Wetting transition of a discrete interface; an approach by random walk^{1}

Nobuo Yoshida^{2}

Contents 1 Introduction 1 2 Proof of Theorem 1.1 (a) 4

2.1 Preliminary 1: Generating functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.2 Preliminary 2: Hitting times . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.3 Asymptotics of the partition functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 2.4 Proof of (1.12) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Proof of Theorem 1.1 (b) 10 3.1 Spectral analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 3.2 Proof of (1.16) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

4 Process level limit theorems 14

1 Introduction We consider a symmetric, nearest neighbor random walk ({S_{n}}_{n≥0}, {P^{x}}_{x∈ℤ}) on ℤ. We formulate the random walk as follows.

• {P^{x}}_{x∈ℤ} are probability measures on a measurable space (Ω, F).

• {X_{n}}_{n≥0} are ℤ-valued random variables (i.e., measurable functions) on (Ω, F) such that

for all n ≥ 1 and x ∈ ℤ; X_{0}, X_{1}, …, X_{n} are P^{x}-independent and

P^{x}{X_{0} = x} = 1, P^{x}{X_{n} = 0} = r < 1, P^{x}{X_{n} = ±1} = v/2,  (1.1)

where r + v = 1.

• S_{n} = X_{0} + X_{1} + ⋯ + X_{n}, n ≥ 1. For a measure μ on (Ω, F), A ∈ F, a random variable F, we will use the following notation;

μ[F] = ∫ F dμ,  (1.2)
μ[F : A] = ∫_{A} F dμ.  (1.3)

1Notes for a course given at Nara Women’s University, November 11–15, 2002. This manuscript is based on a joint work [IsYo01] with Professor Yasuki Isozaki at Osaka University

2[email]nobuo@kusm.kyoto-u.ac.jp [URL] http://www.kusm.kyoto-u.ac.jp/ nobuo

It is well known and is easy to prove that

P^{x}[|S_{n} − x|^{2}] = vn,  (1.4)

which roughly says that S_{n} = O(n^{1/2}); the diffusion of the random walk is of order n^{1/2} at time n. We would like to consider the following situation for fixed n ≥ 1 and x ∈ ℕ;

• The random walk (S_{k})_{k≥0} starting from x stays non-negative for k = 1, …, n.

• A path (S_{k})_{0≤k≤n} which visits the origin 0 many times is more likely to be realized than a path which stays away from the origin.

The situation we want can be formulated by introducing new measures μ_{n}^{x} (n ≥ 1, x ∈ ℕ) as follows. For n ≥ 1, we set

L_{n}^{0} = the number of 0 ≤ j ≤ n − 1 such that S_{j} = 0.  (1.5)

We define a probability measure μ_{n}^{x} on (Ω, F) by

μ_{n}^{x}(dω) = \frac{1}{Z_{n}^{x}} P^{x} [exp (βL_{n}^{0}) : dω ∩ \{S_{1} ≥ 0, …, S_{n} ≥ 0\}] , n ≥ 1  (1.6)

where β ≥ 0 and Z_{n}^{x} is the normalizing constant.

In the case β > 0 which we are mainly interested in, the basic feature of the model can be explained as follows. We refer to the factor exp(βL_{n}^{0}) above as the weak pinning which tries to localize the path (S_{k})_{k=1}^{n} near the origin. The effect of the weak pinning, however, has to compete with the entropic fluctuation of the random walk, which together with the restriction to {S_{1} ≥ 0, …, S_{n} ≥ 0}, has a tendency to push the path away from the origin. We call the latter tendency entropic repulsion. The competetion between weak pinning and entropic repulsion has been well studied at the level of the free energy. We set

β_{c} = ln \frac{2}{1 + r}, .  (1.7)

For β > 0, an equation

q(λ) =^{def.} 4(e^{β} − 1)λ^{2} − 4e^{β}(e^{β} − 1)rλ − e^{2β}v^{2} = 0  (1.8)

has real solutions λ_{±} such that

λ_{−} < 0 < λ_{+}, and λ_{+} = |λ_{−}| + re^{β}.  (1.9)

Since q(1) = − ((1 + r)e^{β} − 2)^{2}, we see that

λ_{+} ≥ 1 with the equality if and only if β = β_{c}.  (1.10)

It is known [Bol99, Fis84] and will also be seen later in this course that the free energy for this model can explicitly be given as follows;

ψ(β) =^{def.} lim_{n↗∞} \frac{1}{n} ln Z_{n}^{x} = \{ \begin{matrix}0 & if β ≤ β_{c}, \\ ln(λ_{+}) > 0 & if β > β_{c}. \end{matrix}  (1.11)

for all x ∈ ℕ. This means in particular that ψ(β) = ψ(0) for β ≤ β_{c} and ψ(β) > ψ(0) for β > β_{c}. We can therefore expect the following. For β < β_{c}, the effect of the weak pinning is smaller than that of the entropic repulsion and the pinning effect does not change the diffusive nature of the path in a qualitative way. On the other hand, if β > β_{c}, the effect of the weak pinning overwhelms that of the entropic repulsion and it confines the path near the origin. The purpose of this course is to justify the above expected phenomena by proving a limit theorem which roughly says that

S_{n} = \{ \begin{matrix}O(n^{1/2}) & if β ≤ β_{c}, \\ O(1) & if β > β_{c}, \end{matrix} as n ↗ ∞.

