在高維度下受波氏分配自我相斥隨機漫步的均場行為 - 政大學術集成
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(2) 謝. 首先要感謝我的指導教授「陳隆奇」教授，在我碩一的時候就直接收我 作為學生，並在很相信我可以做得動 self-avoiding walk 這個主題，雖然我做 的只是佔了整個理論的一小部分，但還是令人覺得頭痛，學了很久，寫了. 政 治 大. 很久，慢慢地終於也到了寫完的這一刻，再一次地感謝老師每一次的提點. 立. 與信任，都讓我充滿了信心來寫這篇論文。. ‧ 國. 學. 再來是要感謝推薦我去找陳隆奇教授的教官「游森棚」教授，在大學時 期，因為教官的教學，讓我對數學產生了高度的興趣，當初要念研究所的. ‧. 時候，教官說了一句：「你一定會喜歡陳隆奇老師的研究的。」因為信任教 官的說詞，不假思索地投注在機率的語言裡，就以結果而言，喜歡是喜歡，. Nat. sit. y. 但是知識真的不簡單，雖說如此我還是努力地學會了一些東西，在這樣的. al. er. io. 過程當中，其實我成長許多，為此也非常感謝教官的推薦，今天沒有教官，. n. 就沒有成長過後的我了。. Ch. i Un. v. 接著是感謝我的實變老師「陳天進」教授，一個喜歡學生叫老大的老. engchi. 師，在修習實變的第一堂課就說：「跟不上不是我的錯，是你們要跟上我， 不是我要放慢速度。」難度很高加上速度很快，讓我繃緊神經每天讀書，時 不時就去敲老大門，最終克服困難，高分過關。 最後是感謝這三年來一直陪伴著我成長的家人、朋友，因為有你們我 才能走到現在，完成屬於我的第一篇著作。. i DOI:10.6814/NCCU202000775.
(3) 中文 要. self-avoiding walk 是線性聚合物的模型。它是機率和統計力學中一個 重要而有趣的模型。一些重要問題已經解決 (c.f. [5]). 然而，許多重要問 題仍未解決，特別是涉及關鍵指數的問題，尤其是遠程模型的關鍵指數。. 政 治 大. 在本文中，我們獲得了對於一個特殊的長域模型，其單步分佈是波松分佈. 立. 的特殊敏感度模型，其敏感性指數滿足均值場行為，且其值大於上臨界值. ‧ 國. 學. （dc = 4）。參數 λ ≥ λd 的類型分佈，其中 λd 取決於維度。為此，我們選擇 一組特殊的 bootstrapping functions，它們類似於 [4]，並使用 lace expansion. ‧. 分析有關 bootstrapping functions 的複雜部分。此外，對於 d > 4，我們得到 λd 的確切值。. n. er. io. sit. y. Nat. al. Ch. engchi. i Un. v. ii DOI:10.6814/NCCU202000775.
(4) Abstract. Self-avoiding walk is a model for linear polymers. It is an important and interesting model in Probability and Statistical mechanics. Some of the important problems had been solved (c.f. [5]). However, many of the important problems. 政 治 大. remain unsolved, particularly those involving critical exponents, especially the. 立. critical exponents for long-range models. In this thesis, we see Lace expansion. ‧ 國. 學. to obtain that the critical exponent of the susceptibility satisfies the mean-field behavior with the dimensions above the upper critical dimension (dc = 4) for a. ‧. special loge-range model in which each one-step distribution is the Poisson-type distribution with parameter λ ≥ λd where λd depends on the dimensions. To. Nat. sit. y. achieve this, we choose a particular set of bootstrapping functions which is similar. al. n. Moreover we get the exactly value of λd for d > 4.. Ch. engchi. er. io. as [4] and using a notoriously complicated part of the lace expansion analysis.. i Un. v. iii DOI:10.6814/NCCU202000775.
(5) Contents 謝. i. 中文 要. ii. 政 治 大. Abstract. 立. Contents. y. Nat. 1 3. sit. ‧ 國. vi. ‧. 1 Introduction. iv. 學. List of Tables. iii. 3. 2 Models and Main Results. 2.2. Main results and their proofs . . . . . . . . . . . . . . . . . . . . . . . . . . .. n. al. er. Notations and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. io. 2.1. Ch. n U engchi. iv. 6. 3 The lace expansion for self-avoiding walk. 10. 4 Diagrammatic bounds estimate. 14. 4.1. Diagrammatic bounds on the lace expansion coefficients . . . . . . . . . . . .. 14. 4.2. Diagramatic bounds on the bootstrapping argument . . . . . . . . . . . . . . .. 25. 5 Random-walk estimate. 29. 5.1. The diagrams bound of random-walk quantities for p = 1 . . . . . . . . . . . .. 29. 5.2. The diagrams bound of random-walk quantities for p > 1 . . . . . . . . . . . .. 35. 6 Proof of Proposition 2.2.7 - 2.2.9 6.1. 40. Proof of Proposition 2.2.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 40. DOI:10.6814/NCCU202000775.
(6) 6.2. Proof of Proposition 2.2.8 - 2.2.9 and Lemma 4.1.1 . . . . . . . . . . . . . . .. 44. Appendix A. 48. Bibliography. 50. 立. 政 治 大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i Un. v. DOI:10.6814/NCCU202000775.
(7) List of Tables Upper bounds A for 6 ≤ d ≤ 9 . . . . . . . . . . . . . . . . . . . . . . . . . .. 立. 46. 政 治 大. 學 ‧. ‧ 國 io. sit. y. Nat. n. al. er. 6.1. Ch. engchi. i Un. v. DOI:10.6814/NCCU202000775.
(8) Chapter 1 Introduction A self-avoiding walk (SAW) is a model for linear polymers. In mathematics, a SAW is a. 政 治 大. sequence of moves on lattices (a lattice path) that does not visit the same point more than once.. 立. This is a special case of the theoretical graphic notion of a path. Very little is known rigorously. ‧ 國. 學. about the SAW from a mathematical perspective, although physicists have provided numerous conjectures that are believed to be true and are strongly supported by numerical simulations.. ‧. In higher dimensions, the SAW is believed to behave much like the corresponding random walk. This behavior is called the mean-field behavior. The dimension dc is called the upper. Nat. sit. y. critical dimension if the behavior is the mean-field behavior for dimensions d > dc . It’s well. er. io. known that dc = 4 for a finite-range SAW model (c.f. [4]) and dc = 2(α ∧ 2) for a long-range model in which one-step distribution decays as |x|−d−α for some α > 0 (c.f. [7]) . The lace. n. al. Ch. i Un. v. expansion is a powerful tool for analyzing the critical behavior for SAW on Zd for d > dc . The. engchi. idea of the lace expansion was initiated by Brydges and Spencer for investigating weakly SAW for d > 4. Later, the lace-expansion was applied to various stochastic-geometrical models, such as SAW for d > 4, (c.f. [5]) , lattice trees/animals for d > 8, (c.f. [9]), percolation for d > 6, (c.f. [6]), oriented percolation for d > 4, (c.f. [13]), contact process for d > 4, (c.f. [11]), and Ising model for d > 4, (c.f. [10]). In this thesis, we consider the mean-field behavior for longrange SAW in which one-step distribution D(x) is symmetric Poisson-type distribution that will be defined in the next chapter. The rest of this thesis is organized as follows. In Chapter 2, we define the symmetry Poisson-type distribution and state the main results of this thesis and their proofs under some key propositions (Proposition 2.2.7-2.2.9). In Chapter 3, we follow the same argument of [3] to. 1 DOI:10.6814/NCCU202000775.
(9) introduce the lace expansion that is the most important ingredient to prove the key propositions, and then we estimate the upper bound of the diagrams on lace expansion coefficients in Chapter 4. In Chapter 5, we evaluate the upper bounds of some random-walk quantities that are useful to estimate the upper bound of diagrams. Finally, in Chapter 6, we can use the previous analysis (Chapter 4-5) to prove the key propositions.. 立. 政 治 大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i Un. v. 2 DOI:10.6814/NCCU202000775.
(10) Chapter 2 Models and Main Results First, we provide precise definitions of the long-range Poisson-type self-avoiding walk on. 政 治 大. Zd , whose translation-invariant 1-step distribution decays as the symmetric Poisson distribution.. 立. Then, we present the main results and explain their proofs according some key propositions.. ‧ 國. 學. 2.1 Notations and Definitions. ‧. xj. j=1. ] ∞ ( ∑ λxj −1 ) λ−xj −1 ) −λ +e . (xj − 1)! (−xj − 1)! =1 −x =1 j. n. al. sit. ∞ (∑. Ch. er. e. −λ. io. 2 =. d [ ∏. Nat. d. y. By Taylor’s formula, we have the following equation. i Un. v. Then, we can define a symmetry Poisson distribution D(x) on Zd with parameter λ > 0 as follows. Since. 1. =. engchi. ) d ( e−λd ∏ ∑ λ|xj |−1 2d j=1 x ̸=0 (|xj | − 1)! j. =. ∑ x=(x1 ,x2 ,...,xd )∈Zd. (∏ d. ) e−λ λ|xj |−1 , 2 (|xj | − 1)! j=1. xk ̸=0,∀k=1,2,...,d. 3 DOI:10.6814/NCCU202000775.
(11) let D(x) = D(x1 , ..., xd ) =. ∏d j=1. . e−λ λ|xj |−1 , 2 (|xj |−1)!. if x1 x2 · · · xd ̸= 0, (2.1.1) if x1 x2 · · · xd = 0.. 0,. ∑ In the rest of the thesis, we denote fˆ(k) ≡ x∈Zd f (x)eik·x where k = (k1 , k2 , ..., kd ) and ∑ (f ∗ g)(x) ≡ y∈Zd f (y)g(x − y) the Fourier transform of function f on Zd and the convolution of two function f and g on Zd , respectively. By (2.1.1), we get ∑. =. D(x)eik·x. x∈Zd −λd. e. =. (xj − 1)!. −xj. ‧ 國. d ∏ ) −ikj e−λ ( λeikj e (cos(kj ) + i sin(kj )) + eλe (cos(kj ) − i sin(kj )) 2 j=1. y. ( ) e−λ+λ cos(kj ) cos(kj ) cos(λ sin(kj )) − sin(kj ) sin(λ sin(kj )). io. sit. d ∏ j=1. d ∏. al. ( ) e−λ+λ cos(kj ) cos kj + λ sin(kj ) .. n. =. λ−xj −1 ikj xj ) e ) (−xj − 1)! =1. d ∞ ∞ ∑ e−λd ∏ ( ∑ λu ikj (u+1) λu ikj (−u−1) ) ( e ) + ( e ) 2d j=1 u=0 u! u! u=0. Nat. =. xj =1. ∞. ikj xj. ‧. =. 立. j=1. xj −1. 學. =. 2d. 政λ 治 大∑ e )+(. d ∞ ∏ ( ∑ (. j=1. Ch. engchi. er. ˆ D(k). i Un. v. Hence ˆ D(k) =. d ∏. ( ) e−λ+λ cos(kj ) cos kj + λ sin(kj ) .. j=1. Let Wn (x, y) be the set of {w0 , w1 , ..., wn } with w0 = x and wn = y, and δ be the Kronecker delta function. Then the two-point function of the random walk (RW) from the origin o to x is denoted by φRW 0 (x) ≡ δ0,x and, for n ≥ 1, φRW n (x). ≡. ∑. n ∏. D(wi − wi−1 ) = D∗n (x),. w∈Wn (o,x) i=1. 4 DOI:10.6814/NCCU202000775.
