}}.$

For the upper bound of $\widetilde{F}_{m+1}$, we use eq. 5.23 to see that
$F_{n}\leqslant F_{n+1}$, so

$\displaystyle\widetilde{F}_{m+1}$
$\displaystyle=3^{-\frac{\beta}{2}(m+1)}F_{0}+\sum_{n=0}^{m}3^{-\frac{\beta}{2}(m-n)}F_{n+1}$
$\displaystyle\leqslant C3^{-\frac{\beta
m}{2}}+\sum_{n=0}^{m}3^{-\frac{\beta}{2}(m-n)}F_{n}.$

Then we apply eq. 5.9 into the result above to get

(5.25) $\begin{split}\widetilde{F}_{m+1}&\leqslant C3^{-\frac{\beta
m}{2}}+\sum_{n=0}^{m}3^{-\frac{\beta}{2}(m-n)}\left(3^{-\beta
n}+\sum_{k=0}^{n}3^{-\beta(n-k)}\tau_{k}\right)\\\ &\leqslant C3^{-\frac{\beta
m}{2}}+3^{-\frac{\beta}{2}m}\sum_{k=0}^{m}\tau_{k}\sum_{n=k}^{m}3^{\frac{\beta}{2}(2k-n)}\\\
&\leqslant C3^{-\frac{\beta
m}{2}}+C\sum_{k=0}^{m}3^{-\frac{\beta}{2}(m-k)}\tau_{k}.\end{split}$

We combine eq. 5.24 and eq. 5.25, to obtain
$C(\widetilde{F}_{m}-\widetilde{F}_{m+1}+\widetilde{C}3^{-\frac{\beta
m}{2}})\geqslant\widetilde{F}_{m+1}$, which implies

$\displaystyle\widetilde{F}_{m+1}\leqslant\theta\widetilde{F}_{m}+C3^{-\frac{\beta
m}{2}},$

for some $\theta(d,\Lambda)\in(0,1)$. We thus conclude for the exponential
decay of $(\widetilde{F}_{m})_{m\geqslant 0}$, and thus also of $F_{m}$, since
$F_{m}\leqslant\widetilde{F}_{m}$. By eq. 4.21, this completes the proof of
Theorem 2.1.

##  Appendix A Some elementary properties of the function spaces

######  Lemma A.1 (Canonical projection).

Let $f:\mathcal{M}_{\delta}({\mathbb{R}^{d}})\to\mathbb{R}$ be a function, and
for every Borel set $U$, measure
$\mu\in\mathcal{M}_{\delta}({\mathbb{R}^{d}})$, and $n\in\mathbb{N}$, let
$f_{n}(\cdot,\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}U^{c})$ denote the (permutation-
invariant) function

$f_{n}(\cdot,\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}U^{c}):\left\\{\begin{array}[]{rcl}U^{n}&\to&\mathbb{R}\\\
(x_{1},\ldots,x_{n})&\mapsto&f\left(\sum_{i=1}^{n}\delta_{x_{i}}+\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}U^{c}\right).\end{array}\right.$

The following statements are equivalent.

(1) The function $f$ is $\mathcal{F}$-measurable.

(2) For every $n\in\mathbb{N}$, the function $f_{n}$ is
$\mathcal{B}_{U}^{\otimes n}\otimes\mathcal{F}_{U^{c}}$-measurable.

###### Proof.

We start from $(1)\Rightarrow(2)$. Because
${\mathcal{F}=\mathcal{F}_{U}\otimes\mathcal{F}_{U^{c}}}$, it suffices to
study the product function

$\displaystyle{f=\mathbf{1}_{\left\\{\mu(V_{1})=n_{1}\right\\}}\mathbf{1}_{\left\\{\mu(V_{2})=n_{2}\right\\}}\mathbf{1}_{\left\\{\mu(U)=n\right\\}}},$

for some Borel sets ${V_{1}\subseteq U},{V_{2}\subseteq U^{c}}$. In this case,
we have

$\displaystyle\\{f_{n}=1\\}$
$\displaystyle=\\{\mu(V_{1})=n_{1}\\}\cap\\{\mu(V_{2})=n_{2}\\}\cap\\{\mu(U)=n\\}$
$\displaystyle=\bigcup_{\sigma\in
S_{n}}\left(\bigcap_{i=1}^{n_{1}}\\{x_{\sigma(i)}\in
V_{1}\\}\bigcap_{j={n_{1}+1}}^{n}\\{x_{\sigma(j)}\in(U\backslash
V_{1})\\}\bigcap\\{\mu(V_{2})=n_{2}\\}\right),$

where $S_{n}$ is the symmetric group. This proves that $f_{n}$ is
$\mathcal{B}_{U}^{\otimes n}\otimes\mathcal{F}_{U^{c}}$-measurable.

We turn to $(2)\Rightarrow(1)$. Let us pick a suitable $f_{n}$ and
$\mu\mathbin{\vrule height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}U=\sum_{i=1}^{n}\delta_{x_{i}}$,
then the main point is to establish the $\mathcal{F}$-measurable property.
Since $f_{n}$ is $\mathcal{B}_{U}^{\otimes
n}\otimes\mathcal{F}_{U^{c}}$-measurable and permutation-invariant, it
suffices to study the function of type

(A.1) $\displaystyle f_{n}=\sum_{\sigma\in
S_{n}}\left(\prod_{i=1}^{n}\mathbf{1}_{\left\\{x_{\sigma(i)}\in
V_{i}\right\\}}\right)\mathbf{1}_{\left\\{\mu\mathbin{\vrule
height=3.01389pt,depth=0.0pt,width=0.3014pt\vrule
height=0.3014pt,depth=0.0pt,width=2.1097pt}U^{c}(V_{0})=n_{0}\right\\}}\mathbf{1}_{\left\\{\mu(U)=n\right\\}},$

for $\\{V_{i}\\}_{0\leqslant i\leqslant n}$ Borel sets. This is still a
complicated function, but we can add one more condition

(A.2) $\displaystyle\forall 1\leqslant i,j\leqslant n,\quad V_{i}=V_{j}\text{
or }V_{i}\cap V_{j}=\emptyset.$

For example, let $\\{\widetilde{V}_{j}\\}_{0\leqslant j\leqslant m}$ be all
the different elements in $\\{V_{i}\\}_{0\leqslant i\leqslant n}$, and
$\widetilde{V}_{j}$ appears $n_{j}$ times. For the functions of type eq. A.1
satisfying the condition eq. A.2, the $\mathcal{F}$-measurable property is
easy to treat since we have

$\sum_{\sigma\in
S_{n}}\left(\prod_{i=1}^{n}\mathbf{1}_{\left\\{x_{\sigma(i)}\in
V_{i}\right\\}}\right)\mathbf{1}_{\left\\{\mu\mathbin{\vrule
height=3.01389pt,depth=0.0pt,width=0.3014pt\vrule
height=0.3014pt,depth=0.0pt,width=2.1097pt}U^{c}(V_{0})=n_{0}\right\\}}\mathbf{1}_{\left\\{\mu(U)=n\right\\}}\\\
=\left(\prod_{j=1}^{m}\mathbf{1}_{\left\\{\mu(\widetilde{V}_{j})=n_{j}\right\\}}\right)\mathbf{1}_{\left\\{\mu\mathbin{\vrule
height=3.01389pt,depth=0.0pt,width=0.3014pt\vrule
height=0.3014pt,depth=0.0pt,width=2.1097pt}U^{c}(V_{0})=n_{0}\right\\}}\mathbf{1}_{\left\\{\mu(U)=n\right\\}},$

which is an $\mathcal{F}$-measurable function.

Finally, let us conclude that for a general $f_{n}$ in eq. A.1, they can be
decomposed into the sum of the functions with the propriety eq. A.2. Let us
see the case $n=2$, where we have the following decomposition

$\displaystyle\mathbf{1}_{\left\\{x_{1}\in
V_{1}\right\\}}\mathbf{1}_{\left\\{x_{2}\in V_{2}\right\\}}$
$\displaystyle=(\mathbf{1}_{\left\\{x_{1}\in(V_{1}\backslash
V_{2})\right\\}}+\mathbf{1}_{\left\\{x_{1}\in(V_{1}\cap
V_{2})\right\\}})(\mathbf{1}_{\left\\{x_{2}\in(V_{2}\backslash
V_{1})\right\\}}+\mathbf{1}_{\left\\{x_{2}\in(V_{1}\cap V_{2})\right\\}})$
$\displaystyle=\mathbf{1}_{\left\\{x_{1}\in(V_{1}\backslash
V_{2})\right\\}}\mathbf{1}_{\left\\{x_{2}\in(V_{2}\backslash
V_{1})\right\\}}+\mathbf{1}_{\left\\{x_{1}\in(V_{1}\backslash
V_{2})\right\\}}\mathbf{1}_{\left\\{x_{2}\in(V_{1}\cap V_{2})\right\\}}$
$\displaystyle\qquad+\mathbf{1}_{\left\\{x_{1}\in(V_{1}\cap
V_{2})\right\\}}\mathbf{1}_{\left\\{x_{2}\in(V_{2}\backslash
V_{1})\right\\}}+\mathbf{1}_{\left\\{x_{1}\in(V_{1}\cap
V_{2})\right\\}}\mathbf{1}_{\left\\{x_{2}\in(V_{1}\cap V_{2})\right\\}}.$

For a general $n$, one can use induction and this concludes the proof. ∎

######  Proposition A.2.

For every $s>0$ and $f\in\mathscr{H}^{1}(Q_{s})$, we have
${\mathsf{A}}_{s}f\in\mathscr{H}^{1}(Q_{s})$, and for every
$x\in\operatorname{supp}(\mu)\cap Q_{s}$

(A.3) $\displaystyle\nabla({\mathsf{A}}_{s}f)(\mu,x)={\mathsf{A}}_{s}(\nabla
f)(\mu,x).$

Moreover, if $s>2$ and $f\in\mathcal{A}(Q_{s})$, then
${\mathsf{A}}_{s}f\in\mathcal{A}(Q_{s-2})$.

