\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 83, pp. 1--36.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/83\hfil Homogenization of a diffusion perturbed by a vector field] {On homogenization of a diffusion perturbed by a periodic reflection invariant vector field} \author[J. G. Conlon\hfil EJDE-2008/83\hfilneg] {Joseph G. Conlon} \address{Joseph G. Conlon \newline Department of Mathematics\\ University of Michigan\\ Ann Arbor, MI 48109-1109, USA} \email{conlon@umich.edu} \thanks{Submitted March 26, 2007. Published May 30, 2008.} \subjclass[2000]{35R60, 60H30, 60J60} \keywords{PDE with periodic coefficients; homogenization} \begin{abstract} In this paper the author studies the problem of the homogenization of a diffusion perturbed by a periodic reflection invariant vector field. The vector field is assumed to have fixed direction but varying amplitude. The existence of a homogenized limit is proven and formulas for the effective diffusion constant are given. In dimension $d=1$ the effective diffusion constant is always less than the constant for the pure diffusion. In $d>1$ this property no longer holds in general. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \section{Introduction} We consider the problem of the homogenization of a diffusion perturbed by a reflection invariant vector field. The general set up we have in mind is to understand the limit as $\varepsilon \to 0$ of the solutions $u_\varepsilon$ to an elliptic equation, $$\label{A1} \begin{gathered} - \frac 1{2d} \Delta u_\varepsilon(x,\omega) - 2b_\varepsilon(x,\omega)\partial u_\varepsilon (x,\omega)/ \partial x_1 + u_\varepsilon(x,\omega) = f(x), \\ x = (x_1,\dots,x_d) \in \mathbb{R}^d, \quad \omega \in \Omega. \end{gathered}$$ Here the function $f : \mathbb{R}^d \to \mathbb{R}$ is smooth of compact support and $\Omega$ is a probability space. For simplicity we have assumed that the vector field is always in the $x_1$ direction and hence can be described by the scalar function $b_\varepsilon$. As $\varepsilon \to 0$ the field becomes rapidly oscillatory and therefore one might expect that $u_\varepsilon(x,\omega)$ converges with probability 1 as $\varepsilon \to 0$ to a homogenized limit $u(x)$ which is the solution to a constant coefficient elliptic equation, $$\label{B1} -q(b)\frac{\partial^2 u}{\partial x^2_1} - \sum^d_{j=2} \frac 1{2d} \frac{\partial^2 u}{\partial x^2_j} + u(x) = f(x), \quad x \in \mathbb{R}^d.$$ The effect of the rapidly oscillating vector field $b_\varepsilon$ is contained in the coefficient $q(b)$ in \eqref{B1}. In order for a limit $u(x)$ satisfying \eqref{B1} to exist it is necessary to make assumptions concerning the rapidly oscillating field $b_\varepsilon$. These are primarily that the distribution functions of the variables $b_\varepsilon(x,\cdot)$, $x \in \mathbb{R}^d$, are translation and reflection invariant. To be specific, we assume that there are translation operators $\tau_x : \Omega \to \Omega$, $x \in \mathbb{R}^d$, which are measure preserving and satisfy the group properties $\tau_x \tau_y = \tau_{x+y}$, $x,y \in \mathbb{R}^d, \ \tau_0=$identity. Suppose $b:\Omega\to\mathbb{R}$ is a bounded function. We then set $b_\varepsilon(x,\omega) = \varepsilon^{-1}b(\tau_{x/\varepsilon} \; \omega)$, $x\in \mathbb{R}^d, \; \omega \in \Omega, \; \varepsilon > 0$. Such a $b_\varepsilon$ has translation invariant distribution functions and is rapidly oscillating as $\varepsilon \to 0$. For $b_\varepsilon$ to satisfy reflection invariance we let $R : \mathbb{R}^d \to \mathbb{R}^d$ be the reflection operator $R(x_1, \dots, x_d) = (-x_1,x_2,\dots,x_d)$, $x = (x_1,\dots,x_d) \in \mathbb{R}^d$. We then require $b : \Omega \to \mathbb{R}$ to satisfy the identities, $$\label{C1} \Big< \prod^n_{i=1} b(\tau_{x_i} \; \cdot) \Big> = (-1)^n \Big< \prod^n_{i=1} b(\tau_{Rx_i} \; \cdot) \Big>, \quad x_i \in \mathbb{R}^d,\; 1\le i \le n, \; n \ge 1,$$ where $\langle \cdot \rangle$ denotes expectation on $\Omega$. Evidently \eqref{C1} implies that $\langle b(\cdot)\rangle = 0$, so the vector field has no net drift. A concrete example of an $\Omega$ and a $b : \Omega \to \mathbb{R}$ which satisfies \eqref{C1} is given by taking $\Omega$ to be a torus, $\Omega = \prod^d_{i=1} [0, L_i]$ with periodic boundary conditions and uniform measure. The operators $\tau_x : \Omega \to \Omega$, $x \in \mathbb{R}^d$, are just translation on $\Omega$ and reflection invariance of \eqref{C1} is guaranteed by the condition, $$\label{D1} b(x_1,x_2,\dots,x_d) = - b(L_1 - x_1,x_2,\dots,x_d), \quad x = (x_1,\dots,x_d) \in \Omega.$$ We shall show that for a discrete version of a periodic $\Omega$ with $b$ satisfying (\ref{D1}) a homogenized limit exists with $q(b)$ satisfying $0 < q(b) < \infty$. For $d=1$ one has $q(b) \le q(0) = 1/2$. For $d > 1$ it is no longer the case that $q(b) \le q(0) =1/2d$ in general although this does hold for $L_1$ sufficiently small. One might wish to understand this difference between $d=1$ and $d > 1$ by observing that only in $d > 1$ can one construct nontrivial divergence free vector fields. The homogenized limit of diffusion perturbed by a divergence free vector field necessarily yields an effective diffusion constant which is larger than the constant for the pure diffusion \cite{fp}. The homogenization problem considered here appears to have only been studied in the case where $\Omega$ is an infinite space for which the variables $b(\tau_x \; \cdot), \ x \in \mathbb{R}$, are uncorrelated on a scale larger than $O(1)$. The problem was introduced by Sinai \cite{s} in a discrete setting. He proved that in dimension $d=1$ a scaling limit of the random walk corresponding to a finite difference approximation to (\ref{A1}) exists with probability 1 in $\Omega$. The limiting process is strongly subdiffusive. In a subsequent paper Kesten \cite{k} obtained an explicit formula for the distribution of the scaling limit. For dimension $d \ge 3$ Fisher \cite{f} and Derrida-L\"{u}ck \cite{dl} predicted that a homogenized limit exists as in \eqref{B1} with $0 < q(b) < \infty$. This was proved for sufficiently small $b$ by Bricmont-Kupiainen \cite{bk} and Sznitman-Zeitouni \cite{sz} using a very difficult induction argument. A formal perturbation expansion for $q(b)$ was obtained in \cite{c2002,c2005} where it was shown that each term of the expansion is finite if $d\ge 3$. One does not expect the series to converge however. For $d=1,2$ there are individual terms in the perturbation expansion which diverge. A main difficulty in the homogenization problem (\ref{A1}), \eqref{B1} is that when $\Omega$ is infinite, good a-priori estimates on the solution to (\ref{A1}) do not hold for all configurations of $b(\cdot)$. In contrast such estimates do hold for divergence form equations with zero drift. The proof of homogenization in these cases is therefore considerably simpler than for the problem (\ref{A1}), \eqref{B1}. The first proofs of homogenization for divergence form equations were obtained by Kozlov \cite{ko} and Papanicolaou-Varadhan \cite{pv1} in the continuous case. K\"{u}nneman \cite{ku} proved a corresponding result for the discrete case. For non-divergence form equations with zero drift the first proofs in the continuous case were given by Papanicolaou-Varadhan \cite{pv2} and Zhikov-Sirazhudinov \cite{zs}. Lawler \cite{l} and Anshelevich et al \cite{aks} proved homogenization for a discrete version. See the books of Bolthausen-Sznitman \cite{bs} for an account of the theory in a discrete setting and of Zhikov et al \cite{zko} for the continuous case. In this paper we shall be concerned with a discrete version of the homogenization problem described by (\ref{A1}), \eqref{B1}, \eqref{C1}. Thus the probability space $\Omega$ is acted upon by translation operators $\tau_x : \Omega \to \Omega$ where now $x \in \mathbb{Z}^d$, the integer lattice in $\mathbb{R}^d$, and satisfy the group properties $\tau_x\; \tau_y = \tau_{x+y}$, $\tau_0 =$ identity. For $i=1,\dots,d$ let ${\bf e}_i \in \mathbb{Z}^d$ be the element with entry 1 in the ith position and 0 in the other positions. the discrete equation corresponding to (\ref{A1}) is given by \label{E1} \begin{aligned} &u_\varepsilon(x,\omega) - \sum^d_{i=1} \frac 1{2d} \left[ u_\varepsilon(x + \varepsilon{\bf e}_i, \omega) + u_\varepsilon(x - \varepsilon{\bf e}_i, \omega)\right] \\ &- b(\tau_{x/\varepsilon}\omega) \left[ u_\varepsilon(x + \varepsilon{\bf e}_1, \omega) - u_\varepsilon(x - \varepsilon{\bf e}_1, \omega)\right] + \varepsilon^2 \; u_\varepsilon(x,\omega) \\ &= \varepsilon^2 f(x), \quad x \in \mathbb{Z}^d = \varepsilon \; \mathbb{Z}^d, \quad \omega \in \Omega. \end{aligned} We assume that $b : \Omega \to \mathbb{R}$ satisfies $\sup_\omega \ |b(\omega)| < 1/2d$, in which case \eqref{E1} is an equation for the expectation value of a function of an asymmetric random walk. Hence \eqref{E1} has a unique bounded solution. We assume that $b$ satisfies the reflection invariant condition \eqref{C1} (with $x_i \in \mathbb{Z}^d, 1 \le i \le n$, now). We also assume that $\Omega$ is finite, in which case one can see (Lemma \ref{lem2.4}) that $\Omega$ is isomorphic to the integer points on a torus and $b$ has the reflection invariance property (\ref{D1}). In ${\S 2}$ we prove the following theorem (with $\lfloor \cdot \rfloor$ denoting the integer part): \begin{theorem} \label{thm1.1} Assume $\Omega$ is a finite probability space and the translation operators $\tau_x : \Omega \to \Omega$ are ergodic, $x \in \mathbb{Z}^d$. Then there exists $q(b), \ 0 < q(b) < \infty$ such that with $u_\varepsilon$ the solution to \eqref{E1} and $u$ the solution to \eqref{B1}, $\lim_{\varepsilon \to 0} \ \sup_{x\in \mathbb{R}^d, \omega \in \Omega} \ |u_\varepsilon(\varepsilon \lfloor x/ \varepsilon \rfloor, \omega) - u(x)| = 0.$ \end{theorem} One should note here that Theorem \ref{thm1.1} may be obtained from a homogenization theorem for divergence form operators. This follows from the representation $$\label{Z1} \mathcal{L}=\Delta+b.\nabla=(1/\phi^*) \mathop{\rm div}(\phi^*(I+H)\nabla\cdot),$$ where $\phi^*$ is the invariant measure for $\mathcal{L}$ on $\Omega$ and $H$ is an antisymmetric matrix. This follows since \eqref{C1} implies that $=0$. Suppose now that $\Omega$ consists of the integer points on the torus $\prod^d_{i=1}[0, L_i] \subset \mathbb{R}^d$ with periodic boundary conditions. The reflection invariant condition corresponding to (\ref{D1}) is given by $$\label{F1} b(x_1,x_2,\dots,x_d) = - b(L_1 - 1-x_1,x_2,\dots,x_d), \ \ x = (x_1,x_2, \dots,x_d) \in \Omega.$$ For $b$ satisfying (\ref{F1}) we prove in $\S$2, $\S$3 the following results concerning the coefficient $q(b)$ of the homogenized equation \eqref{B1}: \begin{theorem} \label{thm1.2} \begin{itemize} \item [(a)] For $d=1$ one has $q(b) \le 1/2$. \item[(b)] If $d \ge 1$ and $L_1 = 2$ one has $q(b) \le 1/2d$. \item[(c)] If $d=2$ and $L_1=4$ one has $q(b) \le 1/4$. \item[(d)] If $d \ge 2$ and $L_1 \ge 6$ is even then there exists $b$ with $q(b) > 1/2d$. \end{itemize} \end{theorem} The proofs of (a), (b), (c) are given in $\S 3$ and are based on applications of the Schwarz inequality to certain representations of $q(b)$ obtained in $\S 2$. The proof of (c) is quite lengthy. Basically one shows that $q(b) \le 1/4$ provided a quadratic form corresponding to a fourth order difference operator is non-negative definite. The non-negative definiteness of the quadratic form depends crucially on actual numerical values for a Green's function associated with standard random walk on the integers. The proof of (d) is given in $\S 2$. One observes that perturbation theory yields $q(b) = 1/2d + O(|b|^2)$ and that the term $O(|b|^2)$ can be positive. In the proof of Theorem \ref{thm1.2} we use a representation for $q(b)$ in terms of invariant measures for random walk on $\Omega$ with drift $b$. Let $\Omega_{d-1}$ consist of the integer points on the $d-1$ dimensional torus $\prod^d_{i=2} [0, L_i] \subset \mathbb{R}^{d-1}$ with periodic boundary conditions. Setting $L_1 = 2L$ with $L$ an integer we define $\hat \Omega$ by $\hat \Omega = \left\{ (n,y) : 1 \le n \le L, \ y \in \Omega_{d-1} \right\},$ whence $\Omega$ is the double of $\hat \Omega$. Observe that the boundary of $\partial \hat \Omega$ is given by $\partial \hat \Omega = \left\{ (1,y), \ (L,y) : y \in \Omega_{d-1} \right\}.$ Let $\varphi^*$ be the invariant measure for random walk on $\hat \Omega$ with drift $b(\cdot)$ in the ${\bf e}_1$ direction and reflecting boundary conditions on $\partial \hat \Omega$. The normalization of $\varphi^*$ is chosen so that $<\varphi^*>_{\hat\Omega}=1$ where $<\cdot>_{\hat\Omega}$ is the uniform probability measure on $\hat\Omega$. We define $\psi : \Omega_{d-1} \to \mathbb{R}$ by $\psi(y) = [1/2d - b(1,y)] \ \varphi^*(1,y), \ y \in \Omega_{d-1},$ Then $q(b)$ is given by the formula, $$\label{G1} q(b) = 8d\big< \psi \left[ -\Delta_{d-1} + 4\right]^{-1} \psi_R \big>_{\Omega_{d-1}},$$ where $\psi_R$ is defined exactly as $\psi$ but with $b$ replaced by $-b$. In (\ref{G1}) the expectation $\langle \cdot \rangle_{\Omega_{d-1}}$ is the uniform probability measure on $\Omega_{d-1}$ and $\Delta_{d-1}$ is the $d-1$ dimensional finite difference Laplacian on functions with domain $\Omega_{d-1}$. The general formula (\ref{G1}) is proven in $\S 4$. \section{Proof of Theorem \ref{thm1.1}} We follow the method introduced in \cite{cn} to obtain homogenized limits. Thus in \eqref{E1} we put $u_\varepsilon(x,\omega) = v_\varepsilon(x, \tau_{x/\varepsilon} \; \omega)$ whence \eqref{E1} becomes \label{A2} \begin{aligned} &v_\varepsilon(x,\omega) - \sum^{2d}_{i=1} \frac 1{2d} \left[ v_\varepsilon(x + \varepsilon{\bf e}_i , \tau_{{\bf e}_i }\; \omega) + v_\varepsilon(x - \varepsilon{\bf e}_i , \tau_{{-\bf e}_i }\; \omega) \right] \\ &- b(\omega) \left[ v_\varepsilon(x + \varepsilon{\bf e}_1 , \tau_{{\bf e}_1 }\; \omega) - v_\varepsilon(x - \varepsilon{\bf e}_1 , \tau_{{-\bf e}_1 }\; \omega) \right] \\ + \varepsilon^2 \ v_\varepsilon(x, \omega) \\ &= \varepsilon^2 \ f(x), \quad x \in \mathbb{Z}^d_\varepsilon=\varepsilon\mathbb{Z}^d, \quad \omega \in \Omega. \end{aligned} Next we wish to take the Fourier transform of (\ref{A2}). To show that this is legitimate we first show that the solution $u_\varepsilon(x,\omega)$ of \eqref{E1} decreases exponentially as $x \to \infty$. \begin{lemma} \label{lem2.1} Suppose $f : \mathbb{Z}^d_\varepsilon \to \mathbb{R}$ has finite support in the set $\{ x = (x_1,\dots,x_d) \in \mathbb{Z}^d_\varepsilon : |x| < R \}.