Feynman-Kac semigroups with discontinuous additive functionals, gauge theorems and applications to dirichlet problems

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Feynman-Kac semigroups with discontinuous additive functionals, gauge theorems and applications to dirichlet problems
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FEYNMAN-KAC SEMIGROUPS WITH DISCONTINUOUS
ADDITIVE FUNCTIONALS, GAUGE THEOREMS AND APPLICATIONS TO DIRICHLET PROBLEMS











By

RENMING SONG











A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY


UNIVERSITY OF FLORIDA


1993













ACKNOWLEDGMENTS

I would like to offer my deepest gratitude to my adviser, Professor Joseph Glover, for his constant guidance and encouragement. His suggestion of a gauge theorem for discontinuous additive functionals was the starting point for the whole thesis. His advice led to numerous improvements.

I am greatly indebted to Professor Murali Rao for all his help. I benefited a lot from collaborating with him on several papers. Working with him is always a great pleasure.

I would like to thank Professor Nicolae Dinculeanu, Professor Malay Ghosh and Professor Zoran Pop-Stojanovic for their service on my supervisory committee.

I would also like to express my gratitude to the following people: Professor Zhichan Li, for introducing me into probability and for supervising my master's thesis; Professor Zhiming Ma, for sending me all his preprints and for collaborating with me on several papers; Professor Rong Wu, for encouraging me to attend the University of Florida to pursue my doctor's degree; and my friend Dr. Qingji Yang, for working, enjoying and suffering together when we were pursuing our master's degrees at Hebei University.

And last, but not the least, I would like to express my gratitude to the members of my family: my parents, my brothers and my sisters, for their love and encouragement; my wife, Junge Guo, for being such a loving, understanding, patient and supportive companion; and my daughter, Linda, for enriching my life.

















TABLE OF CONTENTS




ACKNOWLEDGEMENTS ............................ ii

ABSTRACT .. .. ... .. .. .. . ... .. .. . .. .. . .. iv

CHAPTERS

1 INTRODUCTION ................................ 1


2 PRELIM INARIES ............................... 6

2.1 Symmetric Stable Processes ....................... 6
2.2 Killed Symmetric Stable Processes ................... 13
2.3 Bilinear Forms and Sesquilinear Forms ................. 20

3 THE FEYNMAN-KAC SEMIGROUP .................... 27

3.1 The Kato Class .............................. 27
3.2 The Feynman-Kac Semigroup ...................... 38
3.3 The Bilinear Form ............................ 45
3.4 The Connection between the Semigroup (T) and the Bilinear Form
(S,D) ........ ................................... 56

4 THE GAUGE THEOREM AND ITS APPLICATIONS .............. 64

4.1 The Gauge Theorem ........................... 64
4.2 The Dirichlet Problem for Linear Equations .............. 74
4.3 The Dirichlet Problem for Semilinear Equations ................ 82

5 CONCLUSIONS ........ ................................ 88

REFERENCES ........ ................................... 89

BIOGRAPHICAL SKETCH ................................... 92






iii














Abstract of Dissertation Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy


FEYNMAN-KAC SEMIGROUPS WITH DISCONTINUOUS
ADDITIVE FUNCTIONALS, GAUGE THEOREMS AND APPLICATIONS TO DIRICHLET PROBLEMS By

RENMING SONG

August 1993

Chairman: Dr. Joseph Glover Major Department: Mathematics

Let Xt be a symmetric stable process of index a, 0 < a < 2, in Rd (d > 2), let pi and v be Radon measures on Rd belonging to the Kato class Id, and let F be a Borel function on Rd x Rd satisfying certain conditions. Suppose that A and Bt are the continuous additive functionals with Revuz measures it and v, respectively, and that

At= Au + E F(X,-,X,).
O In this research we first study the following Feynman-Kac semigroup


Ttf(x) = E{eat f(Xt)}

and identify the bilinear form corresponding to (Tt)t>o. Then we prove a criterion, called the gauge theorem, about the boundedness of the following gauge function g(x) = E{eA(rD)}.



iv











Finally we prove that for a suitable exterior function f,

u(x) = Ex (eA(rD)f(X(rD))) + Fx TDA t AdBt

is the unique continuous solution to the Dirichlet problem of + ) u(x) + dIa dy + v = 0
Rd IX -yld+o,

on D with the exterior function f, where G = eF 1.































V

















CHAPTER 1
INTRODUCTION

The study of perturbations of Markov processes created by multiplicative functionals is a classical topic dating back to the early researches of Mark Kac. See [17]. This subject has been revived in the past ten years by both probabilists and mathematical physicists interested in the perturbations of the generators of Markov processes.

If X = (Xt, px) is a Feller process on Rd and if At is a continuous additive functional of X, then the semigroup defined by Ttf(x) = EX{eA, f(X,)}

is called a Feynman-Kac semigroup. Mathematical physicists are mainly interested in, among other things, various properties of the semigroup (Tt)t>o. For literature on this subject, one can consult [1], [2], [4], [5], [7], [31] and the references therein.

Probabilists are interested in, among other things, using the Feynman-Kac semigroup to solve various boundary value problems of second order elliptic partial differential equations. When X is a standard Brownian motion, we know, from probabilistic potential theory, that the function u(x) = EX{eA'(D)f(X(rD))}

(where D is an open set in Rd, TD is the first exit time from D and f is a bounded function on OD) should be a candidate for a solution to the Dirichlet problem of the equation
A
A +,),, = 1
2









2



(where pi is the Revuz measure of At) on D with boundary function f. One of the most important steps in making the sentence above precise is a criterion, generally known as the gauge theorem, about the boundedness of the gauge function g(x) = E'le A(TD)}1.

This was first studied by Chung and Rao in [9] and later generalized by many others. For literature on this and some related problems, see [4], [10], [161, [22], [28], [30] and the references therein.

So far, all the literature on Feynman-Kac semigroups and their applications to various boundary value problems have dealt exclusively with the case when At is a continuous additive functional. It is well known that, when the underlying Markov process X is discontinuous, there are a lot of important discontinuous additive functionals. The most typical example of such an additive functional is as follows: EZ F(X,,X,), (11)
O where F is a Borel function on R d x R d vanishing on the diagonal. Naturally one would also like to study the Feynman-Kac semigroup (Tt)t o with the discontinuous additive functional given in (1.1). As one would expect, there are some difficulties in accomplishing this, since some of the basic tools used in the continuous case are no longer available in the discontinuous case. For example, it is no easy task to determine when

E'le At}

is finite or bounded as a function of x, while the corresponding question in the continuous case can be solved by a simple use of Fubini's theorem.

As for the application to the Dirichlet problem, although the case where X is a diffusion has been thoroughly studied, no one has touched the case when X is discontinuous. Even in the case when no perturbation is involved, discussions about









3


the Dirichiet problem for discontinuous Markov processes are scarce and unsystematic. There is a brief discussion in [20] about solving the Dirichiet problem for the symmetric stable process by using a balayage method. In [24] and [25] the Dirichlet problem for a class of infinitely divisible processes is discussed, but the formulation there is purely probabilistic and is not interpreted in analytic language. Another reference is [21] in which the Dirichlet problem for general discontinuous Markov processes is formulated and solved by using characteristic operators; and when the underlying process is a symmetric stable process an equivalent analytical formulation is also given. References [12] and [13] also contain some related discussions. When compared with the case of a second order differential operator L, the Dirichlet problem in an open set D for the generator A of a discontinuous Markov process has some new features. First, instead of a boundary function as in the diffusion case, we have to use a function which is defined on all of the complement of D (which we call an exterior function) and we have to require that the solution to the Dirichlet problem coincide with the exterior function on the complement of D (because when Xt leaves D it can jump to any point in the complement of D). Secondly, we have to be careful about the phrase "u is a solution of Au =0 in D", especially when we encounter it for the first time. In the historical case where L is a differential operator, the phrase 44u is a solution to the equation Lu = 0 in D" means that u is defined in D and satisfies the equation Lu(x) =0 for every X E D. But in our case, the function u has to be defined everywhere on R'd because the operator A is not a differential operator but is an integral operator and the integration extends over the whole space, not just over D.

Let X =(Xt, Px) be the symmetric stable process of index a, 0 < a < 2 on R d (d > 2), let yi and v be Radon measures on R d belonging to the Kato class Kfd,,, (see Definition 3.1.1) and let F be a Borel function on R d x R d belonging to Ad,,, (see








4


Definition 3.1.3). Suppose that A' and Bt are the continuous additive functionals with Revuz measures it and v, respectively, and that At= A,"+ E F(X_,X.).
O The purposes of this research are (a) to study the Feynman-Kac semigroup Ttf(x) = Ex{eA f(Xt)}

and to identify the bilinear form corresponding to (Tt)t>o; (b) to use Dirichlet form theory to formulate and solve the Dirichlet problem for the equation 2+ Pu) u (x) + I G(x,Y~(~ 12
2x Rd yld+, dy + v = 0 (1.2)
J~Ix -yd+a~
on some bounded open set D, where G = eF 1.

As the title suggests, Chapter 2 serves as preparation for later chapters. In this chapter, we recall some basic properties about symmetric stable processes and killed symmetric stable processes. A section on bilinear forms and sesquilinear forms is also included. Section 2.1 and Section 2.3 contain no new results. Some of the results of Section 2.2 may have never been explicitly stated in the literature, but they are hardly surprising.

In chapter 3, we first introduce the class of additive functionals which we are going to deal with, then we study various properties of the Feynman-Kac semigroup and finally we identify the bilinear form corresponding to the Feynman-Kac semigroup.

In chapter 4, we first prove a criterion, called the gauge theorem, about the boundedness of the gauge function g(x) =ExeA(TD)

Then we formulate the Dirichlet problem of (1.2) on D and prove that for a suitable exterior function f,

u(x) :EX (eA( D)f(X(TD))) + E'jnD eA'dBt








5


is the unique continuous solution to the Dirichlet problem of (1.2) on D with exterior function f. As an application of the gauge theorem and the results in Section 4.2, we prove, in Section 4.3, the existence of a solution to the Dirichlet problem for the semilinear equation
(-(-A)2 +p)u(x)+j !G (X,.)u.).
+ ) u(x) + G(x y )u() dy + ((u) + v = 0 (1.3)


on D, where ( is a continuous differentiable function on R1. The results of Section

4.3 generalize those of [14].

















CHAPTER 2
PRELIMINARIES

2.1 Symmetric Stable Processes

The starting point of our research is a particular symmetric Hunt process X = (Q,,Ft,Xt, Ot,P-): the symmetric stable process of index a, 0 < a < 2, on Rd (d > 2) with the characteristic function e-tilzi

For convenience, we shall assume that Q is the space of all right continuous maps w from [0, 00) to Rd. Set Xt(w) = w(t), and let .Ft and F be the appropriate completions of F = o{X, : s < t} and Jo = r{X: s > 0}. For each t > 0, 0t : Q Q is the shift operator characterized by X, o Ot = Xs+t.

As usual, we use (Pt) to denote the transition semigroup of Xt and U, 7 > 0, to denote the 7-potential of (P), U f (W) =1 e- 'tPtf (x)dt. The following result is well known. See, for instance, [7].

Theorem 2.1.1 There is a function p(t, x, y) on (0, c) x Rd x Rd such that

(i) for any t > 0, any x E R and any nonnegative Borel function f on Rd, Ptf(x) =Rd p(t, x, y)f (y)dy;


(ii) p(t, x, y) is jointly continuous on (0, 00) x Rd x Rd;
6








7


(iii) for any t > 0 and any x,y E Rd, p(t, x, y) = p(t, y, x);


(iv) for any t, s > 0 and any x, y E Rd p(t + s,x,y) = Rd p(t,x,z)p(s,z,y)dz;


(v) for any t >0 and any x, y E Rd, p(t, x, y) = p(t, 0, y X);


(vi) for any t > 0 and any x, y E R ,

0 < p(t,x,y) < p(t,O, 0); (vii) for any bounded continuous function f on Rd, l p(t,x,y)f(y)dy = f(x). Theorem 2.1.1 only guarantees the existence of the density p(t, x, y); it does not provide an explicit formula for p(t, x, y). In fact, except for the cases of a = 1 and a= 1/2, no explicit formula seems to be known for p(t, x, y). But fortunately the following two results (see, for instance, [7]) will provide all the estimates that we are going to use later on. Theorem 2.1.2 For any q > 0, there exist positive constants cl and c2 which depend only on d, a and 7 such that for any x,y E Rd with Ix y 1 7, Cl ) C2
Ix ywi (1. Ix y|d+a









8


Theorem 2.1.3 For any t > 0 and any x,y E R,


p(t,x,y) = t-p(1,t-Ex,t-Ty).

From the two results above we can immediately get the following lemma which will be used in the proof of the next theorem. Lemma 2.1.1 For any i7 > 0, lim sup p(t, X, y) = 0. t0 IX-Y>n

Proof. Fix j > 0 and x,y E Rd such that Ix y > 7. Then by Theorem 2.1.2 and Theorem 2.1.3 we know that there exists a constant c > 0 such that

d C
It- (x- y) d+1a
ct
Ix yld+a
ct
d+aL "

Therefore the conclusion of the lemma is true.

Q.E.D.


Theorem 2.1.4 (Pt)t>o is a strongly continuous semigroup on Co(Rd). Here Co(Rd) is the Banach space of all continuous real- valued functions f on Rd such that liml-_.oo f (x) = 0.


