The Bose Gas in a Box with Neumann Boundary Conditions

Abstract

We consider a gas of n bosonic particles confined in a box $[-\ell /2,\ell /2]^3$ with Neumann boundary conditions. We prove Bose–Einstein condensation in the Gross–Pitaevskii regime, with an optimal bound on the condensate depletion. Moreover, our lower bound for the ground state energy in a small box $[-\ell /2,\ell /2]^3$ implies (via Neumann bracketing) a lower bound for the ground state energy of N bosons in a large box $[-L/2,L/2]^3$ with density $\rho =N/L^3$ in the thermodynamic limit.

A The two-body problem in the Neumann box

This Appendix is devoted to proving Propositions 2.1 and 2.2. We will use the following Lemma.

Lemma A.1

Let $\Omega =[-\ell /2,\ell /2]^6$ and let $\varepsilon$ be such that $0<\varepsilon \ell ^{2}\le 1$. For $y\in \Omega$ let $G_\varepsilon (x,y)$ be the solution of

$$\begin{aligned} \big (-\Delta _x+\varepsilon \big )G_\varepsilon (x,y)=\delta _y(x) \end{aligned}$$

(A.1)

on $\Omega$ with Neumann boundary conditions. There exists a constant $C>0$ (independent of $\varepsilon$ and $\ell$) such that

$$\begin{aligned} G_\varepsilon (x,y)\le C\Big (\frac{1}{|x-y|^4}+\frac{1}{\ell ^6\varepsilon }\Big ) \end{aligned}$$

(A.2)

for every $x,y\in \Omega$. Moreover, let ${\tilde{G}}_\varepsilon$ be the unique solution of

$$\begin{aligned} \big (-\Delta _x+\varepsilon \big ){\tilde{G}}_\varepsilon (x-y)=\delta _y(x) \end{aligned}$$

(A.3)

on $\mathbb {R}^6$ decaying at infinity. Then there exists a constant $C>0$ such that for $1\le i\le 6$

$$\begin{aligned} |\partial _{x^{(i)}}G_\varepsilon (x,y)-\partial _{x^{(i)}}{\tilde{G}}_\varepsilon (x-y)| \le C\left[ \sum _{n}\frac{1}{|x-y_n|^5}+\frac{1}{\varepsilon ^{1/2}\ell ^6}\right] \end{aligned}$$

(A.4)

where the $y_n$ are the (at most $3^6-1$) points obtained by reflecting $y, y_n$ with respect to the planes generated by the sides of the box, whose distance from y is less than $\ell$ (each reflected point is counted only once, and among the $y_n$ we don’t include y itself).

Proof

The solution ${\tilde{G}}_\varepsilon$ to (A.3) can be expressed as

$$\begin{aligned} {\tilde{G}}_\varepsilon (x)=\frac{\varepsilon }{2^3\pi ^{3}}\frac{\text {K}_2(\sqrt{\varepsilon }|x|)}{|x|^{2}} \end{aligned}$$

(A.5)

where $\text {K}_2$ is the modified Bessel function of the third kind of order 2 (see [2]). From the properties of $\text {K}_2$ we deduce that for large $\varepsilon ^{1/2}|x|$ there exists a constant $C_0>0$ such that

$$\begin{aligned} {\tilde{G}}_\varepsilon (x)=C_0\frac{\varepsilon ^{3/4} e^{-\sqrt{\varepsilon }|x|}}{|x|^{2+1/2}}\Big (1+\mathcal {O}((\sqrt{\varepsilon }|x|)^{-1})\Big ) \end{aligned}$$

(A.6)

while for small $\varepsilon ^{1/2}|x|$ there exists a constant $C_1>0$ such that

$$\begin{aligned} {\tilde{G}}_\varepsilon (x)=\frac{C_1}{|x|^4}+\mathcal {O}(\varepsilon |x|^{-2})\,. \end{aligned}$$

(A.7)

We obtain the Green function $G_\varepsilon$ on $\Omega$ with Neumann boundary conditions as follows. For $x,y\in \Omega$,

$$\begin{aligned} G_\varepsilon (x,y)={\tilde{G}}_\varepsilon (x-y)+\sum _{n\in \mathbb {Z}^6\backslash \{0\}} {\tilde{G}}_\varepsilon (x-y_n) \end{aligned}$$

(A.8)

where the positions $y_n$ are all possible reflections (each counted only once) of y and $y_n$ with respect to the infinite planes obtained by extending the sides of the box $\Omega$ and their periodic replicas over all $\mathbb {R}^6$. This operation gives rise to a grid, and each six-dimensional cell contains one and only one $y_n$ (therefore the label $n\in \mathbb {Z}^6\backslash \{0\}$ also identifies the cell where $y_n$ belongs). The positions $y_n$ can be thought as positions of image charges, whose contributions cancels the normal derivative of $G_\varepsilon$ on $\partial \Omega$. Given a point $y=(y^{(1)},\ldots ,y^{(6)})\in \Omega$, the coordinates of its image charges are, for $j=1, \dots , 6$,

$$\begin{aligned} y^{(j)}_n=n^{(j)}\ell +(-1)^{n^{(j)}}y^{(j)}. \end{aligned}$$

In order to estimate (A.8), we deduce from (A.6) and (A.7) that for any $0<\lambda <1$ there exists a $C_\lambda >0$ such that

$$\begin{aligned} {\tilde{G}}_\varepsilon (x)\le \frac{C_\lambda e^{-\lambda \sqrt{\varepsilon }|x|}}{|x|^{4}}. \end{aligned}$$

(A.9)

Using the estimate above, for the charges that are such that $|x-y_n|\ge \ell$ we bound the contribution in the second term on the right-hand side of (A.8) as

$$\begin{aligned} \left| \sum _{n\in \mathbb {Z}^6\backslash \{0\}} \frac{C_\lambda e^{-\lambda \sqrt{\varepsilon }|x-y_n|}}{|x-y_n|^{4}}\right| \le \frac{C_\lambda }{\ell ^{4}} \sum _{n\in \mathbb {Z}^6\backslash \{0\}} \frac{ e^{-\lambda \sqrt{\varepsilon }|n|\ell }}{|n|^{4}} \end{aligned}$$

We estimate the sum with an integral (this can be done since the summand is a continuous decreasing function of n on $\mathbb {R}^6\backslash B_1(0)$, where $B_1(0)$ is the ball or radius one centered in zero), so that

$$\begin{aligned} \sum _{n\in \mathbb {Z}^6\backslash \{0\}} \frac{ e^{-\lambda \sqrt{\varepsilon }|n|\ell }}{|n|^{4}}\le \int _{\mathbb {R}^6\backslash B_1(0)} \textrm{d}n \frac{ e^{-\lambda \sqrt{\varepsilon }|n|\ell }}{|n|^{4}}=\frac{1}{(\lambda \sqrt{\varepsilon }\ell )^{2}}\int _{\mathbb {R}^6\backslash B_1(0)} \textrm{d}n \frac{ e^{-|n|}}{|n|^{4}} \end{aligned}$$

and therefore

$$\begin{aligned} \left| \sum _{n\in \mathbb {Z}^6\backslash \{0\}} \frac{ e^{-\lambda \sqrt{\varepsilon }|x-y_n|}}{|x-y_n|^{4}}\right| \le \frac{C_\lambda }{\varepsilon \ell ^6} \end{aligned}$$

(A.10)

Only a finite number of $y_n$ are such that $|x-y_n|<\ell$, and for those we bound $|x-y|\le |x-y_n|$. We thus obtain (A.2).

We consider now $\partial _{x_i}{\tilde{G}}_\varepsilon (x)$, given by

$$\begin{aligned} \begin{aligned} \partial _{x^{(i)}}{\tilde{G}}_\varepsilon (x)&=-x_i\frac{\varepsilon ^{3/2}}{2^3\pi ^{3}}\frac{\text {K}_3(\sqrt{\varepsilon }|x|)}{|x|^3} \end{aligned} \end{aligned}$$

(A.11)

(see [2, Chapter 3] for properties of the Bessel function of the third kind). For large $\varepsilon ^{1/2}|x|$,

$$\begin{aligned} \begin{aligned} \partial _{x^{(i)}}{\tilde{G}}_\varepsilon (x)&\simeq Cx_i\frac{\varepsilon ^{5/4} }{|x|^{3+1/2}}e^{-\sqrt{\varepsilon }|x|} \end{aligned} \end{aligned}$$

(A.12)

For small $\varepsilon ^{1/2}|x|$,

$$\begin{aligned} \begin{aligned} \partial _{x^{(i)}}{\tilde{G}}_\varepsilon (x)&\simeq C \frac{x_i}{|x|^6}\\ \end{aligned} \end{aligned}$$

(A.13)

The two equations above imply that for any $0<\lambda <1$ there exists a $C_\lambda >0$ such that

$$\begin{aligned} \begin{aligned} | \partial _{x^{(i)}}{\tilde{G}}_\varepsilon (x) |&\le \frac{C_\lambda }{|x|^5}e^{-\lambda \sqrt{\varepsilon }|x|}. \end{aligned} \end{aligned}$$

(A.14)

