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Vol. 5 (2000) Paper no. 5, pages 1–47.
Journal URL
http://www.math.washington.edu/~ejpecp/
Paper URL
http://www.math.washington.edu/~ejpecp/EjpVol5/paper5.abs.html
THE NORM ESTIMATE OF THE DIFFERENCE BETWEEN
THE KAC OPERATOR AND SCHRÖDINGER SEMIGROUP II:
THE GENERAL CASE INCLUDING THE RELATIVISTIC CASE
Takashi Ichinose1
Department of Mathematics, Kanazawa University, Kanazawa, 920-1192, Japan
ichinose@kappa.s.kanazawa-u.ac.jp
Satoshi Takanobu2
Department of Mathematics, Kanazawa University, Kanazawa, 920-1192, Japan
takanobu@kappa.s.kanazawa-u.ac.jp
Abstract More thorough results than in our previous paper in Nagoya Math. J. are given
on the Lp -operator norm estimates for the Kac operator e−tV /2 e−tH0 e−tV /2 compared with the
Schrödinger semigroup e−t(H0 +V ) . The Schrödinger operators H0 + V to be treated in this paper
are more general ones associated with the Lévy process, including the relativistic Schrödinger
operator. The method of proof is probabilistic based on the Feynman-Kac formula. It differs
from our previous work in the point of using the Feynman-Kac formula not directly for these
operators, but instead through subordination from the Brownian motion, which enables us to
deal with all these operators in a unified way. As an application of such estimates the Trotter
product formula in the Lp -operator norm, with error bounds, for these Schrödinger semigroups
is also derived.
Keywords Schrödinger operator, Schrödinger semigroup, relativistic Schrödinger operator,
Trotter product formula, Lie-Trotter-Kato product formula, Feynman-Kac formula, subordination of Brownian motion, Kato’s inequality
AMS subject classification 47D07, 35J10, 47F05, 60J65, 60J35
Submitted to EJP on October 21, 1999. Final version accepted on January 26, 2000.
1
Partially supported by Grant-in-Aid for Scientific Research No. 11440040, Ministry of Education, Science and
Culture, Japan
2
Partially supported by Grant-in-Aid for Scientific Research No. 10440030, Ministry of Education, Science and
Culture, Japan.
1. Introduction
By the Kac operator we mean an operator of the kind K(t) = e−tV /2 e−tH0 e−tV /2 , where H =
H0 + V ≡ −∆/2 + V (x) is the nonrelativistic Schrödinger operator in L2 (R d ) with mass 1 with
scalar potential V (x) bounded from below. This K(t) may correspond to the transfer operator for
a lattice model in statistical mechanics studied by M. Kac [Ka]. There it is one of the important
problems to know asymptotic spectral properties of K(t) for t ↓ 0. To this end, in [H1, H2] Helffer
estimated the L2 -operator norm of the difference between K(t) and the Schrödinger semigroup
e−tH to be of order O(t2 ) for small t > 0, if V (x) satisfies |∂ α V (x)| ≤ Cα (1 + |x|2 )(2−|α|)+ /2
for every multi-index α with a constant Cα . Then such norm estimates may be applied to get
spectral properties of K(t) in comparison with those of H.
In [I-Tak1] and [I-Tak2] we have extended his result to the case of more general scalar potentials V (x) even in the Lp -operator norm, 1 ≤ p ≤ ∞, making a probabilistic approach based
on the Feynman-Kac formula. In [I-Tak2] we have also considered this problem for both the
nonrelativistic Schrödinger
operator H = H0 + V and the relativistic Schrödinger operator
√
H r = H0r + V ≡ −∆ + 1 − 1 + V (x) with light velocity 1. The Lp -operator norm of this
difference is estimated to be of order O(ta ) of small t > 0 with a ≥ 1, though the relativistic
case shows for small t > 0 a slightly different behavior from the nonrelativistic case. As another
application of these results the Trotter product formula for the nonrelativistic and relativistic
Schrödinger operators in the Lp -operator norm with error bounds is obtained. There are also
related L2 results with operator-theoretic methods, for which we refer to [D-I-Tam].
The aim of this paper is to generalize and refine the result of [I-Tak2] in the relativistic case,
√ admitting of more general operators than the free relativistic Schrödinger operator
H0r = −∆ + 1 − 1 as well as relaxing the conditions for the potentials V (x). We use the
probabilistic method with Feynman-Kac formula, though observing everything in a unified way
through subordination from the Brownian motion. In this respect the present method differs
from that in [I-Tak2] used for the relativistic Schrödinger operator H r , which made the best of
r
the explicit expression of the integral kernel of e−tH0 .
The more general operator we have in mind is the following operator
H0ψ = ψ( 12 (−∆ + 1)) − ψ( 12 ),
(1.1)
which will play the same role as the relativistic Schrödinger operator
√
H0r = −∆ + 1 − 1
(1.2)
in [I-Tak2]. Obviously, H0ψ is a selfadjoint operator in L2 (R d ). Here ψ(λ) is a continuous
increasing function on [0, ∞) with ψ(0) = 0 and ψ(∞) = ∞ expressed as
Z
ψ(λ) =
(1 − e−λl )n(dl), λ ≥ 0,
(1.3)
(0,∞)
where n(dl) is a Lévy measure on (0, ∞) (i.e. a measure on (0, ∞) such that
with n((0, ∞)) = ∞. It is clear that
Z
ψ(λ + 12 ) − ψ( 12 ) =
(1 − e−λl )e−l/2 n(dl).
(0,∞)
2
R
(0,∞) l∧1n(dl)
< ∞)
(1.4)
As a special case of H0ψ we have for ψ(λ) = (2λ)α , 0 < α < 1, the operator
(α)
H0
= (−∆ + 1)α − 1,
(1.5)
(1/2)
which reduces to the relativistic Schrödinger operator when α = 1/2: H0
the Lévy measure is n(dl) = {2α α/Γ(1 − α)}l−1−α dl.
= H0r . In this case
To formulate our results we are going to describe what kind of function V (x) is. Let 0 < γ, δ ≤ 1,
0 ≤ κ ≤ 1, 0 ≤ µ, ν, ρ < ∞, 0 ≤ C1 , C2 , c1 , c2 < ∞ and 0 < c < ∞. Let V : R d → [0, ∞) be a
continuous function satisfying one of the following five conditions:
(A)0
|V (x) − V (y)| ≤ C1 |x − y|γ ;
(A)1
V is a C 1 -function such that
(i) |∇V (z)| ≤ C1 (1 + V (z)1−δ ), (ii) |∇V (x) − ∇V (y)| ≤ C2 |x − y|κ ;
(A)2
V is a C 1 -function such that
(i) |∇V (z)| ≤ C1 (1 + V (z)1−δ ),
(ii) |∇V (x) − ∇V (y)|
n
o
≤ C2 V (x)(1−2δ)+ (1 + |x − y|µ ) + 1 + |x − y|ν |x − y|;
(V)1
V is a C 1 -function such that
(i) V (z) ≥ chziρ , (ii) |∇V (z)| ≤ c1 hzi(ρ−1)+ ;
(V)2
V is a C 2 -function such that
(i) V (z) ≥ chziρ , (ii) |∇V (z)| ≤ c1 hzi(ρ−1)+ ,
(iii) |∇2 V (z)| ≤ c2 hzi(ρ−2)+ .
Here hzi :=
p
1 + |z|2 .
Conditions (A)0 , (A)1 and (A)2 on V (x) are used in [Tak] and are more general than in [ITak1,2], while conditions (V)1 and (V)2 are used in [D-I-Tam]. But these conditions may not be
best possible. A simple example of a function which has property (A)0 , (A)1 or (A)2 is, needless
to say, V (x) = |x|r (0 < r < ∞), and a slightly complicated one V (x) = |x|r (2 + sin log |x|),
according as 0 < r ≤ 1, 1 < r < 2 or r ≥ 2. Also V (x) = 1 + |x1 − x2 |r (x = (x1 , x2 , . . . , xd ))
satisfies (A)0 , (A)1 or (A)2 with the same r as above, but neither (V)1 nor (V)2 . To the contrary
R |x|
V (x) = 1 + |x| 0 (1 + sin(θ 2 ))dθ satisfies (V)1 , but neither (V)2 , (A)0 , (A)1 nor (A)2 .
The operator H0ψ +V is essentially selfadjoint on C0∞ (R d ), and so its unique selfadjoint extension
ψ
is also denoted by the same H0ψ + V . The semigroup e−t(H0 +V ) on L2 (R d ) is extended to a
strongly continuous semigroup on Lp (R d ) (1 ≤ p < ∞) and C∞ (R d ), to be denoted by the same
ψ
e−t(H0 +V ) . Here C∞ (R d ) is the Banach space of the continuous functions on R d vanishing at
infinity. To be complete, these and further facts are proved in Appendix.
3
As for the Lévy measure n(dl) introduced in (1.3) and (1.4), we make the following assumption:
(L)
For some α ∈ [0, 1], n((·, ∞)) is regularly varying at zero with exponent −α, i.e., there
exists a slowly varying function L(λ) at infinity such that
n((t, ∞)) ∼ t−α L( 1t )
as t ↓ 0.
(1.6)
Here a positive function L(·) is called slowly varying at infinity if for any c > 0,
lim
λ↑∞
L(cλ)
= 1.
L(λ)
Let φ−1 (·) be the inverse function of φ(λ) := ψ(λ + 1/2) − ψ(1/2). (Note that φ is strictly
increasing.) Under the above assumption, set
(
L1 (λ) :=
Γ(1 − α)L(λ)
if 0 ≤ α < 1
R 1/λ
n((s, ∞))ds if α = 1,
0
L2 (x) := L1 (φ−1 (x))−1/α
if
0 < α ≤ 1.
These two functions are slowly varying at infinity, and we have φ(λ) ∼ λα L1 (λ)Ras λ → ∞ and
φ−1 (x) ∼ x1/α L2 (x) as x → ∞, as will be seen from Fact in Section 6, so that 0· (φ−1 (θ))−α dθ
(0 < α < 1) is also slowly varying at infinity.
Now we state the main results of this paper, which generalize the results in [I-Tak2]. In the
following k · kp→p stands for the Lp -operator norm for 1 ≤ p < ∞ and the supremum norm on
C∞ (R d ) for p = ∞.
Theorem 1. Suppose assumption (L) and let 1 ≤ p ≤ ∞. Then the following estimates (i), (ii)
and (iii) hold for small t > 0.
(i) Under (A)0 ,
ψ
ψ
ke−tV /2 e−tH0 e−tV /2 − e−t(H0 +V ) kp→p ,
ψ
ψ
ke−tV e−tH0 − e−t(H0 +V ) kp→p ,
ψ
ψ
ψ
ke−tH0 /2 e−tV e−tH0 /2 − e−t(H0 +V ) kp→p

if α < γ/2
O(t2 )





R 1/t
=
O(t2 0 (φ−1 (θ))−α dθ) if α = γ/2





O(t1+ γ/2α L2 ( 1t )−γ/2 )
if γ/2 < α.
(ii) Under (A)1 ,
ψ
ψ
ke−tV /2 e−tH0 e−tV /2 − e−t(H0 +V ) kp→p
4

if
O(t1+1∧2δ )





R 1/t
=
O(t1+2δ ) + O(t2 0 (φ−1 (θ))−α dθ)
if





O(t1+2δ ) + O(t1+ (1+κ)/2α L2 ( 1t )−(1+κ)/2 ) if
ψ
α < (1 + κ)/2 or κ = 1
α = (1 + κ)/2 < 1
(1 + κ)/2 < α,
ψ
ke−tV e−tH0 − e−t(H0 +V ) kp→p ,
ψ
ψ
ψ
ke−tH0 /2 e−tV e−tH0 /2 − e−t(H0 +V ) kp→p

if α < 1/2
O(t1+δ )





R 1/t
=
O(t1+δ 0 (φ−1 (θ))−α dθ) if α = 1/2





O(tδ+ 1/2α L2 ( 1t )−1/2 )
if 1/2 < α.
(iii) Under (A)2 ,
ψ
= O(t1+1∧2δ ),
ψ
ke−tV /2 e−tH0 e−tV /2 − e−t(H0 +V ) kp→p
ψ
ψ
ke−tV e−tH0 − e−t(H0 +V ) kp→p ,
ψ
ψ
ψ
ke−tH0 /2 e−tV e−tH0 /2 − e−t(H0 +V ) kp→p

if α < 1/2
O(t1+δ )





R 1/t
=
O(t1+δ 0 (φ−1 (θ))−α dθ) if α = 1/2





O(tδ+ 1/2α L2 ( 1t )−1/2 )
if 1/2 < α.
In fact, the first estimate in (iii) holds independent of (L).
A consequence of Theorem 1 is the following Trotter product formula in the Lp -operator norm
with error bounds.
Theorem 2. Suppose assumption (L) and let 1 ≤ p ≤ ∞. Then the following estimates (i),
(ii), (iii) and (iv) hold uniformly on each finite t-interval on [0, ∞).
(i) Under (A)0 ,
ψ
ψ
k(e−tV /2n e−tH0 /n e−tV /2n )n − e−t(H0 +V ) kp→p ,
ψ
ψ
k(e−tV /n e−tH0 /n )n − e−t(H0 +V ) kp→p ,
ψ
ψ
ψ
k(e−tH0 /2n e−tV /n e−tH0 /2n )n − e−t(H0 +V ) kp→p

if α < γ/2
O(n−1 )





Rn
O(n−1 0 (φ−1 (θ))−α dθ) if α = γ/2
=





O(n−γ/2α L2 (n)−γ/2 )
if γ/2 < α.
5
(ii) Under (A)1 ,
ψ
ψ
k(e−tV /2n e−tH0 /n e−tV /2n )n − e−t(H0 +V ) kp→p ,
ψ
ψ
k(e−tV /n e−tH0 /n )n − e−t(H0 +V ) kp→p ,
ψ
ψ
ψ
k(e−tH0 /2n e−tV /n e−tH0 /2n )n − e−t(H0 +V ) kp→p

if
O(n−1∧2δ )





Rn
O(n−2δ ) + O(n−1 0 (φ−1 (θ))−α dθ)
if
=





O(n−2δ ) + O(n−(1+κ)/2α L2 (n)−(1+κ)/2 ) if
α < (1 + κ)/2 or κ = 1
α = (1 + κ)/2 < 1
(1 + κ)/2 < α.
(iii) Under (A)2 ,
ψ
ψ
k(e−tV /2n e−tH0 /n e−tV /2n )n − e−t(H0 +V ) kp→p ,
ψ
ψ
k(e−tV /n e−tH0 /n )n − e−t(H0 +V ) kp→p ,
ψ
ψ
ψ
k(e−tH0 /2n e−tV /n e−tH0 /2n )n − e−t(H0 +V ) kp→p
= O(n−1∧2δ ).
(iv) Under (V)i (i = 1, 2),
ψ
ψ
k(e−tV /2n e−tH0 /n e−tV /2n )n − e−t(H0 +V ) kp→p ,
ψ
ψ
k(e−tV /n e−tH0 /n )n − e−t(H0 +V ) kp→p ,
ψ
ψ
ψ
k(e−tH0 /2n e−tV /n e−tH0 /2n )n − e−t(H0 +V ) kp→p
= O(n−i/2∨ρ ).
In fact, the asymptotic estimates (iii) and (iv) hold independent of (L).
ψ
Notice here that though the estimates with small t, in Theorem 1, for e−tV e−tH0 and
ψ
ψ
ψ
e−tH0 /2 e−tV e−tH0 /2 are of worse order than that for e−tV /2 e−tH0 e−tV /2 , one has, in Theorem 2,
the same error bounds with large n for these three products.
Finally we give a comment on what kind of operators are to be covered by our H0ψ +V . To this end
(α)
we briefly illustrate how our result reads on the Trotter product formula in the case H0 +V with
(α)
H0 = (−∆ + 1)α − 1, 0 < α < 1, in (1.5). In this case, we have n((t, ∞)) = (2α /Γ(1 − α)) t−α ,
or L2 (·) ≡ 2−1 , so that
Z x
(φ−1 (θ))−α dθ ∼ 2α log x as x → ∞.
0
Therefore Theorem 2 says that for 1 ≤ p ≤ ∞ and uniformly on each finite t-interval in [0, ∞),
(α)
k(e−tV /2n e−tH0
(α)
−tH0 /n
k(e−tV /n e
(α)
−tH0 /2n
k(e
(α)
/n −tV /2n n
) − e−t(H0
e
(α)
−t(H0 +V
)n − e
(α)
−tH0 /2n
e−tV /n e
)
+V )
kp→p ,
kp→p ,
(α)
)n − e−t(H0
6
+V ) k
p→p

if α < γ/2
O(n−1 )