Proof. First, we are going to prove that, for any t > 0, Pt maps Co(Rd) into itself. Fix a t > 0 and an f C Co(Rd). Then for any n > 0, we have


Ptf(x) = j p(t, x, y)f(y)dy + j. p(t, x, y)f(y)dy.








9


The fact that, for any n > 0, the function X lIn p(t, x, y)f(y)dy

is in Co(Rd) follows from the continuity of p and the lemma above. Since

lim sup p(t,x,y)f(y)dy | rlim sup If (y)
n-oo xERd 1, yl>n ptxyf() I n-"o jyl>n = 0,

we know that Ptf is in Co(R d).

From Theorem 2.1.1 we know that for any f E Co(Rd) and any x E Rd, limPtf(x) = f(x),
tjo

thus it follows from [8] that (P)t>o is a strongly continuous semigroup on Co(Rd).
Q.E.D.

For any p E [1, oo), we use LP(Rd) to denote the real Banach space of all the real-valued functions f such that


fd IP(x)dx <00.

The norm on LP(Rd) will be denoted by 1- p. L*(Rd) will stand for the real Banach space of all the essentially bounded real-valued functions on Rd, and the norm on this space will be denoted by I1- 1o. Theorem 2.1.5 If 1 < p < p' < oo, then for any t > 0, Pt is a bounded operator from LP(Rd) into LP'(Rd).

Proof. From Theorem 2.1.1, Theorem 2.1.2 and Theorem 2.1.3 we know that for any t > 0 the function p(t, 0,. -) is in Lq(Rd) for any q E [1, oo]. Applying H6lder's inequality we get that for any f E LP(Rd) and t > 0, IIPtfIlp Ilf lp. Thus for any t > 0, Pt is bounded operator from LP(Rd) into LP(Rd).








10


Applying H6lder's inequality again we get that for t > 0, x E Rd and f E LP(Rd),

IP, f(x)I <_ dp(t, x, y) f |"(x)dy < p(t, 0, 0) lf l.
(J~d

Thus for any t > 0, P is bounded operator from LP(Rd) into L*(Rd).
Now the theorem follows easily from the two paragraphs above.
Q.E.D.

Theorem 2.1.6 For any p E [1, o), (Pt)t>o is a strongly continuous semigroup on LP(Rd).

Proof. For any f E LP(Rd), any t > 0 and almost every x E Rd,

Ptf(x) f(x) = ip(t, z, y)(f(y) f(x))dy = d p(t, 0, y)(f(y + x) f(x))dy. Using H6lder's inequality we get

lIP f- f = JRd Rd P(t, 0o, y)(f(y + x) f(X))dy dx Rd (Rdp(t, 0, y) f(y + X) f (x) IPdy) dx =Rd IIf(y +-) f(-)llp(t,0,y)dy. Since the function y F- if(y + -) f(')jP is bounded and continuous, we get that lim IIPtf f ip = 0.
tl0

Q.E.D.

Theorem 2.1.7 For any t > 0, P maps L"(R d) into bC(Rd). Here bC(Rd) stands for the real Banach space of all the bounded real-valued continuous functions on Rd.

Proof. It follows from the semigroup property that we only need to consider the case when t E (0,1]. Fix a t e (0, 1]. For any f E L(Rd), the boundedness of Ptf








11


is trivial. Let f E L (Rd) and let r be an arbitrary positive number. Then for any n >0,

Ptf(x) = f Yr+n+l p(t, x, y)f (y)dy.

The continuity of the function x Jl
follows from the dominated convergence theorem. From Theorem 2.1.2 and Theorem

2.1.3 we can get

lim sup I p(t,x,y)f(y)dy |- 11lfi lim sup p(t,x,y)dy
n-o ,lr+n+1 n-oo ITlr+n+1

lr IvI>r+n+ x y

< CllfII' lim s-(1+)ds
R-ooloo
= 0,

where C is the constant c2 in Theorem 2.1.2. Therefore Ptf is continuous on {x : IxI < r}. Since r is arbitrary, Ptf is continuous on Rd. The proof is now complete.

Q.E.D.


Theorem 2.1.8 For any t > 0 and any p C [1,oo), Pt maps LP(Rd) into Co(Rd).

Proof. Fix an f E LP(Rd) and at > 0. For any n > 0, put fn(x) = l(o,,n)(x)f(x), with B(0, n) being the the ball of radius n around the origin. Using the dominated convergence theorem we can easily see that the function x Pf(X) = IB(On) p(t,x,y)f(y)dy








12


belongs to Co(Rd). Now by using H6lder's inequality we can get that, for any x E Rd,

IPtf(x) Prfi(x)I = I Rp(t, x, y)(f f.)(y)dy < Ilp(t, x, Y)IIqIIf fAIIP = IIp(tOy)l 1lf f-I1, where q is such that 1 + 1. Since P q
lim If f-ip = 0,
n--oo

we know that Pf must be in C0(Rd).
Q.E.D.

The following simple result will be used in the next section.

Lemma 2.1.2 If A, B are disjoint closed subsets of Rd with B being compact, then p(t, x, y) is uniformly continuous on (0, T] x A x B for any T > 0.

Proof. Follows directly from Lemma 2.1.1 and the joint continuity of p(t, x, y).

Q.E.D.

For any x,y E Rd, define

u(x, y) = p(t, x, y)dt. The function u(., .) is called the Green fuction of X. Although generally we do not have an explicit formula for p(t, x, y), the following nice formula for the Green function is well known. See, for instance, [31]. Theorem 2.1.9 For any x,y E Rd u(x, y)= CIx- yl-d+ where C > 0 is a constant depending on d and a only.








13


2.2 Killed Symmetric Stable Processes

Let D be a bounded open subset of Rd. We are going to use To to denote the first exit time from D, i.e., TD = inf{t > 0 :Xt 4 D}. Then it is well known that the process X(wD() Xt(w), if t < TD; {, if t > TD,

is a symmetric Hunt process on D6 =D U {6}, the one-point compactification of D. This process is called the symmetric stable process of index a killed upon leaving D, or simply the killed symmetric stable process on D. The purpose of this section is to study some of the properties of XD.

We are going to use (PtD)t>o to denote the transition semigroup of XD and

(uP)>o to denote the y-potential of XD: UDf(x) e-tPD f (x)dt.

Lemma 2.2.1 For any I > 0 and x E Rd PX(rD = t) = 0.


Proof. Let n be a positive integer. Then


/liyl<+l P(TrD t)dy < 00, for every I > 0 and hence 0 for only a finite or countably infinite number of values of s. Thus fRd PY(TD = s)dy for only a finite or countably infinite number of values of s.








14


Let t > 0. By the previous paragraph there is an s E (0, t) such that Rd Py(TD = s)dy = 0. Since

{rD = t = D > t s, t s + TD o Ot_, = t }

C {TD 0 Ots, = S}, by the Markov property we get

P"(TD = t) Ex{PX(t-)(rTD = s)}

JRd p(t s,x,y)P"(rD= s)dy =0.


This completes the proof.

Q.E.D.

Lemma 2.2.2 For any t > 0, the function x Px(rD < t) is lower semicontinuous.


Proof. Let 0 < s < t. Then

Px(X, 4 D for some u E (s,t)) = Rdp(s, x, y)P(TD 5 t s)dy, which is continuous in x and increases to P'(-(rD < t)= PZ(TD < t) as s 1 0. Thus the conclusion of this lemma is true.

Q.E.D.

Definition 2.2.1 A bounded open set D in Rd is said to be regular if every point z on OD is regular for Dc, i.e., Pz (TD = 0) = 1.








15


In the following we are going to assume that D is a fixed regular bounded open subset of Rd.

Theorem 2.2.1 (PD)t>o admits an integral kernel pD (t, x,y) defined on (0, oo) x Dx D which satisfies the following properties:

(i) for any x E D and any nonnegative Borel function f on D, PtD f(x) = D pD(t, x, y) f (y)dy;


(ii) pD' (t, x, y) is jointly continuous on (0, oo) x D x D; (iii) for any 1,s > 0 and x,y E D, pD(t + s,x,y)= I pD((t,x,z)PD(s,z,y)dz;


(iv) for any t > 0 and any x,y E D, pD(t,xY) = pD(t,yX);


(v) for any t > 0 and any x,y E D,

0 < pD(tXy) < p(t,x,y);


(vi) if to > 0 and xo (or yo) is a point on aD then lim pD (tn, xn,yn) = 0 whenever (t, ,xn,yn) E (0, 0) x D x D converges to (to, xo, yo).

Proof. For any t > 0 and x, y E D, set

pD(t, xy) = p(t,x,y) E[p(t TrD,XT,y);rTD < t]. (2.1)








16


We are going to show that the pD(t, x, y) defined above satisfies all the requirements of the theorem.

It is clear that for any t > 0 and x, y E D,

p D(t,x, y) < p(t,x, y).

By using the strong Markov property we can prove that pD(t, x, y) satisfies property

(i) of the theorem. From this we can easily derive the following:

(a) for any t > 0 and any x E D, pD(t, x, .) is nonnegative almost everywhere on D;

(b) for any t, s > 0 and any x E D, the following identity is true for almost all y ED,

PD(t + SXy) DD (tXIz)PD(s,z,y)dz;

(c) for any t > 0 and any nonnegative Borel functions f, g on D,


ID ID PD(t, x, y)f (x)g(y)dxdy = I p D(t, y, x)f (x)g(y)dxdy;

(d) for any x E D and any bounded continuous function f on D, lim pD(t,x,y)f(y)dy = f(x).
t-0 JD

Let (to, yo) E (0, 00) x D. Take a closed neighborhood [c, t] x A of (to, yo) such that c > 0 and AC D. For an arbitrary sequence {(t.,y.)},n__ C [E,t] x A such that (ts, y,) converges to (to, yo), we have, by Lemma 2.2.1 and Lemma 2.1.2,

lim Ex[p(t,, 7D, X,,, y,); rD < tn] = Ex[p(to TD, XTD, YO); TD < to],
n--+o0

therefore pD(t, x, y) is jointly continuous in t and y for any fixed x E D. Now applying the continuity of pD in y, (iii) follows from (b) above and the first inequality of (v) follows from (a) above.








17


Since for any (t, y) E (0, c) x D the function pD (t, y, *) is bounded continuous, we get from (d) above

limJ pD(6E z)pD(t, y, z)dz = pD(t, y, x). (2.2)
f10 JD

Since for any x, y E D, pD is continuous in t, it follows from (iii) that

lim[ )p D D(t, z, y)dz = lim pD(t + _,x,y) = pD(t, x, y). (2.3)
0 JD O

Let c, 77 > 0. By (c) above we have


ID ID pD( U, v)pD( x, u)pD(q, y, v)dudv = D pD(t, v, u)pD(, x, u)pD(q, y, v)dudv. First letting c 1 0 and then letting r4 10, we obtain (iv) by applying (2.2) and (2.3) to the identity above. A consequence of (iv) is that pD(t, x, y) is also jointly continuous in (t, x) for any fixed y E D.
Let {(t,,xn,yn)}_ 1 C (0, oo) x D x D and (tn,xn,yn) (t,x,y) E (0, oo) x D x D. Without loss of generality we may assume that for any n > 1, S < t" < T for some constants T > S > 0. From (iii) and (v) we can get that

lim pD(t"' Xn, Yxn,) = li PD (tn 6, En, xZ)PD(E, Z, Y n)dz = pD(txY)
n-+oo n- JD

Thus (ii) is verified.
It remains to prove (vi). Let x0 E D and let {(tn,xn,yn)}n>1 C (2E,T) x D x D be such that (tn, Xn, yn) (t, xo, yo) E (0, oo) x BD x D. Then we have by (iii) and

(v),
pD(t, x~xy) = D PD(, Xz)pD(tn f, z,yn)dz < MP""(E < TD),

where M is a constant. Since the function x P"(E < TD) is upper semicontinuous,








18


we can derive from the above limsuppD(tn),xn,yn) < MPxo( < TD) = 0, n-+OO

which completes the proof of the theorem.
Q.E.D.

From the theorem above we immediately get the following corollary. Corollary 2.2.1 The function uD(x,y) = JpD(t,x,y)dt is finite off the diagonal of D x D. Furthermore, it is jointly continuous off the diagonal of D x D.


Proof. The first assertion follows easily from Theorem 2.1.9 and conclusion (v) of Theorem 2.2.1. We are now going to the second assertion. From (2.1) we have, for any x, y E D,

u(x,y) = UD(X,y) + EX{u(X(TD),y)}, so Ex{u(X(TD), y)} is symmetric in x and y on Dx D. To prove the desired result it suffices to show that EX{u(X(rD), y)} is jointly continuous off the diagonal of D x D. Choose x0, y0 E D with x0 yo. For any e > 0, there is a 6 > 0 such that u(z,y) u(z,yo)l < 2 lu(z,x) u(z, xo)l < for Ix xol < b, ly yol < b, and z E Dc. Choose x,y E D such that Ix xol < b and Il Yol < 6. Then








19


IE'f{u(X(TD),y)} E'o {u(X(TD), yo)}
< E'{ lu(X(rD),y) u(X(rD),yo)I} + Eo {lu(X(rD),x) u(X(rD),xo)I}

<.


Therefore EZ{u(X(TD), y)} is jointly continuous off the diagonal of D x D as desired.
Q.E.D.

By using the theorem above we can also get the following result which will be used in Chapter 4.


Theorem 2.2.2 For any t > 0, pD is a compact operator from L-(D) to L- (D).

