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Vol. 10 (2005), Paper no. 16, pages 577-608.
Journal URL
http://www.math.washington.edu/∼ejpecp/
Large Deviations for Local Times of Stable Processes
and Stable Random Walks in 1 Dimension
Xia Chen1 ,

Wenbo V. Li2

and

Jay Rosen3

Abstract. In Chen and Li (2004), large deviations were obtained for the spatial
Lp norms of products of independent Brownian local times and local times of random
walks with finite second moment. The methods of that paper depended heavily on the
continuity of the Brownian path and the fact that the generator of Brownian motion, the
Laplacian, is a local operator. In this paper we generalize these results to local times of
symmetric stable processes and stable random walks.
Keywords: Large deviation, intersection local time, stable processes, random walks,
self-intersection local time.
AMS 2000 Subject Classifications: Primary 60F10; Secondary 60J30, 60J55.
Submitted to EJP on May 27, 2004. Final version accepted on November 9, 2004.

1

Research partially supported by NSF grant #DMS-0102238.
Research partially supported by NSF grant #DMS-0204513
3
Research partially supported by grants from the NSF and from PSC-CUNY.
2

577

1

Introduction

In the recent paper [3] the first two authors studied large deviations for the spatial L p
norms of Brownian local time Lxt . In particular they showed that there are explicit
constants C1 (p), C2 (p) such that for any p > 1 and λ, h > 0
(1.1)

³
´
1
·
log E eλkLt kp = λ2p/(p+1) C1 (p)
t→∞ t

lim

and
(1.2)

½

¾

1
lim log P kL·t kp ≥ ht = −h2p/(p−1) C2 (p).
t→∞ t
Similar results were obtained for products of independent local times and for local times of
random walks with finite second moment. The methods of that paper depended heavily on
the continuity of the Brownian path and the fact that the generator of Brownian motion,
the Laplacian, is a local operator. The goal of this paper is to generalize these results to
local times of symmetric stable processes and stable random walks, i.e. random walks in
the domain of attraction of a symmetric stable process.
To describe our results let {Xt ; t ≥ 0} denote the symmetric stable process of order
β
β > 1 in R1 , and let Lxt denote its local time. We normalize Xt so that E(eiλXt ) = e−t|λ| .

(Note that when β = 2 this gives a multiple of the standard Brownian motion). Let
Eβ (f, f ) =:

(1.3)

Z

R1

|λ|β |fb(λ)|2 dλ

R
where fb(λ) = R1 f (x)e−2πiλx dx denotes the Fourier transform of f , and

(1.4)

Fβ = {f ∈ L2 (R1 ) | kf k2 = 1 and Eβ (f, f ) < ∞}.

Theorem 1 Let Lxt be the local time for the symmetric stable process of index β > 1 in
R1 . For any p > 1 and λ > 0

(1.5)
where
(1.6)

³
´
pβ
1
·
log E eλkLt kp = λ pβ−(p−1) Mβ,p
t→∞ t

lim

Mβ,p = sup

g∈Fβ

½

kgk22p

¾

− Eβ (g, g) < ∞.

Equivalently, for any h > 0
(1.7)

½

where
(1.8)

¾

1
lim log P kL·t kp ≥ ht = −hpβ/(p−1) Aβ,p
t→∞ t

Aβ,p =

Ã

p−1
pβ

!Ã

pβ − (p − 1)
pβMβ,p

578

! pβ−(p−1)
p−1

.

We will see that the constant Mβ,p can be expressed in terms of the best possible

constant in a Gagliardo-Nirenberg type inequality, see (2.8) and (2.9).
−1/β x
d
By the scaling property of X, for each s ≥ 0 and a ≥ 0, we have Lxas = a1−1/β Las
d
so that kL·t kp = t1−(p−1)/pβ kL·1 kp . Using this, our Theorem is equivalent to the fact that
for any h > 0
n
o
(1.9)
lim t−1 log P kL·1 kppβ/(p−1) ≥ ht = −hAβ,p .
t→∞

Thus

E(e

(1.10)

pβ/(p−1)

λkL·1 kp

)

(

A−1
β,p .

We also note that when p is an integer, kL·t kpp can be expressed as an intersection
local
R
time. To see this, let f ∈ S(Rd ) be a positive, symmetric function with f dx = 1 and
set fǫ (x) = f (x/ǫ)/ǫd . Then it is well known that for β > 1
Lxt

= lim

Z

t

ǫ→0 0

fǫ (Xs − x) ds

where the limit exists a.s. locally uniformly and in all Lp spaces. Thus it is clear, at least
formally, that
Z

R1

(Lxt )p dx = lim

Z

ǫ→0 [0,t]p

Z

p
Y

R1 j=1

fǫ (Xsj − x) dx

p
Y

dsj

j=1

which measures the ‘amount’ of time spent by the path in p−fold intersections. This can
be justified.
There has also been interest in the literature in studying Lp norms for products of
independent local times. Let {Xj,t ; t ≥ 0} j = 1, . . . , m denote m independent copies of
{Xt ; t ≥ 0}. We use Lxj,t to denote the local time at x of {Xj,t ; t ≥ 0} respectively. We
will develop the large deviation principle for the mixed intersection local time
Z

(1.11)

m
Y

R1 j=1

(Lxj,t )p dx,

t≥0

where m ≥ 1 is an integer and real number p > 0 satisfying mp > 1. When p is an
integer, the above quantity measures the ‘amount’ of time that m independent trajectories
intersect together, while each of them intersects itself p times.
−1/β x
By the scaling property of X, for each t ≥ 0 and a ≥ 0, we have Lxat = a1−1/β Lat
so that
Z
Z
m
m

(1.12)

Y

d

R1 j=1

(Lxj,at )p dx = amp(1−1/β)+1/β

Y

R1 j=1

(Lxj,t )p dx.

Theorem 2 For each integer m ≥ 1 and real number p > 0 with mp > 1, and any λ > 0
(1.13)

½ µZ Y
¶1/mp ¾
m
1
log E exp λ
(Lxj,t )p dx
t→∞ t
R1 j=1

lim

= λmpβ/(mp(β−1)+1) c(β, p, m)Mβ,mp .
579

with
(1.14)

c(β, p, m) = m−(mp−1)/(mp(β−1)+1) .

Equivalently, for any h > 0
(1.15)

1
lim log P
t→∞ t

½µ Z

with
(1.16)

Aβ,m,p =

m
Y

R1 j=1

Ã

mp − 1
mpβ

(Lxj,t )p dx
!Ã

¶1/mp

¾

≥ ht = −hmpβ/(mp−1) Aβ,m,p

mpβ − (mp − 1)
mpβc(β, p, m)Mβ,mp

! mpβ−(mp−1)
mp−1

We also have the following law of the iterated logarithm, which is new even in th case
of m = 1.
Theorem 3 For each integer m ≥ 1 and real number p > 0 with mp > 1,
(1.17)

−(mp(1−1/β)+1/β)

lim sup t
t→∞

(log log t)

−(mp−1)/β

Z

m
Y

R1 j=1

(Lxj,t )p dx

−(mp−1)/β

= Aβ,m,p

a.s.

Let now {Sn ; n = 1, 2, . . .} be a symmetric random walk in Z 1 in the domain of
attraction of the symmetric stable process Xt of index β > 1, i.e.
Sn
(1.18)
→ X1
b(n)
in law with b(x) a function of regular variaton of index 1/β. For simplicity we assume
further that our random walk is strongly aperiodic.
We will use
n
lnx =

(1.19)

X

1{Sj =x}

j=1

for the local time of {Sn ; n = 1, 2, . . .} at x ∈ Z 1 .
Let {νn } represents a positive sequence satisfying
(1.20)

νn → ∞ and νn /n → 0.

P

We use k · kp,Z 1 for the norm in lp (Z 1 ), i.e. kf kp,Z 1 = ( x∈Z 1 |f (x)|p ))1/p . We have the
following moderate deviation result for the local times of stable random walks.
Theorem 4 For any positive sequence {νn } satisfying (1.20), any p ≥ 1 and λ > 0,
(1.21)

lim
n→∞

¾

½

pβ
1
b(n/νn )1−1/p ·
log E exp λ
kln kp,Z 1 = λ pβ−(p−1) Mβ,p .
νn
n/νn

Equivalently, for any h > 0
¾
½
n
1
·
1
(1.22)
log
P
kl
k
≥
h
lim
n p,Z
n→∞ ν
b(n/νn )1−1/p
n
= −hpβ/(p−1)

Ã

p−1
pβ

580

!Ã

pβ − (p − 1)
pβMβ,p

! pβ−(p−1)
p−1

.