###### Proof.

At first, we should remark the well-definedness of the right side of eq. A.3.
Notice that the Poisson measure can be decomposed as a sum of the independent
parts ${\mu=\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}+\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c}}$, we have

$\displaystyle{\mathsf{A}}_{s}f=\int_{\mathcal{M}_{\delta}({\mathbb{R}^{d}})}f(\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}+\mu^{\prime}\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c})\,{\mathrm{d}}\mathbb{P}_{\rho}(\mu^{\prime}).$

Thus the right-hand side of eq. A.3 is defined as

(A.4) $\displaystyle{\mathsf{A}}_{s}(\nabla
f)(\mu,x):=\int_{\mathcal{M}_{\delta}({\mathbb{R}^{d}})}\nabla
f(\mu\mathbin{\vrule height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}+\mu^{\prime}\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c},x)\,{\mathrm{d}}\mathbb{P}_{\rho}(\mu^{\prime}).$

We prove eq. A.3 and ${\mathsf{A}}_{s}f\in\mathscr{H}^{1}(Q_{s})$ for the
functions in ${\mathscr{C}^{\infty}(Q_{s})\cap\mathscr{H}^{1}(Q_{s})}$ as they
are dense, and we can focus on the case $\mu(\overline{Q}_{s})=n$ fixed. We
use Lemma A.1 to write

$f\left(\sum_{i=1}^{n}\delta_{x_{i}}+\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c}\right)=f_{n}(x_{1},\cdots,x_{n},\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c}).$

Following the property of product measure, for every
$(x_{1},x_{2},\cdots,x_{n})\in(\overline{Q}_{s})^{n}$, the mapping

$\displaystyle{\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c}\mapsto
f_{n}(x_{1},\cdots,x_{n},\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c})},$

is $\mathcal{F}_{\overline{Q}_{s}^{c}}$-measurable. Thus for every
$(x_{1},x_{2},\cdots,x_{n})\in(Q_{s})^{n}$, the mapping

$\displaystyle{\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c}\mapsto\nabla_{x_{k}}f_{n}(x_{1},\cdots,x_{n},\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c})},$

is also $\mathcal{F}_{\overline{Q}_{s}^{c}}$-measurable because it is the
limit of $\mathcal{F}_{\overline{Q}_{s}^{c}}$-measurable functions. Then we
observe that

$\displaystyle\left\|{\nabla f_{n}}\right\|_{L^{\infty}((Q_{s})^{n})}$
$\displaystyle=\sup_{(Q_{s})^{n}}\left(\sum_{k=1}^{n}|\nabla_{x_{k}}f_{n}(x_{1},\cdots,x_{n},\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c})|^{2}\right)^{\frac{1}{2}}$
$\displaystyle=\sup_{(\mathbb{Q}\cap
Q_{s})^{n}}\left(\sum_{k=1}^{n}|\nabla_{x_{k}}f_{n}(x_{1},\cdots,x_{n},\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c})|^{2}\right)^{\frac{1}{2}},$

as a supremum of a countable number of
$\mathcal{F}_{\overline{Q}_{s}^{c}}$-measurable functions, is finite and
$\mathcal{F}_{\overline{Q}_{s}^{c}}$-measurable. Thus we can define a cut-off
version of $f$ that

$\displaystyle
f^{n,M}=f\mathbf{1}_{\left\\{\mu(\overline{Q}_{s})=n\right\\}}\mathbf{1}_{\left\\{\left\|{\nabla
f_{n}}\right\|_{L^{\infty}((Q_{s})^{n})}\leqslant M\right\\}},$

and we can establish eq. A.3 at first for $f^{n,M}$. For every $x\in
Q_{s}\cap\operatorname{supp}(\mu)$, we have

$\displaystyle\partial_{k}({\mathsf{A}}_{s}f^{n,M})(\mu,x)$
$\displaystyle=\lim_{h\to
0}\int_{\mathcal{M}_{\delta}({\mathbb{R}^{d}})}\frac{f((\mu-\delta_{x}+\delta_{x+h\mathbf{e}_{k}})\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}+\mu^{\prime}\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c})-f(\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}+\mu^{\prime}\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c})}{h}$
$\displaystyle\qquad\qquad\times\mathbf{1}_{\left\\{\left\|{\nabla
f_{n}}\right\|_{L^{\infty}((Q_{s})^{n})}\leqslant
M\right\\}}\,{\mathrm{d}}\mathbb{P}_{\rho}(\mu^{\prime})\mathbf{1}_{\left\\{\mu(\overline{Q}_{s})=n\right\\}},$

for $h$ small enough such that $x+h\mathbf{e}_{k}\in Q_{s}$. Since
$f\in\mathscr{C}^{\infty}(Q_{s})$, we use Lemma A.1 and the mean value theorem

$\displaystyle\frac{f(\mu-\delta_{x}+\delta_{x+h\mathbf{e}_{k}})-f(\mu)}{h}=\partial_{k}f(\mu-\delta_{x}+\delta_{x+\theta\mathbf{e}_{k}},x+\theta\mathbf{e}_{k}),$

for some $\theta\in(0,1)$. With the indicator
$\mathbf{1}_{\left\\{\left\|{\nabla
f_{n}}\right\|_{L^{\infty}((Q_{s})^{n})}\leqslant M\right\\}}$, this term is
bounded by $M$, so we can use the dominated convergence theorem that

$\displaystyle\partial_{k}({\mathsf{A}}_{s}f^{n,M})(\mu,x)$
$\displaystyle=\int_{\mathcal{M}_{\delta}({\mathbb{R}^{d}})}\lim_{h\to
0}\frac{f((\mu-\delta_{x}+\delta_{x+h\mathbf{e}_{k}})\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}+\mu^{\prime}\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c})-f(\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}+\mu^{\prime}\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c})}{h}$
$\displaystyle\qquad\qquad\times\mathbf{1}_{\left\\{\left\|{\nabla
f_{n}}\right\|_{L^{\infty}((Q_{s})^{n})}\leqslant
M\right\\}}\,{\mathrm{d}}\mathbb{P}_{\rho}(\mu^{\prime})\mathbf{1}_{\left\\{\mu(\overline{Q}_{s})=n\right\\}}$
$\displaystyle=\int_{\mathcal{M}_{\delta}({\mathbb{R}^{d}})}\partial_{k}f(\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}+\mu^{\prime}\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c},x)\mathbf{1}_{\left\\{\left\|{\nabla
f_{n}}\right\|_{L^{\infty}((Q_{s})^{n})}\leqslant
M\right\\}}\,{\mathrm{d}}\mathbb{P}_{\rho}(\mu^{\prime})\mathbf{1}_{\left\\{\mu(\overline{Q}_{s})=n\right\\}},$

which establishes the eq. A.3 in the sense eq. A.4. By Jensen’s inequality and
Fubini’s lemma, we observe that

$\displaystyle\mathbb{E}_{\rho}\left[\int_{Q_{s}}|\nabla({\mathsf{A}}_{s}f^{n,M})|^{2}(\mu,x)\,{\mathrm{d}}\mu(x)\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\int_{Q_{s}}\left|\int_{\mathcal{M}_{\delta}({\mathbb{R}^{d}})}\nabla
f^{n,M}(\mu\mathbin{\vrule height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}+\mu^{\prime}\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c},x)\,{\mathrm{d}}\mathbb{P}_{\rho}(\mu^{\prime})\right|^{2}\,{\mathrm{d}}\mu(x)\right]$
$\displaystyle\leqslant\mathbb{E}_{\rho}\left[\int_{Q_{s}}\int_{\mathcal{M}_{\delta}({\mathbb{R}^{d}})}\left|\nabla
f^{n,M}(\mu\mathbin{\vrule height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}+\mu^{\prime}\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c},x)\right|^{2}\,{\mathrm{d}}\mathbb{P}_{\rho}(\mu^{\prime})\,{\mathrm{d}}\mu(x)\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\int_{Q_{s}}\left|\nabla
f^{n,M}\right|^{2}(\mu,x)\,{\mathrm{d}}\mu(x)\right],$

which implies that ${\mathsf{A}}_{s}f^{n,M}\in\mathscr{H}^{1}(Q_{s})$. Then we
use once again Jensen’s inequality for $f^{n,M}$ and $f^{n,M^{\prime}}$ with
$M<M^{\prime}$

$\displaystyle\mathbb{E}_{\rho}\left[\int_{Q_{s}}|\nabla({\mathsf{A}}_{s}f^{n,M})-\nabla({\mathsf{A}}_{s}f^{n,M^{\prime}})|^{2}(\mu,x)\,{\mathrm{d}}\mu(x)\right]$
$\displaystyle\leqslant\mathbb{E}_{\rho}\left[\int_{Q_{s}}\left|\nabla
f^{n,M}-\nabla f^{n,M^{\prime}}\right|^{2}(\mu,x)\,{\mathrm{d}}\mu(x)\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\int_{Q_{s}}\left|\nabla
f\right|^{2}(\mu,x)\,{\mathrm{d}}\mu(x)\mathbf{1}_{\left\\{\mu(\overline{Q}_{s})=n\right\\}}\mathbf{1}_{\left\\{M<\left\|{\nabla
f_{n}}\right\|_{L^{\infty}((Q_{s})^{n})}\leqslant
M^{\prime}\right\\}}\right].$

So $\\{f^{n,M}\\}_{M\geqslant 0}$ gives a Cauchy sequence in
$\mathscr{H}^{1}(Q_{s})$, and the only candidate is
$f\mathbf{1}_{\left\\{\mu(\overline{Q}_{s})=n\right\\}}$ because it is the
limit in $\mathscr{L}^{2}$. By this and a linear combination, we establish eq.
A.3 for $f$ in ${\mathscr{C}^{\infty}(Q_{s})\cap\mathscr{H}^{1}(Q_{s})}$, and
we can then extend to a general function in $\mathscr{H}^{1}(Q_{s})$ by the
density argument.

