The Best Constant of Sobolev Inequality Corresponding to

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Volume 2011, Article ID 875057,17pages doi:10.1155/2011/875057

Research Article

The Best Constant of Sobolev Inequality Corresponding to

Clamped Boundary Value Problem

Kohtaro Watanabe,

¹

Yoshinori Kametaka,

²

Hiroyuki Yamagishi,

³

Atsushi Nagai,

⁴

and Kazuo Takemura

⁴

1Department of Computer Science, National Defense Academy, 1-10-20 Hashirimizu, Yokosuka 239-8686, Japan

2Division of Mathematical Sciences, Graduate School of Engineering Science, Osaka University, 1-3 Machikaneyama-cho, Toyonaka 560-8531, Japan

3Tokyo Metropolitan College of Industrial Technology, 1-10-40 Higashi-ooi, Shinagawa, Tokyo 140-0011, Japan

4Department of Liberal Arts and Basic Sciences, College of Industrial Technology, Nihon University, 2-11-1 Shinei, Narashino 275-8576, Japan

Correspondence should be addressed to Kohtaro Watanabe,[email protected] Received 14 August 2010; Accepted 10 February 2011

Academic Editor: Irena Rach ˚unkov´a

Copyrightq2011 Kohtaro Watanabe et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Green’s functionGx, y of the clamped boundary value problem for the diﬀerential operator

−1^Md/dx^2M on the interval−s, sis obtained. The best constant of corresponding Sobolev inequality is given by max_|y|≤sGy, y. In addition, it is shown that a reverse of the Sobolev best constant is the one which appears in the generalized Lyapunov inequality by Das and Vatsala 1975.

1. Introduction

ForM1,2,3, . . .,s >0, letHH₀^M−s, sbe a SobolevHilbertspace associated with the inner product·,·_M:

HHM

u|u^M∈L²−s, s, uⁱ±s 00≤i≤M−1 , u, v_M

_s

−su^Mxv^Mxdx, u²_M u, u_M.

1.1

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The fact that·,·_Minduces the equivalent norm to the standard norm of the SobolevHilbert space ofMth order follows from Poincar´e inequality. Let us introduce the functionalSuas follows:

Su

sup_|y|≤su y ²

u²_M . 1.2

To obtain the supremum ofSi.e., the best constant of Sobolev inequality, we consider the following clamped boundary value problem:

−1^Mu^2Mfx −s < x < s, uⁱ±s 0 0≤i≤M−1.

BVPM

Concerning the uniqueness and existence of the solution toBVPM, we have the following proposition. The result is expressed by the monomialKjx:

Kjx KjM;x

⎧⎪

⎪⎨

⎪⎪

⎩

x^2M−1−j 2M−1−j

!

0≤j≤2M−1 ,

0

2M≤j .

1.3

Proposition 1.1. For any bounded continuous functionfxon an interval−s < x < s,BVPM has a unique classical solutionuxexpressed by

ux _s

−sG x, y

f y

dy −s < x < s, 1.4

where Green’s functionGx, y GM;x, y −s < x, y < sis given by G

x, y

−1^M 2

K0x−yD⁻¹

Kij2s Ki

s−y Kjsx 0

Kij2s Ki

sy Kjs−x 0

1.5 −1^MD⁻¹

Kij2s Ki

sx∧y Kj

s−x∨y

0

−s < x, y < s

. 1.6

D is the determinant ofM×MmatrixKij2s 0 ≤ i, j ≤ M−1,x∧y minx, y, and x∨ymaxx, y.

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With the aid of Proposition 1.1, we obtain the following theorem. The proof of Proposition 1.1is shown in AppendicesAandB.

Theorem 1.2. iThe supremumCM;−s, s(abbreviated asCMif there is no confusion) of the Sobolev functionalSis given by

CM;−s, s sup

u∈H, u /≡0

Su max

|y|≤sG y, y

G0,0 s^2M−1

2^2M−12M−1{M−1!}² . 1.7 Concretely,

C1,−s, s s

2, C2,−s, s s³

24, C3,−s, s s⁵

640, C4,−s, s s⁷

32256, . . . . 1.8 iiCM;−s, sis attained byuGx,0, that is,SGx,0 CM;−s, s.

Clearly,Theorem 1.2i,iiis rewritten equivalently as follows.

