LATTICE GAS MODELS WITH LONG RANGE INTERACTIONS
arXiv:1604.01337v1 [math-ph] 5 Apr 2016
DAVID ARISTOFF AND LINGJIONG ZHU
Abstract. We study microcanonical lattice gas models with long range interactions, including power law interactions. We rigorously obtain a variational
principle for the entropy. In a one dimensional example, we find a first order
phase transition by proving the entropy is non-differentiable along a certain
curve.
1. Introduction
In this article we study lattice gas models with long range pair interactions. Our
goal is to bridge the gap between mean field and random graph models, in which
all sites or nodes interact with all others with equal order strength, and standard
short range models in statistical mechanics, in which interactions decay fast enough
to be summable.
In Ising or lattice gas models with power law interactions decaying like `−r ,
with ` the distance between sites, it is well known that the thermodynamic limit
exists when r > d, where d is the dimension of the system [21]. We consider
here r < d. We call such interactions long range. For details and discussion, see
the review articles [4, 12]. Due to the inequivalence of ensembles for long range
interactions [3, 5], we consider only the microcanonical ensemble. Though we focus
on lattice gas models, our arguments easily adapt to other long range interacting
models, e.g. the α-Ising model [3].
In the long range setting, the thermodynamic limit does not exist in the traditional sense, and a scaling factor depending on the number of sites must be
multiplied into the interaction to obtain a nontrivial limit. (See the remarks after Assumption 1 below.) Using this scaling and techniques from large deviations
theory, we prove a variational principle for the entropy. See [4] for an outline of
the large deviations approach in the long range setting. See also [3, 9] for similar
work on the α-Ising model, and [6, 7, 8, 14] for rigorous studies of mean field type
models.
The variational principle we obtain is defined on the space of functions on the
unit cube C = [0, 1]d , instead of on a space of measures, as is standard in models
with short range interactions. Variational formulas similar to ours can be found in
random graph models in which the interaction depends on the number of edges and
other subgraphs; see [2, 10, 15, 16, 17, 18, 19]. The variational optimizers describe
the probabilistic structure of the model as the number of sites becomes infinite.
(See the remarks after Theorem 3 below.)
Date: 23 February 2016.
2010 Mathematics Subject Classification. 82B20, 82B26.
Key words and phrases. Statistical mechanics, lattice gas models, long range interactions,
phase transitions.
1
2
DAVID ARISTOFF AND LINGJIONG ZHU
An interesting connection between random graph models and lattice statistical
mechanics models is found in the Curie-Weiss mean field Ising model. The CurieWeiss model is equivalent, up to a continuous change of parameters, to the edge
and 2-star random graph models studied in [1, 20]. Our lattice gas models can be
seen as an extension of such random graph models in which an underlying topology
is responsible for a decay in interaction strength.
Our variational optimizers have an inhomogeneous structure [3], except along a
curve in parameter space. In a one dimensional example, we prove that along this
curve the entropy is non-differentiable, indicating a first order phase transition.
This article is organized as follows. In Section 2 we describe our models in detail.
We present a variational principle for the entropy and the corresponding EulerLagrange equations in Section 3. Using these results, we show a phase transition
in a one dimensional model in Section 4. All proofs are in Section 5.
2. Notation and assumptions
Fix a dimension d ≥ 1, and for n ≥ 1 define Λn = {1, . . . , n}d . We denote
elements of Λn by I, J. We consider particle configurations η : Λn → {0, 1}, sometimes writing
ηn to emphasize dependence on n, with a pair interaction function
√
ψ : [0, d] → R. The interaction between sites I, J ∈ Λn is given by
φn (I, J) := ψ n−1 |I − J| ,
where | · | is the Euclidean norm in Rd . Let C = [0, 1]d and define CI = {x ∈ C :
I − 1 ≤ nx < I}, where the inequalities are componentwise. Throughout, we will
associate a particle configuration η : Λn → {0, 1} with a function f η : C → [0, 1]
defined by
f η (x) = η(I),
if x ∈ CI .
See Figure 1. Let Pn be the uniform probability measure on particle configurations,
d
Pn (η) = 2−n ,
for all η : Λn → {0, 1}.
Under the map η → f η , Pn pushes forward to a probability measure on the space
of measurable functions C → [0, 1], which, abusing notation, we also denote by Pn .