An important application of the large and moderate deviations we establish is to obtain
the law of the iterated logarithm. Indeed, we have
Theorem 5
X

lim sup n−p b(n/ log log n)p−1
n→∞

−(p−1)/β

(lnx )p = Aβ,1,p

a.s.

x∈Z 1

The analogue of Theorem 2 for independent random walks is left to the reader.
Our paper is organized as follows. In Section 2 we develop the Sobolev inequalities
and Feynman-Kac formulae which are used throughout this paper. In Section 3 we study
large deviations for stable local times on the circle, which is then used in section 4 to
prove Theorem 1 on large deviations for stable local times in R 1 . In Section 5 we prove
Theorem 2 involving independent local times and Theorem 3, the law of the iterated
logarithm. Section 6 contains technical material on exponential moments for local times
which is needed in the paper. Section 7 explains how to get Theorems 4 and 5 for random
walks.

2

Sobolev inequalities and Feynman-Kac formulae

Lemma 1 If p > 1 and β > (p − 1)/p then Fβ ⊆ L2p (R1 ), and for any δ > 0
kf k22p ≤ Cδ kf k22 + δEβ (f, f )

(2.1)

for some Cδ < ∞. In particular for any λ > 0
(2.2)

Mp,β (λ) =: sup

f ∈Fβ

µ

λkf k22p

¶

− Eβ (f, f ) < ∞.

Proof of Lemma 1: By the Hausdorff-Young inequality
kf k2p ≤ kfbk2p/(2p−1)

(2.3)

where fb denotes the Fourier transform of f . We also have that for any r > 0

(2.4)

Now
(2.5)
and
(2.6)

2p/(2p−1)

kfbk2p/(2p−1)

Z

(r + |λ|β )p/(2p−1) b
|f (λ)|2p/(2p−1) dλ
β
p/(2p−1)
d
(r
+
|λ|
)
R
≤ k(r + |λ|β )−p/(2p−1) k(2p−1)/(p−1) ·

=

k(r + |λ|β )p/(2p−1) |fb(λ)|2p/(2p−1) k(2p−1)/p .
(2p−1)/p

k(r + |λ|β )p/(2p−1) |fb(λ)|2p/(2p−1) k(2p−1)/p = rkf k22 + Eβ (f, f )
Z

1
dλ
R1 (r + |λ|β )p/(p−1)
which is finite if β > (p − 1)/p, in which case we also have that limr→∞ cr = 0. Summarizing,
³
´
kf k22p ≤ cr(p−1)/p rkf k22 + Eβ (f, f ) .
(2.7)
(2p−1)/(p−1)

cr =: k(r + |λ|β )−p/(2p−1) k(2p−1)/(p−1) =

This gives (2.1) on taking r sufficiently large. This completes the proof of our Lemma.
581

Lemma 2 If p > 1 and β > (p − 1)/p then
(2.8)
and
(2.9)

¯

n
o
1−(p−1)/pβ 1/2
κp,β =: inf C ¯¯ kf k2p ≤ Ckf k2
[Eβ (f, f )](p−1)/pβ < ∞

pβ − (p − 1)
Mp,β (1) =
(p − 1)

Ã

(p − 1)κ2p,β
pβ

!pβ/(pβ−(p−1))

.

Proof of Lemma 2: To see that (2.8) is finite, note that if we set f (x) = s1/2 g(sx), then
kf k2 = kgk2 , kf k22p = s1−1/p kgk22p and Eβ (f, f ) = sβ Eβ (g, g) so that from (2.1) we obtain
(2.10)

³

´

kgk22p ≤ C kgk22 + sβ Eβ (g, g) s−(p−1)/p

and the fact that (2.8) is finite follows on taking sβ = kgk22 /Eβ (g, g). Finally, (2.9) follows
as in the proof of Lemma 8.2 of [2]. This completes the proof of our Lemma.
Let Hβ be the self-adjoint operator associated to the Dirichlet form Eβ . Thus the form
domain Q(Hβ ) = Fβ and for g ∈ Q(Hβ ) we have (g, Hβ g)2 = Eβ (g, g). Let Vf denote
the operator of multiplication by f . Note that if f ∈ Lp/(p−1) (R1 ), by using Hölder’s
inequality and then (2.1) we have that for any g ∈ Q(Hβ )
(2.11)

³

´

(g, Vf g)2 ≤ kgk22p kf kp/(p−1) ≤ kf kp/(p−1) Cδ kgk22 + δEβ (g, g) .

In the terminology of [15], Vf is infinitesimally form bounded with respect to Hβ , written
Vf ≺≺ Hβ . It follows from [15], Theorem X.17 that Hβ −Vf can be defined as a self-adjoint
operator with Q(Hβ − Vf ) = Q(H
) = Fβ .
R β
g
As usual, we write E (·) = g(x)E x (·) dx. Aside from technical integrability issues,
the lemmas below are generalizations of the Feynman-Kac formula. We include the proofs
for lack of a suitable reference.
Lemma 3 If p > 1 and β > (p − 1)/p then for any f ∈ Lp/(p−1) (R1 ) and g, h ∈ L2 (R1 )
(2.12)

(g, e

−t(Hβ −Vf )

h)2 = E

g

µ Rt

e

0

f (Xs ) ds

¶

h(Xt ) .

Proof of Lemma 3: If f ∈ S(R1 ) then using the right-continuity of paths, (2.12) follows
as in the proof of Theorem 6.1 in [16], see also Theorem 1.1 there. Let now fn ∈ S(R1 )
with fn → f in Lp/(p−1) (R1 ). We therefore have for each n
(2.13)

(g, e

−t(Hβ −Vfn )

h)2 = E

g

µ Rt

e

0

fn (Xs ) ds

¶

h(Xt ) .

Using (2.11), it follows from [10], Theorems IX, 2.16 and VIII, 3.6, that
(2.14)

lim (g, e−t(Hβ −Vfn ) h)2 = (g, e−t(Hβ −Vf ) h)2 .

n→∞

582

On the other hand the integrand on the right-hand side of (2.13) converges a.s. to the
corresponding integrand in (2.12). Thus to finish the proof we need only show uniform
integrabilty. We have
µ

E g e(4/3)

(2.15)

≤

Z µ

E

x

µ

fn (Xs ) ds 4/3

h

0

µ

≤ sup E
x0

Rt

4

e

x0

Rt
0

µ

4

e

fn (Xs ) ds

Rt
0

¶

(Xt )

¶¶1/3 ³

fn (Xs ) ds

³

´´2/3

E x h2 (Xt )

¶¶1/3 Z ³

³

g(x) dx

´´2/3

E x h2 (Xt )

g(x) dx

We then use E x (h2 (Xt )) = pt ∗ h2 (x) so that
Z ³

(2.16)

³

´´2/3

E x h2 (Xt )

and
(2.17)

2/3

g(x) dx ≤ kgk2 kpt ∗ h2 k4/3

kpt ∗ h2 k4/3 ≤ kpt k4/3 kh2 k1

which is finite. In fact, for any r > 1
kpt kr ≤ C/t(r−1)/rβ .

(2.18)

This follows from the fact that
pt is a probability
density function so that kpt kr ≤
R iλx
R −tλβ
(r−1)/r
−tλβ
kpt k∞
and kpt k∞ = supx | e e
dλ| ≤ e
dλ ≤ C/t1/β . (Alternatively, one
can use scaling: pt (x) = t−1/β p1 (xt−1/β )).
We now bound
(2.19)

sup E

x0

x0

=

∞
X

k=0

4k

µ

4

e

Z

Rt
0

fn (Xs ) ds

¶

{0≤t1 ≤...≤tk ≤t}

Z

k
Y

Rk j=1

fn (xj )ptj −tj−1 (xj − xj−1 ) dxj dtj .

By Hölder’s inequality
Z

(2.20)

k
Y

Rk j=1

fn (xj )ptj (xj − xj−1 ) dxj

≤ kfn kkp/(p−1) k
≤ kfn kkp/(p−1)

k
Y

j=1

k
Y

j=1

ptj −tj−1 (xj − xj−1 )kp

kptj −tj−1 kp .

Using (2.18) we have
(2.21)

k
Y

j=1

kptj −tj−1 kp = ck

k
Y

(tj − tj−1 )−(p−1)/pβ

j=1

583

Since by assumption (p − 1)/pβ < 1 we have that
(2.22)

Z

k
Y

(tj − tj−1 )−(p−1)/pβ dtj =

{0≤t1 ≤...≤tk ≤t} j=1

ck tk(1−(p−1)/pβ)
.
Γ(k(1 − (p − 1)/pβ))

Thus we have shown that for fixed t
(2.23)

sup E
x0

x0

µ

4

e

Rt
0

fn (Xs ) ds

¶

≤

ck kfn kkp/(p−1)
k=0 Γ(k(1 − (p − 1)/pβ))
∞
X

which is bounded uniformly in n. This completes the proof of Lemma 3.
Fix M > 0 and let TM = R1 /M Z 1 denote the circle of circumference M . We use the
notation kf kp,TM to denote the Lp (TM ) norm with the usual Lebesgue measure on TM .
Set
X
2 1
b
(2.24)
|λ|β |h(λ)|
Eβ,TM (h, h) =:
M
λ∈( 2π )Z 1
M

b denotes the usual Fourier transform for functions on T . Let
where h
M

(2.25)

Fβ,TM = {f ∈ L2 (TM ) |kf k2,TM = 1 and Eβ,TM (f, f ) < ∞}.