Similarly as above, we sum the contribution from charges such that $|x-y_n|>\ell$, so that

$$\begin{aligned} \begin{aligned} \Big |\sum _{|x-y_n|>\ell }\partial _{x^{(i)}} {\tilde{G}}_\varepsilon (x-y_n)\Big |&\le C_\lambda \sum _{n\in \mathbb {Z}^6\backslash \{0\}}\frac{1}{|x-y_n|^{5}}e^{-\lambda \sqrt{\varepsilon }|x-y_n|} \le \frac{C_\lambda }{\varepsilon ^{1/2}\ell ^6} \end{aligned}\nonumber \\ \end{aligned}$$

(A.15)

Therefore

$$\begin{aligned} |\partial _{x^{(i)}} G_\varepsilon (x,y)-\partial _{x^{(i)}} {\tilde{G}}_\varepsilon (x-y)| \le \sum _{n\ne 0, \, |x-y_n|<\ell }\frac{C}{|x-y_n|^5}+\frac{C}{\varepsilon ^{1/2}\ell ^6} \nonumber \\ \end{aligned}$$

(A.16)

for a constant $C>0$. $\square$

Proof of Proposition 2.1

Existence and uniqueness of minimizers can be proved by standard methods. We start by proving (2.12). Let $f_0$ be the zero-energy scattering solution defined in (1.2), and $f(x_1,x_2) = f_0(x_1-x_2)$ for $x_1,\, x_2\in \Lambda _\ell$. We write $\psi = f g$ and integrate by parts. Calling $\Lambda _\ell \times \Lambda _\ell =\Omega$ and writing $\nabla$ for $\nabla _x$, with $x=(x_1,x_2)$, we have

$$\begin{aligned} \int _{\Omega } \left( |\nabla \psi |^2 + \kappa V |\psi |^2 \right) = \int _{\Omega } f^2 |\nabla g |^2 + \int _{\partial \Omega } g^2 f {\hat{n}} \cdot \nabla f \end{aligned}$$

where ${\hat{n}}$ is the unit outward normal vector, and we use the shorthand notation $V(x) = V(x_1-x_2)$ for simplicity. Note that ${\hat{n}}\cdot \nabla f > 0$ since $f_0$ is an increasing function. Moreover, we have $f_0 \ge c_0 > 0$ (for a proof of this property we refer to [13, Lemma 5.1], which easily extends to bounded interaction potentials). Let us write $\tau = \delta _{\partial \Omega } f {\hat{n}} \cdot \nabla f$, so that the second term is simply $\int g^2 \tau$. We thus have

$$\begin{aligned} \int _{\Omega } \left( |\nabla \psi |^2 + \kappa V |\psi |^2 \right) \ge c_0^2 \int _{\Omega } |\nabla g |^2 + \int _{\Omega } g^2 \tau \end{aligned}$$

(A.17)

Let us look for the lowest eigenvalue of the right-hand side, i.e., the largest $\lambda$ such that

$$\begin{aligned} c_0^2 \int _{\Omega } |\nabla g |^2 + \int _{\Omega } g^2 \tau \ge \lambda \int _{\Omega } g^2 \end{aligned}$$

Since $f \le 1$, this is also a lower bound to the eigenvalue we are looking for, i.e.,

$$\begin{aligned} \int _{\Omega } \left( |\nabla \psi |^2 + \kappa V |\psi |^2 \right) \ge \lambda \int _{\Omega } g^2 \ge \lambda \int _{\Omega } \psi ^2 \end{aligned}$$

Clearly $\lambda \le \ell ^{-6} \int \tau$. Using that $f\le 1$ we have

$$\begin{aligned} \int \tau \le \int _{\partial \Omega } {\hat{n}} \cdot \nabla f = \int _{\Omega } \Delta f \le 2 \int _{\Lambda _\ell \times \textbf{R}^3}\textrm{d}x_1\textrm{d}x_2 (\Delta f_0)(x_1-x_2) = 8\pi \mathfrak {a} \ell ^3 \end{aligned}$$

We may assume that g shares the symmetries of $\Omega$, in which case

$$\begin{aligned} \begin{aligned} \int _{\Omega } g^2 \tau&= 12 \int _{\Omega } g^2 \tau _1 = 12 \int _{\Lambda _\ell } \textrm{d}x_2 \int _{[-\ell /2,\ell /2]^2} \textrm{d}x_1^\perp g(-\ell /2,x_1^\perp ,x_2)^2 \\&\qquad \times f(-\ell /2,x_1^\perp , x_2) \frac{\big (-\ell /2-x_2^{(1)}\big )}{|(-\ell /2,x_1^\perp ) - x_2|} f_0'( (-\ell /2,x_1^\perp ) - x_2) \end{aligned} \end{aligned}$$

where we write the vector $x_j$ as $(x_j^{(1)}, x_j^\perp )$, and denote the radial derivative of $f_0$ by $f_0'$. Using the Schur complement formula, we have, with Q the projection orthogonal to the constant function on $\Omega$,

$$\begin{aligned} \lambda \ge \ell ^{-6} \int \tau - \frac{12^2}{\ell ^6} \langle \tau _1, Q[ Q(-c_0^2\Delta - 8\pi \mathfrak {a} \ell ^{-3}) Q ]^{-1} Q\,\tau _1\rangle . \end{aligned}$$

Since the spectral gap of $-\Delta$ equals $(\pi /\ell )^2$ we can further bound

$$\begin{aligned} Q(-c_0^2\Delta - 8\pi a \ell ^{-3}) Q \ge \frac{c_0^2}{2} Q \left( -\Delta + \ell ^{-2} \right) Q\ge \frac{c_0^2}{2} Q \left( -\Delta _{x_1} + \ell ^{-2} \right) Q \end{aligned}$$

as long as $c_0^2(\pi ^2-1)/2 \ge 8\pi \mathfrak {a}/\ell$, which we assume henceforth. In particular,

$$\begin{aligned} Q\left[ Q(-c_0^2\Delta - 8\pi a \ell ^{-3}) Q \right] ^{-1}Q&\le \frac{2}{c_0^2} Q \left[ -\Delta _{x_1} + \ell ^{-2} \right] ^{-1} Q\\&= \frac{2}{c_0^2} \left[ -\Delta _{x_1} + \ell ^{-2} \right] ^{-1} - \frac{2\ell ^2}{c_0^2} P \end{aligned}$$

with $P=1-Q$ the projection onto the constant function. Observing that $-2\ell ^2c_0^{-2}P$ can be dropped for an upper bound, we thus have

$$\begin{aligned} \lambda \ge \ell ^{-6} \int \tau - \frac{2 \cdot 12^2}{c_0^2\ell ^6} \langle \tau _1 , [ -\Delta _{x_1} + 1/\ell ^2 ]^{-1} \tau _1\rangle \end{aligned}$$

An analysis similar to Lemma A.1 shows that the integral kernel of $[ -\Delta _{x_1} + \ell ^{-2} ]^{-1}$ on $[-\ell /2,\ell /2]^3$ is bounded above by $c_1 |x_1-y_1|^{-1}$, hence

$$\begin{aligned} \langle \tau _1 , [ -\Delta _{x_1} + 1/\ell ^2 ]^{-1} \tau _1\rangle \le c_1 \int _{\Lambda _\ell ^{3}}\textrm{d}x_1 \textrm{d}y_1 \textrm{d}x_2 \frac{ \tau _1(x_1,x_2) \tau _1(y_1, x_2) }{|x_1-y_1|} \end{aligned}$$

Using that $f \le 1$ as well as $f_0'(x_1) \le \mathfrak {a}/|x_1|^2$, we have, for fixed $x_2$,

$$\begin{aligned}&\int _{\Lambda _\ell ^{2}} \textrm{d}x_1 \textrm{d}y_1 \frac{ \tau _1(x_1,x_2) \tau _1(y_1, x_2) }{|x_1-y_1|} \\&\quad \le \left( \mathfrak {a} \big (\ell /2+x_2^{(1)}\big )\right) ^2 \int _{[-\ell /2,\ell /2]^4} \textrm{d}x_1^\perp \textrm{d}y_1^\perp \frac{ 1}{|x_1^\perp - y_1^\perp |} \frac{1}{| (-\ell /2,x_1^\perp ) - x_2 |^3} \frac{1}{| (-\ell /2,y_1^\perp ) - x_2 |^3} \\&\quad \le \frac{\mathfrak {a}^2}{\ell /2+x_2^{(1)}} \int _{\mathbb {R}^4} \textrm{d}x_1^\perp \textrm{d}y_1^\perp \frac{ 1}{|x_1^\perp - y_1^\perp |} \frac{1}{ ( 1+ (x_1^\perp )^2 )^{3/2} } \frac{1}{ ( 1 + (y_1^\perp )^2 )^{3/2}} \end{aligned}$$

where $\ell /2+x_2^{(1)}$ has been scaled out after extending the integral to $\mathbb {R}^4$. The final integral is finite by the Hardy–Littlewood–Sobolev inequality. In order to obtain a better bound for $x_2^{(1)}$ close to $-\ell /2$, we use in addition that $f_0'$ is bounded, and hence that $f_0'(x) \le c_2 \mathfrak {a}^{1/2} / |x|^{3/2}$ for some $c_2>0$. Thus

$$\begin{aligned}&\int _{\Lambda _\ell ^{2}} \textrm{d}x_1 \textrm{d}y_1 \frac{ \tau _1(x_1,x_2) \tau _1(y_1, x_2) }{|x_1-y_1|} \\&\quad \le \mathfrak {a}\, c_2^2 (\ell /2+x_2^{(1)} )^2 \int _{[-\ell /2,\ell /2]^4} \textrm{d}x_1^\perp \textrm{d}y_1^\perp \frac{ 1}{|x_1^\perp - y_1^\perp |} \frac{1}{| (-\ell /2,x_1^\perp ) - x_2 |^{5/2}} \frac{1}{| (-\ell /2,y_1^\perp ) - x_2 |^{5/2}} \\&\quad \le \mathfrak {a} \,c_2^2 \int _{\mathbb {R}^4} \textrm{d}x_1^\perp \textrm{d}y_1^\perp \frac{ 1}{|x_1^\perp - y_1^\perp |} \frac{1}{ ( 1+ (x_1^\perp )^2 )^{5/4} } \frac{1}{ ( 1 + (y_1^\perp )^2 )^{5/4}} \end{aligned}$$

where the integral is again finite by the Hardy–Littlewood–Sobolev inequality.