O(n−1 log n) if α = γ/2
=





O(n−γ/2α )
if γ/2 < α


O(n−1∧2δ )







 O(n−1 log n)
=
under (A)0 ,
if
α < (1 + κ)/2
if
α = (1 + κ)/2 and 1/2 ≤ δ ≤ 1


O(n−2δ )
if






 O(n−2δ∧ (1+κ)/2α ) if
under (A)1 .
α = (1 + κ)/2 and 0 < δ < 1/2
(1 + κ)/2 < α
An important remark is the following. In the above example, the case α = 1 is missing. This is
equivalent to the nonrelativistic case H0 + V = −∆/2 + V (x), treated in [Tak] (cf. [I-Tak1,2]).
However we may think that this case is also
in our results, Theorems 1 and
√ implicitly contained
r
2
2
4
2, for α = 1/2. Indeed, by using H0 (c) = −c ∆ + c − c with light velocity c restored in place
of H0r in (1.2), we can obtain the case α = 1/2 so as to involve the parameter c (light velocity).
r
Since, in the nonrelativistic limit c → ∞, the relativistic Schrödinger semigroup e−t(H0 (c)+V ) is
strongly convergent to the nonrelativistic Schrödinger semigroup e−t(H0 +V ) uniformly on each
finite t-interval in [0, ∞) (e.g. [I2]), we can reproduce the nonrelativistic result in [Tak] (cf.
Remark following Theorem 2.3).
In Section 2, we state our results in more general form: we generalize Theorems 1 and 2 to
Theorems 2.1 and 2.2 / 2.3 by introducing the subordinator σt , namely, a time-homogeneous
Lévy process associated with the Lévy measure e−l/2 n(dl). Moreover we state Theorem 2.4 on
asymptotics of the moments of the process σt . Once we know these asymptotics, we can obtain
Theorems 1 and 2 from Theorems 2.1 and 2.2 / 2.3. These four theorems are proved in Sections
3 – 6.
ψ
In Appendix, we give a full study of the semigroups e−t(H0 +V ) , t ≥ 0, on Lp (R d ), 1 ≤ p < ∞
and C∞ (R d ) defined through the Feynman-Kac formula. We show they constitute a strongly
continuous contraction semigroup there. It is also shown that its infinitesimal generator ψ,V
p
ψ
∞
d
has C0 (R ) as a core, by establishing Kato’s inequality for the operator H0 . Some of these
results seem to be new.
G
The authors would like to thank the referee for his / her careful reading of the manuscript and
for a number of comments.
2. General results
In this section we shall prove the theorems in a little more general setting based on probability
theory. To describe it we introduce some notations and notions. For a continuous function
V : R d → [0, ∞), set
ψ
K(t) := e−tV /2 e−tH0 e−tV /2 ,
ψ
G(t) := e−tV e−tH0 ,
7
ψ
ψ
R(t) := e−tH0 /2 e−tV e−tH0 /2
and
ψ
QK (t) := K(t) − e−t(H0 +V ) ,
ψ
QG (t) := G(t) − e−t(H0 +V ) ,
ψ
QR (t) := R(t) − e−t(H0 +V ) .
Suppose we are given the independent random objects N (·) and B(·) on some probability space
(Ω, F, P):
(i) N (dsdl) is a Poisson random measure on [0, ∞)×(0, ∞) such that E [N (dsdl)] = dse−l/2 n(dl);
(ii) (B(t))t≥0 is a d-dimensional Brownian motion starting at 0.
Set
t+Z
Z
σt :=
l N (dsdl).
0
(2.1)
(0,∞)
Then (σt )t≥0 is a time-homogeneous Lévy process with increasing paths such that
E [e−λσt ] = e−t(ψ(λ+ 1/2)−ψ(1/2))
(2.2)
(e.g. Note 1.7.1 in [It-MK]). Note that σt has moments of all order (cf. (6.1)), which is to be
seen at the beginning of Section 6. We use a subordination of B(·) by a subordinator σ· , i.e., a
process (B(σt ))t≥0 on R d . This is a Lévy process such that
E [e
√
−1hp,B(σt )i
] = e−t(ψ((|p|
2 +1)/2)−ψ(1/2)
),
ψ
which corresponds to the semigroup {e−tH0 }t≥0 with generator H0ψ in (1.1).
We prove the following generalization of Theorems 1 and 2.
Theorem 2.1. Let 1 ≤ p ≤ ∞ and t ≥ 0.
(i) Under (A)0 ,
γ/2
kQK (t)kp→p , kQG (t)kp→p , kQR (t)kp→p ≤ const(γ, d) C1 t E [σt
].
(ii) Under (A)1 ,
h
i
2
P
j(1+κ)/2
kQK (t)kp→p ≤ const(δ, κ, d) C12 (t2 + t2δ )E [σt ] +
(C2 t)j E [σt
] ,
j=1
kQG (t)kp→p , kQR (t)kp→p ≤ const(δ, κ, d)
2 n
P
j=1
8
j/2
j(1+κ)/2
C1j (tj + tjδ )E [σt ] + (C2 t)j E [σt
o
] .
(iii) Under (A)2 ,
h
2 n
P
kQK (t)kp→p ≤ const(δ, µ, ν, d) C12 (t2 + t2δ )E [σt ] +
(C2 t)j E [σtj ]
j=1
j(1+ ν/2)
+ (C2 t)j E [σt
] + (C2 t1∧2δ )j E [σtj ] + (C2 t1∧2δ )j E [σt
j(1+ µ/2)
kQG (t)kp→p , kQR (t)kp→p ≤ const(δ, µ, ν, d)
oi
] ,
2 n
P
j/2
C1j (tj + tjδ )E [σt ] + (C2 t)j E [σtj ]
j=1
] + (C2 t1∧2δ )j E [σtj ]
o
j(1+ µ/2)
+ (C2 t1∧2δ )j E [σt
] .
j(1+ ν/2)
+ (C2 t)j E [σt
Theorem 2.2. Let 1 ≤ p ≤ ∞, t ≥ 0 and n ∈ N .
(i) Under (A)0 ,
ψ
ψ
k(e−tV /2n e−tH0 /n e−tV /2n )n − e−t(H0 +V ) kp→p ,
ψ
ψ
k(e−tV /n e−tH0 /n )n − e−t(H0 +V ) kp→p ,
ψ
ψ
ψ
k(e−tH0 /2n e−tV /n e−tH0 /2n )n − e−t(H0 +V ) kp→p
γ/2
≤ const(γ, d) C1 t E [σt/n ].
(ii) Under (A)1 ,
ψ
ψ
k(e−tV /2n e−tH0 /n e−tV /2n )n − e−t(H0 +V ) kp→p
h i
2
P
j(1+κ)/2
≤ const(δ, κ, d) C12 ( nt )2 + ( nt )2δ nE [σt/n ] +
(C2 nt )j nE [σt/n
] ,
j=1
ψ
ψ
k(e−tV /n e−tH0 /n )n − e−t(H0 +V ) kp→p ,
ψ
ψ
ψ
k(e−tH0 /2n e−tV /n e−tH0 /2n )n − e−t(H0 +V ) kp→p
h 1/2
(1+κ)/2
≤ const(δ, κ, d) n1 C1 (t + tδ )E [σt ] + C1 tE [σt
]
+ C1
t
n
i
2
P
1/2
j(1+κ)/2
+ ( nt )δ E [σt/n ] + C12 ( nt )2 + ( nt )2δ nE [σt/n ] +
(C2 nt )j nE [σt/n
] .
j=1
(iii) Under (A)2 ,
ψ
ψ
k(e−tV /2n e−tH0 /n e−tV /2n )n − e−t(H0 +V ) kp→p
h 2 n
P
j
(C2 nt )j nE [σt/n
≤ const(δ, µ, ν, d) C12 ( nt )2 + ( nt )2δ nE [σt/n ] +
]
j=1
j(1+ ν/2)
+ (C2 nt )j nE [σt/n
j
] + (C2 ( nt )1∧2δ )j nE [σt/n
] + (C2 ( nt )1∧2δ )j nE [σt/n
j(1+ µ/2)
9
oi
] ,
ψ
ψ
k(e−tV /n e−tH0 /n )n − e−t(H0 +V ) kp→p ,
ψ
ψ
ψ
k(e−tH0 /2n e−tV /n e−tH0 /2n )n − e−t(H0 +V ) kp→p
h 1/2
1+ µ/2
≤ const(δ, µ, ν, d) n1 C1 (t + tδ )E [σt ] + C2 t1∧2δ (E [σt ] + E [σt
])
1+ ν/2
1/2
+ C2 t(E [σt ] + E [σt
]) + C1 nt + ( nt )δ E [σt/n ] + C12 ( nt )2 + ( nt )2δ nE [σt/n ]
+
2 n
P
j(1+ ν/2)
j
j
(C2 nt )j nE [σt/n
] + (C2 nt )j nE [σt/n
] + (C2 ( nt )1∧2δ )j nE [σt/n
]
j=1
+ (C2 ( nt )1∧2δ )j nE [σt/n
j(1+ µ/2)
oi
] .
Theorem 2.3. Let 1 ≤ p ≤ ∞ and t ≥ 0.
(i) Under (V)1 for n ≥ 22(2∨ρ) ,
ψ
ψ
k(e−tV /2n e−tH0 /n e−tV /2n )n − e−t(H0 +V ) kp→p
h
≤ const(ρ, c, c1 , d) n−1/2∨ρ t2/(ρ∧2)∨1 −1 + (t2 + t2(1∧((ρ∧2)∨1)/2ρ) )nE [σt/n ]
+
2 P
j=1
j(2∨ρ)/2
j
(tj + tj2/2∨ρ )nE [σt/n
] + tj nE [σt/n
ψ
i
] ,
ψ
k(e−tV /n e−tH0 /n )n − e−t(H0 +V ) kp→p ,
ψ
ψ
ψ
k(e−tH0 /2n e−tV /n e−tH0 /2n )n − e−t(H0 +V ) kp→p
h
1/2
≤ const(ρ, c, c1 , d) n−1/2∨ρ t2/(ρ∧2)∨1 −1 + (t + t1∧((ρ∧2)∨1)/2ρ )E [σt ]
(2∨ρ)/2
+ t2/2∨ρ E [σt ] + t(E [σt ] + E [σt
]) + (t2 + t2(1∧((ρ∧2)∨1)/2ρ) )nE [σt/n ]
oi
2 n
P
j(2∨ρ)/2
j
+
(tj + tj2/2∨ρ )nE [σt/n
] + tj nE [σt/n
] .
j=1
(ii) Under (V)2 for n ≥ 1,
ψ
ψ
ke−tV /2n e−tH0 /n e−tV /2n )n − e−t(H0 +V ) kp→p
h
≤ const(ρ, c, c1 , c2 , d) n−2/2∨ρ (t2 + t2/1∨ρ )nE [σt/n ]
+
i
2 P
j(2∨ρ)/2
j
(tj + tj2/2∨ρ )nE [σt/n
] + tj nE [σt/n
] ,
j=1
ψ
ψ
k(e−tV /n e−tH0 /n )n − e−t(H0 +V ) kp→p ,
ψ
ψ
ψ
k(e−tH0 /2n e−tV /n e−tH0 /2n )n − e−t(H0 +V ) kp→p
h
1/2
(2∨ρ)/2
≤ const(ρ, c, c1 , c2 , d) n−2/2∨ρ (t + t1/1∨ρ )E [σt ] + (t + t2/2∨ρ )E [σt ] + tE [σt
]
+ (t2 + t2/1∨ρ )nE [σt/n ] +
2 n
P
j=1
j(2∨ρ)/2
j
(tj + tj2/2∨ρ )nE [σt/n
] + tj nE [σt/n
10
o
]
i
1/2
+ n−1/1∨ρ E [σt/n ](t + t1/1∨ρ ) .
Remark. As noted at the end of Section 1, the nonrelativistic case for H0 + V = −∆/2 + V ,
being equivalent to the case α = 1 which Theorems 1 and 2 fail to cover, can be thought
to be implicitly contained in the relativistic
√ case, of the above three theorems, for the relativistic Schrödinger operator H0r (c) ≡ −c2 ∆ + c4 − c2 with the light velocity c ≥ 1 restored. We have √
H0ψ = H0r (c), where
√ this ψ(λ) is a c-dependent function (1.3) given by
ψ(λ) := ψ(λ; c) = 2c2 λ + c4 − c2 − c4 − c2 associated with the c-dependent Lévy measure
2
e−l/2 n(dl; c) = (2π)−1/2 ce−c l/2 l−3/2 dl. In this case, Theorem 2.1 and Theorems 2.2 / 2.3 hold
with the corresponding c-dependent subordinator σt (c), just as they stand, namely, only with
E [σsa ] replaced by E [σs (c)a ] for each respective s > 0 and a > 0. Then the nonrelativistic case
in question is obtained as the nonrelativistic limit c → ∞ of this c-dependent relativistic case,
turning out to be just Theorems 2.1 and 2.2 / 2.3 with E [σsa ] replaced by sa . This is because
r
one can show that, as c → ∞, the relativistic Schrödinger semigroup e−t(H0 (c)+V ) on the LHS
converges strongly to the nonrelativistic Schrödinger semigroup e−t(H0 +V ) uniformly on each
finite t-interval in [0, ∞) (cf. [I2]), and E [σt (c)a ] on the RHS tends to ta . Then taking the most
dominant contribution on the RHS for small t or large n reproduces the same nonrelativistic
result as in [Tak].
Theorems 1 and 2 follow immediately from Theorems 2.1 and 2.2 / 2.3, if one knows the asymptotics for t ↓ 0 of the moments of σt to investigate which of the terms on the RHS makes a
dominant contribution for small t or large n. These asymptotics are given by the following
theorem.
Theorem 2.4. Suppose assumption (L). Let a > 0.
R
(i) If α < a or a ≥ 1, then (0,∞) la e−l/2 n(dl) < ∞ and
Z
E [σta ]
∼ t
la e−l/2 n(dl)
(0,∞)
as t ↓ 0.
In fact, for a ≥ 1 this always holds independent of (L).
(ii) If α = a and a < 1, then
E [σta ] ∼
1
t
Γ(1 − α)
Z
1/t
(φ−1 (θ))−α dθ
0
as t ↓ 0.
(iii) If 0 < a < α, then
E [σta ] ∼
Γ(1 − αa ) a/α
t L2 ( 1t )−a
Γ(1 − a)
as t ↓ 0.
The proofs of Theorems 2.1, 2.2, 2.3 and 2.4 are given in Sections 3, 4, 5 and 6, respectively. To
show Theorem 2.1, in fact, we prove estimates of the integral kernels of QK (t), QG (t) and QR (t)
by a finite positive linear combination of tc E [|x − y|a σtb p(σt , x − y)], where p(t, x − y) is the heat
kernel (see (A.2)). Such estimates of the integral kernels of the three operators of difference in
Theorems 2.2 / 2.3 also can be obtained (cf. [Tak]), but are omitted.
11
3. Proof of Theorem 2.1
It is easily seen (see (A.6)) that for f ∈ C0∞ (R d )
h
i
K(t)f (x) = E exp − 2t (V (x) + V (x + Xt )) f (x + Xt ) ,
h
i
G(t)f (x) = E exp −tV (x) f (x + Xt ) ,
h
i
R(t)f (x) = E exp −tV (x + Xt/2 ) f (x + Xt )
(3.1)
(3.2)
(3.3)
and generally
h
i
n
t P
K( nt )n f (x) = E exp − 2n
(V (x + X(k−1)t/n ) + V (x + Xkt/n )) f (x + Xt ) ,
G( nt )n f (x)
(3.4)
k=1
h
i
n
t P
= E exp − n
V (x + X(k−1)t/n ) f (x + Xt ) ,
(3.5)
k=1
h
i
n
P
R( nt )n f (x) = E exp − nt
V (x + X(2k−1)t/2n ) f (x + Xt ) .
(3.6)
k=1
Further, for f ∈ C0∞ (R d ) we have (see (A.13))
Z
h
i
QK (t)f (x) =
dyf (y)E σ E B [vK (t, x, y; σ)]p(σt , x − y) ,
Rd
Z
QG (t)f (x) =
Rd
Z
QR (t)f (x) =
Rd
(3.7)
h
i
dyf (y)E σ E B [vG (t, x, y; σ)]p(σt , x − y) ,
(3.8)
h
i
f (y)dy E σ E B [vR (t, x, y; σ)]p(σt , x − y) ,
(3.9)
where E σ and E B are the expectations with respect to σ· and B· , respectively,
Z t
σt ,y
vK (t, x, y; σ) := exp − 2t (V (x) + V (y)) − exp −
V (B0,x
(σs ))ds ,
(3.10)
Z t
σt ,y
vG (t, x, y; σ) := exp −tV (x) − exp −
V (B0,x
(σs ))ds ,
(3.11)
Z t
σt ,y
σt ,y
vR (t, x, y; σ) := exp −tV (B0,x (σt/2 )) − exp −
V (B0,x
(σs ))ds ,
(3.12)
0
0
0
and, for τ > 0, x, y ∈ R d and 0 ≤ θ ≤ τ
τ,y
B0,x
(θ) := x + τθ (y − x) + B0τ (θ)
B0τ (θ)
(3.13)
:= B(θ) − τθ B(τ ).
Since
2
e − e = (a − b)e + (a − b)
a
b
b
Z
1
0
12
(1 − θ)eθa e(1−θ)b dθ,
a, b ∈ R,
we have
vK (t, x, y; σ) = wK (t, x, y; σ) exp − 2t (V (x) + V (y))
Z 1
Z t
σt ,y
2
− wK (t, x, y; σ)
(1 − θ) exp −θ
V (B0,x
(σs ))ds
0
0
t
× exp −(1 − θ) 2 (V (x) + V (y)) dθ
=: vK1 (t, x, y; σ) + vK2 (t, x, y; σ),
vG (t, x, y; σ) = wG (t, x, y; σ) exp −tV (x)
Z 1
Z t
σt ,y
2
− wG (t, x, y; σ)
(1 − θ) exp −θ
V (B0,x
(σs ))ds
0
0
× exp −(1 − θ)tV (x) dθ
(3.14)
=: vG1 (t, x, y; σ) + vG2 (t, x, y; σ),
σt ,y
vR (t, x, y; σ) = wR (t, x, y; σ) exp −tV (B0,x
(σt/2 ))
Z 1
Z t
σt ,y
2
− wR (t, x, y; σ)
(1 − θ) exp −θ
V (B0,x
(σs ))ds
0
0
σt ,y
× exp −(1 − θ)tV (B0,x
(σt/2 )) dθ
(3.15)
=: vR1 (t, x, y; σ) + vR2 (t, x, y; σ),
where
Z
wK (t, x, y; σ) :=
− 2t (V
t
(x) + V (y)) +
0
Z
wG (t, x, y; σ) := −tV (x) +
wR (t, x, y; σ) :=
(3.16)
t
0
σt ,y
V (B0,x
(σs ))ds,
σt ,y
V (B0,x
(σs ))ds,
Z
σt ,y
−tV (B0,x
(σt/2 ))
t
+
0
(3.18)
σt ,y
V (B0,x
(σs ))ds.
When V is further a C 1 -function, since
Z
V (z) − V (w) = h∇V (w), z − wi +
1
0
h∇V (w + θ(z − w)) − ∇V (w), z − widθ,
we have
wK (t, x, y; σ) =
1
2 h∇V
Z
(x) − ∇V (y), y − xi
Z
1
+ 2 h∇V (y), y − xi
t
0
σs
σt ds
t
0
−
σs
σt ds
Z
0
t
σt −σs
σt ds
Z t
D
E
1
+ 2 ∇V (x) + ∇V (y),
B0σt (σs )ds
0
13
(3.17)
(3.19)
Z
Z 1D
∇V (x + θ( σσst (y − x) + B0σt (σs ))) − ∇V (x),
0
0
E
σt
σs
σt (y − x) + B0 (σs ) dθ
Z t Z 1D
1
s
+ 2
∇V (y + θ( σtσ−σ
ds
(x − y) + B0σt (σs ))) − ∇V (y),
t
0
0
E
σt
σt −σs
(x
−
y)
+
B
(σ
)
s dθ
0
σt
+
=:
1
2
t
ds
5
X
wKj (t, x, y; σ),
j=1
D
wG (t, x, y; σ) =
Z
∇V (x),
Z
t
+
t
0
(3.20)
( σσst (y − x) + B0σt (σs ))ds
E
Z 1D
∇V (x + θ( σσst (y − x) + B0σt (σs ))) − ∇V (x),
0
E
σt
σs
(y
−
x)
+
B
(σ
)
s dθ
0
σt
ds
0
=: wG1 (t, x, y; σ) + wG2 (t, x, y; σ),
Z t
D
E
σt ,y
σt ,y
σt ,y
wR (t, x, y; σ) = ∇V (B0,x