Define the energy density En and number density Nn of η : Λn → {0, 1} by
X
X
En (η) = n−2d
η(I)η(J)φn (I, J),
Nn (η) = n−d
η(I).
I,J∈Λn
I∈Λn
For fixed parameters ξ, ρ ∈ R, define the entropy
S(ξ, ρ) = lim+ lim n−d log Pn (En (η) ∈ (ξ − δ, ξ + δ), Nn (η) ∈ (ρ − δ, ρ + δ)) .
δ→0
n→∞
(2.1)
We show below the limit defining S(ξ, ρ) exists under the following assumption.
Assumption 1. The map (x, y) 7→ ψ(|x − y|) is in Lq (C 2 ) for some q > 1. Moreover, it is Riemann integrable.
Assumption 1 will hold throughout the remainder of the paper. We keep in
mind the case of power law interactions in dimension d = 1, where ψ(t) = t−r for
LONG RANGE LATTICE GAS MODELS
3
1
fη
0
0
0.2
0.4
0.6
0.8
1
Figure 1. The graph of the function f η when d = 1, n = 5,
η(1) = η(2) = η(4) = 1 and η(3) = η(5) = 0.
t ∈ (0, 1] and ψ(0) = 0, with r ∈ (0, 1) constant. In this setting,
n
X
−2
En (η) = n
η(i)η(j)φn (i, j)
= n−2
=n
i,j=1
n
X
−r
η(i)η(j) n−1 (i − j)
i,j=1
i6=j
n
X
r−2
(2.2)
η(i)η(j)|i − j|−r .
i,j=1
i6=j
Thus, for a one dimensional power law interaction having strength |i−j|−r between
sites i and j, we must multiply the interaction by nr−2 to obtain a nontrivial limit.
This is in contrast with short range interactions, where the scaling factor is n−1 .
Note that as r → 0 we recover mean field interactions. In Section 4 below, we
consider a modified version of this interaction for which the entropy S is singular.
3. Large deviations, entropy, and Euler-Lagrange equations
Below we write
(
t log t + (1 − t) log(1 − t) + log 2,
Hbin (t) =
∞,
t ∈ [0, 1]
t∈
/ [0, 1]
for the binary entropy function. We will need the following terminology before
proceeding. A sequence Qn of probability measures on a topological space T is said
to satisfy a large deviation principle with speed an and rate function K : T → R
if K is non-negative and lower semicontinuous, and for any measurable set A ⊂ T ,
−1
− inf ◦ K(x) ≤ lim inf a−1
n log Qn (A) ≤ lim sup an log Qn (A) ≤ − inf K(x), (3.1)
x∈A
n→∞
n→∞
x∈Ā
4
DAVID ARISTOFF AND LINGJIONG ZHU
where A◦ denotes the interior of A and Ā the closure of A. We refer to the first
inequality in (3.1) as the lower bound and the last inequality in (3.1) as the upper
bound.
Throughout we fix p ∈ [1, ∞) and q ∈ (1, ∞] with p−1 + q −1 = 1. We prove a
large deviation principle for Pn on the Banach space of functions in Lp (C) endowed
with the weak topology. We denote this space by X . We will also consider the
subset Y = {f ∈ X : f (x) ∈ [0, 1] for a.e. x} ⊂ X . Unless otherwise specified, we
endow Y with the subspace topology.
Theorem 2. The sequence Pn satisfies a large deviation principle on X with speed
nd and rate function
Z
H(f ) =
Hbin (f (x)) dx.
C
Consider the following constrained subset of Y,
Z
Z
Yξ,ρ := f ∈ Y :
f (x)f (y)ψ(|x − y|) dx dy = ξ,
f (x) dx = ρ .
C2
C
Theorem 2 leads to the following variational expression for the entropy.
Theorem 3. We have
S(ξ, ρ) = − inf H(f ) = sup [−H(f )] ,
f ∈Yξ,ρ
(3.2)
f ∈Yξ,ρ
with the infimum over the empty set equal to ∞ by convention.