Proof. Let { f,} be a bounded sequence in LO(D), i.e., there exists a C1 > 0 such that for any n > 0, Ifs|Io < C1. Then {f,} is also bounded in Lk(D) for any k > 1. Since for any k > 1, the unit ball in Lk(D) is weakly compact, by applying diagonal argument we see that there is a subsequence of {fJ}, say, {f} itself, which is weakly convergent to an f E L-(D). Hence for any x E D and any t > 0, lim PtDf(x) = PDf(x)


Consequently for any t > 0, {PDfn},>l converges in measure with respect to the Lebesgue measure m on D. Therefore for any e > 0, there exists an integer N such that

m({y E D: PD fn(y) P f(y)l > e}) < e,
2 2

whenever n > N. Thus for any x E D and any t > 0,








20



IPtDf.(x) PtD f(x) (, x, Y) P n(y) PP f(y)) dyl

JpD( ,y)IPF fn(y) -P f(y)|dy D 2 2
< I fpDt P Pf PD dy
fl~-I}2 2 2
+ -p X, y)PF f(y) P1 f (y)dy
Dflq _nP l <} 2 2 2


< C+2e Ce,

whenever n > N, where C2= 2max(Cl, If I0 PD-, I ", ")).
2

The proof is now complete.

Q.E.D.

2.3 Bilinear Forms and Sesquilinear Forms

Let E be a locally compact Hausdorff space with a countable base. We are going to fix a positive Radon measure m on E such that for any nonempty open subset O of E, m(0) > 0. L2(E, m) will always denote the real L2-space with inner product (u, v) = u(x)v(x)m(dx) and L 2(E, m) will always denote the complex L2-space with inner product (u, v) = u(x)v(x)m(dx). Definition 2.3.1 Let D be a linear subspace of L2(E,m). A map T from D x D into R1 is said to be a bilinear form if

(i) for any ul, u2,v ED and any a,b E R1, T(aul + bu2, v) = aT(ul, v) + bT(u2, v);








21


(ii) for any u, v1, v2 E D and any a, b E R1, T(u, av, + bv2) = aT(u, vi) + bT(u, v2).


D is said to be the domain of T. A bilinear form (T, D) is said to be symmetric if for any u, v E D

T(u, v) = T(v, u).

A bilinear form (T,D) is said to be densely defined if D is dense in L2(E,m).


In the sequel we shall always deal with densely defined bilinear forms; so from now on, whenever we talk about a bilinear form, we mean a densely defined bilinear form.

To any bilinear form (T, D) we can associate the following bilinear form (T, D): T(u, v) {7T(u, v) + T(v, u)}.


It is clear that is a symmetric bilinear form.

For any bilinear form (T, D) and any 3 > 0, (T, D) will always denote the following bilinear form 7O(u, v) = T(u, v) + /3(u, v). Definition 2.3.2 A bilinear form (T, D) is said to satisfy the sector condition if there exists a 3o > 0 and a constant K > 0 such that

(i) for any u E D,

T3o(u, u) > 0;

(ii) for any u, v E D, TO. o(u, v)1 K< KT(u, u) 2-TO(V,V) 2.








22


Definition 2.3.3 A bilinear form (T, D) is said to be closed if there exist a 3o > 0 such that

(i) for any u E D, T 0(u,u) > 0;

(ii) for any 3 > Po, D is a real Hilbert space with respect to the inner product j'6.


From the definition it is easy to see that any closed symmetric bilinear form satisfies the sector condition.

The following result is proved by Kunita. See [19].

Theorem 2.3.1 If(T, D) is a closed bilinear form satisfying the sector condition, then there exists a uniquely determined strongly continuous semigroup (Tt)t>o on L2(E, m) with resolvent

V f (x ) j e-3tTtf (x) dt satisfying

T(Vf,u) = (f,u) for any f E L2(E,m), u E D and /3 > 3o. Furthermore, when (T,D) is symmetric,

(Tt)t>o is a strongly continuous semigroup of self-adjoint operators on L2(E, m). Definition 2.3.4 A bilinear form (T, D) is said to be a Dirichlet form if

(i) for any u E D, T(u, u) > 0;

(ii) (T, D) is a closed bilinear form; (iii) (T, D) satisfies the sector condition;








23


(iv) the strongly continuous semigroup (Tt)t>o determined by (T, D) is Markovian, that is, for any t > 0 we have 0 < Ttf 5 1 whenever f E L2(E,m), 0 < f <
1 m-a.e..

De fnition 2.3.5 Let (T, D) be a Dirichlet form and let Co,o(E) be the family of all continuous functions with compact supports. We shall say that (T, D) is regular if D n Co,0(E) is dense in D with the Ti-norm and dense in Co,o(E) with the uniform norm.

De fnition 2.3.6 A Dirichlet form (T, D) is said to be transient if there exists an m-a.e. strictly positive bounded function f E L2(E, m) such that for any u E D, u|(x) f(x)m(dz) :5 T1(uu).

We know that for a Dirichlet form (T, D), D is a Hilbert space with inner product Tp for any 3 > 0, but D is not even a pre-Hilbert space with respect to T in general. When (T, D) is transient, D is a pre-Hilbert space with inner product T. Definition 2.3.7 Suppose that (T, D) is a transient symmetric Dirichlet form. If we use De to denote the completion of the pre-Hilbert space D with respect to T, then (T,D,) is called an extended Dirichlet form associated with (T,D). Example 2.3.1 (See [13].) Assume that 0 < a < 2, E = Rd(d > 2) and m is the Lebesgue measure on Rd. Put

T(u,v) = il (xv(x)lxldx,

D =u EL 2(Rd) I (x)2xdx < o},

where

i(x) = (27r JRd e (x)u(y)dy.








24


Then (T7, D) is a transient, regular symmetric Dirichlet form and the strongly continuous semigroup determined by (T7, D) is nothing but the transition semigroup of the symmetric stable process of index a on Rd. Another way of writing (T7-, D) is as follows.

T(u, V dI1 (u(x) u(y))(v(x) v(y))
Ixuyd+v)dxdy, 12 naI nad Iz y |d+ae

D {uEL2(Rd):J (u(x) u(y))2dd
Ix yld+ .

The extended Dirichlet form associated with (T, D) can be characterized as follows.

D = {u E L oc(Rd): u is a tempered distribution and J 1(x)12 xlIadx < oo Another characterization of De is as follows.

De = {u = Rf : f E L2(Rd)} where R(O) denotes the Riesz kernel of index 6: R() x =20rF(') XId


Example 2.3.2 (See [13].) Assume that D is a bounded open subset of Rd and (T, D) is the Dirichlet form given in Example 2.3.1. Put DD = {u E D : i = 0 q.e. on DC} where i& denotes a quasi-continuous version of the function u and "q.e." stands for quasi-everywhere. Then (T7-, DD) is again a regular symmetric Dirichlet form. The strongly continuous semigroup determined by (T, DD) is nothing but the transition semigroup of the killed symmetric stable process on D. Definition 2.3.8 Let D be a linear subspace of LC(E,m). A map T from Dx D into C is said to be a sesquilinear form if








25


(i) for any u1, u2, v E D and any a, b E C, T(au1 + bu2, V) = aT(u, v) + bT(U2, v);


(ii) for any u, v1, v2 ED and any a, b E C, T(u, av1 + by2)= -T(u, vi) + bT(u, v2).


D is said to be the domain of T. A sesquilinear form (T, D) is said to be symmetric if for any u, v E D, T(u, v) = T(v, u). A sesquilinear form (T,D) is said to be densely defined if D is dense in L 2(E,m).

In the sequel we shall always deal with densely defined sesquilinear forms; so from now on, whenever we talk about a sesquilinear form we mean a densely defined sesquilinear form.

With each sesquilinear form (T, D) is associated another sesquilinear form (T*, D) defined by T*(u, v) = T(v, u).

(T*, D) is called the adjoint form of (T, D). For any sesquilinear form (T, D), the two sesquilinear forms (1, D) and (, D) defined by

1
1 = (T + T*)
2
1
I = (T T*)
2iZ

are symmetric and T = I + il. For a sesquilinear form (T, D), the set of values of T(u, u) for u C D with (u, u) =

1 is called the numerical range of T and is denoted by 0(T).








26


Definition 2.3.9 A sesquilinear form (T, D) is said to be sectorial if there exist real numbers -7 > 0 and 0 < 0 < such that for any ( E O(T),
2

Sarg(( + )-< 0.


A symmetric sesquilinear form (T, D) is said to be nonnegative if for any u E D, T(u, u) > 0.


From the definition above it is easy to see that a nonnegative symmetric sesquilinear form is always sectorial.

Let (T, D) be a sectorial sesquilinear form. A sequence {u,} C D is said to be Tconvergent to u E L 2(E, m) if u, -- u in L 2(E, m) and T(u, u, u,, Um) 0 as n, m -00.


Definition 2.3.10 A sectorial form is said to be closed if the T-convergence of {un) C D to u E L2(E, m) implies that u E D and T(u, u, u, u) -*0 as n 0.


With any bilinear form (T, D) is associated a sesquilinear form (Tc, Dc) defined as follows.

DC = D iD,


TC(Ul + iu2, v, + iv2) T(ui, vl) + T(u2, v2)

+iT(u2, v1I) iT(ui, v2), VU1, U2, v1, v2 E D. The following result is proved in [18] and will be used in the next chapter.


Theorem 2.3.2 Let (T,D) be a bilinear form. If (TC, DC) is a closed sectorial form, then (T, D) is a closed bilinear form satisfying the sector condition.
















CHAPTER 3
THE FEYNMAN-KAC SEMIGROUP

3.1 The Kato Class

From now on we shall always assume that X = (f,.F, .Ft, Xt, Ot, P") is the symmetric stable process of index a, 0 < a < 2, on Rd (d > 2) with the following characteristic function c-t z 0

and that (T, D) is the regular symmetric Dirichlet form defined below

T(u, v) 1 d R1 (u(x) u(y))(v(x) v(y))dxdy,
2 nIx yd+ D= u L2(Rd):/nd (u(x) u(y))2dxd <4 L2--) :dzy < yd+ .

As usual, we use (Pt) to denote the transition semigroup of Xt and U.,, 7 > 0, to denote the y-potential of (Pt),


Ulf f) = 1 e-tPt f(x)dt. Definition 3.1.1 A signed Radon measure p on Rd is said to be in the Kato class Kd,a if
lim sup I I*(dy)
him sup -yda= 0
rio xERd I-xyl



27








28


A Borel function q is said to be in the Kato class Kd,, if the measure q(x)dx is in Kd,,.

Lemma 3.1.1 Let p be a Radon measure on Rd. If for some r > 0,

sup p(dy)= C XERd fIs-yl then
/x ,(dy)
sup < (dy) .
ERdj Ix-yl>r IX yld+< Proof. For any x E Rd, we can cover Rd with nonoverlapping cubes of side length r with one of the cubes centered at x. Then counting from the cube at x, on the n-th layer,
2n-1 2 2n-1
Ix y > r r.
2 7rd d

There are (2n + 1)d (2n 1)d cubes on the n-th layer. Each cube can be covered by a ball of radius r. Thus

p(dy) )d )d ((2n 1)r -(d+c)
y [(2n + 1) (2n 1) C
-yl>r Ix Y-d+a 0 C
(r ) -(d+a) o
< C 2d(2n + 1)dl(2n 1)-(d


=2dC( -(d+a) o (2n + 1)d-1
7d -o (2n 1)(d+a)
< 00.

The proof is now complete.

Q.E.D.

Lemma 3.1.2 Let V E Kd,,. Then there exists an ro > 0 such that for each fixed

0 < r < ro we have

sup sup p(t,x,y)p*(dy) = C(r) < oo.
t<1 XERd JIy-x>r








29


Proof. From the definition of Kd,o we know that there exists an ro > 0 such that / P*(dy)
supi < 1.
xERd I--X:ro |y Therefore for any 0 < r < r0, I .rda"p*(dy) xERdjY -xl Applying Theorem 2.1.2 and Theorem 2.1.3 we know that for any t < 1,

sup p(t,x,y)p*(dy) = sup t-ap(1,t-ax,t-i y) *(dy)
xERd ly-xl>r xERd y-xl>r
< /2 SUP P*(dy)
tC2 sup Id+a'
z-Rd l-xl>r ly I Now applying Lemma 3.1.1 we immediately get the conclusion of this lemma.
Q.E.D.

Theorem 3.1.1 Let p be a signed Radon measure on Rd. Then i E Kd,, if and only if
lim sup p(s, x,y)p*(dy)ds 0.
t tERd 0 Rd

Proof. Suppose first that y E Kd,a. Lemma 3.1.2 guarantees that we can choose an r > 0 such that for any t < 1, nt
sup p(s,x,y)p*(dy)ds < tC(r).
xERd 0 Iy-zl>r
On the other hand we have, by Theorem 2.1.9, for all t > 0,

1 xl< p(s,x,y)*(dy)ds < <-r p(s,x,y)p*(dy)ds

C (dy)
Cf l tro XERd 0 Rd








30


Now suppose that

lim sup J, p(s,x,y)u*(dy)ds = 0.
tjO XERd JRd

We know that


lo IRd p(s, x, y)p*(dy)ds

Id*(dy) j p(s, x, y)ds

d*(d) -a p(1, s- I Yy)ds

Rd
= R = yd- p -1 -1y)du.