(σt/2 )), (B0,x
(σs ) − B0,x
(σt/2 ))ds
Z
t
+
Z 1D
(3.21)
0
σt ,y
σt ,y
σt ,y
∇V (B0,x
(σt/2 ) + θ(B0,x
(σs ) − B0,x
(σt/2 )))
0
E
σt ,y
σt ,y
σt ,y
− ∇V (B0,x
(σt/2 )), B0,x
(σs ) − B0,x
(σt/2 ) dθ
ds
0
=: wR1 (t, x, y; σ) + wR2 (t, x, y; σ).
(3.22)
In the following we shall prove Theorem 2.1 only in Cases (A)2 and (A)0 . The proof of Case
(A)1 is omitted; it is similar to that of (A)2 .
3.1. Case (A)2
In this subsection, we suppose condition (A)2 on V (x).
Claim 3.1.
h
i
E σ E B [vK1 (t, x, y; σ)]p(σt , x − y) h
≤ const(δ, µ, ν, d) C2 t1∧2δ E σ [|x − y|2 p(σt , x − y)] + E σ [σt p(σt , x − y)]
1+ µ/2
+ E σ [|x − y|2+µ p(σt , x − y)] + E σ [σt
p(σt , x − y)]
+ t E σ [|x − y|2 p(σt , x − y)] + E σ [σt p(σt , x − y)] + E σ [|x − y|2+ν p(σt , x − y)]
i
1+ ν/2
+ E σ [σt
p(σt , x − y)] .
14
Proof. In view of (3.14) and (3.20), we set
5
X
vK1 (t, x, y; σ) =
wKj (t, x, y; σ)e−t(V (x)+V (y))/2
j=1
=:
5
X
vK1j (t, x, y; σ).
(3.23)
j=1
Clearly
E B [wK3 (t, x, y; σ)] =
1
2
D
Z
∇V (x) + ∇V (y),
t
0
E B [B0σt (σs )]ds
E
= 0,
h
i
L
and hence E B vK13 (t, x, y; σ) = 0. By the fact (σt − σt−s )0≤s≤t ∼ (σs )0≤s≤t ,
h
i
E σ wK2 (t, x, y; σ)p(σt , x − y)
=
1
2 h∇V
hZ
(y), y − xi E σ
t
0
σs
σt ds p(σt , x
hZ
i
− y) − E σ
t
0
σt −σt−s
σt −σt−t ds p(σt
i
− σt−t , x − y)
= 0,
h
h
i
i
and hence E σ E B [vK12 (t, x, y; σ)]p(σt , x − y) = E σ vK12 (t, x, y; σ)p(σt , x − y) = 0. By (A)2 (ii)
|vK11 (t, x, y; σ)| = |wK1 (t, x, y; σ)|e−t( V (x)+V (y))/2
≤
1
2 |∇V
≤
C2
2
n
(x) − ∇V (y)||x − y| t e−t(V (x)+V (y))/2
o
V (x)(1−2δ)+ (1 + |x − y|µ ) + 1 + |x − y|ν |x − y|2 t e−tV (x)/2
n
o
V (x)(1−2δ)+ e−tV (x)/2 t(|x − y|2 + |x − y|2+µ ) + t(|x − y|2 + |x − y|2+ν )
n
o
2
2+µ
2
2+ν
+ (1−2δ)+ 1∧2δ
≤ C22 ( 2(1−2δ)
)
t
(|x
−
y|
+
|x
−
y|
)
+
t(|x
−
y|
+
|x
−
y|
)
.
e
≤
C2
2
(3.24)
Here (and hereafter) the following inequality has been (will be) used:
tb e−t ≤ ( eb )b ,
t ≥ 0, b ≥ 0,
(3.25)
where for b = 0 we understand (0/e)0 := 1. By (A)2 (ii) and (3.25) again
|vK14 (t, x, y; σ)| = |wK4 (t, x, y; σ)|e−t(V (x)+V (y))/2
Z t Z 1
1
≤ 2
ds
|∇V (x + θ( σσst (y − x) + B0σt (σs ))) − ∇V (x)|
0
≤
C2
2
0
Z tn
0
× | σσst (y − x) + B0σt (σs )|dθ e−tV (x)/2
V (x)(1−2δ)+ e−tV (x)/2 | σσst (y − x) + B0σt (σs )|2 + | σσst (y − x) + B0σt (σs )|2+µ
15
o
+ | σσst (y − x) + B0σt (σs )|2 + | σσst (y − x) + B0σt (σs )|2+ν ds
≤
C2
2
Z tn
0
+ (1−2δ)+ −(1−2δ)+
( 2(1−2δ)
)
t
e
× | σσst (y − x) + B0σt (σs )|2 + | σσst (y − x) + B0σt (σs )|2+µ
o
+ | σσst (y − x) + B0σt (σs )|2 + | σσst (y − x) + B0σt (σs )|2+ν ds.
(3.26)
Similarly
|vK15 (t, x, y; σ)|
Z tn
C2
+ (1−2δ)+ −(1−2δ)+
≤ 2
( 2(1−2δ)
)
t
e
0
σt
σt
2
2+µ
σt −σs
s
× | σtσ−σ
(x
−
y)
+
B
(σ
)|
+
|
(x
−
y)
+
B
(σ
)|
s
s
0
0
σt
t
o
σt
σt
2
2+ν
σt −σs
s
+ | σtσ−σ
ds.
(x
−
y)
+
B
(σ
)|
+
|
(x
−
y)
+
B
(σ
)|
s
s
0
0
σt
t
Note that for a > 0 and 0 ≤ θ ≤ τ (τ > 0)
h
i
E B | τθ z + B0τ (θ)|a ≤ 3(a−1)+ |z|a + 2C(a, d)τ a/2 ,
h
i
τ
a
(a−1)+
a
a/2
≤
3
|z|
E B | τ −θ
z
+
B
(θ)|
+
2C(a,
d)τ
0
τ
(3.27)
(3.28)
R
where C(a, d) := E B [|B(1)|a ] = Rd |y|a p(1, y)dy. Thus, taking expectation E B in (3.26) and
(3.27), we have
h
h
i
i
E B |vK14 (t, x, y; σ)| + E B |vK15 (t, x, y; σ)|
n
+ (1−2δ)+ 1∧2δ
≤ C2 ( 2(1−2δ)
)
t
e
1+ µ/2
× 3|x − y|2 + 6C(2, d)σt + 31+µ |x − y|2+µ + 31+µ 2C(2 + µ, d)σt
o
1+ ν/2
+ t 3|x − y|2 + 6C(2, d)σt + 31+ν |x − y|2+ν + 31+ν 2C(2 + ν, d)σt
.
Collecting all the above into (3.23) yields the estimate in Claim 3.1 and the proof is complete.
Claim 3.2.
h
i
E σ E B [|vK2 (t, x, y; σ)|]p(σt , x − y)
h
≤ const(δ, µ, ν, d) C12 (t2 + t2δ ) E σ [|x − y|2 p(σt , x − y)] + E σ [σt p(σt , x − y)]
+ C22 t2(1∧2δ) E σ [|x − y|4 p(σt , x − y)] + E σ [σt2 p(σt , x − y)]
2+µ
4+2µ
+ E σ [|x − y|
p(σt , x − y)] + E σ [σt p(σt , x − y)]
16
+ C22 t2 E σ [|x − y|4 p(σt , x − y)] + E σ [σt2 p(σt , x − y)]
i
+ E σ [|x − y|4+2ν p(σt , x − y)] + E σ [σt2+ν p(σt , x − y)] .
Proof. By (A)2 (i)
Z t
P
D
E
3
1
wKj (t, x, y; σ) = 2 ∇V (x),
( σσst (y − x) + B0σt (σs ))ds
0
j=1
Z t
D
E
σt
1
s
+ 2 ∇V (y),
( σtσ−σ
(x
−
y)
+
B
(σ
))ds
s
0
t
≤
C1
2
n
0
(1 + V (x)1−δ )
Z
t
| σσst (y − x) + B0σt (σs )|ds
0
+ (1 + V (y)1−δ )
Z
t
0
o
σt
s
| σtσ−σ
(x
−
y)
+
B
(σ
)|ds
.
s
0
t
This estimate together with (3.26) and (3.27) gives us that
|wK (t, x, y; σ)|e−θt(V (x)+V (y))/4
≤ |
3
P
wKj (t, x, y; σ)|e−θt(V (x)+V (y))/4 +
j=1
5
P
j=4
|wKj (t, x, y; σ)|e−θt(V (x)+V (y))/4
1−δ −1+δ −1+δ
≤ C21 1 + ( 4(1−δ)
)
θ
t
e
Z t
σt
s
×
| σσst (y − x) + B0σt (σs )| + | σtσ−σ
(x
−
y)
+
B
(σ
)|
ds
s
0
t
0
Z
+
C2
2
0
t
n
+ (1−2δ)+
ds θ −(1−2δ)+ t−(1−2δ)+ ( 4(1−2δ)
)
e
× | σσst (y − x) + B0σt (σs )|2 + | σσst (y − x) + B0σt (σs )|2+µ
s
s
+ | σtσ−σ
(x − y) + B0σt (σs )|2 + | σtσ−σ
(x − y) + B0σt (σs )|2+µ
t
t
+ | σσst (y − x) + B0σt (σs )|2 + | σσst (y − x) + B0σt (σs )|2+ν
o
σt
σt
2
2+ν
σt −σs
s
.
+ | σtσ−σ
(x
−
y)
+
B
(σ
)|
+
|
(x
−
y)
+
B
(σ
)|
s
s
0
0
σt
t
By the Schwarz inequality, it follows that
2
|wK (t, x, y; σ)|e−θt(V (x)+V (y))/4
h
≤ 12 ( C21 )2 (t + ( 4(1−δ)
)2(1−δ) θ −2+2δ t−1+2δ )
e
Z t
Z t
σt
2
s
×
| σσst (y − x) + B0σt (σs )|2 ds +
| σtσ−σ
(x
−
y)
+
B
(σ
)|
ds
s
0
t
0
0
n
+ 2(1−2δ)+ −2(1−2δ)+ 2(1∧2δ)−1
+ ( C22 )2 ( 4(1−2δ)
)
θ
t
e
17
(3.29)
Z
×
t
| σσst (y − x) + B0σt (σs )|4 ds +
0
Z
t
+
0
Z
t
+t
0
s
| σtσ−σ
(x
t
| σσst (y
Z
t
+
0
− x) +
s
| σtσ−σ
(x
t
− y)
Z
t
+
0
B0σt (σs )|4 ds
t
| σσst (y − x) + B0σt (σs )|4+2µ ds
0
+ B0σt (σs )|4 ds
B0σt (σs )|4 ds
− y) +
Z
Z
+
0
t
s
| σtσ−σ
(x
t
− y) +
B0σt (σs )|4+2µ ds
| σσst (y − x) + B0σt (σs )|4+2ν ds
Z
+
0
t
s
| σtσ−σ
(x − y) + B0σt (σs )|4+2ν ds
t
oi
.
Take expectation E B above, and integrate in θ. Then
E B [|vK2 (t, x, y; σ)|]
Z
h
2
≤ E B wK (t, x, y; σ)
Z
1
=
0
θE B
1
θe−θt(V (x)+V (y))/2 dθ
i
0
h
2 i
|wK (t, x, y; σ)|e−θt(V (x)+V (y))/4
dθ
h
≤ 12 ( C21 )2 3(t2 + ( 4(1−δ)
)2(1−δ) 1δ t2δ )(|x − y|2 + 2C(2, d)σt )
e
n
2(1∧2δ)
+ 2(1−2δ)+
1
+ ( C22 )2 ( 4(1−2δ)
)
e
1∧2δ t
× [33 (|x − y|4 + 2C(4, d)σt2 ) + 33+2µ (|x − y|4+2µ + 2C(4 + 2µ, d)σt2+µ )]
oi
+ t2 [33 (|x − y|4 + 2C(4, d)σt2 ) + 33+2ν (|x − y|4+2ν + 2C(4 + 2ν, d)σt2+ν )] ,
whence follows immediately the estimate in Claim 3.2.
Claim 3.3.
h
h
i
i
E σ E B [|vG (t, x, y; σ)|]p(σt , x − y) , E σ E B [|vR (t, x, y; σ)|]p(σt , x − y)
≤ const(δ, µ, ν, d)
2 h
P
j=1
j/2
C1j (tj + tjδ ) E σ [|x − y|j p(σt , x − y)] + E σ [σt p(σt , x − y)]
+ C2j tj(1∧2δ) E σ [|x − y|2j p(σt , x − y)] + E σ [σtj p(σt , x − y)]
j(1+ µ/2)
+ E σ [|x − y|j(2+µ) p(σt , x − y)] + E σ [σt
p(σt , x − y)]
+ C2j tj E σ [|x − y|2j p(σt , x − y)] + E σ [σtj p(σt , x − y)]
i
j(1+ ν/2)
+ E σ [|x − y|j(2+ν) p(σt , x − y)] + E σ [σt
p(σt , x − y)] .
Proof. Similarly to what is done in (3.29), (3.26) and (3.27), we have
|wG1 (t, x, y; σ)|e−rtV (x)
≤ C1 (1 +
1−δ
( 1−δ
(rt)−1+δ )
e )
Z
t
0
| σσst (y − x) + B0σt (σs )|ds,
18
(3.30)
|wG2 (t, x, y; σ)|e−rtV (x)
h
+ (1−2δ)+
≤ C2 ( (1−2δ)
)
(rt)−(1−2δ)+
e
Z t
| σσst (y − x) + B0σt (σs )|2 + | σσst (y − x) + B0σt (σs )|2+µ ds
×
0
Z t
i
+
| σσst (y − x) + B0σt (σs )|2 + | σσst (y − x) + B0σt (σs )|2+ν ds ,
(3.31)
0
σt ,y
|wR1 (t, x, y; σ)|e−rtV (B0,x
≤ C1 (1 +
(σt/2 ))
1−δ
( 1−δ
(rt)−1+δ )
e )
Z
t
0
σt ,y
σt ,y
|B0,x
(σs ) − B0,x
(σt/2 )|ds,
(3.32)
σt ,y
|wR2 (t, x, y; σ)|e−rtV (B0,x (σt/2 ))
h
+ (1−2δ)+
≤ C2 ( (1−2δ)
)
(rt)−(1−2δ)+
e
Z t
σt ,y
σt ,y
σt ,y
σt ,y
|B0,x
×
(σs ) − B0,x
(σt/2 )|2 + |B0,x
(σs ) − B0,x
(σt/2 )|2+µ ds
0
Z t
i
σt ,y
σt ,y
σt ,y
σt ,y
+
|B0,x
(σs ) − B0,x
(σt/2 )|2 + |B0,x
(σs ) − B0,x
(σt/2 )|2+ν ds .
(3.33)
0
By (3.15), (3.16), (3.21) and (3.22), note that
|vG (t, x, y; σ)|
≤ |wG1 (t, x, y; σ)|e−tV (x) + |wG2 (t, x, y; σ)|e−tV (x)
Z 1 2
+
θ |wG1 (t, x, y; σ)|e−θtV (x)/2 + |wG2 (t, x, y; σ)|e−θtV (x)/2 dθ,
(3.34)
0
|vR (t, x, y; σ)|
σt ,y
σt ,y
≤ |wR1 (t, x, y; σ)|e−tV (B0,x (σt/2 )) + |wR2 (t, x, y; σ)|e−tV (B0,x (σt/2 ))
Z 1 σt ,y
+
θ |wR1 (t, x, y; σ)|e−θtV (B0,x (σt/2 ))/2
0
σt ,y
+ |wR2 (t, x, y; σ)|e−θtV (B0,x
(σt/2 ))/2
2
dθ.
(3.35)
Also note that for a > 0 and 0 ≤ θ1 , θ2 ≤ τ (τ > 0) (cf. (3.28))
h
i
τ,y
τ,y
E B |B0,x
(θ1 ) − B0,x
(θ2 )|a ≤ 3(a−1)+ (|x − y|a + 2C(a, d)τ a/2 ).
(3.36)
Collecting all the above yields the estimate in Claim 3.3 immediately.
We are now in a position to prove Theorem 2.1(iii). To do so, we need the following lemma.
R
Lemma 3.1. Let 1 ≤ p ≤ ∞. Then, for a, b ≥ 0 with C(a, d) = Rd |y|a p(1, y)dy,
Z
fa,b (t) := |f (y)| E σ [| · −y|a σtb p(σt , · − y)]dy Rd
p
19
a/2 +b
≤ C(a, d)E σ [σt
] kf kp ,
f ∈ Lp (R d ).
Proof. For p = ∞, the described estimate is obvious. So let 1 ≤ p < ∞. First we note the
Minkowski inequality for integrals: If h(x, y) is a measurable function on a σ-finite product
measure space (X × Y, α(dx) × β(dy)), then
Z Z
Z Z
p
1/p
1/p
|h(x, y)|α(dx) β(dy)
≤
|h(x, y)|p β(dy)
α(dx).
Y
X
X
Note also that for c ≥ 0
Z
Rd
Y
|f (y)|| · −y|c p(τ, · − y)dy ≤ C(c, d)τ c/2 kf kp .
p
By these inequalities, the estimate is obtained as follows:
Z
i
hZ
a b
|f (y)| E σ [| · −y| σt p(σt , · − y)]dy ≤ E σ |f (y)|| · −y|a σtb p(σt , · − y)dy Rd
Rd
p
≤
p
a/2 +b
C(a, d)E σ [σt
] kf kp .
Proof of Theorem 2.1(iii). By Claims 3.1, 3.2 with (3.7)
Z
kQK (t)f kp ≤ |f (y)| |E σ [E B [vK1 (t, ·, y; σ)]p(σt , · − y)]|dy
Rd
Z
+
|f (y)| E σ [E B [|vK2 (t, ·, y; σ)|]p(σt , · − y)]dy Rd
h
p
≤ const(δ, µ, ν, d) C12 (t2 + t2δ )(f2,0 (t) + f0,1 (t))
+
2 n
X
C2j tj(1∧2δ) (f2j,0 (t) + f0,j (t) + fj(2+µ),0 (t) + f0,j(1+ µ/2) (t))
j=1
oi
+ C2j tj (f2j,0 (t) + f0,j (t) + fj(2+ν),0 (t) + f0,j(1+ ν/2) (t)) .
By Claim 3.3 with (3.8), (3.9)
Z
kQ G (t)f kp ≤ |f (y)| E σ [E B [|v G (t, ·, y; σ)|]p(σt , · − y)]dy R
Rd
≤ const(δ, µ, ν, d)
R
p
2 h
X
C1j (tj + tjδ )(fj,0 (t) + f0,j/2 (t))
j=1
+ C2j tj(1∧2δ) (f2j,0 (t) + f0,j (t) + fj(2+µ),0 (t) + f0,j(1+ µ/2) (t))
i
+ C2j tj (f2j,0 (t) + f0,j (t) + fj(2+ν),0 (t) + f0,j(1+ ν/2) (t)) .
Combining these with Lemma 3.1 we have the assertion of Theorem 2.1(iii).
20
3.2. Case (A)0
In this subsection, we suppose condition (A)0 on V (x). In this case
|vK (t, x, y; σ)| ≤ |wK (t, x, y; σ)|
Z t
C1
≤ 2
| σσst (y − x) + B0σt (σs )|γ ds +
0
Z
C1
2
t
0
s
| σtσ−σ
(x − y) + B0σt (σs )|γ ds,
t
|vG (t, x, y; σ)| ≤ |wG (t, x, y; σ)|
Z t
≤ C1
| σσst (y − x) + B0σt (σs )|γ ds,
0
|vR (t, x, y; σ)| ≤ |wR (t, x, y; σ)|
Z t
σt ,y
σt ,y
≤ C1
|B0,x
(σs ) − B0,x
(σt/2 )|γ ds.
0
Here taking expectation E B , we have by (3.28) or (3.36),
E B [|vK (t, x, y; σ)|], E B [|vG (t, x, y; σ)|], E B [|vR (t, x, y; σ)|]
γ/2
≤ C1 t(|x − y|γ + 2C(γ, d)σt
)
and hence, by (3.7), (3.8) and (3.9)
|QK (t)f (x)|, |QG (t)f (x)|, |QR (t)f (x)|
h
nZ
i
≤ C1 t
|f (y)| E σ |x − y|γ p(σt , x − y) dy
Rd
Z
h
i o
γ/2
+ 2C(γ, d)
|f (y)| E σ σt p(σt , x − y) dy .
Rd
From this and Lemma 3.1 the assertion of Theorem 2.1(i) follows immediately.
4. Proof of Theorem 2.2
For notational simplicity we set H0 := H0ψ and H := H0 + V , in the following, so that K(t) =
e−tV /2 e−tH0 e−tV /2 , G(t) = e−tV e−tH0 and R(t) = e−tH0 /2 e−tV e−tH0 /2 .
4.1. Proof of Theorem 2.2 for K(t)
Since K(t) and e−sH are contractions, we have
kK( nt )n − e−tH kp→p = k
n−1
X
K( nt )n−1−k (K( nt ) − e−tH/n )e−ktH/n kp→p
k=0
≤
n−1
X
kK( nt ) − e−tH/n kp→p
k=0
= nkQK ( nt )kp→p .
Combined with the estimates for QK (t) in Theorem 2.1, the desired bound for K(t/n)n − e−tH
in Case (A)0 , (A)1 or (A)2 is obtained immediately.
21
4.2. Proof of Theorem 2.2 for G(t) and R(t) in Case (A)0
In the same way as above
kG( nt )n − e−tH kp→p ≤ nkQG ( nt )kp→p ,
kR( nt )n − e−tH kp→p ≤ nkQR ( nt )kp→p ,
from which together with Theorem 2.1(i), the desired bounds follow immediately.
4.3. Proof of Theorem 2.2 for G(t) and R(t) in Case (A)1 or (A)2
In this subsection we suppose that V (x) satisfies (A)1 or (A)2 .
We first observe that for t ≥ 0 and n ∈ N
1 n−1
G( nt )n − e−tH = e−tV /2n (K( n−1
− e−(n−1)tH/n )e−tV /2n e−tH0 /n
n t n−1 )
+ [e−tV /2n , e−(n−1)tH/n ]e−tV /2n e−tH0 /n + e−(n−1)tH/n QG ( nt ),
1 n−1
R( nt )n − e−tH = e−tH0 /2n e−tV /2n (K( n−1
− e−(n−1)tH/n )e−tV /2n e−tH0 /2n
n t n−1 )
+ e−tH0 /2n [e−tV /2n , e−(n−1)tH/n ]e−tV /2n e−tH0 /2n
+ [e−tH0 /2n , e−(n−1)tH/n ]e−tV /n e−tH0 /2n + e−(n−1)tH/n QR ( nt ),
where [A, B] = AB − BA. Hence
1 n−1
kG( nt )n − e−tH kp→p ≤ kK( n−1
− e−(n−1)tH/n kp→p
n t n−1 )
+ k[e−tV /2n , e−(n−1)tH/n ]kp→p + kQG ( nt )kp→p ,
(4.1)
1 n−1
kR( nt )n − e−tH kp→p ≤ kK( n−1
− e−(n−1)tH/n kp→p
n t n−1 )
+ k[e−tV /2n , e−(n−1)tH/n ]kp→p + k[e−tH0 /2n , e−(n−1)tH/n ]kp→p
+ kQR ( nt )kp→p .
(4.2)
As for the first term on the RHS of (4.1) and (4.2), we see by Theorem 2.2 which was proved in
Section 4.1
1 n−1
kK( n−1
− e−(n−1)tH/n kp→p
n t n−1 )