More precisely, we have;

Theorem 1.1 a) If β ≤ β_{c}, then there are positive constants {c(p)}_{p∈ℕ} such that

μ_{n}^{x} [S_{n}^{p}] ∼ c(p)n^{p/2} as n ↗ ∞  (1.12)

for all x, p ∈ ℕ, where

c(p) = \{ \begin{matrix}(2v)^{p/2}Γ(\frac{p}{2} + 1) & if β < β_{c}, \\ \sqrt{(2v)^{p}/π}Γ(\frac{p+1}{2}) & if β = β_{c}, \end{matrix}  (1.13)

b) Suppose that^{3} β > β_{c} and r > 0. We define a probability measure m_{β} on ℕ by

m_{β} = \frac{1}{e^{−β} + ρ(1 − ρ)^{−1}} (e^{−β}δ_{0} + ∑_{x≥1} ρ^{x}δ_{x}) .  (1.14)

where

ρ = \frac{2(e^{−β}λ_{+} − r)}{v} ∈ (0, 1).  (1.15)

Then,

lim_{n↗∞} μ_{n}^{x} [S_{n}^{p}] = ∑_{y∈ℕ}y^{p}m_{β}{y} for all x, p ∈ ℕ.  (1.16)

Exercise 1.1 Prove that c(p) = v^{p/2} ∫_{0}^{∞} t^{p}μ(dt) where

μ(dt) = \{ \begin{matrix}e^{−t^{2}/2}tdt & if β < β_{c}, \\ \sqrt{2/π}e^{−t^{2}/2}dt & if β = β_{c}. \end{matrix}  (1.17)

Exercise 1.2 Show that

c(p) = \{ \begin{matrix}vpc(p − 2) & if β < β_{c} and p ≥ 2, \\ v(p − 1)c(p − 2) & if β = β_{c} and p ≥ 2. \end{matrix}  (1.18)

Remark 1.1 By (1.12) and Exercise 1.1, we see that probability measures μ_{n}^{x}\{S_{n}/\sqrt{vn} ∈ ·}, n ≥ 1 converge weakly to a limit μ on [0, ∞), which is the Rayleigh distribution if β < β_{c} and is the one-sided Gaussian distribution if β = β_{c}. On the other hand, we see from (1.16) that for β < β_{c}, probability measures μ_{n}^{x}{S_{n} ∈ ·}, n ≥ 1 converge weakly to the probability measure m_{β}.

3The assumption r > 0 is not essential. It is assumed here just to make the statement simpler.

2 Proof of Theorem 1.1 (a) 2.1 Preliminary 1: Generating functions For a function φ : ℕ → ℝ, a formal power series

φ̂(s) = ∑_{n∈ℕ}s^{n}φ(n), −1 ≤ s ≤ 1,  (2.1)

is called the generating function of φ. Recall the following fact which is known as the Cauchy-Hadamard theorem; The radius of convergence R(φ) for the series (2.1) is given by the formula;

R(φ)^{−1} = \overline{lim_{n↗∞}}|φ(n)|^{1/n}.  (2.2)

If φ ≥ 0 and R(φ) = 1, then the following important relation between φ and φ̂ is known as a Tauberian theorem [Fel70, page 447];

Lemma 2.1 For α ∈ [0, ∞) and C > 0,

φ̂(s) ∼^{s↗1} C(1 − s)^{−α} if and only if φ(0) + ⋯ + φ(n − 1) ∼^{n↗∞} \frac{Cn^{α}}{Γ(α + 1)}.  (2.3)

Moreover, if α > 0 and φ is monotone, then (2.3) is equivalent to

φ(n) ∼^{n↗∞} \frac{Cn^{α−1}}{Γ(α)} .  (2.4)

2.2 Preliminary 2: Hitting times We set

f^{x}(n) = \{ \begin{matrix}0 & if n = 0, \\ P^{x}\{T_{1} = n\}, & if n ≥ 1, \end{matrix}  (2.5)

where T_{m} denotes the m^{th}-hitting time to the origin;

T_{m} = inf {n ≥ 1 | ∑_{k=1}^{n} 1{S_{k} = 0} = m} .  (2.6)

It is then well known that the generating function of f^{x} is given by

f̂^{x}(s) = \{ \begin{matrix}(\frac{1−rs−γ(s)^{1/2}}{vs} )|x| & if x ≠ 0, |s| < 1, \\ 1 − γ(s)^{1/2} & if x = 0, |s| < 1, \end{matrix}  (2.7)

where γ(s) = (1 − rs)^{2} − s^{2}v^{2}.

Exercise 2.1 Proof of (2.7) will be partially given in the lecture. Complete it.