Applying Theorem 2.1.2 we get
Ot/R y /p* (dyd)_ 1o a c1
p(sxy)p*(dy)ds > t*(dy) U d u
lu-cl 1 p*(dy)
0 Iv-el

Q.E.D.

Definition 3.1.2 A Radon measure p on Rd is said to be of finite energy integral if there exists a constant C > 0 such that for v E D n Co,0(Rd),


Rd Iv(x)| p(dx) < C i(vv). We shall use So to denote the collection of measures of finite energy integrals.

By Riesz's representation theorem, a Radon measure P E So if and only if there exists an element V1p E D such that for any u E D n Co,0(Rd), Tl(V1I ,u) =IRd u(x)p(dx).








31


If for any 7 > 0 and any nonnegative measure v, we use U.,v to denote the following function

U~v(x) = Jn e-Y'p(t,x,y)v(dy)dt = nu(x,y)v(dy). Then we can give the following characterization of So. Lemma 3.1.3 If v is a Radon measure on Rd, then V E So if and only if for some

7 >0,

< U ,V, >:= d Uv(x)v(dx) < 00. Proof. If v E So, then by Lemma 5.1.3 and Theorem 3.2.2 of [13] we know that < U, V v >< 00.

Now let us assume that < Uyv, v > is finite. Then v charges no polar sets, thus by [13] there exists an increasing sequence {F,} of closed sets such that

(a) v(Rd U-,Fn) = 0;

(b) for any n > 1, 1& v E So. From (a) above we get that Uj(1F, v) converges to Uyv everywhere on Rd. From

(b) above we get that for any v G D n Co,o(Rd), "Tj(U(1F4 V), v) = I v(x)v(dx). Since for each n, U,(1F. v ) is quasi-continuous, we know by Theorem 3.3.2 of [13] that for each n,


7(U,(1F v), U(1F. v)) = J U(1f. v)(x)v(dx) < .








32


Therefore there exists a subsequence {U-,(1F.k v)} of {U.y(1F v)} which converges weakly with respect to the ,-norm to some function w E D. Combining this with the fact that U,(1F, v) converges everywhere to U.,v, we get that w = U.,v, hence U.y(1Fk v) converges weakly with respect to the T-norm to U.,v. Now letting k tend to infinity in

(U v), v) v(x)v(dx), vE D Co,o(R ),
F.k

we get that for any v E D n Co,o(Rd), T(UV, v) = v()v(d), which implies that v E So.

Q.E.D.

An immediate consequence of the result above is that for any Radon measure v, if < U.,v, v > is finite, then Uv is quasi-continuous. From this we can easily get that for any f E L2(R d) and any -y > 0, U, f is quasi-continuous. Theorem 3.1.2 Let p be a Radon measure which is in Kd,,. Then for any compact subset F of Rd, the restriction of p to F, 1F p, is of finite energy integral.

Proof. From Theorem 3.1.1 we can see that
OO
sup e~' p(s,x,y)p(dy)ds < 0o. xERd JRd

In particular, the following function U1(1F p)(x) = jo e- p(s, x, y)(dy)ds
0

is bounded, say, by a constant C. Now


R (Ul(1F p)(x))2 dx < C j e-" p(s, x,y)p(dy)dsdx = C(F),








33


so Ul(1F I) E L2(Rd). And since

lim 1 (U(1F )- eP-tP(U(1f -L p)), UI(1F P)) tjo t
  • tI0 t R' 1
    < Cp(F) <00,

    we know by Lemma 1.3.4 of [13] that UI(1F t) E D. Therefore, for any u E D n Co,o(Rd),

    T1(Ul(1F ),u) = lim I (Ul(1F .P) e-tPt(Ul(1F ')),u) tio t
    = lim e-"p(s, x, y)p(dy)ds, u
    tjo( lltI
    u fu(x!y (dx).
    FoI


    Thus 1F pE So.
    Q.E.D.

    Corollary 3.1.1 If a Radon measure p is in Kd,,, then i does not charge polar sets.

    Therefore it follows from [27] that for any Radon measure p in Kd,,,, there exists a unique positive continuous additive functional A = (At)t>o (in the sense of [6]) of X such that for all -y-excessive functions (y7 > 0) h and all nonnegative Borel functions f on Rd,
    1 r
    liml R h(z)E": fot(X,)dAsdx =fd h(s) f (x)(ds). (3.1)


    We shall always deal with additive functionals in the sense of [6], so, from now on, whenever we talk about an additive functional we mean an additive functional in the sense of [6].

    Suppose now that p E Kd,. Let (A+) and (A-) be the positive continuous additive functionals associated with y+ and y- respectively in the manner of (3.1).








    34


    If we set At = A' At-, then (At) is a continuous additive functional of bounded variation. We shall call (At) the continuous additive functional associated with p.

    Now here is an easy corollary to Theorem 3.1.1.

    Theorem 3.1.3 If y E Kd,a and (At) is the continuous additive functional associated with p, then

    lim sup E'At = 0.
    tjO ZERd

    Proof. Without loss of generality we can assume that p is nonnegative. From Lemma 5.1.4 of [13] and Theorem 3.1.2 above we know that if we denote by B, the ball of radius n around the origin then for any n > 0, t > 0, and any nonnegative Borel function f,


    (f, E B,(X,)dA, = fI p(s,, y)p(dy)ds consequently for any t > 0 and n > 0, EXJ1fl(X,)dA, = IBp(sxy)p(dy)ds
    0 Bn
    is true for almost all x E R Multiplying both sides of the equality above by p(e, x0o, x) and integrating over x, we get from the additivity of (At),

    Eoj 1B,(X,)dA, = p(s, xo, y)(dy)ds.

    Letting ~. 0 and n T c, by the monotone convergence theorem we get that

    E'OAt = J p(s, o, ) p(dy)ds

    Since xo is arbitrary, we get that for any x E Rd and any t > 0, ExAt = lRd p(Sxy)y(dy)"ds.

    Now the conclusion of the theorem follows immediately from Theorem 3.1.1.

    Q.E.D.








    35


    Theorem 3.1.4 If p E Kd,a and (At) is the continuous additive functional associated with pi, then lim sup E'(At)2 = 0. tO xERd

    Proof. Without loss of generality we can assume that p is nonnegative. Since for any t > 0,

    (At)2 = 2 (At A,)dA,, do
    we know that for any x E Rd E"(At)2 = 2Ex 0 Ex AtdA, < 2 sup (E'At)2.
    XERd

    Now the conclusion of the theorem follows immediately from Theorem 3.1.3.

    Q.E.D.


    Definition 3.1.3 A bounded Borel function F(., .) on Rd x Rd is said to be admissible with respect to X if

    (i) F vanishes on the diagonal;

    (ii) the function I IF(z,y)l z dy Rd 1z y d+a is in Kd,,.

    We are going to use Ad,0 to denote the collection of all the admissible functions with respect to X.


    It is easy to see from the definition that if F1, F2 E Ad,. and c E R, then cF, F + F2 and FIF2 all belong to Ad,0. Furthermore we have the following








    36


    Lemma 3.1.4 If F E Ad,,, then eF 1 E Ad,a .


    Proof. It is easy to see that the bounded function G = eF 1 satisfies the first condition in the definition above, so we need only to check the second condition. Let M > 0 be such that for any x, y E Rd,


    IF(x,y)I < M. Since on the interval [-M, M], the function ea 1
    a i---
    a

    is bounded, there is aC > 0 such that for any a E [-M, M], e < 1 C
    a

    Consequently we have eF("') 11 < C F(x,y)l. for any x, y E R Therefore for any z Rd,


    I IG(z, y)I
    Rd Iz yd+a
    < IGo Iz dy + C IF(z+, y)I
    fly-z I Z y I-z<) Iz y ,

    hence the second condition in the definition above is also satisfied.

    Q.E.D.

    Theorem 3.1.5 If F e Ad,a, then At = E F(X._, X.),
    O A* = E IFI(X ,,X,)
    0







    37


    are finite almost surely. Furthermore, (At) and (At) are additive functionals such that

    lim sup E' IAt = lim sup E"A* = 0.
    tjo XERd tjO xERd

    Proof. Without loss of generality we can assume that F is nonnegative. From [3] we know that for any t > 0 and any x E Rd,
    F(X.,,y)
    E'At = E' ds I F(XS Y) dy.
    RdX yd+a

    From the third condition in the definition above and Theorem 3.1.1 we immediately see that

    lim sup E'At = 0.
    tjO xERd

    The rest of the conclusions follow easily.

    Q.E.D.


    Theorem 3.1.6 Suppose that F E Ad,a. If we put At= > F(X,_, X,),
    O
    then

    lim sup E'(At)2 = 0. tIO xERd

    Proof. Without loss of generality we can assume that F is nonnegative. From the fact

    (At)2 = 2 (A, A,)dA, + E F2(X,_, X,) O we get that for any x E Rd

    Ex(At)2 = 2E" E'(A_,_)dA, + E" ds F2(X Xy)dy
    10 10 J~d IX., ydc

    < 2 sup (ExAt)2 + Ex ds F2(X, Y) dy.
    xERd IX. yd+








    38


    Since F E Ad,,, F2 is also in Ad,a, thus by Theorem 3.1.1 we have u. E F2(X, y)
    lim sup E' ds 0.
    to nERd Rd |X, yld+c,

    The proof is now complete.

    Q.E.D.

    Combining Theorem 3.1.4 and Theorem 3.1.6 we immediately get the following result.

    Theorem 3.1.7 Suppose that y E Kd,,, F e Ad,a and that A' is the continuous additive functional of X associated with p. If we put At = A + F(X,_, X),
    O then

    lim sup E'(At)2 = 0. tjO xeRd
    3.2 The Feynman-Kac Semigroup

    Before we get to the Feynman-Kac semigroup, some preliminary results are in order. The following result is well known, see, for instance, [10]. A proof is given here for the sake of completeness. Theorem 3.2.1 Suppose that y E Kd,, and (At) is the continuous additive functional of X associated with p. Then there exist constants C > 1 and p > 0 such that for any t > 0,

    sup Ex {supeA,} < Cet.
    xERd s Proof. Without loss of generality we can assume that p is nonnegative. From the assumption p E Kd,, we know, by Theorem 3.1.1, that there exists a to > 0 such that

    sup ExAto < 0 < 1. xERd








    39


    Hence for all x E R, 0 < t < to and n > 2,

    1E{(A)"} = Exf" (At -A,)d(A,)n-1 = Ex j A,(O)d(A,)n-1 = E x Ex[At_,]d(A, )n-1 < OE{(At )n-I}. Therefore by induction we have for all x E Rd E"{(At)"} < n!0", hence,
    1EX{eAt< 1-0
    n=o
    Using the additivity of (At) we get that for any positive integer n and any (n 1)to < t < nto,
    1
    Ex {eAt} < (1 0)n

    This is equivalent to the conclusion of the theorem.

    Q.E.D.

    Theorem 3.2.2 Suppose that F E Ad,a and At = F(X,_, X,).
    O Then there exist constants C > 1 and # > 0 such that for any t > 0, sup Ex{supeA, } < Cet. xERd s O







    40


    Then it follows from [3] that Bt:= f F (X,,y) dyds
    0 IX, yd+a is the dual predictable projection of Bt. Therefore Bt B{ is a P"-martingale for every x E Rd. Now it follows from the exponential formula (see, for instance, [11]) that

    Nt = eB'-B" rI (1 + Fi(X,_, X.)) e-F (X.-,X,) s = e-B g (1 + FI(Xs-, X,)) s e2At-BPt

    is a local martingale under Px for any x E Rd, where the convergence of the infinite product in the first line above is also guaranteed by the exponential formula. (N) is clearly a multiplicative functional of X, so it is a supermartingale multiplicative functional of X. Thus E e2A-BP
    Ex{e2Ax-~'} < 1. Applying the Schwarz inequality we get EX{eA,} EX{eAt-BeB
    -xeEl 2 t ,2
    11
    < (Ex{e2A-B (E"{e}

    < (E{eB }) .

    Now the conclusion of the theorem follows from Theorem 3.2.1 and the fact that F E Ad,a.

    Q.E.D.

    Theorem 3.2.3 Suppose that y E Kd,a, F E Ad,a and that A' is the continuous additive functional of X associated with p. If we put At= A+ Z F(X_,X),
    O







    41


    then there exist constants C > 1 and / > 0 such that for any t > 0, sup E'{supeA.} < Cet.
    xERd s
    Proof. Direct consequence of the two results above and the Schwarz inequality.

    Q.E.D.


    Lemma 3.2.1 Suppose that it E Kd,, F E Ad,, and that A' is the continuous additive functional of X associated with yu. If we put At = A" + E F(X,-,X.), O
    then for any p> 1 lim sup E {IleAt iI 0. tlo xERd

    Proof. Noticing that for any a > 0, ea 1 > 1 e-, we can assume, without loss of generality, that both p and F are nonnegative. Observe that for any a > 0, e' 1
    Ex {leAt -l1} < Ex (ePAt-I)



    p(Ex (At )2) 2 (Ex e2PAt) 2 From Theorem 3.2.3 we know that there exist constants C > 1 and )3 > 0 such that for any t > 0, sup E'e2pAt < Ce O, xERd
    and from Theorem 3.1.7 we get that








    42


    lim sup E'(At)2 = 0, tiO xERd
    therefore the conclusion of the lemma is true.

    Q.E.D.