(12) where D∗n is the n time convolution. Similarily, we define the two-point function of SAW from the origin o to x by φ0 (x) ≡ δ0,x and, for n ≥ 1, ∑. φn (x) ≡. n ∏. ∏. D(wi − wi−1 ). w∈Wn (o,x) i=1. (1 − δwi ,wj ).. 0≤i<j≤n. Since the green function of RW is defined as follows Sp (x) ≡. ∞ ∑. n φRW n (x)p. = δ0,x +. n=0. ∞ ∑. D∗n (x)pn ,. p ∈ (0, 1).. n=1. We can define the time-independent two-point function of SAW as follows Gp (x) ≡. ∞ ∑ n=0. 立. p ∈ (0, p ), 政 治 大. φn (x)pn ,. c. χp ≡. ∑. Gp (x),. p ∈ (0, pc ).. (2.1.2). ‧. x∈Z d. 學. ‧ 國. where pc is the radius of convergence. Moreover, the susceptibility of SAW is defined as follows. al. er. io. χp ≍ (pc − p)−r ,. sit. y. Nat. In Physics, there is a conjecture about the susceptibility of SAW as follows,. n. iv n Cishbounded away from zero and infinity, and r is a critical engchi U exponent that is not statistically distinguishable from a SAW instead on the integer lattice, on the. where f (p) ≍ g(p) means that. f (p) g(p). hexagonal lattice, the triangular lattice, or indeed on any one of a wide variety of d-dimensional lattices. This feature is called universality. Given any dimension d > 4, r = 1 was proved by Slade (c.f. [6]) for spread-out model with sufficiently large L, and was proved by Hara and Slade (c.f. [5]) for nearest-neighbor model. In a general finite-range model it is predict that r =. 43 32. for d = 2, r = 1.162 for d = 3, r = 1 with logarithmic term for d = 4, and r = 1 for d ≥ 5. However, up to now, there is no rigorous mathematical proof. For the long-range model which the one-step distribution decays as |x|−d−α for some α > 0, given any dimension d > 2(α ∧ 2), r = 1 was proved by Heydenreich (c.f. [7]). In particular, for α = 2 and d = 4, r = 1 with logarithmic term was proved by Chen and Sakai (c.f. [3]). For more background and related. 5 DOI:10.6814/NCCU202000775.
(13) results, we refer to [8] and [1]. It is interesting to get r for other long-range models. In our thesis, we discuss a special long-range model.. 2.2 Main results and their proofs Now, we present the main theorems in this thesis. Theorem 2.2.1. (Infrared bound) For SAW on Zd whose one step distribution D(x) is defined in (2.1.1) for any d > 4, there is a λd ,. λd =. if d = 5, 6,. 政 治 大. 35,. 立. 24, 18,. if d = 8,. +. if d = 9,. 1 2 3 )e d−1 [ln(3125000)−1+ln(d(d−4) )] , d−4. if d ≥ 10,. ‧. 2e (1 3. if d = 7,. 學. . 60,. ‧ 國. . Nat. er. io. sit. y. such that for all λ ≥ λd ,. ˆ G ˆ p ||∞ ≡ sup (1 − D(k))| ˆ ˆ p (k)| ≤ 3, ||(1 − D) G. n. al. uniformly in p ∈ [1, pc ) .. k∈Zd. Ch. engchi. i Un. Remark 2.2.2. According Theorem 2.2.1, we obtain that λd → for 0 < λ <. 2e , 3. v. 2e 3. as d goes to infinity. However,. we don’t obtain what we want.. Remark 2.2.3. The λd of Theorem 2.2.1 is not the best estimate for our Poisson-type. Define a bubble function Bp = ||(pD)∗2 ∗ G∗2 p ||∞ . By Theorem 2.2.1, we can estimate χp that is defined in (2.1.2) as follows. Proposition 2.2.4. Under the same assumption of Theorem 2.2.1, we have χ2p. χ2p d ≥ (pχp ) ≥ and Bpc < ∞. dp 1 + Bp. (2.2.1). 6 DOI:10.6814/NCCU202000775.
(14) We will use the same argument of [7] to prove this proposition in Chapter 5.. By. Proposition 2.2.4, we get the second main result of this thesis directly. Theorem 2.2.5. Under the same assumption of Theorem 2.2.1, we have pc pc (1 + Bpc ) ≤ χp ≤ . pc − p pc − p. (2.2.2). By Theorem 2.2.5, we obtain the third main result in our thesis obviously. Corollary 2.2.6. Under the same assumption of Theorem 2.2.1, we have χp ≍ (pc − p)−1 .. 政 治 大. In Corollary 2.2.6, we get the ciritical exponent r = 1 in our model. We use the. 立. Proposition 2.2.4 and Theorem 2.2.1 to prove Theorem 2.2.5 as follow.. ‧ 國. 學. Proof of Theorem 2.2.5 according to the Proposition 2.2.4 and Theorem 2.2.1. Since Bp is increasing as p is increasing and from Proposition 2.2.4, we get Bpc < ∞, by the first inequality. ‧. of (2.2.1), we have. 1 1 1 d(pχp ) ≥ dp ≥ dp, for p ∈ (0, pc ]. 2 χp 1 + Bp 1 + Bpc. n. al. er. io. Then,. sit. y. Nat. dp ≥. ∫ p. pc. 1 dt ≥ t2. ∫. p. pc. 1C h. t2 χ2t. ∫. pc. i U1 n. e n g c ht2i(1 + B. d(tχt ) ≥. p. v. pc ). dt.. This implies pc (1 + Bpc ) pc ≤ χp ≤ . pc − p pc − p □ The proof of Theorem 2.2.1 is rather straightforward, assuming the following three propositions. To state those propositions, we first define ˆ G ˆ p ||∞ , g1 (p) ≡ p, and g2 (p) ≡ ||(1 − D). (2.2.3). 7 DOI:10.6814/NCCU202000775.
(15) where the supremum near k = 0 should be interpreted as the supremum over the limit as |k| → 0, and g3 (p) ≡ sup k,l. 1 Uˆ (k, l). ˆ kG ˆ p (l)|, × |∆. (2.2.4). where ( ) ( ) Sˆ1 (l + k) + Sˆ1 (l − k) ˆ ˆ ˆ ˆ ˆ U (k, l) ≡ 1 − D(k) S1 (l) + 4S1 (l + k)S1 (l − k) , 2 ˆ k fˆ is defined as follows and the notation ∆ ˆ ˆ ˆ k fˆ(l) = f (l + k) + f (l − k) − fˆ(l). ∆ 2. 政 治 大 Now, we state the three 立 aforementioned propositions and show that they indeed imply. ‧ 國. 學. Theorem 2.2.1. To prove these three propositions, we need to use lace expansion and diagram estimates that will be explained in Chapter 3. Then we will show these three propositions in. ‧. Chapter 6.. sit. y. Nat. Proposition 2.2.7. (Continuity). The functions {gi (p)}3i=1 are continuous in p ∈ [1, pc ).. io. d ≥ 6. Then, we have the following Propositions.. al. er. Let K1 = 1.01, K2 = 1.02, K3 = 2.1 for d = 5, and K1 = 1.1, K2 = 1.1, K3 = 2.1 for. n. iv n C U gi(1) < Ki for i = 1, 2, 3. hFor Proposition 2.2.8. (Initial conditions). e nd g> c4, hwei have Proposition 2.2.9. (Bootstrapping argument). For d > 4, and p ∈ (1, pc ), assume gi (p) ≤ Ki , i = 1, 2, 3, where {Ki }3i=1 are the same constants as in Proposition 2.2.8. Then, the stronger inequlities gi (p) < (1 − ϵ)Ki for some ϵ > 0, and i = 1, 2, 3, hold. We only use g1 (p) and g2 (p) in Proposition 2.2.7 - 2.2.9 to prove Theorem 2.2.1 as follow. As for g3 (p), it is used to prove g2 (p) in Proposition 2.2.9, and we will show that in later chapter. Proof of Theorem 2.2.1 according to Proposition 2.2.7-2.2.9. First, by Proposition 2.2.7, we have g1 (p) is continuous in p ∈ (1, pc ), and by Proposition 2.2.8 we have g1 (1) < K1 , and by Proposition 2.2.9 we have g1 (p) < K1 in p ∈ (1, pc ). Therefore, we can conclude that g1 (p) < K1 in p ∈ [1, pc ). In other words, pc ≤ K1 . 8 DOI:10.6814/NCCU202000775.
(16) Then, we use same argument to obtain that g2 (p) < K2 in p ∈ [1, pc ). This complete prove Theorem 2.2.1. □. 立. 政 治 大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i Un. v. 9 DOI:10.6814/NCCU202000775.
(17) Chapter 3 The lace expansion for self-avoiding walk The lace expansion was derived by Brydges and Spencer in [2]. It was later noted that the. 政 治 大. lace expansion can also be seen as a result from repeated application of the inclusion-exclusion. 立. relation. For a more combinatorial description of the lace expansion, see [14]. We use the. ‧ 國. 學. inclusion-exclusion approach to the time-independent two-point function (2.1.2) as follows. By application of the inclusion-exclusion relation, we have. ∑. (φ1 ∗ φn−1 )(x)pn − Rp(1) (x),. n=1. y. sit. al. n. ∞ ∑ ∑. io. Rp(1) (x) =. ∞ ∑. φ1 (y)I[o ∈ ω (1) ]. n=0 y∈Zd ω (1) ∈Wn−1 (y,x). Ch. n−1 ∏. er. where. φn (x)p = δ0,x +. Nat. n=0. n. ‧. Gp (x) ≡. ∞ ∑. v ni. ∏. D(ωi − ωi−1 ). e n g ci=1 hi U. (1 − δωi ,ωj )pn ,. 0≤i<j≤n−1. and I is an indicator function. Observe that. ∞ ∑. =. (φ1 ∗ φn−1 )(x)pn. n=1 ∞ ∑. ∑. φ1 (y)φn−1 (x − y)pn. n=1 y∈Zd. =. ∑. φ1 (y). φn (x − y)pn+1. n=0. y∈Zd. =. ∞ ∑. p(D ∗ Gp )(x).. 10 DOI:10.6814/NCCU202000775.
(18) Then, we get Gp (x) = δo,x + p(D ∗ Gp )(x) − Rp(1) (x). (1). For convenience, we can use Feynman diagram to analyze Rp (x). Here we first introduce some notations that will be needed. We define ” b a. b, and ”. ” as a self-avoiding path from a to. ” as two independent self-avoiding paths and their intersection is only at. c a. b. ” to represent one step from a. (1). Using inclusion-exclusion, we can rewrite Rp (x) as follows. Rp(1) (x). =. o. 政 治 大 x. − o. x. | ∑ y∈Z. }. ‧. =. {z. (2) :=Rp (x). Gp (x − y) − Rp(2) (x). o=y. io. sit. y. Nat. Then, we define. o. 學. ‧ 國. 立=. x. er. b, ”. a. n. a lπp(1)(y) ≡ . iv n Ch engchi U o=y. (2). Continue the process, and we can rewrite Rp (x) as follows. 11 DOI:10.6814/NCCU202000775.
(19) x. Rp(2) (x). = o. x y. −. y. = o. o x. |. {z. }. (3) :=Rp (x). =. ∑. y. y∈Zd. o. Gp (x − y) − Rp(3) (x),. 政 治 大. where y is a first hitting point of two independent self-avoiding paths. Then, we define. 立π. y. (2) p (y). ≡. . o. ‧ 國. 學. By repeating the application of inclusion-exclusion relations, we obtain the lace expansion. represents at least one step, and ”. a. b. ‧. we redefine Feynman diagram of the lace-expansion coefficients depicted as (i.e. the solid line ” can be allowed to merge into one point, and all. y. Nat. sit. solid lines are independent. Their intersection points are only at the vertices in the following. πp(3) (x) ≡. , o. x. al. n. sum of all possible points.). er. io. Feynman diagrams, and some vertices have no English code, e.g. ” ”, which represents the. Ch. πp(4) (x) ≡. e n g c h(5)i x. ,. i Un. πp (x) ≡. o. v. , o. .... x. In summary, we can get the equation as follow ) N +1 (N +1) Gp (x) = δ0,x + ((pD + Π(N Rp (x), p ) ∗ Gp )(x) + (−1). where ) Π(N p (x). =. N ∑. (−1)n πp(n) (x),. n=1. 12 DOI:10.6814/NCCU202000775.
