Discrete approximation

In this last subsection, we give a concrete approximation of non-symmetric diffusions in diver-gence form.

For a matrix a= (aij)^d_i,j=1 we denote by ˜a the symmetric and by ˆa the antisymmetric part.

Also forξ ∈R^d, let

≠ξ, aξÆ

= Xd

i,j

ξiaijξj = Xd

i,j

ξia˜ijξj and kak= max

i

X

j

|aij|.

For≤, M1, M2 >0 be denote by

M^d(≤, M1.M2) ={a= ˜a+ ˆa: ≠ ξ, aξÆ

≥≤kξk² and k˜ak ≤M1, kˆak ≤M2}

the set of uniformly elliptic, bounded matrices. Clearly a is symmetric if and only if M2 = 0.

Given a measurable map a : R^d → M^d(≤, M1, M2), our goal is to find a sequence of Markov chains that approximate the diffusion process whose divergence form is determined bya. Thanks to Theorem 4.6, all we need is to find a sequence (Γn, αn) where Γnis a collection of cyclesγ_iⁿ, i∈ I in Sⁿ with weights αn(γ_iⁿ)≥0 such that (2.2) and (2.8) are satisfied and the corresponding F_ijⁿ(·) converges locally inL¹(R^d) to aij, that is, for allK compact subset of R^d

nlim→∞kF_ijⁿ−aijk^K = lim

n→∞

Z

K|F_ijⁿ(x)−aij(x)|dx= 0, ∀i, j = 1, .., d, (5.22) where as usual we write F_ijⁿ(x) = F_ijⁿ([x]_n) for x ∈ R^d. In Theorem 5.4, which is our main theorem, we will prove that it is possible to find such a sequence.

Our construction will be based on a two scale procedure: we will discretize the matrix a(·) at intermediate scale r_n = [n^1−β]/n, for some β ∈ (0,1) and then construct the corresponding chain on Sⁿ at microscopic scale 1/n.

Clearly, if aⁿ:R^d→ M^d(≤, M1, M2) is a sequence such that

nlim→∞kaⁿ_ij −aijk^K = 0 and lim

n→∞kF_ijⁿ−aⁿ_ijk^K = 0, (5.23) then by the triangle inequality (5.22) holds.

We start with a trivial observation: let (Γ¹_n, α¹_n) and (Γ²_n, α²_n) be two such collections and consider the merged collection Γ_n = Γ¹_n∪Γ²_n with weightsα_n(γ_n) = αⁱ_n(γ_n) if γ_n∈Γⁱ_n, then the corresponding Fⁿ satisfies the additive rule

F_ijⁿ=F_ij^1,n+F_ij^2,n. (5.24) Also if both (Γⁱ_n, αⁱ_n) satisfy (2.2) then of course (Γn, αn) satisfy (2.2).

This additive rule will be a very useful tool for our construction and we will proceed itera-tively. Let us introduce the set of strongly uniformly elliptic bounded symmetric matrices:

M⁽³⁾d (≤, M1) = {a= ˜a :aii− X

j:j6=i

|aij| ≥≤, i= 1, ..., d, kak ≤M1}, and for N ∈N the set of “almost” antisymmetric bounded matrices:

M⁽¹⁾d (N, M2) ={a :aij =−aji, 1≤i < j≤d, aii = 1 2N

X

j:j6=i

|aij|,kak ≤M2}. The appearance of the diagonal term will be explained below.

Next for given L∈N, let

M⁽²⁾d (L, M1) = {a = ˜a :aij = Xd

k=1

νkV_i^k·V_j^k, i, j ∈ {1, ..., d},

for νk≥0, V^k ∈[−L, L]^d∩Z^d, k= 1, ..., d, kak ≤M1},

be the set of symmetric matrices with non-negative eigenvalues and integer valued eigenvectors.