We introduce TM to deal with two technical problems in the proof of the upper bound
in Theorem 1. First, the stable infinitesimal generator is not a local operator when β < 2.
As a consequence, we will not have the upper bound for the Feymann-Kac large deviation
estimate which corresponds to the lower bound given in (4.2) below. Second, as pointed
out in [3] (p. 225-226), the family {t−1 L·t } is not exponentially tight as a stochastic process
taking values in the Banach space Lp (R1 ). To fix these two problems we map the process
Xt into the compact space TM . It is crucial that the image process is Markovian.
An almost identical proof gives the following analogue of Lemma 1.
Lemma 4 If p > 1 and β > (p − 1)/p then Fβ,TM ⊆ L2p (TM ), and for any δ > 0
(2.26)

kf k22p,TM ≤ Cδ kf k22,TM + δEβ,TM (f, f )

for some Cδ < ∞. In particular for any λ > 0
(2.27)

Mp,β,TM (λ) =:

sup
f ∈Fβ,TM

µ

λkf k22p,TM

¶

− Eβ,TM (f, f ) < ∞.

Let Yt be the image of Xt under the quotient map x ∈ R1 7→ x̄ ∈ TM . It is easily
seen that Yt is a Markov process with independent increments. Yt is called the symmetric
stable process of order β on TM .
As before, let Hβ,TM be the self-adjoint operator associated to the Dirichlet form Eβ,TM .
Thus Q(Hβ,TM ) = Fβ,TM and for g ∈ Q(Hβ,TM ) we have (g, Hβ,TM g)2 = Eβ,TM (g, g). Let Vf
denote the operator of multiplication by f . Using Lemma 4, we see that Vf ≺≺ Hβ,TM so
that, as before we can define Hβ,TM − Vf as a self-adjoint operator with Q(Hβ,TM − Vf ) =
Q(Hβ,TM ) = Fβ,TM . An almost identical proof gives the following analogue of Lemma 3.
584

Lemma 5 If p > 1 and β > (p − 1)/p then for any f ∈ Lp/(p−1) (TM ) and g, h ∈ L2 (TM )
µ Rt

(g, e−t(Hβ −Vf ) h)2 = E g e

(2.28)

0

f (Ys ) ds

¶

h(Yt ) .

We next present an important large deviation result. It is possible to derive this result
from the methods of Donsker and Varadhan, see in particular [7], but we prefer to give a
simple self-contained proof.
Lemma 6 If p > 1 and β > (p−1)/p then for any non-negative function f ∈ Lp/(p−1) (TM )
(2.29)

µ

¶

µ

¶

Rt
1
(g, f g)2,TM − Eβ,TM (g, g) .
log E e 0 f (Ys ) ds = sup
t→∞ t
g∈Fβ,TM

lim

Note that using Hölder’s inequality and Lemma 4, the sup on the right-hand side is
finite.
Proof of Lemma 6: Let p̄t denote the density of Yt . Fix t0 > 0, and recall from (2.18),
(more precisely the analogue for TM ), that p̄t0 ∈ L2 (TM ). Then using the non-negativity
of f , the Markov property and (2.28) we have
(2.30)

µ Rt

E e

0

f (Ys ) ds

¶

≥ E

p̄t0

µ R t−t
0

e

0

f (Ys ) ds

¶

= (p̄t0 , e−(t−t0 )(Hβ,TM −Vf ) 1)2,TM .

By (2.24) we can see that σ(Hβ,TM ), the spectrum of Hβ,TM , is purely discrete. In fact
σ(Hβ,TM ) = {

µ

2πj
M

¶β

| j = 0, 1, . . .}

with a complete set of corresponding eigenvectors
e±(2πi)jx/M
1
√
| j = 1, 2, . . .}.
{√ } ∪ {
M
M
Hence, using the fact that Vf ≺≺ Hβ,TM and [15], Theorem XIII.64, (iv), (v), see also
Theorem XIII.68, we find that Hβ,TM − a′ Vf also has purely discrete spectrum for any a′ .
(We note for later that these Theorem’s show that Hβ,TM − a′ Vf has compact resolvent).
′
From Lemma 5 it follows that e−t(Hβ,TM −a Vf ) is positivity preserving and ergodic. It
follows from [15], Theorem XIII.43 that inf σ(Hβ,TM − a′ Vf ) is a simple eigenvalue and the
associated eigenvector is strictly positive. Since p̄t0 is also strictly positive, we find from
(2.30) that
µ Rt
¶
1
f (Ys ) ds
≥ − inf σ(Hβ,TM − Vf ).
(2.31)
lim inf log E e 0
t→∞ t
The lower bound for (2.29) follows by the Rayleigh-Ritz principle, see [15].

585

For the upper bound use the Markov property to see that for any a, a′ with 1/a+1/a′ =
1,
(2.32)

µ Rt

E e

0

f (Ys ) ds

µ Rt
0

=E e
½

0

µ

a

≤ E e
½

µ

a

≤ E e
½

µ

a

= E e

¶

f (Ys ) ds

R t0
0

R t0
0

R t0
0

µ R t−t
0

E Yt0 e

f (Ys ) ds

f (Ys ) ds

f (Ys ) ds

f (Ys ) ds

0

¶¾1/a (

E

¶¾1/a ½

¶¾1/a ½

õ

E E

E

Yt0

E

µ

p̄t0

¶¶

Yt0

µ

e

µ

a′

e

µ R t−t
0
a′

e

0

R t−t0

R t−t0
0

0

f (Ys ) ds

f (Ys ) ds

f (Ys ) ds

¶¶a′ !)1/a′
¶¶¾1/a′

¶¾1/a′

.

By (2.23), (more precisely the analogue for TM ), we have that the first factor on the
right hand side is bounded for any fixed t0 and a. On the other hand, by (2.28)
(2.33)

E

p̄t0

µ

a′

e

Hence

R t−t0
0

f (Ys ) ds

µ

¶

′

= (p̄t0 , e−(t−t0 )(Hβ,TM −a Vf ) 1)2 .
¶

Rt
1
inf σ(Hβ,TM − a′ Vf )
log E e 0 f (Ys ) ds ≤ −
.
a′
t→∞ t
Using once more the fact that Vf ≺≺ Hβ,TM we find that Hβ,TM − zVf is an analytic
family of type (B). We have noted in the last paragraph that Hβ,TM − a′ Vf has compact
resolvent. It follows from [10], VII, Remark 4.22, that

(2.34)

(2.35)

lim sup

lim inf σ(Hβ,TM − a′ Vf ) = inf σ(Hβ,TM − Vf ).

a′ →1

The upper bound for (2.29) follows by taking a′ → 1 and then applying the Rayleigh-Ritz
principle.

3

Large deviations for stable local times on the circle

We use the notation of the previous section. M > 0 is fixed throughout.
Theorem 6 Let L̄xt be the local time for the symmetric stable process of index β > 1 in
TM . For any p > 1 and λ > 0
(3.1)

½

¾

³
´
1
·
λkgk22p,TM − Eβ,TM (g, g) .
lim log E eλkL̄t kp,TM = sup
t→∞ t
g∈Fβ,TM

Proof of Theorem 6: We first establish the lower bound for (3.1). We claim that for
any λ > 0
(3.2)

½

¾

´
³
1
·
lim inf log E eλkL̄t kp,TM ≥ sup
λkgk22p,TM − Eβ,TM (g, g) .
t→∞ t
g∈Fβ,TM

586

Indeed, if we take any f ∈ L(p−1)/p (TM ) with kf k(p−1)/p,TM = 1 then
kL̄·t kp,TM

(3.3)

Z

≥

TM

L̄xt f (x)dx

=

Z

t

0

f (Ys )ds.

Consequently, taking f non-negative, by Lemma 6
(3.4)

lim inf
t→∞

≥

³
´
1
·
log E eλkL̄t kp,TM
t ½

¾

λ(g, f g)2,TM − Eβ,TM (g, g) .

sup

g∈Fβ,TM

Taking supremum on the right hand side over such f we obtain (3.2).
To establish the upper bound and complete the proof of (3.1) we shall prove that for
any λ > 0
(3.5)

½

¾

³
´
1
·
lim sup log E eλkL̄t kp,TM ≤ sup
λkgk22p,TM − Eβ,TM (g, g) .
t→∞ t
g∈Fβ,TM

By (3.3) and Lemma 6, for any non-negative f ∈ L(p−1)/p (TM )
−1

lim t

(3.6)

t→∞

=

sup
g∈Fβ,TM

½ Z

log E exp λ
½

TM

L̄xt f (x)dx

¾

¾

λ(g, f g)2,TM − Eβ,TM (g, g) .