(20) (N ). and we can do a simple analytical method for Rp (x) to get the following inequality 0 ≤ Rp(N ) (x) ≤ (πp(N ) ∗ Gp )(x).. (3.0.1). For example, given N = 5, we have Rp(5) (x). = ≤. o. x. o. x. (πp(5) ∗ Gp )(x).. =. Similarly, we can use the same argument to get the inequality (3.0.1) for N ∈ N. Due to. 政 治 大 Proposition 3.0.1. (Lace expansion, 立 c.f. [4]). For any p < p and N ∈ Z. the construction explained above, we have the lace expansion as follow c. ‧ 國. Gp (x) = δ0,x + (Jp(N ) ∗ Gp )(x) + (−1)N +1 Rp(N +1) (x),. and the remainder. sit. y. n. al. er. io. ) Jp(N ) (x) = pD(x) + Π(N p (x),. (N ) Rp. (3.0.2). ‧. Nat. where. ≡ {0} ∪ N, then we. 學. obtain the recursion equation. +. ni Ch obeys the bound U engchi 0 ≤ Rp(N ) (x) ≤ (πp(N ) ∗ Gp )(x).. (3.0.3). v. (3.0.4). 13 DOI:10.6814/NCCU202000775.
(21) Chapter 4 Diagrammatic bounds estimate 4.1 Diagrammatic bounds on 治 the lace expansion coefficients. 政. 立. 大. In this chapter, we need to evaluate the upper bound of gi (p) for i = 1, 2, 3 in (2.2.3) and. ‧ 國. 學. (2.2.4). First, we need the following lemma to help us get the important result of our thesis. Lemma 4.1.1. For p ∈ [1, pc ), we have. n=1. y. ∞ (n) ∑ ˆ kπ −∆ ˆp < 1. < ∞, and sup ˆ 1 − D(k) k. io. sit. Nat. n=1. π ˆp(n) (0). ‧. ∞ ∑. n. al. er. The proof of Lemma 4.1.1 will be showed that in Chapter 6. Hence, by Lemma 4.1.1 we have. Ch. lim. N →∞. engchi. π ˆp(N ) (0). i Un. v. = 0,. and by (3.0.4) we obtain 0≤. ∑. ˆ p (0) −→ 0 as N → ∞. Rp(N ) (x) ≤ π ˆp(N ) (0)G. x∈Zd. Recall ˆ p (k) = Π. ∞ ∑ ˆ ˆ p (k). (−1)n π ˆp(n) (k), and Jˆp (k) = pD(k) +Π n=1. 14 DOI:10.6814/NCCU202000775.
(22) Then, (3.0.2) and (3.0.4) yield ˆ p (k) = 1 + Jˆp (k)G ˆ p (k) + 0. G. (4.1.1). Through these simple calculations, we can get 1 . 1 − Jˆp (0). χp =. ˆ Since χp ≥ 0, we can get Jˆp (0) ≤ 1. Using D(0) = 1, we have ˆ p (0). p≤1−Π. 立1. 1 − Jˆp (k). =. ˆ ˆ p (k) 1 − pD(k) −Π. 1. ,. sit. ˆ p (0) 1−p−Π. ,. ‧. Nat. ˆ p (0) = χp = G. 1. 學. hence, we have. ‧ 國. ˆ p (k) = G. 政 治 大. y. By (3.0.3) and (4.1.1), we obtain. (4.1.2). n. al. er. io. ˆ p (0) is a positive continuous and increasing function on p ∈ [0, pc ), and pc is a radius of since G convergence. Then, we have. Ch. engchi. i Un. v. ˆ pc (0) = 0. 1 − pc − Π. 15 DOI:10.6814/NCCU202000775.
(23) As above,we have ˆ p (k) G = = = = =. 1 ˆ ˆ ˆ p (k) (pc + Πpc (0)) − pD(k) −Π 1 ˆ pc (0) − Π ˆ p (0)) + p(1 − D(k)) ˆ ˆ p (0) − Π ˆ p (k) (p − pc ) − 1 + 1 + (Π +Π 1 ˆ pc (0)) + (1 − p − Π ˆ p (0)) + p(−∆ ˆ k D(0)) ˆ ˆ kΠ ˆ p (0)) −(1 − pc − Π + (−∆ − χ1p c 1 χp. +. 1 χp. 1 ˆ ˆ ˆ kΠ ˆ p (0)) + p(−∆k D(0)) + (−∆. 1 . ˆ k Jˆp (0)) + (−∆. 政 治 大. Then, we have. ˆ p (k) ≤ 0≤G. 1 . ˆ k Jˆp (0) −∆. 學. ‧ 國. 立. (4.1.3). In order to evaluate the upper bound of gi (p) for i = 1, 2, 3, we must know the upper bound. ‧. io. m=1. π ˆp(2m+1) (0),. m=0. n. al. ∞ ∑. sit. π ˆp(2m) (0) −. (4.1.4). er. Nat. ∞ ∑. ˆ p (0) = Π. y. (n) (n) ˆ kπ of |ˆ πp (0)| and |∆ ˆp (0)|. Since. i Un. v. ∑∞ ∑∞ (2m+1) (2m) ˆ even ˆ odd and we define Π (k) ≡ ˆp (k) and Π ˆp (k) where k ∈ p (k) ≡ p m=0 π m=1 π. Ch. engchi. [−π, π]d . Hence, we can rewrite (4.1.4) as follow. ˆ p (0) = Π ˆ even ˆ odd Π (0) − Π p p (0), and ˆ kΠ ˆ odd ˆ kΠ ˆ p (0) = −∆ ˆ kΠ ˆ even (0) + ∆ −∆ p (0). p In the begining, we define Lp = ||(pD)∗2 ∗ Gp ||∞ = sup(. o. x. ),. x. 16 DOI:10.6814/NCCU202000775.
(24) and Bp = ||(pD)∗2 ∗ G∗2 p ||∞ = sup(. o. ),. x. x. and γp = p||D||∞ + Lp + Bp = sup(. o. x. ), + sup(. x. o. x. ) + sup(. x. o. x. ).. x (n). (n). ˆ kπ Then, we can obtain the bounds of π ˆp (0) and |∆ ˆp (0)| in Lemma 4.1.2 and Lemma 4.1.3. Lemma 4.1.2. (Diagrammatic bounds on the expansion coefficients.) The expansion coefficients ∑ (n) (n) π ˆp (0) ≡ x πp (x), obey the following bound. . Lp ,. if n = 1,. Bp (p||D||∞ + Lp )γp n−2 ,. if n ≥ 2.. ‧. ‧ 國. 立. 學. π ˆp(n) (0) ≤. . 政 治 大. Proof of Lemma 4.1.2. First, we have the following inequality. sit. y. Nat. io. n. al. (4.1.5). er. Gp (x)I[x ̸= 0] ≤ (pD ∗ Gp )(x).. i Un. v. (n). In the following, we will use the above inequality, and we can analyze π ˆp (0) for each. Ch. engchi. n ∈ N to prove Lemma 4.1.2. For n = 1, we have π ˆp(1) (0) ≡. ( ) ≤ (pD)∗2 ∗ Gp (0) ≤ Lp .. ≤ o=y. o=y. (n). For n ≥ 2, we first decompose π ˆp (0) by using subadditivity and then repeatedly apply. 17 DOI:10.6814/NCCU202000775.
(25) (4.1.5) to obtain Lemma 4.1.2. For example, π ˆp(2) (0). =. ∑. x. x̸=o. o. ≤. (∑. ≤. (. 2. Gp (x). )(. ∑ x̸=o. x. x. )(sup. ). x̸=o. o. o. ) sup Gp (x) x̸=0. x̸=0. ( ) ( ) (pD)∗2 ∗ G∗2 p (0) sup(pD ∗ Gp )(x) , | {z } x̸=0. (4.1.5). ≤. ≤Bp. and ∑. =. o. x. x̸=o. ≤. x̸=o. x. )(sup x̸=o. o. ∑. x x y. o. y̸=x. )(sup x̸=o. ∑ ∑ ( )( ) ( Gp (x)2 ) sup Gp (y)Gp (x − y) sup Gp (x). 政 治 大 )( x̸=o. x̸=o. y̸=x. 立 ( B sup (pD) ∗ G (. (4.1.5). ≤. ∗2 p. p. ) ) sup (pD) ∗ Gp (x) ,. ) (x). ‧ 國. Nat. x̸=o. x̸=o. x. ∑ (. ≤. o. x̸=o. x. )(sup o. x̸=o. ‧. =. (. 學. x̸=o. ∑. ∑. x x. 2. y. o. y̸=x. ) (sup x̸=o. io. x̸=o. x̸=o. sit. ∑ ∑ ( )2 ( ) Gp (x)2 ) sup Gp (y)Gp (x − y) sup Gp (x) (. =. y̸=x. ) o. x̸=o. y. =. π ˆp(4) (0). ∑ (. er. π ˆp(3) (0). ) o. x̸=o. n. ( )2 ( ) ( ) ( ) a v ∗2 Bp l sup (pD) ∗ Gp (x) supi (pD) ∗ Gp (x) . n=o Co h x̸= x̸ U engchi. (4.1.5). ≤. By induction, we have ( π ˆp(n) (0). (. ≤ Bp sup (pD) ∗. G∗2 p. ). )n−2 ( ) ( ) (x) sup (pD) ∗ Gp (x) . x̸=o. x̸=o. ( ) In the following step, we will analyze the bound of supx̸=o (pD) ∗ Gp (x) and ( ) supx̸=o (pD) ∗ G∗2 p (x). Notice that, by omitting the spatial variables, we have ( ) (4.1.5) pD ∗ Gp = pD ∗ δGp + (1 − δ)Gp ≤ pD + (pD)∗2 ∗ Gp ,. 18 DOI:10.6814/NCCU202000775.
(26) where δ is the Kronecker delta. Therefore, sup(pD ∗ Gp )(x) ≤ p||D||∞ + ||(pD)∗2 ∗ Gp ||∞ = p||D||∞ + Lp . x. Similarly, we have pD ∗ G∗2 p. ( ) pD ∗ Gp ∗ δGp + (1 − δ)Gp. = (4.1.5). pD ∗ Gp + (pD)∗2 ∗ G∗2 p ( ) pD ∗ δ + (1 − δ)Gp + (pD)∗2 ∗ G∗2 p. ≤ =. (4.1.5). pD + (pD)∗2 ∗ Gp + (pD)∗2 ∗ G∗2 p ,. ≤. 政 治 大. hence. 立 ≤. sup(pD ∗ G∗2 p )(x). ‧ 國. 學. x. p||D||∞ + ||(pD)∗2 ∗ Gp ||∞ + ||(pD)∗2 ∗ G∗2 p ||∞ p||D||∞ + Lp + Bp ≡ γp .. =. ‧. □. Nat. (n). ˆ p (k) = sup(1 − cos k · x)Gp (x). W. n. al. x. Ch. engchi. er. io. sit. y. ˆ kπ Then, we define the following equation to get the bound of |∆ ˆp (0)|.. i Un. v. Lemma 4.1.3. (Diagrammatic bounds on the expansion coefficients.) The expansion coefficients ∑ (n) (n) ˆ kπ |∆ ˆp (0)| ≡ x (1 − cos k · x)πp (x) obey the following bound. ˆ kπ |∆ ˆp(n) (0)| ≤. . ˆ p (k)m2 γ 2m−2 , Bp2 W p. . ˆ p (k)mγp2m−2 , ˆ p (k)m(m − 1)γp2m−3 + Bp W Bp2 W. if n = 2m + 1, if n = 2m.. To prove this lemma, we need the following lemma which can be proved by following the argument in [4].. 19 DOI:10.6814/NCCU202000775.