$ Let $u_\varepsilon(x,\omega)$ be a bounded solution to $\eqref{E1}$. Then there are constants $C,K(\varepsilon) > 0$ such that $$\label{B2} |u_\varepsilon(x,\omega)| \le C \exp [K(\varepsilon)(R - |x|) ] \|f\|_\infty, \ \ x \in \mathbb{Z}^d_\varepsilon.$$ \end{lemma} \begin{proof} We write $u_\varepsilon(x,\omega) = e^{-kx_1} \; u_{\varepsilon,k}(x,\omega)$. Then from \eqref{E1} the function $u_{\varepsilon,k}$ satisfies \label{E2} \begin{aligned} & \frac 1{2d} \sum^{d}_{i=2} \left[ 2u_{\varepsilon,k}(x,\omega) - u_{\varepsilon,k}(x + \varepsilon{\bf e}_i ,\omega) - u_{\varepsilon,k}(x - \varepsilon{\bf e}_i , \omega)\right]/\varepsilon^2 \\ &+ e^{-k\varepsilon} \big[ \frac 1{2d} + b(\tau_{x/\varepsilon} \omega)\big] \left\{ u_{\varepsilon,k}(x,\omega) - u_{\varepsilon,k}(x + \varepsilon{\bf e}_1 , \omega)\right\}\big/\varepsilon^2 \\ &+ e^{k\varepsilon} \big[ \frac 1{2d} - b(\tau_{x/\varepsilon} \omega)\big] \left\{ u_{\varepsilon,k}(x,\omega) - u_{\varepsilon,k}(x - \varepsilon{\bf e}_1 , \omega)\right\}\big/\varepsilon^2 \\ &+ \left\{ 1 - [\cosh k\varepsilon - 1]/d\varepsilon^2 + 2b(\tau_{x/\varepsilon}\;\omega) \sinh \; k\varepsilon/\varepsilon^2\right\} u_{\varepsilon,k}(x,\omega) \\ &= e^{kx_1} \ f(x), \ \ \ \ x \in \mathbb{Z}^d_\varepsilon. \end{aligned} We may assume without loss of generality (wlog) that $f$ is nonnegative, whence $u_{\varepsilon,k}$ is also nonnegative. Suppose $u_{\varepsilon,k}$ attains its maximum at a point $\bar x \in \mathbb{Z}^d_\varepsilon$. Then we have that \label{C2} \begin{aligned} &\left\{ 1 - [\cosh k\varepsilon - 1]/d\varepsilon^2 + 2b(\tau_{\bar x/\varepsilon} \;\omega)\sinh \; k\varepsilon/\varepsilon^2\right\} u_{\varepsilon,k}(\bar x,\omega) \\ &\le \exp \left[ k(\bar x \cdot {\bf e}_1) \right] \|f\|_\infty, \quad \ |\bar x| < R. \end{aligned} Observe that the coefficient of $u_{\varepsilon,k}(\bar x,\omega)$ on the LHS of (\ref{C2}) is positive if $k$ is sufficiently small. Hence it follows that $$\label{D2} u_\varepsilon(x,\omega) \le C \exp \left[ k(\bar x - x) \cdot {\bf e}_1 \right] \|f\|_\infty, \ x \in \mathbb{Z}^d_\varepsilon.$$ We need to show that the point $\bar x$ exists for sufficiently small $k$. To see this assume for contradiction that it does not exist. Then (\ref{E2}) implies that $a_N=\sup_{|x| \le N} u_{\varepsilon,k}(x,\omega)$ grows exponentially in $N$ as $N \to \infty$. This follows since we see from (\ref{E2}) that if $N>R$ then there is a positive $\alpha$ such that $a_{N+1}\ge (1+\alpha)a_N$. The rate of exponential growth remains bounded away from 0 as $k \to 0$. Hence, taking $k$ sufficiently small, we conclude that the function $u_\varepsilon$ is unbounded, contradicting our assumption on $u_\varepsilon$. The inequality (\ref{B2}) now follows from (\ref{C2}), (\ref{D2}) on generalizing to all directions ${\bf e}_j, 1 \le j \le d$. \end{proof} For $\xi \in [-\pi/\varepsilon, \; \pi/\varepsilon]^d$ we put $$\label{F2} \hat v_\varepsilon(\xi, \omega) = \int_{\mathbb{Z}^d_\varepsilon} v_\varepsilon(x,\omega)e^{ix\cdot \xi} dx = \sum_{x\in \mathbb{Z}^d_\varepsilon}\varepsilon^d v_\varepsilon(x,\omega)e^{ix\cdot\xi}.$$ Then from (\ref{A2}) we have that \label{G2} \begin{aligned} &\hat v_\varepsilon(\xi, \omega) - \sum^d_{j=1} \frac {1}{ 2d} \left[ e^{-i\varepsilon\xi_j} \hat v_\varepsilon(\xi, \tau_{{\bf e}_j} \omega) + e^{i\varepsilon\xi_j} \hat v_\varepsilon(\xi, \tau_{-{\bf e}_j} \omega)\right] \\ &- b(\omega) \left[ e^{-i\varepsilon\xi_1} \hat v_\varepsilon(\xi, \tau_{{\bf e}_1} \omega) - e^{i\varepsilon\xi_1} \hat v_\varepsilon(\xi, \tau_{{-\bf e}_1} \omega)\right] + \varepsilon^2 \; \hat v_\varepsilon(\xi, \omega) \\ &= \varepsilon^2 \hat f_\varepsilon(\xi), \quad \xi \in [-\pi/\varepsilon, \pi/\varepsilon]^d, \; \omega \in \Omega, \end{aligned} where $\hat f_\varepsilon$ denotes the discrete Fourier transform (\ref{F2}) of $f$. To solve (\ref{G2}) we define for $\zeta \in [-\pi, \pi]^d$ an operator $\mathcal{L}_\zeta$ on functions $\Psi : \Omega \to \mathbb{C}$ defined by \label{H2} \begin{aligned} \mathcal{L}_\zeta \Psi(\omega) &= \Psi(\omega) - \sum^d_{j=1} \frac 1{2d} \left[ e^{-i\zeta_j} \Psi(\tau_{{\bf e}_j}\omega) + e^{i\zeta_j} \Psi(\tau_{-{\bf e}_j}\omega) \right] \\ &\quad - b(\omega) \left[ e^{-i\zeta_1} \Psi(\tau_{{\bf e}_1}\omega) - e^{i\zeta_1} \Psi(\tau_{-{\bf e}_1}\omega) \right]. \end{aligned} Next we define an operator $T_{\eta,\zeta}, \ \eta > 0$, $\zeta \in [-\pi, \pi]^d$ on $L^\infty(\Omega)$ by $$\label{I2} T_{\eta,\zeta} \; \varphi(\omega) = \eta \left[ \mathcal{L}_\zeta + \eta\right]^{-1} \varphi(\omega), \quad \omega \in \Omega.$$ It is easy to see that $T_{\eta,\zeta}$ is a bounded operator on $L^\infty(\Omega)$ with norm at most 1. In fact the RHS of (\ref{I2}) is the expectation for a continuous time random walk on $\Omega \times \mathbb{Z}^d$. The walk is defined as follows: \begin{itemize} \item[(a)] The waiting time at $(\omega, x) \in \Omega \times \mathbb{Z}^d$ is exponential with parameter 1. \item[(b)] For $j=2,\dots,d$ the particle jumps from $(\omega, x)$ to $(\tau_{{\bf e}_j}\omega, x+{\bf e}_j)$ with probability $1/2d$ and to $(\tau_{-{\bf e}_j}\omega, x-{\bf e}_j)$ with probability $1/2d$. \item[(c)] The particle jumps from $(\omega, x)$ to $(\tau_{{\bf e}_1}\omega, x+{\bf e}_1)$ with probability $1/2d + b(\omega)$, and to $(\tau_{-{\bf e}_1}\omega, x-{\bf e}_1)$ with probability $1/2d - b(\omega)$. \end{itemize} If $[\omega(t), X(t)] \in \Omega \times \mathbb{Z}^d$ is the position of the walk at time $t$ then $$\label{J2} T_{\eta,\zeta} \varphi(\omega) = \eta E \Big[ \int^\infty_0 dt \ e^{-\eta t} \varphi(\omega(t)) \exp [-i X(t) \cdot \zeta] \Big| \omega(0) = \omega, \ X(0) = 0 \Big].$$ It is clear from the representation (\ref{J2}) that $\| T_{\eta,\zeta}\|_\infty \le 1$. We conclude from this that (\ref{G2}) is solvable with solution given by $$\label{K2} \hat v_\varepsilon(\xi,\omega) = \hat f_\varepsilon(\xi)\; T_{\varepsilon^2,\varepsilon \xi}(1)(\omega), \ \omega \in \Omega, \ \ \xi \in [-\pi/\varepsilon, \pi/\varepsilon]^d.$$ To obtain the homogenization theorem we need then to obtain the limit of the RHS of (\ref{K2}) as $\varepsilon \to 0$. To facilitate this we observe from (\ref{H2}) that $$\label{V2} [\mathcal{L}_\zeta + \eta]1 = 1 + \eta - \frac 1 d \sum^d_{j=1} \cos \zeta_j + 2ib(\omega) \sin \zeta_1.$$ It follows therefore that $$\label{L2} T_{\eta,\zeta}(1)(\omega) = \eta \Big/ \Big[ 1 + \eta - \frac 1 d \sum^d_{j=1} \cos \zeta_j \Big] - \Big\{ 2i \sin \zeta_1 \Big/ \Big[ 1 + \eta - \frac 1 d \sum^d_{j=1} \cos \zeta_j \Big] \Big\} T_{\eta,\zeta}b(\omega).$$ Setting $\eta = \varepsilon^2, \ \zeta = \varepsilon \; \xi$ for some fixed $\xi \in \mathbb{R}^d$ we see from (\ref{L2}) that $$\label{M2} \lim_{\varepsilon \to 0}T_{\varepsilon^2,\varepsilon\xi}(1)(\omega) = 1\Big/ \Big[ 1 + \frac {1}{2d} \sum^d_{j=1} \xi^2_j \Big] - \Big\{ 2i\xi_1 \Big/ \Big[ 1 + \frac 1 {2d} \sum^d_{j=1} \xi^2_j \Big] \Big\} \lim_{\varepsilon \to 0}\varepsilon^{-1} T_{\varepsilon^2,\varepsilon\xi}b(\omega).$$ We shall show that under the assumption of \eqref{C1} the limit on the RHS of (\ref{M2}) exists. To do this we define two subspaces of the space $L^\infty(\Omega)$. We define $L^\infty_R(\Omega)$ as all functions $\Phi \in L^\infty(\Omega)$ such that $\Big< \Phi(\tau_x \cdot) \prod^n_{i=1} b(\tau_{x_i} \cdot) \Big> = (-1)^{n+1} \Big< \Phi(\tau_{Rx}\cdot) \prod^n_{i=1} b(\tau_{Rx_i} \cdot) \Big>,$ with $x, x_i \in \mathbb{Z}^d$, $1 \le i \le n$, $n=0,1,2,\dots$. Evidently \eqref{C1} implies that $b \in L^\infty_R(\Omega)$. We also see that if $\Phi \in L^\infty_R(\Omega)$ then $\Phi(\tau_{\bf e_j }\cdot)$ and $\Phi(\tau_{-{\bf e_j}} \cdot)$ are also in $L^\infty_R(\Omega), j=2,\dots,d$. For $j=1$ one has that $\Phi \in L^\infty_R(\Omega)$ implies both $[\Phi(\tau_{\bf e_1} \cdot)+ \Phi(\tau_{-{\bf e_1}} \cdot)]$ and $b(\cdot) [\Phi(\tau_{\bf e_1} \cdot)- \Phi(\tau_{-{\bf e_1}} \cdot)]$ are in $L^\infty_R(\Omega)$. The space $\hat L^\infty_R(\Omega)$ is defined similarly as all functions $\Phi \in L^\infty(\Omega)$ such that $\Big< \Phi(\tau_x \cdot) \prod^n_{i=1} b(\tau_{x_i} \cdot) \Big> = (-1)^{n} \Big< \Phi(\tau_{Rx} \cdot) \prod^n_{i=1} b(\tau_{Rx_i} \cdot) \Big>,$ with $x, x_i \in \mathbb{Z}^d$, $1 \le i \le n$, $n = 0,1,2,\dots$. >From \eqref{C1} we see that the function $\Phi \equiv 1$ is in $\hat L^\infty_R(\Omega)$. As for the space $L^\infty_R(\Omega)$, if $\Phi \in \hat L^\infty_R(\Omega)$ then $\Phi(\tau_{\bf e_j} \cdot)$ and $\Phi(\tau_{-{\bf e}_j} \cdot)$ are also in $\hat L^\infty_R(\Omega), j=2,\dots,d$. Similarly both $[\Phi(\tau_{\bf e_1} \cdot)+ \Phi(\tau_{-{\bf e}_1} \cdot)]$ and $b(\cdot) [\Phi(\tau_{\bf e_1} \cdot)- \Phi(\tau_{-{\bf e}_1} \cdot)]$ are in $\hat L^\infty_R(\Omega)$. We note that the mapping $\Phi(\cdot) \to b(\cdot) \Phi(\cdot)$ maps $L^\infty_R(\Omega)$ into $\hat L^\infty_R(\Omega)$ and vice-versa. We denote the operator $\mathcal{L}_\zeta$ of (\ref{H2}) for $\zeta = 0$ by $\mathcal{L}$. It is evident that $\mathcal{L}$ is the generator of a random walk on $\Omega$. Hence the kernel of the operator $\mathcal{L}$ is just the constant function. Furthermore $\mathcal{L}$ leaves the space $L^\infty_R(\Omega)$ invariant. Since the constant function is not in $L^\infty_R(\Omega)$ it follows that there is a unique function $\varphi \in L^\infty_R(\Omega)$ such that $$\label{N2} \mathcal{L}\varphi = b.$$ Let $\varphi^*$ be the invariant measure for the walk on $\Omega$ generated by $\mathcal{L}$. Thus $\varphi^* > 0$, $$\label{O2} \mathcal{L}^* \varphi^* = 0, \quad \langle \varphi^* \rangle = 1,$$ where $\mathcal{L}^*$ is the adjoint of $\mathcal{L}$. Since $\mathcal{L}$ is non singular on the space $L^\infty_R(\Omega)$ it follows that $\varphi^*$ is orthogonal to $L^\infty_R(\Omega)$. We can also see that $\varphi^* \in \hat L^\infty_R(\Omega)$. One simply notes that both $\mathcal{L}$ and $\mathcal{L}^*$ leave the space $\hat L^\infty_R(\Omega)$ invariant and that the constant function is in $\hat L^\infty_R(\Omega)$. We obtain the limit on the RHS of (\ref{M2}) in terms of the functions $\varphi, \varphi^*$ defined by \eqref{N2}, \eqref{O2}. \begin{lemma} \label{lem2.2} Let $\psi \in L^\infty(\Omega)$ be defined by $$\label{W2} \psi(\cdot) = \big\{ \frac 1{2d} + b(\cdot) \big\} \varphi(\tau_{{\bf e}_1} \cdot) - \big\{ \frac 1{2d} - b(\cdot) \big\} \varphi(\tau_{-{\bf e}_1} \cdot),$$ where $\varphi$ is given by \eqref{N2}. Then if $\varphi^*$ is as in \eqref{O2} there is the limit, $$\label{X2} \lim_{\varepsilon \to 0}\varepsilon^{-1} T_{\varepsilon^2,\varepsilon\xi}b(\omega) = -i \xi_1 <\varphi^*\psi>\Big/ \Big[ 1 +\frac 1 {2d} \sum^d_{j=1} \xi^2_j + 2\xi^2_1 <\varphi^*\psi> \Big],$$ for all $\omega \in \Omega$, provided $\xi_1$ is sufficiently small. \end{lemma} \begin{proof} For $\eta > 0, \ \zeta \in [-\pi, \pi]^d$, let $\varphi(\eta,\zeta)$ be the unique solution to the equation $$\label{P2} [\mathcal{L}_\zeta + \eta] \varphi(\eta, \zeta) = b.$$ It is clear then that $$\label{Q2} \varepsilon^{-1} T_{\varepsilon^2,\varepsilon\xi} b = \varepsilon \varphi(\varepsilon^2, \varepsilon \xi).$$ We define operators $A_\zeta, B_\zeta$ by $A_\zeta = [\mathcal{L}_\zeta + \mathcal{L}_{R\zeta}]/2$, $B_\zeta = [\mathcal{L}_\zeta - \mathcal{L}_{R\zeta}]/2$, where $R(\zeta_1, \dots, \zeta_d) = (-\zeta_1, \zeta_2,\dots,\zeta_d)$. The operator $A_\zeta$ leaves the spaces $L^\infty_R(\Omega), \hat L^\infty_R(\Omega)$ invariant whereas $B_\zeta$ takes $L^\infty_R(\Omega)$ into $\hat L^\infty_R(\Omega)$ and vice versa. Equation (\ref{P2}) is now equivalent to $$\label{T2} [(A_\zeta + \eta) + B_\zeta] \ \varphi(\eta,\zeta) = b \ ,$$ and we may write the solution of this formally as a power series, $$\label{R2} \varphi(\eta,\zeta) = \sum^\infty_{n=0} \left\{ -(A_\zeta + \eta)^{-1} B_\zeta\right\}^n \ (A_\zeta + \eta)^{-1} \ b.$$ The operators $B_\zeta$, $(A_\zeta + \eta)^{-1}$ on $L^\infty(\Omega)$ have norms satisfying $\| B_\zeta\| \le C_2|\zeta_1|$, $\| (A_\zeta + \eta)^{-1} \| \le 1/\eta$, for some constant $C_2$. Since $A_0 = \mathcal{L}$ is invertible on $L^\infty_R(\Omega)$ it follows that for $(\eta, \zeta)$ sufficiently small the operator norm of $(A_\zeta + \eta)^{-1}$ acting on $L^\infty_R(\Omega)$ satisfies $\| (A_\zeta + \eta)^{-1} \| \le C_1$ for some constant $C_1$. We conclude therefore for $(\eta,\zeta)$ sufficiently small that $$\label{S2} \| \left\{ (A_\zeta + \eta)^{-1}B_\zeta \right\}^n (A_\zeta + \eta)^{-1}b\|_\infty \le C^n_2 |\zeta_1|^n C^{n+1-r}_1 \eta^{-r} \|b\|_\infty \ ,$$ where $r = n/2$ if $n$ is even, $r = (n+1)/2$ if $n$ is odd. Hence if $(\eta, \zeta)$ and $|\zeta_1|^2/\eta$ are small then the series in (\ref{R2}) converges in $L^\infty(\Omega)$ to the solution of $(\ref{T2})$. It follows that for $\xi \in \mathbb{R}^d$ fixed with $\xi_1$ sufficiently small we may construct the function $\varphi(\varepsilon^2, \varepsilon \xi)$ by means of (\ref{R2}) as $\varepsilon \to 0$. To obtain the limit in (\ref{X2}) we write $$\label{AF2} \varphi(\eta, \zeta) = \varphi_1(\eta, \zeta) + \varphi_2(\eta, \zeta)$$ where $\varphi_1(\eta, \zeta)$ is the sum on the RHS of (\ref{R2}) over odd powers of $n$. It is evident from (\ref{S2}) that for $|\xi_1| < 1 \big/ C_2\sqrt{{C_1}}$ one has $$\label{Y2} \lim_{\varepsilon \to 0} \ \varepsilon \varphi_2(\varepsilon^2, \varepsilon \xi) = 0.