    From now on we are going to fix a E Kd,a and an F E Ad,a. Let A' be the continuous additive functional associated with I and put At = A + F(X,_,X,).
    O For any t > 0, x E Rd and any nonnegative Borel function f on Rd, define Ttf(x) = E' {eAt f(Xt)}. Then (T)t>o is called a Feynman-Kac semigroup. Theorem 3.2.4 If1 < p < p' < oo, then for any t > 0, Tt is a bounded operator from LP(Rd) into LP'(Rd).

    Proof. For any t > 0, x e Rd and f E LP(Rd), a simple use of Holder's inequality gives us

    ITt f(x)I < (El f (Xt)]P) (ExeqAt) (3.2)

    where 1+ 1.
    P q
    From Theorem 3.2.3 we know that there exist C > 1 and 3 > 0 such that

    sup E'{eqAt} Cet, (3.3)
    zxERd
    thus
    .1 1
    ITl" f (Ce>)P (~d Ex"I f (Xt) "dz) =(Ce"') f .

    Hence Tt is a bounded operator from LP(Rd) into LP(Rd). From (3.2), (3.3) and Theorem 2.1.5 we can easily see that T is bounded operator from LP(Rd) into L*(Rd). By using interpolation the theorem follows easily.

    Q.E.D.









    43


    Theorem 3.2.5 For any p E [1, oo), (Tt)t o is a strongly continuous semigroup on LP(Rd).

    Proof. For any t > 0 and f E LP(Rd),

    IlTtf f lip IlTtf Ptf lip + llPtf fip.

    From Theorem 3.1.6 we know that lrn IlPtf f ip = 0

    Thus to complete the proof, we need only to show that lim lITtf Ptf lIp = 0. (3.4)
    t 10

    Now by H61der's inequality we can get

    IITtf FtfI =l (Ip IE-{(eAt )f(Xt)}IPdx)P


    S(Id (Exe- qI) (EXrlfIP(Xt))dx)P <5 sup (Exle At q (IRd Exf P(Xtdx)P

    xERd


    Since by Lemma 3.2.1, lim sup (ExeAt _-~)I = 0,
    tjo XERd E
    we know that (3.4) is true. The proof is now complete.

    Q.E.D.

    Theorem 3.2.6 For any t > 0, Tt maps L'~(Rd) into bC(Rd).

    Proof. For any t > 0 and f E L'(Rd), we have

    Ttf (x) = Ex eA-Af(Xt)} + E'jeAtAc(eAr 1)f (Xt)1








    44


    The first term on the right hand side of this equation is equal to P, [T_,f](x), therefore it is continuous in x because of Theorem 2.1.7 and Theorem 3.2.6. The second term on the right hand side of the last identity is bounded by
    2
    If l1 (EIeAE 112) sup Ex(sup e2A)
    XERd s
    which tends to 0 uniformly as e 0 because of Theorem 3.2.3 and Lemma 3.2.1. Therefore Ttf is a continuous function on Rd.

    Q.E.D.


    Theorem 3.2.7 For any t > 0 and any p E [1, oc), T maps LP(Rd) into Co(Rd).


    Proof. For any t > 0 and f E LP(Rd), we have

    Ttf(x) = Ex{eAt-At f(Xt)} + Ex{eAt-A'(eAt 1)f(Xt)}

    The first term on the right hand side of the last identity is equal to P,[T_Cf](x) which belongs to Co(Rd) because of Theorem 2.1.4 and Theorem 3.2.6. The second term on the right hand side of the last identity is bounded by

    1 q
    (ExlfIP(Xt)) Esup E(supe2qAt Ele ,_ 2q 2q
    xERd s
    (where + = 1) which tends to 0 uniformly in x as e 1 0 because of Theorem 2.1.8, Theorem 3.2.3 and Lemma 3.2.1. Therefore Ttf E Co(R d).

    Q.E.D.

    Similar to the proof of Theorem 3.2.5 we can get the following result.


    Theorem 3.2.8 (Tt)t>o is a strongly continuous semigroup on Co(Rd).








    45


    3.3 The Bilinear Form

    It is well known that any function in D admits a quasi-continuous version. We are going to use D) to denote the family of all the quasi-continuous functions belonging to D. It is clear that the equivalence classes of D (in the sense of q.e.) are identical with the equivalence classes of D (in the sense of a.e.).

    In this section we are going to deal frequently with integrals of the following type:


    IRd u2(x)v(dx)

    where u is in D and v is a nonnegative measure. The following theorem provides an estimate on the above integral.


    Theorem 3.3.1 There exists an Md, > 0 depending only on a and the dimension d such that for any v E D and any nonnegative Radon measure v charging no polar sets we have
    Iv'(x)(dx) M,0 sup v(dy ) (ivI1 + T(v,v))
    Rd ,xERdRd Ix- y-Id

    Proof. Put

    ui(x,y) = jet p(t,x,y)dt; then ul is the integral kernel of U1. Since v does not charge polar sets, we know that for every Borel set A,

    1A(x) < P'(TA = 0), v -a.e., where TA is the first hitting time of A by the symmetric stable process killed with exponential rate 1.

    For any set A, let K.A be the equilibrium measure of A. Using properties of the capacity we can get








    46


    fndv (x)v(dx) = 2 tv(lvl > t)dt
    = 21 t 1(IIt)(x)v(dx)dt

    2 I P"(T(lviat) = 0)v(dx)dt

    21 I PX(T(ivIet) < C)v(dx)dt
    o
    = 2JOf IRd ui(x, y)(lvlt)(dy)v(dx)dt

    2 sup ui(x,Y)v(dY)) j-tCl(v > t)dt

    MdSup I v(dy) \1-oo~
    2M, Isup X dc tCi(lvI t)dt,

    where ( is the life time of the symmetric stable process killed with exponential rate 1, C1 is the capacity of this killed symmetric stable process and Mi, is a constant depending only on d and a.
    From Theorem 1.6 of [15] we get that OO
    J tC1( v > t)dt 5 Md",,inf{< UIA, A >} where M"'a depends only on d and a, and the infimum is taken over all nonnegative measures A such that UIA > lvI up to a set of zero C1 capacity. The infimum in the second line above can be replaced by infimum over all measures v such that Ulv > lvi a.e., since U1A and v are both quasi-continuous. Therefore, to finish the proof it suffices to show that there exists a p E D of the form U1 A for some positive measure A such that

    p > Iv| a.e.,and T1(p,p) T(v,v). Consider now the set P of potentials of finite energy:

    P = {U1 A : A is a nonnegative measure and < U1 A, A >< oo}.








    47


    It follows from Lemma 3.1.3 that every U1A E 7 is in D and satisfies TW(AUA) =< U1AA >.

    Hence 7 is a closed, convex subset of D, since the set of positive measures of finite energy is complete in the energy norm(see, for instance, [20]). It follows that there exists a projection p of IvI on the set 7. Using the standard manipulation by quadratic equations (see, for example, Chapter 7 of [26]) we can prove that Ti(p- Iv1,q) > o, Vq E P,

    and

    Tir(p- Iv1, p) 0.

    It follows that p > Ivi a.e. and that Ti(IvI p,p) = 0. Hence T1(p,p) = T(p, Iv1) _< T(p,p) Ti(V,v) which finishes the proof.

    Q.E.D.
    The following result is elementary. A proof is given here for the sake of completeness.

    Lemma 3.3.1 For every r > 0, there exists a sequence {cj : j > 1} of points in Rd such that

    (i) {B(cj, 3r) : i > 1} covers Rd.

    (ii) for every x E Rd, B(x, r) intersects at most 5d balls from {B(cj, 3r) j 1}, at most 8d balls from {B(cy,6r) j 1} and at most 9d balls from {B(ci,7r)

    j>1}.








    48


    where for any c E Rd and any y > 0, B(c, 17) stands for the open ball of radius q around c.

    Proof. Let us start with a countable cover of Rd by balls of radius r and let us denote this cover by {Bj = B(aj,r): j > 1}. We can take a maximal subset of {Bj} such that any two balls in the subset are disjoint. First, we take j1 = 1. Then we take j2 = min{j > J, : Bj n Bj, = 0} and we continue inductively in the same way. Therefore we get a sequence {Bjk : k E N} with the following two properties:

    (1) for any k $ 1,

    Bjk n Bj, =

    (2) for any j > 1 there exists a jk >1 such that Bj and Bjk have a nonempty intersection.

    Obviously {jk : k > 1} must be infinite. We claim that Ck = ajk satisfies all the requirements of this lemma.
    Let x E Rd. Since {Bj} is a cover, there exists Bj such that x E Bj. By (2) above there exists Bjk such that Bj n Bj, $ . Hence, Ix- ckl Ix- aj I

    < Ix-ajl+lajk-aI

    < r+2r=3r.

    It follows that {B(ck,3r) : k > 1} is a cover of Rd.
    Let x E Rd, and let B(x,r) intersect B(ck,3r). Then Ix ckl < 4r, i.e., B(ck, r) is contained in B(x,5r). Since the balls in {B(c,,r) : I > 1} are pairwise disjoint, there can be no more than m(B(x, 5r))
    m(B(cj, r))








    49


    balls of the form B(ct, r) inside B(x, 5r), where m denotes the Lebesgue measure in R d. Since
    _(5r)dwi
    m(B(x,5r)) =(5r) 1 (+ 1)
    and

    m(B(ct, r)) d ( +1)'
    there are no more than 5d balls of the form B(ct, r) intersecting B(x, r).
    The assertion about {B(ci, 6r) : i > 1} can be proved by starting with the radius 6r in the paragraph above and going over the same argument. And the assertion about {B(ci, 7r) :i > 1} can be proved similarly.

    Q.E.D.

    Using the two results above we can get the following theorem.

    Theorem 3.3.2 If v e Kd,a is nonnegative, then for any u E D,


    JRd u'(x)v(dx) < oo.

    Furthermore, for every e > 0, there exists a C(e) > 0 such that for every u E D,


    u(x)v(dx) < eT(u, u) + C(c) Iull Proof. Fix an arbitrary e > 0. Since v E Kd, there exists an r0 > 0 such that sup v(dy)
    sup]x xnd Ir-y<0 IX y d-o Let r = ro/6 and let {c : j > 1 } be the corresponding sequence obtained in Lemma 3.3.1. Let 9 : Rd [0, 1] be an infinitely differentiable function such that p(x) 1 on B(0,3r) and p(x) = 0 outside of B(0,6r). For every j > 1, we denote by j the following translate of V: ,j(x) = p(x cj). Notice that there exists a constant M = M(E) > 0 such that IV I2 < M for every j.








    50
    Since JB(cj, 3r) > 11 is a cover of R d and (x), we know
    1B(cj,3r) ::::::: 1B(cj,3r)VJ that


    Rd U'(x)v(dx) Rd IB(c.,,3r)(x)u'(x)v(dx) E IB(cj,3r) U2(X) O, (x)v(dx).
    j=j j=1

    Notice that u pj E f), so we can apply Lemma 3.3.1 to every integral under the above sum:


    B(c,,3r) U2 (X)V, (x)v(dx)

    dc, sup B(c),3r) IX yld-c,) V(wj, UWA + 11U OjI12) (XERd
    Recall that for 0 < a < 2, the Riesz potential satisfies the maximum principle; therefore

    sup v(dy) SU v(dy)
    xERd B (r.,,3r) IX y1d-c, B( .3r) yjd-a
    xE J, B(cj,3r) IX < sup v(dy)
    XEB(cj,3r) 1--yl <

    It follows that

    U2(X)CP2(X) v(dx)
    IB(cj,3r) j
    (T(U Oj, U Oj) + UW, 11 2) Mdr, IE 2 (3-5)

    It is easy to see that


    R, I
    7-(Upj, wj) :5 (U2(X)(Wj(X) pj(y))2 dxdy
    R d ix yld+ci
    y))2 0 (Y)
    + Rd Rd (U(X) _U ( 3 dxdy. (3-6)
    Ix yld+a
    The first double integral on the right hand side of the inequality above can be estimated as follows:







    51
    (U' W GOj W Oj (y)), 'dY R d IX yld+III U'(X)(IRd (vj W Oj (y)), dy) dx
    1X y1d+a
    U2(X)(4 GOj W Oj(y))2 dy)dx
    + Rd (c.,,6r) ix yld+a
    U 2 (X) (I ((Pj(x) (pj (y)) dy)dx
    IB(cj,6r) ,(Cj,8r) ix yld+a
    + U2(X)(4 (Vj(X) Vj(y))2 dy)dx
    (cj,8r)c Ix yld+a U2(X)(4 (10i (X) Vj(y))2 dy)dx
    + 4(,,Ir) (cj,6r) ix Y ld+a
    U2(X)(I (10i (X) Vj(y))2 dy)dx
    + JB(,,Ir) B(cj,6r) Ix yld+ct
    < 2 U, W (I (Vj(X) Vj(y))2 dy)dx
    IB(c,,,7r) ", (c.,,8,) ix yld+a
    U2(X)(I (Vj(X) Vj(y))2 dy)dx
    + B(cj,6,) B(c,,8,)c ix yld+a
    U2(X)(4 (Vi (X) (pj (y)) 2 dy)dx (3.7)
    + 4(,,7r)c (cj,6r) ix yld+a
    It is easy to see that there exists a constant C > 0 depending only on r such that for any > I and any x, y E R d ,

    ((Pj (X) Vj (y)) 2 < C I X Y 12.