(27) Lemma 4.1.4. (the telescopic inequality) d d ∑ ∑ 1 − cos( tj ) ≤ d (1 − cos tj ). j=1. j=1. Proof of Lemma 4.1.4. First, take the real part of the telescopic identity 1−e. i. ∑d. j=1 tj. =. d ∑. (1 − eitj )ei. ∑j−1. h=1 th. .. (4.1.6). j=1. Then, using the inequalities respectively | sin. ∑j−1. h=1 th |. ≤. ∑j−1 h=1. | sin th |, | sin tj || sin th | ≤. (sin2 tj + sin2 th )/2 and sin2 tj ≤ 2(1 − cos tj ), and by (4.1.6), we have. ‧ 國. d ∑. tj −. d ∑ (1 − cos tj ). j=1. −. j=1. d ∑. (1 − cos tj )(1 − cos. Nat. =. j=1. −. sin tj sin 2. n. j−1. j=1 h=1. = = =. al. 2. ∑j−1. d−1 ∑. th ) +. d ∑. h=1. h=1 th. +. h=1. j−1 d ∑ ∑. sin tj sin. j−1 ∑. th. j=1. h=1. | sin tj sin th |. iv n C j−1 d ∑ d ∑ ∑ h e∑ (sin2itj U + sin2 th ) sin tj sin |th | ≤ n g c h j=1. ≤. 2. io. ≤. d ∑. j=1. h=1. ‧. 1 − cos. 立. 學. Then. j=1. y. j=1. 政 治 大. j−1 j−1 d ∑ ∑ ∑ th ) + th ). (1 − cos tj ) cos( sin tj sin(. sit. tj =. d ∑. er. 1 − cos. d ∑. 2. j=1 h=1. j=1 h=1. 2. 1 d−1 ∑ ∑ ) 1( 2 2 2 2 sin t2 + 2 sin t3 + ... + (d − 1) sin td + sin th + ... + sin2 th 2 h=1 h=1. d−1 (sin2 t1 + sin2 t2 + ... + sin2 td ) 2 d d ∑ d−1∑ 2 sin tj ≤ (d − 1) (1 − cos tj ). 2 j=1 j=1 □. Proof of Lemma 4.1.3 according to Lemma 4.1.4. First, we prove Lemma 4.1.3 for n = 2m+1. (1) (1) ˆ kπ Since πp (x) is proportional to δo,x and therefore ∆ ˆp (0) ≡ 0, we can assume m ≥ 1.. 20 DOI:10.6814/NCCU202000775.
(28) ∑ (2m+1) (2m+1) ˆ kπ To bound |∆ ˆp (0)| ≡ (x) for m ≥ 1, we first identify the x (1 − cos k · x)πp diagram vertices along the lowest diagram path from o to x, say y1 , ..., ym−1 , and then split x into {yj − yj−1 }m j=1 , where y0 = o and ym = x. For m = 1, ∑ ( ) 1 − cos k · (y1 − y0 ). ˆ kπ |∆ ˆp(3) (0)| =. y1. ∈Zd. y0 = o. .. y1. Then, by subadditivity, we obtain ˆ kπ |∆ ˆp(3) (0)|. ∑. ≤. y1. (1 − cos k · y1 )Gp (y1 ). ∈Zd. ( ˆ p (k) sup W. ≤. 政 治 大 y1. 立. For m = 2, we have. (1 − cos. y1 ,y2 ∈Zd. ∑. o. y1. k · (yj − yj−1 )). y0 = o. io. n. al. ≤. y. ∑. ((1 − cos k · y1 ) + (1 − cos k · (y2 − y1 ))). y1 ,y2 ∈Zd. ×(Gp (y1 ). Ch. .. sit. 2. y2. er. Nat ≤. ≤. y1. j=1,2. Then, using Lemma 4.1.4 and subadditivity, we obtain ˆ kπ |∆ ˆp(5) (0)|. ˆ p (k)Bp2 . ≤W. ‧. ‧ 國. ∑. ∈Zd. 學. ˆ kπ |∆ ˆp(5) (0)| =. ). y1. o. o y1. y2. engchi. ( ˆ p (k) sup 2W. y1 ∈Zd. o. y1. ˆ p (k)B 2 γ 2 , 4W p p. v. + Gp (y2 − y1 ). i Un. + sup y2 ∈Zd. y1. o. y2. o. ). ). y2. (4.1.7). where the last inequality holds by the same argument of the proof in Lemma 4.1.2. By induction, we have ˆ kπ |∆ ˆp(2m+1) (0)|. ≤. ˆ p (k) × (m diagrams, each bounded by Bp2 γp2m−2 ) mW. ≤. ˆ p (k)m2 γ 2m−2 . Bp2 W p. In the following, we prove Lemma 4.1.3 for n = 2m, and we follow the same lines as. 21 DOI:10.6814/NCCU202000775.
(29) ∑ (2m) (2m) ˆ kπ above for n = 2m + 1. To bound |∆ ˆp (0)| ≡ x (1 − cos k · x)πp (x), we first identify the diagram vertices along the lowest diagram path from o to x, say, y1 , ..., ym−1 , and then split x into {yj − yj−1 }m j=1 , where y0 = o and ym = x. For m = 1, ∑ ( ) 1 − cos k · (y1 − y0 ). ˆ kπ |∆ ˆp(2) | =. y1. y1. . y0 = 0. ∈Zd. Then, by subadditivity, we obtain. ˆ kπ |∆ ˆp(2) |. ∑ (. ≤. y1. ) 1 − cos k · (y1 − y0 ) Gp (y1 ). y0 = 0. y1 ∈Zd. ( 治 ) 政 ˆ (k) sup W 大 y1. ≤. p. 立 ≤. y1 ∈Zd. y0 = 0. ˆ p (k). Bp W. ‧ 國. 學. For m = 2, we have. (1 − cos. y1 ,y2 ∈Zd. ∑. k · (yj − yj−1 )). ‧. y2. y0 = o. n ˆ kπ |∆ ˆp(4) (0)|. ≤. sit. io. Then, by using the same argument of (4.1.7), we obtain. al. .. y1. y. j=1,2. er. ∑. Nat. ˆ kπ |∆ ˆp(4) (0)| =. iv ∑ ( ) n C 2 h (1 − cos k ·U y1 ) + (1 − cos k · (y2 − y1 )) engchi y ,y ∈Z 1. 2. d. ( × Gp (y1 ). y2 o. + Gp (y2 − y1 ). y1. ≤. ( ˆ p (k) 2W. ≤. ˆ p (k)(Bp2 γp 2W. ). + o. y2. ). o y1. o. + Bp γp2 ).. 22 DOI:10.6814/NCCU202000775.
(30) By induction, we have (. ˆ kπ |∆ ˆp(2m) (0)|. ( ) (m − 1) diagrams, each bounded by Bp2 γp2m−3 ) 2m−2 +(1 diagram, bounded by Bp γp ). ≤. ˆ p (k) × mW. ≤. ˆ p (k)m(m − 1)γp2m−3 + Bp W ˆ p (k)mγp2m−2 . Bp2 W □ (n). From Lemma 4.1.2 and Lemma 4.1.3, we can simply get that if γp < 1 then π ˆp (0) and (n). ˆ kπ |∆ ˆp | approaches zero as n goes to infinity. Hence, we need to suppose γp < 1 to achieve our purpose.. 政 治 大. Lemma 4.1.5. Suppose that γp ≡ p||D||∞ + Lp + Bp < 1. We have. 立. γp ˆ odd 0≤Π , p (0) ≤ Lp + Bp (p||D||∞ + Lp ) 1 − γp2. ˆ even 0≤Π (0) ≤ Bp (p||D||∞ + Lp ) p. ‧. ‧ 國. 學. (4.1.8). (4.1.9). er. io. sit. y. Nat. 1 , 1 − γp2. ˆ kΠ ˆ odd ˆp |∆ Bp2 (1 + γp2 ) W p (0)| || sup ≤ || , ˆ ˆ ∞ (1 − γp2 )3 1 − D 1 − D(k) k. n. al. sup k. Ch. engchi. i Un. v. (4.1.10). ˆ kΠ ˆ even ˆp ˆp |∆ (0)| W 2γp 1 W p ≤ Bp2 || ||∞ ||∞ + Bp || . 2 3 ˆ ˆ ˆ (1 − γp ) (1 − γp2 )2 1 − D(k) 1−D 1−D. (4.1.11). Proof. Note that (4.1.8) and (4.1.9) are very easy to get from Lemma 4.1.2. Hence, we explain (4.1.10) and (4.1.11) as follows. By lemma 4.1.3, we obtain ∞ ˆ kΠ ˆ odd (0)| ∑ ˆp |∆ W p 2 sup ≤ Bp || ||∞ m2 γp2m−2 , ˆ ˆ 1 − D(k) 1−D k m=0. 23 DOI:10.6814/NCCU202000775.
(31) where. ∑∞ m=0. m2 γp2m−2 , and we can consider ∞ ∑. γp2m =. m=0. 1 . 1 − γp2. Then, we have ∞ ∑. m2 γp2m−2. m=1. 1 + γp2 . = (1 − γp2 )3. Therefore, we get (4.1.10) as follows. sup k. ˆ kΠ ˆ odd ˆp Bp2 (1 + γp2 ) W |∆ p (0)| ≤ || || . ˆ ˆ ∞ (1 − γp2 )3 1 − D 1 − D(k). 政 治 大. Similarly, by lemma 4.1.3, we obtain. 立. ∞ ∑. m(m − 1)γp2m−3 + Bp ||. γp2 , 1−γp2. we have. al. n. ∞ ∑. ˆ 1−D. ||∞. ∞ ∑. mγp2m−2 .. m=1. y. m=1. ˆp W. sit. ‧ 國 γp2m =. io. m=1. ˆ 1−D. ||∞. er. ∑∞. ˆp W. Nat. Since. Bp2 ||. ‧. ≤. 學. ˆ kΠ ˆ even (0)| |∆ p sup ˆ 1 − D(k) k. Ch. m(m − 1)γp2m−3 =. m=1. i Un. 2γp , (1 − γp2 )3. engchi. v. and ∞ ∑ m=1. mγp2m−2 =. 1 . (1 − γp2 )2. Hence, we can obtain (4.1.11) as follows. sup k. ˆ kΠ ˆ even ˆp ˆp (0)| |∆ W 2γp 1 W p ≤ Bp2 || ||∞ || + B || . ∞ p ˆ ˆ ˆ (1 − γp2 )3 (1 − γp2 )2 1 − D(k) 1−D 1−D. 24 DOI:10.6814/NCCU202000775.
(32) 4.2 Diagramatic bounds on the bootstrapping argument In Section 4.1, we have already estimated the coefficients of lace expansion. In this section, we use them to estimate the upper bounds of gi (p) for i = 1, 2, 3. For g1 (p), by (4.1.2), we can easily get ˆ p (0). g1 (p) ≤ 1 − Π. (4.2.1). ˆ p (k) (i.e. Jˆp (k) = Jˆp (−k)), we obtain And then, by (4.1.3) and using the symmetry of G ˆ k Jˆp (0) = Jˆp (0) − Jˆp (k). Since p ≥ 1, we have −∆ 1 1 = ˆ k Jˆp (0) −∆ Jˆp (0) − Jˆp (k) 1 ˆ ˆ p (0) − pD(k) ˆ ˆ p (k) pD(0) +Π −Π. ≤. ˆ p (k) G. 政 治 大. =. 立. ‧ 國. ] ˆ 1 − D(k) sup ˆ ˆ kΠ ˆ p (0)) + (−∆ k∈[−π,π]d p(1 − D(k)) [ ] ˆ 1 − D(k) sup . ˆ ˆ kΠ ˆ p (0)) + (−∆ k∈[−π,π]d (1 − D(k)). sit. y. ≤. (p≥1). n. al. er. io. ≤. Then, we have. Ch. engchi. [ g2 (p) ≤. sup k∈[−π,π]d. Suppose −. ‧. [. Nat. g2 (p). 學. Therefore. 1 . ˆ ˆ kΠ ˆ p (0)) p(1 − D(k)) + (−∆. =. ˆ kΠ ˆ p (0) −∆ 1+ ˆ 1 − D(k). i Un. v. ]−1 .. (4.2.2). ∑∞ ˆ (n) ∑∞ ∑ (n) ˆ ˆp (0) = n=0 n=0 ∆k π x (1 − cos(k · x))πp (x) is smaller than 1 − D(k).. Then, we can ensure the g2 (p) to be uniformly bounded. To evaluate g3 (p), we need the following proposition which can be found in ( [12], Lemma 5.7 ). ˆ Proposition 4.2.1. ( [12], Lemma 5.7 ) Suppose A(k) = (1 − a ˆ(k))−1 , where a ˆ is the Fourier transform of a symmetric a(x) = a(−x) for all x ∈ Zd , then we have the following identity. 25 DOI:10.6814/NCCU202000775.