Corollary 1.3. Let u ∈ H, then the best constant of Sobolev inequality (corresponding to the embedding ofHintoL^∞−s, s)

sup

|y|≤s

u y₂

≤C _s

−s

u^Mx²dx, 1.9

isCM;−s, s. Moreover the best constantCM;−s, sis attained byux cGx,0, wherecis an arbitrary complex number.

Next, we introduce a connection between the best constant of Sobolev- and Lyapunov- type inequalities. Let us consider the second-order diﬀerential equation

upxu0 −s≤x≤s, 1.10

wherepx ∈ L¹−s, s∩C−s, s. If the above equation has two pointss1 ands2 in−s, s satisfyingus1 0us2, then these points are said to be conjugate. It is wellknown that if there exists a pair of conjugate points in−s, s, then the classical Lyapunov inequality

_s

−spxdx > 2

s, 1.11

holds, where px : maxpx,0. Various extensions and improvements for the above result have been attempted; see, for example, Ha 1, Yang 2, and references there in.

Among these extensions, Levin3and Das and Vatsala4extended the result for higher order equation

−1^Mu^2M−pxu0 −s≤x≤s. 1.12

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For this case, we again call two distinct pointss1and s2 conjugate if there exists a nontrivial C^2M−s, s∩C^M−1−s, ssolution of1.12satisfying

uⁱs1 0uⁱs2 i0, . . . , M−1. 1.13 We point out that the constant which appears in the generalized Lyapunov inequality by Levin3and Das and Vatsala4is the reverse of the Sobolev best embedding constant.

Corollary 1.4. If there exists a pair of conjugate points on−s, swith respect to1.12, then _s

−spxdx > 1

CM;−s, s, 1.14

whereCM;−s, sis the best constant of the Sobolev inequality1.9.

Without introducing auxiliary equation u^2M −1^M−1pu 0 and the existence result of conjugate points as2,4, we can prove this corollary directly through the Sobolev inequalitythe idea of the proof origins to Brown and Hinton5, page 5.

Proof ofCorollary 1.4. Consider _s₂

s1

u^Mx ²dx _s₂

s1

pxux²dx≤

sup

s1≤x≤s2

|ux|

₂_s₂

s1

pxdx

≤CM;s1, s2 _s₂

s1

u^Mx ²dx _s₂

s1

pxdx.

1.15

In the second inequality, the equality holds for the function which attains the Sobolev best constant, so especially it is not a constant function. Thus, for this function, the first inequality is strict, and hence we obtain

1

CM;s1, s2 <

_s₂

s1

pxdx. 1.16

Since

1

CM;−s, s ≤ 1

CM;s1, s2 <

_s₂

−s1

pxdx≤ _s

−spxdx, 1.17

we obtain the result.

Here, at the end of this section, we would like to mention some remarks about 1.12. The generalized Lyapunov inequality of the form 1.14 was firstly obtained by Levin 3 without proof; see Section 4 of Reid 6. Later, Das and Vatsala 4 obtained the same inequality1.14 by constructing Green’s function forBVPM. The expression of the Green’s function of Proposition 1.1 is diﬀerent from that of 4. The expression of

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4, Theorem 2.1is given by some finite series ofxandyon the other hand, the expression ofProposition 1.1 is by the determinant. This complements the results of7–9, where the concrete expressions of Green’s functions for the equation −1^Mu^2M f but diﬀerent boundary conditions are given, and all of them are expressed by determinants of certain matrices asProposition 1.1.

2. Reproducing Kernel

First we enumerate the properties of Green’s functionGx, yofBVPM.Gx, yhas the following properties.

Lemma 2.1. Consider the following:

1

∂^2M_x G x, y

0

−s < x, y < s, x /y

, 2.1

2

∂ⁱ_xG x, y

x±s0

0≤i≤M−1, −s < y < s

, 2.2

3

∂ⁱ_xG x, y

yx−0−∂ⁱ_xG x, y

yx0

⎧⎪

⎨

⎪⎩

0 0≤i≤2M−2,

−1^M i2M−1 −s < x < s,

2.3

4

∂ⁱ_xG x, y

xy0−∂ⁱ_xG x, y

xy−0

⎧⎪

⎨

⎪⎩

0 0≤i≤2M−2,

−1^M i2M−1

−s < y < s .