Exercise 2.2 Prove that

1 − f̂^{x}(s) ∼ \{ \begin{matrix}\sqrt{2v(1 − s)} & if x = 0, \\ x\sqrt{2(1 − s)/v} & if x ≠ 0, \end{matrix} as s ↗ 1.  (2.8)

2.3 Asymptotics of the partition functions For x, n ∈ ℕ = {0, 1, …}, we introduce the following random variables;

h_{n} = 1{S_{1} ≥ 0, …, S_{n} ≥ 0}, n ≥ 1,  (2.9)
e_{n} = exp (βL_{n}^{0}) , β ≥ 0, n ≥ 1,  (2.10)
z_{n} = e_{n}h_{n}, n ≥ 1,  (2.11)
z_{0} = e_{0} = h_{0} = 1.  (2.12)

Here and in what follows,

1{…} = 1 if “....” is satisfied, = 0 if otherwise

The measure μ_{n}^{x} can be rewritten as the following concise form;

μ_{n}^{x}(dω) = \frac{1}{Z_{n}^{x}} P^{x}[z_{n} : dω], x, n ∈ ℕ,  (2.13)

For x, y, n ∈ ℕ, we also define

z_{n}^{y} = z_{n}1{S_{n} = y},  (2.14)
Z_{n}^{x,y} = P^{x}[z_{n}^{y}].  (2.15)

We now introduce

θ_{±}(s) = 1 − \frac{e^{β}}{2} (rs + 1 ± \sqrt{γ(s)}) for s ∈ [−1, 1].  (2.16)

Note that θ_{−}(s) is strictly concave on [−1, +1] and is strictly decreasing on [0, 1]. Since, θ_{−}(0) = 1 and θ_{−}(1) = 1 − \frac{e^{β}}{2} (1 + r), we have that

\{s ∈ [0, 1] ; θ_{−}(s) = 0\} = \{ \begin{matrix}∅ & if β < β_{c}, \\ \{1\} & if β = β_{c}. \end{matrix}  (2.17)

Exercise 2.3 Prove that

θ_{−}(s) ∼^{s↗1} \frac{\sqrt{2v(1 − s)}}{1 + r} if β = β_{c}.  (2.18)

The purpose of this subsection is to prove the following lemma;

Lemma 2.2

∑_{k≥0} Z_{k}^{x,0} = 1/θ_{−}(1) if β < β_{c},  (2.19)
∑_{k=0}^{n−1} Z_{k}^{x,0} ∼ (1 + r)\sqrt{\frac{2n}{πv}} as n ↗ ∞ if β = β_{c}  (2.20)
Z_{n}^{x} ∼ \sqrt{\frac{2}{πvn}} (x + \frac{ve^{β}}{2θ_{−}(1)}) as n ↗ ∞ if β < β_{c}.  (2.21)

Proof of (2.19) and (2.20): We will prove that

Ẑ^{x,0}(s) = \frac{1\{x = 0\} + 1\{x ≥ 1\}f̂^{x}(s)}{θ_{−}(s) ∨ 0} for 0 ≤ s < 1,  (2.22)

where we agree with 1/0 = +∞. Let us first see that (2.22) implies (2.19) and (2.20). In fact, (2.19) follows from (2.22) by taking s = 1. For (2.20), we see from (2.22) and (2.18) that

Ẑ^{x,0}(s) ∼^{s↗1} \frac{1 + r}{\sqrt{2v}}(1 − s)^{−1/2}, if β = β_{c}.  (2.23)

This implies (2.20) by the Tauberian theorem.

The equality (2.22) is essentially the same as the representation of Green function of the random walk in terms of its hitting times. We follow [Bol99]. Note first that

Ẑ^{x,0}(s) = δ^{x,0} + ∑_{n≥1} s^{n}P^{x}[z_{n}^{0}].  (2.24)

To compute the right-hand-side, let us introduce

g^{x}(n) = \{ \begin{matrix}0 & if n = 0, \\ P^{x}\{h_{n} : T_{1} = n\} & if n ≥ 1. \end{matrix}  (2.25)

It is then, easy to see that

g^{x}(n) = \{ \begin{matrix}f^{x}(n) & if x ≥ 1, \\ f^{x}(n) = r & if x = 0 and n = 1, \\ f^{x}(n)/2 & if x = 0 and n ≥ 2. \end{matrix}  (2.26)

and consequently that for 0 ≤ s ≤ 1,

ĝ^{x}(s) = \{ \begin{matrix}(rs + f̂^{0}(s))/2 & if x = 0, \\ f̂^{x}(s) & if x ≥ 1. \end{matrix}  (2.27)

We have in particular that

θ_{−}(s) = 1 − e^{β}g^{0}(s).  (2.28)

We compute the right-hand-side of (2.24), first for the case x ≥ 1;

∑_{n≥1} s^{n}P^{x}[z_{n}^{0}]

= ∑_{n≥1} s^{n} ∑_{m≥1} e^{(m−1)β} ∑_{0<i_{1}<...<i_{m}=n} P^{x}[h_{n} : T_{1} = i_{1}, …, T_{m} = i_{m}]
= ∑_{n≥1} s^{n} ∑_{m≥1} e^{(m−1)β} ∑_{\begin{matrix}j_{1} ≥ 1, …, j_{m} ≥ 1 \\ j_{1} + … + j_{m} = n \end{matrix}} g^{x}(j_{1})g^{0}(j_{2}) ⋯ g^{0}(j_{m})