For the part of $\mathbf{a}$-harmonic function, we suppose
$f\in\mathcal{A}(Q_{s})$ and test $\phi\in\mathscr{H}^{1}_{0}(Q_{s-2})$ with
eq. A.3,

$\displaystyle\mathbb{E}_{\rho}\left[\int_{Q_{s-2}}(\nabla{\mathsf{A}}_{s}f)(\mu,x)\cdot\mathbf{a}(\mu,x)\nabla\phi(\mu,x)\,{\mathrm{d}}\mu(x)\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\int_{Q_{s-2}}{\mathsf{A}}_{s}(\nabla
f)(\mu,x)\cdot\mathbf{a}(\mu,x)\nabla\phi(\mu,x)\,{\mathrm{d}}\mu(x)\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\int_{Q_{s-2}}\left(\int_{\mathcal{M}_{\delta}({\mathbb{R}^{d}})}\nabla
f(\mu\mathbin{\vrule height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}+\mu^{\prime}\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c},x)\,{\mathrm{d}}\mathbb{P}_{\rho}(\mu^{\prime})\right)\cdot\mathbf{a}(\mu,x)\nabla\phi(\mu,x)\,{\mathrm{d}}\mu(x)\right].$

Restricted on $x\in Q_{s-2}$, we have $\mathbf{a}(\mu,x),\nabla\phi(\mu,x)$
are $\mathcal{F}_{Q_{s}}\otimes\mathcal{B}_{Q_{s}}$-measurable, so we have

$\displaystyle\forall x\in\operatorname{supp}(\mu)\cap
Q_{s-2},\qquad\mathbf{a}(\mu,x)\nabla\phi(\mu,x)=\mathbf{a}(\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s},x)\nabla\phi(\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s},x).$

We can enter the part in the integration, and then use Fubini’s lemma

$\displaystyle\mathbb{E}_{\rho}\left[\int_{Q_{s-2}}(\nabla{\mathsf{A}}_{s}f)(\mu,x)\cdot\mathbf{a}(\mu,x)\nabla\phi(\mu,x)\,{\mathrm{d}}\mu(x)\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\int_{Q_{s-2}}\left(\int_{\mathcal{M}_{\delta}({\mathbb{R}^{d}})}\nabla
f(\mu\mathbin{\vrule height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}+\mu^{\prime}\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s}^{c},x)\cdot\mathbf{a}(\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s},x)\nabla\phi(\mu\mathbin{\vrule
height=6.02773pt,depth=0.0pt,width=0.55974pt\vrule
height=0.55974pt,depth=0.0pt,width=4.30554pt}\overline{Q}_{s},x)\,{\mathrm{d}}\mathbb{P}_{\rho}(\mu^{\prime})\right)\,{\mathrm{d}}\mu(x)\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\int_{Q_{s-2}}\nabla
f(\mu,x)\cdot\mathbf{a}(\mu,x)\nabla\phi(\mu,x)\,{\mathrm{d}}\mu(x)\right]$
$\displaystyle=0.$

In the last step, we use $f\in\mathcal{A}(Q_{s})$ and this finishes the proof.
∎

##  Appendix B Equivalent definitions of the effective diffusion matrix

Recall that we defined ${\overbracket{\mathbf{a}}}(U)$ and
${\overbracket{\mathbf{a}}}_{*}(U)$ according to eq. 4.1-eq. 4.3. The proof of
Theorem 2.1 ensures the existence of a constant $C<\infty$ and an exponent
$\alpha>0$ such that for every $m\in\mathbb{N}$,

(B.1)
$\left|{\overbracket{\mathbf{a}}}({\scaleobj{1.2}{\square}}_{m})-{\overbracket{\mathbf{a}}}\right|+\left|{\overbracket{\mathbf{a}}}_{*}({\scaleobj{1.2}{\square}}_{m})-{\overbracket{\mathbf{a}}}\right|\leqslant
C3^{-\alpha m}.$

Throughout this appendix, we will only rely on the qualitative statement that

(B.2)
${\overbracket{\mathbf{a}}}=\lim_{m\to\infty}{\overbracket{\mathbf{a}}}({\scaleobj{1.2}{\square}}_{m})=\lim_{m\to\infty}{\overbracket{\mathbf{a}}}_{*}({\scaleobj{1.2}{\square}}_{m}).$

The first main goal of this appendix is to demonstrate that the definition we
chose for the bulk diffusion matrix indeed coincides with the “stationary”
definition appearing in works such as [69, 34]. Adapted to our context, this
alternative definition takes the following form. For

(B.3)
$\displaystyle\Gamma:=\left\\{\mu\mapsto\int_{\mathbb{R}^{d}}\tau_{x}g(\mu)\,{\mathrm{d}}x,\quad
g\in\mathscr{C}_{c}^{\infty}({\mathbb{R}^{d}})\cap\mathscr{H}^{1}_{0}({\mathbb{R}^{d}})\right\\},$

we let $\widetilde{{\overbracket{\mathbf{a}}}}$ be the $d$-by-$d$ matrix such
that for every $p\in{\mathbb{R}^{d}}$,

(B.4) $\displaystyle
p\cdot\widetilde{{\overbracket{\mathbf{a}}}}p:=\inf_{u\in\Gamma}\mathbb{E}_{\rho}\left[(p+\nabla
u(\mu+\delta_{0},0))\cdot\mathbf{a}(\mu+\delta_{0},0)(p+\nabla
u(\mu+\delta_{0},0))\right],$

where in eq. B.3, we used the notation $\tau_{x}g(\mu):=g(\tau_{-x}\mu)$.
Notice that for
$g\in\mathscr{C}_{c}^{\infty}({\mathbb{R}^{d}})\cap\mathscr{H}^{1}_{0}({\mathbb{R}^{d}})$
and $u:\mu\mapsto\int_{\mathbb{R}^{d}}\tau_{x}g(\mu)\,{\mathrm{d}}x\in\Gamma$,
the function $\mu\mapsto u(\mu)$ is typically not well-defined unless $\mu$ is
of finite support. However, the quantity $\nabla u(\mu,\cdot)$ makes sense
whenever the measure $\mu$ is $\sigma$-finite, since in the sum $\nabla
u(\mu,y)=\int_{\mathbb{R}^{d}}\nabla(\tau_{x}g)(\mu,y)\,{\mathrm{d}}x$, the
function $g$ is local, and thus the integrand $\nabla(\tau_{x}g)(\mu,y)$ is
non-zero only for $x$ in a bounded set. With this interpretation of $\nabla
u$, the right side of eq. B.4 is well-defined.

######  Theorem B.1.

We have ${\overbracket{\mathbf{a}}}=\widetilde{{\overbracket{\mathbf{a}}}}$.

The second main goal of this appendix is to demonstrate that the infimum in
eq. B.4 is achieved in a suitable completion of the space $\Gamma$. We also
show that the optimizer, which we call the (stationary) corrector, can be
obtained as a limit of approximations based on the finite-volume optimizers
for $\nu$ or $\nu^{*}$. Denoting

$\mathcal{M}_{\bullet}({\mathbb{R}^{d}}):=\left\\{(\mu,x)\in\mathcal{M}_{\delta}({\mathbb{R}^{d}})\times{\mathbb{R}^{d}}\
:\ x\in\operatorname{supp}\mu\right\\},$

we introduce the space

$\mathscr{L}^{2}_{\bullet}:=\left\\{f:\mathcal{M}_{\bullet}({\mathbb{R}^{d}})\to\mathbb{R}\
:\ f\text{ is measurable and
}\mathbb{E}_{\rho}\left[\int_{\mathbb{R}^{d}}|f(\mu,x)|^{2}\,{\mathrm{d}}\mu(x)\right]<\infty\right\\},$

and its local version

(B.5)
$\mathscr{L}^{2}_{\bullet,\mathrm{loc}}:=\Big{\\{}f:\mathcal{M}_{\bullet}({\mathbb{R}^{d}})\to\mathbb{R}\
:\ \mbox{$f$ is measurable and }\\\ \mbox{for every compact
$K\subseteq{\mathbb{R}^{d}}$, }\
\mathbb{E}_{\rho}\left[\int_{K}|f(\mu,x)|^{2}\,{\mathrm{d}}\mu(x)\right]<\infty\Big{\\}}.$

In these definitions, we say that
$f:\mathcal{M}_{\bullet}({\mathbb{R}^{d}})\to\mathbb{R}$ is measurable
provided that the mapping

$\left\\{\begin{array}[]{rcl}\mathcal{M}_{\delta}({\mathbb{R}^{d}})\times{\mathbb{R}^{d}}&\to&\mathbb{R}\\\
(\mu,x)&\mapsto&f(\mu,x)\mathbf{1}_{\\{x\in\operatorname{supp}\mu\\}},\end{array}\right.$

is $\mathcal{F}\otimes\mathcal{B}$-measurable. The space
$\mathscr{L}^{2}_{\bullet,\mathrm{loc}}$ is naturally endowed with the family
of seminorms

$\left\\{\begin{array}[]{rcl}\mathscr{L}^{2}_{\bullet,\mathrm{loc}}&\to&\mathbb{R}\\\
f&\mapsto&\mathbb{E}_{\rho}\left[\int_{K}|f(\mu,x)|^{2}\,{\mathrm{d}}\mu(x)\right]^{\frac{1}{2}},\end{array}\right.$

indexed by all the compact sets $K\subseteq{\mathbb{R}^{d}}$. This family of
seminorms turns $\mathscr{L}^{2}_{\bullet,\mathrm{loc}}$ into a complete
space.

For every $p\in{\mathbb{R}^{d}}$ and $m\in\mathbb{N}$, we let
$\phi(\cdot,{\scaleobj{1.2}{\square}}_{m},p)$ be such that the minimizer in
the definition of $\nu({\scaleobj{1.2}{\square}}_{m},p)$ is
$\ell_{p,{\scaleobj{1.2}{\square}}_{m}}+\phi(\cdot,{\scaleobj{1.2}{\square}}_{m},p)$,
where we recall that $\ell_{p,{\scaleobj{1.2}{\square}}_{m}}$ was defined in
eq. 4.1. Similarly, we let $\phi^{*}(\cdot,{\scaleobj{1.2}{\square}}_{m},p)$
be such that
$\ell_{p,{\scaleobj{1.2}{\square}}_{m}}+\phi^{*}(\cdot,{\scaleobj{1.2}{\square}}_{m},p)$
is the maximizer in the definition of
$\nu^{*}({\scaleobj{1.2}{\square}}_{m},{\overbracket{\mathbf{a}}}_{*}({\scaleobj{1.2}{\square}}_{m})p)$.
More precisely, using the notation introduced in Proposition 4.1, we have

$v(\cdot,{\scaleobj{1.2}{\square}}_{m},p)=\ell_{p,U}+\phi(\cdot,{\scaleobj{1.2}{\square}}_{m},p),$

and

$u(\cdot,{\scaleobj{1.2}{\square}}_{m},{\overbracket{\mathbf{a}}}_{*}({\scaleobj{1.2}{\square}}_{m})p)=\ell_{p,{\scaleobj{1.2}{\square}}_{m}}+\phi^{*}(\cdot,{\scaleobj{1.2}{\square}}_{m},p).$

Finally, we define $\nabla\widetilde{\phi}_{p,m}$ according to the formula

(B.6)
$\widetilde{\phi}_{p,m}:\mu\mapsto\frac{1}{|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\mathbb{R}^{d}}}\tau_{x}\phi(\mu,{\scaleobj{1.2}{\square}}_{m},p)\,{\mathrm{d}}x=\frac{1}{|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\mathbb{R}^{d}}}\phi(\mu,x+{\scaleobj{1.2}{\square}}_{m},p)\,{\mathrm{d}}x.$

Notice that, while $\widetilde{\phi}_{p,m}(\mu)$ is ill-defined when
$\mu\sim\text{Poi}(\rho)$, the quantity
$\nabla\widetilde{\phi}_{p,m}(\mu,\cdot)$ is still well-defined, for the same
reason as in the discussion following eq. B.4. Our second main result is as
follows.

######  Theorem B.2.

The following statements hold for every $p\in{\mathbb{R}^{d}}$.