i
h 2
P

j(1+κ)/2
2
t 2
t 2δ
t j

(C2 n ) (n − 1)E [σt/n
] ,
 const(δ, κ, d) C1 ( n ) + ( n ) (n − 1)E [σt/n ] +


j=1






in Case (A)1 ,




h 2 n
≤
P
j
2
t 2
t 2δ
t j

const(δ,
µ,
ν,
d)
C
(
(n
−
1)
(C
(n − 1)E [σt/n
)
+
(
)
E
[σ
]
+
)
]
2

t/n
1
n
n
n


j=1


oi


j(1+ ν/2)
j(1+ µ/2)
j

+ (n − 1)E [σt/n
] + (C2 ( nt )1∧2δ )j (n − 1)E [σt/n
] + (n − 1)E [σt/n
]
,






in Case (A)2 .
22
As for the third term on the RHS of (4.1) and the fourth term of (4.2), we see by Theorem 2.1
kQG ( nt )kp→p , kQR ( nt )kp→p

o
2 n
P
j/2
j(1+κ)/2

j t j
t j
t jδ

C
((
)
+
(
)
)
E
[σ
]
+
(C
)
E
[σ
]
,
const
(δ,
κ,
d)

2n
1 n
n
t/n
t/n


j=1







2 n
P
j/2
const
(δ,
µ,
ν,
d)
C1j (( nt )j + ( nt )jδ )E [σt/n ]
≤

j=1



j(1+ µ/2)
j


+ (C2 ( nt )1∧2δ )j (E [σt/n
] + E [σt/n
])



o


j(1+ ν/2)

+ (C t )j (E [σ j ] + E [σ
]) ,
2n
t/n
t/n
in Case (A)1 ,
in Case (A)2 .
Therefore we need to estimate the middle terms of (4.1) and (4.2).
Claim 4.1. Let s ≥ 0 and t > 0. Then
k[e−sV , e−tH ]kp→p , k[e−sH0 , e−tH ]kp→p
i
h

−1+δ )E [σ 1/2 ] + C E [σ (1+κ)/2 ] ,

(1
+
t
in Case (A)1 ,
const
(δ,
κ,
d)s
C

1
2
t
t




h
1/2
1+ µ/2
≤
const (δ, µ, ν, d)s C1 (1 + t−1+δ )E [σt ] + C2 t−(1−2δ)+ (E [σt ] + E [σt
])



i


1+ ν/2

+ C2 (E [σt ] + E [σt
]) ,
in Case (A)2 .
Proof. First we estimate the Lp -operator norm of [e−sV , e−tH ]. We have (by (A.13)) that for
f ∈ C0 (R d )
[e−sV , e−tH ]f (x)
Z
h
Z t
i
σt ,y
−sV (x)
−sV (y)
=
f (y)(e
−e
)E exp −
V (B0,x
(σr ))dr p(σt , x − y) dy.
Rd
0
Hence we have
|[e−sV , e−tH ]f (x)|
Z
h
Z t
i
σt ,y
≤ s
|f (y)|E |V (y) − V (x)| exp −
V (B0,x
(σr ))dr p(σt , x − y) dy.
Rd
0
To estimate the integrand in (4.3), note by Taylor’s theorem that
Z
V (y) − V (x) =
t
0
σt ,y
h∇V (B0,x
(σr )), y − xi drt
Z
+
Z
1
dθ
0
0
t
σt ,y
h∇V (x + θ(y − x)) − ∇V (B0,x
(σr )), y − xi dr
t .
23
(4.3)
In Case (A)1 , it follows that
Z
|V (y) − V (x)| ≤
t
0
σt ,y
C1 (1 + V (B0,x
(σr ))1−δ ) drt |x − y|
Z
+
Z
1
t
dθ
0
0
C2 |( σσrt − θ)(y − x) + B0σt (σr )|κ dr |x−y|
t
Z t
σt ,y
−1+δ
≤ C1 1 + t
( V (B0,x
(σr ))dr)1−δ |x − y|
+ C2 1t
Z
Z
1
0
t
dθ
0
0
|( σσrt − θ)(y − x) + B0σt (σr )|κ dr|x − y|
(4.4)
where the last inequality is due to Jensen’s inequality. In Case (A)2
|V (y) − V (x)|
Z t
σt ,y
≤
C1 (1 + V (B0,x
(σr ))1−δ ) dr
t |x − y|
0
Z
1
+
Z
n
σt ,y
C2 V (B0,x
(σr ))(1−2δ)+ (1 + |( σσrt − θ)(y − x) + B0σt (σr )|µ )
0
o
+ 1 + |( σσrt − θ)(y − x) + B0σt (σr )|ν |( σσrt − θ)(y − x) + B0σt (σr )| drt |x − y|
t
dθ
0
Z t
σt ,y
−1+δ
≤ C1 1 + t
(
V (B0,x
(σr ))dr)1−δ |x − y|
0
+ C2 t
−(1−2δ)+
Z
t
V
0
σt ,y
(B0,x
(σr ))dr
(1−2δ)+ Z 1 0
max |( σσt − θ)(y − x) + B0σt (σ)|
0≤σ≤σt
+ max |( σσt − θ)(y − x) + B0σt (σ)|1+µ dθ|x − y|
+ C2 1t
Z
0≤σ≤σt
Z t
|( σσrt − θ)(y − x) + B0σt (σr )|
0
σr
+ |( σt − θ)(y − x) + B0σt (σr )|1+ν dr|x − y|.
1
dθ
0
By (3.25), (4.4) and (4.5) imply the desired estimate:
Z t
σt ,y
|V (y) − V (x)| exp −
V (B0,x
(σr ))dr
0
24
(4.5)






