= ∑_{m≥1} ∑_{j_{1}≥1} … ∑_{j_{m}≥1} s^{j_{1}}g^{x}(j_{1}) (e^{β}s^{j_{2}}g^{0}(j_{2})) ⋯ (e^{β}s^{j_{m}}g^{0}(j_{m}))

= ∑_{m≥1} ĝ^{x}(s) (e^{β}ĝ^{0}(s))^{m−1}
= ∑_{m≥1} f̂^{x}(s) (1 − θ_{−}(s))^{m−1} , by (2.27) and (2.28)  (2.29)

which, implies (2.22) for the case x ≥ 1. Equality (2.22) for the case x = 0 can be seen similarly as above just by replacing e^{(m−1)β} and g^{x}(j_{1}) respectively by e^{mβ} and g^{0}(j_{1}). □

To prove (2.21), we prepare one more lemma which will also be used in Section 2.4. To state and prove the lemma, we will use the following notation. For a sequence (a_{n})_{n≥0} (random or non-random), we set Δa_{n} = a_{n} − a_{n−1} for n ≥ 1.

Lemma 2.3 Let Φ_{0} be a constant and Φ_{n} = φ_{n}(S_{1}, …, S_{n}) for n ≥ 1, where each φ_{n} is a function on ℕ^{n}. Then,

P^{x}[Φ_{n}z_{n}] = Φ_{0} + ∑_{k=1}^{n} P^{x}[z_{k−1}ΔΦ_{k}] + ∑_{k=1}^{n} P^{x}[z_{k−1}^{0}(e^{β}1{ΔS_{k} ≥ 0} − 1)ΔΦ_{k}]

−θ_{−}(1) ∑_{k=1}^{n} P^{x}[z_{k−1}^{0}Φ_{k−1}].  (2.30)

In particular,

Z_{n}^{x} = 1 − θ_{−}(1) ∑_{k=0}^{n−1} Z_{k}^{x,0},  (2.31)

P^{x}[S_{n}^{p}z_{n}] = x^{p} + v ∑_{1≤j≤p/2} \frac{p!}{(2j)!(p − 2j)!} ∑_{k=0}^{n−1} P^{x}[S_{k}^{p−2j}z_{k}]

+\frac{v}{2}\{e^{β} − 1 − (−1)^{p}\} ∑_{k=0}^{n−1} Z_{k}^{x,0}.  (2.32)

Proof: We observe that

Δz_{n} = (e^{β}1{ΔS_{n} ≥ 0} − 1) z_{n−1}^{0}.

We therefore have

Δ(z_{n}Φ_{n}) = z_{n}ΔΦ_{n} + Φ_{n−1}Δz_{n}
= (z_{n−1} + Δz_{n})ΔΦ_{n} + Φ_{n−1}Δz_{n}
= z_{n−1}ΔΦ_{n} + (e^{β}1{ΔS_{n} ≥ 0} − 1) z_{n−1}^{0}ΔΦ_{n} + (e^{β}1{ΔS_{n} ≥ 0} − 1) z_{n−1}^{0}Φ_{n−1}

This implies (2.30), since P^{x}[Φ_{n}z_{n}] = Φ_{0} + ∑_{k=1}^{n} P^{x}[Δ(z_{k}Φ_{k})] and P^{x}[e^{β}1{ΔS_{n} ≥ 0} − 1] = −θ_{−}(1). The other two equalities are obtained easily from (2.30). □

Exercise 2.4 Prove (2.31) and (2.32) by (2.30).

Proof of (2.21): We will prove that for 0 ≤ s < 1,

Ẑ^{x}(s) = 1\{x ≥ 1\}\frac{1 − f̂^{x}(s)}{1 − s} + (\frac{1}{2}e^{β} \frac{1 − f̂^{0}(s)}{1 − s} − \frac{1}{2}e^{β} + 1) Ẑ^{x,0}(s).  (2.33)

Let us first conclude (2.21) from (2.33). We see from (2.33) and (2.8) that

Ẑ^{x}(s) ∼^{s↗1} \sqrt{\frac{2}{v(1 − s)}} (x + \frac{ve^{β}}{2θ_{−}(1)}) if β < β_{c}.  (2.34)

Since Z_{n}^{x} is decreasing in n by (2.31), we obtain (2.21) from (2.34) and Tauberian theorem. To prove (2.34), we begin by showing that

Z_{n}^{x} = \frac{1}{2}e^{β} ∑_{m=0}^{n−1} Z_{m}^{x,0}F_{n−m}^{0} + Z_{n}^{x,0} + 1\{x ≥ 1\}F_{n}^{x},  (2.35)

where F_{n}^{x} = P^{x}{T_{1} > n}. To prove this, we set

k_{m,n} = 1{S_{m+1} > 0, …, S_{n} > 0}, 0 ≤ m ≤ n,
d_{n} = max D_{n} if D_{n} =^{def.} {0 ≤ j ≤ n ; S_{j} = 0} ≠ ∅.