$$ We consider the first term in the sum for $\varphi_1$. Setting $\eta = \varepsilon^2$, $\zeta = \varepsilon \xi$ and multiplying the term by $\varepsilon$ as in (\ref{Q2}) we see that \begin{align*} &-\lim_{\varepsilon \to 0} \varepsilon \left(A_{\varepsilon\xi} + \varepsilon^2\right)^{-1} B_{\varepsilon\xi} \left(A_{\varepsilon\xi} + \varepsilon^2\right)^{-1} b\\ &=- \lim_{\varepsilon \to 0} i \varepsilon^2 \xi_1 \left(A_{\varepsilon\xi} + \varepsilon^2\right)^{-1} \Big[ \Big\{ \frac 1{2d} + b(\cdot)\Big\} \varphi(\tau_{{\bf e}_1}\cdot) - \Big\{ \frac 1{2d} - b(\cdot)\Big\} \varphi(\tau_{-{\bf e}_1}\cdot) \Big], \end{align*} where $\varphi$ is the solution of \eqref{N2}. Observe now that for any $\psi \in L^\infty(\Omega)$ we have $- \lim_{\varepsilon \to 0} i \varepsilon^2 \xi_1 \left(A_{\varepsilon\xi} + \varepsilon^2\right)^{-1} \psi = -i\xi_1\big<\varphi^* \psi\big> \lim_{\varepsilon \to 0} \varepsilon^2 \left(A_{\varepsilon\xi} + \varepsilon^2\right)^{-1} \; 1 \,,$ where $\varphi^*$ is the solution of \eqref{O2}. From (\ref{V2}) we have that $\varepsilon^2 \left(A_{\varepsilon\xi} + \varepsilon^2\right)^{-1} \; 1 = \varepsilon^2 \Big/ \Big[ 1 + \varepsilon^2 - \frac 1 d \sum^d_{j=1} \cos \varepsilon \xi_j \Big],$ whence we conclude that $\lim_{\varepsilon \to 0} \varepsilon^2 \left(A_{\varepsilon\xi} + \varepsilon^2\right)^{-1} \; 1 = 1 \Big/ \Big[ 1 + \frac{1}{2d} \sum^d_{j=1} \xi^2_j \Big].$ We have therefore obtained a formula for the limit as $\varepsilon \to 0$ of the first term in the series representation of $\varepsilon \varphi_1(\varepsilon^2, \varepsilon \xi)$. Using the same argument we can obtain a formula for all the terms. For the rth term corresponding to $r = (n+1)/2$ with $n$ as in (\ref{R2}) we see that the limit is given by the formula $$\label{AB2} \frac i{2\xi_1} \Big\{ -2 \xi^2_1 \big<\varphi^* \psi\big> \Big/ \Big[ 1 + \frac{1}{2d} \sum^d_{j=1} \xi^2_j \Big]\Big\}^r,$$ where $\psi$ is the function \eqref{W2}. Evidently $\psi \in \hat L^\infty_R(\Omega)$. We have already observed that $\varphi^*$ is also in $\hat L^\infty_R(\Omega)$. We conclude that $\lim_{\varepsilon \to 0} \varepsilon \varphi_1(\varepsilon^2, \varepsilon \xi) = -i\xi_1 \big<\varphi^* \psi\big> \Big/ \Big[ 1 + \frac{1}{2d} \sum^d_{j=1} \xi^2_j + 2\xi^2_1 \big<\varphi^* \psi\big> \Big].$ Then (\ref{X2}) follows from this and (\ref{Y2}). \end{proof} Lemma \ref{lem2.1} enables us to compute the limit (\ref{M2}) when $\xi_1$ is small. We have $$\label{Z2} \lim_{\varepsilon \to 0} T_{\varepsilon^2,\varepsilon\xi}(1)(\omega) = 1\Big/ \Big[ 1 + \frac{1}{2d} \sum^d_{j=1} \xi^2_j + 2\xi^2_1 \big<\varphi^* \psi\big> \Big].$$ We wish now to extend the identity (\ref{Z2}) to all $\xi \in \mathbb{R}^d$. \begin{lemma} \label{lem2.3} Let $K \subset \mathbb{R}^d$ be a compact set. Then the limit (\ref{X2}) is uniform for $\xi \in K, \ \omega \in \Omega$. \end{lemma} \begin{proof} Since the LHS of (\ref{Z2}) does not exceed 1 in absolute value we conclude that $$\label{AA2} \frac 1{2d} + 2 \big<\varphi^* \psi\big> \ge 0.$$ The inequality (\ref{AA2}) in turn implies that the expression (\ref{AB2}) is the rth power of a number strictly less than 1 provided we also assume that $$\label{AC2} 2 \; \big<\varphi^* \psi\big> \le 1/2d.$$ We show that the power series methods of Lemma \ref{lem2.1} apply to prove the result under the additional assumption (\ref{AC2}). We shall see in $\S$3 that $\big<\varphi^* \psi\big> \le 0$ for dimension $d=1$, in which case (\ref{AC2}) certainly holds. Since the constant function is the unique eigenvector of $A_0 = \mathcal{L}$ with eigenvalue 0 and it is also an eigenvector of $A_\zeta$ it follows that there exists $\delta > 0$ such that if $|\zeta| < \delta$ then the adjoint $A^*_\zeta$ of $A_\zeta$ has a unique eigenvector $\varphi^*_\zeta$ with eigenvalue equal to the eigenvalue of $A_\zeta$ for the constant function. Normalizing $\varphi^*_\zeta$ so that $<\varphi^*_\zeta> = 1$, it is easy to see that there is a constant $C_1$ such that $$\label{AD2} \| \varphi^*_\zeta - \varphi^*\| _\infty \le C_1|\zeta|, \quad |\zeta| < \delta.$$ For $|\zeta| < \delta$, $\eta > 0$ we define a projection $P_{\eta,\zeta}$ by $P_{\eta,\zeta} \; \psi = \left< \bar \varphi^*_\zeta \; \psi \right> \left( A_\zeta + \eta\right)^{-1} 1, \quad \psi \in L^\infty(\Omega).$ Then there is a constant $C_2$ such that $$\label{AE2} \| (A_\zeta + \eta)^{-1} -P_{\eta,\zeta}\| \le C_2, \ |\zeta| < \delta.$$ The uniform convergence of (\ref{X2}) for $\xi \in K$ follows now from(\ref{AD2}), (\ref{AE2}) just as in Lemma \ref{lem2.1}. Finally we consider the situation where (\ref{AC2}) is violated. As in Lemma \ref{lem2.1} we decompose the solution $\varphi(\eta,\zeta)$ of (\ref{P2}) into a sum (\ref{AF2}). The function $\varphi_1(\eta, \zeta)$ is a solution to the equation, $$\label{AG2} \left[ A_\zeta + \eta - B_\zeta \left( A_\zeta + \eta\right)^{-1} B_\zeta \right] \varphi_1(\eta, \zeta) = -B_\zeta (A_\zeta + \eta)^{-1} b.$$ The function $\varphi_2(\eta, \zeta)$ is a solution to the equation, $$\label{AH2} \left[ A_\zeta + \eta - B_\zeta \left( A_\zeta + \eta\right)^{-1} B_\zeta \right] \varphi_2(\eta, \zeta) = b.$$ It is easy to see that if $\varphi_2(\eta, \zeta)$ is a solution of (\ref{AH2}) then the function $\varphi_1(\eta, \zeta) = \varphi(\eta, \zeta) - \varphi_2(\eta, \zeta)$, where $\varphi(\eta, \zeta)$ solves (\ref{P2}), is a solution to (\ref{AG2}). Hence if (\ref{AG2}),(\ref{AH2}) have unique solutions $\varphi_1(\eta, \zeta), \varphi_2(\eta, \zeta)$ then the identity (\ref{AF2}) holds. We show that (\ref{AH2}) has a unique solution in $L^\infty_R(\Omega)$ provided $\eta > 0$ and $(\eta, \zeta)$ are sufficiently small. To see this we write (\ref{AH2}) as $$\label{AI2} \left[ A_\zeta + \eta - L_{\eta,\zeta} - B_\zeta P_{\eta,\zeta} B_\zeta \right] \varphi_2(\eta, \zeta) = b,$$ where by (\ref{AE2}) the operator $L_{\eta,\zeta}$ is invariant on $L^\infty_R(\Omega)$ and satisfies $\| L_{\eta,\zeta}\| \le C|\zeta|^2$ for some constant $C$. Next let $\varphi_3(\eta,\zeta)$ be the solution to $$\label{AJ2} \left[ A_\zeta + \eta - L_{\eta,\zeta} \right] \varphi_3(\eta, \zeta) = b.$$ For $(\eta,\zeta)$ small there is a unique solution to (\ref{AJ2}) in $L^\infty_R(\Omega)$ which satisfies $$\label{AK2} \|\varphi - \varphi_3(\eta,\zeta)\|_\infty \le C[|\eta| + |\zeta|^2],$$ for some constant $C$, where $\varphi$ is the solution of \eqref{N2}. Now it is easy to see that the solution $\varphi_2(\eta, \zeta)$ of (\ref{AI2}) is given in terms of $\varphi_3(\eta,\zeta)$ by the formula \label{AL2} \begin{aligned} \varphi_2(\eta, \zeta) &= \Big[ 1 + \eta - \frac 1 d \sum^d_{j=1} \cos \zeta_j \Big] \varphi_3(\eta, \zeta) \\ &\quad \div \Big[ 1 + \eta - \frac 1 d \sum^d_{j=1} \cos \zeta_j + 2 \left< \bar{\varphi}^*_\zeta B\varphi_3(\eta, \zeta) \right> \sin^2 \zeta_1\Big], \end{aligned} where the operator $B$ is defined by $B_\zeta = i \sin \zeta_1 B$. In view of (\ref{AK2}) and the fact that $B\varphi = \psi$ and we are assuming (\ref{AC2}) is violated, it follows that the denominator in (\ref{AL2}) is positive for $(\eta, \zeta)$ sufficiently small. We have shown a solution $\varphi_2(\eta, \zeta)$ of (\ref{AI2}) exists in $L^\infty_R(\Omega)$. The uniqueness of the solution follows from the uniqueness of the solution to (\ref{AJ2}). Evidently the limit (\ref{Y2}) follows from (\ref{AK2}), (\ref{AL2}) for all $\xi$ and is uniform for $\xi$ restricted to a compact subset of $\mathbb{R}^d$. Next we show that (\ref{AG2}) has a unique solution in $\hat L^\infty_R(\Omega)$ provided $\eta > 0$ and $(\eta,\zeta)$ are sufficiently small. First note that for $(\eta, \zeta)$ small the operator $B_\zeta(A_\zeta + \eta)^{-1} B_\zeta$ leaves $\hat L^\infty_R(\Omega)$ invariant and there is a constant $C$ such that $$\label{AM2} \| B_\zeta(A_\zeta + \eta)^{-1} B_\zeta\| \le C|\zeta|^2.$$ We define the subspace $\mathcal{E}_\zeta$ of $\hat L^\infty_R(\Omega)$ by $\mathcal{E}_\zeta = \left\{ \psi \in \hat L^\infty_R(\Omega): \left< \psi \ \bar \varphi^*_\zeta \right> = 0 \right\} .$ Let $P_\zeta$ be the projection operator on $\hat L^\infty_R(\Omega)$ orthogonal to $\mathcal{E}_\zeta$, whence $P_\zeta \psi = \left< \psi \ \bar \varphi^*_\zeta \right>, \quad \psi \in \hat L^\infty_R(\Omega).$ Consider now the equation related to (\ref{AG2}) given by $$\label{AN2} \Big[ A_\zeta + \eta - (I-P_\zeta)B_\zeta (A_\zeta + \eta)^{-1} B_\zeta \Big] \varphi_4(\eta,\zeta) = -(I-P_\zeta)B_\zeta (A_\zeta + \eta)^{-1} b.$$ In view of (\ref{AM2}) it is clear that for $(\eta,\zeta)$ sufficiently small the equation (\ref{AN2}) has a unique solution $\varphi_4(\eta,\zeta)$ in $\mathcal{E}_\zeta$. Furthermore, if we define $\varphi_1(\eta,\zeta)$ by \label{AO2} \begin{aligned} \varphi_1(\eta,\zeta) &= \Big\{ \Big[ 1 + \eta - \frac 1 d \sum^d_{j=1} \cos \zeta_j \Big] \varphi_4(\eta,\zeta) - i \sin \zeta_1 \left< \bar \varphi^*_\zeta B(A_\zeta + \eta)^{-1} \; b \right> \\ &\quad - \sin^2 \zeta_1 \left< \bar \varphi^*_\zeta B(A_\zeta + \eta)^{-1} \; B \; \varphi_4(\eta,\zeta)\right> \Big\}\\ &\quad \div \Big\{ 1 + \eta - \frac 1 d \sum^d_{j=1} \cos \zeta_j + 2\sin^2 \zeta_1 \left< \bar\varphi^*_\zeta B(A_\zeta + \eta)^{-1} \; b \right> \\ &\quad -2i \sin^3 \zeta_1 \left< \bar\varphi^*_\zeta B(A_\zeta + \eta)^{-1} \; B \varphi_4 \right> \Big\} , \end{aligned} then one sees that the formula (\ref{AO2}) yields a solution to (\ref{AG2}). Conversely, since we are assuming (\ref{AC2}) is violated, it follows that for $(\eta, \zeta)$ small (\ref{AO2}) is the unique solution in $\hat L^\infty_R(\Omega)$ to (\ref{AG2}). It is easy to see now from (\ref{AO2}) that the limit $\lim_{\varepsilon \to 0} \varepsilon\varphi_1(\varepsilon^2,\varepsilon\xi)$ exists and is uniform for $\xi$ in a compact subset of $\mathbb{R}^d$. Furthermore, the limit is given by the RHS of (\ref{X2}). Finally we show that if $\varphi_1(\eta,\zeta), \varphi_2(\eta,\zeta)$ are solutions to (\ref{AG2}), (\ref{AH2}) then (\ref{AF2}) holds. To see this we put $\varphi(\eta,\zeta) =\varphi_1(\eta,\zeta) + \varphi_2(\eta,\zeta)$ and note that (\ref{AG2}), (\ref{AH2}) imply that $\varphi(\eta,\zeta)$ satisfies the equation $\left[ A_\zeta + \eta - B_\zeta (A_\zeta + \eta)^{-1} B_\zeta \right] \varphi(\eta,\zeta) = b - B_\zeta (A_\zeta + \eta)^{-1} b.$ We can rewrite this equation as $\left[ A_\zeta + B_\zeta + \eta - B_\zeta (A_\zeta + \eta)^{-1}(A_\zeta+ B_\zeta+\eta) \right] \varphi(\eta,\zeta) = b - B_\zeta (A_\zeta + \eta)^{-1} b,$ which is the same as $[\mathcal{L}_{R\zeta} + \eta ](A_\zeta + \eta)^{-1} [\mathcal{L}_\zeta + \eta] \varphi(\eta, \zeta) = [\mathcal{L}_{R\zeta} + \eta ](A_\zeta + \eta)^{-1} \; b.$ Now using the fact that the operator $\mathcal{L}_{R\zeta} + \eta$ is invertible we obtain (\ref{P2}). \end{proof} Next we show that there is strict inequality in (\ref{AA2}). In order to do this we shall first obtain a concrete representation of spaces $\Omega$ which satisfy \eqref{B1}. \begin{lemma} \label{lem2.4} Let $\Omega$ be a finite probability space and $b : \Omega \to \mathbb{R}$ satisfy \eqref{C1}. Then $\Omega$ may be identified with a rectangle in $\mathbb{Z}^d$ with periodic boundary conditions. The operators $\tau_x, \; x \in \mathbb{Z}^d$, act on $\Omega$ by translation and the measure $\langle \cdot \rangle$ is simple averaging. Let $R : \Omega \to \Omega$ be the reflection operator defined as reflection in the hyperplane through the center of $\Omega$ with normal ${\bf e}_1$. Then there is the identity $b(\omega) = -b(R\omega), \ \omega \in \Omega$. \end{lemma} \begin{proof} Since $\Omega$ has no nontrivial invariant subsets under the action of the $\tau_{{\bf e}_j}, \; 1 \le j \le d$, it is isomorphic to a rectangle in $\mathbb{Z}^d$ with periodic boundary conditions. Thus we may assume $\Omega$ is given by $$\label{AP2} \Omega = \left\{ x = (x_1, \dots, x_d) \in \mathbb{Z}^d : 0 \le x_i \le L_i - 1, \ 1 \le i \le d \right\},$$ where $L_1,\dots,L_d$ are positive integers. The action of the $\tau_{{\bf e}_j}$ is translation, $\tau_{{\bf e}_j} x = x + {\bf e}_j$ with periodic boundary conditions. The measure on $\Omega$ is averaging, $$\label{AQ2} \left< \Psi(\cdot) \right> = \frac 1{L_1L_2 \dots L_d} \sum_{0\le x_i \le L_i-1,\; 1 \le i\le d} \Psi(x_1,\dots,x_d).$$ Functions $\Psi: \Omega \to \mathbb{R}$ are isomorphic to periodic functions $\Psi : \mathbb{Z}^d \to \mathbb{R}$. Next we consider the condition \eqref{C1}. We define a function $b^R : \Omega \to \mathbb{R}$ by $b^R(\omega) = -b(R\omega)$, $\omega \in \Omega$. It is easy to see that $b^R(\tau_x \omega) = -b(R\tau_x\omega)$ $= -b(\tau_{Rx} R\omega), \omega \in \Omega$. Since $R$ leaves the measure (\ref{AQ2}) invariant \eqref{C1} implies that for any $\theta_1,\dots,\theta_k \in \mathbb{R}$, $x_1,\dots,x_k \in \mathbb{Z}^d$, there is the identity, $\Big< \exp \Big[ \sum^k_{j=1} \theta_j \ b(\tau_{x_j} \cdot ) \Big] \Big> = \Big< \exp \Big[ \sum^k_{j=1} \theta_j \ b^R(\tau_{x_j} \cdot) \Big] \Big>.$ We conclude that $b \equiv b^R$. \end{proof} Next we wish to construct the solutions $\varphi, \varphi^*$ of \eqref{N2}, \eqref{O2} on the domain $\Omega$ defined by (\ref{AP2}). First observe that since $\Omega$ is the fundamental region for the homogenization problem we can assume that $L_1$ is an even integer by simply doubling $\Omega$ if $L_1$ is odd. In that case the function $b$ is determined by its values $b(x), \ x \in \Omega$, $0 \le x_1 \le L_1/2 - 1$. Hence we define a new fundamental region $\hat \Omega$ by $$\label{AR2} \hat \Omega = \{ x \in \Omega : 0 \le x_1 \le L_1/2 - 1 \}.$$ We can extend functions $\Psi : \hat \Omega \to \mathbb{R}$ to $\Omega$ by either symmetric or antisymmetric extension. For a symmetric extension we define $\Psi$ on $\Omega - \hat \Omega$ by $$\label{AS2} \Psi(x_1,\dots,x_d) = \Psi(L_1 - 1 - x_1, x_2, \dots, x_d), \ L_1/2 \le x_1 \le L_1 - 1.$$ For an antisymmetric extension we define $\Psi$ by $$\label{AT2} \Psi(x_1,\dots,x_d) = -\Psi(L_1 - 1 - x_1, x_2, \dots, x_d), \ L_1/2 \le x_1 \le L_1 - 1.