Lemma 5.1 Leta ∈ M^d(≤, M1, M2) and chooseL >[9dM1/≤]and N >[3M2/2≤], then we can find b⁽¹⁾ ∈ M⁽¹⁾d (N, M₂), b⁽²⁾ ∈ M⁽²⁾d (L, M₁), b⁽³⁾ ∈ M⁽³⁾d (≤/3, M₁) such that

a=b⁽¹⁾+b⁽²⁾+b⁽³⁾. (5.25)

Proof.Setb =a−≤I, and writeb= ˜b+ˆbwhere ˜bis the symmetric part and ˆbthe antisymmetric part of b. Set

b⁽¹⁾_ij =−b⁽¹⁾_ji = ˆbij, i6=j, b⁽¹⁾_ii = 1 2N

X

j:j6=i

|ˆbij|. Note that

b⁽¹⁾_ii ≤ M2

2N ≤≤/3.

Next let U¹, ..., U^d∈ R^d and ν¹, ..., ν^d ∈R+ be the eigenvectors and eigenvalues of ˜b. We may assume that the vectors are orthonormal: U^k·U^l =δk,l and therefore

˜b_ij = Xd

k=1

ν_kU_i^k·U_j^k. Let

V_i^k = [LU_i^k], λk = νk

L². then V^k ∈Z^d∩[−L, L]^d and writing ¯V_i^k = [LU_i^k]/L, we have

|U_j^k−V¯_j^k| ≤ 1

L, |V¯_j^k| ≤|U_j^k| ≤1.

Define

b⁽²⁾_ij = Xd

k=1

λkV_i^k·V_j^k= Xd

k=1

νkV¯_i^k·V¯_j^k and

b⁽³⁾ =a−b⁽¹⁾−b⁽²⁾=≤I+ ˜b−b⁽²⁾+ ˆb−b⁽¹⁾. Note thatb⁽³⁾ is symmetric by construction with

b⁽³⁾_ij = ˜b_ij −b⁽²⁾_ij , 1≤i < j ≤d, b⁽³⁾_ii =≤+ ˜b_ii−b⁽²⁾_ii −b⁽¹⁾_ii , i= 1, ..., d.

We claim that

k˜b−b⁽²⁾k ≤ 3dM₁

L ≤≤/3, (5.26)

which implies

b⁽³⁾ ∈ M⁽³⁾d (≤/3, M₂).

By the triangle inequality we have kb⁽²⁾−˜bk ≤

Xd

k=1

maxi

Xd

j=1

νk|U_i^k·U_j^k−V¯_i^kV¯_j^k|

where

|U_i^k·U_j^k−V¯_i^kV¯_j^k| ≤|V¯_i^k||U_j^k−V¯_j^k|+|V¯_j^k||U_i^k−V¯_i^k|+|U_i^k−V¯_i^k||U_j^k−V¯_j^k| ≤ 2 + _L¹

L ≤ 3

L Thus

kb⁽²⁾−˜bk ≤ 3d L

Xd

k=1

ν_k ≤ 3dM1

L .

§ In view of the additive rule, it thus suffices to find a collection of cycles Γ^k_nsuch thatF_ij^k,n(·) converges locally in L¹(R^d) to b^(k)_ij , for each k = 1,2,3.

Examples 1 and 2 of the previous section imply the following.

Lemma 5.2 Let M, N and L be fixed and bⁿ:Sⁿ→[0, M] be such that

nlim→∞kbⁿ(· +y/n)−bⁿk^K = 0, ∀y∈Z^d, ∀K ⊂R^d compact. (5.27) a) Referring to Example 1, for given fixed V ∈[−L, L]^d∩Z^d, take

Γn={γ_xⁿ= (x, x+V /n, x), x∈ Sⁿ}, with weights αn(γ_xⁿ) =bⁿ(x), x∈ Sⁿ, and set aⁿ∈ M⁽³⁾d (L, M) by

aⁿ_ij(x) =bⁿ([x]n)Vi·Vj, 1≤i, j ≤d, x∈R^d. Then for every K ⊂R^d compact, limn→0kF_ijⁿ−aⁿ_ijk^K = 0.