(1) The sequence $(\nabla\widetilde{\phi}_{p,m})_{m\in\mathbb{N}}$ is a Cauchy
sequence in $(\mathscr{L}^{2}_{\bullet,\mathrm{loc}})^{d}$. Its limit, which
we denote by $\nabla\phi_{p}$, satisfies

(B.7) $\displaystyle\forall
v\in\mathscr{H}^{1}_{0}({\mathbb{R}^{d}}),\qquad\mathbb{E}_{\rho}\left[\int_{{\mathbb{R}^{d}}}\nabla
v\cdot\mathbf{a}(p+\nabla\phi_{p})\,{\mathrm{d}}\mu\right]=0.$

(2) We have

(B.8)
$\displaystyle\lim_{m\to\infty}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}|\nabla\phi(\mu,\cdot,{\scaleobj{1.2}{\square}}_{m},p)-\nabla\phi_{p}(\mu,\cdot)|^{2}\,{\mathrm{d}}\mu\right]=0,$

as well as

(B.9)
$\displaystyle\lim_{m\to\infty}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}|\nabla\phi^{*}(\mu,\cdot,{\scaleobj{1.2}{\square}}_{m},p)-\nabla\phi_{p}(\mu,\cdot)|^{2}\,{\mathrm{d}}\mu\right]=0.$

(3) The effective diffusion matrix ${\overbracket{\mathbf{a}}}$ satisfies

(B.10)
$p\cdot{\overbracket{\mathbf{a}}}p=\mathbb{E}_{\rho}\left[(p+\nabla\phi_{p}(\mu+\delta_{0},0))\cdot\mathbf{a}(\mu+\delta_{0},0)(p+\nabla\phi_{p}(\mu+\delta_{0},0))\right],$

as well as

(B.11)
${\overbracket{\mathbf{a}}}p=\mathbb{E}_{\rho}\left[\mathbf{a}(\mu+\delta_{0},0)(p+\nabla\phi_{p}(\mu+\delta_{0},0))\right].$

As a preparation towards the proof of these results, we state in the following
proposition a number of elementary properties about the function space
$\Gamma$.

######  Proposition B.3.

Let
$g\in\mathscr{C}_{c}^{\infty}({\mathbb{R}^{d}})\cap\mathscr{H}^{1}_{0}({\mathbb{R}^{d}})$
be an $\mathcal{F}_{{\scaleobj{1.2}{\square}}_{n}}$-measurable function, and
let $u:=\int_{\mathbb{R}^{d}}\tau_{x}g\,{\mathrm{d}}x\in\Gamma$. The following
properties hold:

  1. (1)

$y\mapsto\nabla u(\mu+\delta_{y},y)$ is a stationary field, i.e. $\nabla
u(\mu+\delta_{y},y)=\nabla u(\tau_{-y}\mu+\delta_{0},0)$.

  2. (2)

$\nabla u$ has mean zero, that is,

(B.12) $\displaystyle\mathbb{E}_{\rho}\left[\nabla
u(\mu+\delta_{0},0)\right]=0.$

  3. (3)

$\nabla u$ satisfies the estimate

(B.13) $\mathbb{E}_{\rho}\left[\nabla
u(\mu+\delta_{0},0)\cdot\mathbf{a}(\mu+\delta_{0},0)\nabla
u(\mu+\delta_{0},0)\right]\\\ \leqslant
3^{2dn}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\int_{{\scaleobj{1.2}{\square}}_{n}}\nabla
g(\mu,y)\cdot\mathbf{a}(\mu,y)\nabla g(\mu,y)\,{\mathrm{d}}\mu(y)\right].$

###### Proof.

(1) We use the definition to write

$\displaystyle\partial_{k}u(\mu+\delta_{y},y)$ $\displaystyle=\lim_{h\to
0}\frac{1}{h}\left(\int_{\mathbb{R}^{d}}\tau_{x}g(\mu+\delta_{y+h\mathbf{e}_{k}})-\tau_{x}g(\mu+\delta_{y})\,{\mathrm{d}}x\right)$
$\displaystyle=\lim_{h\to
0}\frac{1}{h}\left(\int_{y+{\scaleobj{1.2}{\square}}_{n}}\tau_{x}g(\mu+\delta_{y+h\mathbf{e}_{k}})-\tau_{x}g(\mu+\delta_{y})\,{\mathrm{d}}x\right)$
$\displaystyle=\lim_{h\to
0}\frac{1}{h}\left(\int_{y+{\scaleobj{1.2}{\square}}_{n}}\tau_{x-y}g(\tau_{-y}\mu+\delta_{h\mathbf{e}_{k}})-\tau_{x-y}g(\tau_{-y}\mu+\delta_{0})\,{\mathrm{d}}x\right).$

From the first line to the second line, we used the fact that $g$ is
$\mathcal{F}_{{\scaleobj{1.2}{\square}}_{n}}$-measurable, so if the transport
vector $x$ does not belong to $y+{\scaleobj{1.2}{\square}}_{n}$, then the
integrand vanishes (up to a boundary layer that vanishes in the limit $h\to
0$). We then do the change of variables $z=x-y$ to get

$\displaystyle\partial_{k}u(\mu+\delta_{y},y)$
$\displaystyle=\lim_{h\to\infty}\frac{1}{h}\left(\int_{{\scaleobj{1.2}{\square}}_{n}}\tau_{z}g(\tau_{-y}\mu+\delta_{h\mathbf{e}_{k}})-\tau_{z}g(\tau_{-y}\mu+\delta_{0})\,{\mathrm{d}}z\right)$
$\displaystyle=\partial_{k}u(\tau_{-y}\mu+\delta_{0},0),$

which means that $y\mapsto\nabla u(\mu+\delta_{y},y)$ is a stationary gradient
field.

(2) We use the equations developed in the last question. Since
$g\in\mathscr{C}_{c}^{\infty}({\mathbb{R}^{d}})$, we can exchange the
integration and derivative and get

(B.14) $\displaystyle\nabla u(\mu+\delta_{y},y)$
$\displaystyle=\int_{{\mathbb{R}^{d}}}\nabla\tau_{z}g(\tau_{-y}\mu+\delta_{0},0)\,{\mathrm{d}}z.$

We evaluate this gradient at $y=0$

$\displaystyle\mathbb{E}_{\rho}\left[\nabla u(\mu+\delta_{0},0)\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\int_{{\mathbb{R}^{d}}}\nabla\tau_{z}g(\mu+\delta_{0},0)\,{\mathrm{d}}z\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\int_{{\mathbb{R}^{d}}}\nabla
g(\tau_{-z}\mu+\delta_{-z},-z)\,{\mathrm{d}}z\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\int_{{\mathbb{R}^{d}}}\nabla
g(\mu+\delta_{z},z)\,{\mathrm{d}}z\right]$ $\displaystyle=0.$

From the second line to the third line, we used the stationarity of the
Poisson point process. Because
$g\in\mathscr{C}_{c}^{\infty}({\mathbb{R}^{d}})$, then
$\int_{{\mathbb{R}^{d}}}\nabla g(\mu+\delta_{z},z)\,{\mathrm{d}}z=0$ and the
integration in the third line vanishes.

(3) We pick a cube $Q_{L}=\left(-\frac{L}{2},\frac{L}{2}\right)^{d}$ with
$L>0$, make use of the stationarity of ${y\mapsto\nabla u(\mu+\delta_{y},y)}$
and Mecke’s identity (see [52, Theorem 4.1])

$\displaystyle\mathbb{E}_{\rho}\left[\nabla
u(\mu+\delta_{0},0)\cdot\mathbf{a}(\mu+\delta_{0},0)\nabla
u(\mu+\delta_{0},0)\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\frac{1}{|Q_{L}|}\int_{Q_{L}}\nabla
u(\mu+\delta_{y},y)\cdot\mathbf{a}(\mu+\delta_{y},y)\nabla
u(\mu+\delta_{y},y)\,{\mathrm{d}}y\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\frac{1}{\rho|Q_{L}|}\int_{Q_{L}}\nabla
u(\mu,y)\cdot\mathbf{a}(\mu,y)\nabla u(\mu,y)\,{\mathrm{d}}\mu(y)\right].$

We put the definition $u=\int_{\mathbb{R}^{d}}\tau_{x}g\,{\mathrm{d}}x$ into
the equation. For the gradient at $y$, as $g$ is
$\mathcal{F}_{{\scaleobj{1.2}{\square}}_{n}}$-measurable, thus only the term
$\nabla(\tau_{x}g)(\mu,y)$ for $x\in y+{\scaleobj{1.2}{\square}}_{n}$
contributes. This gives

$\mathbb{E}_{\rho}\left[\nabla
u(\mu+\delta_{0},0)\cdot\mathbf{a}(\mu+\delta_{0},0)\nabla
u(\mu+\delta_{0},0)\right]\\\
=\mathbb{E}_{\rho}\left[\frac{1}{\rho|Q_{L}|}\int_{Q_{L}}\left(\int_{y+{\scaleobj{1.2}{\square}}_{n}}\nabla(\tau_{x}g)(\mu,y)\,{\mathrm{d}}x\right)\cdot\mathbf{a}(\mu,y)\left(\int_{y+{\scaleobj{1.2}{\square}}_{n}}\nabla(\tau_{x}g)(\mu,y)\,{\mathrm{d}}x\right)\,{\mathrm{d}}\mu(y)\right].$

We next apply Jensen’s inequality to obtain that

(B.15) $\begin{split}&\mathbb{E}_{\rho}\left[\nabla
u(\mu+\delta_{0},0)\cdot\mathbf{a}(\mu+\delta_{0},0)\nabla
u(\mu+\delta_{0},0)\right]\\\
&=\mathbb{E}_{\rho}\left[\frac{|{\scaleobj{1.2}{\square}}_{n}|^{2}}{\rho|Q_{L}|}\int_{Q_{L}}\left(\strokedint_{y+{\scaleobj{1.2}{\square}}_{n}}\nabla(\tau_{x}g)(\mu,y)\,{\mathrm{d}}x\right)\cdot\mathbf{a}(\mu,y)\left(\strokedint_{y+{\scaleobj{1.2}{\square}}_{n}}\nabla(\tau_{x}g)(\mu,y)\,{\mathrm{d}}x\right)\,{\mathrm{d}}\mu(y)\right]\\\
&\leqslant\mathbb{E}_{\rho}\left[\frac{|{\scaleobj{1.2}{\square}}_{n}|^{2}}{\rho|Q_{L}|}\int_{Q_{L}}\left(\strokedint_{y+{\scaleobj{1.2}{\square}}_{n}}\nabla(\tau_{x}g)(\mu,y)\cdot\mathbf{a}(\mu,y)\nabla(\tau_{x}g)(\mu,y)\,{\mathrm{d}}x\right)\,{\mathrm{d}}\mu(y)\right]\\\
&\leqslant\mathbb{E}_{\rho}\left[\frac{|{\scaleobj{1.2}{\square}}_{n}|}{\rho|Q_{L}|}\int_{Q_{L+3^{n}}}\left(\int_{x+{\scaleobj{1.2}{\square}}_{n}}\nabla(\tau_{x}g)(\mu,y)\cdot\mathbf{a}(\mu,y)\nabla(\tau_{x}g)(\mu,y)\,{\mathrm{d}}\mu(y)\right)\,{\mathrm{d}}x\right].\end{split}$