1−δ t−1+δ )|x − y|
C1 (1 + ( 1−δ
e )
R
R
1
t
+ C2 1t 0 dθ 0 |( σσrt − θ)(y − x) + B0σt (σr )|κ dr|x − y|,
in Case (A)1 ,
1−δ t−1+δ )|x − y|
C1 (1 + ( 1−δ
e )
+ (1−2δ)+ −(1−2δ)+
+ C2 ( (1−2δ)
)
t
e
R
1
≤
× 0 max |( σσt − θ)(y − x) + B0σt (σ)|


0≤σ≤σt




σt
σ
1+µ dθ|x − y|

+
max
|(
−
θ)(y
−
x)
+
B
(σ)|

0

0≤σ≤σt σt



R
R 
1

 + C2 1t 0 dθ 0t |( σσrt − θ)(y − x) + B0σt (σr )|





σt
σr
1+ν

+ |( σt − θ)(y − x) + B0 (σr )|
dr|x − y|,
in Case (A)2 .
We take expectation E B in the above. This time we use the following moment estimate: For
a > 0, τ > 0, 0 ≤ θ ≤ 1 and z ∈ Rd
h
i
E B |( τt − θ)z + B0τ (t)|a ≤ 3(a−1)+ (|z|a + 2C(a, d)τ a/2 ),
h
i
e d)τ a/2 )
E B max |( τt − θ)z + B0τ (t)|a ≤ 3(a−1)+ (|z|a + 2C(a,
(4.6)
0≤t≤τ
e d) = E B [ max |B(t)|a ], and thereby we have
where C(a, d) = E B [|B(1)|a ] and C(a,
0≤t≤1
h
Z t
i
σt ,y
E B |V (y) − V (x)| exp −
V (B0,x
(σr ))dr
0
≤

κ/2
1−δ −1+δ

C1 (1 + ( 1−δ
t
)|x − y| + C2 (|x − y|1+κ + 2C(κ, d)σt |x − y|),

e )





in Case (A)1 ,







1−δ 1−δ −1+δ

)|x − y|

 C1 (1 + ( e ) t
+ (1−2δ)+ −(1−2δ)+
(4.7)
+ C2 ( (1−2δ)
)
t
e




e d)σ 1/2 + 3µ (|x − y|1+µ + 2C(1
e + µ, d)σ (1+µ)/2 ) |x − y|

× |x − y| + 2C(1,
t
t





1/2
ν (|x − y|1+ν + 2C(1 + ν, d)σ (1+ν)/2 ) |x − y|,

+
C
|x
−
y|
+
2C(1,
d)σ
+
3

2
t
t





in Case (A)2 .
Hence follows the desired bound for [e−sV , e−tH ] by Lemma 3.1 with (4.3).
Next we estimate the Lp -operator norm of [e−sH0 , e−tH ].
First we suppose that V : R d → [0, ∞) is in C ∞ and all its derivatives have polynomial growth.
Then it is easily verified that (cf. Claim A.2 and its Remark)
(i) e−tH (S(R d )) ⊂ S(Rd ), in particular, e−tH0 (S(R d )) ⊂ S(R d ), and
T
T
ψ,V
(ii) S(R d ) ⊂
( ψ,V
( ψ,0
= ψ,0
− V on S(R d ).
p ) ∩
p ) and
p
p
G
1≤p≤∞
DG
1≤p≤∞
DG
G
G
G
Here ψ,V
(1 ≤ p < ∞) is the infinitesimal generator of {e−t(H0 +V ) } on Lp (R d ) and ψ,V
p
∞ the
one on C∞ (R d ). By these facts the following formula holds in Lp (R d ) (1 ≤ p < ∞) and C∞ (R d ):
25
For each f ∈ S(R d )
−sH0
[e
−tH
,e
Z
s
]f =
e−uH0 [V, e−tH ]e−(s−u)H0 f du.
0
Hence, taking Lp -norm in the above yields that for each f ∈ S(R d )
Z s
−sH0 −tH
k[e
,e
]f kp ≤
k[V, e−tH ]e−(s−u)H0 f kp du.
(4.8)
0
Now let V satisfy (A)1 or (A)2 . In this case V is not necessarily
smooth. So, take a nonnegative
R
h ∈ C0∞ with support in {x ∈ R d ; |x| ≤ 1} and Rd h(x)dx = 1. Set V ε = V ∗ hε with
hε (x) = (1/ε)d h(x/ε). Then V ε is in C ∞ (R d → [0, ∞)), and satisfies condition (A)1 or (A)2
with the same const’s as V does. Further, by (A)1 (i) or (A)2 (ii) all the derivatives of V ε have
polynomial growth. Hence, by (4.7) and Lemma 3.1 it holds that for g ∈ S(R d )
k[V ε , e−t(H0 +V ) ]gkp
i
h

−1+δ )E [σ 1/2 ] + C E [σ (1+κ)/2 ] kgk ,

(1
+
t
in Case (A)1 ,
const
(δ,
κ,
d)
C

1
2
p
t
t




h
1/2
1+ µ/2
≤
const
(δ,
µ,
ν,
d)
C1 (1 + t−1+δ )E [σt ] + C2 t−(1−2δ)+ (E [σt ] + E [σt
])



i


1+ ν/2

+ C2 (E [σt ] + E [σt
]) kgkp ,
in Case (A)2 .
ε
Since (4.8) holds with V = V ε , by combining this with the above we have
k[e−sH0 , e−t(H0 +V ) ]f kp
i
h

−1+δ )E [σ 1/2 ] + C E [σ (1+κ)/2 ] kf k ,

(1
+
t
in Case (A)1 ,
const
(δ,
κ,
d)s
C

1
2
p
t
t




h
1/2
1+ µ/2
≤
const (δ, µ, ν, d)s C1 (1 + t−1+δ )E [σt ] + C2 t−(1−2δ)+ (E [σt ] + E [σt
])



i


1+ ν/2

+ C2 (E [σt ] + E [σt
]) kf kp ,
in Case (A)2 .
ε
Finally let ε ↓ 0. Since V ε → V compact uniformly, we see by the Feynman-Kac formula
ε
ε
(A.6) that e−t(H0 +V ) f → e−t(H0 +V ) f boundedly pointwise, so that [e−sH0 , e−t(H0 +V ) ]f →
[e−sH0 , e−t(H0 +V ) ]f pointwise. Hence the desired bound for [e−sH0 , e−t(H0 +V ) ] follows immediately by the Fatou inequality.
We return to estimate G(t/n)n − e−tH and R(t/n)n − e−tH . By Claim 4.1
k[e−tV /2n , e−(n−1)tH/n ]kp→p , k[e−tH0 /2n , e−(n−1)tH/n ]kp→p
h
i

1
δ )E [σ 1/2 ] + C tE [σ (1+κ)/2 ] ,

C
(t
+
t
in Case (A)1 ,
const
(δ,
κ,
d)

1
2
t
t
n




h
1/2
1+ µ/2
≤
const (δ, µ, ν, d) n1 C1 (t + tδ )E [σt ] + C2 t1∧2δ (E [σt ] + E [σt
])