Below we will refer to functions f∗ ∈ Yξ,ρ with S(ξ, ρ) = −H(f∗ ) as optimizers
of the variational problem (3.2). Optimizers represent the most likely structure of
large particle configurations. For instance, if f∗ is the unique optimizer of (3.2)
and n is large, then f∗ (n−1 I) is roughly the probability that ν(I) = 1, i.e., there is
a particle at site I ∈ Λn . If (ξ, ρ) is in the space of parameters defined by
Ω = {(ξ, ρ) : Yξ,ρ 6= ∅} ,
then optimizers exist; moreover optimizers in
F := {f ∈ X : ∃ > 0 s.t. f (x) ∈ [, 1 − ] for a.e. x}
satisfy the appropriate Euler-Lagrange equations.
Theorem 4. Optimizers of (3.2) exist whenever (ξ, ρ) ∈ Ω. If f∗ ∈ F ∩ Yξ,ρ is an
optimizer, then for a.e. x, either
R
exp µ + β C f∗ (y)ψ(|x − y|) dy
,
R
(3.3)
f∗ (x) =
1 + exp µ + β C f∗ (y)ψ(|x − y|) dy
for some β, µ ∈ R with (β, µ) 6= (0, 0), or
Z
f∗ (y)ψ(|x − y|) dy ≡ ξ/ρ.
(3.4)
C
Theorem 4 immediately leads to the following.
Corollary 5. If the Euler-Lagrange equation (3.3) holds, then f∗ is continuous.
Corollary 5 has the following
interesting consequence. Suppose that (ξ, ρ) ∈ Ω
R
and ξ 6= ρ2 λ, where λ := C 2 ψ(|x − y|) dx dy. Then if an optimizer of (3.2) is in
F, it cannot be constant or piecewise constant. Rather, it has a curved shape;
see [3]. This is in contrast with short range interactions, where large configurations
are spatially homogeneous.
LONG RANGE LATTICE GAS MODELS
5
4. Singularity of the entropy in a one dimensional example
Here we consider an example in dimension d = 1 in which the entropy S is
singular. We will consider ψ with the following structure.
Assumption 6. For some constants r ∈ (0, 1) and M > 0,
(
t−r , 0 < t < 1/4
,
ψ(t) =
M, 1/4 ≤ t ≤ 1/2
and ψ(0) = 0. Also, ψ is symmetric: for each t ∈ [0, 1], ψ(t) = ψ(1 − t).
Note that symmetry of ψ corresponds to periodic boundary conditions for the
particle configurations, i.e., particle configurations on a circle. Clearly, ψ satisfies
Assumption 1. Note that our ψ is not continuous at 1/4 in general. However, it
will be clear that all the results below also hold for a smoothed version of ψ; see
Figure 2 and the remarks below the proof of Lemma 7 in Section 5. We choose this
form of the interaction for simpler arguments. Below we write
Z
λ=
ψ(|x − y|) dx dy
[0,1]2
for the integrated interaction function, and we define
Z
ξ(f ) =
f (x)f (y)ψ(|x − y|) dx dy.
[0,1]2
We begin by making a claim about the parameter space Ω.
Lemma 7. For each r ∈ (0, 1), there is an interaction ψ satisfying Assumption 6
with the following property. There is > 0 such that the curve
{(ξ, ρ) ∈ Ω : ξ = λρ2 , ρ ∈ (1/4 − , 1/4)}
is in the interior of Ω.
Lemma 7 is proved by exhibiting functions f which integrate to ρ and have values
of ξ(f ) both larger and smaller than ξ(ρ) = λρ2 . Such functions can be found for
suitable M . We do not attempt to find the complete interior or boundary of Ω.
Fortunately, Lemma 7 suffices for the following.
Theorem 8. Let ψ be as in Lemma 7 with r < 1/2. Then the entropy S is nondifferentiable along the curve {(ξ, ρ) ∈ Ω : ξ = λρ2 , ρ ∈ (1/4 − , 1/4)}.
Note that we needed Lemma 7 to show that the curve ξ = λρ2 is in the interior
of Ω for ρ ∈ (1/4 − , 1/4). Theorem 8 shows there is a first order phase transition,
i.e., a discontinuity in the first derivative of the entropy, across this curve. The
curve corresponds to optimizers that are constant valued. (Recall from Corollary 5
that optimizers are non-constant off this curve.) The first order transition suggests
a qualitative change in the structure of the non-constant optimizers across the
singularity. Indeed, the proof of Lemma 7 below suggests that above the curve
optimizers are bimodal, while below it they are unimodal.