Altogether, we have thus shown that

$$\begin{aligned} \int _{\Omega } \textrm{d}x_1 \textrm{d}y_1 \frac{ \tau _1(x_1,x_2) \tau _1(y_1, x_2) }{|x_1-y_1|} \le C \mathfrak {a}^2 \min \left\{ \frac{1}{\ell /2+x_2^{(1)}} , \frac{1}{\mathfrak {a}} \right\} \end{aligned}$$

Integrating this over $x_2$ yields

$$\begin{aligned} \int _{\Lambda _\ell } \textrm{d}x_2 \int _{\Omega } \textrm{d}x_1 \textrm{d}y_1 \frac{ \tau _1(x_1,x_2) \tau _1(y_1, x_2) }{|x_1-y_1|} \le C \mathfrak {a}^2 \ell ^2 \ln \frac{\ell }{\mathfrak {a}} \end{aligned}$$

and thus

$$\begin{aligned} \lambda \ge \ell ^{-6} \int \tau - C \frac{\mathfrak {a}^2}{\ell ^4} \ln \frac{\ell }{\mathfrak {a}} \end{aligned}$$

To complete the lower bound on $\lambda$, we need a lower bound on $\int \tau$. We have

$$\begin{aligned} \int \tau= & {} \frac{1}{2} \int _{\partial \Omega } {\hat{n}} \cdot \nabla f^2 = \frac{1}{2} \int _{\Omega } \Delta f^2 = \int _{\Omega } ( |\nabla f|^2 + \kappa V f^2 )\nonumber \\= & {} 8\pi \mathfrak {a} \ell ^3 - \int _{\Lambda _\ell } \textrm{d}x_1 \int _{\Lambda _\ell ^c} ( |\nabla f|^2 + \kappa V f^2 ) \end{aligned}$$

(A.18)

Using that $\Delta f^2 = 2 ( |\nabla f|^2 + \kappa V f^2 ) \le C \min \{ \mathfrak {a}^{-2} ,\mathfrak {a}^2 /|x_1-x_2|^4\}$, the error term is bounded by

$$\begin{aligned} C \mathfrak {a}^2 \int _{\Lambda _\ell } \textrm{d}x_1 \min \{ (\ell /2 + x_1^1)^{-1} , 1 /\mathfrak {a} \} = C\mathfrak {a}^2 \ell ^2 \ln \frac{\ell }{\mathfrak {a}} \end{aligned}$$

We thus conclude that

$$\begin{aligned} \lambda \ge 8\pi \mathfrak {a} \ell ^3 - C\mathfrak {a}^2 \ell ^2 \ln \frac{\ell }{\mathfrak {a}} \end{aligned}$$

and from (A.17)

$$\begin{aligned} \int _{\Omega } \left( |\nabla \psi |^2 + \kappa V |\psi |^2 \right) \ge \lambda \int _{\Omega } |g|^2 \ge \lambda \int _\Omega |\psi |^2 \end{aligned}$$

(A.19)

since $f_0 \le 1$. In particular, $\lambda _\ell \ge \lambda$, and this concludes the lower bound. The upper bound follows by taking the trial function $\psi =f$ corresponding to $g=1$ and using again (A.18) together with

$$\begin{aligned} \Vert f\Vert ^2_2\ge \ell ^6-C \mathfrak {a} \ell ^5, \end{aligned}$$

where the latter follows from (1.3). This completes the proof of (2.12).

The estimate (2.13) in point i) clearly follows from (2.12). We proceed with point ii). The minimizer satisfies the eigenvalue equation on $\Omega$ with Neumann boundary conditions

$$\begin{aligned} \begin{aligned} \Big [-\Delta _x+\kappa V(x)\Big ]f( x)=\lambda _\ell f( x), \end{aligned} \end{aligned}$$

(A.20)

with $\lambda _\ell =8\pi \mathfrak {a}\ell ^{-3}\big (1+\mathcal {O}(\mathfrak {a}\ell ^{-1}\ln (\ell /\mathfrak {a}))\big )$. As before $x=(x_1,x_2)\in \Lambda _\ell \times \Lambda _\ell =\Omega$ and $\Delta _x=\Delta _{x_1}+\Delta _{x_2}$. Abusing notation we wrote $V(x)=V(x_1-x_2)$. It is useful to introduce a parameter $0<\varepsilon \le \ell ^{-2}$ and write (A.20) as

$$\begin{aligned} \begin{aligned} \big (-\Delta _x+\varepsilon \big )f( x)=\big (\lambda _\ell +\varepsilon -\kappa V(x)\big ) f( x). \end{aligned} \end{aligned}$$

(A.21)

We can express the solution to (A.21) as

$$\begin{aligned} f(x)=\int _\Omega \textrm{d}y\,G_\varepsilon (x,y)\big (\lambda _\ell +\varepsilon -\kappa V(y)\big ) f( y) \end{aligned}$$

with $G_\varepsilon (x,y)$ defined in (A.1). Lemma A.1 and the positivity of the minimizer f, of $G_\varepsilon (x,y)$ and of the potential V imply that

$$\begin{aligned} f(x)\le C(\lambda _\ell +\varepsilon )\int _\Omega \textrm{d}y\,\frac{ f( y)}{|x-y|^4}+\frac{C(\lambda _\ell +\varepsilon )}{\ell ^6\varepsilon } \int _\Omega \textrm{d}y\,f( y) \end{aligned}$$

(A.22)

The last term can be bounded as

$$\begin{aligned} \int _\Omega \textrm{d}y\,f( y)\le \Vert f\Vert _2\Vert \chi _\Omega \Vert _2= \ell ^3 \end{aligned}$$

We split the first integral in (A.22) as

$$\begin{aligned} \int _\Omega \textrm{d}y\,\frac{ f( y)}{|x-y|^4}=\int _{\Omega \cap B_\delta (x)} \textrm{d}y\,\frac{ f( y)}{|x-y|^4}+\int _{\Omega \backslash B_\delta (x)} \textrm{d}y\,\frac{ f( y)}{|x-y|^4} \end{aligned}$$

(A.23)

for $0<\delta \le \ell$ and $B_\delta (x)=\{y\in \mathbb {R}^6:|x-y|\le \delta \}$. We have

$$\begin{aligned} \int _{B_\delta (x)} \textrm{d}y\,\frac{ f( y)}{|x-y|^4}\le C\delta ^2\Vert f\Vert _\infty \end{aligned}$$

(A.24)

and

$$\begin{aligned} \int _{\Omega \backslash B_\delta (x)} \textrm{d}y\,\frac{ f( y)}{|x-y|^4} \le \Vert f\Vert _2\left( \int _{\mathbb {R}^6\backslash B_\delta (x)} \textrm{d}y\,\frac{ 1}{|x-y|^8}\right) ^{1/2}= \frac{C}{\delta } \end{aligned}$$

(A.25)

Hence

$$\begin{aligned} \Vert f\Vert _\infty \le C(\lambda _\ell +\varepsilon )\Big [\delta ^2\Vert f\Vert _\infty +\frac{1}{\delta }+ \frac{1}{\ell ^3\varepsilon }\Big ]. \end{aligned}$$

(A.26)

We set $\varepsilon =\ell ^{-2}$ and $\delta ^2=\big (2C(\lambda _\ell +\varepsilon )\big )^{-1}$, so that $\Vert f\Vert _\infty \le C'\ell ^{-3}$, proving (2.14).

