The Parisi formula has a unique minimizer
Antonio Auffinger
∗
Wei-Kuo Chen
University of Chicago
University of Chicago
arXiv:1402.5132v2 [math.PR] 27 Feb 2014
†
February 28, 2014
Abstract
In 1979, G. Parisi [14] predicted a variational formula for the thermodynamic limit
of the free energy in the Sherrington-Kirkpatrick model and described the role played
by its minimizer. This formula was verified in the seminal work of Talagrand [19] and
later generalized to the mixed p-spin models by Panchenko [12]. In this paper, we prove
that the minimizer in Parisi’s formula is unique at any temperature and external field
by establishing the strict convexity of the Parisi functional.
1
Introduction and main results
The Sherrington-Kirkpatrick (SK) model was introduced in [16]. For any N ≥ 1, its Hamiltonian at (inverse) temperature β > 0 and external field h ∈ R is given by
N
N
X
1 X
√
gij σi σj + h
σi
N i,j=1
i=1
for σ = (σ1 , . . . , σN ) ∈ ΣN := {−1, +1}N , where gij ’s are independent standard Gaussian
random variables. It is arguably the most well-known model of disordered mean field spin
glasses. Over the past few decades, its study has generated hundreds of papers in both
theoretical physics and mathematics communities. We refer readers to the book of MézardParisi-Virasoro [9] for physics’ methodologies and predictions and the books of Talagrand [17]
and Panchenko [13] for its recent rigorous treatments.
This paper is concerned with a generalization of the SK model, the so-called mixed p-spin
model, which corresponds to the Hamiltonian
HN (σ) = HN′ (σ) + h
N
X
i=1
∗
†
[email protected]
[email protected]
1
σi
(1)
for σ ∈ ΣN , where
HN′ (σ)
∞
X
=
βp HN,p (σ)
p=2
is the linear combination of the pure p-spin Hamiltonian,
HN,p (σ) =
1
N (p−1)/2
N
X
gi1 ,...,ip σi1 . . . σip .
(2)
i1 ,...,ip =1
Here gi1 ,...,ip ’s are independent standard Gaussian random variables for all p ≥ 2 and all
(i1 , . . . , ip ). The nonnegative real sequence (βp )p≥2 is called the temperature parameters and
h ∈ R denotes the strength of the external field. We P
assume that βp > 0 for at least one
p 2
p ≥ 2 and (βp )p≥2 decreases fast enough, for instance, ∞
p=2 2 βp < ∞. The SK model can
be recovered by choosing βp = 0 for all p ≥ 3. A direct computation gives
EHN′ (σ1 )HN′ (σ 2 ) = Nξ(R1,2 ),
where R1,2 := N −1
PN
i=1
σi1 σi2 is the overlap between spin configurations σ 1 and σ 2 and
ξ(s) :=
∞
X
p=2
βp2 sp , ∀s ∈ [0, 1].
(3)
Define the Gibbs measure as GN (σ) = ZN−1 exp(−HN (σ)) for σ ∈ ΣN , where the normalizing
factor ZN is known as the partition function.
Let M be the collection of all probability measures on [0, 1] endowed with the metric
R1
d(µ, µ′) := 0 |µ([0, s]) − µ′ ([0, s])|ds. Denote by Md the collection of all atomic measures
from M. For µ ∈ M, let Φµ be the solution to the Parisi PDE,
∂s Φµ (s, x) = −
ξ ′′ (s)
∂xx Φµ (s, x) + µ([0, s]) (∂x Φµ (s, x))2 , (s, x) ∈ [0, 1] × R
2
(4)
with terminal condition Φµ (1, x) = log cosh x. In the case µ ∈ Md , this PDE can be solved
explicitly by performing the Hopf-Cole transformation. Indeed, suppose that µ has exactly
k + 1 jumps, k ≥ 0, at positions (ql )1≤l≤k+1 satisfying µ([0, ql ]) = ml for 1 ≤ l ≤ k + 1. The
sequences (ml )1≤l≤k+1 and (ql )1≤l≤k+1 satisfy
0 ≤ m1 < m2 < · · · < mk < mk+1 = 1,
0 ≤ q1 < q2 < · · · < qk < qk+1 ≤ 1.
(5)
Set m0 = q0 = 0 and qk+2 = 1. Then for qk+1 ≤ s ≤ qk+2 ,
1
Φµ (s, x) = log cosh x + (ξ ′ (1) − ξ ′ (s))
2
(6)
and for 0 ≤ l ≤ k, one can solve decreasingly,
Φµ (s, x) =
p
1
log E exp ml Φµ (ql+1 , x + z ξ ′ (ql+1 ) − ξ ′ (s))
ml
2
(7)
for ql ≤ s < ql+1 , where z is a standard Gaussian random variable. It is well-known (see
Guerra [7]) that µ 7→ Φµ defines a Lipschitz functional from (Md , d) to (C([0, 1] ×R), k · k∞ ).