    Thus for any x E B(cj, 7r),
    ((pj(X) pj(y))2 dy < C I X Y 12 dy
    (cj,8r) ix yld+cv IB(c,,8r) IX yld+,,
    < C I -1-.
    fx-yl<15r IX yld+a-2 = C4 dy
    (0,15r) jyjd+a-2
    = M, (r) < oo (3.8)







    52

    For any x E B(cj, 6r),
    ( Oj W vj MY dy ( Pj(x)-- Vj(y))'dy < I.-Yl>r I x yld+or
    B (c,,8r)l yld+a < 1 dy

    I.-yl>r I x Yjd+cv
    lyl>r jyj 1 '0' dy M2(r) < oo. (3-9)

    For any x B(cj, 7r),

    ((Pj (x) vj (y)), dy
    (c,,6r) x yld+ct
    1 1 dy
    p6r) Fx y +- lrld+a ) T- ) 1B(c,,6r)(y)dy. (3-10)
    lRd x yld+a r Fa
    Combining (3.6)-(3-10), we get

    7-(uvj, U(pj)
    Y))2V (y)
    (u(x) u( 3 dxdy
    Rd IRd I x Y I d+ce
    +2M, (r) Rd u 2 (X)IB(c,,7r)(x)dx
    + M2 (r) JRd u 2 (X)lB(cj,6r)(x)dx

    + x Y +a
    U2(X) I A F -)lB(c,,6r)(y)dy) dx
    ,,d (IRd (
    y))2V2(y)
    < IRd JRd(U(X) U( ld+a dxdy
    + (2M, (r) + M2 Rd u 2(X) IB(cj,7r)(x)dx
    U2(X) (1 1 1 6r)(y)dy) dx. (3-11)
    +1 A )lB(c,
    ,d F W+- Tr F







    53

    Combining (3.5) and (3.11) we get


    B(c,,3r) U2 W 4(x)v(dx)
    y))2W (y) Md,.-E I IRd Rd (U (X) -U ( dxdy Ix yld+a
    + IRd U2(X)[(l + 2M,(r) + M2(r))IB(c,,7r)(x)]dx U2(X) '-- A T'
    Rd J ,d( ix YF- r F,, B (cj,6r) (y) dy dx

    which implies


    IRd U'(x)v(dx)
    2 00
    5 Md,. 6 1 Rd IRd (U(X) U(Y)) E V (y)dxdy Ix yld+a 0

    IRd U2(X)[(l + 2M,(r) + M2(r)) E 1B(cj,7r)(x)]dx
    0
    U2(X) -) E' IB(cj,6r)(y)dydx
    + Rd 4d( X yp+, A I r +, 0

    For every x E R', B(x, r) intersects at most 8 d balls from I B (cj, 6r) j 1} and
    at Most 9d balls from B(cj, 7r) > I). Thus for any x, y E R',
    00
    (P2 (Y) < 8d,
    0
    00
    E IB(c,,6r)(Y) < 8d,
    0
    00
    E IB(c,,7,)(X) < 9d0
    Hence we obtain that








    54



    JRd u2(x)v(dx)
    < d, c 2.8 8dT(u, u) + Md,c. c[(1 + 2M, (r) M2(r))9d + 8 dM3(r)]IjuII' where

    M3(r) = ~ A~ A dy < o

    The proof is now complete.

    Q.E.D.


    Theorem 3.3.3 If G c Ad,a. is a nonnegative function such that the function Gdefine by

    ?7(x, y) = G(y, x) is also in Ada then for any u E )




    Furthermore, for any f > 0, there exists C(c) > 0 such that for any u E )

    J~IRd u(x)G(xl Y)u(y) dx dy K cT( u, u) + C(c)I~uIIl.


    Proof. Observe that

    IfdIf u (X)G (x, y) u(y) dxdy
    J~dJ~d IX -yld+-f U 2(X) (J G~xy ) d) d JI U2(y) (J G(x, y) dX) dy.

    From the assumptions on G we know that the functions








    55

    X G(x, y) d

    "d Ix yld+c dy
    Id G(x,y)
    R Ix y]d+ x

    are in Kd,,, therefore the conclusion of this theorem follows from Theorem 3.3.2.
    Q.E.D.

    If, in addition, we assume that the function F defined by F(x, y) = F(y, x) is also in Ad,a, then from Lemma 3.1.4, Theorem 3.3.2 and Theorem 3.3.3 we know that E given by

    E(u, v) =T(u, v) u (x v ~ Rd i dx u(x)G(x, Y)v(Y)dxdy
    R Ix yl d+

    with G = eF 1. E(u, v) is finite for all u, v E b and (E, D) is a bilinear form. The following theorem is the main result of this section.


    Theorem 3.3.4 If, in addition, we assume that the function F defined by F(x, y) = F(y,x) is also in Ad,a, then (,D) is a closed bilinear form satisfying the sector condition.

    Proof. From Theorem 3.3.2 and Theorem 3.3.3 we can easily show that for any E > 0 there exists a C(c) > 0 such that for any u E D D iD, I(S- T)c(u, u)j < CTc(U,, u)+ C(e)ju,5 i.e., the relative form bound of (E ')c with respect to Tc is zero. Therefore by Theorem 6.1.33 of [18] we know that ((e T)c, D E iD) is a closed sectorial form. Thus it follows from Theorem 2.3.2 that (S, D) is a closed bilinear form satisfying the sector condition.
    Q.E.D.








    56


    From Theorem 2.3.1 and the theorem above we know that (9, D) uniquely determines a strongly continuous semigroup on L2(Rd). In the next section we are going to show that this semigroup is nothing but the semigroup (Tt)t>o of the Section 3.2.

    3.4 The Connection between the Semigroup (T) and the Bilinear Form (6, D)

    In this section, in addition to assuming that F E Ad,a, we are going to assume that the function F defined by F(x, y) = F(y, x) is also in Ad,a.

    From Theorem 3.2.3 we know that there exist constants C1, C2 > 0 and /1, /2 > 1 such that

    sup EX{eAt} < Cle't (3.12)
    xERd
    sup EX{e2At} C2e2t (3.13)
    xERd
    Put O0 = 01 V 02. Then for any 0 > /o, the mapping f E' e- teAt f(Xt)dt

    maps L'(Rd) into L*(Rd) and maps L2 (Rd) into L2(Rd). Theorem 3.4.1 Iff E L" (Rd), then for any / > #o, the function u(x) = Ex e-'t eAt f(Xt)dt is a bounded continuous function in D such that for any v D,

    T(u,v) + /(u, v) f(x)v(x)dx u(x)v(x)p(dx)
    Rd Rd v(x)G(x, Y)u(y)dxdy
    Ix yld+a
    where G = eF- 1.

    Proof. The fact that u is bounded and continuous follows from the properties of the Feynman-Kac semigroup. Set Mt = e-t eAt u(Xt) + e CSeA' f(X.)ds.








    57


    Then we have for each x E Rd, Mt_ E2 [10 e-08 CAf (X)ds I.F] Therefore (Mt)t 0 is a continuous Prx-martingale for each x E Rd. Applying the integration by parts formula for semnimartingales, we get

    Cot u(Xt) e e AtCo e At u(Xt)

    = PeAt (Mt e-3ssfX j)

    =U(XO) + j0 CAs-dM, J- C e ff(X,)ds + -:e,,(,( eF(X,-,X)) 0 ~eu(X)dAl".

    We know that fol e-A-dMs is a local Px-martingale for each X E Rd, thus ~~t

    10 6-13'u(X,)dA" I e'3su(X3)(1 eF(X,,-X.))
    0 0
    is a local P-T-martingale for each x E Rd. Since u and f are bounded, it is easy to check that (Nt)t 0 is in fact a uniformly integrable Px-inartingale for each x E Rd Taking expectations we get for every x E Rd

    u(x) =E'e-oiu(Xt) + Ex j0 e-03f (Xs)ds

    jE e-'3u(X,)dA" + E' Ite~ G(X., Y)U(Y)dydS.
    10 0 ~R d IX. -yld+ce

    Letting t T oo, we get

    u (x) = jx 0 eOsf (Xs)ds + Ex 1j e-Osu(Xs)dAI'

    jE Fi JO G(X.,, Y)U(y) dyds
    0 JRIX, yld+c.
    Therefore by Lemma 5.1.3 of [13] we know that u E D.








    58


    Applying the integration by parts formula for sernimartingales again we get

    U(Xt) = eote-Ate-ot C At U(Xt)
    ,at -At (Mt t e-j3seA. f (X,
    0
    t t
    = U(XO) + 10 e OS CAdMs 10 f (X,)ds t t
    u(X,)ds u(X,)dAm + 1: U(X,)(I
    0 0<3
    Thus

    t t
    u(X,)ds + u(X.)dA"' Zt U(Xt) U(XO) JO 10

    Z U(X')(I F(X.,-,X,)) + It f (X.)ds O
    is a local Px-martingale for each x E R d Again from the boundedness of u and f we can show that (Zt)t>o is uniformly Px-integrable for each x E R therefore it is d d
    a P'-martingale for each X E R Taking expectation, we get that for each X E R t t
    u(x) Exu(Xt)+ E' 10 f (X.)ds OE' JO u(X,,)ds

    +E- It u(X,)dA" + Ex It j G(X.,, Y)u(y)dyd..
    0 S 0 Rd IX., yld+o,

    Thus from Lemma 5.1.4 of [131 we get that for any -y > 0 and any 9 E L'(R d),

    T(U' U"g) = lim 1 (U Ptu, U'g) tjo t

    = IRd f(x)Ug(x)dx #(U'Ug) + IRd u(x)Ug(x)dx + IRd IRd U-yg(x)G(X, Y)U(Y) dxdy.
    Ix yld+ce

    From Theorem 3.3.2 we can easily see that

    v IRd u(x)v(x)p(dx)

    v IRd IRd v(x)G(x, Y)U(Y)dxdy
    Ix yld+a








    59


    are bounded linear functionals on ) with respect to the Ti-norm. Since U.,(L2 (R d)) is dense in D, we can deduce immediately from the identity above that for any v E D,

    T(u, v) + 0(u, v) = Jf (x)v(x)dx + Rd u(x)v(x)p(dx) Id v(x)G(x, y)u(y)dzdy.
    S Ix yd+a
    Q.E.D.

    Lemma 3.4.1 Iff E L2(Rd) and /3> #o, then the function v(x) = Ex eteAt If I(Xt)dt
    0

    is finite quasi everywhere on Rd.

    Proof. From (3.2) and (3.13) we can easily get that for any x E Rd,

    lo 1
    v(x) < C e-(~-3o)t Ef f2(X,)) dt
    C1 < (o e-(d-~)tEx f2(Xt)dt)

    It is easy to show that for any v E So IIUlIlo < oo implies IIUvlloo < oo for any y > 0, for any v E So. Therefore for any v E So with IIUvllooI < o, we have


    Ia v2(x)v(dx) < e-3- )'E" f(X,)div(dx)
    C u- dRdo(x, y)f2(y)dyu(dx)
    V9 -TO d Rd
    C
    ___ 3---0 TOu-01ooI l Thus from Theorem 3.3.2 of [13] we know that v is finite quasi everywhere.
    Q.E.D.

    Lemma 3.4.2 Suppose that {un} C D and that u, u in L2(Rd). If sup T(u, u~) = sup / IxI-1dI2dx < oo, (3.14)
    n nJRd








    60


    then u E D and there is a subsequence {un,} of {un} which converges weakly to u with respect to the TI -norm.

    Proof. Using Fatou's lemma we get that

    J zd Ixl-lI2(x)dx < lim inf lz Ix "2dx <00,


    thus u E D.

    From (3.14) we know that there is a subsequence {u,} of {un} such that Ixlz?,k converges weakly in L2 (Rd) to some function w E L2 (Rd). Taking convex combinations of { IX 1f, } if necessary, we can assume that Ix 1|2, converges strongly in L2(Rd) to w. Combining this with the fact that un converges to u in L2(R d), we can see that w = x It. Therefore Ix1i2fk converges weakly in L2(Rd) to Iz| I, that is to say, for any f E L2(Rd),

    lim Ix l2ak f(x)dx = R I(x)f(x)dx.
    k-0oo d d

    In particular, we have, for any v E D,

    lim l xf n,(x)V(x)dx = I xIa(x)i(x)dx,
    k--oo JRdJd

    i.e.,

    lim T(un,, v) = T(u, v). k-oo

    The proof is now complete.
    Q.E.D.


    Theorem 3.4.2 If f 2(Rd) and # > /o, then the function u(x) = Ex e -teAt f(Xt)dt
    0








    61


    is in Db and satisfies for any v E D

    T(u, V) +13(u, V) = ./Rd()(x)dx-JdU(~~~~


    IRd IRd vxydx, YUYdxdy, (3.15)

    where GC e F 1.