In order to prove (2.15) in point iii), we decompose f as $f=c+g$, with $\int _\Omega g=0$ and $c=\ell ^{-6}\int f$. We shall show that

$$\begin{aligned} \Vert g\Vert _2\le C\kappa \ell ^{-1} \end{aligned}$$

(A.27)

for a constant $C>0$. Since

$$\begin{aligned} \Vert f-1/\ell ^3\Vert _2^2\le 2\Vert f-c\Vert _2^2+2\Vert c-1/\ell ^3\Vert _2^2=2\Vert g\Vert _2^2+2|\ell ^3c-1|^2 \end{aligned}$$

and, since

$$\begin{aligned} \Vert f-c\Vert _2^2=1-c^2\ell ^6\ge |1-c\ell ^3|^2, \end{aligned}$$

we have

$$\begin{aligned} \Vert f-1/\ell ^3\Vert _2^2\le 4\Vert g\Vert _2^2 \end{aligned}$$

Hence (2.15) follows from (A.27). To prove (A.27) we write Eq. (A.21) as

$$\begin{aligned} \begin{aligned} \big (-\Delta _x+\varepsilon \big )g( x)=\big (\lambda _\ell -\kappa V(x)\big ) f( x)+\varepsilon g( x) \end{aligned} \end{aligned}$$

(A.28)

for some $0<\varepsilon \le \ell ^{-2}$. We have

$$\begin{aligned} g(x)=\int _\Omega \textrm{d}y\,G_\varepsilon (x,y)\big (\lambda _\ell -\kappa V(y)\big ) f( y)+\varepsilon \int _\Omega \textrm{d}y\,G_\varepsilon (x,y)g( y) \end{aligned}$$

(A.29)

By Lemma A.1 and the Hardy–Littlewood–Sobolev and Hölder inequalities we have

$$\begin{aligned} \Big \Vert \lambda _\ell \int _\Omega \textrm{d}y\,G_\varepsilon (\,\cdot \,,y) f( y)\Big \Vert _2\le & {} C\lambda _\ell \Big \Vert \int _\Omega \textrm{d}y\,\Big (\frac{1}{|\,\cdot \,-y|^4}+\frac{1}{\ell ^6\varepsilon }\Big ) f( y)\Big \Vert _2\nonumber \\\le & {} C\lambda _\ell \Vert f\Vert _{6/5}+C\frac{\lambda _\ell }{\ell ^3\varepsilon }\Vert f\Vert _1 \le \frac{C\kappa }{\ell ^3\varepsilon } \end{aligned}$$

(A.30)

To bound the contribution proportional to V in (A.29), we use (2.14) and estimate

$$\begin{aligned} \begin{aligned} \int _\Omega \textrm{d}y\,\Big (\frac{1}{|x-y|^4}+\frac{1}{\ell ^6\varepsilon }\Big )V(y) f( y)&\le \frac{C}{\ell ^3}\int _\Omega \textrm{d}y\,\frac{1}{|x-y|^4}V(y) + \frac{C}{\ell ^9\varepsilon }\int _\Omega \textrm{d}y\,V(y) \end{aligned} \end{aligned}$$

Using the notation $y=(y_1,y_2)\in \Lambda _\ell \times \Lambda _\ell$, we observe that

$$\begin{aligned}{} & {} \int _{\Lambda _\ell \times \Lambda _\ell } \textrm{d}y_1\textrm{d}y_2\,\frac{V(y_1-y_2)}{\big [|x_1-y_1|^2+|x_2-y_2|^2\big ]^2}\nonumber \\{} & {} \qquad \le \int _{\mathbb {R}^3}\textrm{d}y_2\,V(y_2) \int _{\mathbb {R}^3} \textrm{d}y_1\frac{1}{\big [|x_1-y_1|^2+|x_2-y_1+y_2|^2\big ]^2}\nonumber \\{} & {} \qquad = C\int _{\mathbb {R}^3}\textrm{d}y_2\, \frac{V(y_2)}{|x_1-x_2-y_2|} \le \frac{C}{|x_1-x_2|} \int _{\mathbb {R}^3}\textrm{d}y\,V(y) \end{aligned}$$

(A.31)

where we have used Newton’s theorem in the last step. The $L^2$ norm of the last expression is thus bounded by $(\int V) \ell ^2$, and we conclude that

$$\begin{aligned} \begin{aligned} \Big \Vert \int _\Omega \textrm{d}y\,G_\varepsilon (\,\cdot \,,y)V(y) f( y)\Big \Vert _2&\le \frac{C}{\varepsilon \ell ^3}\Vert V\Vert _1 \end{aligned} \end{aligned}$$

(A.32)

We are left with the last contribution in (A.29). Since g is orthogonal to the constant function, we can use the spectral gap $(\pi /\ell )^2$ of the Laplacian to obtain the bound

$$\begin{aligned} \begin{aligned} \varepsilon \Vert G_\varepsilon g\Vert _2 \le \varepsilon \frac{\ell ^2}{\pi ^2}\Vert g\Vert _2 \end{aligned} \end{aligned}$$

(A.33)

By (A.30), (A.32) and (A.33) and with the choice $\varepsilon =\ell ^{-2}$ (so that $\varepsilon \frac{\ell ^2}{\pi ^2}<1$) we therefore arrive at (A.27), proving (2.15). The estimate (2.16) follows by Cauchy–Schwarz.

Next we examine point iv). Again, we decompose f as $f=c+g$, with $\int _\Omega g=0$ and c a positive constant. We observe that

$$\begin{aligned} \begin{aligned} |1-\ell ^3f(x_1,x_2)|\le |1-\ell ^3c|+\ell ^3|g(x_1,x_2)|\le \Vert g\Vert _2+\ell ^3|g(x_1,x_2)| \end{aligned} \end{aligned}$$

Hence, if we show that

$$\begin{aligned} \begin{aligned} \sup _{x\in \Omega }\,(|x_1-x_2|+1)|g(x_1,x_2)|\le C\kappa \ell ^{-3}, \end{aligned} \end{aligned}$$

(A.34)

the bound (2.17) follows. To show (A.34), we multiply (A.29) by $|x_1-x_2|+1$ to obtain

$$\begin{aligned} \begin{aligned}&(|x_1-x_2|+1)g(x_1,x_2)\\&\quad =\int _{\Lambda _\ell \times \Lambda _\ell } \textrm{d}y_1\textrm{d}y_2\, (|x_1-x_2|+1)G_\varepsilon (x_1,x_2,y_1,y_2)\big (\lambda _\ell -\kappa V(y_1-y_2)\big ) f( y_1,y_2)\\&\qquad +\varepsilon \int _{\Lambda _\ell \times \Lambda _\ell } \textrm{d}y_1\textrm{d}y_2\, (|x_1-x_2|+1)G_\varepsilon (x_1,x_2,y_1,y_2)g( y_1,y_2) \end{aligned}\nonumber \\ \end{aligned}$$

(A.35)

We use Lemma A.1 to estimate $G_\varepsilon$ and (2.14) as well as (2.12) to get

$$\begin{aligned} \begin{aligned}&\lambda _\ell \int _{\Lambda _\ell \times \Lambda _\ell } \textrm{d}y_1\textrm{d}y_2\,(|x_1-x_2|+1)G_\varepsilon (x_1,x_2,y_1,y_2)f( y_1,y_2)\\&\quad \le \frac{C\kappa }{\ell ^6}\int _{\Lambda _\ell \times \Lambda _\ell } \textrm{d}y_1\textrm{d}y_2\,(|x_1-x_2|+1)\left[ \frac{1}{\big [|x_1-y_1|^2+|x_2-y_2|^2\big ]^2}+\frac{1}{\ell ^6\varepsilon }\right] \\&\quad \le C\kappa \ell ^{-3}+ C\kappa \varepsilon ^{-1}\ell ^{-5}. \end{aligned} \end{aligned}$$

Moreover, with (A.31) and (2.14), we have

$$\begin{aligned} \begin{aligned} \int _{\Lambda _\ell \times \Lambda _\ell } \textrm{d}y_1\textrm{d}y_2&\,(|x_1-x_2|+1)G_\varepsilon (x_1,x_2,y_1,y_2)\kappa V(y_1-y_2)f( y_1,y_2)\\&\le \frac{C\kappa }{\ell ^3}\int _{\mathbb {R}^3}\textrm{d}y_2\, (|x_1-x_2|+1)\frac{V(y_2)}{|x_1-x_2-y_2|}+\frac{C\kappa }{\varepsilon \ell ^5}\int _{\mathbb {R}^3}V(y_2)\textrm{d}y_2\\ \end{aligned} \end{aligned}$$

By Newton’s theorem we see that

$$\begin{aligned}{} & {} \int _{\mathbb {R}^3}\textrm{d}y_2\, (|x_1-x_2|+1)\frac{\kappa V(y_2)}{|x_1-x_2-y_2|}\nonumber \\{} & {} \quad \le C\int _{\mathbb {R}^3}\textrm{d}y_2\, \kappa V(y_2)+\frac{1}{|x_1-x_2|}\int _{|y_2|\le |x_1-x_2|}\textrm{d}y_2\,\kappa V(y_2)\nonumber \\{} & {} \qquad +\int _{|y_2|> |x_1-x_2|}\textrm{d}y_2\, \frac{\kappa V(y_2)}{|y_2|}\nonumber \\{} & {} \quad \le C\kappa , \end{aligned}$$

(A.36)

where we used that $\int \textrm{d}x\, V(x) |x|^{-1}$ is finite. We conclude that

$$\begin{aligned} \int _{\Lambda _\ell \times \Lambda _\ell } \textrm{d}y_1\textrm{d}y_2\,(|x_1-x_2|+1)G_\varepsilon (x_1,x_2,y_1,y_2)\kappa V(y_1-y_2)f( y_1,y_2)\le C\kappa (\ell ^{-3}+\varepsilon ^{-1}\ell ^{-5}) \end{aligned}$$

We are left with the last contribution in (A.35). We write, using (2.16),

$$\begin{aligned}{} & {} \varepsilon \int _{\Lambda _\ell \times \Lambda _\ell } \textrm{d}y_1\textrm{d}y_2\, (|x_1-x_2|+1)G_\varepsilon (x_1,x_2,y_1,y_2)g( y_1,y_2)\nonumber \\ {}{} & {} \quad \le C\varepsilon \int _{\Lambda _\ell \times \Lambda _\ell } \textrm{d}y_1\textrm{d}y_2\, (|x_1-x_2|+1)\left[ \frac{1}{|x-y|^4}+\frac{1}{\varepsilon \ell ^6}\right] g( y_1,y_2)\nonumber \\{} & {} \quad \le C\varepsilon \int _{\Lambda _\ell \times \Lambda _\ell } \textrm{d}y_1\textrm{d}y_2\, \frac{(|x_1-x_2|+1)}{|x-y|^4}g( y_1,y_2)+\frac{C\kappa }{\ell ^3} \end{aligned}$$