This guarantees the existence of the Parisi PDE solution for arbitrary µ ∈ M. The Parisi
functional is defined as
Pξ,h (µ) = Φµ (0, h)
(8)
for µ ∈ M. With these notations, the thermodynamic limit of the free energy can now be
computed through
Theorem 1 (Parisi formula [19, 12]). We have almost surely,
Z
1 1 ′′
1
log ZN = min log 2 + Pξ,h (µ) −
sξ (s)µ([0, s])ds .
lim
µ∈M
N →∞ N
2 0
(9)
This formula was predicted in the ground-breaking work of Parisi [14, 15] in the setting
of the SK model. It was proved and generalized by Talagrand [19] to the mixed even p-spin
models after the celebrated discovery of the replica symmetric breaking bound by Guerra [7].
Later Panchenko [12] verified its validity in the mixed p-spin models including odd p. We
shall call a minimizer of (9) a Parisi measure throughout this paper. Parisi’s prediction goes
beyond the variational formula. In his picture, the Parisi measure is unique. It also describes
the limiting distribution of the overlap R1,2 under EG⊗2
N and encodes all information of the
model. Mathematically, uniqueness of the Parisi measure was only known in the generic case,
that is, when βp > 0 for all p ≥ 2 (see Theorem 1.2 (c) in Talagrand [18]) and in the spherical
version of the present model [20].
As can be seen from (9), the third term on the right-hand side is linear in µ. Therefore
the proof of the uniqueness of the Parisi measure is related to Talagrand’s conjecture [18, 19]
that the functional Pξ,h is strictly convex. The first partial result along this direction was presented in Panchenko [10] where he established convexity between measures that stochastically
dominate each other; his result was later pushed forward slightly by Chen [5] using a PDE
approach. In this paper, based on a variational representation for the Parisi PDE solution
motivated by the works of Boué-Dupuis [3] and Borell [2], we prove Talagrand’s conjecture:
Theorem 2. For any ξ and h, the Parisi functional Pξ,h is strictly convex.
This directly implies
Corollary 1. For any ξ and h, there exists a unique Parisi measure.
As an immediate consequence of Corollary 1, we remark that one can identify the high
temperature regime of the model as the collection of all (βp )p≥2 and h ∈ R such that the
corresponding Parisi measure is a Dirac measure. We refer the reader to Theorem 13.4.1 in
Talagrand [17] for the characterization of the high temperature regime in the SK model. For
physicists’ predictions and rigorous qualitative properties about the Parisi measure, one may
consult authors’ recent work [1]. We also remark that the proof of Theorem 2 and Corollary
1 can be extended to other models that share a similar characterization of the limiting free
energy. These include, for instance, the Ghatak-Sherrington model [6, 11] and the SK with
multidimensional spins [4]. We do not pursue this direction here.
3
The rest of the paper is organized as follows. In Section 2, we establish a variational
representation for the Parisi PDE solution and give an expression for the optimizer as well as
a criterion for its uniqueness. Using these results, we will establish a general strict convexity
for the Parisi PDE solution in Section 3 and conclude immediately Theorem 2.
Acknowledgements. The second author thanks Dmitry Panchenko for introducing the
problem of the strictly convexity of the Parisi functional to him. We are indebted to LouisPierre Arguin for fruitful discussions during the early stage of the project and to CIRM at
Luminy for the hospitality during the workshop on disordered systems in 2013. Special thanks
are due to Gregory Lawler for explaining the representation formula in Boué-Dupuis [3] to
us.
2
A variational representation
Recall ξ from (3). Define ζ = ξ ′′ . Let (B(r))r≥0 be a standard Brownian motion and P denote
the Wiener measure. For 0 ≤ s ≤ t ≤ 1, we denote by D[s, t] the space of all progressively
measurable processes u on [s, t], with respect to the filtration generated by (B(r))r≥0 , that
satisfy sups≤r≤t |u(r)| ≤ 1. We endow D[s, t] with the norm
Z t
1/2
2
kuk = E
u(r) dr
.
s
The main result of this section is the following characterization.
Theorem 3 (Variational formula). Let µ ∈ M and α be its distribution function. Suppose
that Φ is the Parisi PDE solution corresponding to µ. Let 0 ≤ s < t ≤ 1. For any x ∈ R and
u ∈ D[s, t], define
F s,t(u, x) = E C s,t (u, x) − Ls,t (u) ,
(10)
where
s,t
Z
t
C (u, x) = Φ t, x +
α(r)ζ(r)u(r)dr +
s
Z
1 t
s,t
α(r)ζ(r)u(r)2dr.