6
DAVID ARISTOFF AND LINGJIONG ZHU
30
25
20
15
ψ
10
5
0
0.1
0.2
0.3
0.4
0.5
Figure 2. A example of a smooth interaction ψ for which S is singular as in Theorem 8. Here we take periodic boundary conditions,
i.e., ψ(t) = ψ(1 − t).
5. Proofs
We begin by proving Theorem 2. First, we need the following lemmas.
Lemma 9. For any s ∈ [0, 1],
1 1 t
sup st − log
+ e
= Hbin (s).
2 2
t∈R
(5.1)
Proof. When s ∈
/ [0, 1] the quantity in brackets has no upper bound in t. When
s ∈ [0, 1], the maximum is attained when t = log s − log(1 − s), and plugging this
back into (5.1) yields the result.
Lemma 10. Suppose θ : R → R is Lipschitz continuous and let g ∈ Lp (C). Then
Z
X Z
−d
d
lim n
θ n
g(x) dx =
θ(g(x)) dx.
n→∞
CI
I∈Λn
C
Proof. Consider the operator An : Lq (C) → Lq (C) defined by
X Z
An g =
nd
g(x) dx 1CI .
CI
I∈Λn
Note that
Z
θ(An g(x)) dx = n
C
−d
X
Z
θ nd
g(x) dx .
CI
I∈Λn
Since θ is Lipschitz, for a constant c > 0,
Z
Z
Z
θ(An g(x)) dx − θ(g(x)) dx ≤
|θ(An g(x)) − θ(g(x))| dx
C
C
C
Z
≤ c |An g(x) − g(x)| dx.
C
(5.2)
LONG RANGE LATTICE GAS MODELS
7
Clearly An g → g in norm when g is continuous. Since An is a bounded operator
and continuous functions are dense in Lq (C), we see that An g → g in norm for any
g ∈ Lq (C). Thus, the last expression in (5.2) vanishes as n → ∞.
Proof of Theorem 2. Recall that
Y := {f ∈ X : f (x) ∈ [0, 1] for a.e. x} ,
F := {f ∈ X : ∃ > 0 s.t. f (x) ∈ [, 1 − ] for a.e. x} .
(5.3)
We claim that Y is compact. Note that Y is closed, convex and bounded in Lp (C).
Thus, by the Banach-Alaoglu theorem, Y is compact if 1 < p < ∞. Since the weak
topology in L1 (C) is coarser than the weak topology in Lp (C) for 1 < p < ∞, the
p = 1 case follows. We follow Baldi’s theorem; see Theorem 4.5.3 of [13]. Let En
be expectation associated to Pn . Write f ηn : C → [0, 1] for the function drawn from
Pn associated to ηn : Λn → {0, 1}. Thus, (ηn (I))I∈Λn are iid Bernoulli-1/2 random
variables. For any g ∈ Lq (C),
Z
H ∗ (g) := lim n−d log En exp nd f ηn (x)g(x) dx
n→∞
C
"
!#
Z
X
−d
d
= lim n log En exp n
ηn (I)
g(x) dx
n→∞
CI
I∈Λn
Z
Y
−d
d
= lim n log
En exp ηn (I)n
n→∞
= lim n
n→∞
X
g(x) dx
CI
I∈Λn
−d
Z
log En exp ηn (I)nd
g(x) dx
CI
I∈Λn
Z
1 1
+ exp nd
g(x) dx
n→∞
2 2
CI
I∈Λn
Z
1 1 g(x)
=
log
+ e
dx.
2 2
C
= lim n−d
X
log
The last equality follows from Lemma 10, since θ(t) := log( 12 + 21 et ) is Lipschitz.
Notice Y is compact and the Pn are supported on Y. In particular, Pn is exponentially tight. Thus (see Theorem 4.5.3 (a) of [13]) Pn satisfies the large deviation
upper bound in Lp (C) with rate function
Z
H(f ) =
Hbin (f (x)) dx
C
Z
Z
1 1 g(x)
+ e
dx ,
= sup
f (x)g(x) dx − log
2 2
g∈Lq (C)
C
C
where we used Lemma 9 for the second equality. Since the weak topology is coarser
than the norm topology, Pn also satisfies the large deviation upper bound in X . It
is easy to check that the rate function H is nonnegative and lower semi-continuous.