We have that

P^{x}[k_{0,n}] = \{ \begin{matrix}F_{n}^{x}/2 & if x = 0, \\ F_{n}^{x} & if x ≥ 1. \end{matrix}  (2.36)

Note also that D_{n} = ∅ if and only if x ≥ 1 and T_{1} > n. Since we have either d_{n} ∈ {0, …, n} or D_{n} = ∅, we see that z_{n} can be decomposed as follows;

z_{n} = ∑_{m=0}^{n−1} z_{n}1{d_{n} = m} + z_{n}1{d_{n} = n} + 1{x ≥ 1}z_{n}1{T_{1} > n}

= ∑_{m=0}^{n−1} e^{β}z_{m}^{0}k_{m,n} + z_{n}^{0} + 1{x ≥ 1}k_{0,n}.

We therefore have

Z_{n}^{x} = e^{β} ∑_{m=0}^{n−1} P^{x}[z_{m}^{0}k_{m,n}] + Z_{n}^{x,0} + 1{x ≥ 1}P^{x}[k_{0,n}]

= \frac{1}{2}e^{β} ∑_{m=0}^{n−1} Z_{m}^{x,0}F_{n−m}^{0} + Z_{n}^{x,0} + 1\{x ≥ 1\}F_{n}^{x}, by (2.36).

This proves (2.35).

It is easy to see that

F̂^{x}(s) = \frac{1 − f̂^{x}(s)}{1 − s}.  (2.37)

Therefore, (2.33) is an easy consequence of (2.22), (2.37) and (2.35). □

Exercise 2.5 Check (2.37).

2.4 Proof of (1.12)

We begin with the case β < β_{c}. We set M_{n}(p) = P^{x}[S_{n}^{p}z_{n}]/C, where C = \sqrt{\frac{2}{πv}} (x + \frac{ve^{β}}{2θ_{−}(1)}). Then, by (2.21), (1.12) for β < β_{c} is equivalent to that;

lim_{n↗∞} n^{− \frac{p−1}{2}} M_{n}(p) = c(p).  (2.38)

We prove (2.38) by induction. For p = 0, (2.38) can immediately be seen from (2.21). For p = 1, (2.38) can be seen from (2.19), (2.32) and (2.21). Let us consider the case p ≥ 2. We may now assume (2.38) with p replaced by 0, 1, ..., p − 2. It is then not difficult to see that

n^{− \frac{p−1}{2}} ∑_{k=0}^{n−1} M_{k}(p − 2j)

= n^{− \frac{p−1}{2}} ∑_{k=0}^{n−1} k^{− \frac{p−2j−1}{2}} M_{k}(p − 2j)k^{\frac{p−2j−1}{2}} ⟶^{n↗∞} \{ \begin{matrix}\frac{c(p−2)}{(p−1)/2}, & if j = 1, \\ 0 & if otherwise. \end{matrix}  (2.39)

We have that

M_{n}(p) = v ∑_{1≤j≤p/2} \frac{p!}{(2j)!(p − 2j)!} ∑_{k=0}^{n−1} M_{k}(p − 2j) + O(1) as n ↗ ∞

by (2.19) and (2.32). We see from this and (2.39) that

lim_{n↗∞} n^{− \frac{p−1}{2}} M_{n}(p) = v \frac{p!}{2!(p − 2)!} \frac{c(p − 2)}{(p − 1)/2}
= vpc(p − 2) = c(p), by (1.18).

which completes the proof of (2.38).

Proof of (1.12) for β = β_{c} is similar as above. Note that Z_{n}^{x} ≡ 1 by (2.31). Therefore, (1.12) is equivalent to that

lim_{n↗∞} n^{− \frac{p}{2}} M_{n}(p) = c(p).  (2.40)

We prove (2.40) by induction. For p = 0, (2.40) can immediately be seen from (2.31). For p = 1, (2.20) and (2.32) imply (2.40).

Let us consider the case p ≥ 2. We may now assume (2.40) with p replaced by 0, 1, ..., p − 2. It is then not difficult to see that

n^{− \frac{p}{2}} ∑_{k=0}^{n−1} M_{k}(p − 2j)

= n^{− \frac{p}{2}} ∑_{k=0}^{n−1} k^{− \frac{p−2j}{2}} M_{k}(p − 2j)k^{\frac{p−2j}{2}} ⟶^{n↗∞} \{ \begin{matrix}\frac{c(p−2)}{p/2} & if j = 1, \\ 0 & if otherwise. \end{matrix}  (2.41)

We have

M_{n}(p) = v ∑_{1≤j≤p/2} \frac{p!}{(2j)!(p − 2j)!} ∑_{k=0}^{n−1} M_{k}(p − 2j) + O(n^{1/2}) as n ↗ ∞

by (2.32) and (2.20). We see from this and (2.41) that

lim_{n↗∞} n^{− \frac{p}{2}} M_{n}(p) = v \frac{p!}{2!(p − 2)!} \frac{c(p − 2)}{p/2}
= v(p − 1)c(p − 2) = c(p), by (1.18),

which completes the proof of (2.40). □

Exercise 2.6 Let c_{n} ⟶ c and 0 ≤ α ≤ γ. Prove then that

lim_{n↗∞} n^{−(1+α)} ∑_{k=1}^{n} k^{γ}c_{k} = \{ \begin{matrix}\frac{c}{1+α} & if γ = α, \\ 0 & if γ < α. \end{matrix}  (2.42)

Then, use this to verify (2.39) and (2.41)