L (u) =
2 s
Z
t
ζ(r)
1/2
dB(r) ,
s
(11)
Then we have that
(i)
Φ(s, x) = max F s,t (u, x)|u ∈ D[s, t] .
(12)
u∗ (r) = ∂x Φ(r, X(r)),
(13)
(ii) The maximum in (12) is attained by
where (X(r))s≤r≤t is the strong solution to
dX(r) = α(r)ζ(r)∂xΦ(r, X(r))dr + ζ(r)1/2 dB(r),
X(s) = x.
4
(14)
Remark 1. It is known that |∂xx Φ| ≤ 1 (see (17)), which gives for any s, y1 , y2 ,
|α(s)ζ(s)∂xΦ(s, y1 ) − α(s)ζ(s)∂x Φ(s, y2 )| ≤ ζ(1)|y1 − y2 |.
Therefore, [8, Proposition 2.13] ensures the existence of the strong solution (X(r))s≤r≤t for
any µ.
Before we turn to the proof of Theorem 3, we summarize some properties about the Parisi
PDE in the following proposition.
Proposition 1. Let µ ∈ M. Denote by α the distribution function and by Φ the Parisi PDE
solution associated to µ. Then
(i) For 0 ≤ j ≤ 4,
∂xj Φ exists and is continuous.
(15)
|∂x Φ(s, x)| ≤ 1,
C
≤ ∂xx Φ(s, x) ≤ 1,
cosh2 x
|∂x3 Φ(s, x)| ≤ 4,
(16)
(ii) For all (s, x) ∈ [0, 1] × R,
(17)
(18)
where C > 0 is a constant depending only on ξ.
(iii) If α is continuous on [0, 1], then
Φ, ∂x Φ, ∂xx Φ ∈ C 1,2 ,
(19)
where C 1,2 is the space of all functions f on [0, 1] × R with continuous ∂s f and ∂xx f.
(iv) Suppose that (µn )n≥1 ∈ M converges weakly to µ and Φn is the Parisi PDE solution
associated to µn . For 0 ≤ j ≤ 2, uniformly on [0, 1] × R,
lim ∂xj Φn = ∂xj Φ.
n→∞
(20)
Proof. Statements (15), (16), (17) and (20) are parts of the results of Proposition 1 and 2 in
[1]. As for (18), it follows from (14.272) in [17] and (20). For (19), note that the continuity
of α gives
∂s Φ(s, x) = −
ζ(s)
(∂xx Φ(s, x) + α(s)(∂x Φ(s, x))2 )
2
for all (s, x) ∈ [0, 1] × R. This and (15) give (19).
The first step to prove Theorem 3 is the following lemma about atomic measures.
5
Lemma 1. Let µ ∈ Md and α be its distribution function. Let s, t ∈ [0, 1] with s < t be both
jump points of µ. Using the notations of Theorem 3, we have
Φ(s, x) ≥ sup{F s,t(u, x)|u ∈ D[s, t]}.
(21)
Proof. Suppose that µ has exactly k + 1 jumps at (ql )1≤l≤k+1 with µ([0, ql ]) = ml where
(ql )1≤l≤k+1, and (ml )1≤l≤k+1 satisfy (5) and that for some 1 ≤ a < b ≤ k + 1 we have
qa = s, qb = t.
Set m0 = q0 = 0 and qk+2 = 1. As we have discussed in the Section 1, Φ can be solved
through (6) and (7). Using the standard Brownian motion (B(r))r≥0 , for each a ≤ l ≤ b − 1,
we can write
Z ql+1
1
1/2
ζ(r) dB(r) .
log E exp ml Φ ql+1 , x +
Φ(ql , x) =
ml
ql
Let u ∈ D[s, t]. Set
Z
Z ql+1
1 ql+1 2
2
1/2
Zl = exp −
ml u(r) ζ(r)dr −
ml u(r)ζ(r) dB(r) .
2 ql
ql
Rr
Define dP̃ = Zl dP and B̃(r) = ql ml u(a)ζ(a)1/2 da + B(r). We use Ẽ to denote the expectation
with respect to P̃. The Girsanov theorem [8, Theorem 5.1] says
Z ql+1
1/2
E exp ml Φ ql+1 , x +
ζ(r) dB(r)
ql
Z ql+1
1/2
ζ(r) dB̃(r)
= Ẽ exp ml Φ ql+1 , x +
ql
Z ql+1
Z ql+1
1/2
= E exp ml Φ ql+1 , x +
ml u(r)ζ(r)dr +
ζ(r) dB(r)
ql
ql
Z ql+1
Z
1 ql+1 2
1/2
2
ml u(r)ζ(r) dB(r) .