Let ǫ > 0 and γ > 0 be fixed and let K ⊂ Lp (TM ) be the compact set given in Lemma
11. By the fact that the set of bounded measurable functions on TM is dense in the unit
ball of Lq (TM ), and by the Hahn-Banach Theorem, for each h ∈ γK, there is a bounded
function f such that kf k(p−1)/p,TM = 1 and
Z

TM

µZ

f (x̄)h(x̄)λ(dx̄) >

p

TM

|h(x̄)| λ(dx̄)

¶1/p

− ǫ.

Consequently, there are finitely many bounded functions f1 , · · · , fN in the unit sphere of
Lq (TM ) such that
µZ

TM

p

|h(x̄)| λ(dx̄)

¶1/p

< max

1≤i≤N

Z

TM

fi (x̄)h(x̄)λ(dx̄) + ǫ

Therefore,
(3.7)

µ

E exp
≤ e

ǫt

N
X
i=1

½

λkL̄·t kp,TM

E exp

½Z

TM

587

¾

; t

−1

L̄·t

fi (x)L̄xt

∈ γK
¾

dx .

¶

∀h ∈ γK.

In view of (3.6),
µ

¾

½

lim sup t−1 log E exp λkL̄·t kp,TM ; t−1 L̄·t ∈ γK

(3.8)

t→∞

≤ ǫ + max

sup

1≤i≤N g∈Fβ,T
M

≤ ǫ + sup

g∈Fβ,TM

½

½

λ(g, fi g)2,TM − Eβ,TM (g, g)

λkgk22p,TM

− Eβ,TM (g, g)

¾

¾

¶

where the second step follows from the Hölder’s inequality and the fact kfi k(p−1)/p,TM = 1
for 1 ≤ i ≤ N . Letting ǫ −→ 0 gives
−1

(3.9)

lim sup t
t→∞

≤

sup
g∈Fβ,TM

µ

log E exp

½

½

λkL̄·t kp,TM

¾

; t

−1

L̄·t

¾

∈ γK

¶

o¶1/2

.

λkgk22p,TM − Eβ,TM (g, g) .

By the Cauchy-Schwarz inequality, on the other hand,
µ

¾

½

E exp λkL̄·t kp,TM ; t−1 L̄·t 6∈ γK

(3.10)
≤

µ

E exp

½

2λkL̄·t kp,TM

¾¶1/2 µ n

−1

P t

¶

L̄·t

6∈ γK

Note that Lemma 8 and scaling (1.12) imply that
(3.11)

sup E exp
t≤1

½

2λkL̄·t kp,TM

¾

0 and hence N (γ) can be arbitrarily large. Combining (3.9) and (3.12) we
obtain (3.5) completing the proof of our theorem.

588

Large deviations for stable local times in R 1

4

Proof of Theorem 1: We first establish the lower bound for (1.5). We claim that for
any λ > 0
½
¾
³
´
1
2
λkL·t kp
lim inf log E e
(4.1)
≥ sup λkgk2p − Eβ (g, g) .
t→∞ t
g∈Fβ
This will follow exactly as in the proof of Theorem 6 once we establish that for any
non-negative f ∈ L(p−1)/p (R1 )
(4.2)

½

¶

µ

¾

Rt
1
lim inf log E e 0 f (Xs )ds ≥ sup (g, f g)2 − Eβ (g, g) .
t→∞ t
g∈Fβ

But as in the proof of (2.30) we have, for any bounded g1 , g2
(4.3)

µ Rt

E e

0

f (Xs ) ds

µ R t−t
0

¶

≥ E pt0 e

0

f (Xs ) ds

µ R t−t
0

≥ E g1 pt0 e

0

¶

f (Xs ) ds

¶

g2 (Xt−t0 ) /kg1 k∞ kg2 k∞

= (g1 pt0 , e−(t−t0 )(Hβ −Vf ) g2 )2 /kg1 k∞ kg2 k∞ .
Since pt0 (x) > 0 for all x, by varying g1 , g2 we obtain
µ

¶

Rt
1
lim log E e 0 f (Xs ) ds ≥ − inf σ(Hβ − Vf )
t→∞ t

(4.4)

which gives (4.2) by the Rayleigh-Ritz principle.
We next establish the upper bound for (1.5). We claim that for any λ > 0
½

¾

³
´
1
·
lim sup log E eλkLt kp ≤ sup λkgk22p − Eβ (g, g) .
t→∞ t
g∈Fβ

(4.5)

Fix M > 0 and recall from the last section the symmetric stable process Yt of index β
in TM and its local time L̄xt . It can be easily verified that
L̄xt =

X

Lx+kM
,
t

t ≥ 0, x ∈ R1

k∈Z 1

Consequently, for any p > 1
Z

R

(4.6)

(Lxt ))p
1

dx =
≤

X Z

M

(Lx+kM
))p dx
t

0
k∈Z 1
Z Mµ X
0

Lx+kM
t

k∈Z 1

¶p

dx =

Z

TM

(L̄xt )p dx.

Hence (4.5), and thus our theorem, will follow from Theorem 6 once we verify the following
lemma.
589

Lemma 7 For any p > 1, r > 0
(4.7)

lim sup sup
M →∞ ḡ∈Fβ,TM

½

¾

½

¾

rkḡk22p,TM − Eβ,TM (ḡ, ḡ) ≤ sup rkgk22p − Eβ (g, g) .
g∈Fβ

Proof of Lemma 7: Without loss of generality we will prove this lemma with r = 1.
Recall the definition
Z
Eβ (f, f ) =:
(4.8)
|λ|β |fb(λ)|2 dλ.
R1

Using

β

|λ| = cβ
where c−1
β =
(4.9)

R

R1

1−cos(y)
|y|1+β

Z

1 − cos(λy)
dy
|y|1+β

R1

dy and Parseval’s formula we find that
Eβ (f, f ) = cβ

Z

R1

Z

R1

|f (y) − f (x)|2
dy dx.
|y − x|1+β

Similarly, for any M-periodic function h
(4.10)

Z MZ
1
|h(x + y) − h(x)|2
|λ| h(λ)|
Ēβ (h, h) =
= cβ
dy dx
M
|y|1+β
0
R1
λ∈( 2π )Z 1
X
M

βb

2

where the last equality follows as in the proof of (4.9).
Let ḡ be an M-periodic function in Fβ,TM . We need to construct a function f ∈ Fβ
which is equal to ḡ on [M 1/2 , M − M 1/2 ] and is negligible in some suitable sense on the
1/2
rest of the real line as M gets large. Let E = [0, M 1/2 ] ∪ [M − M
, M ]. By Lemma 3.4
R
2
in Donsker-Varadhan (1975), there is a real number a such that E ḡ (x − a)dx ≤ 2M −1/2 .
We may assume a = 0, i.e,
Z
(4.11)
ḡ 2 (x)dx ≤ 2M −1/2
E

for otherwise we can replace ḡ(·) by ḡ(· + a). Define

xM −1/2





1
M 1/2 − xM −1/2



0

ϕ(x) = 

0 ≤ x ≤ M 1/2
M −1/2 ≤ x ≤ M − M 1/2
M − M 1/2 ≤ x ≤ M
otherwise.

It is straightforward to verify that

0 ≤ ϕ(x) ≤ 1, |ϕ′ (x)| ≤ M −1/2 , |(ϕ2 (x))′ | ≤ 2M −1/2 ,
Define
f (x) = ḡ(x)ϕ(x) ·

µZ

∞

−∞

590

2

2

ḡ (x)ϕ (x)dx

¶−1/2

−∞ < x < ∞.
.