3 Proof of Theorem 1.1 (b)

3.1 Spectral analysis

We introduce matrix Q = (Q_{xy})_{x,y∈ℕ} by

Q = ( \begin{matrix}e^{β}r & e^{β} \frac{v}{2} & 0 & 0 & ⋯ \\ \frac{v}{2} & r & \frac{v}{2} & 0 & ⋯ \\ 0 & \frac{v}{2} & r & \frac{v}{2} & ... \\ 0 & 0 & \frac{v}{2} & r & ... \\ ⋮ & ⋮ & ... & ... & ... \end{matrix} ) .  (3.1)

Exercise 3.1 Check

λ_{+} = e^{β} (r + \frac{v}{2}ρ) = r + \frac{v}{2} (ρ + ρ^{−1}) .  (3.2)

Then, use this to prove that ∑_{y∈ℕ} Q_{xy}ρ^{y} = λ_{+}ρ^{x} for all x ∈ ℕ.

Let us recall that we have defined functions θ_{±}(s) by (2.16). For β > 0, we see by direct computations that

{s ∈ [−1, 1] ; θ_{−}(s) = 0} ⊂ {1/λ_{−}, 1/λ_{+}}.  (3.3)

Since θ_{−}(s) is strictly decreasing on [0, 1], θ_{−}(0) = 1 and θ_{−}(1) < 0 for β > β_{c}, we see that

{s ∈ [0, 1] ; θ_{−}(s) = 0} = {1/λ_{+}} for β > β_{c}.  (3.4)

Lemma 3.1 Let λ ∉ (−1, +1) and s = 1/λ. Then, for β > β_{c}, an equation

Qu = λu,  (3.5)

given the value of u(0), can uniquely be solved as

u(x) = \frac{u(0)}{e^{β}\sqrt{γ(s)}} (θ_{−}(s)f̂^{1}(s)^{−x} − θ_{+}(s)f̂^{1}(s)^{x}) , x ≥ 1.  (3.6)

In particular, the equation (3.5) for |λ| > 1 has a bounded solution only if λ = λ_{±}.

Proof: The equation (3.5) reads

e^{β}ru(0) + \frac{1}{2}e^{β}vu(1) = λu(0),  (3.7)
ru(x) + \frac{1}{2}v (u(x − 1) + u(x + 1)) = λu(x) x ≥ 1.  (3.8)

If s = 1/λ ∈ (−1, +1), then the characteristic equation

rt + \frac{1}{2}v (1 + t^{2}) = λt

for (3.8) has f̂^{1}(s)^{±1} as its two distinct real solutions (f̂^{1}(s) < 1 < f̂^{1}(s)^{−1}). It is therefore elementary to get (3.6). □

Exercise 3.2 Fill in the detail of the proof of (3.6)

We now consider a bilinear form

⟨ f_{1}, f_{2} ⟩_{β} = e^{−β}f_{1}(0)f_{2}(0) + ∑_{x≥1} f_{1}(x)f_{2}(x)  (3.9)

for f_{i} : ℕ → ℝ (i = 1, 2) when the summation on the right-hand-side converges absolutely. For β > β_{c}, we set

ρ_{2}(x) = \frac{ρ^{x}}{\sqrt{e^{−β} + ρ^{2}(1 − ρ^{2})^{−1}}} x ∈ ℕ,  (3.10)

so that ⟨ ρ_{2}, ρ_{2} ⟩_{β} = 1 and m_{β}[f] = ⟨ f, ρ_{2} ⟩_{β}/⟨ 1, ρ_{2} ⟩_{β}. Let p ≥ 0, m ≥ 1 and f : I^{m} → ℝ, where I is either [0, ∞) or ℕ. We say f is p-bounded if there is p ≥ 0 such that

sup_{(x_{j})_{j=1}^{m}∈I^{m}}(1 + |x_{1}| + … + |x_{m}|)^{−p}|f(x_{1}, …, x_{m})| < ∞.  (3.11)

We say f is polynomially bounded if it is p-bounded for some p ≥ 0.

Lemma 3.2 For β > β_{c}, there is a constant ε_{3.12} = ε_{3.12}(β, v) ∈ (0, 1) such that

|λ_{+}^{−n}P^{x}[f(S_{n})z_{n}] − ⟨ f, ρ_{2} ⟩_{β}ρ_{2}(x)| ≤ (1 + |x|)^{p}B_{3.12}(f)ε_{3.12}^{n}  (3.12)

for all x, n ∈ ℕ and a p-bounded function f : ℕ → ℝ, where B_{3.12}(f) is a constant which depends only on β, v and f.