In the last line, we use Fubini’s lemma and exchange
${\int(\cdots)\,{\mathrm{d}}x}$ with ${\int(\cdots)\,{\mathrm{d}}\mu(y)}$. In
this procedure, we have to enlarge the domain from $Q_{L}$ to $Q_{L+3^{n}}$,
because for the gradient at $y\in Q_{L}$, $\nabla(\tau_{x}g)(\mu,y)$
contributes for the transport $x\in Q_{L+3^{n}}$ (see Figure 4 as an
illustration). Using the stationarity of the Poisson point process, we have

$\mathbb{E}_{\rho}\left[\int_{x+{\scaleobj{1.2}{\square}}_{n}}\nabla(\tau_{x}g)(\mu,y)\cdot\mathbf{a}(\mu,y)\nabla(\tau_{x}g)(\mu,y)\,{\mathrm{d}}\mu(y)\right]\\\
=\mathbb{E}_{\rho}\left[\int_{{\scaleobj{1.2}{\square}}_{n}}\nabla
g(\mu,y)\cdot\mathbf{a}(\mu,y)\nabla g(\mu,y)\,{\mathrm{d}}\mu(y)\right],$

which helps us conclude that

$\displaystyle\mathbb{E}_{\rho}\left[(\xi+\nabla
u(\mu+\delta_{0},0))\cdot\mathbf{a}(\mu+\delta_{0},0)(\xi+\nabla
u(\mu+\delta_{0},0))\right]$ $\displaystyle\leqslant
3^{2dn}\frac{|Q_{L+3^{n}}|}{|Q_{L}|}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\int_{{\scaleobj{1.2}{\square}}_{n}}\nabla
g(\mu,y)\cdot\mathbf{a}(\mu,y)\nabla g(\mu,y)\,{\mathrm{d}}\mu(y)\right].$

We take $L\to\infty$ and obtain the desired result. ∎

######  Remark B.4.

The inequality eq. B.13 is essentially sharp when $\nabla g$ itself is close
to a stationary field. Indeed, if $g$ is close to a stationary field, then

$\displaystyle\nabla\tau_{x}g(\mu,y)=\nabla g(\tau_{-x}\mu,y-x)\simeq\nabla
g(\mu,y),$

which implies that the application of Jensen’s inequality in eq. B.15 is
essentially sharp. The error introduced by a boundary layer in a subsequent
step of the proof disappears as we take $L\to\infty$ at the end.

As a corollary of Proposition B.3, we can also propose the following
equivalent definition of $\widetilde{{\overbracket{\mathbf{a}}}}$.

######  Corollary B.5.

For any open set $U\subseteq{\mathbb{R}^{d}}$, we have

(B.16)
$\displaystyle\xi\cdot\widetilde{{\overbracket{\mathbf{a}}}}\xi=\inf_{u\in\Gamma}\mathbb{E}_{\rho}\left[\frac{1}{\rho|U|}\int_{U}(\xi+\nabla
u)\cdot\mathbf{a}(\xi+\nabla u)\,{\mathrm{d}}\mu\right].$

###### Proof.

It is a direct result of Mecke’s identity (see [52, Theorem 4.1]) and the
stationarity of ${y\mapsto\nabla u(\mu+\delta_{y},y)}$. ∎

Figure 4. The red cube represents the cube $Q_{L}$ and the blue cubes
represent the transport of small cubes $x+\square_{n}$. It is clear that for
$x\in Q_{L+3^{n}}$, the blue cubes and the red cube intersect.

With the help of Proposition B.3, we can now prove the first main theorem of
this appendix.

###### Proof of Theorem B.1.

We decompose the proof into two steps.

Step 1: Bound from below
$\widetilde{{\overbracket{\mathbf{a}}}}\geqslant{\overbracket{\mathbf{a}}}$.
We fix $m\in\mathbb{N}$ and a sequence of approximate minimizers
$\\{\phi_{{p}}^{(i)}\\}_{i\geqslant 1}$ for the variational problem in eq.
B.4, which we write in the form

$\displaystyle\phi_{{p}}^{(i)}(\mu)=\int_{{\mathbb{R}^{d}}}\tau_{x}g_{i}(\mu)\,{\mathrm{d}}x$

for some
$g_{i}\in\mathscr{C}_{c}^{\infty}({\mathbb{R}^{d}})\cap\mathscr{H}^{1}_{0}({\mathbb{R}^{d}})$.
Now we propose a modified version in
$\mathscr{H}^{1}({\scaleobj{1.2}{\square}}_{m})$ defined by

$\displaystyle\widetilde{\phi}_{{p}}^{(i)}(\mu):=\int_{K_{i}}\tau_{x}g_{i}(\mu)\,{\mathrm{d}}x,$

with $K_{i}\subseteq{\mathbb{R}^{d}}$ a large compact set so that
$\widetilde{\phi}_{{p}}^{(i)}\in\mathscr{H}^{1}({\scaleobj{1.2}{\square}}_{m})$
and

$\displaystyle\forall
y\in{\scaleobj{1.2}{\square}}_{m},\qquad\nabla\widetilde{\phi}_{{p}}^{(i)}(\mu,y)=\nabla\phi_{{p}}^{(i)}(\mu,y).$

Then we test ${p}+\nabla\widetilde{\phi}_{{p}}^{(i)}$ in the optimization
problem for $\nu^{*}({\scaleobj{1.2}{\square}}_{m},q)$ to get that

$\displaystyle\frac{1}{2}q\cdot{\overbracket{\mathbf{a}}}_{*}^{-1}({\scaleobj{1.2}{\square}}_{m})q$
$\displaystyle\geqslant\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\left(-\frac{1}{2}({p}+\nabla\widetilde{\phi}_{{p}}^{(i)})\cdot\mathbf{a}({p}+\nabla\widetilde{\phi}_{{p}}^{(i)})+q\cdot({p}+\nabla\widetilde{\phi}_{{p}}^{(i)})\right)\,{\mathrm{d}}\mu\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\left(-\frac{1}{2}({p}+\nabla\phi_{{p}}^{(i)})\cdot\mathbf{a}({p}+\nabla\phi_{{p}}^{(i)})+q\cdot({p}+\nabla\phi_{{p}}^{(i)})\right)\,{\mathrm{d}}\mu\right].$

We use the stationarity of $y\mapsto\nabla\phi_{{p}}^{(i)}(\mu+\delta_{y},y)$,
eq. B.12 and let $i\to\infty$ to obtain

$\displaystyle\frac{1}{2}q\cdot{\overbracket{\mathbf{a}}}_{*}^{-1}({\scaleobj{1.2}{\square}}_{m})q\geqslant-\frac{1}{2}{p}\cdot\widetilde{{\overbracket{\mathbf{a}}}}{p}+{p}\cdot
q.$

Taking $q={\overbracket{\mathbf{a}}}_{*}({\scaleobj{1.2}{\square}}_{m}){p}$
leads to

$\displaystyle{p}\cdot\widetilde{{\overbracket{\mathbf{a}}}}{p}\geqslant{p}\cdot{\overbracket{\mathbf{a}}}_{*}({\scaleobj{1.2}{\square}}_{m}){p}.$

Finally, we let $m\to\infty$ and conclude that
$\widetilde{{\overbracket{\mathbf{a}}}}\geqslant{\overbracket{\mathbf{a}}}$.

Step 2: Bound from above
$\widetilde{{\overbracket{\mathbf{a}}}}\leqslant{\overbracket{\mathbf{a}}}$.
We hope to prove
$\widetilde{{\overbracket{\mathbf{a}}}}\leqslant{\overbracket{\mathbf{a}}}$ by
testing the variational formula eq. B.4 with a suitable candidate, namely the
function $\widetilde{\phi}_{p,m}$ introduced in eq. B.6. Since
$\phi(\cdot,{\scaleobj{1.2}{\square}}_{m},p)\in\mathscr{H}^{1}_{0}({\scaleobj{1.2}{\square}}_{m})$,
and using eq. B.13, we can approximate $\nabla\widetilde{\phi}_{p,m}$ in
$(\mathscr{L}^{2}_{\bullet,\mathrm{loc}})^{d}$ arbitrarily closely with
elements of $\Gamma$. It thus follows that we can use $\widetilde{\phi}_{p,m}$
as a candidate in the variational problem in eq. B.4, and use the comparison
inequality eq. B.13 to get that

(B.17) $\displaystyle{p}\cdot\widetilde{{\overbracket{\mathbf{a}}}}{p}$
$\displaystyle\leqslant\mathbb{E}_{\rho}\left[({p}+\nabla\widetilde{\phi}_{{p},m}(\mu+\delta_{0},0))\cdot\mathbf{a}(\mu+\delta_{0},0)({p}+\nabla\widetilde{\phi}_{{p},m}(\mu+\delta_{0},0))\right]$
$\displaystyle\leqslant\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}({p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p}))\cdot\mathbf{a}({p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p}))\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle={p}\cdot{\overbracket{\mathbf{a}}}({\scaleobj{1.2}{\square}}_{m}){p}.$

Finally, we let $m\to\infty$ and conclude that
$\widetilde{{\overbracket{\mathbf{a}}}}\leqslant{\overbracket{\mathbf{a}}}$. ∎

In the proof above, we used $\\{\widetilde{\phi}_{{p},m}\\}_{m\geqslant 1}$ as
a sequence of approximate minimizers for the variational problem in eq. B.4.
This already gives us a good hint for the validity of at least some of the
statements in Theorem B.2. We now turn to the proof of the first part of this
result.

###### Proof of part (1) of Theorem B.2.

We decompose the proof into four steps.