i


1+ ν/2

+ C2 t(E [σt ] + E [σt
]) ,
in Case (A)2 .
Therefore, collecting all the estimates above yields the desired bounds for G(t/n)n − e−tH and
R(t/n)n − e−tH .
26
5. Proof of Theorem 2.3
As in the previous section, we are setting H0 = H0ψ and H = H0 + V .
5.1. Case (V)2
Condition (V)2 implies (A)2 with δ = 1 ∧ 1/ρ, C1 = c1 c−(1−1∧ 1/ρ) , C2 = c2 2(ρ−3)+
((1/2)c−(1−2(1∧ 1/ρ))+ ∨ 1), µ = 0 and ν = (ρ − 2)+ . So this case follows immediately from
Theorem 2.2(iii).
5.2. Case (V)1
In this subsection we suppose condition (V)1 on V (x).
Let us adopt an Ridea in [D-I-Tam]. Take again a nonnegative h ∈ C0∞ with support in {x ∈
R d ; |x| ≤ 1} and Rd h(x)dx = 1. For 0 < ε ≤ 1/4, set
1 d Z
x − y
Vε (x) :=
V (y)dy,
h
εhxiη
Rd εhxiη
where η := ((ρ − 1) ∨ 0) ∧ 1. Then Vε is a smooth function and it satisfies the following:
Lemma 5.1. (i) Vε (x) ≥ c0 hxiρ where c0 = c/4ρ .
(ii) |Vε (x) − V (x)| ≤ C 0 εhxi(ρ−1)+ +η where C 0 = c1 (5/4)(ρ−1)+ .
(iii) |∇Vε (x)| ≤ c01 hxi(ρ−1)+ where c01 = c1 (5/4)ρ∨1 .
(iv) |∇Vε (x) − ∇Vε (y)| ≤ (1/ε)c02 {hxi(ρ−2λ)+ + |x − y|(ρ−2λ)+ }|x − y| where λ := (1 + η)/2 and
c02 = c1 (5/4)(ρ−1)+ 2(ρ−3)+ (5d/16 + 2).
The proof is not difficult, so is omitted (cf. [Tak]).
As a consequence of Lemma 5.1, it is easily seen that Vε satisfies condition (A)2 , i.e.
(A)2,ε
|∇Vε (x)| ≤ C10 Vε (x)1−1∧ λ/ρ ,
n
o
|∇Vε (x) − ∇Vε (y)| ≤ 1ε C20 Vε (x)(1−2(1∧ λ/ρ))+ + |x − y|(ρ−2)+ |x − y|
where C10 = c01 c0−(1−1∧ λ/ρ) and C20 = c02 (c0−(1−2(1∧ λ/ρ))+ ∨ 1). Indeed, by the definition of λ, we
have ρ − ρ ∧ λ ≥ (ρ − 1)+ , (ρ − 2(ρ ∧ λ))+ = (ρ − 2λ)+ = (ρ − 2)+ . Hence (A)2,ε follows, because,
by (i) with hxi ≥ 1,
Vε (x)1−1∧ λ/ρ ≥ (c0 )1−1∧ λ/ρ hxiρ−ρ∧λ ≥ (c0 )1−1∧ λ/ρ hxi(ρ−1)+ ,
Vε (x)(1−2(1∧ λ/ρ))+ ≥ (c0 )(1−2(1∧ λ/ρ))+ hxi(ρ−2(ρ∧λ))+ = (c0 )(1−2(1∧ λ/ρ))+ hxi(ρ−2λ)+ .
In what follows we write c, C, c1 , c2 , C1 and C2 simply for c0 , C 0 , c01 , c02 , C10 and C20 .
Now let Kε (t) := e−tVε /2 e−tH0 e−tVε /2 , Gε (t) := e−tVε e−tH0 and Rε (t) := e−tH0 /2 e−tVε e−tH0 /2 .
27
Claim 5.1. Let t ≥ 0 and n ∈ N . Then with H ε = H0 + Vε
kKε ( nt )n − e−tH kp→p
h
2 n
P
j
(C2 1ε n1 t)j nE [σt/n
≤ const(ρ, d) C1 2 (( nt )2 + ( nt )2(1∧ λ/ρ) )nE [σt/n ] +
]
ε
j=1
+ (C2 1ε n1 t)j nE [σt/n
j(2∨ρ)/2
oi
j
] + (C2 1ε ( n1 )1∧ 2λ/ρ t1∧ 2λ/ρ )j nE [σt/n
] ,
kGε ( nt )n − e−tH kp→p , kRε ( nt )n − e−tH kp→p
h
1/2
≤ const(ρ, d) n1 C1 (t + t1∧ λ/ρ )E [σt ] + C2 1ε n1 t1∧ 2λ/ρ E [σt ]
ε
ε
(2∨ρ)/2
+ C2 1ε n1 t(E [σt ] + E [σt
1/2
]) + C1 ( nt + ( nt )1∧ λ/ρ )E [σt/n ]
+ C1 2 (( nt )2 + ( nt )2(1∧ λ/ρ) )nE [σt/n ] +
2 n
P
j(2∨ρ)/2
j
(C2 1ε n1 t)j nE [σt/n
] + (C2 1ε n1 t)j nE [σt/n
]
j=1
oi
j
+ (C2 1ε ( n1 )1∧ 2λ/ρ t1∧ 2λ/ρ )j nE [σt/n
] .
This is obvious from (A)2,ε and Theorem 2.2(iii).
Claim 5.2. Let t ≥ 0 and n ∈ N . Then
ke−tH − e−tH kp→p ,
ε
kK( nt )n − Kε ( nt )n kp→p , kG( nt )n − Gε ( nt )n kp→p , kR( nt )n − Rε ( nt )n kp→p
≤ const(C, c, ρ) ε t2/((ρ∧2)∨1)−1 .
Proof. Let f ∈ C0∞ (R d ). By (3.4), (3.5) and (3.6) with (A.6),
|(e−tH − e−tH )f (x)|
h
Z t
Z t
i
≤ E exp −
V (x + Xs )ds − exp −
Vε (x + Xs )ds |f (x + Xt )| ,
ε
0
(5.1)
0
|(K( nt )n − Kε ( nt )n )f (x)|
n
h
X
t
≤ E exp − 2n
(V (x + X(k−1)t/n ) + V (x + Xkt/n ))
k=1
n
i
X
t
− exp − 2n
(Vε (x + X(k−1)t/n ) + Vε (x + Xkt/n )) |f (x + Xt )| ,
(5.2)
k=1
|(G( nt )n − Gε ( nt )n )f (x)|
n
h
X
t
≤ E exp − n
V (x + X(k−1)t/n )
k=1
n
i
X
t
− exp − n
Vε (x + X(k−1)t/n ) |f (x + Xt )| ,
k=1
28
(5.3)
|(R( nt )n − Rε ( nt )n )f (x)|
n
h
X
≤ E exp − nt
V (x + X(2k−1)t/2n )
k=1
n
i
X
t
− exp − n
Vε (x + X(2k−1)t/2n ) |f (x + Xt )| .
k=1
By a formula
−a
−b
−e
e
Z
1
=
0
(b − a)e−θa e−(1−θ)b dθ,
a, b ∈ R
and Lemma 5.1, we have
Z t
Z t
V (x + Xs )ds − exp −
Vε (x + Xs )ds exp −
Z
≤
0
1
0
Z
t
dθ
0
0
|Vε (x + Xs ) − V (x + Xs )|ds
Z t
Z t
× exp −θ
V (x + Xs )ds exp −(1 − θ)
Vε (x + Xs )ds
Z
≤ Cε
0
t
0
hx + Xs i(ρ−1)+ +η ds exp −c
Z
0
0
t
hx + Xs iρ ds ,
n
X
t
(V (x + X(k−1)t/n ) + V (x + Xkt/n ))
exp − 2n
k=1
n
X
t
− exp − 2n
(Vε (x + X(k−1)t/n ) + Vε (x + Xkt/n )) Z
≤
k=1
1
dθ
0
t
2n
n X
|Vε (x + X(k−1)t/n ) − V (x + X(k−1)t/n )|
k=1
+ |Vε (x + Xkt/n ) − V (x + Xkt/n )|
n
X
t
× exp −θ 2n
(V (x + X(k−1)t/n ) + V (x + Xkt/n ))
k=1
n
X
t
× exp −(1 − θ) 2n
(Vε (x + X(k−1)t/n ) + Vε (x + Xkt/n ))
k=1
n
X
t
≤ Cε 2n
hx + X(k−1)t/n i(ρ−1)+ +η +
k=1
t
2n
n
X
hx + Xkt/n i(ρ−1)+ +η
k=1
n
n
X
X
t
t
× exp −c 2n
hx + X(k−1)t/n iρ exp −c 2n
hx + Xkt/n iρ .
k=1
k=1
29
(5.4)
Similarly
n
n
X
X
t
t
V (x + X(k−1)t/n ) − exp − n
Vε (x + X(k−1)t/n ) exp − n
≤ Cε nt
k=1
n
X
k=1
n
X
hx + X(k−1)t/n i(ρ−1)+ +η exp −c nt
hx + X(k−1)t/n iρ ,
k=1
exp − nt
≤ Cε nt
k=1
n
X
n
X
t
V (x + X(2k−1)t/2n ) − exp − n
Vε (x + X(2k−1)t/2n ) k=1
n
X
k=1
n
X
hx + X(2k−1)t/2n i(ρ−1)+ +η exp −c nt
hx + X(2k−1)t/2n iρ .
k=1
k=1
By Jensen’s inequality and (3.25),
Z t
Z t
V (x + Xs )ds − exp −
Vε (x + Xs )ds ,
exp −
0
0
n
n
X
X
t
t
exp
−
V
(x
+
X
)
−
exp
−
V
(x
+
X
)
ε
(k−1)t/n
(k−1)t/n ,
n
n
k=1
k=1
n
n
X
X
V (x + X(2k−1)t/2n ) − exp − nt
Vε (x + X(2k−1)t/2n ) exp − nt
k=1
≤ Cε t1−((ρ−1)+ +η)/ρ
k=1
(ρ−1)+ +η 1
ρ
ec
((ρ−1)+ +η)/ρ
,
n
X
t
(V (x + X(k−1)t/n ) + V (x + Xkt/n ))
exp − 2n
k=1
n
X
t
− exp − 2n
(Vε (x + X(k−1)t/n ) + Vε (x + Xkt/n )) k=1
((ρ−1)+ +η)/ρ
1
≤ Cε( 2t )1−((ρ−1)+ +η)/ρ 2 (ρ−1)ρ+ +η ec
,
where for ρ = 0 we understand ((ρ − 1)+ + η)/ρ = 0. Substituting these into (5.1), (5.2), (5.3)
and (5.4), respectively, we have
|(e−tH − e−tH )f (x)|,
ε
|(K( nt )n − Kε ( nt )n )f (x)|, |(G( nt )n − Gε ( nt )n )f (x)|, |(R( nt )n − Rε ( nt )n )f (x)|
≤ const(C, c, ρ) ε t2/((ρ∧2)∨1) −1 E [|f (x + Xt )|],
which imply the estimates in Claim 5.2 and the proof is complete.
Proof of Theorem 2.3(i). By Claims 5.1 and 5.2
kK( nt )n − e−tH kp→p
30
≤ kK( nt )n − Kε ( nt )n kp→p + kKε ( nt )n − e−tH kp→p + ke−tH − e−tH kp→p
h
≤ const(ρ, C, c, d) ε t2/((ρ∧2)∨1) −1 + C1 2 (( nt )2 + ( nt )2(1∧ λ/ρ) )nE [σt/n ]
ε
+
2 n
P
j=1
j
(C2 1ε n1 t)j nE [σt/n
] + (C2 1ε n1 t)j nE [σt/n
ε
j(2∨ρ)/2
]
oi
j
+ (C2 1ε ( n1 )1∧ 2λ/ρ t1∧ 2λ/ρ )j nE [σt/n
] ,
kG( nt )n − e−tH kp→p , kR( nt )n − e−tH kp→p
h
1/2
≤ const(ρ, C, c, d) ε t2/((ρ∧2)∨1) −1 + n1 C1 (t + t1∧ λ/ρ )E [σt ]
(2∨ρ)/2
+ C2 1ε n1 t1∧ 2λ/ρ E [σt ] + C2 1ε n1 t(E [σt ] + E [σt
])
1/2
+ C1 ( nt + ( nt )1∧ λ/ρ )E [σt/n ] + C1 2 (( nt )2 + ( nt )2(1∧ λ/ρ) )nE [σt/n ]
2 n
P
j(2∨ρ)/2
j
+
(C2 1ε n1 t)j nE [σt/n
] + (C2 1ε n1 t)j nE [σt/n
]
j=1
oi
j
+ (C2 1ε ( n1 )1∧ 2λ/ρ t1∧ 2λ/ρ )j nE [σt/n
] .
Now let n ≥ 22(2∨ρ) and ε := n−(1/2)∧(λ/ρ) = n−1/2∨ρ . Then ε ≤ 1/4, ε−1 n−1∧ 2λ/ρ = n−1/2∨ρ
and ε−1 n−1 ≤ n−1/2∨ρ . Therefore we have
kK( nt )n − e−tH kp→p
h
≤ const(ρ, C, c, C1 , C2 , d) ( n1 )1/2∨ρ t2/((ρ∧2)∨1) −1 + (t2 + t2(1∧((ρ∧2)∨1)/2ρ) )nE [σt/n ]
+
oi
2 n
P
j(2∨ρ)/2
j
(tj + tj2/2∨ρ )nE [σt/n
] + tj nE [σt/n
] ,
j=1
kG( nt )n − e−tH kp→p , kR( nt )n − e−tH kp→p
h
1/2
≤ const(ρ, C, c, C1 , C2 , d) ( n1 )1/2∨ρ t2/((ρ∧2)∨1) −1 + (t + t1∧((ρ∧2)∨1)/2ρ )E [σt ]
(2∨ρ)/2
+ t2/2∨ρ E [σt ] + t(E [σt ] + E [σt
]) + (t2 + t2(1∧((ρ∧2)∨1)/2ρ) )nE [σt/n ]
oi
2 n
P
j(2∨ρ)/2
j
+
(tj + tj2/2∨ρ )nE [σt/n
] + tj nE [σt/n
] ,
j=1
and the proof is complete.
6. Proof of Theorem 2.4
For a > 0, the proof will be given, divided into the three cases a = 1, a > 1 and 0 < a < 1.
First we note that for every a > 0
E [σta ] < ∞.
31
(6.1)
In fact, it is enough to show when
√ a = ν is a positive integer. To do so, let ϕt be the characteristic
function of σt , i.e., ϕt (ξ) = E [e −1 ξσt ]. We have ϕt (ξ) = e−tf (ξ) , where
Z
√
f (ξ) =
(1 − e −1 ξl )e−l/2 n(dl).
(0,∞)
Since smoothness of ϕt (ξ) near ξ = 0 implies existence of moments of σt (cf. Exercise 2.6(viii)
∞
in [It]), we have only to show that ϕt or f is in C
ξ = 0. But this is obvious, because, by
R near
ν
a property of the Lévy measure n, the integral (0,∞) l e−l/2 n(dl) is convergent, so that by the
Lebesgue convergence theorem
Z
√
√
d ν
( dξ ) f (ξ) = −
( −1 l)ν e −1 ξl e−l/2 n(dl).
(0,∞)
By Itô’s formula (e.g. [Ik-Wa]),
t+Z
Z
σta
n
=
Z
0
t+Z
(0,∞)
o
a
(σs− + l)a − σs−
N (dsdl)
Z
=
1
a
0
0
(0,∞)
and hence, by taking expectation E
Z t Z
a
E [σt ] =
ds
0
(σs− + θl)a−1 dθ l N (dsdl),
−l/2
le
Z
1
n(dl) a
0
(0,∞)
h
i
E (σs + θl)a−1 dθ.
(6.2)
This is further, by the change of variable r = st , rewritten as
1
t
Z
E [σta ]
Z
1
=
−l/2
dr
0
le
Z
1
n(dl) a
0
(0,∞)
h
i
E (σtr + θl)a−1 dθ.
(6.3)
6.1. The case a = 1
By (6.3), it is clear that
1
t
Z
E [σt ] =
(0,∞)
le−l/2 n(dl) ∈ (0, ∞).
(6.4)
6.2. The case a > 1
By (6.1) and (6.3), E [(σr + θl)a−1 ] is of course integrable on (0, ∞) × [0, 1] × [0, 1] w.r.t.
le−l/2 n(dl)drdθ. Since σt is increasing in t with σ0+ = σ0 = 0 and a − 1 > 0, we have
(σtr + θl)a−1 ↓ θ a−1 la−1 as t ↓ 0. It follows by the Lebesgue convergence theorem that
Z
a
1
la e−l/2 n(dl) ∈ (0, ∞).
(6.5)
t E [σt ] ↓
(0,∞)
32
6.3. The case 0 < a < 1
By the same reason as above (but in this case, a − 1 < 0), we have (σtr + θl)a−1 ↑ θ a−1 la−1 as
t ↓ 0, and hence, by the monotone convergence theorem
Z
a
1
la e−l/2 n(dl) ∈ (0, ∞].
(6.6)
t E [σt ] ↑
(0,∞)
This time the integral on the RHS is not always convergent. To find the exact asymptotics we
suppose assumption (L).
We start with a remark on (L) and ψ(λ) defined by (1.3):
Fact. (i) If 0 ≤ α < 1, then
ψ(λ) ∼ Γ(1 − α)λα L(λ)
(ii) If α = 1, then
and
R·
0
as λ ↑ ∞.
Rt
n((s, ∞))ds is slowly varying at zero, L(1/t) = o( 0 n((s, ∞))ds) as t ↓ 0
Z
ψ(λ) ∼ λ
1/λ
0
n((s, ∞))ds
as λ ↑ ∞.
Proof. First of all note that
Z
∞ >
Z
(0,∞)
Z
l ∧ 1 n(dl) =
∞
ψ(λ) = λ
0
1
0
n((t, ∞))dt,
(6.7)
Z t
e−λt d
n((s, ∞))ds .
(6.8)
0
By (1.6), n((1/y, ∞)) ∼ y α L(y) as y ↑ ∞, and by (6.7),
Z ∞
Z 1/x
1
1
n(( y , ∞))dy =
n((s, ∞))ds < ∞
y2
for any x > 0.
0
x
R∞
Let us apply Lemma and Theorem 1 of §VIII.9 in [Fe]. These say that · 1/y 2 n((1/y, ∞))dy is
regularly varying with exponent −1 + α and
(1/x)n((1/x, ∞))
R∞
−→ 1 − α
2
x 1/y n((1/y, ∞))dy
as x ↑ ∞.
Combining these with (1.6), we see that when 0 ≤ α < 1
Z t
1
1
n((s, ∞))ds ∼ 1−α
t n((t, ∞)) ∼ 1−α
t1−α L( 1t )
0
and that when α = 1,
as t ↓ 0.
R·
0
as t ↓ 0,
Rt
n((s, ∞))ds is slowly varying at zero and L(1/t) = o( 0 n((s, ∞))ds)
By virtue of (6.8), if we apply the Abelian
theorem (cf. Theorem 2 of §XIII.5 in [Fe]), the
R
asymptotics of ψ follow from those of 0· n((s, ∞))ds.
33
Remark. Conversely, when 0 ≤ α < 1, we have (1.6) by Fact (i) by the Tauberian theorem.
Recall functions φ, L1 and L2 around assumption (L) in Section 1. By Fact, L1 is slowly varying
at infinity and
φ(λ) ∼ λα L1 (λ)
as λ ↑ ∞.
(6.9)
As ψ is strictly increasing with ψ(0) = 0 and ψ(∞) = ∞, so is φ, so that the inverse φ−1 exists.
By (6.9), if 0 < α ≤ 1,
φ−1 (x) ∼ x1/α L2 (x)
as x ↑ ∞.
(6.10)
Since, by (6.9)Ragain, φ is regularly varying with exponent α, so is φ−1 with exponent 1/α, and
hence L2 and 0· (φ−1 (θ))−α dθ (0 < α < 1) are also slowly varying at infinity.
Now we are in a position to show the asymptotics of E [σta ] for 0 < a < 1.
Claim 6.1. (i) If 0 < a < α,
E [σta ] ∼
Γ(1 − a/α) a/α
Γ(1 − a/α) −1 1 −a
t L2 ( 1t )−a ∼
φ (t)
Γ(1 − a)
Γ(1 − a)
as t ↓ 0.
(ii) If a = α,
E [σtα ]
(iii) If α < a < 1, then
1
∼
t
Γ(1 − α)
Z
1/t −α
φ−1 (θ)
dθ
0
R∞
λ−1−a φ(λ)dλ ∈ (0, ∞) and
Z ∞
a
a
E [σt ] ∼ t
λ−1−a φ(λ)dλ
Γ(1 − a) 0
as t ↓ 0.
0
as t ↓ 0.
Proof. To rewrite (6.2), we see first with (2.2)
h
a−1
E (σs + θl)
i
1
=
Γ(1 − a)
=
1
Γ(1 − a)
Z
∞
0
Z
∞
h
i
λ−a e−λθl E e−λσs dλ
λ−a e−λθl e−sφ(λ) dλ,
0
and then we have
E [σta ]
a
=
Γ(1 − a)
a
=
Γ(1 − a)
Z
Z
t
∞
ds
Z
0
λ
e
0
Z
t
ds
0
−1−a −sφ(λ)
Z
dλ
(0,∞)
∞
λ−1−a φ(λ)e−sφ(λ) dλ.
0
34
(1 − e−λl )e−l/2 n(dl)
The λ-integral in the last line is further computed by the change of variable λ = φ−1 (x) as
follows:
Z ∞
λ−1−a φ(λ)e−sφ(λ) dλ
0
Z
∞
(φ−1 (x))−1−a xe−sx (φ−1 )0 (x)dx
0
Z ∞
h
i∞
−a −sx
−1
1
1
= − a (φ (x)) xe
+ a
(φ−1 (x))−a (e−sx − sxe−sx )dx
0
0
Z ∞
nZ ∞
o
= a1
(φ−1 (x))−a e−sx dx − s
(φ−1 (x))−a xe−sx dx
0
0
Z ∞
nZ ∞
R
R
o
x −1
x
−a
−sx
1
= a
e d 0 (φ (θ)) dθ − s
e−sx d 0 (φ−1 (θ))−a θdθ
=
0
0
n R
R
o
= a1 L s, 0· (φ−1 (θ))−a dθ − sL s, 0· (φ−1 (θ))−a θdθ .
Here L(·, G) denotes Rthe Laplace transform of a right-continuous increasing function G : [0, ∞) →
∞
[0, ∞): L(s, G) := 0 e−sx dG(x). The last fourth and third equalities are respectively bea 1−a −sx
cause
0 ≤ (φ−1 (x))−a xe−sx ≤ (ψ 0 (1/2))
x ↓ 0, and because for b > a − 1,
R ∞ −1
R ∞ xb−a e−sx → 0 as
−a
b
−sx
0
a
0 (1/2))a sa−b−1 Γ(b−a+1) < ∞. Hence
(φ
(x))
x
e
dx
≤
(ψ
(1/2))
x
e
dx
=
(ψ
0
0
(6.2) is rewritten as follows:
Z t
R
R
1
a
L(s, 0· (φ−1 (θ))−a dθ) − sL(s, 0· (φ−1 (θ))−a θdθ) ds.
E [σt ] =
(6.11)
Γ(1 − a) 0
1◦ The case 0 < a < α. Then 0 < α ≤ 1. By (6.10), (φ−1 (·))−a is regularly varying with
exponent −a/α ∈ (−1, 0). By Theorem 1 of §VIII.9 in [Fe],
x2 (φ−1 (x))−a
Rx
−→ 2 −
−1
−a
0 (φ (θ)) θdθ
a
α
> 0,
x(φ−1 (x))−a
Rx
−→
−1
−a
0 (φ (θ)) dθ
α−a
α
> 0