(33) ˆ |∆k A(l)|. 1 ˆ ˆ + k)]A(l)[ˆ ˆ aav (0) − a [A(l − k) + A(l ˆav (k)] 2 ˆ − k)A(l) ˆ A(l ˆ + k)[ˆ +4A(l aav (0) − a ˆav (k)][ˆ aav (0) − a ˆav (l)],. ≤. where aav (x) = |a(x)|. ˆ p (k) = (1 − Jˆp (k))−1 Then, we use Proposition 4.2.1, and substitute A to Gp , noting that G ,we obtain the upper bound of g3 (p) as follows g3 (p). ≡. sup k,l. 1. ˆ kG ˆ p (l)|, × |∆. ˆ p (k) = G. where. Uˆ (k, l) {[ ˆ ˆ ˆ p (l − k) ] Gp (l + k) + G 1 − D(k). 1 1 − Jˆp (k). ˆ ˆ ˆ p (l) |∆k Jp (l)| G ˆ 2 Uˆ (k, l) 1 − D(k) k,l } ˆ ˆ ˆ ˆ ˆ p (l + k)G ˆ p (l − k) −∆l |Jp |(0) −∆k |Jp |(0) , (4.2.3) +4G ˆ 1 − Jˆp (l) 1 − D(k). ≤. sup. 政 治 大. 學. ‧ 國. 立. ∑ where |Jˆp |(k) = x∈Zd eik·x |Jˆp (x)|.. ‧. Applying these bounds in (4.2.1), (4.2.2), and (4.2.3), we obtain the following bounds on the bootstrapping functions gi (p) for i = 1, 2, 3.. sit. y. Nat. io ≤. g2 (p). ≤. g3 (p). ≤. al. n. g1 (p). er. Lemma 4.2.2. p ∈ [1, pc ), we have. v. Bp (p||D||∞ + Lp )γp , 1 − γp2 ˆp Bp2 (1 + γp2 ) W (1 − || ||∞ )−1 , 2 3 ˆ (1 − γp ) 1−D ( ˆp W 2γp max{g2 (p), 1}3 p + Bp2 || ||∞ ˆ (1 − γp2 )3 1−D )2 ˆp ˆp Bp2 (1 + γp2 ) W 1 W || || . +Bp || + || ˆ ∞ (1 − γp2 )2 ˆ ∞ (1 − γp2 )3 1 − D 1−D 1 + Lp +. Ch. engchi. i Un. Proof. Since ˆ odd ˆ p (0) = Π ˆ even (0) − Π Π p (0), p. 26 DOI:10.6814/NCCU202000775.
(34) and ˆ kΠ ˆ p (0) = ∆ ˆ kΠ ˆ even (0) − ∆ ˆ kΠ ˆ odd (0). ∆ p p We obtain (4.2.1). ≤. g1 (p). ˆ p (0) ≤ 1 + Π ˆ odd (0) 1−Π p. (4.1.8). ≤. 1 + Lp + Bp (p||D||∞ + Lp ). γp , 1 − γp2. and ˆ kΠ ˆ odd ˆ kΠ ˆ p (0) |∆ −∆ p (0)| −1 −1 sup(1 + ) ≤ sup(1 − ) ˆ ˆ 1−D 1−D k k ˆp Bp2 (1 + γp2 ) W (1 − ||∞ )−1 . || 2 3 ˆ (1 − γp ) 1−D. (4.2.2). ≤. (4.1.10). 立. ≤. ≤. sup k,l. ˆ ˆ p (l + k) + G ˆ p (l − k) ˆ ˆ 1 − D(k) G ˆ p (l) |∆k Jp (l)| {[ ]G ˆ 2 Uˆ (k, l) 1 − D(k). y. Nat. ˆ ˆ ˆ ˆ ˆ p (l + k)G ˆ p (l − k) −∆l |Jp |(0) −∆k |Jp |(0) } +4G ˆ 1 − Jˆp (l) 1 − D(k). io. sit. g3 (p). ‧. ‧ 國. 學. For g3 (p),. 政 治 大. al. er. g2 (p). v. n. ˆ p (k) ≡ 1/(1 − Jˆp (k)) and |G ˆ p (k)| ≤ g2 (p)Sˆ1 (k) ≡ g2 (p)/(1 − D(k)), ˆ since G we obtain g3 (p). ≤. (2.2.5). ≤. Ch. engchi. i Un. { ˆ ˆ ˆ k Jˆp (0)| ( S1 (l + k) + Sˆ1 (l − k) ) 1 − D(k) |∆ Sˆ1 (l)g2 (p)2 sup ˆ 2 Uˆ (k, l) 1 − D(k) k,l } ˆ l |Jˆp |(0) −∆ ˆ k |Jˆp |(0) −∆ ) +4Sˆ1 (l + k)Sˆ1 (l − k)g2 (p)3 ˆ ˆ 1 − D(l) 1 − D(k) { ˆ k Jˆp (l)| ( ˆ k |Jˆp |(0) )2 } |∆ −∆ 3 max{g2 (p), 1} max sup , sup . ˆ ˆ 1 − D(k) k,l 1 − D(k) k. 27 DOI:10.6814/NCCU202000775.
(35) Since Jp = pD + Πp , we have ˆ k Jˆp (l)| |∆ ˆ 1 − D(k). ∑ 1 | (1 − cos k · x)eil·x (pD(x) + Πp (x))| ˆ 1 − D(k) x ∑ 1 even (1 − cos k · x)(pD(x) + Πodd (x)) p (x) + Πp ˆ 1 − D(k). = ≤. x. ≤. p+. ˆ kΠ ˆ even (0)| |∆ p ˆ 1 − D(k). ˆ kΠ ˆ odd (0)| |∆ p + , ˆ 1 − D(k). ˆ k |Jˆp |(0)/(1 − D(k)) ˆ which is larger than 1, since p ≥ 1. It is easy to check that −∆ obeys the same bound. Therefore, by using (4.1.10) - (4.1.11), we obtain. g3 (p). ˆ kΠ ˆ odd ˆ kΠ ˆ even |∆ (0)| |∆ p p (0)| 2 + ) ˆ ˆ 1 − D(k) 1 − D(k) ( ˆp 2γp W 3 ||∞ max{g2 (p), 1} · p + Bp2 || ˆ (1 − γp2 )3 1−D )2 ˆp ˆp Bp2 (1 + γp2 ) W 1 W +Bp || || + || || , ˆ ∞ (1 − γp2 )2 ˆ ∞ (1 − γp2 )3 1 − D 1−D. ≤. max{g2 (p), 1}3 (p +. 立. ≤. ‧. ‧ 國. 學. as required.. 政 治 大. y. Nat. Therefore, in order to prove the main result of our thesis, we also need to evaluate the upper. io. n. al. er. ˆ. sit. Wp bound of Lp , Bp , || 1− ˆ ||∞ , and γp for p ∈ [1, pc ) in the next chapter. D. Ch. engchi. i Un. v. 28 DOI:10.6814/NCCU202000775.
(36) Chapter 5 Random-walk estimate In Section 5.1, we use the inequlity G1 (x) ≤ S1 (x) to evaluate the diagrams (i.e.. 政 治 大. Wp Lp , Bp , || 1− ˆ ||∞ ) at p = 1. In Section 5.2, we use bootstrapping argument to deal with the D ˆ. 立. diamgrams for p ∈ (1, pc ).. ‧ 國. 學. 5.1 The diagrams bound of random-walk quantities for p = 1. ‧. W1 To estimate the upper bound of L1 , B1 , || 1− ˆ ||∞ , we can use two RW quantities, such as D ˆ. Nat. al. ≡. (D ∗. S1∗1 )(0). =. sit. ∞ ∑. er. ≡. n. ε2. io. ε1. ∗2. y. the RW loop ε1 , the RW bubble ε2 , defined as. D∗n (0),. n=2 iv n ∞ C h∗2 ∗2 ∑ (D e∗nS1g)(0) c h=i U D∗n(0)(n − 1).. (5.1.1). n=2. ∑∞. Since the symmetry Poisson distribution is long-range, we can easy to get (D∗2 ∗S1∗1 )(0) =. n=2. D∗n (0). Similarly, the analysis of (D∗2 ∗ S1∗2 )(0) can be written as follows ∗2. (D ∗. S1∗2 )(0). =. ∞ ∑ ∞ ∑. ∗n. D (x)D. ∗m. (−x) =. m=1 n=1. = =. ∞ ∑ ∞ ∑. D∗(n+m) (0). m=1 n=1. ∞ ∑ ∞ ∑ ∞ ∑ u=2 m=1 n=1 ∞ ∑. D∗u (0)I{n + m = u}. (u − 1)D∗u (0) =. u=2. ∞ ∑ (n − 1)D∗n (0). n=2. 29 DOI:10.6814/NCCU202000775.
(37) And then, we evaluate the diagrams for p ∈ [1, pc ). First, by the Fourier transform of the −1 ˆ RW Green function Sˆ1 (k) ≡ (1 − D(k)) , we have. ≡. ε1. ∫. ∗2. (D ∗ S1 )(0) = [−π,π]d. ˆ 2 dd k D(k) , ˆ (2π)d 1 − D(k). and ∗2. ≡. ε2. S1∗2 )(0). (D ∗. ∫ = [−π,π]d. ˆ 2 D(k) dd k . 2 (2π)d ˆ (1 − D(k)). Hence we can estimate the upper bound as follows. [−π,π]d. 政 治 大. 立 ˆ |D(k)| ≤. j=1. ∫. ≤. io. n. =. [−π,π]d. a1dl. π. ∫ e 2 dd k e 2 dd k D(k) d D(k) = 2 e e (2π)d (2π)d 1 − D(k) 1 − D(k) [0,π]d. y. Nat. ε1. e e−λ+λ cos(kj ) ≡ D(k).. ‧. We get. d ∏. 學. ‧ 國. Since. ∫. e 2 D(k) dd k. e 1 − D(k). Ch. [0,π]d. As above, we have 1 ε1 ≤ d π. sit. ε1 =. ∫ 2 ˆ 2 dd k ˆ |D(k)| dd k D(k) ≤ , ˆ ˆ (2π)d (2π)d 1 − D(k) [−π,π]d 1 − |D(k)|. engchi. ∫ [0,π]d. er. ∫. i Un. v. e 2 D(k) dd k. e 1 − D(k). (5.1.2). Then, we do the analysis and calculation for this formula as follows. First, we divide the scope into three parts, namely I1 , I2 and I3 , 1 1 1 I1 = {k : ||k||2 ∈ [0, √ ]}, and I2 = {k : ||k||2 ∈ [ √ , λ− 3 ]}, λ λ. 30 DOI:10.6814/NCCU202000775.