Step 1: $\\{\nabla\widetilde{\phi}_{{p},m}\\}_{m\geqslant 1}$ is a Cauchy
sequence in ${(\mathscr{L}^{2}_{\bullet,\mathrm{loc}})^{d}}$. We fix $n<m$
and, recalling the notation
$\mathcal{Z}_{m,n}:=3^{n}{\mathbb{Z}^{d}}\cap{\scaleobj{1.2}{\square}}_{m}$,
we observe that

$\displaystyle\nabla\widetilde{\phi}_{{p},n}(\mu,y)$
$\displaystyle=\frac{1}{|{\scaleobj{1.2}{\square}}_{n}|}\int_{{\mathbb{R}^{d}}}\nabla\phi(\mu,y,x+{\scaleobj{1.2}{\square}}_{n},{p})\,{\mathrm{d}}x$
$\displaystyle=\frac{1}{|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\mathbb{R}^{d}}}\sum_{z\in\mathcal{Z}_{m,n}}\nabla\phi(\mu,y,x+z+{\scaleobj{1.2}{\square}}_{n},{p})\,{\mathrm{d}}x$
$\displaystyle=\frac{1}{|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\mathbb{R}^{d}}}\nabla\tau_{x}\phi_{{p},m,n}(\mu,y)\,{\mathrm{d}}x,$

where the function $\phi_{{p},m,n}$ is defined as

(B.18)
$\displaystyle\phi_{{p},m,n}(\mu):=\sum_{z\in\mathcal{Z}_{m,n}}\phi(\mu,z+{\scaleobj{1.2}{\square}}_{n},{p}).$

For any compact set $K$, we use Mecke’s identity (see [52, Theorem 4.1]) and
the stationarity of $\nabla\widetilde{\phi}_{{p},m}$

$\displaystyle\mathbb{E}_{\rho}\left[\int_{K}|\nabla\widetilde{\phi}_{{p},m}-\nabla\widetilde{\phi}_{{p},n}|^{2}(\mu,y)\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle=\rho\mathbb{E}_{\rho}\left[\int_{K}|\nabla\widetilde{\phi}_{{p},m}-\nabla\widetilde{\phi}_{{p},n}|^{2}(\mu+\delta_{y},y)\,{\mathrm{d}}y\right]$
$\displaystyle=\rho|K|\mathbb{E}_{\rho}\left[|\nabla\widetilde{\phi}_{{p},m}-\nabla\widetilde{\phi}_{{p},n}|^{2}(\mu+\delta_{0},0)\right].$

Then we use the comparison inequality eq. B.13 and obtain that

$\displaystyle\mathbb{E}_{\rho}\left[\int_{K}|\nabla\widetilde{\phi}_{{p},m}-\nabla\widetilde{\phi}_{{p},n}|^{2}(\mu,y)\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle\leqslant\rho|K|\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\left|\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})-\nabla\phi_{{p},m,n}(\mu,y)\right|^{2}\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle\leqslant\rho|K|(\nu({\scaleobj{1.2}{\square}}_{n},{p})-\nu({\scaleobj{1.2}{\square}}_{m},{p})).$

By eq. B.2, this shows that $\\{\nabla\widetilde{\phi}_{{p},m}\\}_{m\geqslant
1}$ is a Cauchy sequence in ${(\mathscr{L}^{2}_{\bullet,\mathrm{loc}})^{d}}$.

Step 2: Harmonic property - setting up. Denote the limit by $\nabla\phi_{p}$,
we set things up to prove the harmonic property eq. B.7 by approximation. We
fix $s>0$ and
${v\in\mathscr{C}^{\infty}_{c}({\mathbb{R}^{d}})\cap\mathscr{H}^{1}_{0}({\mathbb{R}^{d}})}$
which is $\mathcal{F}_{Q_{s}}$-measurable, and observe that

$\displaystyle\mathbb{E}_{\rho}\left[\int_{{\mathbb{R}^{d}}}\nabla
v\cdot\mathbf{a}({p}+\nabla\phi_{{p}})\,{\mathrm{d}}\mu\right]$
$\displaystyle=\lim_{m\to\infty}\mathbb{E}_{\rho}\left[\int_{Q_{s}}\nabla
v\cdot\mathbf{a}({p}+\nabla\widetilde{\phi}_{{p},m})\,{\mathrm{d}}\mu\right]$
$\displaystyle=\lim_{m\to\infty}\mathbb{E}_{\rho}\left[\int_{Q_{s}}\nabla
v(\mu,y)\cdot\mathbf{a}(\mu,y)\left(\strokedint_{y+{\scaleobj{1.2}{\square}}_{m}}{p}+\nabla\phi(\mu,y,x+{\scaleobj{1.2}{\square}}_{m},{p})\,{\mathrm{d}}x\right)\,{\mathrm{d}}\mu(y)\right].$

We then use Fubini’s lemma to exchange the order of integration,

$\displaystyle\mathbb{E}_{\rho}\left[\int_{{\mathbb{R}^{d}}}\nabla
v\cdot\mathbf{a}({p}+\nabla\phi_{{p}})\,{\mathrm{d}}\mu\right]=\lim_{m\to\infty}\frac{1}{|{\scaleobj{1.2}{\square}}_{m}|}$
$\displaystyle\times\mathbb{E}_{\rho}\left[\int_{Q_{3^{m}+s}}\left(\int_{x+{\scaleobj{1.2}{\square}}_{m}}\nabla
v(\mu,y)\cdot\mathbf{a}(\mu,y)\left({p}+\nabla\phi(\mu,y,x+{\scaleobj{1.2}{\square}}_{m},{p})\right)\mathbf{1}_{\left\\{y\in
Q_{s}\right\\}}\,{\mathrm{d}}\mu(y)\right)\,{\mathrm{d}}x\right].$

For $m$ sufficiently large, we can decompose the domain of integration in $x$
in the expression above into $Q_{3^{m}-s}$ and $Q_{3^{m}+s}\backslash
Q_{3^{m}-s}$. We analyse the contribution of each of these quantities in each
of the following two steps.

Step 3: Integration in $Q_{3^{m}-s}$. Notice that for $x\in Q_{3^{m}-s}$, we
have $Q_{s}\subseteq x+{\scaleobj{1.2}{\square}}_{m}$ (see Figure 5 for an
illustration), thus we can drop the indicator $\mathbf{1}_{\left\\{y\in
Q_{s}\right\\}}$ in the inner integral and use the $\mathbf{a}$-harmonic
property of ${{p}+\nabla\phi(x+{\scaleobj{1.2}{\square}}_{m},{p})}$ to get
that

$\displaystyle\frac{1}{|{\scaleobj{1.2}{\square}}_{m}|}\mathbb{E}_{\rho}\left[\int_{Q_{3^{m}-s}}\left(\int_{x+{\scaleobj{1.2}{\square}}_{m}}\nabla
v(\mu,y)\cdot\mathbf{a}(\mu,y)\left({p}+\nabla\phi(\mu,y,x+{\scaleobj{1.2}{\square}}_{m},{p})\right)\,{\mathrm{d}}\mu(y)\right)\,{\mathrm{d}}x\right]$
$\displaystyle\quad=\frac{1}{|{\scaleobj{1.2}{\square}}_{m}|}\int_{Q_{3^{m}-s}}\mathbb{E}_{\rho}\left[\int_{x+{\scaleobj{1.2}{\square}}_{m}}\nabla
v(\mu,y)\cdot\mathbf{a}(\mu,y)\left({p}+\nabla\phi(\mu,y,x+{\scaleobj{1.2}{\square}}_{m},{p})\right)\,{\mathrm{d}}\mu(y)\right]\,{\mathrm{d}}x$
$\displaystyle\quad=0.$

Figure 5. The red cube represents $Q_{s}$, and the blue and green cubes
represent respectively $Q_{3^{m}+s}$ and $Q_{3^{m}-s}$. For $x\in
Q_{3^{m}-s}$, we have $Q_{s}\subseteq x+\square_{m}$; for $x\notin
Q_{3^{m}+s}$, ${(x+\square_{m})\cap Q_{s}=\emptyset}$; for $x\in
Q_{3^{m}+s}\backslash Q_{3^{m}-s}$, $x+\square_{m}$ and $Q_{s}$ have non-empty
intersection but $Q_{s}$ is not totally contained in $x+\square_{m}$. These
three cases are represented by $z_{1},z_{2},z_{3}$.

Step 4: Boundary layer $Q_{3^{m}+s}\backslash Q_{3^{m}-s}$. We use Young’s
inequality to bound the term with the integral over $x\in
Q_{3^{m}+s}\backslash Q_{3^{m}-s}$ by

$\displaystyle\frac{\Lambda}{|{\scaleobj{1.2}{\square}}_{m}|}\mathbb{E}_{\rho}\left[\int_{Q_{3^{m}+s}\backslash
Q_{3^{m}-s}}\left(\int_{(x+{\scaleobj{1.2}{\square}}_{m})\cap Q_{s}}|\nabla
v(\mu,y)|^{2}+\left|{p}+\nabla\phi(\mu,y,x+{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\right)\,{\mathrm{d}}x\right]$
$\displaystyle\leqslant\underbrace{\frac{\Lambda|Q_{3^{m}+s}\backslash
Q_{3^{m}-s}|}{|{\scaleobj{1.2}{\square}}_{m}|}\mathbb{E}_{\rho}\left[\int_{Q_{s}}|\nabla
v(\mu,y)|^{2}\,{\mathrm{d}}\mu(y)\right]}_{\mathbf{(A)}}$
$\displaystyle\quad+\underbrace{\frac{\Lambda}{|{\scaleobj{1.2}{\square}}_{m}|}\mathbb{E}_{\rho}\left[\int_{Q_{3^{m}+s}\backslash
Q_{3^{m}-s}}\left(\int_{(x+{\scaleobj{1.2}{\square}}_{m})\cap
Q_{s}}\left|{p}+\nabla\phi(\mu,y,x+{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\right)\,{\mathrm{d}}x\right]}_{\mathbf{(B)}}.$

For the term $\mathbf{(A)}$, we have, for a constant $C$ that may depend on
$s$,

$\displaystyle\mathbf{(A)}\leqslant C\Lambda
3^{-m}\mathbb{E}_{\rho}\left[\int_{Q_{s}}|\nabla
v(\mu,y)|^{2}\,{\mathrm{d}}\mu(y)\right]\xrightarrow[m\to\infty]{}0.$

For the term $\mathbf{(B)}$, we use the stationarity to observe that

$\mathbb{E}_{\rho}\left[\int_{(x+{\scaleobj{1.2}{\square}}_{m})\cap
Q_{s}}\left|{p}+\nabla\phi(\mu,y,x+{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\right]\\\
=\mathbb{E}_{\rho}\left[\int_{{\scaleobj{1.2}{\square}}_{m}\cap(-x+Q_{s})}\left|{p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\right].$