    Proof. For any n > 0, let

    f (x), if If(x) I n; {-n, if f(x) <-n

    and define

    un(x) =Ex 0e,te Atfn(x)dt. By Theorem 3.4.1 we know that for each n, Un is a continuous function in D such that for any v E D,

    T(u,, v) + ,3(u,, v) =Jd f (x)v(x)dx Jd Un(x)v(x)1u(dx)
    I~ dv(x)G(x, y)un(Y) dxdy (3.16) J~JdIx y d+cv We can easily show that un, tends to u strongly in L2(Rd), thus Sup{InII~2} < 00n

    Now from (3.16), Theorem 3.3.2 and Theorem 3.3.3 we obtain

    T(Un, Un) = -/3(Un, Un) + Jd f (x) u,(x) dx

    Rdu(xp(dx)-J I ~xGx ~ny dxdy

    < 1~ 2~I~H IIfI2HIUnhI2 + 2 T(Un, Un) + CIIUnII2, where C is a positive constant. Therefore T(Un, Un) 2(1 + C) IjUn112 + 211f1 1jIUn11,








    62


    consequently
    sup T(u,, un) < oo00.
    n

    Hence by Lemma 3.4.2 we know that u E D and there is a subsequence {uk } of {un converging to u weakly with respect to the Tl-norm. Thus there is a sequence {vk} with vk being finite convex combinations of {u,} such that vk converges to u strongly with respect to the Tl-norm. Now from Lemma 3.4.1 and the dominated convergence theorem we can see that un converges to u quasi everywhere on Rd, thus vk converges to u quasi everywhere on Rd. For each k, vk is continuous since Un is continuous for each n. Therefore by Lemma 3.1.4 of [13] we know that u is quasi-continuous.

    Now we are going to prove (3.15). From the facts that u, converges to u strongly in L2(R d) and that unk converges to u weakly with respect to the Tl-norm, we can easily see that for any v E D,


    lim T(uk,unk) = 7T(u,u),
    k-,oo
    lim(unk,uUk) = (u,u),
    k-oo
    lim nf(x)u,(x)dx = f(x)u(x)dx.
    k-oo d d

    Therefore by (3.16) we know that in order to prove (3.15), we need only to prove that for any v E D,


    lim R unk,(x)v(x)p(dx) = u(x)v(x)p(dx), (3.17)
    k-oo JRd JRd
    lim v(x)G(x'y)unk(Y)dxdy = [v(x)G(x, y)u(Y)dxdy. (3.18)
    h m d z y = z d y ( 3 1 8 )
    k-oo Rd -Rd Ix yld+a- YId+a We are only going to show (3.17). (3.18) can be shown similarly.

    Fix a v D h. By Schwarz's inequality and Theorem 3.3.2 we know that for any r > 0 there exists a C(e) > 0 such that








    63








    1
    Me + C() u() u (x) v (x),(dx) ,





    where

    M =d ()Q 2 sup {T(uk u k

    Since us tends to u strongly in L2(Rd), we can take a K > 0 such that Iu,<- Ul2 (E) ( v (x)*(dx)1 ,) whenever k > K. Therefore

    SIRd u,(x)v(x)[(dx) u(x)v(x)p(dx) < (M + 1)e, whenever k > K, and consequently lim / u(x)v(x)(dx) = f u(x)v(x)(dx),
    k-ooJ d Rd

    which is (3.17).

    Q.E.D.

    From Theorem 3.4.2 we immediately get the main result of this section.

    Theorem 3.4.3 (T)t>o is the unique strongly continuous semigroup on L2(Rd) determined by the bilinear form (, D).

    From Theorem 3.3.4 and the result above we obtain the following result which generalizes the main result of [29]. Corollary 3.4.1 If F 0 and tt = 0, then the Markov process (Xt, eAt) satisfies the sector condition.

















    CHAPTER 4
    THE GAUGE THEOREM AND ITS APPLICATIONS

    4.1 The Gauge Theorem

    Recall that p E Kd,, and F E .Ad,a are fixed and

    At= A" + S F(X_,X,).
    O
    Let A+ and A'- be the continuous additive functionals associated with + and j-, respectively, and put A+t = Al++ F+(X-, X);
    O A = A- + F-(X,_,X,);
    O A* = A+ + At; Lt = eA
    It = A


    Then {Lt, t > O} and {Kt, I > O} are multiplicative functionals of X and furthermore for all I > 0, eAt = Kt -. Lt, Kt < 1, Lt > 1.

    From now on we are going to fix a regular bounded open domain D in Rd and define

    g(x) = E'(eA(TDo

    The function g is called the gauge function of (A, D) with respect to the process X, or simply the gauge function of (A, D) if there is no confusion. The purpose of this section is to study the boundedness property of g.
    64








    65


    Lemma 4.1.1 Suppose that v E Kd,a is nonnegative and A' is the continuous additive functional associated with v. Then the function x ExA"(rD)


    is bounded on Rd


    Proof. From the definition of Kd,, we see that the function x u(x,y)v(dy)


    is bounded on Rd. From [23] we know that ATD is the continuous additive functional of XD associated with 1D v. Therefore by repeating the argument in the proof of Theorem 3.1.3, we can get that for any x E D, EXAV(T.D) (4.1)
    EuA(rD) = D(x, y)v(dy). (4.1)

    Hence for any x E D,

    EXAv(TD)
    and thus the conclusion of this lemma is true.

    Q.E.D.


    Theorem 4.1.1 infxEndg(x) > 0.


    Proof. From the definition of Ad,, and Lemma 4.1.1 we get that the function

    x E (A"+ + + A + f [F(Xt, y) dydt
    is bounded by a constant C on Rd. Hence by Jensen's inequality we et
    is bounded by a constant C on R d. Hence by Jensen's inequality we get









    66


    g(x) = ExleAT) Ex ,'(D)

    > exp(-Ex{A*(,TD)})

    -exp {Ex (Am +(TD) + Am- (TD) + fI f~ IF(Xt,Y)I1 dydt)

    > e-C>O0.


    The proof is now complete.

    Q.E.D.


    Lemma 4.1.2 There exists a -y> 0 such that sup Ex /T eOtt <00o. xE Rd JO

    whenever /3> -y.


    Proof. By using the integration by parts formula for semnimartingales we can get that

    jdLt Lt dA"'+ + E e exp( E F+ (X,,X,)) (eF+X~t )
    10 0 O
    Therefore for any x E R,


    Exj1 dLt



    ExJ LtdA't' + ExJ IA+~< EX ]1Ld0~ X L IRd Ixt yld+a dd

    ExF/ LtdBt

    (Ex (B1 )2 (Ex eA') where

    -A' + t eF+(Xt,y) -_dydt.
    t 1tiRd IXt -y d+oe








    67


    Since p and the function x eF+(xY) dy
    aIx yld+a

    are both in Kd,a, it follows from Theorems 3.1.4 and 3.2.3 that


    sup (Ex(B)2) (Exe2At) < C, xERd

    for some constant C1.

    Now for any x E Rd, we have


    E J"TD e-tdLt

    EX{~ ~e- tdLt}


    = _,Ez: n< TD;i,' e-IBtdL}

    0
    Ex T, IDA(nfl) Ct~
    0 DAn

    oo n D A(n+1)



    C1 1 6-mnEXLn.
    = Ex n < 7rD; e6- nE ()tdAJd
    0 (n

    <, E" e-n < DI-pnxs ~

    0

    From Theorem 3.2.3 we know that there exist constants C2 and 7 > 0 such that for any n > 0, sup ExL, < C2e'. Rd
    Therefore if 3> 7, then


    sup EX TD e-tdLt xERd 0
    00
    <- C C2 E e (y-)0)n
    o
    < 00.
    <..E.D.
    Q.E.D.








    68


    Theorem 4.1.2 (The Gauge Theorem) The gauge function g is either identically infinite on D or bounded on D.


    Proof. Lemma 4.1.2 implies that eA' is compatible in the following sense:

    fTD
    sup e-OtKtdLt < c,
    XED0

    for some 3> 0. It follows from Theorem 2.2.2 that U1 is compact as a map from L'(D) to L (D) for any -y > 0. The operator U'g defined by f E j e-Ytf (Xt)KtdLt


    is dominated by U4, consequently UK is compact from L-(D) to L-(D) for any

    -y > 0 by Remark (3.5.b)of [28]. It follows again from Remark (3.5) of [28] that for any -y > 0, UD',K is non-degenerate and irreducible in the following sense: UDK1 is not identically zero and for any nonnegative f, UDgf is either identically zero or strictly positive on D. Theorem 4.1.1 above guarantees that g is strictly positive. Thus all the conditions of the general Gauge Theorem (Theorem (3.4)) of [28] are satisfied, and so the gauge function is either identically infinite or bounded on D.
    Q.E.D.

    Now we are going to derive some important consequences of the boundedness of the gauge function which will be very useful in the next two sections. But first we state two lemmas which we will use to derive the results mentioned above. Lemma 1.1.3 There exists an 77 > 0 such that for any Borel subset B of D whose Lebesgue measure is less than 717,

    sup Ex{eA(TB)} < cC).
    XERd








    69


    Proof. The proof is similar to that of Lemma 4 of [10], we give it here for the sake of completeness.

    If P"{rB = 0} = 1, then E{eA('TB)} = 1. Otherwise P{rB = 0} = 0 and we have


    E"{eA(-B)} < E' {L(rB) 00
    Z E I {n < -B; LE X{0 < 7B < 1;L(TB)}
    n=O
    oo
    00
    < E {n < TB; L ,Ex { L}} n=o
    OO
    SC1 E E" {n < Trs;exp(A+)} n=o

    where

    C1 = sup E'{L1,} < co
    xERd
    by Theorem 3.2.3. The sum above is bounded by

    00
    -(Px{n < TB}) (Ex{exp(2A+)})1. n=O

    Applying Theorem 3.2.3 again we know that there exist constants C2 and b such that


    sup E'{exp(2A)} < C2enb, xERd

    Hence the last sum is bounded by OO
    2 Z(Px{n < rI})12e (4.2)
    n= o

    It is easy to see that there exists an 7 > 0 such that for any Borel subset B of D whose Lebesgue measure is less than ri we have


    sup Px{1 < TB} < e-2b. xERd

    It follows from the Markov property that








    70


    P"{n < rsB} < e-2b


    Using the above inequality in (4.2) we see that the infinite series in (4.2) is convergent.

    Q.E.D.


    Lemma 4.1.4 If the gauge function g is bounded, then for any i > 0, there exists a constant C(r1) such that for any x E D,
    00
    IlgIIl-lg(X) < Ex{k?7 < TD; e A~k,,} < C')
    k=0

    Proof. Since the proof is the same for any 77 > 0 we take 7 = 1. For each x E D the quantity Ex{n < TD; eA(rD)} decreases to zero as k -* co. Hence for any e > 0, we can choose an integer N and a set H such that the Lebesgue measure of D \ H is less than e and

    sup Ex{N < TD; eA(rD)} < C. xEH
    By choosing a compact subset of H if necessary we may assume that H itself is compact. Here e is such that sup Ex{e 2A(TD\H)} := b < o. XERd

    This is possible by Lemma 4.1.2. Now


    Ex{2N < TD; eA(TD)}

    < Ex{N < TD\H; eA(TD)} + Ex{N > TD\H, 2N < TD; eA(rD)}

    5 gooExf{N < TD\H;e A(rD\H)}

    +E'f{TD\H < TD; eA(D\H)Ex(D\H){N < TD; eA(D)}}.

    The expectation in the third line above can be estimated by Ex{e2A(-D\H)} P{rD\H >








    71


    which is uniformly small if N is large, by the transience of X. On the set {TD\H < TD}, XD\H e H, so the expectation in the fourth line above is bounded by V/b. Thus for large N,

    sup EX{N < TD; eA(D)I
    XERd
    is small. Now from Theorem 4.1.1 g is bounded from below, say g a > 0. Then EX{N < TD; eA(N)} < 1EX{N < TD; eA(N)g(XN)}
    a
    < 1Ex{N < TD; CA(TD)}.
    a

    Therefore we can see that for N large enough sup EX{N < TD; eA(N)} < <1. xERd
    By the Markov property we have for any k,

    sup Ex{kN < TD; eA(kN)} ik. xERd
    Further if j < N, Ex{kN + j < TD; eA(kN+j) = E{j < TD; EA(j){kN
    fk sup sup E'{eA(j) j! N xERd
    All these inequalities give

    SEx{n < TD; CA(n) n=0
    o N
    = 5 E'{kN + j < TDr; eA(kN+j)}
    k=o j=1

    k=w
    Finally we have








    72


    OO
    g(x) = E {n < TD < n 1;A(-D)}
    n=O

    < E({n < TD; eA(n)g(XI)}
    n=o

    SIghlo E EX{n < TD; eA(n) n=o

    which proves the result.

    Q.E.D.

    Now here is one consequence of the boundedness of the gauge function.


    Theorem 4.1.3 Let v E Kd,o be nonnegative and Bt be the continuous additive functional associated with v. If the gauge function g is bounded on D, then


    sup Ex D eA(t)dBt < oo. xED 0

    Proof. Obviously we have


    Ex 1eAtdBt < Ex eA *dBt
    < Exfe Bt <_ E": eA B1} < Ex {eA+B1}. By Theorem 3.2.3 we know that there exists a constant C such that


    sup E'x{eA+B1} < C. xERd

    Thus we have


    sup E eTD A'dBt
    xED 0

    = sup Ex JnDA(fl) eAt dB A
    xED n=0 Dn

    = sup t Ex n < TD; DA tdB
    zGDa= n= III









    73


    00 An X~n TDA1 \t
    sup E n < TD; eAEX( A dBet
    zED nJO=0
    00 / 1l
    < sup E n < TD; An Ex(n) eAt dBt
    XED n=0
    OO
    < Csup Z E"{n < TD; CeAn} xED n=O
    < 00,


    where the last inequality follows from Lemma 4.1.4.

    Q.E.D.

    Now here is another consequence of the boundedness of the gauge function. It looks much stronger than the assumption that the gauge function is bounded.