2.4

Proof. Fork 1≤k≤2Mand−s < x, y < s,x /y, we have from1.5

∂^k_xG x, y

−1^M 2

sgn

x−y_k

Kkx−y D⁻¹

Kij2s Ki

s−y Kkjsx 0

Kij2s Ki

sy

−1^kKkjs−x 0

. 2.5

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Fork 2M, noting the factKjx 0 2M≤j, we have1. Next, for 0≤k ≤M−1 and

−s < y < s, we have from2.5

∂^k_xG x, y

x−s

−1^M 2

−1^kKk

sy

D⁻¹

Kij2s Ki

s−y Kkj0 0

Kij2s Ki

sy

−1^kKkj2s 0

. 2.6 SinceKk0, . . . , KkM−10 0, . . . ,0, we have

−1^Mk2 ∂^k_xG x, y

x−sKk

sy D⁻¹

Kij2s Ki

sy Kkj2s 0

Kk

sy D⁻¹

Kij2s Ki

sy 0 · · · 0 −Kk

sy 0.

2.7

Note that subtracting the kth row from Mth row, the second equality holds. Equation

∂^k_xGx, y|_xs 0 is shown by the same way. Hence, we have2. For 0 ≤ k ≤ 2M−1, we have

∂^k_xG x, y

yx−0−∂^k_xG x, y

yx0

−1^M 2

1−−1^k Kk0

⎧⎪

⎨

⎪⎩

0 0≤k≤2M−2,

−1^M k2M−1 −s < x < s,

2.8

where we used the factKk0 0k /2M−1, 1 k2M−1. So we have3, and4follows from3.

UsingLemma 2.1, we prove that the functional spaceHassociated with inner norm

·,·_Mis a reproducing kernel Hilbert space.

Lemma 2.2. For anyu∈H, one has the reproducing property

u y

u·, G

·, y

M

_s

−su^Mx∂^M_x G x, y

dx

−s≤y≤s

. 2.9

Proof. For functionsuuxandvvx Gx, ywithyarbitrarily fixed in−s≤y≤s, we have

u^Mv^M−u−1^Mv^2M

⎛

⎝^M−1

j0

−1^M−1−j u^jv^2M−1−j

⎞

⎠. 2.10

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Integrating this with respect toxon intervals−s < x < yandy < x < s, we have _s

−su^Mxv^Mxdx− _s

−sux−1^Mv^2Mxdx

⎡

⎣^M−1

j0

−1^M−1−ju^jxv^2M−1−jx

⎤

⎦^xy−0_x−s ^xs_xy0

^M−1

j0

−1^M−1−j

u^jsv^2M−1−js−u^j−sv^2M−1−j−s

^M−1

j0

−1^M−1−ju^j y

v^2M−1−j y−0

−v^2M−1−j y0

.

2.11

Using1,2, and4inLemma 2.1, we have2.9.

3. Sobolev Inequality

In this section, we give a proof ofTheorem 1.2andCorollary 1.3.

Proof ofTheorem 1.2andCorollary 1.3. Applying Schwarz inequality to2.9, we have u

y²≤ _s

−s

∂^Mx G

x, y²dx _s

−s

u^Mx²dxG y, y

s

−s

u^Mx²dx. 3.1

Note that the last equality holds from2.9; that is, substituting2.9,u· G·, y. Let us assume that

CM;−s, s CM max

|y|≤sG y, y

G0,0, 3.2

holdsthis will be proved in the next section. From definition ofCM, we have

sup

|y|≤s|u y

| ₂

≤CM _s

−s

u^Mx²dx. 3.3

Substitutingux Gx,0∈Hin to the above inequality, we have

sup

|y|≤s|G y,0

| ₂

≤CM _s

−s

∂^M_x Gx,0²dx CM². 3.4

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Combining this and trivial inequalityCM² G0,0²≤sup_|y|≤s|Gy,0|², we have

CM²≤

sup

|y|≤s

G

y,02

≤CM _s

−s

∂^M_x Gx,0²dx CM². 3.5

Hence, we have

sup

|y|≤s|G y,0

| ₂

CM _s

−s

∂^M_x Gx,0²dx, 3.6

which completes the proof ofTheorem 1.2andCorollary 1.3.

Thus, all we have to do is to prove3.2.