We apply once again Fubini’s lemma to $\mathbf{(B)}$ and get that

$\displaystyle\mathbf{(B)}$
$\displaystyle=\frac{\Lambda}{|{\scaleobj{1.2}{\square}}_{m}|}\mathbb{E}_{\rho}\left[\int_{Q_{3^{m}+s}\backslash
Q_{3^{m}-s}}\left(\int_{{\scaleobj{1.2}{\square}}_{m}}\left|{p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\mathbf{1}_{\left\\{y\in(-x+Q_{s})\right\\}}\,{\mathrm{d}}\mu(y)\right)\,{\mathrm{d}}x\right]$
$\displaystyle=\frac{\Lambda}{|{\scaleobj{1.2}{\square}}_{m}|}\mathbb{E}_{\rho}\left[\int_{{\scaleobj{1.2}{\square}}_{m}}\left|{p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\left(\int_{Q_{3^{m}+s}\backslash
Q_{3^{m}-s}}\mathbf{1}_{\left\\{y\in(-x+Q_{s})\right\\}}\,{\mathrm{d}}x\right)\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle\leqslant\frac{\Lambda|Q_{s}|}{|{\scaleobj{1.2}{\square}}_{m}|}\mathbb{E}_{\rho}\left[\int_{{\scaleobj{1.2}{\square}}_{m}}\left|{p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\mathbf{1}_{\left\\{\operatorname{dist}(y,\partial{\scaleobj{1.2}{\square}}_{m})\leqslant
s\right\\}}\,{\mathrm{d}}\mu(y)\right].$

That this term converges to zero is a consequence of the stronger estimate
given by Lemma B.6 below. This concludes the proof for
${v\in\mathscr{C}^{\infty}_{c}({\mathbb{R}^{d}})\cap\mathscr{H}^{1}_{0}({\mathbb{R}^{d}})}$,
and then we can use density argument to extend to general
${v\in\mathscr{H}^{1}_{0}({\mathbb{R}^{d}})}$. ∎

In the proof above, we appealed to the following boundary layer estimate,
which we state as a separate lemma for future reference (and which is stronger
than what was needed for the purpose of the proof above, since the boundary
layer size is allowed to increase with $m$).

######  Lemma B.6 (Boundary layer estimate).

For every sequence $(s_{m})_{m\in\mathbb{N}}$ such that $s_{m}\leqslant 3^{m}$
and $\lim_{m\to\infty}3^{-m}s_{m}=0$, we have

(B.19)
$\displaystyle\lim_{m\to\infty}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\left|{p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\mathbf{1}_{\left\\{\operatorname{dist}(y,\partial{\scaleobj{1.2}{\square}}_{m})\leqslant
s_{m}\right\\}}\,{\mathrm{d}}\mu(y)\right]=0.$

###### Proof.

The idea is to make use of the renormalization argument. We define a
mesoscopic scale $n$ associated to $m$ such that $s_{m}\leqslant 3^{n}$,
$n\to\infty$ and $m-n\to\infty$. Then we immediately have

$\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\left|{p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\mathbf{1}_{\left\\{\operatorname{dist}(y,\partial{\scaleobj{1.2}{\square}}_{m})\leqslant
s_{m}\right\\}}\,{\mathrm{d}}\mu(y)\right]\\\
\leqslant\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\left|{p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\mathbf{1}_{\left\\{\operatorname{dist}(y,\partial{\scaleobj{1.2}{\square}}_{m})\leqslant
3^{n}\right\\}}\,{\mathrm{d}}\mu(y)\right].$

We propose to compare $\phi(\cdot,{\scaleobj{1.2}{\square}}_{m},p)$ with
$\phi_{{p},m,n}\in\mathscr{H}^{1}_{0}({\scaleobj{1.2}{\square}}_{m})$ defined
as in eq. B.18:

$\displaystyle\phi_{{p},m,n}(\mu)=\sum_{z\mathcal{Z}_{m,n}}\phi(\mu,z+{\scaleobj{1.2}{\square}}_{n},{p}).$

Then we have

(B.20)
$\begin{split}&\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\left|{p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\mathbf{1}_{\left\\{\operatorname{dist}(y,\partial{\scaleobj{1.2}{\square}}_{m})\leqslant
3^{n}\right\\}}\,{\mathrm{d}}\mu(y)\right]\\\
&\leqslant\underbrace{2\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{Q_{3^{m}}\backslash
Q_{3^{m}-2\times
3^{n}}}\left|{p}+\nabla\phi_{{p},m,n}(\mu,y)\right|^{2}\,{\mathrm{d}}\mu(y)\right]}_{\text{\lx@cref{creftype~refnum}{eq:BdDecom}-a}}\\\
&\qquad+\underbrace{2\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{Q_{3^{m}}\backslash
Q_{3^{m}-2\times
3^{n}}}\left|\nabla\phi_{{p},m,n}(\mu,y)-\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\right]}_{\text{\lx@cref{creftype~refnum}{eq:BdDecom}-b}}.\end{split}$

For the first term eq. B.20-a, we do partition of sum into cubes of size
$3^{n}$, then we have

$\displaystyle\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{Q_{3^{m}}\backslash
Q_{3^{m}-2\times
3^{n}}}\left|{p}+\nabla\phi_{{p},m,n}(\mu,y)\right|^{2}\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle=\frac{|{\scaleobj{1.2}{\square}}_{n}|}{|{\scaleobj{1.2}{\square}}_{m}|}\sum_{z\in\mathcal{Z}_{m,n}\cap(Q_{3^{m}}\backslash
Q_{3^{m}-2\times
3^{n}})}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{n}|}\int_{z+{\scaleobj{1.2}{\square}}_{n}}\left|{p}+\nabla\phi(\mu,y,z+{\scaleobj{1.2}{\square}}_{n},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle=\frac{|Q_{3^{m}}\backslash Q_{3^{m}-2\times
3^{n}}|}{|Q_{3^{m}}|}\nu({\scaleobj{1.2}{\square}}_{n},{p})$
$\displaystyle\leqslant 3^{-(m-n)}\Lambda|{p}|^{2}.$

For the second term eq. B.20-b, we have

$\displaystyle\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{Q_{3^{m}}\backslash
Q_{3^{m}-2\times
3^{n}}}\left|\nabla\phi_{{p},m,n}(\mu,y)-\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle\leqslant\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{Q_{3^{m}}}\left|\nabla\phi_{{p},m,n}(\mu,y)-\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle=\nu({\scaleobj{1.2}{\square}}_{n},{p})-\nu({\scaleobj{1.2}{\square}}_{m},{p}).$

Therefore, when we take $m,n\to\infty$, both eq. B.20-a and eq. B.20-b go to
$0$, so the boundary layer in a mecroscopic scale can be neglected. ∎

Now that the gradient of the whole-space corrector $\nabla\phi_{p}$ is well-
defined, we can proceed to complete the proof of Theorem B.2.

###### Proof of parts (2) and (3) of Theorem B.2.

We start by discussing the validity of the identities eq. B.10 and eq. B.11.
We use the stationary approximate corrector $\widetilde{\phi}_{{p},n}$ defined
in eq. B.6, and observe from eq. B.17 and Theorem B.1 that

$\displaystyle{p}\cdot{\overbracket{\mathbf{a}}}{p}$
$\displaystyle=\lim_{m\to\infty}\mathbb{E}_{\rho}\left[({p}+\nabla\widetilde{\phi}_{{p},m}(\mu+\delta_{0},0))\cdot\mathbf{a}(\mu+\delta_{0},0)({p}+\nabla\widetilde{\phi}_{{p},m}(\mu+\delta_{0},0))\right]$
$\displaystyle=\lim_{m\to\infty}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{0}|}\int_{{\scaleobj{1.2}{\square}}_{0}}({p}+\nabla\widetilde{\phi}_{{p},m})\cdot\mathbf{a}({p}+\nabla\widetilde{\phi}_{{p},m})\,{\mathrm{d}}\mu\right].$

The identity eq. B.10 then follows from the convergence of
${\nabla\widetilde{\phi}_{{p},m}}$ to ${\nabla\phi_{{p}}}$ in
$(\mathscr{L}^{2}_{\bullet,\mathrm{loc}})^{d}$. For the second identity, we
can use the fact that

$\displaystyle\lim_{m\to\infty}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\mathbf{a}\left({p}+\nabla\phi(\mu,\cdot,{\scaleobj{1.2}{\square}}_{m},{p})\right)\,{\mathrm{d}}\mu\right]={\overbracket{\mathbf{a}}}{p},$

the estimate eq. B.8, and the stationarity of $\nabla\phi_{p}$. The identity
eq. B.9 can also be deduced from eq. B.8 because

$\displaystyle\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}|\nabla\phi^{*}(\mu,\cdot,{\scaleobj{1.2}{\square}}_{m},{p})-\nabla\phi_{{p}}(\mu,\cdot)|^{2}\,{\mathrm{d}}\mu\right]$
$\displaystyle\leqslant
2\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}|\nabla\phi(\mu,\cdot,{\scaleobj{1.2}{\square}}_{m},{p})-\nabla\phi_{{p}}(\mu,\cdot)|^{2}\,{\mathrm{d}}\mu\right]$
$\displaystyle\qquad+2\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}|\nabla\phi^{*}(\mu,\cdot,{\scaleobj{1.2}{\square}}_{m},{p})-\nabla\phi(\mu,\cdot,{\scaleobj{1.2}{\square}}_{m},{p})|^{2}\,{\mathrm{d}}\mu\right].$

By eq. 4.26 and eq. 4.27, the second term can be bounded by
$J({\scaleobj{1.2}{\square}}_{m},{p},{\overbracket{\mathbf{a}}}_{*}({\scaleobj{1.2}{\square}}_{m}){p})$;
and by eq. 4.19 and eq. B.2, this quantity converges to $0$ as $m\to\infty$.
From now on, we thus focus on the proof of eq. B.8.