ml u(r) ζ(r)dr −
· exp −
2 ql
ql
From Jensen’s inequality m−1 log E exp mA ≥ EA for any random variable A and m > 0 and
noting that ml = α(r) for ql ≤ r < ql+1 , it follows
Z ql+1
Z ql+1
1/2
Φ(ql , x) ≥ E Φ ql+1 , x +
α(r)u(r)ζ(r)dr +
ζ(r) dB(r)
ql
ql
Z
1 ql+1
2
α(r)u(r) ζ(r)dr
−
2 ql
6
for all a ≤ l ≤ b − 1. Using this, an iteration argument on l from a to b − 1 gives
Φ(s, x) = Φ(qa , x)
"
≥ E Φ qb , x +
−
qa ,qb
=F
(u, x)
s,t
= F (u, x).
b−1 Z
X
1
2
ql+1
α(r)u(r)ζ(r)dr +
l=a ql
b−1 Z ql+1
X
l=a
b−1 Z
X
l=a
α(r)u(r)2ζ(r)dr
ql
ql+1
ζ(r)1/2 dB(r)
ql
!
#
Since this is true for arbitrary u ∈ D[s, t], this finishes our proof.
Proof of Theorem 3. First, we claim that for any µ ∈ M,
Φ(s, x) ≥ sup F s,t(u, x)|u ∈ D[s, t] .
(22)
Pick a sequence (µn )n≥1 ∈ Md that converges weakly to µ and have jumps at s and t. Denote
by αn the distribution function and by Φn the Parisi PDE solution associated to µn . Since
(αn )n≥1 converges almost everywhere to α, the uniform boundedness of u ∈ D[s, t] and the
dominated convergence theorem give
Z t
Z t
αn (r)ζ(r)u(r)dr →
α(r)ζ(r)u(r)dr,
s
s
Z t
Z t
2
αn (r)ζ(r)u(r) dr →
α(r)ζ(r)u(r)2dr.
s
s
From (20), it implies that the sequence of functionals (Fns,t )n≥1 associated to (αn )n≥1 converges
uniformly to F s,t and therefore Lemma 1 gives (22).
With the help of this claim, our proof will be finished if we show that u∗ is a maximizer
and equality of (22) holds. We check the case that α is continuous first. Define for s ≤ r ≤ t,
Z
Z r
1 r
∗
2
α(v)ζ(v)u (v) dv −
u∗ (v)ζ(v)1/2 dB(v).
Y (r) = Φ(r, X(r)) −
2 s
s
Since Φ ∈ C 1,2 by (19), we obtain from Itô’s formula [8, Theorem 3.6] and (4),
1
dΦ = (∂s Φ)dr + (∂x Φ)dX + (∂xx Φ)dr
2
1
2
= ∂s Φ + αζ (∂x Φ) + ζ(∂xx Φ) dr + ζ 1/2 (∂x Φ)dB
2
1
= αζ (∂x Φ)2 dr + ζ 1/2 (∂x Φ)dB
2
1
= αζ(u∗)2 dr + ζ 1/2 u∗ dB,
2
7
which implies
1
dY = dΦ − αζ(u∗)2 dr − ζ 1/2 u∗ dB = 0.
2
(23)
Since
Y (s) = Φ(s, X(s)) = Φ(s, x)
and
1
Y (t) = Φ(t, X(t)) −
2
Z
s
t
∗
2
α(v)ζ(v)u (v) dv −
Z
t
u∗ (v)ζ(v)1/2 dB(v),
s
if we take expectation in the equation above, it follows that from (23),
Z t
s,t ∗
F (u , x) = EY (t) = EY (s) + E
dY (r) = Φ(s, x).
(24)
s
This means that u∗ is a maximizer. As for arbitrary α, let us pick a sequence of probability measures (µn )n≥1 ⊂ M such that the distribution function of each µn is continuous and
(µn )n≥1 has weak limit µ. Denote by (αn )n≥1 , (Φn )n≥1 , (Xn )n≥1 and (u∗n )n≥1 the distribution functions, the Parisi PDE solutions, the SDE solutions (14) and the maximizers (13)
associated to (αn )n≥1 . Write
|Xn (r) − X(r)|
Z r
Z r
= αn (v)ζ(v)∂x Φn (v, Xn (v))dv −
α(v)ζ(v)∂xΦ(v, X(v))dv s
Z rs
≤
ζ(v)|αn (v) − α(v)||∂x Φn (v, X(v))| + α(v)ζ(v)|∂xΦn (v, Xn (v)) − ∂x Φ(v, X(v))|dv
s
Z r
≤C
|αn (v) − α(v)||∂x Φn (v, X(v))| + |∂x Φn (v, Xn (v)) − ∂x Φ(v, X(v))|dv
s
for some C > 0, where in the last line we used the fact that ζ is bounded above by ζ(1)
and α ≤ 1. From (16) and the almost everywhere convergence of (αn )n≥1 , the dominated
convergence theorem tells us that the first term in the last line converges to zero. As for the
second term, the mean value theorem, (17) and (20) imply that for given ε > 0, if n is large
enough,
Z r
α(v)|∂x Φn (v, Xn (v)) − ∂x Φ(v, X(v))|dv
s
Z r
Z r
≤
|∂x Φ(v, Xn (v)) − ∂x Φn (v, Xn (v))|dv +
|∂x Φ(v, X(v)) − ∂x Φ(v, Xn (v))|dv
s
s
Z r
≤ε+
|Xn (v) − X(v)|dv
s
for all s ≤ r ≤ t. Apply the Gronwall inequality, we conclude that for large enough n,
|Xn (r) − X(r)| ≤ εeCr for all s ≤ r ≤ t. Thus,
lim sup |Xn (r) − X(r)| = 0.