Set α =
(4.12)

R∞

−∞

ḡ 2 (x)ϕ2 (x)dx. Note that
|ḡ(y)ϕ(y) − ḡ(x)ϕ(x)|2
= |(ḡ(y) − ḡ(x))ϕ(y) + ḡ(x)(ϕ(y) − ϕ(x))|2
= |ḡ(y) − ḡ(x)|2 ϕ2 (y) + ḡ 2 (x)|ϕ(y) − ϕ(x)|2
+2|ḡ(y) − ḡ(x)|ϕ(y)ḡ(x)|ϕ(y) − ϕ(x)|

Now
Z

Z

|ḡ(y) − ḡ(x)|2 ϕ2 (y)
dy dx
|y − x|1+β
R1 R1
Z Z M
|ḡ(y) − ḡ(x)|2
≤ cβ
dy dx
|y − x|1+β
R1 0
= Ēβ (ḡ, ḡ).
cβ

(4.13)

Using
ḡ 2 (x)|ϕ(y) − ϕ(x)|2
≤ ḡ 2 (x)|ϕ(y) − ϕ(x)|2 (1[0,M ] (x) + 1[0,M ] (y))

(4.14)

≤ 2M −1/2 ḡ 2 (x)(1[0,M ] (x) + 1[0,M ] (y))(|y − x| ∧ |y − x|2 )

we have
Z

Z

ḡ 2 (x)|ϕ(y) − ϕ(x)|2
dy dx
|y − x|1+β
R1 R1
Z MZ
|ḡ(x)|2 (|y − x| ∧ |y − x|2 )
−1/2
≤ 2M
cβ
dy dx
|y − x|1+β
0
R1
Z Z M
|ḡ(x)|2 (|y − x| ∧ |y − x|2 )
dy dx
+2M −1/2 cβ
|y − x|1+β
R1 0
≤ cM −1/2

(4.15)

cβ

where for the first integral we used the change of variables y → y + x and for the second
we used the change of variables x → y + x and the periodicity of ḡ(x). Finally we use
(4.16)

2|ḡ(y) − ḡ(x)|ϕ(y)ḡ(x)|ϕ(y) − ϕ(x)|
≤ 4M −1/4 |ḡ(y) − ḡ(x)|ḡ(x)(|y − x|1/2 ∧ |y − x|)1[0,M ] (y).

the Cauchy-Schwarz inequality and (4.13), (4.15) to see that
(4.17)

Z

Z

|ḡ(y) − ḡ(x)|ϕ(y)ḡ(x)|ϕ(y) − ϕ(x)|
dy dx
|y − x|1+β
µZ Z
¶1/2
|ḡ(y) − ḡ(x)|2 1[0,M ] (y)
−1/4
≤ 8M
cβ
dy dx
|y − x|1+β
R1 R1
¶1/2
µZ Z
ḡ 2 (x)(|y − x|2 ∧ |y − x|)1[0,M ] (y)
dy
dx
|y − x|1+β
R1 R1
2cβ

R1

R1

1/2

≤ cM −1/4 Ēβ (ḡ, ḡ)

591

Putting this all together we find that
(4.18)

αEβ (f, f )
=

Z

R1

Z

|ḡ(y)ϕ(y) − ḡ(x)ϕ(x)|2
dy dx
|y − x|1+β

R1

1/2

≤ Ēβ (ḡ, ḡ) + cM −1/4 Ēβ (ḡ, ḡ) + cM −1/2
On the other hand,
µZ

M

0

2p

ḡ (x)dx

¶1/p

µ

≤

(4.19)

α

p

Z

≤ α

µZ

¶1/p

≤

∞

2p

−∞
∞

−∞

|f (x)| dx +
2p

|f (x)| dx

Z

¶1/p

2p

|ḡ(x)| dx

E

+

µZ

E

¶1/p
2p

|ḡ(x)| dx

¶1/p

and from (4.11)
(4.20)

µZ

|ḡ(x)|2p dx

E

sup |ḡ(x)|2/q

0≤x≤M

≤ (2M

−1/2 1/p

µZ

E

|ḡ(x)|2 dx

¶1/p

sup |ḡ(x)|2/q .

)

0≤x≤M

Observe that since β > 1
(4.21)

sup |ḡ(x)| ≤
x

≤
≤

X

q

λ∈( 2π
)Z 1
M

µ

)Z 1
λ∈( 2π
M

q

1 + |k|β

X

λ∈( 2π
)Z 1
M

µ

X

1 + |k|β

(4.22)

µZ

E

Let J = supg∈Fβ
(4.23)

½

|ḡ(x)|2p dx

λkgk22p

µZ

0

M

µ

1
M
1
M

¶1/2 µ

X

β

λ∈( 2π
)Z 1
M

2
b̄
(1 + |k| )|g(k)|

¶1/2

.

¶1/p

≤ (2M −1/2 )1/p c 1 + Ēβ (ḡ, ḡ)

µ

¾

1
M

¶1/q

¶1/2

.

− Eβ (g, g) . By combining (4.18), (4.19), (4.22), we see

2p

ḡ (x)dx

− cM

b̄
|g(k)|

1
1
β
1 + |k| M

≤ c 1 + Ēβ (ḡ, ḡ)
Therefore,

b̄
|g(k)|

−1/4

µ

¶1/p

− (1 − ǫ)Ēβ (ḡ, ḡ)

Ēβ (ḡ, ḡ)

≤ c(1 − ǫ)M −1/2 + α

¶1/2

½µ Z

∞

−∞

+ (2M

−1/2 1/p

|f (x)|2p dx

592

)

¶1/p

µ

c 1 + Ēβ (ḡ, ḡ)

¶1/q ¶

− (1 − ǫ)Eβ (f, f )

¾

≤ cM

−1/2

+ sup
f ∈Fβ

³

= cM −1/2 +

½µ Z

∞

−∞

2p

|f (x)| dx

p−1
1 ´ p(β−1)+1
J
1−ǫ

¶1/p

− (1 − ǫ)Eβ (f, f )

¾

where the second inequality follows from the fact that α ≤ 1 and the final step from the
substitution
³ 1 ´p/2(p(β−1)+1) ³³ 1 ´p/(p(β−1)+1) ´
h
x .
f (x) =
1−ǫ
1−ǫ
Since q > 1, there is a sufficiently large M = M (ǫ) > 0 such that M −1/2 ≤ ǫ and that
cM −1/4 x1/2 + (2M −1/2 )1/p c(1 + x)1/q ≤ ǫ(x + 1)
for all x ≥ 0. Note that the choice of M is independent of the function g!. For such M ,
µZ

(4.24)

M

0

2p

ḡ (x)dx

¶1/p

− Ēβ (ḡ, ḡ) ≤ ǫ +

³

This completes the proof of Lemma 7.

5

p−1
1 ´ p(β−1)+1
J
1−ǫ

Large deviations for independent stable local times
and the law of the iterated logarithm

Proof of Theorem 2: The upper bound for (1.13) follows exactly as in [3]. Given
Lemma 11 and our proof of the lower bound in Theorem 1 the lower bound for (1.13) will
follow exactly as in [3].
Proof of Theorem 3: Using (1.15) and the scaling (1.12) we see that for any h > 0
(5.1)

¾
½Z Y
m
1
x p
(mp−1)/β
lim log P
= −hβ/(mp−1) Aβ,m,p .
(Lj,1 ) dx ≥ ht
t→∞ t
R1 j=1

Replacing t by log log t we get
¾
½Z Y
m
1
x p
(mp−1)/β
(5.2) lim
= −hβ/(mp−1) Aβ,m,p .
(Lj,1 ) dx ≥ h(log log t)
log P
t→∞ log log t
R1 j=1

Let tk = θk for θ > 1. Using (5.2) and the scaling (1.12) we see that
(5.3)

∞
X

k=1


Z

P

m
Y

R1 j=1

−(mp−1)/β

(Lxj,tk )p dx

≥



(mp(1−1/β)+1/β)
ctk
(log log tk )(mp−1)/β 

Aβ,m,p
. Borel-Cantelli and interpolation then give the upper bound in
Theorem 3.
To prove the lower bound, write sk = k 2k , k ≥ 1 and notice that
(5.4)

x−Xj,s

Lxj,sk+1 − Lxj,sk = Lj,k,sk+1k−sk
593

where Lxj,k,t is the local time of Xj,k,t :
Xj,k,t = Xj,sk +t − Xj,sk

t ≥ 0, k ≥ 1, 1 ≤ j ≤ m.

Hence, using |a1/m − b1/m | ≤ 21/m |a − b|1/m
(5.5)

¯m
¯ Y
m
Y
¯ m
1/m ¯¯
1/m
x
x
x
¯k
[Lj,sk+1 − Lj,sk ]kp,R1 − k
Lj,k,sk+1 −sk kp,R1 ¯
¯
j=1

=

j=1

¯ Y
¯m
m
Y
¯ m x−Xj,sk
1/m
1/m ¯¯
x
¯k
Lj,k,sk+1 −sk kp,R1 − k
Lj,k,sk+1 −sk kp,R1 ¯
¯
j=1

j=1

≤ 21/m k
Q

Then using m
j=1 aj −
inequality we see that
(5.6)

m
Y

j=1

m
Y

x−Xj,s

Lj,k,sk+1k−sk −

Qm

j=1 bj

=

Pm

j=1

Qj−1

j=1 (

Lxj,k,sk+1 −sk kp,R1 .

ai )(aj − bj )(

i=1

Qm

k=j+1 bk )

followed by Hölder’s

¯m
¯ Y
m
Y
¯ m
1/m
1/m ¯¯
x
x
x
¯k
Lj,k,sk+1 −sk kp,R1 ¯
[Lj,sk+1 − Lj,sk ]kp,R1 − k
¯
j=1

≤ 21/m

m
X

j=1




j=1

Y

i6=j



x−Xj,s

kLxi,k,sk+1 −sk kmp,R1  kLj,k,sk+1k−sk − Lxj,k,sk+1 −sk kmp,R1 .