4. Diagonal Value of Green’s Function

In this section, we consider the diagonal value of Green’s function, that is,Gx, x. From Proposition 1.1, we have forM1,2,3

G1;x, x

s²−x²

2s , G2;x, x

s²−x²₃

24s³ , G2;x, x

s²−x²₅

650s⁵ . 4.1 Thus, we can expect thatGx, xtakes the formGM;x, x const. K0M; 1xK0M; 1−x.

Precisely, we have the following proposition.

Proposition 4.1. Consider Gx, x −1^MD⁻¹

Kij2s Kis−x Kjsx 0

2M−1 M−1

1

K02sK0sxK0s−x

2M−1 M−1

1 K02s

s²−x²_2M−1 {2M−1!}².

4.2

Hence,

CM;−s, s sup

|x|≤sGx, x G0,0 −1^MD⁻¹

Kij2s Kis Kjs 0

s^2M−1 2^2M−12M−1!

2M−1 M−1

s^2M−1

2^2M−12M−1M−1!²,

4.3

wherei, jsatisfy 0≤i, j≤M−1.

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To prove this proposition, we prepare the following two lemmas.

Lemma 4.2. Letux c1Gx, x, where

c⁻¹₁ −1^M

22M−1 2M−1

D⁻¹

1 Kij2s 0 ... 0 1 0 · · · 0 0

, 4.4

(i, jsatisfy 0≤i, j≤M−1), then it holds that

−u^22M−11 −s < x < s, 4.5

uⁱ±s 0 0≤i≤2M−2, 4.6

u^2M−1s −

2M−1 M−1

c1. 4.7

Lemma 4.3. Letux c2K0sxK0s−x −s < x < s, wherec₂⁻¹

22M−1

2M−1 , then it holds that4.6andu^2M−1s −K02sc2.

Proof ofProposition 4.1. From Lemmas 4.2 and 4.3, ux c1Gx, x and ux c2K0s xK0s−xsatisfy BVP2M−1 in case offx 1−s < x < s. So we have

c1Gx, x c2K0sxK0s−x −s < x < s, 4.8 2M−1

M−1

c1 K02sc2. 4.9

Inserting4.9into4.8, we haveProposition 4.1.

Proof ofLemma 4.2. Let

ux c1Gx, x c1−1^MD⁻¹vx, vx

Kij2s Kis−x Kjsx 0

, 4.10

then diﬀerentiatingvxktimes we have

v^kx ^k

l0

−1^l k

l

wk,lx, wk,lx

Kij2s Klis−x Kk−ljsx 0

. 4.11

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At first, fork22M−1, we have

v^22M−1x ^22M−1

l0

−1^l

22M−1 l

w22M−1,lx

^2M−2

l0

−1^l

22M−1 l

w22M−1,lx−

22M−1 2M−1

w22M−1,2M−1x

^22M−1

l2M

−1^l

22M−1 l

w22M−1,lx.

4.12

The first term vanishes because

K22M−1−ljsx K2M2M−2−ljsx 0 0≤l≤2M−2. 4.13

The third term also vanishes because

Klis−x 0 2M≤l≤22M−1. 4.14 Thus, we have

v^22M−1x −

22M−1 2M−1

w22M−1,2M−1x,

w22M−1,2M−1x

K_ij2s K_2M−1is−x

K2M−1jsx 0

1 K_ij2s 0 ... 0 1 0 · · · 0 0 .

4.15

Hence, we have

−u^22M−1x −c1−1^MD⁻¹ v^22M−1x 1, 4.16

by which we obtain4.5. Next, for 0≤k≤M−1, we have

v^ks ^k

l0

−1^l k

l

wk,ls, wk,ls

Kij2s Kli0 Kk−lj2s 0

. 4.17

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Since 0≤li≤2M−2, we havewk,ls 0. Thus, we havev^ks 00 ≤k≤M−1. For M≤k≤2M−2, we have

v^ks ^M−1

l0

−1^l k

l

wk,ls ^k

lM

−1^l k

l

wk,ls. 4.18

The first term vanishes becauseKli0 00 ≤l ≤ M−1. Next, we show that the second term also vanishes. Let