(A.37)

We bound the first term above as follows

$$\begin{aligned} \begin{aligned}&C\varepsilon \int _{\Lambda _\ell \times \Lambda _\ell } \textrm{d}y\,\frac{(|x_1-x_2|+1)}{|x-y|^4}\frac{(|y_1-y_2|+1)g(y)}{(|y_1-y_2|+1)}\\&\quad \le C\varepsilon \Big [\sup _{y\in \Lambda _\ell \times \Lambda _\ell }(|y_1-y_2|+1)g(y)\Big ]\int _{\Lambda _\ell \times \Lambda _\ell } \textrm{d}y\,\frac{|x_1-x_2|+1}{|y_1-y_2|+1}\frac{1}{|x-y|^4} \end{aligned}\nonumber \\ \end{aligned}$$

(A.38)

Similarly as we did in (A.31) we estimate

$$\begin{aligned}{} & {} \int _{\Lambda _\ell \times \Lambda _\ell } \textrm{d}y\,\frac{1}{|y_1-y_2|+1}\frac{1}{|x-y|^4}\nonumber \\{} & {} \quad \le \int _{[-\ell ,\ell ]^3}\textrm{d}y_2\,\frac{1}{|y_2|+1} \int _{\mathbb {R}^3} \textrm{d}y_1\frac{1}{\big [|x_1-y_1|^2+|x_2-y_1+y_2|^2\big ]^2}\nonumber \\{} & {} \quad \le C\int _{|y_2|\le \sqrt{3}\ell } \textrm{d}y_2\, \frac{1}{|y_2|}\frac{1}{|x_1-x_2-y_2|} \le C \frac{ \ell ^2 }{ |x_1 - x_2| + \ell } \end{aligned}$$

(A.39)

where we again applied Newton’s theorem in the last step. Thus

$$\begin{aligned} \begin{aligned} \int _{\Lambda _\ell \times \Lambda _\ell } \textrm{d}y\,\frac{|x_1-x_2|+1}{|y_1-y_2|+1}\frac{1}{|x-y|^4}&\le C\ell ^2 \end{aligned} \end{aligned}$$

(A.40)

In conclusion, we have

$$\begin{aligned} \begin{aligned} (|x_1-x_2|+1)|g(x_1,x_2)|&\le \frac{C\kappa }{\varepsilon \ell ^5}\int _{\mathbb {R}^3}\textrm{d}y\, V(y)+\frac{C\kappa }{\ell ^3}\\&\quad +C\varepsilon \ell ^2\Big [\sup _{y\in \Lambda _\ell \times \Lambda _\ell }(|y_1-y_2|+1)|g(y)|\Big ] \end{aligned} \end{aligned}$$

therefore, by setting $\varepsilon =(2C\ell ^2)^{-1}$, we obtain (A.34).

Finally we investigate point v). As above, we decompose $f=c+g$ with $c=\ell ^{-6}\int f$. We shall prove that

$$\begin{aligned} \begin{aligned} \big [d(\tfrac{x_1+x_2}{2} )^{5/3}+1\big ]|\nabla _{x_1+x_2}g(x)|\le C\kappa \ell ^{-3} \end{aligned} \end{aligned}$$

(A.41)

where d(x) is the distance of x from the boundary of the box $\Lambda _\ell$. By (A.29), we have

$$\begin{aligned} \nabla _{x_1+x_2}g(x)= & {} -\int _\Omega \textrm{d}y\,\nabla _{y_1+y_2}{\tilde{G}}_\varepsilon (x-y)\big (\lambda _\ell -V(y)\big ) f( y)\nonumber \\{} & {} -\varepsilon \int _\Omega \textrm{d}y\,\nabla _{y_1+y_2}{\tilde{G}}_\varepsilon (x-y)g( y)\nonumber \\{} & {} +\int _\Omega \textrm{d}y\,\nabla _{x_1+x_2}\Big [G_\varepsilon (x,y)-{\tilde{G}}_\varepsilon (x-y)\Big ]\big (\lambda _\ell -\kappa V(y)\big ) f( y)\nonumber \\{} & {} +\varepsilon \int _\Omega \textrm{d}y\,\nabla _{x_1+x_2}\Big [G_\varepsilon (x,y)-{\tilde{G}}_\varepsilon (x-y)\Big ]g( y) \end{aligned}$$

(A.42)

We integrate by parts in the first line, and obtain

$$\begin{aligned} \nabla _{x_1+x_2}g(x)= & {} \int _\Omega \textrm{d}y\,{\tilde{G}}_\varepsilon (x-y)\big (\lambda _\ell -\kappa V(y)\big ) \nabla _{y_1+y_2}f( y)\nonumber \\{} & {} +\varepsilon \int _\Omega \textrm{d}y\,{\tilde{G}}_\varepsilon (x-y)\nabla _{y_1+y_2}g( y)\nonumber \\{} & {} +\int _{\partial \Omega } \textrm{d}\sigma _y\,\hat{n}\,{\tilde{G}}_\varepsilon (x-y)\big (\lambda _\ell -\kappa V(y)\big ) f( y)\nonumber \\{} & {} +\varepsilon \int _{\partial \Omega } \textrm{d}\sigma _y\,\hat{n}\,{\tilde{G}}_\varepsilon (x-y)g( y)\nonumber \\{} & {} +\int _\Omega \textrm{d}y\,\nabla _{x_1+x_2}\Big [G_\varepsilon (x,y)-{\tilde{G}}_\varepsilon (x-y)\Big ]\big (\lambda _\ell -\kappa V(y)\big ) f( y)\nonumber \\{} & {} +\varepsilon \int _\Omega \textrm{d}y\,\nabla _{x_1+x_2}\Big [G_\varepsilon (x,y)-\tilde{G}_\varepsilon (x-y)\Big ]g( y)=\sum _{j=1}^6\text {D}_j(x)\nonumber \\ \end{aligned}$$

(A.43)

where $\textrm{d}\sigma _y$ is the surface element of the boundary of the box $\partial \Omega$ and ${\hat{n}}$ is the unit vector pointing outwards. We start by considering $\text {D}_2$. Using (A.9), we can bound, for every $x\in \Omega$,

$$\begin{aligned} \begin{aligned}&|\big [d( \tfrac{x_1+x_2}{2})^{5/3}+1\big ]\text {D}_2(x)|\\&\quad \le C\varepsilon \big [d( \tfrac{x_1+x_2}{2})^{5/3}+1\big ]\int _{\Omega } \textrm{d}y\,\frac{\big |\big [d(y_1+y_2)^{5/3}+1\big ]\nabla _{y_1+y_2}g(y)\big |}{|x-y|^4\big [d(\tfrac{y_1+y_2}{2})^{5/3}+1\big ]}\\&\quad \le C\varepsilon \sup _{y\in \Omega }\big |\big [d(\tfrac{y_1+y_2}{2})^{5/3}+1\big ]\nabla _{y_1+y_2}g(y)\big | \int _{\Omega } \textrm{d}y\,\frac{\textrm{d}(\tfrac{x_1+x_2}{2})^{5/3}+1}{|x-y|^4\big [d(\tfrac{y_1+y_2}{2})^{5/3}+1\big ]} \end{aligned}\nonumber \\ \end{aligned}$$

(A.44)

In the following we shall show that

$$\begin{aligned} \int _{\Omega } \textrm{d}y\,\frac{1}{|x-y|^4\big [d(\tfrac{y_1+y_2}{2})^{5/3}+1\big ]} \le \frac{ C \ell ^2}{ d(\tfrac{x_1+x_2}{2})^{5/3} +1} \end{aligned}$$

Since

$$\begin{aligned} \frac{ 1}{ d(\tfrac{x_1+x_2}{2})^{5/3} +1} \le \sum _{i=1}^3 \sum _{j=1}^2 \frac{ 1}{2^{-5/3} |x_1^{(i)}+x_2^{(i)}- (-1)^j \ell |^{5/3} +1} \le \frac{ 6}{ d(\tfrac{x_1+x_2}{2})^{5/3} +1}\nonumber \\ \end{aligned}$$

(A.45)

it is sufficient to prove that

$$\begin{aligned} \int _{\Omega } \textrm{d}y\,\frac{1}{|x-y|^4\big [|y_1^{(1)}+y_2^{(1)}-\ell |^{5/3}+1\big ]} \le \frac{ C \ell ^2}{ |x_1^{(1)}+x_2^{(1)}-\ell | ^{5/3} +1}\qquad \quad \end{aligned}$$

(A.46)

For this purpose, we shall write

$$\begin{aligned} \begin{aligned} |x_1- y_1|^2+|x_2- y_2|^2=\frac{1}{2}\big |(x_1+x_2)-( y_1+ y_2)\big |^2+\frac{1}{2}\big |(x_1-x_2)-( y_1- y_2)\big |^2; \end{aligned}\nonumber \\ \end{aligned}$$

