Praca semestralna nr 2

Sebastian Sydor
Uniwersytet Wrocławski
Gronwall inequality for semigroups
Praca semestralna nr 2
(semestr zimowy 2010/11)
Opiekun pracy: Krzysztof Bogdan
Gronwall inequality for semigroups
Sebastian Sydor
February 2, 2011
Abstract
We give a new proof of an estimate for Schrödinger semigroups
given in [1]. The proof is motivated by Gronwall inequality.
1
Introduction
Consider an arbitrary set X with a σ -algebra M and a (nonnegative) measure
m dened on M. To simplify the notation we will write dz for m(dz) in what
follows. Consider the σ -algebra B of Borel subsets of R, and the Lebesgue
measure, du, dened on B . The space-time, R × X , will be equipped with
the σ -algebra B × M and the product measure du dz = du m(dz).
Let p be a B ×M×B ×M-measurable function dened on R×X ×R×X .
We will call p a transition density on X if
p(s, x, t, y) = 0,
for s ≥ t ,
(1)
for s < t , x, y ∈ X ,
0 < p(s, x, t, y) < ∞,
and the following Chapman-Kolmogorov equations hold for s < u < t,
Z
p(s, x, u, z)p(u, z, t, y) dz = p(s, x, t, y) , x, y ∈ X .
(2)
(3)
X
Let function q be nonnegative and measurable with respect to B × M. For
s, t ∈ R and x, y ∈ X , we let p0 (s, x, t, y) = p(s, x, t, y) and
Z tZ
pn (s, x, t, y) =
pn−1 (s, x, u, z)p(u, z, t, y) q(u, z) dzdu ,
(4)
s
for n ≥ 1. We dene
pq (s, x, t, y) =
X
∞
X
pn (s, x, t, y) ,
n=0
x, y ∈ X ,
s, t ∈ R .
Some properties of pn and pq are collect in section 3(see also [1]).
1
(5)
2
Main result
We will consider the condition
Zt Z
s
p(s, x, u, z)q(u, z)p(u, z, t, y)dzdu ≤ [η + β(t − s)]p(s, x, t, y)
(6)
X
where s < t ∈ R, x, y ∈ X and β and η are nonnegative numbers.
We will give a new, shorter proof of the following result of [1].
Theorem 1 If q satises (6) with η < 1, then, for all s < t and x, y ∈ X ,
1
pq (s, x, t, y) ≤
exp
1−η
β
(t − s) p(s, x, t, y).
1−η
(7)
Proof. The proof is based on the elementary identity
exp
Zu
β
β
β
(u − s) = 1 +
exp
(v − s) dv, u > s.
1−η
1−η
1−η
(8)
s
We note that
pq (s, x, t, y) = p(s, x, t, y) +
Zt Z
s
pq (s, x, u, z)q(u, z)p(u, z, t, y)dzdu.
(9)
X
Let function f be nonegative and measurable with respect to B ×M×B ×M.
For s < t and x, y ∈ X we dene
T f (s, x, t, y) = p(s, x, t, y) +
Zt Z
s
f (s, x, u, z)q(u, z)p(u, z, t, y)dzdu.
X
The Chapman-Kolmogorov equations imply that for s < v < t and x, y ∈ X ,
Zt Z
v
=
X
Zt Z Z
v
=
p(s, x, u, z)q(u, z)p(u, z, t, y)dzdu
Z
X
p(s, x, v, w)p(v, w, u, z)q(u, z)p(u, z, t, y)dwdzdu
X X
p(s, x, v, w)
Zt Z
v
p(v, w, u, z)q(u, z)p(u, z, t, y)dudzdw
X
2
Z
≤
p(s, x, v, w)[η + β(t − v)]p(v, w, t, y)dw
X
(10)
= [η + β(t − v)]p(s, x, t, y).
Assume that for all s < t, x, y ∈ X ,
1
exp
f (s, x, t, y) ≤
1−η
Then we claim that
1
T f (s, x, t, y) ≤
exp
1−η
β
(t − s) p(s, x, t, y).
1−η
(11)
β
(t − s) p(s, x, t, y),
1−η
for all s < t, x, y ∈ X . Indeed, by (8) and (10),
T f (s, x, t, y) = p(s, x, t, y) +
Zt Z
s
1
≤ p(s, x, t, y) +
1−η
Zt
s
Z
X
exp
f (s, x, u, z)q(u, z)p(u, z, t, y)dzdu
X
β
(u − s)
1−η
×p(s, x, u, z)q(u, z)p(u, z, t, y)dzdu