The idea of the proof of eq. B.8 is very close to that for eq. B.7 and eq.
B.19. We fix a mescroscopic scale $n=\lfloor\frac{m}{3}\rfloor$, and then use
the decomposition

(B.21)
$\begin{split}&\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}|\nabla\phi_{{p}}(\mu,\cdot)-\nabla\phi(\mu,\cdot,{\scaleobj{1.2}{\square}}_{m},{p})|^{2}\,{\mathrm{d}}\mu\right]\\\
&\leqslant\underbrace{2\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}|\nabla\phi_{{p}}(\mu,\cdot)-\nabla\widetilde{\phi}_{{p},n}(\mu,\cdot)|^{2}\,{\mathrm{d}}\mu\right]}_{\text{\lx@cref{creftype~refnum}{eq:AppCorrectorDecom}-a}}\\\
&\qquad+\underbrace{2\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}|\nabla\widetilde{\phi}_{{p},n}(\mu,\cdot)-\nabla\phi(\mu,\cdot,{\scaleobj{1.2}{\square}}_{m},{p})|^{2}\,{\mathrm{d}}\mu\right]}_{\text{\lx@cref{creftype~refnum}{eq:AppCorrectorDecom}-b}}.\end{split}$

For the first term eq. B.21-a, we use the stationarity to transform to the
integration on unit cube ${\scaleobj{1.2}{\square}}_{0}$, and then use the
fact that ${\nabla\widetilde{\phi}_{{p},n}}$ converges to ${\nabla\phi_{{p}}}$
in $(\mathscr{L}^{2}_{\bullet,\mathrm{loc}})^{d}$ to get that

$\displaystyle\lim_{m\to\infty}\text{\lx@cref{creftype~refnum}{eq:AppCorrectorDecom}-a}=\lim_{n\to\infty}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{0}|}\int_{{\scaleobj{1.2}{\square}}_{0}}|\nabla\widetilde{\phi}_{{p},n}(\mu,\cdot)-\nabla\phi_{{p}}(\mu,\cdot)|^{2}\,{\mathrm{d}}\mu\right]=0.$

Thus it suffices to finish the second term eq. B.21-b. We use the definition
in eq. B.6 and Jensen’s inequality to get that

eq. B.21-b
$\displaystyle=2\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\left|\left(\strokedint_{y+{\scaleobj{1.2}{\square}}_{n}}\nabla\phi(\mu,y,x+{\scaleobj{1.2}{\square}}_{n},{p})\,{\mathrm{d}}x\right)-\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle\leqslant
2\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}\strokedint_{y+{\scaleobj{1.2}{\square}}_{n}}\left|\nabla\phi(\mu,y,x+{\scaleobj{1.2}{\square}}_{n},{p})-\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}x\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle\leqslant 2\times
3^{-dn}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{Q_{3^{m}+3^{n}}}\int_{x+{\scaleobj{1.2}{\square}}_{n}}\left|\nabla\phi(\mu,y,x+{\scaleobj{1.2}{\square}}_{n},{p})-\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\,{\mathrm{d}}x\right]$
$\displaystyle\leqslant
2\times\left(\text{\lx@cref{creftype~refnum}{eq:AppCorrectorDecom}-b1}+\text{\lx@cref{creftype~refnum}{eq:AppCorrectorDecom}-b2}+\text{\lx@cref{creftype~refnum}{eq:AppCorrectorDecom}-b3}\right),$

where in the last line we decompose once again the integration into three
terms with respect to the domain

eq. B.21-b1
$\displaystyle=3^{-dn}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{Q_{3^{m}-10\times
3^{n}}}\int_{x+{\scaleobj{1.2}{\square}}_{n}}\left|\nabla\phi(\mu,y,x+{\scaleobj{1.2}{\square}}_{n},{p})-\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\,{\mathrm{d}}x\right],$
eq. B.21-b2
$\displaystyle=3^{-dn}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{Q_{3^{m}+3^{n}}\backslash
Q_{3^{m}-10\times
3^{n}}}\int_{x+{\scaleobj{1.2}{\square}}_{n}}\left|{p}+\nabla\phi(\mu,y,x+{\scaleobj{1.2}{\square}}_{n},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\,{\mathrm{d}}x\right],$
eq. B.21-b3
$\displaystyle=3^{-dn}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{Q_{3^{m}+3^{n}}\backslash
Q_{3^{m}-10\times
3^{n}}}\int_{x+{\scaleobj{1.2}{\square}}_{n}}\left|{p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\,{\mathrm{d}}x\right].$

The terms eq. B.21-b2 and eq. B.21-b3 are easy to treat as they are boundary
layer terms. For eq. B.21-b2 we can use the energy bound

$\displaystyle\text{\lx@cref{creftype~refnum}{eq:AppCorrectorDecom}-b2}\leqslant
C3^{-(m-n)}\nu({\scaleobj{1.2}{\square}}_{n},{p})\xrightarrow[m\to\infty]{}0.$

For eq. B.21-b3, since the function
$\left|{p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}$
does not involve $x$, we use Fubini’s lemma that

eq. B.21-b3
$\displaystyle=3^{-dn}\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{Q_{3^{m}+3^{n}}}\left(\int_{Q_{3^{m}+3^{n}}\backslash
Q_{3^{m}-10\times 3^{n}}}\mathbf{1}_{\left\\{x\in
y+{\scaleobj{1.2}{\square}}\right\\}}\,{\mathrm{d}}x\right)\left|{p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle\leqslant\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{Q_{3^{m}+3^{n}}\backslash
Q_{3^{m}-20\times
3^{n}}}\left|{p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle=\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{Q_{3^{m}}\backslash
Q_{3^{m}-20\times
3^{n}}}\left|{p}+\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\right]+\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{Q_{3^{m}+3^{n}}\backslash
Q_{3^{m}}}\left|p\right|^{2}\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle\xrightarrow[m\to\infty]{}0.$

Here from the second line to the third line, we use the fact that the gradient
contributes only on $Q_{3^{m}+3^{n}}\backslash Q_{3^{m}-20\times 3^{n}}$. Then
we do a decomposition: the integration on $Q_{3^{m}+3^{n}}\backslash
Q_{3^{m}}$ can be calculated directly, since
$\phi(\mu,{\scaleobj{1.2}{\square}}_{m},{p})$ is
$\mathcal{F}_{{\scaleobj{1.2}{\square}}_{m}}$-measurable and the gradient
vanishes; the integration on $Q_{3^{m}+3^{n}}\backslash Q_{3^{m}-20\times
3^{n}}$ can be bounded by the boundary layer estimate in Lemma B.6.

Finally, we focus on eq. B.21-b1. We rewrite the integration
${\int_{Q_{3^{m}-10\times 3^{n}}}}$ as

$\text{\lx@cref{creftype~refnum}{eq:AppCorrectorDecom}-b1}\\\
\leqslant\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\strokedint_{{\scaleobj{1.2}{\square}}_{n}}\left(\sum_{\begin{subarray}{c}z\in\mathcal{Z}_{m,n}\\\
\operatorname{dist}(z,\partial{\scaleobj{1.2}{\square}}_{m})\geqslant 5\times
3^{n}\end{subarray}}\int_{x+z+{\scaleobj{1.2}{\square}}_{n}}\left|\nabla\phi(\mu,y,x+z+{\scaleobj{1.2}{\square}}_{n},{p})-\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\,\right){\mathrm{d}}x\right].$

For each fixed $x\in{\scaleobj{1.2}{\square}}_{n}$, we can propose a sub-
minimizer $\ell_{{p},{\scaleobj{1.2}{\square}}_{m}}+w_{x}$ for
$\nu({\scaleobj{1.2}{\square}}_{m},{p})$ defined by (see Figure 6 for an
illustration)

$\displaystyle w_{x}:=\phi(\cdot,{\scaleobj{1.2}{\square}}_{m}\backslash
U_{x},{p})+\sum_{\begin{subarray}{c}z\in\mathcal{Z}_{m,n}\\\
\operatorname{dist}(z,\partial{\scaleobj{1.2}{\square}}_{m})\geqslant 5\times
3^{n}\end{subarray}}\phi(\cdot,x+z+{\scaleobj{1.2}{\square}}_{n},{p}),$

where

$\displaystyle
U_{x}:=\overline{\bigcup_{\begin{subarray}{c}z\in\mathcal{Z}_{m,n}\\\
\operatorname{dist}(z,\partial{\scaleobj{1.2}{\square}}_{m})\geqslant 5\times
3^{n}\end{subarray}}(x+z+{\scaleobj{1.2}{\square}}_{n})}.$

The gradient of $w_{x}$ and
$\phi(\cdot,x+z+{\scaleobj{1.2}{\square}}_{n},{p})$ coincides on every cube
$x+z+{\scaleobj{1.2}{\square}}_{n}$, so we can write

$\displaystyle\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\left(\sum_{\begin{subarray}{c}z\in\mathcal{Z}_{m,n}\\\
\operatorname{dist}(z,\partial{\scaleobj{1.2}{\square}}_{m})\geqslant 5\times
3^{n}\end{subarray}}\int_{x+z+{\scaleobj{1.2}{\square}}_{n}}\left|\nabla\phi(\mu,y,x+z+{\scaleobj{1.2}{\square}}_{n},{p})-\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})\right|^{2}\,{\mathrm{d}}\mu(y)\,\right)\right]$
$\displaystyle\leqslant\mathbb{E}_{\rho}\left[\frac{1}{\rho|{\scaleobj{1.2}{\square}}_{m}|}\int_{{\scaleobj{1.2}{\square}}_{m}}|\nabla
w_{x}(\mu,y)-\nabla\phi(\mu,y,{\scaleobj{1.2}{\square}}_{m},{p})|^{2}\,{\mathrm{d}}\mu(y)\right]$
$\displaystyle\leqslant\left(\sum_{\begin{subarray}{c}z\in\mathcal{Z}_{m,n}\\\
\operatorname{dist}(z,\partial{\scaleobj{1.2}{\square}}_{m})\geqslant 5\times
3^{n}\end{subarray}}\frac{|{\scaleobj{1.2}{\square}}_{n}|}{|{\scaleobj{1.2}{\square}}_{m}|}\nu({\scaleobj{1.2}{\square}}_{n},{p})+\frac{|{\scaleobj{1.2}{\square}}_{m}\backslash
U_{x}|}{|{\scaleobj{1.2}{\square}}_{m}|}\nu({\scaleobj{1.2}{\square}}_{m}\backslash
U_{x},{p})\right)-\nu({\scaleobj{1.2}{\square}}_{m},{p})$
$\displaystyle\leqslant\nu({\scaleobj{1.2}{\square}}_{n},{p})-\nu({\scaleobj{1.2}{\square}}_{m},{p})+5\times
3^{-\frac{2m}{3}}\Lambda|{p}|^{2},$

where we used the quadratic response (4.9) from the second line to the third
line. This implies that
$\lim_{m\to\infty}\text{\lx@cref{creftype~refnum}{eq:AppCorrectorDecom}-b1}=0$,
and thus completes the proof of eq. B.8. ∎

Figure 6. The function $\ell_{{p},\square_{m}}+w_{x}$ is a sub-minimizer for
the problem $\nu(\square_{m},{p})$, which combines the minimizer in cubes of
scale $3^{n}$ biased by a vector $x$ (the cubes in blue), and a minimizer in
$U_{x}$ (the domain in red).

### Acknowledgements

Part of this project was developed while AG was affiliated to the University
of Bonn and supported through the CRC 1060 (The Mathematics of Emergent
Effects) that is funded through the German Science Foundation (DFG), and the
Hausdorff Center for Mathematics (HCM). CG was supported by a PhD scholarship
from Ecole Polytechnique. Part of this project was developed while CG was an
academic visitor at the Courant Institute, NYU. JCM was partially supported by
the NSF grant DMS-1954357. Part of this project was developed while JCM was
affiliated at CNRS and ENS Paris, PSL University, and was partially supported
by the ANR grants LSD (ANR-15-CE40-0020-03) and Malin (ANR-16-CE93-0003).

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