b) Referring to Example 2, for fixed N and l6=m∈ {1, ..., d}, take Γn={γ_N,x^n,(l,m), x∈ Sⁿ}, with weights αn(γ_N,x^n,(l,m)) = bⁿ(x)

4N² , x∈ Sⁿ, and set aⁿ∈ M⁽¹⁾d (N, M) by

aⁿ_l,l(x) = aⁿ_m,m(x) = bⁿ([x]_n)

2N , aⁿ_l,m(x) =−aⁿ_m,l(x) =bⁿ([x]n), aⁿ_ij(x) = 0, i, j /∈ {l, m}, for x∈R^d. Then for every K ⊂R^d compact, limn→0kF_ijⁿ−aⁿ_ijk^K = 0.

Proof.Since V is constant, we first see that R^2,n_ij (z) = R^3,n_ij (z) = 0, cf. (5.17). Next using the fact that |bⁿ(x)| ≤M, |V_i| ≤L we get in view of (5.11) and (5.21) writing kyk1 =Pd

i=1|y_i|,

|R^1,n_ij (z)| ≤L² max

y:kyk1≤L|bⁿ(z+y/n)−bⁿ(z)| ≤L² X

y:kyk1≤L

|bⁿ(z+y/n)−bⁿ(z)|, and

|R^4,n_ij (z)| ≤4N² max

y:kyk1≤4N|bⁿ(z+y/n)−bⁿ(z)| ≤4N² X

y:kyk1≤4N

|bⁿ(z+y/n)−bⁿ(z)|.

Using our assumption this yields

nlim→0kR_ij^k,nk^K = 0, k= 1,4,

and implies the result. §

Take β ∈(0,1), set rn= [n^1−β]/n∈(1/n)Z⁺, Jrn =rnZ^d⊂ Sⁿ, and let Q(x, r_n) = {y∈R^d: 0≤min

i (y_i−x_i)≤max

i (y_i−x_i)< r_n}, x∈J_rn, be a partition of disjoint cubes of R^d. For a measurable mapa :R^d → M^d(≤, M1, M2), set

aⁿ_ij(y) = X

x∈J_rn

≥ 1 r^d_n

Z

Q(x,rn)

a_ij(z)dz¥

1_Q(x,rn)(y), y∈R^d. Then

aⁿ(x)∈ M^d(≤, M1, M2), ∀x∈ Sⁿ, and for every compactK ⊂R^d we have

nlim→∞kaⁿ_ij −aijk^K = 0. (5.28) Furthermore,aⁿ_ij(x) is a rn-piecewise constant function, that is

aⁿ_ij(x) = aⁿ_ij≥ (rn[x1

rn

],· · · , rn[xd

rn

])¥

, x= (x1,· · ·xd)∈ Sⁿ.

As we see, rn is an “intermediate” scale and aⁿ_ij is an approximation of aij which is constant in each cell of Jrn.

Lemma 5.3 a) Let bⁿ : Sⁿ → [−M, M] be rn-piecewise constant. Then for each compact K ⊂R^d and fixed y∈Z^d,

kbⁿ(·)−bⁿ(·+y/n)k^K ≤ kyk¹CKM

[n^1−β] (5.29)

for some CK < ∞ depending on the diameter of K, where n is taken large enough so that [n^1−β]≥2kyk¹.

b) Referring to Example 1, let bⁿ : Sⁿ → [0, M] and Vⁿ : Sⁿ → [−L, L]^d∩Z^d be rn-piecewise constant and take

Γn={γ_xⁿ= (x, x+Vⁿ(x)/n, x), x∈ Sⁿ}, with weights αn(γ_xⁿ) =bⁿ(x), x∈ Sⁿ, and set aⁿ∈ M⁽²⁾d (L, M) by

aⁿ_ij(x) = bⁿ([x]_n)V_iⁿ([x]_n)·V_jⁿ([x]_n), 1≤i, j ≤d, x∈R^d. Then for every K ⊂R^d compact, limn→0kF_ijⁿ−aⁿ_ijk^K = 0.