$$ \begin{lemma} \label{lem2.5} The solution $\varphi : \Omega \to \mathbb{R}$ of \eqref{N2} is an antisymmetric extension of its restriction to $\hat \Omega$. The solution $\varphi^* : \Omega \to \mathbb{R}$ of \eqref{O2} is a symmetric extension of its restriction to $\hat \Omega$. \end{lemma} \begin{proof} This follows easily from the fact that $b : \Omega \to \mathbb{R}$ is an antisymmetric extension of its restriction to $\hat \Omega$ and the uniqueness of the solution to \eqref{N2}, \eqref{O2}. \end{proof} Lemma \ref{lem2.4} implies that we can find the functions $\varphi, \varphi^*$ by solving \eqref{N2}, \eqref{O2} on $\hat \Omega$ with antisymmetric and symmetric boundary conditions respectively. Thus $\mathcal{L}$ acting on functions $\Psi : \hat \Omega \to \mathbb{R}$ with antisymmetric boundary conditions is defined by \label{AU2} \begin{aligned} \mathcal{L} \Psi(x) &= \Psi(x) - \sum^d_{j=1} \frac 1{2d} \left[ \Psi(x + {\bf e}_j) + \Psi(x - {\bf e}_j) \right] \\ &\quad - b(x) \left[ \Psi(x + {\bf e}_1) - \Psi(x - {\bf e}_1) \right] , \quad x \in \hat \Omega, \end{aligned} where the boundary conditions are $$\label{AV2} \begin{gathered} \Psi(-1, x_2,\dots,x_d) = -\Psi(0, x_2,\dots,x_d), \\ \Psi(L_1/2, x_2,\dots,x_d) = -\Psi(L_1/2 - 1, x_2,\dots,x_d) , \end{gathered}$$ where $0 \le x_j \le L_j - 1$, $j=2,\dots,d$; and periodic boundary conditions in the directions ${\bf e}_j, \ 2 \le j \le d$. Evidently (\ref{AV2}) is derived from (\ref{AT2}). It is easy to see that the operator $\mathcal{L}$ is invertible on the space $L^\infty(\hat \Omega)$ if the boundary conditions (\ref{AV2}) are imposed. In fact the solution to the equation $$\label{AW2} \mathcal{L} \Psi(x) = f(x), \quad x \in \hat \Omega,$$ with boundary conditions (\ref{AV2}) can be represented as an expectation for a continuous time Markov chain $X(t), t \ge 0$, on $\hat \Omega$. For the chain the transition probabilities at a site $x \in \hat \Omega$ satisfying $0 < x_1 < L_1/2 - 1$ are given by $x \to x + {\bf e}_j, x \to x - {\bf e}_j$, $2 \le j \le d$, each with probability $1/2d$, with periodic boundary conditions in direction ${\bf e}_j, 2 \le j \le d$. In the direction ${\bf e}_1$ then $x \to x + {\bf e}_1$ with probability $1/2d + b(x)$ and $x \to x - {\bf e}_1$ with probability $1/2d - b(x)$. The waiting time at site $x$ is exponential with parameter 1. If $x_1 = 0$ then $x \to x \pm {\bf e}_j$, $2 \le j \le d$, with probability $1/2d[1+1/2d - b(x)] < 1/2d$, and $x\to x + {\bf e}_1$ with probability $[1/2d + b(x)] / [1+1/2d - b(x)] < 1/d$. The waiting time is exponential with parameter $[1+1/2d - b(x)]$. Note that there is a positive probability that the walk will be killed at a site $x$ with $x_1 = 0$. A similar situation occurs at a site $x$ with $x_1 = L_1/2 - 1$. Now $x \to x - {\bf e}_1$ with probability $[1/2d - b(x)] / [1+1/2d + b(x)] < 1/d$ and the waiting time is exponential with parameter $[1+1/2d + b(x)].$ The solution $\Psi$ of \eqref{AW2} with boundary conditions (\ref{AV2}) has the representation $$\label{BA2} \Psi(x) = E \Big[ \int^\tau_0 f (X(t))dt | X(0) = x \Big], \quad x \in \hat \Omega,$$ where $\tau$ is the killing time for the chain. We may also consider the operator $\mathcal{L}$ of (\ref{AU2}) with symmetric boundary conditions, $$\label{AX2} \begin{gathered} \Psi(-1, x_2,\dots,x_d) = \Psi(0, x_2,\dots,x_d), \\ \Psi(L_1/2,x_2,\dots,x_d) = \Psi(L_1/2 - 1, x_2,\dots,x_d) \end{gathered}$$ where $0 \le x_j \le L_j - 1$, $j=2,\dots,d$, corresponding to (\ref{AS2}). This is also associated with a continuous time Markov chain $X(t)$ on $\hat \Omega$. The transition probabilities and waiting time at a site $x \in \hat \Omega$ with $0 < x_1 < L_1/2 - 1$ are as for the chain defined in the previous paragraph. For $x \in \hat \Omega$ with $x_1=0$ reflecting boundary conditions corresponding to (\ref{AX2}) are imposed. Thus the waiting time at $x$ is exponential with parameter $[1 - 1/2d + b(x)], x \to x \pm {\bf e}_j$, $2 \le j \le d$, with probability $1/2d[1-1/2d + b(x)]$ and $x\to x+{\bf e}_1$ with probability $[1/2d+b(x)] / [1-1/2d + b(x)]$. A similar situation occurs at $x \in \hat \Omega$ with $x_1 = L_1/2 - 1$. The formal adjoint $\mathcal{L}^*$ of the operator $\mathcal{L}$ of (\ref{AU2}) is given by \label{AY2} \begin{aligned} \mathcal{L}^* \Psi(x) &= \Psi(x) - \sum^d_{j=1} \frac 1{2d} \left[ \Psi(x + {\bf e}_j) + \Psi(x - {\bf e}_j) \right] \\ &\quad - b(x-{\bf e}_1) \Psi(x - {\bf e}_1) + b(x + {\bf e}_1) \Psi(x + {\bf e}_1), \quad x \in \hat \Omega. \end{aligned} It is easy to see that for functions $\Phi, \Psi$ on $\hat \Omega$ satisfying the symmetric boundary conditions (\ref{AX2}) there is the identity $$\label{AZ2} \left< \Phi \; \mathcal{L}^* \; \Psi \right>_{\hat \Omega} = \left< \Psi \; \mathcal{L} \; \Phi\; \right>_{\hat \Omega},$$ where $\langle \cdot \rangle_{\hat \Omega}$ is the uniform probability measure on $\hat \Omega$. Note that to show (\ref{AZ2}) one has to use the fact that the function $b$ satisfies the antisymmetric conditions ( \ref{AV2}). Hence the adjoint of the operator $\mathcal{L}$ acting on functions $\Psi : \hat \Omega \to \mathbb{R}$ with symmetric boundary conditions (\ref{AX2}) is the operator $\mathcal{L}^*$ of (\ref{AY2}) also acting on functions with symmetric boundary conditions. In particular, it follows from Lemma \ref{lem2.4} that the solution $\varphi^*$ of \eqref{O2}, restricted to $\hat \Omega$, is the unique invariant measure for the Markov chain $X(t)$. Next let $\psi_0 : \hat \Omega \to \mathbb{R}$ be the solution of the homogeneous equation \eqref{AW2} i.e. $f \equiv 0$, with the non-homogeneous antisymmetric boundary conditions $$\label{BB2} \begin{gathered} \psi_0(-1, x_2,\dots,x_d) = -\psi_0(0, x_2,\dots,x_d), \\ \psi_0(L_1/2, x_2,\dots,x_d) = 1 -\psi_0(L_1/2 - 1, x_2,\dots,x_d) \end{gathered}$$ where $0 \le x_j \le L_j - 1$, $j=2,\dots,d$. One can see that $\psi_0$ is a positive function since it has a representation given by (\ref{BA2}), where $f$ is the function $f(x) =\begin{cases} \frac 1{2d} + b(x), & x \in \hat \Omega, \; x_1 = L_1/2 - 1, \\ 0, &\text{otherwise}. \end{cases}$ The following lemma now shows that there is strict inequality in (\ref{AA2}). \begin{lemma} \label{lem2.6} Let $\varphi^*$ be the solution of \eqref{O2} and $\psi$ be given by \eqref{W2}. Then there is the identity, $$\label{BC2} \frac 1{2d} + 2 \left<\varphi^* \psi\right> = L^2_1 \big< \varphi^*(\cdot) \big[ \frac 1{2d} - b(\cdot)\big]\psi_0(\cdot) \chi_0(\cdot)\big>_{\hat \Omega},$$ where $\chi_0 : \hat \Omega \to \mathbb{R}$ is defined by $\chi_0(x) = 1$ if $x_1=0$, $\chi_0(x) = 0$, otherwise. \end{lemma} \begin{proof} Since both $\varphi^*$ and $\psi$ are symmetric on $\Omega$ in the sense of (\ref{AS2}) we may regard them as functions on $\hat \Omega$ with symmetric boundary conditions (\ref{AX2}). We define a function $\psi_1 : \hat \Omega \to \mathbb{R}$ by $\psi_1(x) = [L_1/2 - 1/2 - x_1]\varphi(x), \ x \in \hat{\Omega}$, where $\varphi$ is the solution to \eqref{N2}. It is easy to see that \label{BD2} \mathcal{L} \psi_1(x) = [L_1/2 - 1/2 - x_1]b(x) + \psi(x), \quad x \in \hat \Omega, \; 0 < x_1 &= 2 \left<\varphi^* \psi\right> _{\hat \Omega} \\ & = 2L_1 \big< \varphi^*(\cdot) \big[ \frac 1{2d} - b(\cdot)\big] \varphi(\cdot) \chi_0(\cdot) \big>_{\hat \Omega} - \left< \varphi^*(\cdot) \left[ L_1 - 1 -2x_1\right] b(\cdot) \right>_{\hat \Omega}. \end{aligned} Next we define a function $\psi_2 : \hat \Omega \to \mathbb{R}$ by $\psi_2(x) = (L_1 - 1 -2x_1)^2, x \in \hat \Omega$. Then we have \label{BG2} \mathcal{L}\psi_2(x) = 8[L_1 - 1 -2x_1] b(x) - 4/d, \quad x \in \hat \Omega, \; 0 < x_1 _{\hat \Omega} = - 1/2d + \frac{L_1}2 \big< \varphi^*(\cdot) \big[ \frac 1{2d} - b(\cdot)\big] \chi_0(\cdot) \big>_{\hat \Omega}, where we have used the fact that $\left<\varphi^*\right>_{\hat \Omega} = 1$. It follows now from (\ref{BF2}), (\ref{BH2}) that $$\label{BJ2} 1/2d + 2 \left< \varphi^*\psi \right> = \frac{L_1}2 \big< \varphi^*(\cdot) \big[ \frac 1{2d} - b(\cdot)\big] [1 + 4\varphi(\cdot)] \chi_0(\cdot) \big>_{\hat \Omega} .$$ We put now $\psi_0(x) = [2x_1 + 1 + 4\varphi(x)]/2L_1$, and it is easy to verify that $\psi_0$ satisfies the homogenous equation \eqref{AW2} with the boundary conditions (\ref{BB2}). The result follows then from (\ref{BJ2}). \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.1}] The proof proceeds identically to the proof of \cite[Theorem 1.1]{cp}, on using lemmas \ref{lem2.1}-\ref{lem2.4}. \end{proof} Finally we wish to show that Theorem \ref{thm1.2} holds to leading order in perturbation theory (see \cite{bes} for an introduction to perturbation theory). \begin{theorem} \label{thm2.1} There exists $\delta > 0$ such that if $b : \Omega \to \mathbb{R}$ satisfies $0 < \sup_{\omega\in\Omega} |b(\omega)| < \delta$ then $q(b) < 1/2d$, provided $d=1$, or $d > 1$ and $L_1 \le 4$. If $d\ge 2$ and $L_1 \ge 6$ then there exists arbitrarily small $b$ with $q(b) >1/2d$. \end{theorem} \begin{proof} We shall use the LHS of (\ref{BC2}) as an expression for $q(b)$. If $b \equiv 0$ then $\varphi^* \equiv 1, \varphi \equiv 0 \Rightarrow \psi \equiv 0$. Thus to obtain an expression for $q(b)$ which is correct to second order in perturbation theory we need to expand $\varphi^*$ to first order in $b$ and $\varphi$ to second order. We consider first $\varphi^*$ which is the solution to \eqref{O2}. Letting $\Delta$ be the finite difference Laplacian acting on functions $\Psi : \Omega \to \mathbb{R}$ with periodic boundary conditions, $\Delta \Psi(x) = \sum^d_{j=1} \left[ \Psi(x + {\bf e}_j) + \Psi(x - {\bf e}_j) - 2\Psi(x) \right] , \ x \in \Omega,$ we have from (\ref{AY2}) that \eqref{O2} is given by $$\label{BK2} -\frac{\Delta}{2d} \varphi^*(x) + b(x + {\bf e}_1) \varphi^*(x + {\bf e}_1) - b(x - {\bf e}_1) \varphi^*(x - {\bf e}_1) = 0, \quad x \in \Omega, \; \left<\varphi^*\right> = 1.$$ Since $\left< \left[ \tau_{-{\bf e}_1} - \tau_{{\bf e}_1} \right] b\right> = 0$ the solution to (\ref{BK2}) is to first order in perturbation theory given by $$\label{BL2} \varphi^* = 1 + (-\Delta/2d)^{-1} \left[ \tau_{-{\bf e}_1} - \tau_{{\bf e}_1} \right] b.$$ From (\ref{AU2}) equation \eqref{N2} is the same as $$\label{BM2} -\frac{\Delta}{2d} \varphi(x) - b(x) \left[ \tau_{{\bf e}_1} - \tau_{-{\bf e}_1} \right] \varphi(x) = b(x), \quad x \in \Omega.$$ Using the fact that $\left = \left< b \left[ \tau_{{\bf e}_1} - \tau_{-{\bf e}_1} \right] (-\Delta/2d)^{-1} b\right> = 0,$ we see that the solution to (\ref{BM2}) correct to second order in $b$ is given by $$\label{BN2} \varphi = (-\Delta/2d)^{-1} b + (-\Delta/2d)^{-1} b \left[ \tau_{{\bf e}_1} - \tau_{-{\bf e}_1} \right] (-\Delta/2d)^{-1} b.$$ From \eqref{W2} and (\ref{BN2}) we can obtain an expression for $\psi$ which is correct to second order in $b$, \label{BO2} \begin{aligned} \psi &= \frac 1{2d} \left[ \tau_{{\bf e}_1} - \tau_{-{\bf e}_1} \right] (-\Delta/2d)^{-1} b + b \left[ \tau_{{\bf e}_1} + \tau_{-{\bf e}_1} \right] (-\Delta/2d)^{-1} b \\ &\quad + \frac 1{2d} \left[ \tau_{{\bf e}_1} - \tau_{-{\bf e}_1} \right] (-\Delta/2d)^{-1} b \left[ \tau_{{\bf e}_1} - \tau_{-{\bf e}_1} \right] (-\Delta/2d)^{-1} b. \end{aligned} From (\ref{BL2}), (\ref{BO2}) we see that the lowest order term in the expansion of $\left<\varphi^* \psi\right>$ in powers of $b$ is second order. Thus correct to second order we have \label{BP2} \begin{aligned} \left<\varphi^*\psi\right> &= \left< b \left[ \tau_{{\bf e}_1} + \tau_{-{\bf e}_1} \right] (-\Delta/2d)^{-1} b\right> \\ &\quad + \frac 1{2d} \left< b\left[ \tau_{{\bf e}_1} - \tau_{-{\bf e}_1} \right] (-\Delta/2d)^{-1} \left[ \tau_{{\bf e}_1} - \tau_{-{\bf e}_1} \right] (-\Delta/2d)^{-1} b \right>. \end{aligned} The RHS of this equation is a translation invariant quadratic form, whence it has eigenvectors $\exp[i\xi \cdot x], \ x \in \Omega$, with corresponding eigenvalue given by the formula, $$\label{BQ2} 2d \Big\{ \cos \xi_1 - \frac{\sin^2 \xi_1}{\sum^d_{j=1}(1-\cos \xi_j)} \Big\} \Big/ \sum^d_{j=1} (1 - \cos \xi_j).$$ We obtain an expression for the quadratic form (\ref{BP2}) by doing an eigenvector decomposition in the $x_1$ direction. Putting $L = L_1/2$ we have that $\xi_1 = \pi k/L$, $k=0, \pm 1,\dots,\pm (L-1), L$. The function $b$ then has a representation, $b(x_1) = \sum_{\xi_1} e^{i\xi_1x_1} \Big( \frac 1{2L} \sum^{2L-1}_{y=0} \ b(y)e^{-i\xi_1y} \Big).$ If we use the antisymmetry property of $b$, $b(y) = -b(2L - 1 - y)$ then one has that $\sum^{2L-1}_{y=0} \ b(y)e^{-i\xi_1y} = -2i e^{i\xi_1/2} \sum^{L-1}_{y=0} b(y) \sin \xi_1(y + 1/2).$ We conclude from this and (\ref{BQ2}) that the expression (\ref{BP2}) is the same as \label{BR2} \begin{aligned} \left<\varphi^*\psi\right> &= -\frac{4d}{L^2} \Big< \Big( \sum^{L-1}_{y=0} (-1)^y b(y) \Big) [-\Delta_{d-1} + 4]^{-1} \Big( \sum^{L-1}_{y=0} (-1)^y b(y) \Big)\Big> \\ &\quad + \frac{8d}{L^2} \sum^{L-1}_{k=1} \Big< \Big( \sum^{L-1}_{y=0} b(y) \sin \pi k(y + \frac 1 2)/L \Big) \\ &\quad \times \Big\{ \cos \frac{\pi k}L - 2 \sin^2 \frac{\pi k}L \left[ -\Delta_{d-1} + 2 \big( 1 - \cos(\pi k/L) \big)\right]^{-1}\Big\} \\ &\quad\times \big[ -\Delta_{d-1} + 2 \big( 1 - \cos(\pi k/L) \big)\big]^{-1} \Big( \sum^{L-1}_{y=0} b(y) \sin \pi k(y + \frac 1 2)/L \Big) \Big>, \end{aligned} where $\Delta_{d-1}$ denotes the $d-1$ dimensional Laplacian acting on the space $\{x_1 = 0\}$. Observe now that the $L$ dimensional vectors $\sin \pi k(y+1/2)/L, 0 \le y \le L-1$, are mutually orthogonal, $k=1,\dots,L$. This is a consequence of the fact that they are the eigenvectors of the second difference operator on the set $\{ 0 \le y \le L-1\}$ with antisymmetric boundary conditions. It follows that the quadratic form (\ref{BR2}) is negative definite if and only if all the eigenvalues (\ref{BQ2}) are negative. This is the case for $d=1$. For $d > 1$ it is still true provided $L \le 2$, but already for $L=3$ it is false. Thus for $L=3$ one can find a $b$ such that the homogenized limit has an effective diffusion constant which is larger than the $b \equiv 0$ case. \end{proof} \section{Proof of Theorem \ref{thm1.2}} We shall use the representation for the effective diffusion constant given by the RHS of (\ref{BC2}). We consider first the $d=1$ case. \begin{lemma} \label{lem3.1} Let $\hat\Omega$ be the space $\hat\Omega = \{ x \in \mathbb{Z} : 1 \le x \le L\}$. If $\varphi^* : \Omega \to \mathbb{R}$ is the solution to \eqref{O2} then $\varphi^*(1)$ is given by the formula, $$\label{A3} \varphi^*(1)\delta_1 = L \prod^L_{k=1} \delta_k \big/ \sum^L_{r=1} \prod^{r-1}_{j=1} \bar\delta_j \prod^L_{j=r+1} \delta_j,$$ where the $\delta_j, \bar\delta_j$, $1 \le j \le L$, are given by $$\label{B3} \delta_j = 1/2 - b(j), \quad \bar\delta_j = 1/2 + b(j).