(A.47)

with the change of variable $y_1+ y_2=b, \, y_1- y_2=a$ we have

$$\begin{aligned} \begin{aligned}&\int _{\Omega } \textrm{d}y\,\frac{|x_1^{(1)}+x_2^{(1)}-\ell |^{5/3}+1}{|x-y|^4\big [|y_1^{(1)}+y_2^{(1)}-\ell |^{5/3}+1\big ]}\\&\quad =\frac{1}{2}\int _{[-\ell ,\ell ]^3} \textrm{d}b\int _{\omega (b)}\textrm{d}a\,\frac{|x_1^{(1)}+x_2^{(1)}-\ell |^{5/3}+1}{\big [\big |(x_1+x_2)-b\big |^2+\big |(x_1-x_2)-a\big |^2\big ]^2\big [|b^{(1)}-\ell |^{5/3}+1\big ]} \end{aligned} \end{aligned}$$

(A.48)

where $\omega (b)=[|b^{(1)}|-\ell ,\ell -|b^{(1)}|]\times [|b^{(2)}|-\ell ,\ell -|b^{(2)}|]\times [|b^{(3)}|-\ell ,\ell -|b^{(3)}|]$. Let us introduce the notation $a= (a^{(1)},a^\perp )$ and $b=(b^{(1)}, b^{\perp })$. To bound (A.48) we bound the numerator with $2 (2\ell )^{5/3}$ (assuming $2\ell \ge 1$) and extend the integration domain of the variable $a^\perp$ to $[-\ell ,\ell ]^2$; dropping the term involving $a^{(1)}$ in the denominator, we can integrate over $a^{(1)}$ to obtain the bound

$$\begin{aligned} \begin{aligned}&\int _{[-\ell ,\ell ]^3} \textrm{d}b\int _{\omega (b)}\textrm{d}a\,\frac{|x_1^{(1)}+x_2^{(1)}-\ell |^{5/3}+1}{\big [\big |(x_1+x_2)-b\big |^2+\big |(x_1-x_2)-a\big |^2\big ]^2\big [|b^{(1)}-\ell |^{5/3}+1\big ]} \\&\quad \le 2 (2\ell )^{5/3}\int _{[-\ell ,\ell ]} \textrm{d}b^{(1)}\frac{1}{|b^{(1)}-\ell |^{2/3}} \int _{[-\ell ,\ell ]^4}db^\perp \textrm{d}a^\perp \,\frac{1}{\big [\big |(x_1+x_2)-b\big |^2+\big |(x_1^\perp -x_2^\perp )-a^\perp \big |^2\big ]^2} \end{aligned} \end{aligned}$$

We estimate

$$\begin{aligned} \begin{aligned} \int _{[-\ell ,\ell ]^4}\textrm{d}b^\perp \textrm{d}a^\perp \,&\frac{1}{\big [\big |(x_1+x_2)-b\big |^2+\big |(x_1^\perp -x_2^\perp )-a^\perp \big |^2\big ]^2}\le C\int _{|(x_1^{(1)}+x_2^{(1)})-b^{(1)}|}^{3\ell }\textrm{d}z\,\frac{1}{z}\\&= C\ln \left( \frac{3\ell }{|(x_1^{(1)}+x_2^{(1)})-b^{(1)}|}\right) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\int _{[-\ell ,\ell ]} \textrm{d}b^{(1)}\frac{1}{|b^{(1)}-\ell |^{2/3}}\ln \left( \frac{3\ell }{|(x_1^{(1)}+x_2^{(1)})-b^{(1)}|}\right) \\&\le (2\ell )^{1/3} \sup _{0<s<1} \int _0^1 \textrm{d}t \, t^{-2/3} \ln \frac{3/2}{ t - s } \le C \ell ^{1/3} \end{aligned} \end{aligned}$$

This proves (A.46). We have thus shown that

$$\begin{aligned} \begin{aligned}&|\big [d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]\text {D}_2(x)|\le C\varepsilon \ell ^2\sup _{y\in \Omega }\big |\big [d(\tfrac{y_1+y_2}{2})^{5/3}+1\big ]\nabla _{y_1+y_2}g(y)\big | \end{aligned}\nonumber \\ \end{aligned}$$

(A.49)

We proceed with $\text {D}_1$, which we write as $\text {D}_1=\text {D}_{11}+\text {D}_{12}$, with

$$\begin{aligned} \begin{aligned} \text {D}_{11}(x)&=\lambda _\ell \int _\Omega \textrm{d}y\,{\tilde{G}}_\varepsilon (x-y) \nabla _{y_1+y_2}f( y) \end{aligned} \end{aligned}$$

(A.50)

and

$$\begin{aligned} \begin{aligned} \text {D}_{12}(x)&=-\int _\Omega \textrm{d}y\,{\tilde{G}}_\varepsilon (x-y)\kappa V(y) \nabla _{y_1+y_2}f( y) \end{aligned} \end{aligned}$$

(A.51)

Using the same method as above we estimate

$$\begin{aligned} \begin{aligned} |\big [d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]\text {D}_{11}(x)|&\le C\lambda _\ell \ell ^2\sup _{y\in \Omega }|\big [d(\tfrac{y_1+y_2}{2})^{5/3}+1\big ]\nabla _{y_1+y_2}g(y)|. \end{aligned}\nonumber \\ \end{aligned}$$

(A.52)

For $\text {D}_{12}$ we have

$$\begin{aligned} \begin{aligned} |\big [&d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]\text {D}_{12}(x)|\\&\le \sup _{y\in \Omega }|\big [d(\tfrac{y_1+y_2}{2})^{5/3}+1\big ]\nabla _{y_1+y_2}g(y)|\int _\Omega \textrm{d}y\,\frac{\kappa V(y)\big [d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]}{|x-y|^4\big [d(\tfrac{y_1+y_2}{2})^{5/3}+1\big ]} \end{aligned}\nonumber \\ \end{aligned}$$

(A.53)

Because of (A.45) it again suffices to bound the last integral with d(z) replaced by $|z^{(1)} - \ell |$ for both $z=x_1+x_2$ and $z=y_1+y_2$. With the same change of variables as in (A.48) we have

$$\begin{aligned} \begin{aligned} \int _{\Omega }&\textrm{d}y\,\frac{\kappa V(y)}{|x-y|^4\big [\big (y_1^{(1)}+y_2^{(1)}\big )^{5/3}+1\big ]}\\&\le \frac{1}{2}\int _{\mathbb {R}^3} \textrm{d}a\int _{\mathbb {R}^3}\textrm{d}b\,\frac{\kappa V(a)}{\big [\big |(x_1+x_2)-b\big |^2+\big |(x_1-x_2)-a\big |^2\big ]^2\big [|b^{(1)}-\ell |^{5/3}+1\big ]} \end{aligned}\nonumber \\ \end{aligned}$$

(A.54)

where we extended integration domain to $\mathbb {R}^6$. Integrating first in the variable $b^\perp$ we have

$$\begin{aligned} \begin{aligned} \int _{\mathbb {R}^3} \textrm{d}a&\int _{\mathbb {R}^3}\textrm{d}b\,\frac{\kappa V(a)}{\big [\big |(x_1+x_2)-b\big |^2+\big |(x_1-x_2)-a\big |^2\big ]^2\big [|b^{(1)}-\ell |^{5/3}+1\big ]}\\&= C\int _{\mathbb {R}} \textrm{d}b^{(1)} \frac{1}{|b^{(1)}-\ell |^{5/3}+1}\int _{\mathbb {R}^3} \textrm{d}a\frac{\kappa V(a)}{\big |\big (x_1^{(1)}+x_2^{(1)}\big )-b^{(1)}\big |^2+\big |(x_1-x_2)-a\big |^2} \end{aligned}\nonumber \\ \end{aligned}$$

(A.55)

Using that V is bounded and of compact support, one readily checks that

$$\begin{aligned} \int _{\mathbb {R}^3} \textrm{d}a \frac{V(a)}{X+|Y-a|^2} \le \frac{C}{X+ |Y|^2+1} \end{aligned}$$

(A.56)

for all $X\ge 0$ and $Y\in \mathbb {R}^3$.