wk,ls

Kj2s 0 ... ... K2M−2−lj2s 0 K2M−1−lj2s 1

K2M−lj2s 0

... ... KM−1j2s 0 Kk−lj2s 0

M≤l≤k≤2M−2. 4.19

Since 0≤k−l ≤2M−2−l, two rows, including the last row, coincide, and hence we have wk,ls 0. Thus, we havev^ks 0M≤k≤2M−2. So we have obtainedu^ks 00≤ k≤2M−2. By the same argument, we haveu^k−s 00≤k≤2M−2. Hence, we have 4.6. Finally, we will show4.7. Fork2M−1, notingKli0 00≤l≤M−1, we have

v^2M−1s ^2M−1

lM

−1^l

2M−1 l

w2M−1,ls, 4.20

where

w2M−1,ls

Kij2s Kli0

K2M−1−lj2s 0

Kj2s 0 ... ... K2M−2−lj2s 0 K2M−1−lj2s 1

K2M−lj2s 0

... ... KM−1j2s 0 K2M−1−lj2s 0

Kj2s 0

... ...

K2M−2−lj2s 0

K2M−1−lj2s 1

K2M−lj2s 0

... ... KM−1j2s 0 0 · · · 0 −1

−D.

4.21

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Thus, we obtainw2M−1,ls −DM≤l≤2M−1. Hence we have

v^2M−11 ^2M−1

lM

−1^l

2M−1 l

w2M−1,ls −D

2M−1

lM

−1^l

2M−1 l

−D^2M−2

lM

−1^l

2M−2 l−1

2M−2 l

D −1^M1D

2M−1 M−1

,

4.22

that is,

u^2M−1s c1−1^MD⁻¹v^2M−1s −

2M−1 M−1

c1. 4.23

This completes the proof ofLemma 4.2.

Proof ofLemma 4.3. Let

ux c2K0sxK0s−x c2

2M−1!²

s²−x² ^2M−1. 4.24

Diﬀerentiatingux ktimes, we have

u^kx c2

k l0

−1^l k

l

Kk−lsxKls−x. 4.25

Fork22M−1, notingK22M−1−lsx 0 0≤l≤2M−2,K2M−1sx K2M−1s−x 1, andKls−x 0 2M≤l≤22M−1, we have

−u^22M−1x c2

22M−1 2M−1

1. 4.26

Thus, we have4.5. If 0≤k≤2M−2, then we have

u^ks c2

k l0

−1^l k

l

Kk−l2sKl0 0. 4.27

Sinceu^k−x −1^ku^kx, we haveu^k−s 00≤k≤2M−2. Hence, we have4.6. If k2M−1, then we have

u^2M−1s c2 2M−1

l0

−1^l

2M−1 l

K2M−1−l2sKl0 −c2K02s. 4.28

This provesLemma 4.3.

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Appendices

A. Deduction of 1.5

In this section,1.5in Proposition 1.1 is deduced. Suppose thatBVPM has a classical solutionux. Introducing the following notations:

u ^tu0, . . . , u_2M−1, ui uⁱ 0≤i≤2M−1, e ^t0, . . . ,0,1 2M×1 matrix,

N

⎛

⎜⎜

⎜⎝ 0 1

0 . ..

. .. 1 0

⎞

⎟⎟

⎟⎠

2M×2Mnilpotent matrix ,

A.1

BVPMis rewritten as

uNue−1^Mfx −s < x < s,

ui±s 0 0≤i≤M−1. A.2

Let the fundamental solutionExbe expressed asEx expNx KxK0⁻¹, where

Kx

Kij

x, K0

⎛

⎝ 1

· · · 1

⎞

⎠K0⁻¹, A.3

theni, jsatisfy 0≤i, j ≤2M−1.Exsatisfies the initial value problemE NE, E0 I. I is a unit matrix. SolvingA.2, we have

ux Exsu−s _x

−sE x−y

e−1^Mf y

dy, ux Ex−sus−

_s

x

E x−y

e−1^Mf y

dy,

A.4

or equivalently, for 0≤i≤2M−1, we have

uix ^2M−1

j0

K_ijxsu_2M−1−j−s _x

−s−1^MKi

x−y f

y dy,

uix ^2M−1

j0

Kijx−su2M−1−js− _s

x

−1^MKi

x−y f

y dy.

A.5

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Employing the boundary conditionsA.2, we have

uix ^M−1

j0

Kijxsu2M−1−j−s _x

−s−1^MKi

x−y f

y dy,

uix ^M−1

j0

Kijx−su2M−1−js− _s

x

−1^MKi

x−y f

y dy.