Zu
Zt Z
β
β
1
1 +
exp
(v − s) dv 
= p(s, x, t, y) +
1−η
1−η
1−η
s
X
s
×p(s, x, u, z)q(u, z)p(u, z, t, y)dzdu
1
[η + β(t − s)] + 1
≤ p(s, x, t, y)
1−η
Z t Z Zu
β
β
+
exp
(v − s)
(1 − η)2
1−η
s
X
s
×p(s, x, u, z)q(u, z)p(u, z, t, y)dvdzdu
1
= p(s, x, t, y)
[η + β(t − s)] + 1
1−η
Zt
β
β
+
exp
(v − s)
(1 − η)2
1−η
s
×
Zt
v
Z
p(s, x, u, z)q(u, z)p(u, z, t, y)dzdudv
X
3
1
≤ p(s, x, t, y)
[η + β(t − s)] + 1
1−η
Zt
β
β
(v − s) [η + β(t − v)]p(s, x, t, y)dv
+
exp
(1 − η)2
1−η
s
1
[η + β(t − s)] + 1
= p(s, x, t, y)
1−η
Zt
β
β
+
exp
(v − s)
(1 − η)2
1−η
s
h
i
× η + β(t − s) − β(v − s) p(s, x, t, y)dv
1
= p(s, x, t, y)
[η + β(t − s)] + 1
1−η
β
1−η
β
+
[η + β(t − s)]
exp
(t − s) − 1
(1 − η)2
β
1−η
Zt
β
β
exp
(v − s) β(v − s)dv
−
(1 − η)2
1−η
s
β(t−s)/(1−η)
Z
h
i
η + β(t − s)
β
= p(s, x, t, y) 1 +
exp
(t − s) −
ew wdw
1−η
1−η
0
h
η + β(t − s)
β
= p(s, x, t, y) 1 +
exp
(t − s)
1−η
1−η
i
β
β
β
− exp
(t − s)
(t − s) + exp
(t − s) − 1
1−η
1−η
1−η
β
η
(t − s)
+1
= p(s, x, t, y) exp
1−η
1−η
1
β
=
exp
(t − s) p(s, x, t, y).
1−η
1−η
Let us notice that f (s, x, t, y) = p(s, x, t, y) satises inequality (11).
By induction we verify that for every n ∈ N we have
. . ◦ T} p(s, x, t, y) =
|T ◦ .{z
n
4
n
X
k=0
pk (s, x, t, y).
Therefore
n
X
1
exp
pk (s, x, t, y) ≤
1−η
k=0
β
(t − s) p(s, x, t, y).
1−η
This yields (7). The above proof is motivated by the following version of Gronwall's
lemma:
If q ≥ 0 is integrable on [a, b] and u is bounded on [a, b] and satises
u(t) ≤ 1 +
Zt
(12)
q(s)u(s)ds,
a
for t ∈ [a, b], then for t ∈ [a, b] we have

u(t) ≤ exp 
Zt
a

q(s)ds .
(13)
Recall that the result may be directly veried by iterating (12).
We consider (12) an analogue of the perturbation formula (9), and then
(13) is an analogue of (7). In fact, if q(u, z) = q(u) (depends only on time),
then
Zt Z
s
p(s, x, u, z)q(u, z)p(u, z, t, y)dzdu = p(s, x, t, y)
Zt
q(u)du,
s
X
(compare (6)) and we have