We now verify the remaining conditions in Baldi’s theorem. Let f ∈ F, and define
hf (x) = log f (x) − log(1 − f (x)) for x ∈ C. Then hf is an exposing hyperplane for
8
DAVID ARISTOFF AND LINGJIONG ZHU
f , since
Z
f (x)hf (x) dx − H(f ) −
g(x)hf (x) dx − H(g)
C
C
Z g(x)
1 − g(x)
=
g(x) log
+ (1 − g(x)) log
dx > 0
f (x)
1 − f (x)
C
Z
(5.4)
whenever f 6= g on a set of positive measure. Clearly, H ∗ (hf ) exists and H ∗ (γhf )
exists and is finite for all γ > 1. If for all open sets U ⊂ X , we have
inf
f ∈U ∩F
H(f ) = inf H(f ),
(5.5)
f ∈U
then (see Theorem 4.5.20 (b)-(c) of [13]) Pn satisfies the large deviation lower bound
in X . The norm topology in Lp (C) is coarser than the uniform topology, since an
-ball in Lp (C) contains the corresponding uniform -ball when < 1. Thus, to
prove (5.5) it suffices to consider a set U open in the uniform topology. If U ∩F = ∅
then H(f ) = ∞ for all f ∈ U and both sides of (5.5) equal ∞. Suppose then that
f ∈ U with H(f ) < ∞, and define


f (x), f (x) ∈ [, 1 − ]
.
f (x) = 1 − , f (x) > 1 − 

,
f (x) < For sufficiently small, f ∈ U ∩ F and H(f ) ≤ H(f ). This shows that
inf
f ∈U ∩F
H(f ) ≤ inf H(f ).
f ∈U
The reverse inequality holds since U ∩ F ⊂ U , so we are done.
Now we turn to the proof of Theorem 3. We will need Lemmas 11 and 12 below.
Lemma 11. The maps Y → R defined by
Z
f 7→
f (x)f (y)ψ(|x − y|) dx dy,
C2
Z
f 7→
f (x) dx
C
are continuous.
Proof. Let {fn } in Y converge to f ∈ Y. From Assumption
1, (x, y) 7→ ψ(|x − y|) is
R
in Lq (C 2 ). By Jensen’s inequality, it follows that x 7→ C ψ(|x − y|) dy is in Lq (C),
since
q
Z Z
Z
ψ(|x − y|) dy
dx ≤
ψ(|x − y|)q dx dy < ∞.
2
C
C
C
R
By boundedness of f , x 7→ C f (y)ψ(|x − y|) dy is also in Lq (C) and thus
Z
Z
lim
[fn (x) − f (x)]
f (y)ψ(|x − y|) dy dx = 0.
(5.6)
n→∞
C
C
Notice also that since (x, y) 7→ ψ(|x − y|) ∈ Lq (C 2 ), y 7→ ψ(|x − y|) is in Lq (C). So
since ψ is integrable and fn , f are uniformly bounded, by dominated convergence
Z Z
lim
[fn (y) − f (y)]ψ(|x − y|) dy dx = 0.
n→∞
C
C
Thus, using uniform boundedness of fn again,
Z
Z
lim
fn (x)
[fn (y) − f (y)]ψ(|x − y|) dy dx = 0.
n→∞
C
C
(5.7)
LONG RANGE LATTICE GAS MODELS
9
Combining (5.6) and (5.7) yields
Z
lim
[fn (x)fn (y) − f (x)f (y)]ψ(|x − y|) dx dy = 0.
n→∞
C2
Continuity of the other map is clear, so the proof is complete.
Next we prove exponential equivalences for the sums defining Nn (η) and En (η).
Lemma 12. For any > 0,
lim sup n
n→∞
−d
!
Z
−d X
ηn
ηn (I) − f (x) dx ≥ = −∞
log P n
C
(5.8)
I∈Λn
and

X
−d

ηn (I)ηn (J)φn (I, J)
lim sup n log P n−2d
n→∞
I,J∈Λn
(5.9)
Z
−
f ηn (x)f ηn (y)ψ(|x − y|) dx dy ≥ = −∞.