Proof. a) Simply note that

bⁿ(z) = X

x∈Jrn

bⁿ(x)1Q(x,rn)(z) and therefore

kbⁿ(·)−bⁿ(·+y/n)k^K ≤M X

x∈Jrn

ØØ Ø

Z

K

(1Q(x,rn)([z]n)−1Q(x,rn)([z]n+y/n))dzØØØ. Now for the interior points of Q(x, rn) such that

IQ(x, rn,kyk¹/n) :={z ∈Q(x, rn) :kyk¹/n≤min

i (zi−xi)≤max

i (zi−xi)≤rn− kyk¹/n}, we clearly have

1Q(x,rn)(z)−1Q(x,rn)(z+y/n) = 0, z ∈IQ(x, rn,kyk¹/n).

So all what remains are the boundary terms

BQ(x, rn,kyk¹/n) := Q(x, rn)\IQ(x, rn,kyk¹/n)

with X

x∈Jr

ØØ Ø

Z

K

(1_{BQ(x,rn,kyk1/n)}([z]_n)dzØØØ≤ kyk1C_K [n^1−β] . b) In view of (5.7), (5.11)–(5.12) we have

|R_ij^1,n(z)| ≤CL^d+1 max

y:kyk1≤L|bⁿ(z+y/n)−bⁿ(z)| ≤CL^d+1 X

y:kyk1≤L

|bⁿ(z+y/n)−bⁿ(z)|,

|R_ij^2,n(z)| ≤CM L^d max

y:kyk1≤L|V_jⁿ(z+y/n)−V_jⁿ(z)| ≤CM L^d X

y:kyk1≤L

|V_jⁿ(z+y/n)−V_jⁿ(z)|, and a) shows that kR^1,n_ij k^K and kR^2,n_ij k^K → 0 as n → ∞. Next note that (5.17) implies R^3,n_ij (z) = 0 if z ∈ ∪^x∈J_rnIQ(x, rn, L/n) and by (5.7), (5.13), |R^3,n_ij (z)| ≤ CM L^d+1 if z ∈

∪^x∈J_rnBQ(x, rn, L/n). As in a) this shows kR_ij^3,nk^K →0 as n → ∞. § Next set

bⁿ(x) =aⁿ(x)−≤I, x∈ Sn,

and denote by ˜bⁿ(x) and ˆbⁿ(x) the symmetric and antisymmetric part of bⁿ(x). Choose N = [3M2/≤] + 1, L= [9dM1/≤] + 1, and define b^(i),n(x) as in Lemma 5.1, that is

aⁿ(x) =b^(1),n(x) +b^(2),n(x) +b^(3),n(x) where b^(1),n(x)∈ M⁽¹⁾d (N, M2), b^(3),n(x)∈ M⁽³⁾d (≤/3, M1), and

b⁽²⁾_ij (x) = Xd

k=1

λⁿ_k(x)V_i^k,n(x)·V_j^k,n(x)∈ M⁽²⁾d (L, M1).

Note that by construction, b^(1),n, b^(3),n, λⁿ_k and V^k,n are bounded and rn-piecewise constant, so by Lemma 5.3 a), for all compact K ⊂R^d and fixedy ∈Z^d,

kb^(1),n_ij (· +y/n)−b^(1),n_ij k^K ≤ kyk¹CKM2

[n^1−β] , kb^(3),n_ij (· +y/n)−b^(3),n_ij k^K ≤ kyk¹CKM1

[n^1−β] , (5.30) kV_j^k,n(· +y/n)−V_j^k,nk^K ≤ kyk¹CKL

[n^1−β] , (5.31)

kλⁿ_k(· +y/n)−λⁿ_kk^K ≤ kyk¹CKM1

[n^1−β] . (5.32)