$$ \end{lemma} \begin{proof} From (\ref{AY2}) we see that $\varphi^* : \hat\Omega \to \mathbb{R}$ satisfies the equation $$\label{C3} \begin{split} &\varphi^*(x) - \frac 1 2 \Big[ \varphi^*(x+1) + \varphi^*(x-1)\Big]\\ & - b(x-1) \varphi^*(x-1)+b(x+1)\varphi^*(x+1)\\ &=0, \quad 1 \le x \le L, \end{split}$$ with the symmetric boundary conditions and normalization given by $$\label{D3} \varphi^*(0) = \varphi^*(1), \ \varphi^*(L+1)= \varphi^*(L), \left<\varphi^*\right>_{\hat\Omega} = 1.$$ We can solve (\ref{C3}), (\ref{D3}) uniquely by standard methods. Thus putting $D\varphi^*(x) = \varphi^*(x) -\varphi^*(x-1)$, $1 \le x \le L$, then we may write (\ref{C3}) as $$\label{E3} \frac 1 2 \big[ D\varphi^*(x) - D\varphi^*(x+1)\big] - b(x-1) \varphi^*(x-1)+b(x+1)\varphi^*(x+1)=0, \quad 1 \le x \le L.$$ If we sum (\ref{E3}) over the set $\{1 \le x \le y\}$ we obtain the equation, \begin{align*} &\frac 1 2 \big[ D\varphi^*(1) - D\varphi^*(y+1)\big] - b(0) \varphi^*(0) - b(1) \varphi^*(1) \\ &+ b(y) \varphi^*(y)+ b(y+1)\varphi^*(y+1) = 0, \quad 1 \le y \le L. \end{align*} Then, using the fact that $\varphi^*(1) = \varphi^*(0)$, $b(1) = -b(0)$, we conclude that $\varphi^*(y+1) = \bar\delta_y \varphi^*(y) \big/ \delta_{y+1}, \quad 1 \le y \le L,$ whence we have $$\label{F3} \varphi^*(y) = \varphi^*(1) \prod^{y-1}_{j=1} \bar\delta_j / \delta_{j+1}, \quad 1 \le y \le L.$$ Formula (\ref{A3}) follows from (\ref{F3}) and the normalization condition in (\ref{D3}). \end{proof} \begin{lemma} \label{lem3.2} Let $\hat\Omega$ be the space $\hat\Omega = \{ x \in \mathbb{Z} : 1 \le x \le L\}$. If $\psi_0 : \hat\Omega \to \mathbb{R}$ is the solution to the homogeneous equation \eqref{AW2} with the boundary conditions (\ref{BB2}) then $\psi_0(1)$ is given by the formula $$\label{G3} 2\psi_0(1) = \prod^L_{k=1} \bar\delta_k \big/ \sum^L_{r=1} \prod^{r-1}_{j=1} \delta_j \prod^L_{j=r+1} \bar\delta_j,$$ where $\delta_j, \bar\delta_j$, $1 \le j \le L$, are as in \eqref{B3}. \end{lemma} \begin{proof} >From \eqref{AU2}, \eqref{AW2}, \eqref{BB2}, we see that $\psi_0(x)$ satisfies the equation $$\label{H3} \psi_0(x) - \frac 1 2 \big[ \psi_0(x+1) + \psi_0(x-1)\big] - b(x)\big[ \psi_0(x+1) - \psi_0(x-1)\big]=0, \quad 1 \le x \le L,$$ with the boundary conditions, $$\label{I3} \psi_0(0) = -\psi_0(1), \quad \psi_0(L+1) = 1 - \psi_0(L).$$ We can solve (\ref{H3}), (\ref{I3}) by standard methods. Thus putting $D\psi_0(x) = \psi_0(x) - \psi_0(x-1)$ equation (\ref{H3}) implies $$\label{J3} D\psi_0(x+1)=\delta_x D\psi_0(x)/\bar\delta_x,\quad 1 \le x \le L.$$ Observing from (\ref{I3}) that $D\psi_0(1)=2\psi_0(1)$ we see from (\ref{J3}) that $$\label{K3} D\psi_0(y+1)= 2\psi_0(1) \prod^y_{j=1} \delta_j / \bar\delta_j, \quad 1 \le y \le L.$$ If we sum (\ref{K3}) we obtain a formula for $\psi_0(y)$ given by $$\label{L3} \psi_0(y)= 2\psi_0(1)\Big\{ 1/2+\sum^y_{r=2} \prod^{r-1}_{j=1} \delta_j / \bar\delta_j\Big\}, \quad 1 \le y \le L.$$ Since $D\psi_0(L+1) = 1-2\psi_0(L)$ from (\ref{I3}) it follows that if we add (\ref{K3}) to twice (\ref{L3}) when $y=L$ we obtain a formula for $\psi_0(1)$ given by $$\label{M3} 2\psi_0(1) = \prod^L_{k=1} \bar\delta_k \Big/ \Big\{ \prod^L_{j=1} \bar\delta_j +2 \sum^L_{r=2} \prod^{r-1}_{j=1} \delta_j \prod^L_{j=r}\bar\delta_j + \prod^L_{j=1} \delta_j \Big\} .$$ One can easily see that the denominator of the expression in (\ref{M3}) can be rewritten as in (\ref{G3}). \end{proof} \begin{remark} \label{rmk1} \rm Observe from (\ref{A3}), (\ref{G3}) that under the reflection $b \to -b$ the expression $\varphi^*(1)\delta_1 / L$ becomes $2\psi_0(1)$. \end{remark} \begin{lemma} \label{lem3.3} There is the inequality $\varphi^*(1)\delta_1 \psi_0(1) \le 1/8L$. \end{lemma} \begin{proof} For $1 \le r \le L$ define $a_r$ by $a_r = \prod^{r-1}_{j=1} \bar\delta_j \prod^{L}_{j=r+1} \delta_j ,$ and $\bar a_r$ the corresponding value of $a_r$ under the reflection $b \to -b$. From Lemma \ref{lem3.1}, \ref{lem3.2} we see that we need to prove that $4L^2 \prod^L_{k=1} \delta_k \bar\delta_k \le \Big\{ \sum^L_{r=1} a_r \Big\} \Big\{ \sum^L_{r=1} \bar a_r \Big\}.$ Using the fact that for $1 \le r,s \le L$, $(a_r \bar a_s + a_s \bar a_r)/2 \ge (a_r \bar a_r a_s \bar a_s)^{1/2},$ we see that $\Big\{ \sum^L_{r=1} a_r \Big\}\Big\{ \sum^L_{r=1} \bar a_r \Big\} \ge \sum^L_{r,s=1} (a_r \bar a_r a_s \bar a_s)^{1/2} \ge \sum^L_{r,s=1} 4 \ \prod^L_{k=1} \delta_k \bar\delta_k = 4L^2 \prod^L_{k=1} \delta_k \bar\delta_k,$ where we have used the fact that for $1 \le j \le L$, one has $\delta_j \bar\delta_j \le 1/4$. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.2} ($d=1$)] This follows from Lemmas \ref{lem3.1}--\ref{lem3.3} and Lemma \ref{lem2.5}, using the RHS of (\ref{BC2}) as the representation for $q(b)$. \end{proof} Next we turn to the $d > 1$ case with $L_1 = 2$. Then we can write $\hat \Omega = \{(0,y):y\in\Omega_{d-1}\}$ where $\Omega_{d-1} \subset \mathbb{Z}^{d-1}$ is a $d-1$ dimensional rectangle. It is easy to see now from (\ref{BK2}), on using the anti-symmetry of $b$ and the symmetry of $\varphi^*$, that $\varphi^* \equiv 1$. Also from (\ref{BM2}), on using the anti-symmetry of $b$ and $\varphi$, we have that $$\label{N3} \varphi(0,y) = 2d \left[ -\Delta_{d-1} +4\right]^{-1} b(0,y),$$ where in (\ref{N3}) the operator $\Delta_{d-1}$ is the discrete Laplacian acting on functions $\Psi : \Omega_{d-1} \to \mathbb{R}$. We have then from \eqref{W2} that $\psi(0,y) = -2b(0,y)\varphi(0,y)$, and so we get the formula for the effective diffusion constant, $$\label{O3} 1/2d+ 2 \langle \phi^*\psi\rangle = 1/2d - 8d \langle b(\cdot)\left[-\Delta_{d-1} + 4\right]^{-1} b \rangle_{\Omega_{d-1}}.$$ It is clear that the RHS of (\ref{O3}) is smaller than $1/2d$. We can alternatively derive the effective diffusion constant formula by using the expression on the RHS of (\ref{BC2}). Thus we have $\psi_0(0,y) = 2d \left[ -\Delta_{d-1} + 4\right]^{-1} [ 1/2d +b(0,y)],$ whence the effective diffusion constant is given by the formula $$\label{P3} \begin{split} &L^2_1 \left< \phi^*(\cdot) \left[ 1/2d - b(\cdot)\right] \psi_0(\cdot) \chi_0(\cdot) \right>_{\hat\Omega} \\ &= 8d \left< \left[ 1/2d - b(\cdot)\right] \left[-\Delta_{d-1} + 4\right]^{-1} \left[ 1/2d + b(\cdot)\right] \right>_{\Omega_{d-1}}. \end{split}$$ We shall use the formula on the LHS of (\ref{P3}) to obtain an expression for the effective diffusion constant in the case $L_1=4$. Here $\hat \Omega$ is the space $\hat \Omega = \{(n,y) : n=0,1, \ y \in \Omega_{d-1}\}$. For $y \in \Omega_{d-1}$ we define $\delta_y, \; \bar\delta_y, \; \varepsilon_y, \; \bar\varepsilon_y$ by $$\label{Q3} \begin{gathered} \delta_y = 1/2d - b(0,y), \ \bar\delta_y = 1/2d + b(0,y), \\ \varepsilon_y = 1/2d + b(1,y), \ \bar\varepsilon_y = 1/2d - b(1,y). \end{gathered}$$ We see then from (\ref{BK2}), (\ref{Q3}) that $\varphi^*$ satisfies the system of equations, $$\label{R3} \begin{gathered} \big( \frac{-\Delta_{d-1} + 2}{2d}\big) \varphi^*(0,y) - \bar\varepsilon_y \; \varphi^*(1,y) - \delta_y\varphi^*(0,y) = 0, \\ \big( \frac{-\Delta_{d-1} + 2}{2d}\big) \varphi^*(1,y) - \varepsilon_y \; \varphi^*(1,y) - \bar\delta_y\varphi^*(0,y) = 0. \end{gathered}$$ Adding the two equations above we conclude that $-\Delta_{d-1}[\varphi^*(0,y) + \varphi^*(1,y)]=0$, $y\in\Omega_{d-1}$, whence on using the normalization $<\varphi^*>_{\hat\Omega} = 1$ we conclude that $\varphi^*(0,y) + \varphi^*(1,y) = 2, \ y \in \Omega_{d-1}$. Hence from (\ref{R3}) we have that $\varphi^*(0,y)$ satisfies the equation, $$\label{S3} \left[ -\Delta_{d-1}/2d + \bar\varepsilon_y +\bar\delta_y \right] \varphi^*(0,y) = 2\bar\varepsilon_y, \ y \in \Omega_{d-1}.$$ Evidently (\ref{S3}) has a unique positive solution. We proceed similarly to obtain a formula for $\psi_0$. Thus $\psi_0$ satisfies the system of equations $$\label{T3} \begin{gathered} \big( \frac{-\Delta_{d-1} + 2}{2d}\big) \psi_0(0,y) - \bar\delta_y \;\psi_0(1,y) + \delta_y \psi_0(0,y) = 0, \\ \big( \frac{-\Delta_{d-1} + 2}{2d}\big) \psi_0(1,y) + \varepsilon_y \psi_0(1,y) - \bar\varepsilon_y \psi_0(0,y) = \varepsilon_y, \quad y \in \Omega_{d-1}. \end{gathered}$$ Adding the two equations in (\ref{T3}) we get $$\label{U3} \left[ -\Delta_{d-1}/2d + \varepsilon_y +\delta_y \right] \left\{ \psi_0(0,y) + \psi_0(1,y)\right\}=\varepsilon_y, \quad y \in \Omega_{d-1}.$$ We may also rewrite the first equation of (\ref{T3}) as, $$\label{V3} \big( \frac{-\Delta_{d-1} + 4}{2d}\big) \psi_0(0,y) = \bar\delta_y \{\psi_0(0,y) + \psi_0(1,y)\}, \quad y \in \Omega_{d-1}.$$ We conclude from (\ref{S3}), (\ref{U3}), (\ref{V3}), and the formula on the LHS of (\ref{P3}) that the effective diffusion constant is $$\label{W3} \begin{split} q(b) &= 2^7d^3 \Big< \left\{ \delta \left[ -\Delta_{d-1} + 2 - V\right]^{-1} \bar\varepsilon \right \} \\ &\quad\times \left[ -\Delta_{d-1} + 4 \right]^{-1} \left\{ \bar\delta \left[ -\Delta_{d-1} + 2 + V\right]^{-1} \varepsilon \right \} \Big>_{\Omega_{d-1}}, \end{split}$$ where $V : \Omega_{d-1} \to \mathbb{R}$ is given by $V(y) = 2d[b(1,y) - b(0,y)]$, $y \in \Omega_{d-1}$. We first show that $q(b) \le 1/2d$ in the case where $V$ is constant. \begin{lemma} \label{lem3.4} Let $q(b)$ be given by \eqref{W3} and assume $V$ is constant. Then there is the inequality, $q(b) \le 1/2d$. \end{lemma} \begin{proof} Since $\varepsilon + \delta= (2 + V)/2d$ there is an $f : \Omega_{d-1} \to \mathbb{R}$ such that $$\label{X3} \begin{gathered} \varepsilon = (2+V)/4d + f, \quad \delta = (2+V)/4d - f, \\ \bar \varepsilon = (2-V)/4d - f, \quad \bar\delta = (2-V)/4d + f. \end{gathered}$$ We rewrite the expression on the RHS of (\ref{W3}) in terms of $f$. To do this we let $w_+, w_-$ be solutions to the equations $$\label{Y3} \begin{gathered} \left[ -\Delta_{d-1} + 2 + V \right]w_+ = f, \quad \left[ -\Delta_{d-1} + 2 - V \right]w_- = f. \end{gathered}$$ It follows that \begin{gather*} \left[ -\Delta_{d-1} + 2 + V \right]^{-1} \varepsilon = 1/4d + w_+ , \\ \left[ -\Delta_{d-1} + 2 - V \right]^{-1} \bar\varepsilon = 1/4d - w_- . \end{gather*} Hence from (\ref{W3}) $q(b)$ is given by the expression, \begin{align*} q(b) &= 2^7d^3 \Big< \Big\{ \big[ \frac{2+V}{4d} - f\big] \big[ \frac 1{4d} - w_-\Big]\Big\} \\ &\big[-\Delta_{d-1} + 4\big]^{-1} \Big\{ \big[ \frac{2-V}{4d} + f\big]\big[ \frac 1{4d} + w_+\big] \Big\} \Big>_{\Omega_{d-1}}. \end{align*} This is a quartic expression in $f$ and the zeroth order term is given by, $$\label{Z3} \text{zeroth order } = \frac 1{2d} \left< (2+V) \left[ -\Delta_{d-1}+4\right]^{-1} (2-V) \right>_{\Omega_{d-1}}.$$ Observe that the expression in (\ref{Z3}) is identical to the RHS of (\ref{P3}) if $\varepsilon = \delta$. For the first order term we have the expression \label{AA3} \begin{aligned} &= 2 \left< (2+V)\left[ -\Delta_{d-1} +4\right]^{-1} f \right>_{\Omega_{d-1}} - 2 \left< f \left[ -\Delta_{d-1}+4\right]^{-1} (2-V) \right>_{\Omega_{d-1}} \\ &\quad + 2 \left< (2+V)\left[ -\Delta_{d-1}+4\right]^{-1} (2-V)w_+ \right>_{\Omega_{d-1}} \\ &\quad -2 \left< (2+V) w_- \left[ -\Delta_{d-1}+4\right]^{-1} (2-V) \right>_{\Omega_{d-1}} . \end{aligned} Now from (\ref{Y3}) we have $$\label{AC3} \begin{gathered} (2-V)w_+ = \left[ -\Delta_{d-1}+4\right] w_+ - f, \\ (2+V)w_- = \left[ -\Delta_{d-1}+4\right] w_- - f. \end{gathered}$$ From this we conclude that the expression in (\ref{AA3}) is the same as $2 \left< (2+V)w_+ \right>_{\Omega_{d-1}} - 2 \left< (2-V)w_- \right>_{\Omega_{d-1}} = 0.$ The second order term in (\ref{W3}) is given by \label{AB3} \begin{aligned} &- 2^3d \left< f \left[ -\Delta_{d-1}+4\right]^{-1} f \right>_{\Omega_{d-1}}\\ &- 2^3d \left< (2+V)w_- \left[ -\Delta_{d-1}+4\right]^{-1} (2-V)w_+ \right>_{\Omega_{d-1}} \\ &- 2^3d \left< f \left[ -\Delta_{d-1}+4\right]^{-1} (2-V)w_+ \right>_{\Omega_{d-1}} \\ &- 2^3d \left< (2+V)w_- \left[ -\Delta_{d-1}+4\right]^{-1} f \right>_{\Omega_{d-1}} \\ &+ 2^3d \left< (2+V) \left[ -\Delta_{d-1}+4\right]^{-1} fw_+ \right>_{\Omega_{d-1}} \\ &+ 2^3d \left< f w_- \left[ -\Delta_{d-1}+4\right]^{-1} (2-V) \right>_{\Omega_{d-1}} . \end{aligned} Observe from (\ref{AC3}) that there is the identity, \label{AD3} \begin{aligned} &\left< (2+V)w_- \left[ -\Delta_{d-1}+4\right]^{-1} (2-V)w_+ \right>_{\Omega_{d-1}}\\ &= \left< f \; \left[ -\Delta_{d-1}+4\right]^{-1} \; f \right>_{\Omega_{d-1}} + \left< w_-\left[ -\Delta_{d-1}+4\right]w_+ \right>_{\Omega_{d-1}} \\ &\quad - \left< f w_+ \right>_{\Omega_{d-1}} - \left< f w_- \right>_{\Omega_{d-1}} . \end{aligned} Similarly, we have $$\label{AE3} \begin{gathered} \big< f \left[ -\Delta_{d-1}+4\right]^{-1} (2-V)w_+ \big>_{\Omega_{d-1}} = - \big< f \left[ -\Delta_{d-1}+4\right]^{-1} \; f \big>_{\Omega_{d-1}} + \left< f w_+ \right>_{\Omega_{d-1}} , \\ \big< (2+V)w_- \left[ -\Delta_{d-1}+4\right]^{-1} f \big>_{\Omega_{d-1}} = - \big< f \; \left[ -\Delta_{d-1}+4\right]^{-1} f \big>_{\Omega_{d-1}} + \left< f w_- \right>_{\Omega_{d-1}} . \end{gathered}$$ Define now $U : \Omega_{d-1} \to \mathbb{R}$ as the solution to the equation, $$\label{AF3} \left[ -\Delta_{d-1}+4\right]U = V.$$ Then from equations (\ref{AB3}) - (\ref{AF3}) we see that the second order term in (\ref{W3}) is given by \label{AG3} \begin{aligned} \text{second order } &= 2^3d \Big\{ - \left< w_- \left[ -\Delta_{d-1}+4\right]w_+ \right>_{\Omega_{d-1}} \\ &\quad + \frac 1 2 \left< f \left[ w_+ + w_-\right]\right>_{\Omega_{d-1}} + \left< fU \left[ w_+ - w_-\right]\right>_{\Omega_{d-1}} \Big\} . \end{aligned} The term of third order is given by \label{AH3} \begin{aligned} &2^5d^2 \left< fw_- \left[ -\Delta_{d-1}+4\right]^{-1} \ f \right>_{\Omega_{d-1}} + 2^5d^2 \left< fw_- \left[ -\Delta_{d-1}+4\right]^{-1} (2-V)w_+ \right>_{\Omega_{d-1}} \\ &- 2^5d^2 \left< f \left[ -\Delta_{d-1}+4\right]^{-1}fw_+\right> - 2^5d^2 \left< (2+V)w_- \left[ -\Delta_{d-1}+4\right]^{-1} \; fw_+ \right>_{\Omega_{d-1}}. \end{aligned} Using (\ref{AC3}) again we see that the expression (\ref{AH3}) is the same as $2^5d^2 \left< fw_- w_+ \right>_{\Omega_{d-1}} - 2^5d^2 \left< fw_- w_+ \right>_{\Omega_{d-1}} = 0.