as x ↑ ∞. Hence, by combining this with (6.10),
Z x
1
1
(φ−1 (θ))−a θdθ ∼
x2 (φ−1 (x))−a ∼
x2− a/α L2 (x)−a ,
2
−
a/α
2
−
a/α
0
Z x
α
α
(φ−1 (θ))−a dθ ∼
x(φ−1 (x))−a ∼
x1− a/α L2 (x)−a
α
−
a
α
−
a
0
as x ↑ ∞. By applying the Abelian theorem (cf. Theorem 2 of §XIII.5 in [Fe]), this implies that
R
Γ(2 − a/α + 1) −2+ a/α
L(s, 0· (φ−1 (θ))−a θdθ) ∼
L2 ( 1s )−a = Γ(2 − αa )s−2+ a/α L2 ( 1s )−a ,
s
2 − a/α
R
L(s, · (φ−1 (θ))−a dθ) ∼ α Γ(2 − a )s−1+ a/α L ( 1 )−a
0
α−a
2 s
α
as s ↓ 0, and hence
R
R
α
L(s, 0· (φ−1 (θ))−a dθ) − sL(s, 0· (φ−1 (θ))−a θdθ) ∼ α−a
− 1 Γ(2 − αa )s−1+ a/α L2 ( 1s )−a
=
35
a
α−a −1+ a/α
L2 ( 1s )−a
α Γ( α )s
as s ↓ 0.
Now if, for simplicity, we set
R
R
Z(x) := L( x1 , 0· (φ−1 (θ))−a dθ) − x1 L( x1 , 0· (φ−1 (θ))−a θdθ),
then, by (6.11)
E [σta ]
1
=
Γ(1 − a)
Z
t
0
Z( 1s )ds
1
=
Γ(1 − a)
Z
∞
x−2 Z(x)dx
1/t
and also,
Z(x) ∼
1− a/α
a
α−a
L2 (x)−a
α Γ( α ) x
as x ↑ ∞.
Therefore, applying Theorem 1 of §VIII.9 in [Fe] again, we have
(1/t)−2+1 Z(1/t)
−→ Γ(1 − a) αa
E [σta ]
as t ↓ 0,
and consequently
E [σta ] ∼
Γ((α − a)/α) a/α
t L2 ( 1t )−a ,
Γ(1 − a)
which is just the assertion (i).
2◦ The case a = α. Then 0 < α < 1 and hence, by (6.10), (φ−1 (·))−α is regularly varying with
exponent −1. Once again, by Theorem 1 of §VIII.9 in [Fe],
as x ↑ ∞, and
x2 (φ−1 (x))−α
Rx
−→ 1,
−1
−α θdθ
0 (φ (θ))
R·
−1
−α dθ
0 (φ (θ))
Z
0
x
x(φ−1 (x))−α
Rx
−→ 0
−1
−α dθ
0 (φ (θ))
is slowly varying at infinity. By combining this with (6.10)
(φ−1 (θ))−α θdθ ∼ x2 (φ−1 (x))−α ∼ xL2 (x)−α ,
Rx
L2 (x)−α ∼ x(φ−1 (x))−α = o( 0 (φ−1 (θ))−α dθ)
as x ↑ ∞, and hence, by the Abelian theorem
R
L s, 0· (φ−1 (θ))−α θdθ ∼ s−1 L2 ( 1s )−α ,
Z
R
·
−1
−α
L s, 0 (φ (θ)) dθ ∼
1/s
(φ−1 (θ))−α dθ
0
as s ↓ 0. Therefore
R
R
Z( 1s ) = L s, 0· (φ−1 (θ))−α dθ − sL s, 0· (φ−1 (θ))−α θdθ
Z
∼
1/s
(φ−1 (θ))−α dθ
as s ↓ 0.
0
36
In exactly the same way as in 1◦ we eventually have
(1/t)−2+1 Z(1/t)
−→ Γ(1 − α)
E [σtα ]
as t ↓ 0,
from which the assertion (ii) is easily seen.
3◦ The case α < a < 1. Then 0 ≤ α < 1. By (6.6), it is enough to show that
Z
Z ∞
a
a −l/2
l e
n(dl) =
λ−1−a φ(λ)dλ < ∞.
Γ(1 − a) 0
(0,∞)
First this identity is seen from the following computation:
Z ∞
Z ∞
λ−1−a φ(λ)dλ =
λ−1−a ψ(λ + 12 ) − ψ( 12 ) dλ
0
0
Z
−l/2
=
e
(0,∞)
=
Γ(1 − a)
a
Z
∞
n(dl)
Z
0
λ−1−a (1 − e−λl )dλ
la e−l/2 n(dl).
(0,∞)
Next this integral is convergent. Indeed, since φ(λ) ≤ ψ 0 (1/2)λ (λ ≥ 0),
Z R
Z R
R1−a
−1−a
0 1
λ
φ(λ)dλ ≤ ψ ( 2 )
λ−a dλ = ψ 0 ( 12 )
< ∞
1−a
0
0
for any R > 0. On the other hand, since φ(λ) ∼ λα L1 (λ) as λ ↑ ∞, and L1 (·) is slowly varying
at infinity, there exists an Rε > 0 for 0 < ε < a − α (cf. Lemma 2 of §VIII.8 in [Fe]) such that
φ(λ) ≤ 2λα L1 (λ) and L1 (λ) < λε for any λ ≥ Rε . Hence
Z ∞
Z ∞
1 a−α−ε
2
λ−1−a φ(λ)dλ ≤
λ−1−a 2λα+ε dλ =
< ∞.
a − α − ε Rε
Rε
Rε
Appendix:
ψ
Semigroups e−t(H0 +V ) and their generators in Lp (R d ) and C∞ (R d )
In this appendix we suppose only that V : R d → [0, ∞) is a continuous function. The main
result is Theorem A.1, which follows from Lemma A.2 (Kato’s inequality).
Let M (dsdx) be a Poisson random measure on [0, ∞)×(R d \{0}) with intensity measure dsJ(dx),
where
Z
J(dx) :=
e−l/2 p(l, x)n(dl)dx,
(A.1)
(0,∞)
1 d/2
|x|2 p(l, x) :=
exp −
.
2πl
2l
(A.2)
This M (·) may be defined on the same probability space (Ω, F, P) as in Section 2. Note that
for p ∈ [1, ∞) the 2p-th order absolute moment of J is finite, i.e.,
Z
|x|2p J(dx) < ∞.
(A.3)
Rd \{0}
37
Following the notation in [Ik-Wa], we set
c(dsdx) := dsJ(dx), M
f(dsdx) := M (dsdx) − M
c(dsdx)
M
and define an R d -valued right-continuous process (Xt )t≥0 by
Z
Xt :=
t+Z
0
Z
Rd \{0}
x1|x|≥1 M (dsdx) +
t+Z
0
Rd \{0}
f(dsdx),
x1|x|<1 M
(A.4)
f. This is a d-dimensional
where the second term on the RHS is a stochastic integral w.r.t. M
time-homogeneous Lévy process starting at the origin such that
√
E [e
−1hp,Xt i
] = e−t(ψ((|p|
2 +1)/2)−ψ(1/2)
),
which is easily seen by Itô’s formula (cf. [Ik-Wa]), so that
L
(Xt )t≥0 ∼ (B(σt ))t≥0 .
(A.5)
We now define a system of operators Ptψ,V , t ≥ 0, by the Feynman-Kac formula:
h
Z t
i
Ptψ,Vf (x) := E exp −
V (x + Xs )ds f (x + Xt ) .
(A.6)
0
From this definition the following is easily seen:
(i) If f is a nonnegative Borel measurable function, so is Ptψ,Vf , and it satisfies
Z
Rd
ψ,V
Ptψ,V(Psψ,Vf ) = Pt+s
f,
Z
|Ptψ,Vf (x)|p dx ≤
|f (x)|p dx,
Rd
(A.7)
1 ≤ p < ∞.
(A.8)
(ii) If f ∈ C∞ (R d ), then Ptψ,Vf ∈ C∞ (R d ) and
kPtψ,Vf k∞ ≤ kf k∞ ,
(A.9)
kPtψ,Vf − f k∞ → 0 as t ↓ 0.
(A.10)
(iii) For two nonnegative Borel measurable functions f, g
Z
Z
ψ,V
Pt f (x)g(x)dx =
f (x)Ptψ,Vg(x)dx.
Rd
Rd
(A.11)
By (i) and (ii), {Ptψ,V }t≥0 is a strongly continuous contraction semigroup on C∞ (R d ). By the
Riesz-Banach theorem there exists a finite measure P ψ,V (t, x, dy) on R d such that
Z
ψ,V
Pt f (x) =
f (y)P ψ,V (t, x, dy), f ∈ C∞ (R d ).
(A.12)
Rd
38
Indeed, by noting (A.5), P ψ,V (t, x, dy) is absolutely continuous w.r.t. the Lebesgue measure dy
on R d and expressed as
h
Z t
i
σt ,y
ψ,V
P (t, x, dy) = E exp −
V (B0,x
(σs ))ds p(σt , x − y) dy,
(A.13)
0
τ,y
where B0,x
(θ) is defined in (3.13).
By (i) and (ii) again Ptψ,V is uniquely extended to a bounded operator on Lp (R d ), which is
denoted by the same Ptψ,V , and thus {Ptψ,V }t≥0 is a strongly continuous contraction semigroup
on Lp (R d ). Clearly, for f ∈ Lp (R d )
h
Z t
i
Ptψ,Vf (x) = E exp −
V (x + Xs )ds f (x + Xt ) a.e. x
0
and, when p = 2, Ptψ,V is symmetric.
G
Let ψ,V
be the infinitesimal generator of {Ptψ,V }t≥0 on Lp (R d ) for 1 ≤ p < ∞, and on C∞ (R d )
p
for p = ∞. Their domains are denoted by ( ψ,V
p ).
DG
Put
Z
H0ψ f (x)
:= −
Rd \{0}
{f (x + y) − f (x) − hy, ∇f (x)i1|y|<1 }J(dy),
H ψ f (x) := H0ψ f (x) + V (x)f (x).
(A.14)
(A.15)
Claim A.1. T
(i) For f ∈ S(R d ), H0ψ f is in S(Rd ), and hence, for f ∈ C0∞ (R d ), H ψ f ∈
C∞ (R d ) ∩
Lp (R d ).
1≤p<∞
(ii) For f ∈ C ∞ (R d ) ∩ Lp (R d ) (where 1 ≤ p < ∞), H0ψ f is well-defined, i.e., the integral in
d
∞ d
d
(A.14) is convergent for a.e. x, and H0ψ f ∈ Lloc
p (R ). Also, for f ∈ C (R ) ∩ L∞ (R ), the
ψ
integral in (A.14) is convergent for every x and H0 f ∈ C(R d ).
For the proof, cf. [I1].
Claim A.2. C0∞ (R d ) ⊂
T
1≤p≤∞
ψ,V
∞ d
ψ
D(Gψ,V
p ), and for f ∈ C0 (R ), Gp f = −H f .
Proof. Let f ∈ C0∞ (R d ).
We start with the proof that
ψ,V
1
t (Pt f
− f ) −→ −H ψ f
t↓0
in C∞ (R d ).
(A.16)
Since H ψ f ∈ C∞ (R d ) by Claim A.1, it is enough to check pointwise convergence (cf. Lemma
31.7 in [Sa]). To do so we apply Itô’s formula for (A.4) to obtain
Z t
exp −
V (x + Xs )ds f (x + Xt )
0
39
Z
= f (x) −
Z
t+Z
+
Z
0
t+Z
+
0
Z tZ
s
0
Rd \{0}
V (x + Xr )dr V (x + Xs )f (x + Xs )ds
Z
exp −
s
Z
exp −
s
0
0
Z
exp −
Rd \{0}
0
Z
exp −
Rd \{0}
0
+
t
V (x + Xr )dr
f (x + Xs− + y) − f (x + Xs− ) 1|y|≥1 M (dsdy)
V (x + Xr )dr
f(dsdy)
f (x + Xs− + y) − f (x + Xs− ) 1|y|<1 M
s
V (x + Xr )dr
0
f (x + Xs + y) − f (x + Xs )
c(dsdy).
− hy, ∇f (x + Xs )i 1|y|<1 M
Note that the third term on the RHS is a martingale, so that the expectation is zero. Taking
expectation and changing the variable s = tσ we have
Z 1 h
Z tσ
i
ψ,V
1
exp
−
(P
f
(x)
−
f
(x))
+
E
V
(x
+
X
)dr
(V
f
)(x
+
X
)
r
tσ dσ
t
t
Z
Z
1
=
dσ
0
Z
|y|≥1
+
Z
=
0<|y|<1
Z
1
dσ
0
Rd \{0}
V (x + Xr )dr
Z
E exp −
h
dσ
0
0
tσ
0
Z
1
0
Z
E exp −
h
i
f (x + Xtσ + y) − f (x + Xtσ ) J(dy)
tσ
V (x + Xr )dr
0
f (x + Xtσ + y) − f (x + Xtσ )
i
− hy, ∇f (x + Xtσ )i J(dy)
Z
E exp −
h
V (x + Xr )dr
0
Z
×
tσ
1
0
i
(1 − θ)hy, ∇2 f (x + Xtσ + θy)yidθ J(dy),
(A.17)
where the second equality is due to Taylor’s theorem with the aid of symmetry of J(dy). On
letting t ↓ 0 in the first equality of (A.17) we have (A.16) pointwise.
Next we prove for 1 ≤ p < ∞ that
ψ,V
1
t (Pt f
− f ) −→ −H ψ f
t↓0
in Lp (R d ).
(A.18)
Since H ψ f ∈ Lp (R d ) by Claim A.1, it is enough to check weak convergence (cf. Lemma 32.3 in
[Sa]).
First of all, we note by (A.17) that
sup k 1t (Ptψ,Vf − f )kp < ∞
t>0
and that
Z
lim lim sup
R→∞
t↓0
|x|>R
for 1 ≤ p < ∞
| 1t (Ptψ,Vf (x) − f (x))|dx = 0.
40
(A.19)
(A.20)
Indeed, by the second equality of (A.17)
Z 1
ψ,V
1
| t (Pt f (x) − f (x))| ≤
E [|(V f )(x + Xtσ )|]dσ
0
Z
Z
1
+
dσ
R
0
d \{0}
|y|2 J(dy)
Z
1
0
(1 − θ)E [|∇2 f (x + Xtσ + θy)|]dθ.
(A.21)
Hence, by Minkowski’s inequality, Jensen’s inequality and Fubini’s theorem
Z
Z
1/p
ψ,V
p
1
1
| t (Pt f (x) − f (x))| dx
≤ kV f kp + 2
|y|2 J(dy) k∇2 f kp ,
Rd
Rd \{0}
which shows (A.19). To show (A.20), take R0 > 0 such that supp f ⊂ {x ∈ R d ; |x| < R0 }, and
let R > R0 . Note that 1|x|>R h(x + y) = 1|x|>R h(x + y)1|y|≥R−R0 for h = f , ∇f or ∇2 f . Hence,
by (A.21),
Z
| 1t (Ptψ,Vf (x) − f (x))|dx
|x|>R
Z
≤
1
0
Z
E
hZ
Z
1
+
|x|>R
dσ
0
×E
Rd \{0}
hZ
≤ kV f k1
Z
2
Z
+
i
|(V f )(x + Xtσ )|dx ; |Xtσ | ≥ R − R0 dσ
|x|>R
0
2
1
2 k∇ f k1
1
|y| J(dy)
1
0
(1 − θ)dθ
|∇2 f (x + Xtσ + θy)|dx ; |Xtσ + θy| ≥ R − R0
i
P(|Xtσ | ≥ R − R0 )dσ
Z
Z
1
dσ
0
R
d \{0}
|y|2 J(dy) P(|Xtσ | + |y| ≥ R − R0 ).
Since lim Xtσ = 0 a.s., by the Lebesgue-Fatou inequality
t↓0
Z
lim sup
t↓0
≤
≤
|x|>R
2
1
2 k∇ f k1
2
1
2 k∇ f k1
| 1t (Ptψ,Vf (x) − f (x))|dx
Z
Z
1
dσ
Z
0
Rd \{0}
|y|≥R−R0
|y|2 J(dy) lim sup P(|Xtσ | + |y| ≥ R − R0 )
t↓0
|y|2 J(dy),
and thus (A.20) follows.
Now we show weak convergence in Lp (R d ) of (A.18). When 1 < p < ∞, let q be the conjugate
exponent of p. For each g ∈ Lq (R d ) and R > 0
Z
|h 1t (Ptψ,Vf − f ) + H ψ f, gi| ≤ k 1t (Ptψ,Vf − f ) + H ψ f k∞
|g(x)|dx
|x|≤R
41
+
Z
− f )kp + kH f kp )
(sup k 1t (Ptψ,Vf
t>0
ψ
|x|>R
|g(x)|q dx
1/q
.
By (A.16), the first term tends to zero as t ↓ 0 for fixed R > 0, and the second term tends to
zero as R ↑ ∞. This shows weak convergence in Lp (R d ). Next, when p = 1, for each g ∈ L∞ (R d )
and R > 0,
Z ψ,V
ψ
1
(P
f
(x)
−
f
(x))
+
H
f
(x)
g(x)dx
t
t
Rd
≤ k 1t (Ptψ,Vf − f ) + H ψ f k∞
Z
+
|x|>R
Z
|x|≤R
|g(x)|dx
| 1t (Ptψ,Vf (x) − f (x))|dx +
Z
|x|>R
|H ψ f (x)|dx kgk∞ .
Therefore, by (A.16) and (A.20), similarly we can show weak convergence in L1 (R d ). The proof
of Claim A.2 is complete.
Remark. When V is further a C ∞ -function and all its derivatives
polynomial growth, it
T have ψ,V
d
can be shown in exactly the same way as above that S(R ) ⊂
( p ) and ψ,V
= −H ψ
p
1≤p≤∞
DG
G
on S(R d ).
By Claim A.2, H ψ on C0∞ (R d ) is closable as an operator in Lp (R d ), 1 ≤ p < ∞, or C∞ (R d ). It
is natural to ask whether or not its smallest closed extension agrees with − ψ,V
p . The following
theorem is an affirmative answer.
G
Theorem A.1. The smallest closed extension of H ψ = −
Gψ,V
p |C
ψ,V
d
Gψ,V
∞ |C (R ) in C∞ (R )) agrees with −Gp
C0∞ (R d ) is a core of Gψ,V
(1 ≤ p ≤ ∞).
p
(resp. H ψ = −
∞
0
d
R
∞
d)
0 (
(resp. −
in Lp (R d ) (1 ≤ p < ∞)
Gψ,V
∞ ).
In other words,
Needless to say, this theorem for p = 2, the L2 -case, says nothing but that H0ψ + V is essentially
selfadjoint on C0∞ (R d ).
In the same way as in [I1] and [I-Tsu]
we prove this theorem. Take ρ ∈ C0∞ (R d ) such that ρ ≥ 0,
R
supp ρ ⊂ {x ∈ Rd ; |x| ≤ 1} and Rd ρ(x)dx = 1. For 0 < δ ≤ 1 set ρδ (x) := (1/δ)d ρ(x/δ). For
d
δ
δ
∞ d
δ
u ∈ Lloc
1 (R ), we denote the convolution u ∗ ρδ by u . Clearly u ∈ C (R ) and u → u in
loc
d
L1 (R ) as δ ↓ 0.
d
Lemma A.1. Let 1 ≤ q ≤ ∞. Suppose u ∈ Lq (R d ) is such that H0ψ u ∈ Lloc
1 (R ), i.e., for some
loc
d
∞
d
f ∈ L1 (R ) it holds that for any ϕ ∈ C0 (R )
Z
Z
f (x)ϕ(x)dx =
u(x)H0ψ ϕ(x)dx.
(A.22)
Rd
Rd
d
Then H0ψ uδ → H0ψ u in Lloc
1 (R ) as δ ↓ 0.
42
d
d
Proof. Since u ∈ Lq (R d ), uδ ∈ C ∞ (R d ) ∩ Lq (R d ). By Claim A.1, H0ψ uδ ∈ Lloc
q (R ) or ∈ C(R )
d
according as 1 ≤ q < ∞ or q = ∞, and hence H0ψ uδ ∈ Lloc
1 (R ). For the proof, it is enough to
ψ δ
ψ δ
check that H0 u = (H0 u) .
By (A.22)
Z
(H0ψ u)δ (x) =
Z
=
Z
=
R
d
Rd
Rd
(H0ψ u)(y)ρδ (x − y)dy
u(y)H0ψ ρδ (x − ·)(y)dy
Z
u(y)dy −
n
Rd \{0}
ρδ (x − y − z) − ρδ (x − y)
o
− hz, ∇ρδ (x − ·)(y)i1|z|<1 J(dz) .
(A.23)
∞ d
The
integral on the RHS is convergent,
R
R because2 with ρδ (x − ·)2 =: g ∈ C0 (R ), it is bounded by
|z|≥1 J(dz) kukq 2kgkq/(q−1) + (1/2) 0<|z|<1 |z| J(dz) kukq k∇ gkq/(q−1) . Here when q = 1 or ∞
we understand k · kq/(q−1) = k · k∞ or k · k1 . Hence by noting that ∇ρδ (x − ·)(y) = −(∇ρδ )(x − y),
Fubini’s theorem gives us that
Z
Z
Z
(H0ψ u)δ (x) = −
J(dz)
u(y)ρδ (x − z − y)dy −
u(y)ρδ (x − y)dy
Rd \{0}
Rd
− 1|z|<1 − z,
Rd
Z
= −
Rd \{0}
Rd
Z
D
u(y)(∇ρδ )(x − y)dy
E
(uδ (x + z) − uδ (x) − 1|z|<1 hz, ∇uδ (x)i)J(dz)
= H0ψ uδ (x),
where the symmetry of J(dz) has been used. The proof is complete.
Lemma A.2. (Kato’s inequality). Let 1 ≤ q ≤ ∞. Suppose u ∈ Lq (R d ) is such that H0ψ u ∈
d
Lloc
1 (R ). Then the following distributional inequality holds:
sgn uH0ψ u ≥ H0ψ |u|,
i.e. for any ϕ ∈ C0∞ (R d ) with ϕ ≥ 0,
Z
Z
(sgn u)(x)H0ψ u(x)ϕ(x)dx ≥
R
R
d
Here sgn u is a bounded function on Rd defined by