(38) and { } 1 I3 = k : ||k||2 ≥ λ− 3 , and |kj | ≤ π for all 1 ≤ j ≤ d . So we can divide the (5.1.2) into three parts. i.e. ≤. ε1. =. 1 πd 1 πd |. ∫ [0,π]d. ∫. I1. e 2 D(k) dd k e 1 − D(k). ∫ ∫ e 2 e 2 e 2 D(k) D(k) D(k) 1 1 d d d k+ d d k+ d dd k . e e e π π 1 − D(k) I2 1 − D(k) I3 1 − D(k) {z } | {z } | {z } (1). (2). (3). 政 治 大. First, we evaluate (1) as follow. Since I1 = {k : ||k||2 ∈ [0, √1λ ]} and cos(kj ) ≤ 1 − ≤1−. 2. (1 −. 1 ), 12λ. we have. 立. e(− 2 kj (λ− 12 )) = e(− 2 ||k|| 1 2. 1. 1. y. Nat. I1. e 2 D(k) dd k e 1 − D(k). n. ∫. al =. Ch. e∫n gD(k) ec h2i. 1 πd. I1. ∫. ≤ =. 1 π d 12 (λ −. =. i Un. v. 1 d k= d e π 1 − D(k). 1 πd. ≤. sit. io. 1 πd. .. j=1. 1 1 1 3 e ]. 1 − D(k) ≥ ||k||22 (λ − )[ + 2 12 4 48λ Therefore. 2 (λ− 1 )) 12. ‧. Then. ≤. d ∏. 學. e D(k). +. er. 4!. kj2. ‧ 國. kj4. kj2 2. ∫. d. ||k||2 ∈[0, √1 ] λ. 1. ||k||2 ∈[0, √1 ] λ. 1 ||k||22 (λ 2. 1 1 )( 3 + 12 4. 1 σd 1 1 d π 2 (λ − 12 )( 34 + 1 σd 1 1 d π 2 (λ − 12 )( 34 +. ∫. 1 ) 48λ. −. 1 )[ 3 12 4. +. ||k||2 ∈[0, √1 ]. ∫. λ. √1 λ. r. 1 d ] 48λ. d. k. 1 d d k ||k||22. d−1. dr r2 1 1 ( √ )d−2 1 )d−2 λ 48λ 1 ) 48λ. e 2 D(k) dd k e 1 − D(k). 0. , d > 2,. 31 DOI:10.6814/NCCU202000775.
(39) where σd is the surface area of the d-dimensional unit ball. To evaluate (2) as follow. Since I2 = {k : ||k||2 ∈ [ √1λ , λ− 3 ]} and cos(kj ) ≤ 1− 1. 1−. kj2 (1 2. −. kj2 ) 12. kj2 (1 2. ≤1−. −. 1 ), 12λ2/3. kj2 2. k4. + 4!j =. we have. (− (1− (− (1− )||k||2 ) (− (1− ) ) )) e D(k) ≤ e 2 12λ2/3 ≤ e 2 12λ2/3 λ = e 2 12λ2/3 . λ. 1. λ. 2. 1. 1. 1. 1. Therefore 1 πd. ∫ I2. e 2 D(k) dd k e 1 − D(k). =. 1 πd. ∫ I2. ∫ e 2 D(k) 1 d d k= d e π ||k||2 ∈[ √1 1 − D(k) −(1−. ≤. 立. ). ∫. d. 1 ) 12λ2/3. (− 12 (1−. ,λ− 3 ]. 1. λ− 3. rd−1 dr 0. 政e 治 σ 大d1 ( λ1 ) ] π [1 − e −(1−. =. 1. 12λ2/3 1 e σd 1 π d 1 − e− 2 (1− 12λ12/3 ). 1. λ. e 2 D(k) dd k e 1 − D(k). d. 1 )) 12λ2/3. d. 1 3. , d ∈ N.. 2. 2. 2. ‧. ‧ 國. 學. { 1 Finally, we evaluate (3) as follow. Since I3 = k : ||k||2 ≥ λ− 3 and |kj | ≤ π, for all 1 ≤ } 2 j ≤ d and cos(kj ) ≤ 1 − 21 kj2 (1 − π12 ), we have 1. 2. − 1 π λ π λ π 2 e D(k) ≤ e(− 2 (1− 12 )||k||2 ) ≤ e(− 2 (1− 12 )λ 3 ) = e(− 2 λ 3 (1− 12 )) .. = ≤ ≤ ≤. ∫. e 2 a lD(k) iv dd k n e C I 1 − D(k) h e n g c h∫ i U ∫. n. 1 πd. er. io. sit. y. Nat. Therefore. 3. e 2 e 2 1 D(k) 1 D(k) d d k = dd k e e π d I3 1 − D(k) π d I3 1 − D(k) ∫ 2 1 1 (−λ(1− π12 )||k||22 ) d e d k 1 π d 1 − e(− 12 (λ 3 (1− π122 ))) I3 ∫ √dπ2 2 σd −λ(1− π12 )r2 d−1 e r dr 1 2 −1 − 21 λ 3 (1− π12 ) d 3 λ π (1 − e ) [√ ( d − 1 ) d−1 1] σd − d−1 2 2 − λ− 3 , 2 e dπ 2 1 2 π 1 2λ(1 − π12 ) π d (1 − e− 2 λ 3 (1− 12 ) ). where the last inequality can be explained in the following inequaily.. 32 DOI:10.6814/NCCU202000775.
(40) π2. e−λ(1− 12 )r rd−1 ≤ e− 2. d−1 2. (. d−1 d−1 ) 2 . π2 2λ(1 − 12 ). (5.1.3). π2. The proof of (5.1.3) is easy, if we let f (r) = e−λ(1− 12 ) rd−1 , using the first order derivative √ d−1 d−1 d−1 law, we can get, when r = , the maximum value of f (r) is e− 2 ( d−1π2 ) 2 . π2 2λ(1− 12 ). 2λ(1− 12 ). As the above estimate, we obtain ≤. ε1. ∫. =. [−π,π]d. ∫ e 2 ˆ 2 1 D(k) D(k) dd k ≤ dd k. 2 (2π)d 2 ˆ e π d [0,π]d (1 − D(k)) (1 − D(k)). n. al. er. sit. ∫ ∫ e 2 e 2 e 2 1 1 D(k) D(k) D(k) d d d k+ d d k+ d dd k . 2 2 2 e e e π π (1 − D(k)) I2 (1 − D(k)) I3 (1 − D(k)) {z } | {z } | {z }. io. I1. y. Nat. ∫. (5.1.5). ‧. ‧ 國. S1∗2 )(0). (5.1.4). We can also divide the (5.1.5) into three parts. i.e. 1 ε2 = d π |. ). 學. ε2 ≡ (D ∗. 1. 政 治 大. 立. Similarly, ∗2. −(1−. 12λ2/3 σ e 1 1 1 d 1 d−2 d √ ) + ( ( 1) 1 1 1 (− (1− )) )d−2 λ π d [1 − e 2 12λ2/3 ] d λ 3 48λ d−1 [√ 1] σd d−1 − d−1 2 − λ− 3 , d > 2. 2 ( 2 + e ) dπ 1 2 2 1 π 2λ(1 − π12 ) π d (1 − e− 2 λ 3 (1− 12 ) ). 1 σd 1 1 d π 2 (λ − 12 )( 43 +. (1′ ). Ch. (2′ ). engchi. i Un. v. (3′ ). And then, using the same arguments of estimating ϵ1 , we get. ε2. ≤. −(1−. 1. ). 12λ2/3 σ 1 1 d−4 e 1 1 d d √ ( ( 1/3 ) ) + 1 1 1 2 − (1− ) ] d−4 λ π d [1 − e 2 12λ2/3 ]2 d λ 48λ [√ ( d − 1 ) d−1 1] σd − d−1 2 2 + e dπ 2 − λ− 3 , d > 4. 2 2 π π 1 1 2λ(1 − 12 ) π d (1 − e− 2 λ 3 (1− 12 ) )2 (5.1.6). 4σd 1 2 3 d π (λ − 12 ) [ 4 +. Note that σd =. 2π d/2 , Γ( d2 ). we can estimate the upper bound of ϵ1 for d > 2 and the upper bound. of ϵ2 for d > 4, which are very important for estimating bootstrapping argument.. 33 DOI:10.6814/NCCU202000775.
(41) Proposition 5.1.1. Let d ≥ 5 and p = 1. For all λ > 0, we have L1 ≤ ϵ1 , B1 ≤ ϵ2 , and ||. ˆ1 W ˆ 1−D. ||∞ ≤ 5(1 + 2ϵ1 + ϵ2 ).. Proof. For p = 1, using the inequality G1 (x) ≤ S1 (x), x ∈ Zd obtain the upper bound of L1 . ∫. ∗2. L1 ≤ ||D ∗ S1 ||∞. = [−π,π]d ∗2. ˆ 2 dd k D(k) ˆ (2π)d 1 − D(k). (D ∗ S1 )(0) ≡ ϵ1 .. = Similarily, we have ∗2. S1∗2 ||∞. B1 ≤ ||D ∗. 立. ˆ 2 D(k) dd k 2 (2π)d ˆ (1 − D(k)). 政 治 大 = (D ∗ S )(0) ≡ ϵ . =. [−π,π]d ∗2. ∗2 1. 2. ‧ 國. 學 ∫. io. sit. Nat. ≤. y. ≤. ‧. (1 − cos k · x)G1 (x). d il·x d l ˆ ˆ (−∆k S1 (l))e (1 − cos k · x)S1 (x) = (2π)d [−π,π]d ∫ dd l . Uˆ (k, l) (2π)d [−π,π]d. n. al. er. Finally, since. ∫. Ch. i Un. v. Then by Lemma 4.2.1 and using the Schwarz inequality, we have ∫. d. d l Uˆ (k, l) (2π)d [−π,π]d. engchi. ∫. dd l Sˆ1 (l)2 (2π)d [−π,π]d. ≤. ˆ 5(1 − D(k)). =. ∗2 ˆ 5(1 − D(k))S 1 (0),. by (5.1.1), we have S1∗2 (0). = =. ∞ ∑ ∞ ∑. D∗n (x)D∗m (−x) =. m=0 n=0 ∞ ∑. ∞ ∑ ∞ ∑ ∞ ∑. D∗u (0)I[n + m = u]. u=0 m=0 n=0. (u + 1)D∗u (0). u=0. =. 1 + 2D(0) + 2ϵ1 + ϵ2 = 1 + 0 + 2ϵ1 + ϵ2 .. 34 DOI:10.6814/NCCU202000775.
(42) Therefore ∫. dd l ˆ ˆ U (k, l) ≤ 5(1 − D(k))(1 + 2ϵ1 + ϵ2 ). (2π)d [−π,π]d. Hence, we have ||. ˆ1 W ˆ 1−D. ||∞ ≤ 5(1 + 2ϵ1 + ϵ2 ).. 5.2 The diagrams bound of random-walk quantities for p > 1. 政 治 大. Wp First, we estimate the upper bound of Lp , Bp , || 1− ˆ ||∞ for p ∈ [1, pc ) under the D ˆ. bootstrapping assumptions.. 立. ‧ 國. 學. Proposition 5.2.1. Let d ≥ 5 and p ∈ (1, pc ) and suppose that gi (p) ≤ Ki , i = 1, 2, 3 for some contants {Ki }3i=1 . For all λ > 0, we have. =. ||(pD). ||∞ ≤ 5K3 (1 + 2ϵ1 + ϵ2 ).. y. ˆ 1−D. sit. al. n. Lp. ˆp W. ∗2. er. io. Proof.. ‧. Nat. Lp ≤ K12 K2 ϵ1 , Bp ≤ K12 K22 ϵ2 , and ||. i v dd k 2n ˆ ˆ C ∗ Gp ||h D(k) |Gp (k)| ∞ ≤ p (2π)d e n g[−π,π] chi U ∫ ∫. 2. d. ∫. ˆ d k D(k) ≤ K12 K2 g2 (p) d ˆ (2π) 1 − D(k) d. 2. ≤. p. =. [−π,π]d 2 K1 K2 ϵ1 ,. 2. [−π,π]d. ˆ 2 dd k D(k) ˆ (2π)d 1 − D(k). and Bp. =. ∗2. ||(pD) ∗ ∫. ≤. p. 2 [−π,π]d. G∗2 p ||∞. ∫ ≤p. 2. d ˆ 2G ˆ p (k)2 d k D(k) (2π)d [−π,π]d. ˆ 2 D(k) dd k g2 (p)2 ≤ K12 K22 ϵ2 . d 2 ˆ (2π) (1 − D(k)). 35 DOI:10.6814/NCCU202000775.