Hence we find that

$$\begin{aligned}{} & {} \int _{\Omega } \textrm{d}y\,\frac{\kappa V(y)}{|x-y|^4\big [|y_1^{(1)}+y_2^{(1)}-\ell |^{5/3}+1\big ]}\nonumber \\{} & {} \quad \le C\kappa \int _{\mathbb {R}} \textrm{d}b^{(1)} \frac{1}{|b^{(1)}-\ell |^{5/3}+1} \frac{1}{|(x_1^{(1)}+x_2^{(1)})-b^{(1)}|^{2}+1}\nonumber \\{} & {} \quad \le \frac{C\kappa }{\big [|x_1^{(1)}+x_2^{(1)}-\ell |^{5/3}+1\big ]} \end{aligned}$$

(A.57)

which is the desired bound, allowing us to conclude that

$$\begin{aligned} \begin{aligned}&|\big [d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]\text {D}_{12}(x)|\le C\kappa \sup _{y\in \Omega }|\big [d(\tfrac{y_1+y_2}{2})^{5/3}+1\big ]\nabla _{y_1+y_2}g(y)|. \end{aligned} \end{aligned}$$

(A.58)

In order to bound $\text {D}_5$ we split it into

$$\begin{aligned} \begin{aligned} \text {D}_{51}(x)&=\lambda _\ell \int _\Omega \textrm{d}y\,\nabla _{x_1+x_2}\Big [G_\varepsilon (x,y)-{\tilde{G}}_\varepsilon (x-y)\Big ]f( y) \end{aligned} \end{aligned}$$

(A.59)

and

$$\begin{aligned} \begin{aligned} \text {D}_{52}(x)&=-\int _\Omega \textrm{d}y\,\nabla _{x_1+x_2}\Big [G_\varepsilon (x,y)-{\tilde{G}}_\varepsilon (x-y)\Big ]\kappa V(y)f( y) \end{aligned} \end{aligned}$$

(A.60)

We easily bound

$$\begin{aligned}{} & {} |\big [d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]\text {D}_{51}(x)|\nonumber \\{} & {} \quad \le \lambda _\ell \big [d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]\int _\Omega \textrm{d}y\,\Big [ \sum _{ n}\frac{1}{|x-y_n|^5}+\frac{1}{\varepsilon ^{1/2}\ell ^6}\Big ]f( y)\nonumber \\{} & {} \quad \le C \kappa (\ell ^{-3-1/3}+\ell ^{-4-1/3}\varepsilon ^{-1/2}) \end{aligned}$$

(A.61)

where we used (A.4), (2.12) and (2.14) and we estimated $d(\tfrac{x_1+x_2}{2})\le \ell /2$. For $\text {D}_{52}$, we use again (A.4) and (2.14) to estimate

$$\begin{aligned} \begin{aligned}&|\big [d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]\text {D}_{52}(x)|\\&\quad \le \frac{C\big [d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]}{\ell ^3}\int _\Omega \textrm{d}y\,\Big [ \sum _{ n}\frac{1}{|x- y_n|^5}+\frac{1}{\varepsilon ^{1/2}\ell ^6}\Big ]\kappa V(y). \end{aligned} \end{aligned}$$

(A.62)

For the second contribution we have

$$\begin{aligned} \begin{aligned} \frac{\big [d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]}{\varepsilon ^{1/2}\ell ^9}\int _\Omega \textrm{d}y\,\kappa V(y)\le \frac{C\kappa }{\varepsilon ^{1/2}\ell ^{4+1/3}}\\ \end{aligned} \end{aligned}$$

(A.63)

For the first contribution in (A.62), among the image charges $y_n$ we start considering the one that has all coordinates equal to those of y except the first one. We rename it as $\tilde{y}$, and we have ${\tilde{y}}_1^{(1)}=-\ell - y_1^{(1)}$. We write

$$\begin{aligned}{} & {} \int _\Omega \textrm{d}y\,\frac{1}{|x-{\tilde{y}}|^5} V(y)=\int _\Omega \textrm{d}y_1\textrm{d}y_2\,\frac{V(y_1-y_2)}{(|x_1-{\tilde{y}}_1|^2+|x_2-{\tilde{y}}_2|^2)^{5/2}}\nonumber \\{} & {} \quad =\int _{[-3\ell /2,-\ell /2]\times [-\ell /2,\ell /2]^5} \textrm{d}\tilde{y}_1d{\tilde{y}}_2\,\frac{ V({\tilde{y}}^{(1)}_1+\tilde{y}^{(1)}_2+\ell ,{\tilde{y}}^{\perp }_1-\tilde{y}^{\perp }_2)}{(|x_1-{\tilde{y}}_1|^2+|x_2-{\tilde{y}}_2|^2)^{5/2}}\qquad \end{aligned}$$

(A.64)

The denominator can be expressed as in (A.47); with the change of variables ${\tilde{y}}_1+{\tilde{y}}_2=b, \, \tilde{y}_1-{\tilde{y}}_2=a$, we have (extending the integration domain to $\mathbb {R}^6$)

$$\begin{aligned} \begin{aligned} \int _\Omega \textrm{d}y\,\frac{1}{|x-{\tilde{y}}|^5}V(y)&\le \frac{1}{\sqrt{2}}\int _{\mathbb {R}^3} \textrm{d}b^{(1)}\textrm{d}a^{\perp }\, V(b^{(1)}+\ell , a^{\perp })\\&\quad \times \int _{\mathbb {R}^3}\textrm{d}a^{(1)} \textrm{d}b^{\perp }\,\frac{1}{\big (\big |x_1+x_2-b\big |^2+\big |x_1-x_2-a\big |^2\big )^{5/2}}\\&= C \int _{\mathbb {R}^3} \textrm{d}b^{(1)}\textrm{d}a^{\perp }\frac{ V(b^{(1)}+\ell , a^{\perp })}{\big (x_1^{(1)}+x_2^{(1)}-b^{(1)}\big )^2+\big (x_1^\perp -x_2^\perp -a^\perp \big )^2} \end{aligned}\nonumber \\ \end{aligned}$$

(A.65)

Using now again (A.56), we arrive at

$$\begin{aligned} \begin{aligned} \frac{1}{\ell ^3}&\left| \int _\Omega \textrm{d}y\,\frac{1}{|x-{\tilde{y}}|^5}\kappa V(y)\right| \le \frac{C}{\ell ^3}\frac{\kappa }{|x_1^{(1)}+x_2^{(1)}-\ell |^2+1} \le \frac{C}{\ell ^3}\frac{\kappa }{d(x_1+x_2)^2+1} \end{aligned}\nonumber \\ \end{aligned}$$

(A.66)

The contribution from the other image charges can be estimated similarly, and we omit the details. We conclude that

$$\begin{aligned} \begin{aligned} |\big [d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]\text {D}_{52}(x)|&\le \frac{C\kappa }{\varepsilon ^{1/2}\ell ^{4+1/3}}+\frac{C\kappa }{\ell ^{3}} \end{aligned} \end{aligned}$$

(A.67)

Next we investigate $\text {D}_6$. With (A.4) and (A.34) we have

$$\begin{aligned} \begin{aligned} |\text {D}_6(x)|&=\Big |\varepsilon \int _\Omega \textrm{d}y\,\nabla _{x_1+x_2}\Big [G_\varepsilon (x,y)-{\tilde{G}}_\varepsilon (x-y)\Big ]g( y)\Big |\\&\le \frac{C\varepsilon ^{1/2}}{\ell ^6}\int _\Omega \textrm{d}y\,|g( y)|+\frac{\varepsilon \kappa }{\ell ^3}\sum _{n}\int _\Omega \textrm{d}y\,\frac{1}{|x- y_n|^5}\frac{1}{|y_1-y_2|+1}\\ \end{aligned}\nonumber \\ \end{aligned}$$

(A.68)

To bound the first term we can also use (A.34), which gives $\int |g|\le C\kappa \ell ^{2}$. To estimate the second term in (A.68) we start, as above, by considering the image charge ${\tilde{y}}$ such that ${\tilde{y}}_1^{(1)}=-\ell - y_1^{(1)}$, ${\tilde{y}}_1^{(i)}= y_1^{(i)}$ for $i=2,3$ and ${\tilde{y}}_2^{(j)}= y_2^{(j)}$ for $j=1,2,3$. We perform again the change of variables ${\tilde{y}}_1+{\tilde{y}}_2=b, \, \tilde{y}_1-{\tilde{y}}_2=a$ and extend the integration domains so that

$$\begin{aligned} \begin{aligned}&\int _\Omega \textrm{d}y\,\frac{1}{|x- {\tilde{y}}|^5}\frac{1}{|y_1-y_2|+1}\\&\le \frac{1}{\sqrt{2}}\int _{[-\ell ,\ell ]^2} \textrm{d}a^{\perp }\,\int _{\mathbb {R}^4}\textrm{d}b^{(1)}\textrm{d}b^{\perp }\textrm{d}a^{(1)}\\ {}&\qquad \quad \times \frac{1}{\big [\big |x_1+x_2-b\big |^2+\big |x_1-x_2-a\big |^2\big ]^{5/2}}\frac{1}{\big [|b^{(1)}+\ell |^2+|a^{\perp }|^2\big ]^{1/2}+1}\\&\le C \int _{ [-\ell ,\ell ]^2} \textrm{d}a^{\perp }\,\frac{1}{ |a^{\perp }|+1}\frac{1}{\big |x_1^\perp -x_2^\perp -a^\perp \big |} \end{aligned} \end{aligned}$$

where we dropped the term $|b^{(1)}+\ell |^2$ in the last step in order to be able to explicitly integrate over $b^{(1)}$. It is easy to see that the remaining integral is bounded by $\ln \ell$, uniformly in $x_1^\perp - x_2^\perp$. The same estimate can be applied to the other imaged charges, with the result that

$$\begin{aligned} \begin{aligned} |\text {D}_6(x)|&\le C\kappa \varepsilon ^{1/2}\ell ^{-4}+{C\kappa \varepsilon \ell ^{-3}\ln (\ell )} \end{aligned} \end{aligned}$$

In particular,

$$\begin{aligned} \begin{aligned} |\big [d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]\text {D}_6(x)|&\le C\kappa \varepsilon ^{1/2}\ell ^{-7/3}+C\kappa \varepsilon \ell ^{-4/3}\ln (\ell ) \end{aligned} \end{aligned}$$

(A.69)