By the already proven upper bound in Theorem 3, taking m = 1, replacing p by mp
and using the abbreviation φ(s) = smp(1−1/β)+1/β ( log log s)(mp−1)/β we have
kLxi,k,sk+1 −sk kmp,R1 = O(φ(sk+1 ))1/mp a.s.

(5.7)

for each 1 ≤ i ≤ m.
It follows from Lemma 10, after rescaling, that for any α > 0
lim lim sup (log log t)−1

δ→0+

t→∞

log P

n

sup
|x−y|≤δ(t/ log log t)1/β

o

1−1/β
kLty+· − Lx+·
log log t1/β = −∞.
t kmp,R1 ≥ αt

Hence by the Borel-Cantelli lemma
(5.8)

−(1−1/β)

lim lim sup sk+1

δ→0+

k→∞

−1/β

log log sk+1

sup
|x−y|≤δ(sk+1 / log log sk+1 )1/β

y+·
kLj,k,s
− Lx+·
j,k,sk+1 −sk kmp,R1
k+1 −sk

=0

a.s.

However, it is easy to see that nk+1 − nk > nk k 2 as k → ∞ so that by the scaling
property of the stable process
P (|Xnk | >

(nk+1 − nk )1/β
k 2/β
) ≤ P (|X1 | >
) ≤ C/k 2−2ǫ
log log(nk+1 − nk )
log k
594

for any ǫ > 0, since X1 has β − ǫ moments. By the Borel-Cantelli lemma, with probability
1 the events
{|(X1,nk , . . . , Xm,nk )| ≤

(nk+1 − nk )1/β
}
log log(nk+1 − nk )

k = 1, 2, · · ·

eventually hold. Therefore, by (5.8)
Xj,s +·

(1−1/β)

kLj,k,skk+1 −sk − L+·
j,k,sk+1 −sk kmp,R1 = o(sk+1

(5.9)

1/β

log log sk+1 )

= o(φ(sk+1 ))1/mp

a.s.

a.s.

Combining (5.6), (5.7) and (5.9), we reach the conclusion that
(5.10)

¯µ Z
¯
¯
¯

∞

m
Y

−∞ j=1

[Lxj,sk+1

= o(φ(sk+1 ))1/mp

−

Lxj,sk ]p dx

¶1/mp

−

µZ

∞

m
Y

−∞ j=1

(Lxj,k,sk+1 −sk )p dx

a.s. (k → ∞).
−(mp−1)/β

On the other hand, by (1.15) in Theorem 2, for any γ < Aβ,m,p
X
k

=

X
k


Z

P

m
Y

−∞ j=1


Z

P

∞

∞

−∞




(Lxj,k,sk+1 −sk )p dx ≥ γφ(sk+1 )

m
³Y

j=1




´p

Lxj,sk+1 −sk dx ≥ γφ(sk+1 ) = ∞.

Then by the Borel-Cantelli lemma and the independence of the sequence
Z

∞

m
Y

−∞ j=1

we have
(5.11)

lim sup
k→∞

(Lxj,k,sk+1 −sk )p dx,

k = 1, 2, · · ·

Z ∞ Y
m
1
−(mp−1)/β
(Lxj,k,sk+1 −sk )p dx ≥ Aβ,m,p
a.s.
φ(sk+1 ) −∞ j=1

In view of (5.10),
(5.12)
Note that

Z ∞ Y
m
1
−(mp−1)/β
[Lxj,sk+1 − Lxj,sk ]p dx ≥ Aβ,m,p
a.s.
lim sup
k→∞ φ(sk+1 ) −∞ j=1
Z

∞

m
Y

−∞ j=1

Hence,

(Lxj,sk+1 )p dx ≥

Z

∞

m
Y

−∞ j=1

[Lxj,sk+1 − Lxj,sk ]p dx,

Z ∞ Y
m
1
−(mp−1)/β
(Lxj,t )p dx ≥ Aβ,m,p
lim sup
t→∞ φ(sk+1 ) −∞ j=1

which finishes the proof of Theorem 3.
595

∀k ≥ 1.
a.s.

¶1/mp ¯
¯
¯
¯

6

Exponential moments for local times

Let Lxt denote the local time for X in R1 , and L̄xt denote the local time for X∗ in TM . We
use k · kp,R1 , k · kp,TM to denote the norms in Lp (R1 , dx) and Lp (TM , dx) respectively.
Lemma 8 Let p > 1. For the symmetric stable process in R1 of index β > 1 and any
γ 0
Ã

Ã

kLy+· − Lx+·
1 kp,R1
sup E exp γ 1
|y − x|ζ
x,y

(6.2)
Furthermore

Ã

!!

Ã

kL1y+· − Lx+·
1 kp,R1
lim sup E exp γ
γ→0 x,y
|y − x|ζ

(6.3)

< ∞.

!!

= 1.

Similar results hold for the symmetric stable process in TM of index β > 1 with
y+·
·
kL·1 kp,R1 , kL1y+· − Lx+·
− L̄x+·
1 kp,R1 replaced by kL̄1 kp,TM , kL̄1
1 kp,TM .
Proof of Lemma 8: We note that the tail estimate in [9] implies a slightly stronger
result than (6.1). The direct proof of (6.1) given here serves as a warmup for (6.2) and
(6.3). We will first assume that p is an integer greater than or equal to 2. Recall the
notation
Z t
x
Lt,ǫ =
(6.4)
fǫ (Xs − x) ds.
0

For any integer m we have


E

(6.5)

m
Y

j=1

=

Z

Rmp



x
(Lt,ǫj )p 

Z

m
Y

[0,1]mp j=1

ei(


Z
=E

Rmp

Pp

k=1

λj,k )xj

Z

p
m Y
Y

[0,1]mp j=1 k=1



E

p
m Y
Y

j=1 k=1



eiλj,k (xj −Xtj,k ) fb(ǫλj,k ) dtj,k dλj,k 


e−iλj,k Xtj,k 

p
m Y
Y

j=1 k=1

Using the Fourier inversion formula in the form
(6.6)

Z

i

Rm

e

Pm Pp−1
j=1

(

k=1

Z

λj,k )xj

i

Rm

e

Pm

j=1

λj,p xj

F (λj,p )

m
Y

fb(ǫλj,k ) dtj,k dλj,k .

dλj,p dxj = F (−

j=1

p−1
X

λj,k )

k=1

we have that
(6.7)

E
=

Z

³

kL·1,ǫ kpm
p,R1

Rm(p−1)

Z

[0,1]mp

´

=

Z

Rm



E

E

p
m Y
Y

j=1 k=1





m
Y
 (Lxt,ǫj )p 
j=1
j=1
m
Y



−iλj,k Xtj,k 

e

596

dxj

p
m Y
Y

j=1 k=1

fb(ǫλ

j,k ) dtj,k

m p−1
Y
Y

j=1 k=1

dλj,k

P

p−1
where in the last line λj,p = − k=1
λj,k . To evaluate the expectation, for each bijection
π of {1 ≤ n ≤ mp} onto {(j, k) ; 1 ≤ j ≤ m , 1 ≤ k ≤ p} we let

(6.8)

Dπ = {tj,k ; tπ(1) < tπ(2) < · · · < tπ(mp) < 1}

and
(6.9)

uπ,n =

mp
X

λπ(l) .

l=n

We use C to denote the set of such bijections π. Then


E

(6.10)

p
m Y
Y

e

j=1 k=1

=

X

µ

=

X

µ

X

−i

E e

π∈C

=

−i

E e

π∈C

e−

π∈C



−iλj,k Xtj,k 

Pmp

Pmp

n=1

Pmp

n=1

n=1

λπ(n) Xtπ(n)

¶

uπ,n (Xtπ(n) −Xtπ(n−1) )

|uπ,n |β (tπ(n) −tπ(n−1) )

¶

.

We will bound this by dropping the last factor in which λj,p appears for each j.
To be more precise, let νπ,j be the unique n such that uπ,n − uπ,n+1 = λj,p and set
Vπ = {νπ,j , 1 ≤ j ≤ m}. Combining the above we have uniformly in ǫ > 0
(6.11)

³

E kL·1,ǫ kpm
p,R1
≤

XZ

´

m(p−1)
π∈C R

≤ cmp

XZ

Z

−

Dπ

Y

π∈C Dπ n∈Vπc

e

P

n∈Vπc

|uπ,n |β (tπ(n) −tπ(n−1) )

mp
Y

dtπ(n)

n=1

mp
Y
1
dtπ(n)
|tπ(n) − tπ(n−1) |1/β n=1

m p−1
Y
Y

dλj,k

j=1 k=1

≤ cmp (mp)!/Γ(m(p − 1)(1 − 1/β) + m) ≤ cmp ((mp)!)(p−1)/pβ .
Since this is true for any integer m we can use uniform integrability to obtain

Then for any integer n
(6.13)

³

´

mp
E kL·1 kpm
((mp)!)(p−1)/pβ .
p,R1 ≤ c

(6.12)

³

´

n

³

E kL·1 knp,R1 ≤ E kL·1 kpn
p,R1

´o1/p

≤ cn (n!)(p−1)/pβ .