(43) Finally, since g3 (p) ≤ K3 , we obtain ∫ 0 ≤ (1 − cos(k · x))Gp (x). = ≤. d il·x d l ˆ ˆ (−∆k Gp (l))e (2π)d [−π,π]d ∫ dd l , K3 Uˆ (k, l) (2π)d [−π,π]d. uniformly in x and k. Then, by the similar of analysis for p = 1, we have ||. ˆp W ˆ 1−D. ||∞ ≤ 5K3 (1 + 2ϵ1 + ϵ2 ).. Finally, we need to estimate the upper bounded of ||D(x)||∞ as follows.. 政 治 大. Lemma 5.2.2. For any λ ≥ 1,. ||D(x)||∞ ≤. d ∏ e−λ. io. al. n. Then ||D(x)||∞. sit. j=1. λ|xj |−1 , if x1 x2 · · · xd ̸= 0. 2 (|xj | − 1)!. er. Nat. D(x) =. e−dλ = d 2. y. ‧. Proof. By (2.1.1). 1 √ . d (2 2π) (⌊λ − 1⌋)d/2. 學. ‧ 國. 立. Ch. max x=(x1 ,x2 ,...,xd )∈Zd. x1 x2 ···xd ̸=0. d (∏. e λn|xg|−1c h) i. i Un. v. d e−dλ ∏ λ|xj | = d ( max ). (|xj | − 1)! 2 j=1 xj ∈Z\{0} |xj |! j=1 j. Using the Stirling’s formula n! =. √. n 1 1 2πn( )n ea , where <a< , e 12n + 1 12n. 36 DOI:10.6814/NCCU202000775.
(44) we have the lower bound of n! is λ|xj | |xj |!. √. 1. 2πn( ne )n e 12n+1 . Therefore 1 1 −x − λ|xj | √ e−xj xj j e 12xj +1 2πxj 1 λe − 1 −1 √ ( )xj e 12xj +1 xj 2 , x ≥ 1 2π xj 1 xj [ln(λe)−ln(xj )] − 21 √ e xj 2π 1 √ f (xj ), 2π. ≤ = ≤ =. where f (t) = et[ln(λe)−ln(t)] t− 2 . 1. 政 治 大 Using the simple analysis of calculus, there is t such that f (t ) = 0, hence f (t ) is 立 maximum value, and t satisfies the identity ln( ) = . However, it not easy to get t exactly, ′. c. ‧ 國. 1 2tc. c. c. 學. λ tc. c. c. so we just evaluate the approximation of tc . Let h(t) = ln( λt ) −. 1 . 2t. Clearly, h(tc ) = 0 and we. want to find the interval of the root of h(t) as follows. If t = λ − 12 , we have. ‧. ∑1 1 λ 1 1 λ 1 ) = ( )k = + 2 + ... ln( ) = ln( ) = − ln(1 − 1 t 2λ k 2λ 2λ 8λ λ− 2 k=1 ∞. n. al. er. io. sit. y. Nat. and. Ch. i Un. v. 1 1 1 1 1 ∑ 1 k 1 1 = = [ ] = ( ) = + 2 + ... 1 2t 2λ − 1 2λ 1 − 2λ 2λ k=0 2λ 2λ 4λ ∞. engchi. We obtain 1 h(λ − ) < 0, and h(t) is decreasing if t > 1, 2. 37 DOI:10.6814/NCCU202000775.
(45) and h(λ − 1) λ 1 1 1 1 ln( )− = − ln(1 − ) − [ ] λ−1 2(λ − 1) λ 2λ 1 − λ1 ∞ ∞ ∑ 1 k1 1 ∑ 1 k ( ) ( ) − λ k 2λ k=0 λ k=1 ) 1 1( 1 1 1 1 1 1 + 3 ( + + 2 + ...) − ( + + 2 + ...) 2λ λ 3 4λ 5λ 2 2λ 2λ ∞ 1 1 ∑ 1 1 1 n ( + − )( ) 2λ λ3 n=0 n + 3 2 λ √ 1 λ2 − λ − 1 1 1 −1 1 1+ 5 + ( )= > 0, if λ > ≈ 1.6 2λ λ3 2 1 − λ1 2 λ2 (λ − 1) 2. = = = = >. 政 治 大. By Intermediate value theorem, there is tc ∈ (λ − 1, λ − 21 ) such that h(tc ) = 0. Hence. 立. f (tc ) is a maximum value of f (t) for t > 1. Since xj ∈ Z,. ‧ 國. xj ∈Z. λ|xj | |xj |!. 學. max. 1 max{ √ f (t)} t∈Z 2π 1 1 1 √ max{f (⌊λ − ⌋), f (⌈λ − ⌉), f (⌊λ − 1⌋)}, 2 2 2π. ≤. ‧. ≤. y. Nat. er. io. sit. for j = 1, 2, ..., d. Then, we evaluate f (⌊λ − 21 ⌋) and f (⌊λ − 1⌋) and f (⌈λ − 12 ⌉) as follows 1 1 1 1 . f (⌊λ − ⌋) = e(⌊λ− 2 ⌋)[ln(λe)−ln(λ− 2 )] √ 2 ⌊λ − 12 ⌋. n. al. Ch. engchi. i Un. v. Since 1 ln(λe) − ln(λ − ) 2. ∞ ∑ λ 1 1 1 ) = 1 + ln( ) = 1 + ( )k 1 1 2λ k λ− 2 1 − 2λ k=1. =. 1 + ln(. ≤. 1 2 ) 1 1∑ 1 k 1 1 ( 2λ 1+ + + ( ) =1+ 1 , 2λ 2 k=2 2λ 2λ 2 1 − 2λ ∞. 38 DOI:10.6814/NCCU202000775.
(46) so we have 1 f (⌊λ − ⌋) 2. 1 (⌊λ− 21 ⌋)(1+ 2λ +. 1 1 ) 2− 1 4λ2 λ. √. ≤. e. =. exp(λ −. 1 1 [1 − 4λ 2−. 1 λ. ≤. exp(λ −. 1 1 [1 − 4λ 2−. 1 λ. 1 ⌊λ − 21 ⌋. +. 1 1 ]) √ . 4λ − 2 ⌊λ − 12 ⌋. +. 1 1 ]) √ . 4λ − 2 ⌈λ − 12 ⌉. Similarily, 1 f (⌈λ − ⌉) 2 and f (⌊λ − 1⌋). ≤. 立. 政 1治 1大 1 ]) √ exp(λ − [1 − + λ 2λ − 2 2− 2 λ. 1 1 1 1 √ max{ √ ,√ ,√ } 2π ⌊λ − 1⌋ ⌊λ − 12 ⌋ ⌈λ − 21 ⌉. ≤. 1 1 √ √ . 2π ⌊λ − 1⌋. io. =. C∏ hd j=1. =. y. iv n d U ∏ e λ|xi | e−λ h n c g max( )) ≤ (. n. al. er. Nat ||D(x)||∞. ≤. sit. ‧ 國. λ|xj | max( ) xj ∈Z |xj |!. Therefore. .. ‧. e. ⌊λ − 1⌋. 學. As above, we obtain −λ. 1. j. |xj |!. 2. 1 1 √ √ 2 2π ⌊λ − 1⌋ j=1. 1 √ . (2 2π)d (⌊λ − 1⌋)d/2. 39 DOI:10.6814/NCCU202000775.
(47) Chapter 6 Proof of Proposition 2.2.7 - 2.2.9 In this chapter, we use the same argument of [4].. 政 治 大. 立 2.2.7 6.1 Proof of Proposition. ‧ 國. 學. First, we prove Proposition 2.2.7, and we have. al. er. io. sit. y. Nat. where. ˆ ˆ ˆ G ˆ p ||∞ , and g3 (p) = sup |∆k Gp (l)| , g2 (p) = ||(1 − D) Uˆ (k, l) k,l. ‧. g1 (p) = p,. v. n. 1 ˆ ˆ Uˆ (k, l) ≡ (1 − D(k))( (S1 (l + k) + Sˆ1 (l − k))Sˆ1 (l) + 4Sˆ1 (l + k)Sˆ1 (l − k)). 2. Ch. engchi. i Un. Obviously, g1 (p) = p is continuous. In order to prove the continuity of g2 (p) and g3 (p), we define ˆ G ˆ p (k), g˜2,k (p) = (1 − D(k) and g˜3,k,l (p) =. 1 Uˆ (k, l). ˆ kG ˆ p (l), ×∆. 40 DOI:10.6814/NCCU202000775.
(48) and, clearly, they are continuous in p ∈ [1, pc ) for every k, l ∈ Zd . However, since g2 (p) = sup |˜ g2,k (p)|, and g3 (p) = sup |˜ g3,k,l (p)|, k∈Zd. k,l∈Zd. and the supremum of continuous functions is not necessarily continuous, we must use the following lemma to provide a sufficient condition for the supremum to be continuous. Lemma 6.1.1. (Lemma 5.13 of [12], in our language). Fix p0 ∈ [1, pc ) and let {fˆk (p)}k∈Zd be an equicontinuous family of functions in p ∈ [1, p0 ]. Suppose that supk∈Zd fˆk (p) < ∞ for every p ∈ [1, p0 ]. Then, supk∈Zd fˆk (p) < ∞ is continuous in p ∈ [1, p0 ]. Therefore, in order to prove the continuity of {gi (p)}i=2,3 .. We want to show that. 治 政 大(i) and (ii) in the following lemma. ∈ [1, p ). To prove this, it suffices to show that 立. {˜ g2,k (p)}k∈[−π,π]d and {g3,k,l }k,l∈[−π,π]d are equicontinuous families of functions in p ∈ [1, p0 ]. 學. Lemma 6.1.2.. c. ‧ 國. for each p0. (i) g˜2,k (p) and ∂p g˜2,k (p) are uniformly bounded in k ∈ [−π, π]d and p ∈ [1, p0 ].. ‧ sit. y. Nat. and. n. al. er. io. (ii) g˜3,k,l (p) and ∂p g˜3,k,l (p) are uniformly bounded in k, l ∈ [−π, π]d and p ∈ [1, p0 ]. Proof.. Ch First, we prove (i) as follows. engchi. i Un. v. ˆ ˆ p (k)| ≤ G ˆ p (0) ≡ χp and the monotonicity of χp in p, we obtain By 0 ≤ 1 − D(k) ≤ 2, |G ˆ G ˆ p | ≤ 2χp ≤ 2χp0 < ∞, so g˜2,k (p) is uniformly bounded in k ∈ [−π, π]d and |˜ g2,k | = |(1 − D) p ∈ [1, p0 ]. By subadditivity and translation-invariance, we have ∂p (pGp (x)) = Gp (x) + p(∂p Gp ),. (6.1.1). 41 DOI:10.6814/NCCU202000775.
(49) where ∂p Gp (x). =. ∂p (. ∞ ∑. n. φn (x)p ) =. n=0. =. ∞ ∑. =. nφn (x)pn−1. n=0. nφn (x)pn−1. ∞ ∑ = (n + 1)φn+1 (x)pn. n=1. ≤. ∞ ∑. n=0. ∞ ∑. (n + 1)D ∗ φn (x)p = n. n=0 ∞ ∑ ∞ ∑. ∞ ∑ n=0. D(y). y. y. =. Nat. y. D∗. φ (x − y)p 政 φ ∗治 大 ∑∑∑ n. n+m. m. D(y). φn (z)φm (x − y − z)pn+m. m=0 n=0 ∞ ∑∑ z. D(y). ∑. z. φn (z)pn. n=0. ∞ ∑. φm (x − y − z)pm. m=0. Gp (z)Gp (x − y − z) =. D(y)G∗2 p (x − y). y. z ∗2 Gp (x).. io. al. ∑. ‧. =. 立. D(y). y. ∑. m=0 n=0 ∞ ∑ ∞ ∑ m=0 n=0 ∞ ∞. y. ∑. φn+m (x − y)pn+m. 學. =. ∑. ‧ 國. =. D(y). y. y. ∑. D(y)φn+m (x − y)pn+m. er. ≤. 1. sit. ∑. m=0 n=0 ∞ ∑ ∞ ∑. n ∑ m=0. ∞ ∑ ∞ ∑ ∑. D ∗ φn+m (x)pn+m =. m=0 n=0. =. D ∗ φn (x)p. n. n. Then, by (6.1.1) we obtain. Ch. engchi. ∂p Gp (x) ≤ D ∗ G∗2 p (x),. i Un. v. hence |∂p g˜2,k (p)|. =. ˆ pG ˆ p (k)| ≤ 2|∂p G ˆ p (k)| |(1 − D)∂. ≤. ˆ G ˆ p (k)2 | ≤ 2|G ˆ p (k)|2 2|D(k). ≤. 2χ2p ≤ 2χ2p0 .. This is the complete proof of (i). Next, we prove (ii) as follows. Using Lemma 4.1.4, and. 42 DOI:10.6814/NCCU202000775.