For b⁽¹⁾(x)∈ M⁽¹⁾d (N, M2) consider (Γ^(1),n, αⁿ) as follows

Γ^(1),n ={γ^n,(i,j)_N,x , γ_N,x^n,(j,i), 1≤i < j ≤d, x∈ Sⁿ} with weights

αⁿ(γ_N,x^n,(i,j)) = (b^(1),n_ij (x))⁺

4N² = (ˆaⁿ_ij(x))⁺

4N² , αⁿ(γ_N,x^n,(j,i)) = (b^(1),n_ij (x))⁻

4N² = (ˆaⁿ_ij(x))⁻ 4N² ,

where a⁺:=a∨0 and a⁻= (−a)∨0 fora∈R. Then, in view of (5.30), the additive rule and Lemma 5.2 b), we see that the corresponding F_ij^(1),n satisfies

nlim→∞kF_ij^(1),n−b^(1),n_ij k^K = 0.

Next consider (Γ^(2),n, αⁿ) of the form

Γ^(2),n ={γ_x^k,n = (x, x+V^k,n(x)/n, x), with weights αⁿ(γ_x^k,n) = λⁿ_k(x), k = 1, ..., d, x∈ Sⁿ}, then in view of (5.31) and (5.32), Lemma 5.3 b) and the additive rule, we see that the corre-sponding F_ij^(2),n satisfies

nlim→∞kF_ij^(2),n−b^(2),n_ij k^K = 0.

Finally for

b^(3),n(x) = aⁿ(x)−b^(1),n(x)−b^(2),n(x)∈ M⁽³⁾d (≤/3, M), take (Γ^(3),n, αⁿ) of the form

Γ^(3),n ={γ_ij^n,±(x) = (x, x+ei/n±ej/n, x),1≤i < j≤d, γ_iⁿ(x) = (x, x+ei/n, x), x∈ Sⁿ} with weights

αⁿ(γ_ij^n,+(x)) = (b^(3),n_ij (x))⁺, αⁿ(γ_ij^n,−(x)) = (b^(3),n_ij (x))⁻, αⁿ(γ_iⁿ(x)) =b^(3),n_ii (x)−X

j:j6=i

|b^(3),n_ij (x)| ≥≤/3, x∈ Sⁿ. We callγ_iⁿ(x) a nearest neighbor cycle and γ_ij^n,±(x) a diagonal cycle.

Then using (5.30), the additive rule and the Lemma 5.2 a), we see that the corresponding F_ij^(3),n satisfies

nlim→∞kF_ij^(3),n−b^(3),n_ij k^K = 0.

Putting things together we have the following.

Theorem 5.4 For any measurable map a : R^d → M^d(≤, M1, M2), we can find a sequence (Γn, αn)that satisfies(2.2) in Assumption 2.1 and(2.8), (2.9) in Assumption 2.3, such that the correspondingF_ijⁿ(x) converges to aij(x)locally in L¹(R^d). Furthermore, writing Γn={γn,i, i∈ I}, each cycle γi,n is either a two cycle or a rotational cycle that satisfies

αn(γn,i)≤max(M1, M2), `(γn,i)≤max(2,8([3M2/≤] + 1)), Range(γn,i)≤max(2,[9dM1/≤] + 1)/n, ∀i, n, and (2.8) is satisfied with N = 1 and δ=≤/3.

Note thatαn(γi,n)≥0 in the above construction. However, by neglecting cycles withαn(γi,n) = 0, we may say that weights of cycles in Γ_n are all positive.

Remark 5.5 (i) Our construction is very explicit. For example, when approximating a sym-metric diffusion matrix in [SZ], they have additional procedure of smoothing the matrix by convolution, whereas we can avoid this procedure. We think that our construction is practical in that it is useful when simulating diffusions in divergence form.

(ii) As we have seen, once the lattice approximation of the symmetric part is computed, the antisymmetric part can be easily dealt with rotational cycles which are just translates of a fixed cycle. In general we need to compute the eigenvalues and eigenvectors of the symmetric part.