This immediately gives (6.1). To obtain (6.2) and (6.3) we begin with p an even integer
y+·
and note that if we replace L·1,ǫ in (6.7)
by L1,ǫ
− Lx+·
, then in the last line of (6.7) we
1,ǫ
³
´
Qm Qp
iλj,k y
iλj,k x
will have an extra factor of j=1 k=1 e
−e
. Using the bound

(6.14)

|eiλj,k y − eiλj,k x | ≤ 2|λj,k |ζ |y − x|ζ
597

which is valid for any 0 ≤ ζ ≤ 1 and proceeding in a manner similar to (6.11) leads to
(6.2) and (6.3).
If q > 1 is not an integer, then for some integer p ≥ 1 we have that q = αp+(1−α)(p+1)
for some 0 < α < 1. By Hölder’s inequality, for any g we have
kgkqq,R1 =

(6.15)
so that
(6.16)

Z

(1−α)(p+1)

R1

|g(x)|αp+(1−α)(p+1) dx ≤ kgkαp
p,R1 kgkp+1,R1
(1−α)(p+1)/q

αp/q

≤ kgkp,R1 + kgkp+1,R1

kgkq,R1 ≤ kgkp,R1 kgkp+1,R1

The general case of q > 2 then follows from case of integral q ≥ 2. Finally, if 1 < q < 2,
since kL·1 k1,R1 = 1 we obtain (6.1) using (6.16). For (6.2) and (6.3), if q = α + 2(1 − α)
y+·
for some 0 < α < 1, we have by (6.15) and the fact that kL1,ǫ
− Lx+·
1,ǫ k1,R1 ≤ 2 that
2(1−α)/q

y+·
y+·
α/q
kL1,ǫ
− Lx+·
kL1,ǫ
− Lx+·
1,ǫ kq,R1 ≤ 2
1,ǫ k2,R1

(6.17)

and (6.2)-(6.3) for 1 < q < 2 now follows from the case of q = 2.
We now turn to the analogue of (6.1) for the symmetric stable process in TM . We
have as above for integer m, p
(6.18)



E

m
Y

j=1





−mp
(L̄t j )p  = E 
M
x

m
Y

X

= M −mp

ei(

Z 1 )mp j=1
λ· ∈( 2π
M

E kL̄·1 kpm
p,TM

(6.19)

´

Pp


Z
=E

k=1

λj,k )xj

m
Y

m
TM
j=1

X

= M −m(p−1)

λ· ∈( 2π
Z 1 )m(p−1)
M

where in the last line λj,p = −
³

E kL̄·1 kpm
p,TM

(6.20)
≤

XZ

π∈C Dπ

≤c

mp



´

Pp−1

 −m(p−1)
M

XZ

π∈C Dπ




Y

n∈Vπc

k=1

(

[0,1]mp

λ· ∈( 2π
Z 1 )mp
M

so that as before
³

Z

X



E

x
(L̄t j )p

Z

[0,1]mp

p
m Y
Y

j=1 k=1

p
m Y
Y

j=1 k=1



eiλj,k (xj −Xtj,k ) dtj,k 



e−iλj,k Xtj,k 

p
m Y
Y

dtj,k

j=1 k=1



dxj 


E

p
m Y
Y

j=1 k=1



e−iλj,k Xtj,k 

p
m Y
Y

dtj,k

j=1 k=1

λj,k . Again as above this leads to

X

−

e

λ· ∈( 2π
Z 1 )m(p−1)
M

P

n∈Vπc



|uπ,n |β (tπ(n) −tπ(n−1) ) 

)

mp
Y
1

dtπ(n)
1+
|tπ(n) − tπ(n−1) |1/β
n=1



mp
Y

dtπ(n)

n=1

and as before this leads to the analogue of (6.1) for the symmetric stable process in T M .
The analogues of (6.2)-(6.3) follows similarly. This completes the roof of Lemma 8.
598

Consider the Young’s function ψ(x) = exp(x)−1, and let k·kψ denotes the Orlicz space
norm with respect to E. By (5.1.9) of [8] we have that for any finite set of non-negative
random variables Y1 , . . . , Yn
k sup Yk kψ ≤ cψ −1 (n) sup k Yk kψ .

(6.21)

k≤n

k≤n

Let D denote the set of dyadic numbers in [0, 1]. Thus D = ∪m Dm where Dm is the
set of numbers in [0, 1] of the form i/2m for some integer i. The next Lemma follows from
a standard chaining argument, see e.g. [12], Chapter 11, which also contains historical
references.
Lemma 9 Let {Zt , t ∈ D ⊆ [0, 1]} be a Banach space valued stochastic process such that
for some finite constants c, ζ > 0
k |Zt − Zs |kψ ≤ c|t − s|ζ ,

(6.22)

∀s, t ∈ D.

Then for any ζ ′ < ζ that
(6.23)

′

k sup |Zt − Zs |/|s − t|ζ kψ ≤ 2
s,t∈D

s6=t

∞
X

m=0

′

C2ζ (m+1) 2−mζ < ∞.

Lemma 10 Let p > 1. For the symmetric stable process in R1 of index β > 1, for some
ζ′ > 0
°
°
kL1y+· − Lx+·
°
1 kp,R1 °
° sup
(6.24)
° < ∞.
ψ
|x − y|ζ ′
x6=y
Furthermore, for any α > 0

(6.25)

1
lim sup log P
δ→0 t≥1 t

Ã

sup

|x−y|≤δ

kLty+·

!

− Lx+·
t kp,R1 > αt = −∞.

Similar results hold for the symmetric stable process in TM of index β > 1 with kLty+· −
y+·
Lx+·
− L̄x+·
t kp,TM .
t kp,R1 replaced by kL̄t
Proof of Lemma 10: Using (6.3) we see that for some c < ∞ and all x, y
ζ
k kL1y+· − Lx+·
1 kp,R1 kψ ≤ c|x − y| .

(6.26)

Using Lemma 9 we see that
°
°
kL1y+· − Lx+·
°
1 kp,R1 °
° sup
° < ∞.
′
ζ

(6.27)

x6=y

x,y∈D

|x − y|

ψ

But since, using Fatou’s Lemma and the continuity of local time
(6.28)

sup
x6=y

x,y∈[0,1]

kL1y+· − Lx+·
kL1y+· − Lx+·
1 kp,R1
1 kp,R1
=
sup
′
|x − y|ζ
|x − y|ζ ′
x6=y
x,y∈D

599

we get

°
°
°

(6.29)
Note that for any z

sup
x6=y

x,y∈[0,1]

°
kL1y+· − Lx+·
1 kp,R1 °
° < ∞.
ψ
|x − y|ζ ′

kL1z+y+· − Lz+x+·
kp,R1 = kL1y+· − Lx+·
1
1 kp,R1

(6.30)
hence

kL1y+· − Lx+·
kL1y+· − Lx+·
1 kp,R1
1 kp,R1
=
sup
.
′
|x − y|ζ
|x − y|ζ ′
x6=y
0 0
sup
x6=y

′
1
||L1·+x − L·+y
||Lt·+x − L·+y
t ||p,R1 d 1+ pβ
− 1+ζ
1 ||p,R1
β
sup
=
t
.
′
ζ
|x − y|
|x − y|ζ ′
x6=y

Hence by (6.24)

y+·

°
kL
sup °° sup t

(6.32)

1≤t≤2

so that

x6=y

(

°
− Lx+·
t kp,R1 °
° ≤ K < ∞,
ψ
|x − y|ζ ′
)

´
kLty+· − Lx+·
t kp,R1
sup E exp sup
(6.33)
≤ 2.
′
K|x − y|ζ
1≤t≤2
x6=y
Using the additivity of local times and the Markov property we then have that for any
t≥1
)
(
´
³
kLty+· − Lx+·
t kp,R1
(6.34)
≤ 2t .
E exp sup
K|x − y|ζ ′
x6=y
Hence by Chebycheff

(6.35)

sup P
t≥1

Ã

sup
|x−y|≤δ

Ã

³

kLty+·

−

Lx+·
t kp,R1

> αt

!

kLty+· − Lx+·
αt
t kp,R1
≤ sup P
sup
>
′
K|x − y|ζ
Kδ ζ ′
t≥1
|x−y|≤δ

!