(50) we obtain. = = =. ˆ kG ˆ p (l)| ≡ | 1 (G ˆ p (l + k) + G ˆ p (l − k)) − G ˆ p (l)| |∆ 2 ∑ ∑ 1 ∑ | ( Gp (x)ei(l+k)·x + Gp (x)ei(l−k)·x ) − Gp (x)eil·x | 2 x x x ∑ ∑ 1 ∑ Gp (x) cos(l + k) · x + Gp (x) cos(l − k) · x) − Gp (x) cos(l · x)| | ( 2 x x x ∑ ∑ | Gp (x) cos(l · x) cos(k · x) − Gp (x) cos(l · x)| x. ≤. |. ∑. x. Gp (x) cos(l · x)(cos(k · x) − 1)|. x. ≤. ∑. Gp (x)(1 − cos(k · x)). x ∞ n ∑ ∑ ∑ ∏ (1 − cos(k · x)) D(ωi − ωi−1 ). 政 治 大. x. n=0 ω∈Wn (x) n ∏ i=1. ∑. io. I[ωi−1 = u and ωi = v]p. i=1. n n ∏ i=1. Ch. ∏. D(ωi − ωi−1 ). (1 − δωi ,ωj ).. 0≤i<j≤n. n. al. n. y. ω∈Wn (x). Nat. n ∑. (1 − δωi ,ωj )pn. 0≤i<j≤n. (1 − cos k · (v − u)). u,v,x. ×. ∏. ‧. ≤. ∑. k · (ωi − ωi−1 )). i=1. D(ωi − ωi−1 ). 0≤i<j≤n. sit. ×. 立 (1 − cos. ∑. (1 − δωi ,ωj )pn. 學. x. n=0 ω∈Wn (x) i=1 n ∑. ‧ 國. =. ∞ ∑∑. ∏. er. =. i Un. v. Ignoring the self-avoiding constraint between η ≡ (ω0 , ω1 , ..., ωi−1 ) and ξ ≡ (ωi , ωi+1 , ..., ωn ). engchi. ˆ kG ˆ p (l)| as and using translation-invariance, we can further bound |∆ ˆ kG ˆ p (l)| |∆. ≤. ∑. (1 − cos k · (v − u))pD(v − u). u,v,x. ∑. (|η| + |ξ| + 1). η∈ωi−1 (0,u). ξ∈ωn−i+1 (v,x). ×p. |η|. |η| ∏. D(ηi − ηi−1 )p. |ξ|. ≤. ∞ ∑∑ ∑ x. ×. n ∏. D(ξj − ξj−1 ). j=1. i=1. ˆ 2p(1 − D(k))χ p. |ξ| ∏. D(ωj − ωj−1 ). j=1. ∏. (1 − δωi ,ωj ). 0≤i<j≤n. (n + 1)pn. n=0 ω∈Wn (x). ∏. (1 − δωi ,ωj ).. 0≤i<j≤n. 43 DOI:10.6814/NCCU202000775.
(51) However, by the identity |ω| + 1 =. ∑ y. I[y ∈ ω] for a self-avoiding path ω, subadditivity. and translation-invariance, the sum in the last line is bounded as ∞ ∑∑ ∑ x. ≤. ∑. (n + 1)p. n. n=0 ω∈Wn (x). n ∏. ∏. D(ωj − ωj−1 ). j=1. (1 − δωi ,ωj ). 0≤i<j≤n. Gp (y)Gp (x − y) = χ2p .. x,y. As a result, we arrive at 3 ˆ kG ˆ p (l)| ≤ 2p0 (1 − D(k))χ ˆ |∆ p0 ,. (6.1.2). which implies that g˜3,k,l (p) is uniformly bounded in k, l ∈ Zd and p ∈ [1, p0 ]. For the derivative. 政 治 大. ˆ k ∂p G ˆ p (k), we note that ∂p g˜3,k,l (p) ≡ Uˆ (k, l)−1 ∆ ∑. (1 − cos k · x)∂p Gp (x). 學. x. ∑ (1 − cos(k · x))(D ∗ G∗2 p )(x). ≤. x. lemma 4.1.4. ≤. 2 ˆ 3((1 − D(k))χ p + 2χp. v. n. al. {z. }. ˆ kG ˆ p (0) ∆. 2 4 ˆ ˆ 3((1 − D(k))χ p0 + 4χp0 p0 (1 − D(k))). er. io =. (1 − cos(k · v))Gp (v)). sit. Nat. |. (6.1.2). ≤. ∑. ‧. ‧ 國. =. 立. y. ˆ k ∂p G ˆ p (k)| |∆. 2 ˆ i v 2p p0). + 4χ n C h 3(1 − D(k))χp (1 engchi U 0. 0. Therefore, ∂p g˜3,k,l (p) is uniformly bounded in k, l ∈ [−π, π]d and p ∈ [1, p0 ].. 6.2 Proof of Proposition 2.2.8 - 2.2.9 and Lemma 4.1.1 First, we prove Proposition 2.2.8. In (5.1.4) and (5.1.6), it is obvious that ϵ1 and ϵ2 are decreasing when d and λ are increasing. To estimate the upper bounds of them, we only need to fix d = 5 and λ = 60. γ1 ≡ ||D||∞ + L1 + B1 ≤. 1 √ + ϵ1 + ϵ2 ≤ 0.0417827. (2 2π)d (⌊λ − 1⌋)d/2 44 DOI:10.6814/NCCU202000775.
(52) Remark 6.2.1. For d ≥ 6 and λ = 60, we get γ1 = 0.00789534870075646. In addition, by (4.1.8) - (4.1.11) and Proposition 5.1.1, we have ∞ ∑. (n) π ˆ1 (0). ≤ 0.0094953729577, and sup k. n=1. ∞ (n) ∑ ˆ kπ −∆ ˆ1 (0) n=1. ˆ 1 − D(k). ≤ 0.176357145340,. which imply that the inequalities in Lemma 4.1.1 hold for p = 1 and d > 4. Moreover, by Lemma 4.2.2, we have g1 (1). ≤. 1.009495372,. g2 (1). ≤. 1.005553012,. g3 (1). ≤. 立. Then Proposition 2.2.8 holds.. 1.406997428. 治 政 大. ‧ 國. 學. Next, we prove Proposition 2.2.9 holds for p ∈ (1, pc ). First, we fix d = 5 and λ = λ5 ≡ 60, according the assumptions gi (p) ≤ Ki for i = 1, 2, 3, we have. ‧. γp ≤ 0.044147.. sit. y. Nat. In addition, by (4.1.8) - (4.1.11) and Proposition 5.2.1, we have. er. ∞ ∑ ˆ (n) a l Cand sup −∆k πˆˆp (0)n ≤i v0.3753353454618. ≤ 0.00988157659877, U h ek nn=1g c1 h− iD(k). n. n=1. io. ∞ ∑. π ˆp(n) (0). (6.2.1). Then, we have g1 (p). ≤. 1.0098815763136844 < K1 ,. g2 (p). ≤. 1.0132463877593048 < K2 ,. g3 (p). ≤. 2.0508909504570774 < K3 .. Hence Proposition 2.2.9 holds for p ∈ (1, pc ) and λ ≥ λ5 . Moreover, this also implies that the inequalities in Lemma 4.1.1 hold for p ∈ (1, pc ) and d = 5. Next, we have to prove that Proposition 2.2.9 holds for d > 5 and λ > λd . To do this, we should estimate ϵ1 and ϵ2 in (5.1.4) and (5.1.6) for different dimensions d, respectively. First, 45 DOI:10.6814/NCCU202000775.
(53) substitute d = 6 and λ = 2d in some small terms of right-hand sides of (5.1.4) and (5.1.6), it yields ϵ1 ≤ 5.10 × 10−5 + 10.88A,. (6.2.2). ϵ2 ≤ 0.0002 + 59.14A,. (6.2.3). and. √. where A =. (. dπ d−1 Γ( d2 ) 3λ. ) d−1 2. .. Using Lemma 4.2.2, we obtain g1 (p) ≤ 1 + Lp +. 政 治 大. Bp γp2 Bp (p||D||∞ + Lp )γp ≤ 1 + L + . p 1 − γp2 1 − γp2. 立. ‧ 國. 學. Since g1 (1) < K1 , g2 (1) < K2 , suppose γp = 0.03 < 1, Proposition 5.2.1, and Lemma 5.2.2 hold, we have. ‧. g1 (p) ≤ 1 + K12 K2 ϵ1 + 0.00091(K12 K22 ϵ2 ) ≤ 1.0001 + 14.6A.. y. Nat. n. al. er. io. sit. Let g1 (p) < (1 − ϵ)K1 for all λ ≥ λd and some ϵ > 0. Then, we obtain λd as follows . λd =. . C 60,h. e nifgd =c h5, 6,i. 35,. if d = 7,. 24,. if d = 8,. 18,. if d = 9,. i Un. v. We use them to obtain the upper bound of A is shown in Table 6.1 as follow, Table 6.1: Upper bounds A for 6 ≤ d ≤ 9. A. d = 6,λ6 = 60. d = 7, λ7 = 35. d = 8, λ8 = 24. d = 9, λd = 18. 0.00028. 0.00027. 0.00024. 0.00023. 46 DOI:10.6814/NCCU202000775.
(54) For d ≥ 10, let A < 0.0002 and using Γ( d2 ) ≥ (⌊ d2 ⌋ − 1)!. λ ≥ 3 )e d−4. 2e (1 3. +. 3 )e d−4. 1 [ln(3125000)−1+ln(d(d−4)2 )] d−1. 1 [ln(3125000)−1+ln(d(d−4)2 )] d−1. .. Then, we obtain. Hence, when d ≥ 10, λd =. 2e (1 3. +. .. We use them to evidence that γp < 0.03 holds as follows. γp = p||D||∞ + Lp + Bp ≤. 2d (λ. K1 + K12 K2 ϵ1 + K12 K22 ϵ2 < 0.03. − 1)d−2. Next, we need to prove g2 (p) < (1 − ϵ)K2 and g3 (p) < (1 − ϵ)K3 for some ϵ > 0. Suppose g1 (1) < K1 , g2 (1) < K2 , g3 (p) < K3 , and γp = 0.03 < 1, Table 6.1, Lemma 5.2.1, and Lemma 5.2.2, we have g2 (p). < ≤. 治 政 1.016 < (1 − ϵ)K , 大 (. 立1.1. 2. K1 + (K12 K22 ϵ2 )2 5K3 (1 + 2ϵ1 + ϵ2 )0.061. 學. ‧ 國. g3 (p). (1 − 1.004(K12 K22 ϵ2 )2 5K3 (1 + 2ϵ1 + ϵ2 ))−1. ≤. +K12 K22 ϵ2 5K3 (1 + 2ϵ1 + ϵ2 )1.002. )2 + 2ϵ1 + ϵ2 )1.004. ‧. +(K12 K22 ϵ2 )2 5K3 (1 1.66 < (1 − ϵ)K3 .. <. sit. y. Nat. al. n. completes the proof of Proposition 2.2.9 and Lemma 4.1.1.. Ch. engchi. er. io. Therefore, This proves Proposition 2.2.9 and Lemma 4.1.1 for d > 5 and p ∈ (1, pc ). This. i Un. v. □. 47 DOI:10.6814/NCCU202000775.
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