However if the symmetric part of aⁿ is strongly irreducible, that is if

˜

aⁿ(x)∈ M⁽³⁾d (≤, M1), ∀x∈ Sⁿ,

then we can avoid the computation of eigenvalues and eigenvectors. In this case, we only use nearest neighbors cycles and diagonal cycles.

(iii) Although we do not investigate the convergence speed of our approximation it is very natural to take β = 1/2, since for “nice” a(x) we expect

kaⁿ_ij −aijk^K =O(n^−β), whereas

kaⁿ_ij −aⁿ_ij(· +y/n)k^K =O(n^−1+β).

Finally we demonstrate that for the two dimensional case, we can make approximation without computing the eigenvalues and eigenvectors of the diffusion matrix.

Example 3: (2-dimensional case.) Consider a measurable mapa⁽⁰⁾_ij :R² → M^d(≤, M1, M2).

For simplicity we use the following notation:

a⁽⁰⁾_1,1(x) =a(x) +≤, a⁽⁰⁾_2,2(x) = c(x) +≤, a⁽⁰⁾_1,2(x) = b(x) +d(x), a⁽⁰⁾_2,1(x) = b(x)−d(x), where

a(x), c(x)≥0, a(x)c(x)≥b²(x). (5.33)

As above we define aⁿ(x) by integration as follows aⁿ(y) = X

x∈J_rn

≥1 r_n²

Z

Q(x,rn)

a(z)dz¥

1Q(x,rn)(y), y∈R²,

and define bⁿ(x), cⁿ(x), dⁿ(x) similarly. Next, for L∈N, define aⁿ_L,a¯ⁿ_L, cⁿ_L,c¯ⁿ_L, and bⁿ_L,¯bⁿ_L by aⁿ_L(x) = [L·aⁿ(x)]

L , ¯aⁿ_L(x) = aⁿ(x)−aⁿ_L(x), cⁿ_L(x) = [L·cⁿ(x)]

L , ¯cⁿ_L(x) = cⁿ(x)−cⁿ_L(x), bⁿ_L(x) = [L·(bⁿ(x))⁺]

L − [L·(bⁿ(x))⁻]

L , ¯bⁿ_L(x) = bⁿ(x)−bⁿ_L(x).

Note that

0≤¯aⁿ_L(x)≤ 1

L, 0≤¯cⁿ_L(x)≤ 1

L, |¯bⁿ_L(x)| ≤ 1

L, |bⁿ_L(x)| ≤|bⁿ(x)|. (5.34) We first deal with the antisymmetric part: for fixed N ∈N, consider a family of rotational cycles {γ_N,x^n,(1,2), γ_N,x^n,(2,1), x∈ Sⁿ}with weights

αⁿ(γ_N,x^n,(1,2)) = (dⁿ(x))⁺

4N² , αⁿ(γ_N,x^n,(2,1)) = (dⁿ(x))⁻ 4N² .

As in Lemma 5.2 b), the corresponding diffusion matrix, denoted by a⁽¹⁾(x), is of the form a⁽¹⁾_1,1(x) =a⁽¹⁾_2,2(x) = |d(x)|

2N , a⁽¹⁾_1,2(x) = −a⁽¹⁾_2,1(x) =d(x).

Next consider the family of cycles {γ_x^n,2 = (x, x+Vⁿ(x)/n, x), x ∈ Sⁿ} where Vⁿ(x) = (V₁ⁿ(x), V₂ⁿ(x))∈Z² is of the form

V₁ⁿ(x) = (Laⁿ_L(x) + 1)1_{aⁿ_L(x)≤cⁿ_L(x)}+Lbⁿ_L(x)1_{aⁿ_L(x)>cⁿ_L(x)}, V₂ⁿ(x) =Lbⁿ_L(x)1_{aⁿ_L(x)≤cⁿ_L(x)}+ (Lcⁿ_L(x) + 1)1_{aⁿ_L(x)>cⁿ_L(x)}, with weights