$ Finally the fourth order term is given by $$\label{AI3} \text{fourth order } = 2^7d^3 \left< fw_- \left[ -\Delta_{d-1}+4\right]^{-1} f w_+ \right>_{\Omega_{d-1}} .$$ Hence from (\ref{Z3}), (\ref{AG3}), (\ref{AI3}) we have the formula \label{AJ3} \begin{aligned} q(b) &= \frac 1{2d} \left< (2+V)\left[ -\Delta_{d-1}+4\right]^{-1} (2-V) \right>_{\Omega_{d-1}} \\ &\quad - 2^3d \left< w_- \left[ -\Delta_{d-1}+4\right]w_+ \right>_{\Omega_{d-1}} \\ &\quad + 2^2d \left< f[w_+ + w_-]\right>_{\Omega_{d-1}} + 2^3d \left< fU [w_+ - w_-]\right>_{\Omega_{d-1}} \\ &\quad+ 2^7d^3 \left< fw_- \left[ -\Delta_{d-1}+4\right]^{-1} f w_+ \right>_{\Omega_{d-1}}. \end{aligned} It is clear from the definitions of $V,f$ that $$\label{AK3} |V(y)| < 2, \ |f(y)| < [2 - |V(y)|] /4d, \ \ \ y \in \Omega_{d-1}.$$ From (\ref{AK3}) it follows that there is the inequality \label{AL3} \begin{aligned} 2^7d^3 \left< fw_- \left[ -\Delta_{d-1} + 4\right]^{-1} fw_+\right>_{\Omega_{d-1}} &\le 2^4d^3 \left< (fw_-)^2\right>_{\Omega_{d-1}} + 2^4d^3 \left< (fw_+)^2\right>_{\Omega_{d-1}} \\ &\le d \left< [2 - |V|]^2 [w^2_- + w^2_+] \right> _{\Omega_{d-1}}. \end{aligned} We define now a quadratic form $Q_V$ depending on $V$ by \label{AM3} \begin{aligned} Q_V(f) &= \left< w_-\left[ -\Delta_{d-1} + 4\right] w_+ \right>_{\Omega_{d-1}} - \frac 1 2 \left< f [w_+ + w_-] \right> _{\Omega_{d-1}} \\ &\quad - \left< fU [w_+ - w_-] \right> _{\Omega_{d-1}} - \frac 1 8 \left< [2 - |V|]^2 [w^2_- + w^2_+] \right> _{\Omega_{d-1}}. \end{aligned} It is evident from (\ref{AJ3}), (\ref{AL3}) that $$\label{AN3} q(b) \le \frac 1{2d} \left< (2+V)\left[ -\Delta_{d-1} + 4\right]^{-1} (2-V) \right>_{\Omega_{d-1}}- 2^3 d Q_V(f).$$ Thus to prove the result it will be sufficient to show that $Q_V$ is nonnegative definite. Since $V$ is constant we can compute the eigenvalues of $Q_V$ explicitly. Thus if $p^2$ denotes the eigenvalue of $-\Delta_{d-1}$, corresponding to the eigenvector $\exp[i\xi \cdot x], x \in \Omega_{d-1}$, then the eigenvalue of $Q_V$ is \label{AO3} \begin{aligned} &\Big\{ 2(4 + V^2)(p^2 + 2 + V)(p^2 + 2 - V) \\ &-\frac 12(2 - |V|)^2 \left[ (p^2 + 2 + V)^2+(p^2 + 2 - V)^2\right] \Big\} \Big/ 4(p^2 + 2 + V)^2 (p^2 + 2 - V)^2. \end{aligned} We can rewrite the numerator of (\ref{AO3}) as $$\label{AP3} \left\{ 2(4 + V^2) - (2 - |V|)^2 \right\} [p^2+2]^2- V^2 \left\{ 2(4 + V^2) + (2 - |V|)^2 \right\}.$$ Since $|V| < 2$ the expression in (\ref{AP3}) is bounded below by its value for $p=0$ which can be written as $(4 + V^2) (2 - |V|) [2 + 3 |V|] \ge 0.$ \end{proof} We proceed now to the general case which will follow from the fact that the quadratic form (\ref{AM3}) is nonnegative definite for any $V$ satisfying (\ref{AK3}). From here on we shall denote $\Delta_{d-1}, \langle\cdot\rangle_{\Omega_{d-1}}$ simply as $\Delta$ and $\langle\cdot\rangle$ respectively. We first note that by using (\ref{Y3}) we can obtain some alternate representations for $Q_V$. Thus if we write $\left< w_-\left[ -\Delta_{d-1} + 4\right] w_+ \right> = 2 \left + \frac 1 2 \left< f[w_+ + w_-] \right>,$ we see that $Q_V$ is given by $$\label{AQ3} Q_V(f) = 2 \left< w_- w_+ \right> - \left< fU [w_+ - w_-] \right> - \frac 1 8 \left< [2 - |V|]^2 [w^2_- + w^2_+]\right> .$$ We also have \begin{align*} \left< f U [w_+ - w_-] \right> &= \left< \left\{ U[-\Delta + 2 - V]w_- \right\} w_+ \right> - \left< \left\{ U[-\Delta + 2 +V]w_+ \right\} w_- \right> \\ &= - 2 \left< UV w_- w_+ \right> - \left< U w_+ \Delta w_- \right> + \left< U w_- \Delta w_+ \right>. \end{align*} Hence from (\ref{AQ3}) we have \label{AR3} \begin{aligned} Q_V(f)&= 2 \left< w_- w_+[1+UV] \right> + \left< U w_+ \Delta w_- \right>\\ &\quad -\left< U w_- \Delta w_+ \right> - \frac 1 8 \left< [2 - |V|]^2 [w^2_- + w^2_+]\right> . \end{aligned} We first show that a simple quadratic form related to $Q_V$ is nonnegative definite. \begin{lemma} \label{lem3.5} Let $V$ satisfy (\ref{AK3}) and $w_+, w_-$ be solutions to (\ref{Y3}) for any $f : \Omega_{d-1} \to \mathbb{R}$. Then there is the inequality $\left \ge 0$. \end{lemma} \begin{proof} Let $w$ be the solution to the equation, $$\label{AS3} [-\Delta + 2 + V] [-\Delta + 2 - V]w = f.$$ Then from (\ref{Y3}) we see that $w_+ = [-\Delta + 2 - V]w$. Hence we have from (\ref{Y3}), (\ref{AS3}) that \begin{align*} \left< w_+ w_-\right> &= \left< [-\Delta + 2 - V]w \ w_- \right> \\ &= \left< w[-\Delta + 2 - V] w_- \right> = \left \\ &= \left<\left\{ \left[ -\Delta + 2 + V \right] w \right\} \left\{ \left[-\Delta + 2 - V \right] w \right\} \right> \\ &=\left< \left\{ (-\Delta + 2 )w \right\}^2 \right> - \big< V^2w^2 \big>\\ &\ge \big< (4-V^2)w^2 \big> \ge 0. \end{align*} \end{proof} To proceed further we need to localize the quadratic form (\ref{AR3}). \begin{lemma} \label{lem3.6} Let ${\mathcal{L}}_+, {\mathcal{L}}_-$ be operators on functions $\Phi : \Omega_{d-1} \to \mathbb{R}$ defined by \begin{gather*} {\mathcal{L}}_+ \Phi = (-\Delta + 2) \Phi / V - \Phi, \\ {\mathcal{L}}_- \Phi = (-\Delta + 2) \Phi / V + \Phi, \end{gather*} where we assume $V$ satisfies (\ref{AK3}) and $V(y) \not= 0, y \in \Omega_{d-1}$. Then ${\mathcal{L}}_+, {\mathcal{L}}_-$ are invertible and there is the identity, $$\label{AT3} [-\Delta + 2 + V]{\mathcal{L}}_+= [-\Delta + 2 - V] {\mathcal{L}}_- \, .$$ \end{lemma} The statement of the above lemma is easy to verify; we omit its proof. It follows from Lemma \ref{lem3.6} that we can choose $f$ in (\ref{Y3}), (\ref{AR3}) as the operator (\ref{AT3}) acting on a function $\Phi : \Omega_{d-1} \to \mathbb{R}$, in which case $w_+ = {\mathcal{L}_+ }\Phi, \ w_- = {\mathcal{L}}_- \Phi$. If we substitute into (\ref{AR3}) we obtain a quadratic form $\tilde Q_V(\Phi)$ which is local in $\Phi$, and it is this quadratic form which we will show is nonnegative definite. First we show that the quadratic form obtained from $\tilde Q_V$ upon replacing $U$ by $V/4$ is nonnegative definite. \begin{lemma} \label{lem3.7} For $\Phi : \Omega_{d-1} \to \mathbb{R}$ and $w_+ = {\mathcal{L}}_+ \Phi$, $w_- = {\mathcal{L}}_- \Phi$, there is for $d=2$ the inequality, $$\label{AU3} 2 \left< w_- w_+ [4 + V^2] \right> + \left< Vw_+ \Delta w_- \right> - \left< Vw_- \Delta w_+ \right> - \frac 1 2 \left< [2 - |V|]^2 [w^2_- + w^2_+]\right> \ge 0.$$ \end{lemma} \begin{proof} We first note that the first term in (\ref{AU3}) is nonnegative. Thus we have \begin{align*} \left< w_- w_+ [4 + V^2] \right> &= \left< [(-\Delta + 2)\Phi]^2 [4/V^2 + 1] \right> - \left<\Phi^2 [4 + V^2] \right> \\ &\ge 2 \left< [(-\Delta + 2)\Phi]^2 \right> - \left< \Phi^2[4 + V^2] \right> \ge \left< \Phi^2[4 - V^2] \right> \ge 0, \end{align*} where we have used (\ref{AK3}). The second and third terms of (\ref{AU3}) are given by the formula \begin{align*} &\left< Vw_+ \Delta w_- \right> - \left< Vw_- \Delta w_+ \right> \\ &= 2\left< [(-\Delta + 2)\Phi](\Delta \; \Phi) \right> - 2\big< [\Delta(V\Phi)] \big[ \frac{(-\Delta +2)\Phi}{V} \big]\big>, \end{align*} on summation by parts. From the last two equations we therefore have that \label{AV3} \begin{aligned} & 2\left< w_- w_+ [4 + V^2] \right> + \left< Vw_+ \Delta w_- \right> - \left< Vw_- \Delta w_+ \right> \\ & = 8 \big< \frac{[ (-\Delta +2)\Phi]^2} {V^2} \big> - 2 \left< V^2 \Phi^2\right> + 4 \left< (\nabla \Phi)^2 \right> - 2 \big< [ \Delta(V \Phi)] \big[ \frac {(-\Delta +2)\Phi]}{V} \big] \big> . \end{aligned} Using the fact that \begin{align*} &-\frac{1} {V(x)} \Delta(V \Phi)(x)\\ &= 2(d-1) \Phi(x) - \sum^d_{j=2} \Big[ \frac{V(x+{\bf e}_j)} {V(x)} \Phi(x+{\bf e}_j) + \frac{V(x-{\bf e}_j)}{V(x)} \Phi(x-{\bf e}_j)\Big], \end{align*} we conclude from (\ref{AV3}) that \label{AW3} \begin{aligned} &\left< w_- w_+ [4 + V^2] \right> + \frac 1 2 \left< Vw_+ \Delta w_- \right> - \frac 1 2 \left< Vw_- \Delta w_+ \right> \\ &= 4 \big< \frac{[ (-\Delta +2)\Phi]^2}{V^2} \big> + \left< [4(d-1) - V^2]\Phi^2 \right> + 2d \left< (\nabla \Phi)^2 \right> \\ &\quad - \Big< [(-\Delta + 2)\Phi(\cdot)] \sum^d_{j=2} \Big[ \frac{V(\cdot +{\bf e}_j)}{V(\cdot)} \Phi(\cdot +{\bf e}_j) + \frac{V(\cdot-{\bf e}_j)}{V(\cdot)} \Phi(\cdot -{\bf e}_j) \Big] \Big>. \end{aligned} Now the Schwarz inequality yields \begin{align*} \big| \left[ (-\Delta+2)\Phi(x)\right] \frac{V(y)\Phi(y)}{V(x)} \big| &\le V(y)^2 \Big[ \alpha \; \frac{\Phi(y)^2}{V(y)^2} + \frac 1{4\alpha} \frac{[(-\Delta + 2)\Phi(x)]^2}{V(x)^2} \Big] \\ &\le \alpha \Phi(y)^2 + \frac 1\alpha \frac{[(-\Delta + 2)\Phi(x)]^2}{V(x)^2}, \quad x,y \in \Omega_{d-1}, \end{align*} for any $\alpha > 0$. Hence there is from (\ref{AW3}) the inequality, \label{AVV3} \begin{aligned} &\left< w_- w_+ [4 + V^2] \right> + \frac 1 2 \left< Vw_+ \Delta w_- \right> - \frac 1 2 \langle Vw_- \Delta w_+ \rangle \\ &\ge \big[ 4 - \frac{2(d-1)}{\alpha} \big] \big< \frac{[ (-\Delta +2)\Phi]^2}{V^2} \big> \\ &\quad + \left< \left[2(d-1)(2-\alpha)-V^2\right]\Phi^2\right> + 2d\left<(\nabla\Phi)^2\right>. \end{aligned} For $d=3$ and $\alpha = 1$ the RHS of (\ref{AVV3}) is evidently nonnegative but this is no longer the case when $d>3$. For $\alpha = 1, d=2$ the RHS of (\ref{AVV3}) is bounded below by the nonnegative quantity, $$\label{AWW3} \frac 1 2 \big< \big( \frac{4-V^2}{V^2}\big) \left[ (-\Delta + 2)\Phi\right]^2\big> + \left< (4-V^2)\Phi^2 \right>.$$ We consider now the last term in the expression (\ref{AU3}). We have $$\label{AX3} \frac 1 4 \left< [2 - |V|]^2 [w^2_- + w^2_+] \right> = \frac 1 2 \big< [2-|V|]^2 \big\{ \frac{[ (-\Delta +2)\Phi]^2}{V^2} + \Phi^2\big\} \big>.$$ If we now use the inequality $[2 - |V|]^2 \le 4 - V^2$ we see that the expression (\ref{AX3}) is less than (\ref{AWW3}). Hence the inequality (\ref{AU3}) is established for $d=2$. \end{proof} Next we turn to showing that $\tilde Q_V$ is nonnegative definite for $d=2$. To do this we write the solution $U$ of (\ref{AF3}) as $$\label{AZ3} U(x) = \frac 1 4 \sum_y G(y) V(x+y),$$ where $G(y)$ is the Green's function for $[-\Delta/4 + 1]^{-1}$, whence $G(y)$ is nonnegative for all $y$ and $$\label{AY3} \sum_y G(y) = 1, \quad G(y) = G(-y), \quad y = 1,2,\dots \,.$$ We consider the first three terms in the expression (\ref{AR3}) for $Q_V$. Using Lemma \ref{lem3.6} and setting $w_+ = {\mathcal{L}}_+ \Phi, w_- = {\mathcal{L}}_- \Phi$ we have that \label{BA3} \begin{aligned} & \left< w_- w_+ [1 + UV] \right> + \frac 1 2 \left< Uw_+ \Delta w_- \right> -\frac 1 2 \left< Uw_- \Delta w_+\right> \\ &= \big< \frac 1{V^2} \left[ (-\Delta + 2) \Phi \right]^2 [1 + UV] \big> - \left< \Phi^2[1 + UV] \right> \\ &\quad + \big< \frac U V [\Delta\Phi] \left[ (-\Delta + 2) \Phi\right]\big> - \big< \Delta(U\Phi) \frac 1 V [-\Delta+2]\Phi \big> \\ &= \big< \frac 1{V^2} \left[ (-\Delta + 2) \Phi \right]^2\big> - \langle \Phi^2 \rangle - \big<\Phi^2 UV\big> \\ &\quad + 2 \big< \frac U V \Phi (-\Delta + 2) \Phi \big> - \big< \Delta(U\Phi) \frac 1 V[-\Delta + 2]\Phi \big> \\ &= \big< \frac 1{V^2} \left[ (-\Delta + 2) \Phi \right]^2 \big> - \big< \Phi^2 \big> - \big<\Phi^2 UV\big> \\ &\quad + 4 \big< \frac U V \Phi (-\Delta + 2) \Phi \big> - \big< (\tau_1 U)(\tau_1 \Phi) \frac 1 V[-\Delta + 2]\Phi\big> \\ &\quad- \big< (\tau_{-1} U)(\tau_{-1} \Phi) \frac 1 V[-\Delta + 2]\Phi\big>, \end{aligned} where $\tau_x \varphi(y) = \varphi(x + y)$, $y \in \mathbb{Z}$. We consider the last three terms in the previous expression. We write using \eqref{AZ3}, \begin{gather*} 4 \big< \frac U V \Phi (-\Delta + 2) \Phi \big> = G(0) \left< (\nabla \Phi)^2 +2\Phi^2 \right> + \sum_{y \not= 0} G(y) \big< \frac{\tau_y V}{V} \ \Phi(-\Delta + 2) \Phi \big>, \\ \begin{aligned} \big< (\tau_1 U)(\tau_1 \Phi) \frac 1 V[-\Delta + 2]\Phi\big> &= \frac 1 4 G(-1) \left\{ \left< \nabla(\tau_1\Phi) \nabla\Phi \right> + 2 \left<( \tau_1\Phi) \Phi \right> \right\} \\ &\quad + \frac 1 4 \sum_{y \not= 0} G(y-1) \big< \frac{\tau_y V}{V} \tau_1 \Phi(-\Delta + 2) \Phi \big> , \end{aligned} \end{gather*} with a similar expression for the last term in (\ref{BA3}). We conclude from this that the last three terms of (\ref{BA3}) are given by, \label{BB3} \begin{aligned} &4 \big< \frac U V \Phi (-\Delta + 2) \Phi \big> - \big< (\tau_1 U)(\tau_1 \Phi) \frac 1 V[-\Delta + 2]\Phi\big> - \big< (\tau_{-1} U)(\tau_{-1} \Phi) \frac 1 V[-\Delta + 2]\Phi\big> \\ &= G(0) \left< (\nabla \Phi)^2 + 2\Phi^2 \right> - \frac 1 4 G(-1) \left\{ \left< \nabla(\tau_1\Phi)\nabla \Phi \right> + 2 \left< (\tau_1\Phi )\Phi \right> \right\}\\ &\quad - \frac 1 4 G(1) \left\{ \left< \nabla(\tau_{-1}\Phi)\nabla \Phi \right> + 2 \left< (\tau_{-1}\Phi) \Phi \right> \right\} \\ &\quad + \sum_{y \ge 1} \big[ G(y) - \frac 1 4 \; G(y-1) - \frac 1 4 \; G(y+1) \big] \big< \frac{\tau_y V}{V} \ \tau_1 \Phi(-\Delta + 2) \Phi \big> \\ &\quad +\quad \sum_{y \le -1} \big[ G(y) - \frac 1 4 \; G(y-1) - \frac 1 4 \; G(y+1) \big] \big< \frac{\tau_y V}{V} \ \tau_{-1} \Phi(-\Delta + 2) \Phi \big> \\ &\quad+ \sum_{y \ge 1} G(y) \big< \frac{\tau_y V}{V} \left[ \Phi - \tau_1 \Phi\right] (-\Delta + 2) \Phi \big>\\ &\quad - \frac 1 4 \sum_{y \ge 1} G(y+1) \big< \frac{\tau_y V}{V} \left[ \tau_{-1}\Phi - \tau_1 \Phi\right] (-\Delta + 2) \Phi \big> \\ &\quad + \sum_{y \le -1} G(y) \big< \frac{\tau_y V}{V} \left[ \Phi - \tau_{-1} \Phi\right] (-\Delta + 2) \Phi \big> \\ &\quad - \frac 1 4 \sum_{y \le -1} G(y-1) \big< \frac{\tau_y V}{V} \left[ \tau_1\Phi - \tau_{-1} \Phi\right] (-\Delta + 2) \Phi \big>. \end{aligned} We shall use the representation (\ref{BB3}) to show that the quadratic form $Q_V$ is non-negative definite. \begin{lemma} \label{lem3.8} Suppose the function $G(y)$ of \eqref{AZ3} is decreasing, non-negative for $y \ge 1$, satisfies (\ref{AY3}) and the inequalities, $$\label{CB3} (-\Delta + 2) G(y) \le 0, \quad y \ge 1; \quad 1 - G(0) - 2G(1) < G(1)/2; \quad G(2) < G(1)/5.