C2
Proof. By the definitions of ηn and f νn ,
Z
X
n−d
ηn (I) =
f ηn (x) dx,
C
I∈Λn
which implies (5.8). Define
φI,J
n =
Z
ψ(|x − y|) dx dy
CI ×CJ
and observe that
Z
−2d X
νn
νn
n
η
(I)η
(J)φ
(I,
J)
−
f
(x)f
(y)ψ(|x
−
y|)
dx
dy
n
n
n
C2
I,J∈Λn
X
= ηn (I)ηn (J) n−2d φn (I, J) − φI,J
n
I,J∈Λn
X n−2d φn (I, J) − φI,J
.
≤
n
(5.10)
I,J∈Λn
Using Riemann integrability of (x, y) 7→ ψ(|x − y|), it is easy to see that the last
expression in (5.10) is less than for sufficiently large n. This implies (5.9).
Proof of Theorem 3. This is an immediate consequence of the contraction principle
(see Theorem 4.2.1 of [13]) along with Theorem 2, Lemma 11, and Lemma 12. Now we are ready to prove Theorem 4.
Proof of Theorem 4. Since Y is compact and
Z
f ∈ Y 7→
f (x)f (y)ψ(|x − y|) dx dy,
C2
Z
f ∈ Y 7→
f (x) dx
C
are continuous, Yξ,ρ is compact. Thus, optimizers of (3.2) exist when (ξ, ρ) ∈ Ω.
Suppose now that f∗ ∈ F is an optimizer of (3.2) for some (ξ, ρ). For the remainder
of the proof we will equip Y with the topology induced by the uniform norm. Thus,
10
DAVID ARISTOFF AND LINGJIONG ZHU
f∗ is in the interior of Y. To obtain the Euler-Lagrange equations (3.3) we follow
Theorem 9.1 of [11]. The multiplier rule there states that there exist β, µ ∈ R and
ν ∈ {0, 1} such that (β, µ, ν) 6= (0, 0, 0) and for f = f∗ and all δf ∈ L∞ (C),
Z Z
0=β
f (y)ψ(|x − y|) dy δf (x) dx
C
C
(5.11)
Z
Z
0
+ µ δf (x) dx − ν Hbin
(f (x))δf (x) dx,
C
C
provided the Frechét derivatives in (5.11) are continuous for f ∈ F. Continuity of
the second Frechét derivative is obvious. Continuity of the first Frechét derivative
follows from integrability of (x, y) 7→ ψ(|x − y|), and continuity of the third Frechét
0
derivative follows from uniform continuity of Hbin
on [, 1 − ] for each > 0. Thus,
Z
0
β f∗ (y)ψ(|x − y|) dy + µ − νHbin
(f∗ (x)) = 0
(5.12)
C
for a.e. x. When ν = 1, this is a rearrangement of (3.3). If ν = 0 then β 6= 0 and
Z
f∗ (y)ψ(|x − y|) dy = γ
C
for a.e. x, where γ = −µ/β. Note that
Z
Z
ξ=
f∗ (x)f∗ (y)ψ(|x − y|) dy dx = γ f∗ (x) dx = γρ,
C2
C
so in fact γ = ξ/ρ.
Proof of Corollary 5. Let f satisfy (3.3). Since γ(t) := ψ(|t|) is locally integrable,
Z
(f ∗ γ)(x) ≡
f (y)ψ(|x − y|) dy
C
is continuous. Now continuity of f follows from the Euler-Lagrange equation (3.3).
Proof of Lemma 7. Let 0 < ρ ≤ 1/4 and define
f1 (x) = 1[0,ρ] (x), f2 (x) ≡ ρ, and f3 (x) = 1[0,ρ/2] + 1[1/2−ρ/2,1/2] .
R
Then ρ = [0,1] fi (x) dx for i = 1, 2, 3, and
2ρ2−r
,
(1 − r)(2 − r)
4r−1 ρ2
M ρ2
ξ(f2 ) = λρ2 = 2
+
1−r
2
2−r
4 ρ2
M ρ2
ξ(f3 ) =
+
.