Proof: Since Q_{xy} = P^{x}[z_{1}^{y}], we have

P^{x}[f(S_{n})z_{n}] = ∑_{y∈ℕ}P^{x}[z_{n}^{y}]f(y)
= ∑_{x_{1},x_{2},...,x_{n−1},y∈ℕ}Q_{xx_{1}}Q_{x_{1}x_{2}} ⋯ Q_{x_{n−1}y}f(y)
= Q^{n}f(x).  (3.13)

We will prove (3.12) by studying the matrix Q. To do so, let us introduce a Hilbert space

ℓ_{β}^{2} = {f : ℕ → ℝ ; ⟨ f, f ⟩_{β} < ∞},

equipped with the inner product defined by (3.9). Let us first prove that

∥λ_{+}^{−n}Q^{n}f − ⟨ f, ρ_{2} ⟩_{β}ρ_{2}∥_{β} ≤ ∥f − ⟨ f, ρ_{2} ⟩_{β}ρ_{2}∥_{β}ε_{3.14}^{n} for f ∈ ℓ_{β}^{2},  (3.14)

where ∥ · ∥_{β} denots the ℓ_{β}^{2}-norm. Consider Q as a bounded operator on ℓ_{β}^{2}. Then, Q is symmetric and bounded. Therefore, σ(Q): the set of ℓ_{β}^{2}-spectra of Q, is a compact set in ℝ. We claim

σ(Q) ⊂ [−1, 1] ∪ {λ_{−}, λ_{+}},  (3.15)
λ_{+} > sup{|λ| ∈ σ(Q)\{λ_{+}}}.  (3.16)

We let σ^{disc.}(Q) and σ^{ess.}(Q) denote the discrete and the essential spectra of Q. Now, consider an operator

P = ( \begin{matrix}r & \frac{v}{2} & 0 & 0 & ⋯ \\ \frac{v}{2} & r & \frac{v}{2} & 0 & ⋯ \\ 0 & \frac{v}{2} & r & \frac{v}{2} & ... \\ 0 & 0 & \frac{v}{2} & r & ... \\ ⋮ & ⋮ & ... & ... & ... \end{matrix} ) .

Then, P is ℓ_{0}^{2}-symmetric and has the operator norm one on ℓ_{0}^{2}. This implies σ(P) ⊂ [−1, 1], since the set of ℓ_{β}^{2}-spectra of a bounded operator is independent of the choice of β. Note also that Q − P is of finite rank and hence is a compact operator. This implies that σ^{ess.}(Q) = σ^{ess.}(P) ⊂ [−1, 1], cf. [ReSi78, page 111, Lemma 3]. On the other hand, we see from Lemma 3.1 that σ^{disc.}(Q)\[−1, +1] ⊂ {λ_{−}, λ_{+}}. This, together with (1.9), implies (3.15) and (3.16). The claim (3.14) can now easily be seen by (3.15), (3.16) and and the spectral decomposition.

The desired estimate (3.12) for polynomially bounded f can be seen by applying (3.14) to a cut-off function f_{x,n}(y) = f(y)1{|x − y| ≤ n} as follows. Since Q_{n}f_{x,n}(x) = Q_{n}f(x), we have

|λ_{+}^{−n}Q^{n}f(x) − ⟨ f, ρ_{2} ⟩_{β}ρ_{2}(x)|
≤ |λ_{+}^{−n}Q^{n}f_{x,n}(x) − ⟨ f_{x,n}, ρ_{2} ⟩_{β}ρ_{2}(x)| + |⟨ f_{x,n} − f, ρ_{2} ⟩_{β}|ρ_{2}(x)  (3.17)

Since f_{x,n} ∈ ℓ_{β}^{2}, we see from (3.14) that the first term of (3.17) is not greater than

e^{β/2} ( ∑_{y∈ℕ:|y−x|≤n}|f(y)|^{2})^{1/2} ε_{3.14}^{n},

which has a bound in the form of the right-hand-side of (3.12). On the other hand, it is easy to bound the second term of (3.17) by B(f)ρ^{n}, where B(f) is a constant which depends only on β, v and f. □

Exercise 3.3 Fill in the details of the estimates of (3.17) to conclude (3.12).

3.2 Proof of (1.16)

It is easy to see from (3.12) that there is a constant C_{3.18} = C_{3.18}(β, v) > 0 such that

1/2 ≤ \frac{λ_{+}^{−n}Z_{n}^{x}}{⟨ 1, ρ_{2} ⟩_{β}ρ_{2}(x)} ≤ 3/2 if n ≥ (1 + x)C_{3.18}.  (3.18)

We assume n ≥ (1 + x)C_{3.18}. We set f(y) = y^{p} and rewrite the left-hand-side of (1.16) as follows;

|μ_{n}^{x}[f(S_{n})] − m_{β}[f]|
= |\frac{λ_{+}^{−n}P^{x}[f(S_{n})z_{n}]}{λ_{+}^{−n}Z_{n}^{x}} − \frac{⟨ f, ρ_{2} ⟩_{β}}{⟨ 1, ρ_{2} ⟩_{β}} |
= \frac{|λ_{+}^{−n}P^{x}[f(S_{n})z_{n}]⟨ 1, ρ_{2} ⟩_{β} − λ_{+}^{−n}Z_{n}^{x}⟨ f, ρ_{2} ⟩_{β}|}{λ_{+}^{−n}Z_{n}^{x}⟨ 1, ρ_{2} ⟩_{β}}  (3.19)