n→∞ s≤r≤t
8
and so from (20), (u∗n )n≥1 converges to u∗ uniformly on [s, t]. This combining with (20) and
the almost everywhere convergence of (αn )n≥1 to α implies that (24) is also true for µ. In
other words, u∗ is a maximizer. This ends our proof.
Proposition 2 (Uniqueness of u∗ ). Let µ ∈ M and α be its distribution function. Let
0<
R ts < t ≤ 1 with α(s) > 0. Suppose that Φ is the Parisi PDE solution corresponding to µ.
If s α(r)ζ(r)dr < 1, then the maximizer u∗ given by (13) for the variational representation
(12) is unique.
Rt
Proof. Suppose that s α(r)ζ(r)dr < 1. It suffices to prove that F s,t(·, x) defines a strictly
concave functional on D[s, t]. Let u0 , u1 ∈ D[s, t] with u0 6= u1 . This implies that
Z t
2
ku0 − u1 k = E
|u0 (r) − u1 (r)|dr > 0.
(25)
s
Note that u0 and u1 are uniformly bounded above by one. Define uλ = λu0 + (1 − λ)u1 for
λ ∈ [0, 1]. A direct computation using the dominated convergence theorem gives
"
Z
2 #
t
∂λλ F s,t (uλ, x) = E C s,t (uλ, x)
s
−E
Z
α(r)ζ(r)(u0(r) − u1 (r))dr
t
s
2
α(r)ζ(r)(u0(r) − u1 (r)) dr .
Note that from the Cauchy-Schwarz inequality,
Z
t
s
α(r)ζ(r)(u0(r) − u1 (r))dr
2
≤
Z
t
s
α(r)ζ(r)dr ·
Z
t
s
α(r)ζ(r)(u0(r) − u1 (r))2 dr.
Using this and (16), it follows that
∂λλ F s,t(uλ , x)
"Z
#
2 Z t
t
≤E
α(r)ζ(r)(u0(r) − u1 (r))dr −
α(r)ζ(r)(u0(r) − u1 (r))2 dr
s
Z
t
s
Z
t
2
Z
t
2
≤E
α(r)ζ(r)dr ·
α(r)ζ(r)(u0(r) − u1 (r)) dr −
α(r)ζ(r)(u0(r) − u1 (r)) dr
s
s
s
Z t
Z t
2
=
α(r)ζ(r)dr − 1 E
α(r)ζ(r)(u0(r) − u1 (r)) dr .
s
s
Note that since α is nondecreasing and α(s) > 0, we have α(r) > 0 for
R t all s ≤ r ≤ t. Also
note that ζ(r) > 0 for all 0 < r < t. Consequently, using (25) and s α(r)ζ(r)dr < 1, we
conclude that ∂λλ F s,t(uλ , x) < 0 and this gives the strict concavity of F s,t(·, x).
9
3
Strict convexity of the Parisi PDE solution
Suppose that µ0 , µ1 ∈ M and x0 , x1 ∈ R. For λ ∈ [0, 1], we set
µλ = λµ0 + (1 − λ)µ1 ,
xλ = λx0 + (1 − λ)x1 .
Denote by α0 , αλ , α1 the distribution functions and by Φ0 , Φλ , Φ1 the Parisi PDE solutions
corresponding to µ0 , µλ, µ1 , respectively. Let τ be the last time that α0 and α1 are different,
that is,
τ = min {s ∈ [0, 1] : α0 (r) = α1 (r), ∀r ∈ [s, 1]} .
Note that since α0 (1) = α1 (1) = 1 and α0 , α1 are right continuous, τ is well-defined and that
if µ0 6= µ1 , then τ > 0. The following general result immediately implies Theorem 2 by letting
s = 0 and x0 = x1 = h in (ii) below.