 u(x)
|u(x)|
(sgn u)(x) :=
 0
43
d
|u(x)|H0ψ ϕ(x)dx.
if
u(x) 6= 0
if
u(x) = 0.
d
d
Proof. First let u ∈ C ∞ (R d ) ∩ Lq (R d ). By Claim A.1 H0ψ u ∈ Lloc
q (R ) or ∈ C(R ) according as
p
1 ≤ q < ∞ or q = ∞. For ε > 0, set uε (x) := |u(x)|2 + ε2 . Clearly uε ∈ C ∞ (R d ) and uε ≥ ε.
Since |u(x)| |u(x + y)| ≤ uε (x) uε (x + y) − ε2 , we have
−|u(x)| |u(x + y)| + u(x)2 ≥ −uε (x) uε (x + y) + uε (x)2 .
By noting that 2u(x)∇u(x) = ∇|u(x)|2 = ∇uε (x)2 = 2uε (x)∇uε (x), this inequality gives us
that
−u(x){u(x + y) − u(x) − hy, ∇u(x)i1|y|<1 }
= −u(x)u(x + y) + u(x)2 + hy, u(x)∇u(x)i1|y|<1
≥ −uε (x)uε (x + y) + uε (x)2 + hy, uε (x)∇uε (x)i1|y|<1
= −uε (x){uε (x + y) − uε (x) − hy, ∇uε (x)i1|y|<1 }.
Integrating both sides by J(dy), we have u(x)H0ψ u(x) ≥ uε (x)H0ψ uε (x), or
u(x)
uε (x)
H0ψ u(x) ≥ H0ψ uε (x).
(A.24)
d
δ
∞ d
d
Second let u ∈ Lq (R d ) be such that H0ψ u ∈ Lloc
1 (R ). Since u = u ∗ ρδ ∈ C (R ) ∩ Lq (R ), it
holds by (A.24) that
Z
Z
uδ (x)
ψ δ
H0 u (x)ϕ(x)dx ≥
H0ψ (uδ )ε (x) ϕ(x)dx
(uδ )ε (x)
Rd
Rd
Z
=
R
d
(uδ )ε (x)H0ψ ϕ(x)dx
(A.25)
for any nonnegative ϕ ∈ C0∞ (R d ). In (A.25) let δ ↓ 0 first and ε ↓ 0 second. As δ ↓ 0,
d
δ
loc d
H0ψ uδ → H0ψ u in Lloc
1 (R ) by Lemma A.1, and u → u in L1 (R ). By taking a subsequence if
δ
δ
necessary we may suppose that u → u a.e. Since |(u )ε − uε | ≤ |uδ − u| and |uδ /(uδ )ε | ≤ 1,
uδ /(uδ )ε → u/uε boundedly a.e. Hence, letting δ ↓ 0 in (A.25), we have
Z
Z
u(x)
ψ
uε (x)H0ψ ϕ(x)dx.
(A.26)
uε (x) H0 u(x)ϕ(x)dx ≥
Rd
Rd
Finally, by |uε − |u|| ≤ ε and |u/uε | ≤ 1, we obtain that u/uε → sgn u boundedly as ε ↓ 0.
Consequently, letting ε ↓ 0 in (A.26) yields that
Z
Z
ψ
(sgn u)(x)H0 u(x)ϕ(x)dx ≥
|u(x)|H0ψ ϕ(x)dx
Rd
Rd
and the proof is complete.
Proof of Theorem A.1. First consider the Lp -case, 1 ≤ p < ∞. It suffices to show that Im (H0ψ +
V + 1) = (H0ψ + V + 1)(C0∞ (R d )) is dense in Lp (R d ). By the Hahn-Banach theorem, this is
further reduced to show the following: Let q be the conjugate exponent of p. If v ∈ Lq (R d )
satisfies that hv, (H0ψ + V + 1)ϕi = 0 for any ϕ ∈ C0∞ (R d ), then v = 0 in Lq (R d ).
44
d
Let v ∈ Lq (R d ) be as above. Then H0ψ v = −(V + 1)v and hence H0ψ v ∈ Lloc
1 (R ). By Lemma
∞
d
A.2, it is seen that for any nonnegative ϕ ∈ C0 (R )
Z
Z
ψ
|v(x)|H0 ϕ(x)dx ≤
(sgn v)(x)H0ψ v(x)ϕ(x)dx
Rd
Rd
Z
= −
and hence
Rd
(V (x) + 1)|v(x)|ϕ(x)dx,
Z
R
d
|v(x)|(H0ψ + 1)ϕ(x)dx ≤ 0.
(A.27)
∞ d
Each ϕ ∈ S(R d ) can be approximated by a sequence {ϕn }∞
n=1 of C0 (R ) in the sense that
ϕn → ϕ and (H0ψ + 1)ϕn → (H0ψ + 1)ϕ in Lp (R d ). If ϕ is moreover nonnegative, so are {ϕn }∞
n=1 .
Therefore (A.27) is valid for nonnegative ϕ ∈ S(Rd ).
Now note that the resolvent (1 −
(1 −
−1 is expressed as
Gψ,0
p )
G
ψ,0 −1
p ) f (x)
Z
∞
=
0
e−t E [f (x + Xt )]dt.
G
−1
d
Then it is not difficult to check that if f ∈ S(R d ), then (1 − ψ,0
p ) f ∈ S(R ) and further, if
ψ,0 −1
f is nonnegative, so is (1 − p ) f . Also, by Remark following Claim A.2 (with V (x) ≡ 0)
ψ,0 −1
ψ
ψ,0 −1
f = (1 − ψ,0
p )(1 − p ) f = (H0 + 1)(1 − p ) f . Hence, by (A.27)
Z
|v(x)|f (x)dx ≤ 0,
G
G
G
G
Rd
whence it immediately follows that v = 0 and the proof in the Lp -case is complete.
Next let us consider the C∞ -case. In the same reason as above we have only to show that
(H0ψ + V + 1)(C0∞ (R d )) is dense in C∞ (R d ). For this let ν ∈ C∞ (R d )∗ , the dual of C∞ (R d ),
be such that hν, (H0ψ + V + 1)ϕi = 0 for any ϕ ∈ C0∞ (R d ). By the Riesz-Banach theorem, ν is
regarded as a finite signed Borel measure on R d , and thus
Z
(H0ψ + V + 1)ϕ(x)ν(dx) = 0
for any ϕ ∈ C0∞ (R d ).
(A.28)
Rd
R
Let ν δ = ν ∗ ρδ , i.e., ν δ (x) := Rd ρδ (x − y)ν(dy), x ∈ Rd . Then ν δ ∈ Cb∞ (R d ) ∩ L1 (R d ). It follows
by Claim A.1 and by (A.28) that H0ψ ν δ ∈ C(Rd ) and
Z
ψ δ
δ
H0 ν = −ν (1 + V ) −
(V (y) − V (·))ρδ (· − y)ν(dy).
Rd
By Lemma A.2, this implies that for nonnegative ϕ ∈ C0∞ (R d )
Z
|ν δ (x)|H0ψ ϕ(x)dx
Rd
≤
Z
Rd
(sgn ν δ )(x)H0ψ ν δ (x)ϕ(x)dx
Z
≤ −
Z
|ν (x)|ϕ(x)dx +
δ
Rd
Rd
Z
ϕ(x)dx
45
Rd
|V (y) − V (x)|ρδ (x − y)|ν|(dy),
where |ν| is the total variation of ν, and hence we have that for nonnegative ϕ ∈ C0∞ (R d )
Z
Z
Z
ψ
δ
|ν (x)|(H0 + 1)ϕ(x)dx ≤
ϕ(x)dx
|V (y) − V (x)|ρδ (x − y)|ν|(dy).
(A.29)
Rd
Rd
Rd
G
−1
d
Let f ∈ S(Rd ) be nonnegative. Then, ϕ := (1 − ψ,0
∞ ) f is in S(R ) and nonnegative, and
ψ
2
2
∞
(H0 + V )ϕ = f . Set ϕn (x) := ϕ(x)χ(|x| /n ) with a χ ∈ C ([0, ∞) → R) such that 0 ≤ χ ≤ 1,
χ(t) = 1 (0 ≤ t ≤ 1) and χ(t) = 0 (t ≥ 2). Clearly ϕn ∈ C0∞ (R d ), 0 ≤ ϕn ≤ ϕ and
√
supp ϕn ⊂ {x; |x| ≤ 2n}. Moreover kϕn − ϕk∞ and kH0ψ ϕn − H0ψ ϕk∞ → 0 as n → ∞. From
(A.29) and this observation it follows that
Z
Z
δ
|ν (x)|f (x)dx =
|ν δ (x)|(H0ψ + 1)ϕ(x)dx
Rd
Rd
Z
=
R
Z
≤
d
Rd
Z
|ν
δ
(x)|(H0ψ
+ 1)ϕn (x)dx +
R
d
Z
ϕn (x)dx
Rd
|ν δ (x)|(H0ψ + 1)(ϕ − ϕn )(x)dx
|V (y) − V (x)|ρδ (x − y)|ν|(dy)
+ k(H0ψ + 1)(ϕ − ϕn )k∞ |ν|(R d )
Z
Z
≤ kϕk∞
dx
|V (y) − V (x)|ρδ (x − y)|ν|(dy)
√
Rd
|x|≤ 2n
+ k(H0ψ + 1)(ϕ − ϕn )k∞ |ν|(R d ).
Here, recalling that ρδ (z) has support in {z; |z| ≤ ρ}, we see that for each n ∈ N
Z
Z
|V (y) − V (x)|ρδ (x − y)|ν|(dy)
√ dx
|x|≤ 2n
Z
≤
Z
=
−→ 0
Rd
√
|y|≤ 2n+δ
√
|y|≤ 2n+δ
Z
|ν|(dy)
Z
|ν|(dy)
Rd
|V (y) − V (x)|ρδ (x − y)dx
|x|≤1
|V (y) − V (y + δx)|ρ(x)dx
as δ ↓ 0.
On the other hand, noting that ν δ (x)dx → ν(dx) weakly, we see that
Z
Z
Z
δ
δ
|ν (x)|f (x)dx ≥ f (x)ν (x)dx −→ f (x)ν(dx)
Rd
Rd
Rd
as δ ↓ 0.
R
Therefore it follows that Rd f (x)ν(dx) = 0 for f ∈ S(Rd ), f ≥ 0, which implies that ν = 0, and
the proof in the C∞ -case is complete.
ψ
ψ
In this paper we have denoted the semigroups Ptψ,0 and Ptψ,V by e−tH0 and e−t(H0 +V ) , respectively, taking Theorem A.1 into account. With the general theory ([Trot], [Ch]) we have taken
for granted that the Trotter product formula holds in the strong topology of Lp (R d ) or C∞ (R d ):
ψ
ψ
s- lim (e−tH0 /n e−tV /n )n = s- lim (e−tV /2n e−tH0 /n e−tV /2n )n
n→∞
n→∞
ψ
ψ
ψ
= s- lim (e−tH0 /2n e−tV /n e−tH0 /2n )n = e−t(H0 +V ) .
n→∞
46
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