pq (s, x, t, y) = p(s, x, t, y) exp 
as can be directly veried.
3
Zt
s

q(u)du ,
Algebra of perturbation series
The identities we intend to prove below rely merely on changing the order of
integration, which is justied if the integrals involved are absolutely convergent or nonnegative. We shall rst consider the latter situation and we will
assume that q ≥ 0.
5
Lemma 2 For all s < u < t, x, y ∈ X , and n = 0, 1, . . .,
n Z
X
pm (s, x, u, z)pn−m (u, z, t, y) dz = pn (s, x, t, y) .
(14)
X
m=0
Proof. We note that (14) is true for n = 0 by (3). Assume that n ≥ 1 and
(14) holds for n − 1. The sum of the rst n terms in (14) can be dealt with
by induction:
n−1 Z
X
pm (s, x, u, z)pn−m (u, z, t, y) dz
=
m=0 X
n−1 Z
X
m=0
=
=
pm (s, x, u, z)
X
Z tZ
u
X
u
X
Z tZ
Z tZ
u
n−1 Z
X
m=0
pn−1−m (u, z, r, w)p(r, w, t, y) q(r, w)dwdr dz
X
pm (s, x, u, z)pn−1−m (u, z, r, w) dz
X
!
p(r, w, t, y) q(r, w)dwdr
(15)
pn−1 (s, x, r, w)p(r, w, t, y) q(r, w)dwdr .
By (4), the (n + 1)-st term is
Z
pn (s, x, u, z)p0 (u, z, t, y) dz
ZX Z u Z
=
pn−1 (s, x, r, w)p(r, w, u, z) q(r, w)dwdr p(u, z, t, y) dz
X s
X
Z uZ
pn−1 (s, x, r, w)p(r, w, t, y) q(r, w)dwdr .
(16)
=
s
X
and (14) follows adding (15) and (16). P
We next prove the Chapman-Kolmogorov equation for pq = ∞
n=0 pn .
Lemma 3 For all s < u < t and x, y ∈ X ,
Z
pq (s, x, u, z)pq (u, z, t, y) dz = pq (s, x, t, y) .
X
Proof. By Lemma 2,
Z
pq (s, x, u, z)pq (u, z, t, y) dz =
X
=
Z X
∞
pi (s, x, u, z)
X i=0
∞ X
n
X
n=0 m=0
=
∞
X
n=0
6
Z
∞
X
pj (u, z, t, y) dz
j=0
pm (s, x, u, z)pn−m (u, z, t, y) dz
X
pn (s, x, t, y) = pq (s, x, t, y) . The extension of (4) is the following fact
Lemma 4 For all n = 1, 2, . . ., m = 0, 1, . . . , n − 1, s, t ∈ R and x, y ∈ X ,
pn (s, x, t, y) =
Z tZ
s
pn−1−m (s, x, u, z)pm (u, z, t, y) q(u, z) dzdu .
(17)
X
Proof. For m = 0, equality (17) holds by denition of pn . In particular,
this proves our claim for n = 1. If n ≥ 1 such that (17) holds, then, for every
m = 1, 2, . . . , n,
Z Z
pn+1 (s, x, t, y) =
pn (s, x, u, z)p(u, z, t, y) q(u, z) dzdu
R
X
Z Z Z Z
=
pn−1−(m−1) (s, x, w, v)pm−1 (w, v, u, z) q(v, w)dwdv
R
=
X
Z Z
R
X
R
X
p(u, z, t, y) q(u, z) dzdu
pn−m (s, x, w, v)pm (w, v, t, y) q(v, w)dwdv . References
[1] K. Bogdan, W. Hansen and T. Jakubowski.
Time-dependent
Schrödinger perturbations of transition densities.
Studia Math.,
189(3):235254, 2008.
[2] T. Jakubowski. On combinatorics of Schrödinger perturbations. Potential Anal., 31:45-55, 2009.
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