$$ Then the quadratic form $Q_V$ of (\ref{AR3}) is nonnegative definite. \end{lemma} \begin{proof} We estimate the terms in (\ref{BB3}) by applying the Schwarz inequality. Before doing this we make one further simplification of terms in (\ref{BB3}). We write \label{BC3} \begin{aligned} & \sum_{y \ge 1} G(y) \big< \frac{\tau_y V}{V} \left[ \Phi - \tau_1 \Phi\right] (-\Delta + 2) \Phi \big> + \sum_{y \le -1} G(y) \big< \frac{\tau_y V}{V} \left[ \Phi - \tau_{-1} \Phi\right] (-\Delta + 2) \Phi \big> \\ & = \sum_{y \ge 2} G(y) \big< \frac{\tau_y V}{V} \left[ \Phi - \tau_1 \Phi\right] (-\Delta + 2) \Phi \big> + \sum_{y \le -2} G(y) \big< \frac{\tau_y V}{V} \left[ \Phi - \tau_{-1} \Phi\right] (-\Delta + 2) \Phi \big> \\ &\quad + G(1) \big< \tau_1 V \left[ \frac 1 V - \frac V 4\right] \left[ \Phi - \tau_1 \Phi\right] (-\Delta + 2) \Phi \big>\\ &\quad +G(-1) \big< \tau_{-1} V \left[ \frac 1 V - \frac V 4\right] \left[ \Phi - \tau_{-1} \Phi\right] (-\Delta + 2) \Phi \big> \\ &\quad - \frac{G(1)}4 \left< (\tau_1 V) V[\Phi - \tau_1 \Phi] \Delta \Phi \right> - \frac{G(-1)}4 \left< (\tau_{-1} V) V[\Phi - \tau_{-1} \Phi] \Delta \Phi \right> \\ &\quad + \frac{G(1)} 2 \left< (\tau_1 V)V(\nabla \Phi)^2 \right>, \end{aligned} where we have used the fact that $G(1) = G(-1)$. We also rewrite the first two terms on the RHS of (\ref{BC3}) as \label{BD3} \begin{aligned} &\sum_{y \ge 2} G(y) \big< \frac{(\tau_y V-V)}{V} \left[ \Phi - \tau_1 \Phi\right] (-\Delta + 2) \Phi \big> \\ &+ \sum_{y \le -2} G(y) \big< \frac{(\tau_y V-V)}{V} \left[ \Phi - \tau_{-1} \Phi\right] (-\Delta + 2) \Phi \big> \\ & +\sum_{y \ge 2} G(y) \left< (\Delta\Phi)^2 + 2 (\nabla \Phi)^2 \right> . \end{aligned} We similarly rewrite the sum of the last and third last terms of (\ref{BB3}) as \label{BE3} \begin{aligned} &- \frac 1 4 \sum_{y \ge 1} G(y+1) \big< \frac{(\tau_y V-V)}{V} \left[ \tau_{-1}\Phi - \tau_1 \Phi\right] (-\Delta + 2) \Phi \big> \\ &- \frac 1 4 \sum_{y \le -1} G(y-1) \big< \frac{(\tau_y V-V)}{V} \ \left[ \tau_1\Phi - \tau_{-1} \Phi\right] (-\Delta + 2) \Phi \big> . \end{aligned} Consider now the first three terms on the RHS of (\ref{BA3}). These can be written as \label{BF3} \begin{aligned} &\big< \big( \frac 1{V^2} - \frac 1 4 \big) [(-\Delta + 2) \Phi]^2 \big> + \frac 1 4 \big< \left(\Delta \Phi\right)^2 \big> + \big<\left(\nabla \Phi\right)^2 \big> \\ &-\frac 1 4 G(0) \left< V^2 \Phi^2 \right> + \frac 1 8 \sum_{y\not= 0} G(y) \left[ \left< (\tau_y V-V)^2 \Phi^2\right> - \left< \left\{ (\tau_y V)^2 + V^2 \right\}\Phi^2 \right> \right]. \end{aligned} Next we apply the Schwarz inequality to terms in (\ref{BB3}). Thus we estimate $$\label{BG3} \big| \big< \frac{\tau_y V}{V} \; \tau_1\Phi(-\Delta + 2)\Phi \big> \big| \le \left <\Phi^2\right> + \big< \frac 1{V^2} [(-\Delta +2)\Phi]^2\big>,$$ with a similar estimate when $\tau_1 \Phi$ is replace by $\tau_{-1} \Phi$. Observe now that $\sum_{y \ge 1} \big[ G(y) - \frac 1 4 G(y-1) - \frac 1 4 G(y+1)\big] = -\frac{1}4 [2G(0)-1-G(1)],$ where we have used (\ref{AY3}). Hence on using the fact that $[-\Delta + 2]G(y) < 0$, $y \ge 1$, we see from (\ref{BG3}) that the sum of the first five terms on the RHS of (\ref{BB3}) are bounded below by the expression, $$\label{BH3} [G(0) - \frac 1 2 \; G(1)] \left< (\nabla \Phi)^2 + 2\Phi^2 \right> -\big\{ \left<\Phi^2\right> + \big< \frac 1{V^2} [(-\Delta +2)\Phi]^2\big> \big\} [G(0) - \frac 1 2 - \frac 1 2 \; G(1)].$$ If we combine the estimate (\ref{BH3}) with (\ref{BF3}) and use the fact that $|V| < 2$ we get a lower bound for the sum of the first three terms of (\ref{BA3}) and the first five terms of (\ref{BB3}). It is given by, \label{BI3} \begin{aligned} &\big< \big( \frac 1{V^2} - \frac 1 4 \big) \left[ (-\Delta + 2)\Phi\right]^2 \big> \big\{ \frac 3 2 + \frac 1 2 \; G(1) - G(0) \big\} \\ &+ \frac 1 4 \left< (\Delta \Phi)^2 \right> \big\{ \frac 3 2 + \frac 1 2 \; G(1) - G(0) \big\} + \frac 3 2 \langle (\nabla \Phi)^2 \rangle + \frac 1 4 G(0) \left<(4-V^2) \Phi^2 \right> \\ & + \frac 1 8 \sum_{y\not= 0} G(y) \left< (\tau_y V-V)^2 \Phi^2 \right> + [1-G(0)] \langle \Phi^2 \rangle\\ & - \frac 1 8 \sum_{y\not= 0} G(y) \left< \left\{ (\tau_y V)^2 + V^2\right\} \Phi^2 \right>. \end{aligned} Observe that all terms in (\ref{BI3}) except for the final one are nonnegative. Furthermore, the sum of the last two terms is nonnegative. Next we estimate the terms on the RHS of (\ref{BC3}) which involve $G(1)$ and $G(-1)$. To do this we use the Schwarz inequalities \label{BJ3} \begin{gathered} \begin{aligned} & \big| \big< \tau_1 V \big[ \frac 1 V - \frac V 4 \big] \left[ \Phi - \tau_1 \Phi \right] (-\Delta + 2)\Phi \big> \big| \\ &\le \alpha_1 \big< \big[ \frac 1 {V^2} - \frac 1 4 \big] \left[ (-\Delta + 2)\Phi\right]^2\big> + \frac 1{\alpha_1} \left< (\nabla\Phi)^2\right>, \end{aligned} \\ \frac 1 2 \left| \left< (\tau_1 V)V \left[ \Phi - \tau_1 \Phi \right] \Delta\Phi \right> \right| \le \alpha_2 \left< (\Delta\Phi)^2\right> +\frac 1{\alpha_2} \left< (\nabla\Phi)^2\right> , \end{gathered} for any constants $\alpha_1, \alpha_2 > 0$. Hence on using the fact that $|V| < 2$ we see that the expression (\ref{BI3}) plus the terms in $G(1), G(-1)$ of (\ref{BC3}) is bounded below by the expression, \label{BK3} \begin{aligned} &\big< \big( \frac 1{V^2} - \frac 1 4 \big) [ (-\Delta + 2)\Phi]^2 \big> \big\{ \frac 3 2 + \frac 1 2 \; G(1) - G(0) - 2\alpha_1 G(1) \big\}\\ &+ \frac 1 4 \left< (\Delta \Phi)^2 \right> \big\{ \frac 3 2 + \frac 1 2 G(1) - G(0) - 4\alpha_2 G(1)\big\}\\ &+ \left< (\nabla \Phi)^2 \right> \big\{ \frac 3 2 - 2G(1) - \frac 2{\alpha_1} G(1) - \frac 1{\alpha_2} G(1) \big\} \\ &+ \frac 1 4 \ G(0) \left<(4-V^2) \Phi^2 \right> + \frac 1 8 \sum_{y\not= 0} G(y) \left< (\tau_y V-V)^2 \Phi^2 \right>. \end{aligned} Let us assume for the moment that $G(y) = 0$ for $y \ge 2$. Then (\ref{BA3}) is bounded below by (\ref{BK3}). We write $G(0) = 1 - \gamma$ whence $G(1) = \gamma/2$. Since $(-\Delta + 2) G(y) \le 0$, $y \ge 1$, we must have $\gamma < 1/3$. We choose $\alpha_1$ such that $$\label{BL3} \frac 3 2 + \frac 1 2 \; G(1) - G(0) - 2\alpha_1 G(1) = \frac 1 2,$$ which yields $\alpha_1 = 5/4$. We choose $\alpha_2$ so that $$\label{BM3} \frac 3 2 + \frac 1 2 \; G(1) - G(0) - 4\alpha_2 G(1) = 0,$$ which yields $\alpha_2 = 5/8 + 1/4\gamma$. The coefficient of $<(\nabla \Phi)^2>$ in (\ref{BK3}) is therefore bounded below by $1.5 - 2.6\gamma > 0$ since $\gamma < 1/3$. Hence from (\ref{AX3}) the quadratic form $Q_V/2$ of (\ref{AR3}) is bounded below by twice the expression, \label{BN3} \begin{aligned} &\frac 1 2 \big< \big( \frac 1{V^2} - \frac 1 4 \big) \left[ (-\Delta + 2)\Phi \right]^2 \big> + \frac 1 4 G(0) \left< (4-V^2)\Phi^2 \right> \\ &- \frac 1 8 \big< [2 - |V|]^2 \big\{ \frac {[(-\Delta+2)\Phi]^2} {V^2} + \Phi^2 \big\} \big>. \end{aligned} If we now use the fact that $[2 - |V|]^2 \le 4 - V^2$ we see that (\ref{BN3}) is nonnegative. To complete the proof of the lemma we need to estimate the sum of the terms in (\ref{BD3}), (\ref{BE3}). We rewrite these as \label{BO3} \begin{aligned} &- \frac 1 4 G(2) \big< \frac{(\tau_1 V-V)}{V} \left[ \tau_{-1}\Phi - \tau_1 \Phi\right] (-\Delta + 2) \Phi \big> \\ &- \frac 1 4 G(-2) \big< \frac{(\tau_{-1} V-V)}{V} \left[ \tau_1\Phi - \tau_{-1} \Phi\right] (-\Delta + 2) \Phi \big> \\ &+ \sum_{y \ge 2} \big[ G(y) - \frac 1 4 G(y+1)\big] \big< \frac{(\tau_y V-V)}{V} \left[ \Phi -\tau_1\Phi \right] (-\Delta + 2) \Phi \big> \\ &+ \frac 1 4 \sum_{y \ge 2}G(y+1)\big< \frac{(\tau_y V-V)}{V} \left[ \Phi - \tau_{-1} \Phi\right] (-\Delta + 2) \Phi \big> \\ &+ \sum_{y \le -2} \big[ G(y) - \frac 1 4 \;G(y-1)\big] \big< \big(\frac{\tau_y V-V}{V}\big) \left[ \Phi -\tau_{-1}\Phi \right] (-\Delta + 2) \Phi \big> \\ &+ \frac 1 4 \sum_{y \le -2}G(y-1) \big< \frac{(\tau_y V-V)}{V} \left[ \Phi - \tau_1\Phi\right] (-\Delta + 2) \Phi \big> \\ &+ \frac 1 2 [1 - G(0) - 2G(1)] \left< (\Delta \Phi)^2 + 2(\nabla \Phi)^2 \right> . \end{aligned} We first estimate the third term in (\ref{BO3}). Thus we write \label{BP3} \begin{aligned} &\big< \frac{(\tau_y V-V)}{V} \ \left[ \Phi -\tau_1\Phi \right] (-\Delta + 2) \Phi \big> \\ &= \big< (\tau_y \; V-V) \big[ \frac 1 V - \frac V 4 \big] [\Phi - \tau_1\Phi] (-\Delta + 2) \Phi \big> \\ &\quad - \frac 1 4 \left< (\tau_y \; V-V) V[\Phi - \tau_1\Phi] \Delta \Phi \right> + \frac 1 2 \left< (\tau_y \; V-V) V[\Phi - \tau_1\Phi] \Phi \right> . \end{aligned} We estimate the first two terms on the RHS of (\ref{BP3}) similarly to (\ref{BJ3}). For the third term we use $$\label{BQ3} | \left< (\tau_y \; V-V) V[\Phi - \tau_1\Phi] \Phi \right> | \le \alpha \left< (\nabla \Phi)^2 \right> + \frac 1 \alpha \left< (\tau_y \; V-V)^2 \Phi^2 \right>,$$ for any $\alpha > 0$. Choosing $\alpha = 4$ in (\ref{BQ3}) it follows that the sum of the third, fourth, fifth and sixth terms of (\ref{BO3}) is bounded below by \label{BR3} \begin{aligned} &- \sum_{|y| \ge 2} G(y) \Big\{ 2\alpha_3 \big< \big( \frac 1{V^2} - \frac 1 4 \big) [(-\Delta + 2)\Phi]^2 \big> \\ &+ \alpha_4 \left< (\Delta\Phi)^2 \right> + \big[ \frac 2{\alpha_3} + \frac 1{\alpha_4} + 2 \big] \left< (\nabla \Phi)^2 \right> + \frac 1 8 \left< (\tau_y \; V-V)^2 \Phi^2 \right> \Big\}, \end{aligned} for any $\alpha_3, \alpha_4 > 0$. We estimate the sum of the first two terms in (\ref{BO3}) from below similarly. Choosing now $\alpha = 2G(2) / G(1)$ in (\ref{BQ3}) we obtain the lower bound, \label{BS3} \begin{aligned} &- G(2) \Big\{ 2\alpha_5 \big< \big( \frac 1{V^2} - \frac 1 4 \big) [(-\Delta + 2)\Phi]^2 \big> \\ &+ \alpha_6 \left< (\Delta\Phi)^2 \right> + \big[ \frac 2{\alpha_5} + \frac 1{\alpha_6} + \frac{G(2)}{G(1)} \big] \left< (\nabla \Phi)^2 \right>\Big\} \\ &- \frac {G(1)} 8 \left< (\tau_1 \; V-V)^2 \Phi^2 \right> - \frac {G(-1)} 8 \left< (\tau_{-1} \; V-V)^2 \Phi^2 \right> , \end{aligned} for any $\alpha_5, \alpha_6 > 0$. We may now obtain a lower bound for (\ref{BA3}) by adding (\ref{BK3}) to the final term in (\ref{BO3}) and the expressions of (\ref{BR3}) and (\ref{BS3}). We obtain the lower bound, \label{BT3} \begin{aligned} & \big< \big( \frac 1{V^2} - \frac 1 4 \big) \left[ (-\Delta + 2)\Phi\right]^2 \big> \Big\{ \frac 3 2 + \frac 1 2 \; G(1) - G(0) \\ &- 2\alpha_1 G(1) - 2\alpha_5 G(2) - 2\alpha_3 [1 - G(0) - 2G(1)] \Big\} \\ &+ \frac 1 4 \left< (\Delta \Phi)^2 \right> \Big\{ \frac 3 2 + \frac 1 2 \; G(1) - G(0) - 4\alpha_2 G(1) - 4\alpha_6 G(2) \\ &- 4\alpha_4 [1 - G(0) - 2G(1)] + 2 [1 - G(0) - 2G(1)] \Big\} \\ &+ \left< (\nabla \Phi)^2 \right> \Big\{ \frac 3 2 - 2G(1) - \frac 2{\alpha_1} G(1) - \frac 1{\alpha_2} G(1) - G(2)^2/G(1) \\ &- \big[ \frac 2{\alpha_5} + \frac 1 {\alpha_6}\big] G(2) - \big[ \frac 2{\alpha_3} + \frac 1{\alpha_4} + 2\big] [1-G(0)-2G(1)] \\ &+ [1 - G(0) - 2G(1)] \Big\} + \frac 1 4 G(0) \left< (4-V^2) \Phi^2 \right> . \end{aligned} We may rewrite the coefficient of the first term in (\ref{BT3}) as \label{BU3} \begin{aligned} &\frac 3 2 + \frac 1 2 \; G(1) - G(0) - 2\alpha_1 G(1) - 2\alpha_5 G(2) -2\alpha_3 [1 - G(0) - 2G(1)] \\ &= \frac 1 2 + (1-2\alpha_3) [1 - G(0) - 2G(1)] + \big[ \frac 5 2 - 2\alpha_1 - 2\alpha_5 \frac{G(2)}{G(1)} \big] G(1). \end{aligned} If we set now $$\label{BV3} \alpha_3 = 1/2, \ \ \alpha_1 + \alpha_5 G(2) /G(1) = 5/4,$$ we see as in (\ref{BL3}) that the coefficient of the first term in (\ref{BT3}) is $1/2$. We similarly rewrite the coefficient of the second term as \label{BW3} \begin{aligned} &\frac 3 2 +\frac 1 2 G(1) - G(0) - 4\alpha_2 G(1) - 4\alpha_6 G(2) \\ &- 4\alpha_4 [1 - G(0) - 2G(1)] + 2 [1 - G(0) - 2G(1)] \\ &=\frac 1 2 +(3-4\alpha_4 )[1 - G(0) - 2G(1)] + \big[ \frac 5 2 - 4\alpha_2 - 4\alpha_6 \frac{G(2)}{G(1)} \big] G(1). \end{aligned} Hence if we set $$\label{BX3} \alpha_2 = 5/8, \quad \alpha_4 = 3/4, \quad \alpha_6 = 1/8 \; G(2)$$ then the second term in (\ref{BT3}) is zero. We consider the third term in (\ref{BT3}). This can be written as \label{BY3} \begin{aligned} & \frac 3 2 - 2G(1) - \frac 2{\alpha_1} G(1) - \frac 1{\alpha_2} G(1) - G(2)^2/G(1) \\ &- \big[ \frac 2{\alpha_5} + \frac 1{\alpha_6}\big] G(2) - \big[ \frac 2{\alpha_3} + \frac 1{\alpha_4} + 1\big] [1-G(0)-2G(1)] \\ &= - \frac 3 2(-\Delta + 2)G(1) + \big[ \frac 1 2 - \frac 2{\alpha_3} - \frac 1{\alpha_4}\big][1 - G(0) - 2G(1)] \\ &\quad + \big[ 7 - \frac 2{\alpha_1} - \frac 1{\alpha_2}\big]G(1)-G(2)^2/G(1) - \big[ \frac 3 2 + \frac 2{\alpha_5} + \frac 1{\alpha_6}\big] G(2). \end{aligned} Using the inequalities (\ref{CB3}) we see this is bounded below by the expression, \label{BZ3} \begin{aligned} &\left\{ 7 - \frac 2{\alpha_1} - \frac 1{\alpha_2} + \frac 1 2 \Big[ \frac 1 2 - \frac 2{\alpha_3} - \frac 1{\alpha_4}\Big] - \frac 1{25} - \frac 1 5 \Big[ \frac 3 2 + \frac 2{\alpha_5} + \frac 1{\alpha_6}\Big] \right\}G(1) \\ &= \left\{ 6.91 - \frac 2{\alpha_1} - \frac 1{\alpha_2} - \frac 1{\alpha_3} - \frac 1{2\alpha_4} - \frac 2{5\alpha_5} - \frac 1{5\alpha_6} \right\}G(1). \end{aligned} If we substitute the values (\ref{BV3}), (\ref{BX3}) for $\alpha_3, \alpha_4, \alpha_2, \alpha_6$ into (\ref{BT3}) we see that the coefficient of $G(1)$ is bounded below by $$\label{CA3} 2.64 - 1.6\; G(2) - g_a(\alpha_1), \quad a = G(2)/G(1),$$ where the function $g_a(z)$ is defined by $g_a(z) = \frac 2 z + \frac{8a}{5[5 - 4z]}, \quad 0 < z < 5/4, \; a > 0.$ We can easily compute the minimum of $g_a$ to be $\inf_{0 0. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.2} (d=2, \; L_1 = 4)] We need only verify that the function G defined by (\ref{AF3}), \eqref{AZ3} satisfies the inequalities (\ref{CB3}). Since G(y), \ y \ge 1, decays exponentially one can verify these inequalities with aid of a computer. In particular we see that \[ G(0) = .7071, \quad G(1) = .1213, \quad G(2) = .0208,$ correct to 4 decimal places, whence (\ref{CB3}) holds. \end{proof} \section{Formula for Effective Diffusion Constant} In this section we obtain the formula (\ref{G1}) of $\S$1 for the effective diffusion constant which generalizes the formulas obtained in $\S$3. We take $L_1 = 2L$ with $L \ge 2$ in Lemma \ref{lem2.6}. Then $\hat \Omega = \{ (n,y) : 1 \le n \le L, \ y \in \Omega_{d-1}\}$. For $y \in \Omega_{d-1}$, $1 \le j \le L$ we define $\delta_j(y), \bar\delta_j(y)$ by $$\label{A4} \delta_j(y) = \frac 1{2d} - b(j,y), \quad \bar\delta_j(y)=\frac 1{2d} + b(j,y).