Theorem 4. We have that
(i) For any µ0 , µ1 ∈ M,
Φλ (s, xλ ) ≤ λΦ0 (s, x0 ) + (1 − λ)Φ1 (s, x1 )
(26)
for all λ, s ∈ [0, 1] and x0 , x1 ∈ R.
(ii) Suppose that µ0 , µ1 are distinct. Then for any 0 ≤ r < τ, the inequality (26) is strict
for all λ ∈ (0, 1) and x0 , x1 ∈ R.
As one shall see, while the statement (26) follows directly from our representation theorem,
the proof for the strict inequality of (26) is more delicate and is based on subtle properties of
the maximizers that we summarize as follows.
Lemma 2. Let µ ∈ M. Denote by α the distribution function and by Φ the Parisi PDE
solution corresponding to µ. Let 0 ≤ s ≤ t ≤ 1 and x ∈ R. Suppose that (X(r))s≤r≤t satisfy
dX(r) = α(r)ζ(r)∂xΦ(r, X(r))dr + ζ(r)1/2 dB(r),
X(s) = x.
(27)
Then for any s ≤ a ≤ b ≤ t, we have
∂x Φ(b, X(b)) − ∂x Φ(a, X(a)) =
Z
b
ζ(r)1/2 ∂xx Φ(r, X(r))dB(r)
(28)
a
and
∂xx Φ(b, X(b)) − ∂xx Φ(a, X(a))
Z b
Z b
2
=−
α(r)ζ(r)(∂xxΦ(r, X(r))) dr +
ζ(r)1/2 ∂x3 Φ(r, X(r))dB(r),
a
a
where the last Itô’s integral is well-defined by (18).
10
(29)
Proof. By an approximation argument as we did in the proof of Theorem 3, it suffices to assume that α is continuous on [0, 1]. From (19), this assumption ensures that Φ, Φx , Φxx ∈ C 1,2 .
Recall that Φ satisfies
ζ
∂s Φ = − (∂xx Φ + α(∂x Φ)2 ).
2
A direct computation yields
1
∂xs Φ = − ζ(∂x3 Φ) − αζ(∂xx Φ)(∂x Φ)
2
1
∂xxs Φ = − ζ(∂x4 Φ) − αζ(∂x3 Φ)(∂x Φ) − αζ(∂xx Φ)2 .
2
Now, using Itô’s formula [8, Theorem 3.6] and these two equations,
1
d(∂x Φ) = (∂xs Φ)dr + (∂xx Φ)dX + ζ(∂x3 Φ)dr
2
1
= − ζ(∂x3 Φ) − αζ(∂xx Φ)(∂x Φ) dr
2
1
+ (∂xx Φ) αζ(∂x Φ)dr + ζ 1/2 dB + ζ(∂x3 Φ)dr
2
1/2
= ζ (∂xx Φ)dB
and
1
d(∂xx Φ) = (∂xxs Φ)dr + (∂x3 Φ)dX + ζ(∂x4 Φ)dr
2
1
2
= − ζ(∂x4 Φ) − αζ(∂x3 Φ)(∂x Φ) − αζ(∂xx Φ) dr
2
1
+ (∂x3 Φ) αζ(∂x Φ)dr + ζ 1/2 dB + ζ(∂x4 Φ)dr
2
2
1/2
= −αζ(∂xx Φ) dr + ζ (∂x3 Φ)dB.
These two equations complete our proof of Lemma 2.
Proof of Theorem 4. Let µ0 , µ1 ∈ M, x0 , x1 ∈ R, λ ∈ [0, 1], 0 ≤ s ≤ t ≤ 1 and u ∈ D[s, t].
Recall Theorem 3 and set µλ = λµ0 + (1 − λ)µ1 . For θ = 0, λ, 1, denote by
Fθs,t, Cθs,t , Ls,t
θ ,
the functionals defined in the variational formulas corresponding respectively to µθ :
Φθ (s, xθ ) = max{Fθs,t(u, xθ )|D[s, t]}.
To show (i), let 0 ≤ s ≤ 1 and take t = 1. Suppose that u ∈ D[s, 1]. Note that
Z 1
s,1
Lλ (u) =
αλ (r)ζ(r)u(r)2dr
s
Z 1
Z 1
2
=λ
α0 (r)ζ(r)u(r) dr + (1 − λ)
α1 (r)ζ(r)u(r)2dr
=
s
s,1
λL0 (u)
(30)
s
+ (1 −
λ)Ls,1
1 (u).