A.6

In particular, ifi0, then we have

u0x ^M−1

j0

Kjxsu2M−1−j−s _x

−s−1^MK0

x−y f

y dy,

u0x ^M−1

j0

Kjx−su_2M−1−js− _s

x

−1^MK0

x−y f

y dy.

A.7

On the other hand, using the boundary conditionsA.2again, we have

0uis ^M−1

j0

Kij2su2M−1−j−s _s

−s−1^MKi

s−y f

y dy,

0ui−s ^M−1

j0

Kij−2su2M−1−js− _s

−s−1^MKi

−s−y f

y dy.

A.8

Solving the above linear system of equations with respect to u_2M−1−i−s, u2M−1−is 0≤i≤M−1, we have

u_2M−1−i−s −

_s

−s−1^M Kij₋₁

2sKi s−y

f y

dy,

u2M−1−is

_s

−s−1^M Kij₋₁

−2sKi

−s−y f

y dy.

A.9

SubstitutingA.9intoA.7, we have

u0x − _s

−s−1^M Kj

xs Kij₋₁

2sKi s−y

f y

dy

_x

−s−1^MK0x−yf y

dy,

u0x _s

−s−1^M Kj

x−s Kij₋₁

−2sKi

−s−y f

y dy

_s

x

−1^MK0x−yf y

dy.

A.10

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Taking an average of the above two expressions and notingux u0x, we obtain1.4, where Green’s functionGx, yis given by

G x, y

−1^M 2

K0x−y− Kj

xs Kij₋₁

2sKi s−y

Kj

x−s Kij₋₁

−2sKi

−s−y .

A.11

Using propertiesKi−x −1ⁱ¹Kix, we have Kj

x−s − Kj

s−x

−1ⁱδij , Kij

−2s

−1^ij1Kij 2s −

−1ⁱδij Kij 2s

−1ⁱδij , Ki

−s−y

−1ⁱ¹Ki sy −

−1ⁱδij Ki sy

,

A.12

where δij is Kronecker’s delta defined by δij 1i j, 0 i /j. Inserting these three relations intoA.11, we have

G x, y

−1^M 2

K0x−y− Kj

sx Kij₋₁

2sKi s−y

− Kj

s−x Kij₋₁

2sKi sy

.

A.13

Applying the relation

ta A⁻¹b −

A b

ta 0

|A| , A.14

whereA is anyN×Nregular matrix anda and b are anyN×1 matrices, we have1.5.

B. Deduction of 1.6

To prove1.6, we show

K0

x−y

−D⁻¹

Kij2s Ki

s−y Kjsx 0

−

Kij2s Ki

sy Kjs−x 0

−s < x, y < s . B.1

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Letx≥y. IfB.1holds, substituting it to1.5, replacingxwithx∨y,ywithx∧y, then we obtain1.6. The casex≤ yis shown in a similar way. Lety−s ≤ y ≤ sbe fixed, and let ux K0x−y. Thenusatisfies

u^2M0 −s < x < s, uⁱ−s −1ⁱ¹Ki

sy

, uⁱs Ki

s−y

0≤i≤M−1. B.2 On the other hand, let

vx −D⁻¹

Kij2s Ki

s−y Kjsx 0

−

Kij2s Ki

sy Kjs−x 0

. B.3

Diﬀerentiatingv ktimes with respect tox, we have

v^kx −D⁻¹

Kij2s Ki

s−y Kkjsx 0

−−1^k

Kij2s Ki

sy Kkjs−x 0

. B.4

Fork2M, noticingKkjsx Kkjs−x 0, we havev^2Mx 0. For 0≤k≤M−1, we have

v^k−s −D⁻¹

Kij2s Ki

s−y Kkj0 0

−−1^k

Kij2s Ki

sy Kkj2s 0

−1^kD⁻¹

Kij2s Ki

sy 0 · · · 0 −Kk

sy

−1^k1Kk

sy ,

B.5

where we usedKkj0 0. Similarly, for 0 ≤ k ≤ M−1, we havev^ks Kks−y. So vxsatisfies

v^2M 0 −s < x < s, vⁱ−s −1ⁱ¹Ki

sy

, vⁱs Ki

s−y

0≤i≤M−1. B.6 which is the same equation asB.2. Hence, we havevx ux.

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