We are left with considering $\text {D}_3$ and $\text {D}_4$. With the aid of (A.9) and (A.34) we can bound

$$\begin{aligned} \begin{aligned} |\text {D}_4(x)|&=\Big |\varepsilon \int _{\partial \Omega } \textrm{d}\sigma _y\,{\hat{n}}\,{\tilde{G}}_\varepsilon (x-y) g( y)\Big |\le \frac{C\varepsilon \kappa }{\ell ^3}\int _{\partial \Omega } \textrm{d}\sigma _y \frac{1}{|x-y|^4} \frac{1}{|y_1-y_2|+1} \end{aligned}\nonumber \\ \end{aligned}$$

(A.70)

It clearly suffices to consider the contribution to the boundary integral coming from $y_1^{(1)}= -\ell /2$. With the change of variables $y_1^{\perp }+ y_2^{\perp }=b^{\perp }, \, y_1^{\perp }- y_2^{\perp }=a^{\perp }$ we have, similarly as above,

$$\begin{aligned} \begin{aligned}&\int _{\partial \Omega } \textrm{d}\sigma _y\,\frac{1}{|x-y|^4}\frac{1}{|y_1-y_2|+1}\\&\quad \le \int _\mathbb {R}\textrm{d}y_2^{(1)}\int _{[-\ell ,\ell ]^2}\textrm{d}a^\perp \,\frac{1}{\big [|y_2^{(1)}+\ell /2|^2+\big | a^{\perp }\big |^2\big ]^{1/2}+1}\\&\qquad \times \int _{\mathbb {R}^2} \textrm{d}b^\perp \,\frac{1}{\Big [\big |x_1^\perp +x_2^\perp -b^\perp \big |^2+\big |x_1^\perp -x_2^\perp -a^\perp \big |^2+ 2\big |x_1^{(1)}+\ell /2\big |^2 + 2 \big |x_2^{(1)}-y_2^{(1)}\big |^2\Big ]^2}\\&\quad \le C \int _{[-\ell ,\ell ]^2}\textrm{d}a^\perp \,\frac{1}{ \big | a^{\perp }\big |+1} \frac{1}{\big |x_1^\perp -x_2^\perp -a^\perp \big |} \le C \ln \ell \end{aligned} \end{aligned}$$

and thus

$$\begin{aligned} \begin{aligned} |\big [d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]\text {D}_4(x)|&\le C\kappa \varepsilon \ell ^{-4/3}\ln (\ell ) \end{aligned} \end{aligned}$$

(A.71)

In $\text {D}_3$ we estimate the contribution proportional to $\lambda _\ell$ as

$$\begin{aligned} \begin{aligned} \lambda _\ell \int _{\partial \Omega } \textrm{d}\sigma _y\,{\tilde{G}}_\varepsilon (x-y)f( y)\le C\frac{\lambda _\ell }{\ell ^3}\int _{\partial \Omega } \textrm{d}\sigma _y \frac{1}{|x-y|^4}\le \frac{C \kappa }{\ell ^{5}} , \end{aligned} \end{aligned}$$

where we used (2.14) and (2.12). For the contribution proportional to V, we use again (2.14) to bound it as

$$\begin{aligned} \begin{aligned}&\int _{\partial \Omega } \textrm{d}\sigma _y\,{\tilde{G}}_\varepsilon (x-y)\kappa V(y)f( y)\le \frac{C}{\ell ^3}\int _{\partial \Omega } \textrm{d}\sigma _y\, \frac{\kappa V(y) }{|x-y|^4} \\ \end{aligned} \end{aligned}$$

To estimate the first term on the right-hand side, we perform the same change of variables as in $\text {D}_4$. Extending the domain of integration to $\mathbb {R}^5$ and doing the integration over $b^\perp$ we have

$$\begin{aligned} \begin{aligned}&\int _{\partial \Omega ,\, y_1^{(1)}=-\ell /2} \textrm{d}\sigma _y\,\frac{V(y)}{|x-y|^4}\\&\quad \le C \int _{\mathbb {R}^3} \textrm{d}y_2^{(1)}\textrm{d}a^{\perp } \frac{V(y_2^{(1)}+\ell /2,a^{\perp })}{\big |x_1^{\perp }-x_2^{\perp }-a^{\perp }\big |^2+ \big |x_1^{(1)} + x_2^{(1)}+\ell /2 -y_2^{(1)}\big |^2} \\&\quad \le C\frac{1}{|x_1^{(1)}+x_2^{(1)}-\ell |^2+1} \end{aligned} \end{aligned}$$

where we used again (A.56) in the last step. Hence

$$\begin{aligned} \begin{aligned} |\big [d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]\text {D}_3(x)|&\le C\kappa \ell ^{-3} \end{aligned} \end{aligned}$$

(A.72)

By combining (A.43), (A.49), (A.52), (A.58), (A.61), (A.67), (A.69), (A.71) and (A.72) we have thus shown that

$$\begin{aligned} \begin{aligned}&\big [d(\tfrac{x_1+x_2}{2})^{5/3}+1\big ]|\nabla _{x_1+x_2}g(x)|\\&\quad \le C\kappa \left( \ell ^{-3} + \varepsilon \ell ^{-4/3} \ln (\ell ) \right) \\&\qquad +C\left( \varepsilon \ell ^2 +\lambda _\ell \ell ^2 + \kappa \right) \sup _{y\in \Omega }\big |\big [d(\tfrac{y_1+y_2}{2})^{5/3}+1\big ]\nabla _{y_1+y_2}g(y)\big |. \end{aligned} \end{aligned}$$

We choose $\varepsilon =c\ell ^{-2}$ with small enough c so that the factor $C( \varepsilon \ell ^2 + \lambda _\ell \ell ^2 + \kappa )$ is smaller than one for large $\ell$ and small $\kappa$, concluding the proof of (2.18). $\square$

Proof of Proposition 2.2

From (2.13) it follows that

$$\begin{aligned} \begin{aligned} \int _{\Lambda _1\times \Lambda _1} \textrm{d}x\textrm{d}y\,\Big [|\nabla _xw_\ell (x, y)|^2+|\nabla _yw_\ell (x, y)|^2\Big ]\le \frac{C\kappa }{\ell }, \end{aligned} \end{aligned}$$

(A.73)

estimate (2.14) implies

$$\begin{aligned} \begin{aligned} |w_\ell (x, y)|\le C \end{aligned} \end{aligned}$$

(A.74)

and from (2.15) it follows that

$$\begin{aligned} \begin{aligned} \int _{\Lambda _1\times \Lambda _1} \textrm{d}x\textrm{d}y\,\big |w_\ell (x, y)\big |^2\le \frac{C\kappa ^2}{\ell ^2}, \end{aligned} \end{aligned}$$

(A.75)

while (2.16) shows that

$$\begin{aligned} \begin{aligned} \int _{\Lambda _1\times \Lambda _1} \textrm{d}x\textrm{d}y\,|w_\ell (x,y)|\le \frac{C\kappa }{\ell }. \end{aligned} \end{aligned}$$

(A.76)

By Eq. (2.24) and bounds (A.75), (A.73) we find

$$\begin{aligned} \begin{aligned} \int _{\Lambda _1\times \Lambda _1} \textrm{d}x\textrm{d}y\,\big |\mu (x, y)\big |^2&\le C\kappa \frac{n^2}{\ell ^2}\\ \int _{\Lambda _1\times \Lambda _1} \textrm{d}x\textrm{d}y\,\Big [\big |\nabla _x\mu (x, y)\big |^2+\big |\nabla _y\mu (x, y)\big |^2\Big ]&\le C\kappa \frac{n^2}{\ell } \end{aligned} \end{aligned}$$

which imply (2.27) and (2.28). By (2.14) we have

$$\begin{aligned} |\eta (x,y)|\le n|w_\ell (x,y)|+|\mu (x,y)|\le C n \end{aligned}$$

(A.77)

which proves (2.29). Estimate (2.30) follows from (2.17). Point ii) follows from (2.30).

We consider now point iii). From the definition of r, we find

$$\begin{aligned} \begin{aligned} r(x,y)&= \sum _{n=1}^\infty \frac{1}{(2n+1)!} \eta ^{2n+1} (x,y) \\&= \sum _{n=1}^\infty \frac{1}{(2n+1)!} \int \textrm{d}z \textrm{d}w \, \eta (x,z)\,\eta ^{2n-1}(z,w)\, \eta (w,y); \end{aligned} \end{aligned}$$

(A.78)

using (2.27), which implies $\Vert \eta \Vert _2\le C$, we arrive at

$$\begin{aligned} |r (x,y)|\le & {} \sum _{n=1}^\infty \frac{1}{(2n+1)!} \left[ \int \textrm{d}w \textrm{d}z |\eta (x,z)|^2 |\eta (w,y)|^2 \right] ^{1/2} \left[ \int \textrm{d}w \textrm{d}z \, |\eta ^{2n-1}(z,w)|^2 \right] ^{1/2} \nonumber \\\le & {} \sum _{n=1}^\infty \frac{1}{(2n+1)!} \Vert \eta \Vert _2^{2n-1} \, \Vert \eta (x,\cdot ) \Vert _2 \Vert \eta (\cdot ,y) \Vert _2 \le C \Vert \eta \Vert _2 \Vert \eta (x,\cdot ) \Vert _2 \Vert \eta (\cdot ,y) \Vert _2\nonumber \\ \end{aligned}$$

(A.79)

for every $x,y \in \Lambda _1$. The bound for p can be proven analogously. This proves (2.33) and consequently (2.34) and (2.32). $\square$