11
(31)
Write
Z
1
Z
1
αλ (r)ζ(r)u(r)dr +
ζ(r)1/2 dB(r)
s
s
Z 1
Z 1
1/2
= λ x0 +
α0 (r)ζ(r)u(r)dr +
ζ(r) dB(r)
s
s
Z 1
Z 1
1/2
+ (1 − λ) x1 +
α1 (r)ζ(r)u(r)dr +
ζ(r) dB(r) .
xλ +
s
(32)
s
Since Φ0 (1, x) = Φλ (1, x) = Φ1 (1, x) = log cosh x is a convex function, we obtain
Z 1
Z 1
s,1
1/2
Cλ (u, xλ ) = Φλ 1, xλ +
αλ (r)ζ(r)u(r)dr +
ζ(r) dB(r)
s
s
Z 1
Z 1
1/2
≤ λΦλ 1, x0 +
α0 (r)ζ(r)u(r)dr +
ζ(r) dB(r)
s
s
Z 1
Z 1
1/2
+ (1 − λ)Φλ 1, x1 +
α1 (r)ζ(r)u(r)dr +
ζ(r) dB(r)
s
=
λC0s,1 (u, x0 )
+ (1 −
s
λ)C1s,1 (u, x1 ).
Combining with (31) and taking expectation, one has
Fλs,1 (u, xλ ) ≤ λF0s,1 (u, x0 ) + (1 − λ)F1s,1 (u, x1 ).
Since this is true for any u ∈ D[s, 1], the representation formula (30) gives (i).
Next, we turn to the proof of (ii). Suppose that α0 6= α1 and λ ∈ (0, 1). We will show that
first, there exists some τ ′ ∈ (0, τ ) such that (26) is strict for all s ∈ [τ ′ , τ ) and then, prove (26)
is Ralso strict for Rall s ∈ [0, τ ′ ). The way of finding such τ ′ can be argued as follows. Note that
τ
τ
if s α0 (r)dr = s α1 (r)dr = 0 for all s ∈ [0, τ ), then α0 = α1 = 0 on [0, τ ) since α0 and α1
are nondecreasing. This implies that α0 = α1 on [0, τ ], which contradicts
of τ.
R τ the definition
Rτ
′
Therefore, there exists τ < τ such that at least one of the integrals τ ′ α0 (r)dr, τ ′ α1 (r)dr is
nonzero. Without loss of generality this implies that α0 (τ ′ ) > 0. Making τ ′ bigger if necessary
we see that we can choose 0 < τ ′ < τ such that the following statement holds:
Rτ
α0 (τ ′ ) > 0 and τ ′ α0 (r)ζ(r)dr < 1.
(33)
Now we argue by contradiction. Suppose equality in (26) holds for some s ∈ [τ ′ , τ ) and
x0 , x1 . Take t = τ . Let u∗0 , u∗λ , u∗1 be the corresponding maximizers of (30) generated by (13).
As in (31),
s,t ∗
s,t ∗
∗
Ls,t
λ (uλ ) = λL0 (uλ ) + (1 − λ)L1 (uλ ).
Also writing
xλ +
Z
t
s
αλ (r)ζ(r)u∗λ(r)dr
in the same away as (32), (i) gives
+
Z
t
ζ(r)1/2dB(r)
s
Cλs,t (u∗λ, xλ ) ≤ λC0s,t(u∗λ , x0 ) + (1 − λ)C1s,t (u∗λ , x1 ).
12
They together imply that
Fλs,t (u∗λ, xλ ) ≤ λF0s,t(u∗λ , x0 ) + (1 − λ)F1s,t(u∗λ , x1 ).
(34)
Note that
F0s,t (u∗λ, x0 ) ≤ Φ0 (s, x0 ), F1s,t (u∗λ, x1 ) ≤ Φ1 (s, x1 ), Fλs,t(u∗λ , xλ ) = Φλ (s, xλ ).
Consequently, from (34) and the assumption
Φλ (s, xλ ) = λΦ0 (s, x0 ) + (1 − λ)Φ1 (s, x1 ),
we obtain
F0s,t(u∗λ , x0 ) = Φ0 (s, x0 ),
F1s,t(u∗λ , x1 ) = Φ1 (s, x1 ).
In other words, u∗λ realizes the maxima of the representations for Φ0 (s, x0 ) and Φ1 (s, x1 ). Now
from (33) and Proposition 2, we conclude uniqueness of the maximizer for Φ0 (s, x0 ), that is,
u∗0 = u∗λ with respect to the norm k · k. Since u∗0 and u∗λ are continuous on [s, t], we have
∂x Φ0 (r, X0 (r)) = u∗0 (r) = u∗λ (r) = ∂x Φλ (r, Xλ (r))
(35)
for all s ≤ r ≤ t, where (X0 (r))s≤r≤t and (Xλ (r))s≤r≤t satisfy respectively,
dX0 (r) = α0 (r)ζ(r)∂x Φ0 (r, X0 (r))dr + ζ(r)1/2 dB(r),
X0 (s) = x0 ,
dXλ (r) = αλ (r)ζ(r)∂x Φλ (r, Xλ (r))dr + ζ(r)1/2dB(r),
Xλ (s) = xλ .