$$ We see from (\ref{BK2}), (\ref{A4}) that $\varphi^*$ satisfies the system of equations, $$\label{B4} \begin{gathered} \frac{(-\Delta + 2)}{2d} \varphi^*(1,y) - \delta_2(y) \varphi^*(2,y) - \delta_1(y)\varphi^*(1,y) = 0, \\ \frac{(-\Delta + 2)}{2d} \varphi^*(2,y) - \delta_3(y) \varphi^*(3,y) - \bar\delta_1(y)\varphi^*(1,y) = 0, \\ \dots - \dots - \dots = 0, \\ \frac{(-\Delta + 2)}{2d} \varphi^*(L-1,y) - \delta_L(y) \varphi^*(L,y) - \bar\delta_{L-2}(y)\varphi^*(L-2,y) = 0, \\ \frac{(-\Delta + 2)}{2d} \varphi^*(L,y) - \bar\delta_L(y) \varphi^*(L,y) - \bar\delta_{L-1}(y)\varphi^*(L-1,y) = 0, \end{gathered}$$ where $\Delta = \Delta_{d-1}$ is the $d-1$ dimensional Laplacian. If we add all the equations in (\ref{B4}) we obtain the equation $-\Delta \sum^L_{j=1} \varphi^*(j,y) = 0, \ \ y \in \Omega_{d-1}.$ On using the normalization $\langle \varphi^*\rangle _{\hat \Omega} = 1$ we conclude that $$\label{C4} \sum^L_{j=1} \varphi^*(j,y) = L, \quad y \in \Omega_{d-1}.$$ Evidently we can rewrite the first equation of (\ref{B4}) as $$\label{D4} \big[ -\frac{\Delta}{2d} + \bar \delta_1(y) \big] \varphi^*(1,y) - \delta_2(y)\varphi^*(2,y) = 0.$$ If we add (\ref{D4}) to the second equation of (\ref{B4}) we obtain the equation $$\label{E4} \big( -\frac{\Delta}{2d}\big)\varphi^*(1,y) + \big[- \frac{\Delta}{2d} + \bar \delta_2(y) \big] \varphi^*(2,y) - \delta_3(y)\varphi^*(3,y) = 0.$$ Adding (\ref{E4}) to the third equation of (\ref{B4}) and proceeding similarly with subsequent equations we obtain the system \label{F4} \begin{aligned} &\big(- \frac{\Delta}{2d}\big)\varphi^*(1,y) + \big(- \frac{\Delta}{2d}\big)\varphi^*(2,y)\\ &+ \big[ -\frac{\Delta}{2d} + \bar \delta_3(y) \big] \varphi^*(3,y) - \delta_4(y)\varphi^*(4,y) = 0, \\ &\dots \\ &\big( -\frac{\Delta}{2d}\big)\varphi^*(1,y) + \dots + \big(- \frac{\Delta}{2d}\big)\varphi^*(L-2,y)\\ &+ \big[- \frac{\Delta}{2d} + \bar \delta_{L-1}(y) \big] \varphi^*(L-1,y) - \delta_L(y)\varphi^*(L,y) = 0, \end{aligned} where we have omitted the final equation of (\ref{B4}). >From (\ref{D4}), (\ref{E4}), (\ref{F4}) we can write $\varphi^*(j,y)$, $2 \le j \le L$, in terms of $\varphi^*(1,y)$. Substituting these into (\ref{C4}) we obtain an equation for $\varphi^*(1,y)$ of the form $$\label{G4} \mathcal{L} \; \varphi^*(1,y) = L, \quad y \in \Omega_{d-1},$$ where $\mathcal{L}$ is an operator on functions on $\Omega_{d-1}$. Next we consider the equations \eqref{AW2}, (\ref{BB2}) for the function $\psi_0$ on $\hat \Omega$. Thus $\psi_0$ satisfies the system of equations, $$\label{H4} \begin{gathered} \frac{(-\Delta + 2)}{2d} \psi_0(1,y) - \bar\delta_1(y) \psi_0(2,y) + \delta_1(y)\psi_0(1,y) = 0, \\ \frac{(-\Delta + 2)}{2d} \psi_0(2,y) - \bar\delta_2(y) \psi_0(3,y) - \delta_2(y)\psi_0(1,y) = 0, \\ \dots - \dots - \dots = 0, \\ \frac{(-\Delta + 2)}{2d} \psi_0(L-1,y) - \bar\delta_{L-1}(y) \psi_0(L,y) - \delta_{L-1}(y)\psi_0(L-2,y) = 0, \\ \frac{(-\Delta + 2)}{2d} \psi_0(L,y) + \bar\delta_L(y) \psi_0(L,y) - \delta_L(y)\psi_0(L-1,y) = \bar\delta_L(y). \end{gathered}$$ We can rewrite the first equation of (\ref{H4}) as $$\label{J4} \frac{(-\Delta + 4)}{2d} \psi_0(1,y) = \bar\delta_1(y) \left[ \psi_0(1,y) + \psi_0(2,y)\right] = \bar\delta_1(y) u(1,y), \quad y \in \Omega_{d-1},$$ where $u(1) = \psi_0(1) + \psi_0(2)$. If we put now $u(2) = \psi_0(3) - \psi_0(1)$ then on using $(\ref{J4})$ we see that the second equation of (\ref{H4}) is the same as $$\label{K4} \big[ -\frac{\Delta}{2d} + \delta_1\big]u(1) - \bar \delta_2u(2) = 0.$$ Observe that (\ref{K4}) is identical to (\ref{D4}) under the reflection $b \to -b$. We can similarly obtain the reflection of the equations (\ref{E4}), (\ref{F4}) by defining the variables $u(j), j = 3,\dots$ by $u(j) = \psi_0(j+1) - \psi_0(j-1), \quad j = 3,\dots, L-1.$ Let us assume that the $u(j)$, $j=1,\dots,J-1$, satisfy the reflection of the first $J-2$ of the equations (\ref{D4}), (\ref{E4}), (\ref{F4}). We show that $u(J)$ then satisfies the $(J-1)$st equation provided $J \le L-1$. To see this we consider the $J$th equation of (\ref{H4}) which we may write as $$\label{L4} \big( \frac{-\Delta + 2}{2d}\big) \psi_0(J) - \bar\delta_J \left[ u(J) + \psi_0 (J-1) \right] - \delta_J \psi_0(J-1) = 0.$$ We may rewrite (\ref{L4}) as \label{M4} \begin{aligned} &\big[- \frac{\Delta}{2d} + \delta_{J-1}\big] u(J-1) - \bar\delta_J u(J) +\bar \delta_{J-1} u(J-1) \\ &+ \frac{(-\Delta + 2)}{2d} \psi_0(J-2) - \frac 2{2d} \psi_0(J-1)=0. \end{aligned} If $J=3$, then \begin{align*} &\bar \delta_{J-1} u(J-1) + \frac{(-\Delta + 2)}{2d} \psi_0(J-2) - \frac 2{2d} \psi_0(J-1) \\ &= \bar \delta_2 u(2) + \frac{(-\Delta + 4)}{2d} \psi_0(1) - \frac 2{2d} \left[ \psi_0(1) + \psi_0(2) \right] \\ &= \bar \delta_2 u(2) + \bar \delta_1 u(1) - \frac 2{2d} u(1) \\ &= \bar \delta_2 u(2) - \delta_1 u(1) \\ &= \big[ -\frac{\Delta}{2d} + \delta_1\big] u(1) - \delta_1u(1)\\ & = -\frac{\Delta}{2d} u(1). \end{align*} We have shown that the result holds for $J=3$. More generally we have \begin{align*} &\bar \delta_{J-1} u(J-1) + \frac{(-\Delta + 2)}{2d} \psi_0(J-2) - \frac 2{2d} \psi_0(J-1)\\ &= \big[- \frac{\Delta}{2d} + \delta_{J-2}\big] u(J-2) + \sum^{J-3}_{j=1} - \frac{\Delta}{2d} u(j) + \frac{(-\Delta + 2)}{2d} \psi_0(J-2) - \frac 2{2d} \psi_0(J-1) \\ &= \sum^{J-2}_{j=1} - \frac{\Delta}{2d} u(j) - \bar\delta_{J-2} u(J-2) - \frac 2{2d}\psi_0(J-3) + \frac{(-\Delta + 2)}{2d} \psi_0 (J-2). \end{align*} To complete the proof we need then to show that $-\bar\delta_{J-2} \; u(J-2) - \frac2{2d} \psi_0(J-3) + \frac{(-\Delta + 2)}{2d} \psi_0(J-2) = 0.$ This last equation is however simply the $(J-2)$nd equation of (\ref{H4}). We have shown that $u(j), j=1,\dots,L-1$ satisfies the reflection of the first $L-2$ equations of (\ref{D4}), (\ref{E4}), (\ref{F4}). Define now $u(L) = 1 - \psi_0(L) - \psi_0(L-1)$, whence there is the identity $$\label{N4} \sum^L_{j=1} \ u(j) = 1.$$ We shall show that the $u(j), j=1,\dots,L$ satisfy the reflection of the final equation of (\ref{F4}). To see this we write the final equation of (\ref{H4}) as $\big( \frac{-\Delta + 2}{2d}\big) \psi_0(L) + \bar\delta_L \left[ 1-u(L) - \psi_0 (L-1) \right] - \delta_L \psi_0(L-1) = \bar\delta_L,$ whence we have $\big( \frac{-\Delta + 2}{2d}\big) \left[ u(L-1) + \psi_0 (L-2) \right] - \bar \delta_L u(L) - \frac 2{2d} \psi_0(L-1) = 0.$ We may rewrite the previous equation as \begin{align*} &\big[ -\frac{\Delta}{2d} + \delta_{L-1}\big] u(L-1) - \bar \delta_L u(L) + \bar\delta_{L-1} u(L-1)\\ &+ \frac{(-\Delta + 2)}{2d} \psi_0 (L-2) - \frac 2{2d} \psi_0(L-1) = 0. \end{align*} Now if we use the identity already established, $\bar \delta_{L-1} u(L-1) = \big[ -\frac{\Delta}{2d} + \delta_{L-2}\big] u(L-2) + \sum^{L-3}_{j=1} - \frac{\Delta}{2d} u(j),$ we see that it is sufficient to show that $\delta_{L-2} u(L-2) + \frac{(-\Delta + 2)}{2d} \psi_0(L-2) - \frac 2{2d} \psi_0(L-1) = 0.$ This last equation is just the $(L-2)$nd equation of (\ref{H4}). Let $\mathcal{L}_R$ be the reflection of the operator $\mathcal{L}$ of (\ref{G4}) obtained by replacing $b$ by $-b$. Then on comparing (\ref{C4}), (\ref{N4}) we see that $u(1)$ satisfies the equation $$\label{O4} \mathcal{L}_R \ u(1) = 1.$$ We are able now to come up with a new formula for the effective diffusion constant. On using (\ref{BC2}), (\ref{G4}), (\ref{J4}), (\ref{O4}) we have that the effective diffusion constant is given by $$\label{P4} 8 L^2 d \left< \left[\delta_1 \mathcal{L}^{-1} \; 1\right] (-\Delta + 4)^{-1} \left[\bar\delta_1 \mathcal{L}^{-1}_R \; 1\right] \right>,$$ where $\langle\cdot\rangle$ is the uniform probability measure on $\Omega_{d-1}$. The formula (\ref{G1}) follows from (\ref{P4}). In order for (\ref{P4}) to be valid we need to show that $\mathcal{L}$ is invertible. \begin{lemma} \label{lem4.1} Let $\mathcal{L}$ be the matrix defined by (\ref{G4}). Then $\mathcal{L}$ is invertible and the matrix $\mathcal{L}^{-1}$ has all positive entries. \end{lemma} \begin{proof} We proceed by induction. For $k = 2,3\dots$, let $\mathcal{L}_k$ be the operator (\ref{G4}) when $L = k$. It is easy to see from (\ref{C4}) - (\ref{F4}) that the $\mathcal{L}_k$ satisfy the recurrence relation, $$\label{Q4} \mathcal{L}_{k+1} = \frac 1{\delta_{k+1}} \big[ -\frac{\Delta}{2d} + \bar \delta_k + \delta_{k+1} \big] \mathcal{L}_k - \frac{\bar\delta_k}{\delta_{k+1}} \mathcal{L}_{k-1}, \quad k \ge 1; \; \mathcal{L}_0 = 0, \; \mathcal{L}_1 = 1.$$ The result will follow by showing that the matrices $A_k = \mathcal{L}_{k-1} \mathcal{L}_k^{-1}, k \ge 2$, have all positive entries and principal eigenvalue strictly less than 1. Evidently this is the case for $k=2$. Now from (\ref{Q4}) we see that the $A_k$ satisfy the recurrence relation, $$\label{R4} A_{k+1} = \big\{ -\frac{\Delta}{2d} + \bar \delta_k + \delta_{k+1} - \bar \delta_k A_k\big\}^{-1} \delta_{k+1}.$$ If $A_k$ has all positive entries with principal eigenvalue strictly less than 1 then the matrix $[-\Delta/2d + \bar \delta_k + \delta_{k+1}]^{-1}\bar \delta_k A_k$ has the same property and the matrix $A_{k+1}$ defined by (\ref{R4}) has all positive entries. To conclude the induction step we need therefore to show that $A_{k+1}$ has principal eigenvalue strictly less than $1$. To see this note that if $1$ denotes the vector with all entries $1$ then $\big\{- \frac{\Delta}{2d} + \bar \delta_k + \delta_{k+1} - \bar \delta_k A_k\big\} 1 > \delta_{k+1},$ whence we conclude that $\big\{ -\frac{\Delta}{2d} + \bar \delta_k + \delta_{k+1} - \bar \delta_k A_k\big\}^{-1} \delta_{k+1} (1) < 1.$ \end{proof} \subsection*{Acknowledgements} The author would like to thank the anonymous referee for bringing his attention to the formula (\ref{Z1}). This research was partially supported by grant DMS-0500608 from the NSF. \begin{thebibliography}{00} \bibitem{aks}V. V. Anshelevich, K. M. Khanin and Ya. G. Sinai, \newblock \textit{Symmetric random walks in random environments}, \newblock Commun. Math. Phys. \textbf{85} (1982), 449-470, MR 84a:60082. \bibitem{bes} D. Bes, \newblock Quantum mechanics. A modern and concise introductory course. \newblock \textit{Springer-Verlag, Berlin}, 2004, 204 pp. MR 2064081. \bibitem{bs} E. Bolthausen and A. Sznitman, \newblock Ten lectures on random media. \newblock \textit{ DMV Seminar} {\bf 32} \textit{ Birkhauser Verlag, Basel}, 2002, 116 pp. MR 2003f:60183. \bibitem{bk} J. Bricmont and A. Kupiainen, \newblock \textit{Random walks in asymmetric random environments}, \newblock Commun. Math. Phys. \textbf{142} (1991), 345-420, MR 1137068. \bibitem{c2002} J. Conlon, \newblock \textit{ Homogenization of random walk in asymmetric random environment}, \newblock New York J. Math.\textbf{8} (2002), 31-61, MR 1887697. \bibitem{c2005} J. Conlon, \newblock \textit{ Perturbation theory for random walk in asymmetric environment }, \newblock New York J. Math.\textbf{11} (2005), 465-476, MR 2188251. \bibitem{cp} J. Conlon and I. Pulizzotto, \newblock \textit{On homogenization of non-divergence form partial difference equations}, \newblock Electron. Commun. Probab. \textbf{10} (2005) 125-136, MR 2150701. \bibitem{cn} J.~Conlon and A.~Naddaf, \newblock \textit{On homogenisation of elliptic equations with random coefficients}, \newblock Electronic Journal of Probability \textbf{5} (2000) paper 9, 1-58, MR 1768843. \bibitem{dl} B. Derrida and J. Luck, \newblock \textit{ Diffusion on a random lattice: Weak-disorder expansion in arbitrary dimension}, \newblock Phys. Rev. B \textbf{28} (1983), 7183-7190. \bibitem{fp} A. Fannjiang and G. Papanicolaou, \newblock \textit{Convection-enhanced diffusion for random flows}, \newblock J. Statist. Phys. \textbf{88} (1997), 1033-1076, MR 1478061. \bibitem{f} D. Fisher, \newblock \textit{Random walks in random environment}, \newblock Phys, Rev. A \textbf{30} (1984), 960-964, MR 85h:82028. \bibitem{k} H. Kesten, \newblock \textit{The limit distribution of Sinai's random walk in random environment}, \newblock Phys. A. \textbf{138} (1986), 299-309, MR 88b:60165. \bibitem{ko} S. Kozlov, \newblock Averaging of random structures, \newblock \textit{Dokl. Akad. Nauk. SSSR} \textbf{241}, 1016-1019 (1978), MR 80e:60078. \bibitem{ku} R. K\"unnemann, \newblock \textit{The diffusion limit for reversible jump processes on $\mathbb{Z}$ with ergodic random bond conductivities}, \newblock Commun. Math. Phys. \textbf{90} (1983), 27-68, MR0714611. \bibitem{l} G. Lawler, \newblock \textit{Weak convergence of a random walk in a random environment}, \newblock Commun. Math. Phys. \textbf{87} (1982), 81-87, MR 84b:60093. \bibitem{pv1} G. Papanicolaou and S.R.S. Varadhan, \newblock \textit{Boundary value problems with rapidly oscillating random coefficients}, \newblock Volume 2 of Coll. Math. Soc. Janos Bolyai \textbf{27}, Random Fields, Amsterdam, North Holland Publ. Co. 1981, pp. 835-873, MR 84k:58233. \bibitem{pv2} G. Papanicolaou and S.R.S. Varadhan, \newblock Diffusions with random coefficients. \newblock \textit{Statistics and probability: essays in honor of C.R. Rao}, pp. 547-552, \textit{North-Holland, Amsterdam}, 1982. MR 85e:60082. \bibitem{s} Y. Sinai, \newblock \textit{ Limiting behavior of a one-dimensional random walk in a random medium}, \newblock Theory Prob. Appl. \textbf{27} (1982), 256-268, MR 83k:60078. \bibitem{sz1} A. Sznitman, \newblock \textit{Slowdown and neutral pockets for a random walk in random environment}, \newblock Probab. Theory Related Fields \textbf{115} (1999), 287-323, MR 2001a:60035. \bibitem{sz2} A. Sznitman, \newblock \textit{Slowdown estimates and central limit theorem for random walks in random environment}, \newblock J. Eur. Math. Soc. \textbf{2} (2000), 93-143, MR 1763302. \bibitem{sz} A. Sznitman and O. Zeitouni, \newblock \textit{An Invariance principle for Isotropic Diffusions in Random Environment}, \newblock Invent. Math. \textbf{164} (2006), 455-567, MR 2221130. \bibitem{zko} V. V. Zhikov, S. M. Kozlov and O. A. Oleinik, \newblock Homogenization of Differential Operators and Integral Functionals. Translated from the Russian by G.A. Yosifian. \newblock \textit{Springer-Verlag, Berlin}, 1994, 570 pp. MR 96h:35003b. \bibitem{zs} V. V. Zhikov and M. M. Sirazhudinov, $G$ compactness of a class of second-order nondivergence elliptic operators. (Russian) \textit{Izv. Akad. Nauk SSSR Ser. Mat.} {\bf45} (1981), 718-733. MR 83f:35041. \end{thebibliography} \end{document}