≤ e−αtK

−1 δ −>ζ ′

2t .

Our lemma then follows.
For the next lemma let Ra L·t denote the restriction of Lxt to x ∈ [−a, a].
Lemma 11 Let p > 1. For the symmetric stable process in R1 of index β > 1, for any
0 < a < ∞ we can find a compact K ⊆ Lp ([−a, a]) such that
(6.36)

lim lim sup
γ→∞
t→∞

³
´
1
log P t−1 Ra L·t ∈
/ γK = −∞.
t

For the symmetric stable process in TM of index β > 1, we can find a compact K ⊆
L (TM ) such that
³
´
1
−1 ·
lim
lim
sup
(6.37)
log
P
t
L
∈
/
γK
= −∞.
t
γ→∞ t→∞ t
p

600

Proof of Lemma 11: By (6.25), for any k ≥ 1, there is a δk > 0 such that
sup
t≥1

1
1
log P { sup ||Lt·+h − L·t ||p,R1 ≥ t} ≤ −k
t
k
|h|≤δk

Similarly we can take N > 0 such that
sup
t≥1

1
log P {||L·t ||p,R1 ≥ N t} ≤ −1.
t

If
AN,δk = {f | kf kp,R1 ≤ N }

∞ n
\

f|

k=1

sup kf (x + h) − f (x)kp,R1 ≤

|h|≤δk

1o
k

we take to be the closure of Ra AN,δk in Lp ([−a, a]). By Lemma 15, K is a compact subset
of Lp ([−a, a]) and we have for any γ ≥ 1
(6.38)

P {t−1 Ra L·t 6∈ γK}
≤ P {||L·t ||p,R1 ≥ N γt} +
h

−γ −1

≤ 1 + (1 − e )
Lemma 11 follows immediately.

7

i

−tγ

e

∞
X

k=1

P { sup ||L·+h
− L·t ||p,R1 ≥
t
|h|≤δk

1
γt}
k

.

Random walks

We begin by studying exponential integrabilty for local times of random walks. We will
use the notation
(7.1)
lenx = b(n)1−1/p n−1 lnx .
Lemma 12 Let p > 1. For any γ < ∞

(7.2)
and
(7.3)
For some ζ > 0
(7.4)
and
(7.5)

³

³

sup E exp γklen· kp,Z 1
n

³

³

´´

lim sup E exp γklen· kp,Z 1

γ→0

n

Ã

Ã

´´

kle· − leny+· kp,R1
sup E exp γ n
|y/b(n)|ζ
n,y
Ã

Ã

0 we can find a
c > 0 such that
β−ζ
(7.8)
|φ(u/b(n))| ≤ e−c|u| /n ,
1 ≤ |u| ≤ πb(n).
Hence for any s ≤ n

Z

(7.9)

R1

|φ(u/b(n))|s |u|a du

≤C+

Z

s

R1

e−c n |u|

β−ζ

|u|a du

µ ¶−(1+a)/(β−ζ)

s
≤C +C
n

≤C

µ ¶−(1+a)/(β−ζ)

s
n

.

The proof of (7.2) is then completed by following the proof of Lemma 8 and (7.3)
follows similarly.
For (7.4) we see as in the derivation of (7.7) that when p > 1 is an integer.
(7.10)

klen· − len·+y kpp,Z 1 =
·Z

1
p
n (2π)p−1
i

b(n)[−π,π](p−1)

e

Pp

j=1

n
X

n1 ,...,np =0

λj ·Snj /b(n)

p
Y

j=1

(1 − e

iλj ·y/b(n)

)

p−1
Y
j=1

¸

dλj .

Using |1 − eiλj ·y/b(n) | ≤ |λj · y/b(n)|ζ and (7.9) the proof of Lemma 12 is then completed
by following the proof of Lemma 8.
Set
(7.11)
lbnx = b(n)n−1 ln[b(n)x] .
Lemma 13 Let p > 1. For any γ < ∞

(7.12)

³

³

sup E exp γklbn· kp,R1
n

602

´´

0.
(7.18)


X Z

klbn· − lbny+· kp,R1 = 

j∈Z 1

and

Z

(7.19)

{x: [b(n)x]=j}

=

Z

{x: [b(n)x]=j}

1/p


( lbnx − lbnx+y )p dx

( lbnx − lbnx+y )p dx

{x: [b(n)x]=j}

( b(n)n−1 (lnj − ln[b(n)(x+y)] ) )p dx.

Let u = b(n)x − [b(n)x], v = b(n)y − [b(n)y]. Then 0 ≤ u, v ≤ 1 and [b(n)(x + y)] =
[b(n)x] + [b(n)y] if u < 1 − v while [b(n)(x + y)] = [b(n)x] + [b(n)y] + 1 if u ≥ 1 − v. Thus
(7.20)

Z

{x: [b(n)x]=j}
−1

( b(n)n−1 (lnj − ln[b(n)(x+y)] ) )p dx

= ( b(n)n (lnj − lnj+[b(n)y] ) )p (1 − v)/b(n)
+( b(n)n−1 (lnj − lnj+[b(n)y]+1 ) )p v/b(n)
= ( (lenj − lenj+[b(n)y] ) )p (1 − v) + ( (lenj − lenj+[b(n)y]+1 ) )p v.
603

Thus by (7.18)-(7.20) for any ζ ≤ 1/p
klbn· − lbny+· kp,R1

(7.21)

≤ (1 − v)1/p klen· − len[b(n)y]+· kp,Z 1 + v 1/p klen· − len[b(n)y]+1+· kp,Z 1
≤ (1 − v)ζ klen· − len[b(n)y]+· kp,Z 1 + v ζ klen· − len[b(n)y]+1+· kp,Z 1 .

Note that if [b(n)y] = 0 then the first term on the right is 0 and v = b(n)y so that we
have the bound
klbn· − lbny+· kp,R1 ≤ (b(n)y)ζ klen· − len1+· kp,Z 1 = y ζ

(7.22)

klen· − len1+· kp,Z 1
.
(1/b(n))ζ

If [b(n)y] ≥ 1 we obtain

klbn· − lbny+· kp,R1

(7.23)

≤

(1 − v)ζ [b(n)y]ζ klen· − len[b(n)y]+· kp,Z 1
b(n)ζ
([b(n)y]/b(n))ζ
+

v ζ ([b(n)y] + 1)ζ klen· − len[b(n)y]+1+· kp,Z 1
.
b(n)ζ
(([b(n)y] + 1)/b(n))ζ

Since
(1 − v)[b(n)y] + v([b(n)y] + 1) = b(n)y
we see from (7.23) that
klbn· − lbny+· kp,R1
yζ
kle· − len[b(n)y]+· kp,Z 1
klen· − len[b(n)y]+1+· kp,Z 1
≤ n
+
.
([b(n)y]/b(n))ζ
(([b(n)y] + 1)/b(n))ζ

(7.24)

Using (7.4) then completes the proof of (7.14). (7.15) follows similarly.
(7.16) follows from the proof of Lemma 10, using the fact that lbnx is right continuous
in x.
Lemma 14 Let p > 1. Then
(7.25)

d

lbn· −→ L·1

as Lp (R1 ) valued random variables. In particular,
d

klen· kp,Z 1 −→ kL·1 kp,R1 .

(7.26)

Proof of Lemma 14: Let us first show that the measures induced by the sequence lbn·
on Lp (R1 ) are tight. To do this, let K be the closure of the precompact set
A=

∞ n
\

k=1

f ; f ≡ 0 outside [−a, a], ||f ||p ≤ M and sup ||f (· + h) − f (·)||p ≤
|h|≤δk

604

1o
k

with a, δ1 , δ2 , . . . to be chosen. Then for any ǫ > 0, using (1.18) with a < ∞ sufficiently
large and Lemma 13 as in the proof of Lemma 11 with δk → 0 sufficiently rapidly
(7.27)

P {lbn· 6∈ K} ≤ P {||lbn· ||p ≥ M } + P {max |S(k)| ≥ ab(n)}
k≤n

+

∞
X

k=1

1
}≤ǫ
k

P { sup ||lbn·+h − lbn· ||k ≥
|h|≤δk

which establishes tightness. Hence by Prohorov’s criterion every subsequence lbn· j has a
subsequence which converges in distribution. It only remains to identify the limit with
the measure induced by L·1 on Lp (R1 ). To this end it suffices to show that
Z

(7.28)

∞

−∞

d
f (x)lbnx dx −→

for each f ∈ S(R1 ). But for such f
(7.29)

Z

∞

−∞

1
f (x)lbnx dx =

n

Z

∞

−∞

f

Z

∞

−∞

³ x ´

b(n)

f (x)Lx1 dx

ln[x] dx

n
³ S(k) ´
n
1X