We see from (3.18) that the denominator of (3.19) is bounded from below by \frac{1}{2}⟨ 1, ρ_{2} ⟩_{β}^{2}ρ_{2}(x)^{2}. We now use (3.12) to give the following upper bound of the numerator of (3.19);

|λ_{+}^{−n}P^{x}[f(S_{n})z_{n}] − ⟨ f, ρ_{2} ⟩_{β}ρ_{2}(x)| ⟨ 1, ρ_{2} ⟩_{β}
+ |λ_{+}^{−n}Z_{n}^{x} − ⟨ 1, ρ_{2} ⟩_{β}ρ_{2}(x)| |⟨ f, ρ_{2} ⟩_{β}|
≤ B_{3.20}(1 + |x|)^{p}ε_{3.12}^{n},  (3.20)

where B_{3.20} = B_{3.20}(β, v, f) > 0 and p = p(f) ≥ 0. Consequently, (3.19) is bounded from above by

B_{3.21}(1 + |x|)^{p}ρ^{−x}ε_{3.12}^{n}  (3.21)

for some B_{3.21} = B_{3.21}(β, v, f) > 0. We therefore get (1.16). □

Exercise 3.4 Prove (3.18).

4 Process level limit theorems

Following [IsYo01], we present process level limit theorems.

Theorem 4.1 For β ≤ β_{c}, consider probability measures μ̃_{n}^{x}, n ≥ 1, x ∈ ℕ on C([0, 1]) defined by

μ̃_{n}^{x} = μ_{n}^{x} \{(S_{nt}/\sqrt{vn})_{0≤t≤1} ∈ ·\} ,  (4.1)

where (S_{t})_{t≥0} is the linear interpolation of the random walk (S_{n})_{n∈ℕ}. Then, for any fixed x ∈ ℕ, as n ↗ ∞, μ̃_{n}^{x} converges weakly to the law of a process W = (W_{t})_{0≤t≤1}. The process W is described as follows;

W_{t} = \{ \begin{matrix}(1 − τ)^{−1/2}B_{τ+t(1−τ)} & if β < β_{c}, \\ |B_{t}| & if β = β_{c}, \end{matrix}  (4.2)

where B = (B_{t})_{0≤t≤1} is a one-dimensional Brownian motion starting from 0 and τ = sup{t ∈ [0, 1] ; B_{t} = 0}.

Remark 4.1 Similar limit theorems are discussed in the literatures. Bolthausen et al consider the case of one-dimensional weakly pinned Gaussian field in the delocalized phase [BDZ00, Section4]. In [Upt99], Upton discusses one-dimensional SOS interfaces and the phase boundaries of two-dimensional Ising model, where the thermodynamic parameters are assumed to be sufficiently inside delocalized region.

We have a process level limit theorem also for β > β_{c}. We will have a positively recurrent Markov chain as the limit process. To describe the Markov chain obtained as the limit, we introduce a matrix R = (R_{xy})_{x,y∈ℕ} by

R_{xy} = λ_{+}^{−1}ρ^{−x}Q_{xy}ρ^{y}.  (4.3)

By Exercise 3.1, R defines a transition probability for an ℕ-valued Markov chain. We denote this Markov chain by ({S_{n}}_{n∈ℕ}, {ν^{x}}_{x∈ℕ}). It is easy to check that the Markov chain {ν^{x}}_{x∈ℕ} is reversible with respect to the probability measure m_{β} on ℕ defined by (1.14). In particular, {ν^{x}}_{x∈ℕ} is positively recurrent.

Theorem 4.2 For β > β_{c}, there are constants C_{4.4} = C_{4.4}(β, v) > 0 and ε_{4.4} = ε_{4.4}(β, v) ∈ (0, 1) as follows; if n ≥ k + (1 + x)C_{4.4}, f : ℕ^{k} → ℝ is polynomialy bounded and F = f(S_{1}, …, S_{k}), then

|μ_{n}^{x}[F] − ν^{x}[F]| ≤ B_{4.4}(f)ε_{4.4}^{n−k},  (4.4)

where B_{4.4}(f) is a constant which depends only on β, v and f.

By Theorem 4.2, we get the following strong mixing properties for both μ_{n}^{x} and ν^{x} for β > β_{c};

Theorem 4.3 For β > β_{c}, there are constants C_{4.5} = C_{4.5}(β, v) > 0 and ε_{4.5} = ε_{4.5}(β, v) ∈ (0, 1) as follows; if n ≥ l+m, l ≥ k+(1+x)C_{4.5}, f : ℕ^{k} → ℝ and g : ℕ^{m} → ℝ are polynomially bounded, F = f(S_{1}, …, S_{k}) and G = g(S_{l+1}, …, S_{l+m}) then

|μ_{n}^{x}(F G) − μ_{n}^{x}(F)μ_{n}^{x}(G)| ≤ B_{4.5}(f, g)ε_{4.5}^{l−k},  (4.5)
|ν^{x}(F G) − ν^{x}(F)ν^{x}(G)| ≤ B_{4.5}(f, g)ε_{4.5}^{l−k},  (4.6)

where B_{4.5}(f, g) is a constant which depends only on β, v, f and g.

Remark 4.2 In [BDZ00, Section4], it is mentioned that a similar statement to (4.5) is true for the one-dimensional weakly pinned Gaussian field in the localized phase. Localization of two-dimensional weakly pinned surfaces are discussed in [DeVe00, DMRR92, IoVe00], however without the positivity constraint.

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