(1 − r)(2 − r)
2
ξ(f1 ) =
(5.13)
Observe that when ρ = 1/4,
4r−5/2
M ρ2
43r/2−2
M ρ2
+
<
+
= ξ(f3 ),
(5.14)
1−r
2
(1 − r)(2 − r)
2
where the inequality can be checked by straightforward calculus. If > 0 is sufficiently small, ξ(f2 ) < ξ(f3 ) whenever ρ ∈ (1/4 − , 1/4). Moreover, when M
is sufficiently large, ξ(f1 ) < ξ(f2 ). All values of ξ between ξ(f1 ) and ξ(f3 ) are
attainable by, for example, taking convex combinations of f1 and f3 .
ξ(f2 ) =
LONG RANGE LATTICE GAS MODELS
11
To see that Lemma 7 holds for a smoothed function of ψ, let γ be a bounded
function supported on (1/4 − δ, 1/4 + δ) such that γ + ψ is smooth. Then for
sufficiently small δ > 0, the arguments above still go through.
Proof of Theorem 8. Let ψ be as in Lemma 7. Take ρ ∈ (1/4 − , 1/4), and let f
be an optimizer of (3.2) at (ξ, ρ) ∈ int(Ω). We can write f (x) = ρ + δf (x), where
Z
δf (x) dx = 0.
(5.15)
[0,1]
Observe that
Z
H(f ) − H(ρ) =
Hbin (ρ + δf (x)) dx − Hbin (ρ)
[0,1]
Z
0
[Hbin (ρ + δf (x)) − Hbin
(ρ)δf (x) − Hbin (ρ)] dx
=
[0,1]
Z
≥c
δf (x)2 dx,
[0,1]
where by convexity,
c=
0
Hbin (ρ + t) − Hbin
(ρ)t − Hbin (ρ)
> 0.
t2
t∈[−ρ,1−ρ]\{0}
min
Note that H(ρ) ≤ H(f ) with equality if and only if f ≡ ρ a.e. It follows that the
optimizer of (3.2) at (λρ2 , ρ) is the constant function with value ρ. Thus,
δS := S(ξ, ρ) − S(λρ2 , ρ) = H(ρ) − H(f )
Z
≤ −c
δf (x)2 dx.
[0,1]
Now note that
δξ := ξ(f ) − ξ(ρ)
Z
Z
=
(ρ + δf (x))(ρ + δf (y))ψ(|x − y|) dx dy −
ρ2 ψ(|x − y|) dx dy
[0,1]2
[0,1]2
Z
Z
= 2ρ
δf (x)ψ(|x − y|) dx dy +
δf (x)δf (y)ψ(|x − y|) dx dy
[0,1]2
[0,1]2
Z
=
δf (x)δf (y)ψ(|x − y|) dx dy,
[0,1]2
(5.16)
with the last equality coming from (5.15) and the fact that for each x ∈ [0, 1],
Z
ψ(|x − y|) dy = λ.
[0,1]
Since r < 1/2, the integral kernel Ψ defined by
Z
Ψf (y) =
f (x)ψ(|x − y|) dx
[0,1]
is a Hilbert-Schmidt operator on L2 [0, 1]. Thus,
Z
Z
δf (x)δf (y)ψ(|x − y|) dx dy ≤ σ
δf (x)2 dx,
[0,1]2
[0,1]
12
DAVID ARISTOFF AND LINGJIONG ZHU
where σ is the spectral radius of Ψ. Putting this in (5.16) yields
Z
|δξ| ≤ σ
δf (x)2 dx.
[0,1]
Combining the estimates for δS and δξ, we get
c
δS ≤ − |δξ|.
σ
Thus, S is not differentiable at (λρ2 , ρ) ∈ Ω.
Acknowledgements
The authors would like to thank R. Mark Bradley, Olivier Pinaud, Dan Pirjol, Charles Radin and Clayton Shonkwiler for helpful comments and suggestions.
D. Aristoff gratefully acknowledges support from the National Science Foundation
via the award NSF-DMS-1522398.
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Department of Mathematics
Colorado State University
1874 Campus Delivery
Fort Collins, CO-80523
United States of America
E-mail address:
aristoff@math.colostate.edu
Department of Mathematics
Florida State University
1017 Academic Way
Tallahassee, FL-32306
United States of America
E-mail address:
zhu@math.fsu.edu