From (28) and (35),
Z t
ζ(r)1/2 ∂xx Φ0 (r, X0 (r))dB(r) = ∂x Φ0 (t, X0 (t)) − ∂x Φ0 (s, X0 (s))
s
= ∂x Φλ (t, Xλ (t)) − ∂x Φλ (s, Xλ (s))
Z t
=
ζ(r)1/2 ∂xx Φλ (r, Xλ (r))dB(r).
s
This gives by Itô’s isometry,
Z t
2
ζ(r)E ∂xx Φ0 (r, X0 (r)) − ∂xx Φλ (r, Xλ (r)) dr
s
Z t
2
1/2
=E
ζ(r)
∂xx Φ0 (r, X0 (r)) − ∂xx Φλ (r, Xλ (r)) dB(r) = 0
s
and therefore, by the continuity of ∂xx Φ0 (·, X0(·)) and ∂xx Φλ (·, Xλ (·)) on [s, t], we obtain
∂xx Φ0 (r, X0 (r)) = ∂xx Φλ (r, Xλ (r)), ∀r ∈ [s, t].
13
(36)
Next we use (29) and (36) to get that for all a, b satisfying s ≤ a ≤ b ≤ t,
Z b
−
α0 (r)ζ(r)E(∂xxΦ0 (r, X0 (r)))2 dr
a
Z b
Z b
2
1/2
=E −
α0 (r)ζ(r)(∂xx Φ0 (r, X0 (r))) dr +
ζ(r) ∂x3 Φ0 (r, X0 (r))dB(r)
a
a
= E ∂xx Φ0 (b, X0 (b)) − ∂xx Φ0 (a, X0 (a))
= E ∂xx Φλ (b, Xλ (b)) − ∂xx Φλ (a, Xλ (a))
Z b
Z b
2
1/2
=E −
αλ (r)ζ(r)(∂xx Φλ (r, Xλ (r))) dr +
ζ(r) ∂x3 Φλ (r, Xλ (r))dB(r)
a
a
Z b
=−
αλ (r)ζ(r)E(∂xx Φλ (r, Xλ (r)))2dr.
a
Note that by (17), ∂xx Φ0 (r, X0 (r)) and ∂xx Φλ (r, Xλ (r)) are positive continuous functions on
[s, t] and that ζ > 0 on [s, t]. Let A0 and Aλ be the sets of all points of continuity of α0 and
αλ in [s, t], respectively. We then deduce from the fundamental theorem of calculus and (36)
that α0 (r) = αλ (r) for all r ∈ A0 ∩ Aλ . Using αλ = λα0 + (1 − λ)α1 and the assumption
λ ∈ (0, 1), we get α0 (r) = α1 (r) for all r ∈ A0 ∩ Aλ . Since Ac0 and Acλ are at most countable,
by the right continuity of α0 and α1 and noting α0 (s) = αλ (s), we reach α0 = α1 on [s, t],
which contradicts the definition of τ. Thus, the inequality (26) must be strict for all s ∈ [τ ′ , τ )
and x0 , x1 . This finishes our first part of the argument.
In the second part, we will prove that (26) is also strict for all s ∈ [0, τ ′ ) and x0 , x1 .
Let s ∈ [0, τ ′ ) and x0 , x1 ∈ R. Take t = τ ′ . Recall the representation formula for Φλ (s, u∗λ)
from (30). We observe that from the first part of our proof for (ii), for any y0 , y1 ∈ R and
yλ = λy0 + (1 − λ)y1 ,
Φλ (t, yλ ) < λΦ0 (t, y0 ) + (1 − λ)Φ1 (t, y1 ).
Therefore, we get the following strict inequality,
Cλs,t (u∗λ , xλ ) < λC0s,t (u∗λ , x0 ) + (1 − λ)C1s,t (u∗λ, x1 )
and using
s,t ∗
s,t ∗
∗
Ls,t
λ (uλ ) = λL0 (u0 ) + (1 − λ)L1 (u1 )
gives
Fλs,t (u∗λ, xλ ) < λF0s,t(u∗λ , x0 ) + (1 − λ)F1s,t (u∗λ, x1 ).
Thus, since u∗λ is the maximizer for Φλ (s, xλ ),
Φλ (s, xλ ) = Fλs,t(u∗λ , xλ )
< λF0s,t(u∗λ , x0 ) + (1 − λ)F1s,t(u∗λ , x1 )
≤ λΦ0 (s, x0 ) + (1 − λ)Φ1 (s, x1 ).
This completes our proof.
14
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