αⁿ(γ_x^n,2) = 1

L²aⁿ_L(x) +L1_{aⁿ_L(x)≤cⁿ_L(x)}+ 1

L²cⁿ_L(x) +L1_{aⁿ_L(x)>cⁿ_L(x)}. This yields the following corresponding diffusion matrixa⁽²⁾(x) (cf. Lemma 5.3 b))

a⁽²⁾_1,1(x) = (aL(x) + 1

L)1_{a_L(x)≤c_L(x)}+ b²_L(x)

cL(x) + _L¹1_{a_L(x)>c_L(x)}, a⁽²⁾_2,2(x) = b²_L(x)

aL(x) + _L¹1_{aL(x)≤cL(x)}+ (cL(x) + 1

L)1_{aL(x)>cL(x)},

a⁽²⁾_1,2(x) = a⁽²⁾_2,1(x) =b_L(x).

Here and in the following, we omit the super-suffix n.

The third family of cycles is of the form

{γ_x^n,3,+ = (x, x+e1/n+e2/n, x), γ_x^n,3,−= (x, x+e1/n−e2/n, x), x∈ Sⁿ} with weights

αⁿ(γ_x^n,3,+) = (¯bL(x))⁺, αⁿ(γ_x^n,3,−) = (¯bL(x))⁻, yields the diffusion matrix (cf. Lemma 5.2 a))

a⁽³⁾_1,2(x) =a⁽³⁾_2,1(x) = ¯bL(x) a⁽³⁾_1,1(x) =a⁽³⁾_2,2(x) = |¯bL(x)|. Putting things together we see that

a⁽¹⁾_1,1(x) +a⁽²⁾_1,1(x) +a⁽³⁾_1,1(x)

= |d(x)|

2N + (a_L(x) + 1

L)1_{{aL(x)≤cL(x)}}+ b²_L(x)

cL(x) + _L¹1_{{aL(x)>cL(x)}}+|¯b_L(x)|, a⁽¹⁾_2,2(x) +a⁽²⁾_2,2(x) +a⁽³⁾_2,2(x)

= |d(x)|

2N + (cL(x) + 1

L)1_{cL(x)<aL(x)}+ b²_L(x)

aL(x) + _L¹1_{cL(x)≥aL(x)}+|¯bL(x)|, a⁽¹⁾_1,2(x) +a⁽²⁾_1,2(x) +a⁽³⁾_1,2(x) = d(x) +b_L(x) + ¯b_L(x) =a⁽⁰⁾_1,2(x),

a⁽¹⁾_2,1(x) +a⁽²⁾_2,1(x) +a⁽³⁾_2,1(x) = −d(x) +bL(x) + ¯bL(x) =a⁽⁰⁾_2,1(x).

What remains is the matrix

a⁽⁴⁾(x) = a⁽⁰⁾(x)−a⁽¹⁾(x)−a⁽²⁾(x)−a⁽³⁾(x) which is a diagonal matrix. Now if we choose, L, N so large that

≤≥ 3 L + M2

2N. Note that for a.e. x and all δ, by taking n large, we have

(a_L(x) + 1

L)(c_L(x) + 1

L)≥(bⁿ(x))²−δ≥(b_L(x))²−δ, which is due to (5.33) and (5.34). So, for large n, we see that

a⁽⁴⁾_1,1(x)≥0 and a⁽⁴⁾_2,2(x)≥0. (5.35) Now we choose the last family of cycle

{γ_x^n,4,i = (x, x+ei/n, x), i= 1,2, x∈ Sⁿ} with weights

αⁿ(γ_x^n,4,i) =a⁽⁴⁾_ii (x), i= 1,2, which simply gives the diffusion matrixa⁽⁴⁾(x).

Acknowledgment. The authors thank M.T. Barlow and T. Shirai for stimulating discussions, and I. Shigekawa for valuable comments on Lemma 3.1.

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