    Theorem 4.1.4 If the gauge function g is bounded on D, then


    sup E"' sup eAt xED O
    Proof. We have

    eAt = 1+ eA dA"
    e0
    + eA exp( F(Xr-,xr)) ex-,x,) -1)
    O
    < 1+ eA'dA"8
    + e At exp( F(Xr_,X))(elFI(X,-x) 1)
    O
    Therefore,


    sup eAt 1+ eArdA"t
    O SeA exp( E F(X,_,X,))(eIFl(xt-x) i)
    O
    Hence for any x E D,








    74



    Ex su e At
    0 < + EX eA'dAF"
    0
    'D A e]FI(Xt,y) 1
    +E eA exp(o
    1 + Ex [ eA'dAu"
    +Ex 7D eAt j F (t yJ dydt.
    ot JRd IXt yld+0dt

    Now the conclusion of this theorem follows Theorem 4.1.3.

    Q.E.D.

    4.2 The Dirichlet Problem for Linear Equations

    In addition to p, F and D, we are going to fix another Radon measure v on Rd. We are going to assume that v is in Kd,, and Bt is the continuous additive functional associated with v. The purpose of this section is to formulate and solve the Dirichlet problem of the following linear equation (-(-A) + u~x G (x, y)u(y).
    + P) u(x) + I G(xY)u(Y)dy + v = 0 (4.3)
    ax yjd+cw

    on D, where G = eF 1 as in the previous chapter.

    Before we discuss the Dirichlet problem of (2.1), we are going to recall some useful facts first.


    Definition 4.2.1 A function u e De is said to be harmonic with respect to X on D if T(u, v) = 0 for any v E De n Co,o(D) = D n Co,o(D). Here Co,o(D) stands for the family of continuous functions with compact support in D.


    In the sequel a function which is harmonic with respect to X on D will simply be said to be harmonic on D. Here we need to keep in mind that a function that is








    75


    harmonic on D is required to be defined on all of Rd and even the part of the function on the complement of D is involved in the above definition. Lemma 4.2.1 ff E D, is bounded, then the function u(x) Exf(X(7-D))

    is harmonic and continuous on D. Here f denotes a quasi-continuous version of f. Furthermore, if f is continuous on DC, then u is continuous everywhere and coincides with f on Dc.

    Proof. That u is harmonic follows from the transient version of Theorem 4.4.1 of [13]. The rest are proven in [24] and [25].
    Q.E.D.

    Lemma 4.2.2 If a bounded function u E De is harmonic on D, then ul(x):- Exii(X(TD))

    is a version of u which is continuous in D.

    Proof. Follows immediately from the transient version of Theorem 4.4.1 of [13] and Lemma 4.2.1 above.
    Q.E.D.

    Lemma 4.2.3 If f is a bounded function on Rd, then u(x) := Ex 0 f0(X)dBt is a bounded continuous function which belongs to DD, vanishes on Dc and T(u, v) =IRd v(x)f(x)v(dx) for any v E D, n C0,O(D) = D n C0,0(D).








    76


    Proof. That u vanishes on DC is trivial from the regularity of D. The boundedness of u follows from Lemma 4.1.1. Now we are going to prove the continuity of u. It follows from (4.1) that for any x E D, U()= I~D '(x, y)f (y) v(dY)
    u(x)


    From Corollary 2.2.1 and Theorem 2.2.1 we know that for any r > 0,




    is continuous on D and that, as x tends to any point z on the boundary of D, ur(x) tends to zero. By the definition of Kd,,c and the fact that for any x, y E D, U(x, y) < Ix Yj-d+a,

    we can easily see that u, tends to u uniformly on D, therefore u is continuous on D and u(x) tends to zero as x tends to any point z on the boundary of D. Hence u is continuous on Rd.

    The rest of the conclusions follow from the transient version of Lemma 5.1.3 of [13].
    Q.E.D.


    Definition 4.2.2 A bounded function u E De is called a solution to the equation (4.3) in D if
    T(u, v) u(x)v(x)p(dx) v(x)G(x, Y)u(Y)dd

    Ix yV

    Ij d v(x)v(dx), Vv E D n C,(D). (4.4)

    Again we need to bear in mind that a solution to the equation (4.3) in D is required to be defined on all of Rd and even the part of u on the complement of D plays a role in the definition above.








    77


    The definition above is a generalization of a solution in distributional sense to the following Schr6dinger equation (- + P)u(x) = V. (4.5)
    2

    In fact, when the underlying process X is the standard Brownian motion, then for any F vanishing on the diagonal of R x Rd, the additive functional S F(X,_, X,)
    O
    is identically zero, or equivalently, F can be replaced by the function which is zero everywhere, thus G can be replaced by the function which is zero everywhere and so the third term in the first line of (4.4) disappears and (4.4) becomes 7(u, v) Rd u(x)v(x)p(dx) = Rd v(x)v(dx), VvE Dn Co,0(D), which is exactly the definition of a solution of (4.5) in the distributional sense. Theorem 4.2.1 Any solution u of (4.3) in D has a version which is continuous in D.

    Proof. Put

    u,(x) = E"B(rD) + u(Xt)dAY
    0
    +Ex jD ( G(Xt, y)u(y)dy dt.
    o na XuXt y d

    Then it follows from Lemma 4.2.3 that ul E DD is a bounded continuous function on Rd such that for any v E D n Co,0(D),

    T(u1, v) = v(x)v(dx) + J u(x)v(x)(dx)

    d id v(x)G(x, Y)U(y)dxdy ixI ld+a







    78


    Consequently the function u2 := u u1 E De satisfies the following condition T(u2, v) = 0,

    for any v E D n Co,o(D), i.e., u2 is harmonic on D. Therefore, by Lemma (4.2.2) above, we know that u3(x) := Ex t2(X ) is a version of u2 which is continuous in D. Thus the function u3 + UI is a version of u which is continuous in D.
    Q.E.D.

    Definition 4.2.3 Suppose f is a bounded function in De. We say that a bounded function u in De is a solution to the Dirichlet problem of (4.3) with exterior function f if the following three conditions are satisfied:

    (i) u is a solution of (4.3) in D;

    (ii) UID: = fID:; (iii) for every z E lim u(x) = f(z). D3x -z

    Theorem 4.2.2 If the gauge function g is bounded, then for any f E D, which is bounded and continuous on DC, u(x) :E (eA(D)f(X(rD))) + E-fI eA0dB, is the unique continuous solution to the Dirichlet problem of (4.3) in D with exterior function f.








    79


    Proof. It follows from the regularity of D, the boundedness of f, the boundedness of the gauge function and Theorem 4.1.3 that the function u defined above is bounded on Rd. It follows easily from the definition of u that M : eA(tArD) U(XtAT-D) + j0tr e A'dBt is a Px-martingale for each x E D. Using the integration by parts formula we get

    U(XtATD) = ecA(tATD) (M, IjtATrD eAtdBt)

    =u (XO) + jOtr e-A(s)dM, B(t A TD)

    JtArD u(X,)dAs + E u(X8) ( I FX-,.
    0<8 tArp
    Thus

    Nt U(XtA'D)B(t A 7D)

    + JtATD u(X,)dA, u(X8,) (1 eF(XX.))

    is a local martingale with respect to Px for each x E D. Now, for any t > 0,

    Nt : B(7-D)

    +I1U1-Io ( + A*(TrD) + E (elFI(X.-,Xs)-1) and for each x E D,

    Ex S (IFI(X,-X,) 1) = Ej 17D eIFI(X,,y) -1 ds E 0 RdIXS yd+add

    Thus by Lemma 4.1.1, we know that Nj is uniformly Px-integrable for any x E D and consequently Nt is a uniformly integrable Px-martingale for any x E D. Hence

    u(x) =Exu(XD) + E'B(TrD)

    +I'Du(x,)dAs.Ex 1 u(X,)G(X8-,X,) =Exf (X,,,) + ExB(TrD) + I'D u(X,)dA, + E-'D I G(XS, Y)u(Y)dds
    0 0 Rd ~JX8 la








    80


    Now it follows from Lemma 4.2.3 that

    ui(x) := EXB(rD) + Exj u(X,)dA, +Ex d(a dyds
    o IX, yld+ satisfies the following:

    (a) u1 E DD is a bounded continuous function which vanishes outside D;

    (b) for any v E D n Co,0(D), T(ui, v) = JRd v(x)v(dx) + Rd u(x)v(x)p,(dx) v(x)G(x, y)u(y)dzdy.
    Id Rd x- yId+ d

    And by Lemma 4.2.1 we know that U2(x) = Ex f(X(rD))

    satisfies the following:

    (c) U2 E De is a bounded function which is continuous in D;

    (d) u2 Dc = f De;

    (e) for any z E OD, limu(x) = f(z);
    X- Z

    (f) for any v E D n Co,o(D), T(u2, v) = 0. Therefore u = u1 + u2 is a continuous solution to the Dirichlet problem of (4.3) with exterior function f.








    81


    Now let us assume that W is a continuous solution to the Dirichlet problem of (4.3) with exterior function f. Then by Lemma 4.2.3 we know that the function
    TrD
    (x) := EB(rD) + E" U(X,)dAs +E3 TD G(X,, y)9(y) dd E JRdX, y-da Y satisfies the following

    (g) ui e DD is a bounded continuous function which vanishes on De;

    (h) for any v E D n C0,0(D), T(Wx, v) = Rd v(x)v(dx) + I (x)v(x)p(dx) + d v(x)G(x, y)U(y) ddy.
    S I| yld+ dxdy.

    Therefore uz is a bounded continuous function in De such that T(U U1, v) = 0, for any v E D n Co,o(D), i.e., U Ux is harmonic on D. Hence by Lemma (4.2.2) we know that for any x E D,

    (U Ul)(x) = Ex( Ul)(X,,rD) = E f(XD), consequently

    U(x) = EXf(XrD) + EB(TD)
    TD TDG(X,y)(y)d + U(X,)dA, + Ex J" D IXsYf dyds.
    -0-0 R d IX. y1 +
    From this we can easily get that H(XtATD)+B(tA7D) + AU(X,)dA, + S U(X,)G(X._, X)
    O O







    82


    is a Pz-martingale for each x E D. Using the integration by parts formula we get eA(tArD)U(XtAv )

    = U(Xo) + eA(s-)dYs -tATD eAt dBt.

    Therefore

    Zt := eA(tAD)U(XtA) + tATD eAdBt is a Px-local martingale for each x E D. It follows from Theorem 4.1.3 and Theorem 4.1.4 that {Zt} is uniformly Px-integrable for each x E D, so {Zt} is a uniformly integrable Px-martingale for each x E D, and consequently

    U(x) = Ex eA(rD)U(X(TD))} + Ex eAtdBt = {e eA(TD)f(X(TD))} + E JIT eAtdB,
    = t(X),


    for each x E D. The proof is now complete.

    Q.E.D.

    4.3 The Dirichlet Problem for Semilinear Equations

    In this section, D, u, F, G and v are fixed as before. For convenience, we are going to use B+, Bt and B,* to denote the continuous additive functionals associated with v+, v- and v* = v + v-, respectively.

    The purpose of this section is to prove the existence of solutions to the Dirichlet problem for the following semilinear equation
    (-(A) + P) u(x) + I (x, uy)
    + y) u(x) + )dy + ((u) + v = 0 (4.6)

    on D, where ( is a continuous differentiable function on R'.

    In order to prove the existence of solutions to the Dirichlet problem for (4.6), we need the following result.







    83


    Lemma 4.3.1 Suppose that jT E Ka,o is negative and that At is the continuous additive functional associated with 7i. If the gauge function g is bounded, then for any f E De which is bounded continuous on DC, the function defined below u(x) = Ex (eA(1D)f(X(TD))) + Ex TD eA'dB (4.7)

    satisfies the following relation

    u(x) = Ex (e A(D)+(rD)f(X(rD))) + Ex j eAt+AttdBt

    -Ex TD eAt+Atu(X,)dAt. (4.8)

    Proof. It follows from Theorem 4.2.2 that the function u defined in (4.7) is the unique continuous solution to the Dirichlet problem for (4.3) on D with exterior function f. Therefore u is a continuous solution to the Dirichlet problem for the following equation
    (_(_+ P + +-) u(x) + I,, G(x, y)u(y).
    ( ) + +) u() + J y)UY dy + (v u(x)-) = 0 (4.9) on D with exterior function f. Now applying Theorem 4.2.2 to equation (4.9) we know that u satisfies (4.8).
    Q.E.D.

    Definition 4.3.1 Suppose f is a bounded function in De. We say that a bounded function u in De is a solution to the Dirichlet problem of (4.6) with exterior function f if the following three conditions are satisfied:

    (i) for any v E D n Co,o(D),

    T(u, v) J u(x)v(x)pu(dx) I ((u(x))v(x)dx / / v(x)G(x, y)u(y) dxdy + IRd v(x)v(dx); ..d.dy + Xv()-(d); Rd x -. _yzd+a a








    84


    (ii) uDc = fIDc;

    (iii) for every z E 9D, lim u(x)= f(z).
    Dax-z

    Theorem 4.3.1 If the gauge function g is bounded and if satisfies s(s) 0, Vs E R, then for any f E De which is continuous on Dc, the Dirichlet problem for (4.6) on D with exterior function f has at least one continuous solution.

    Proof. For any r > 0, define

    p(r) = 1 + sup{ '(s) : -r < s < r}. Then p is a continuous function on [0, c) such that for any r > 0, p(r)t (t) is increasing as a function of t on [-r, r].
    For any x E Rd, define

    v(x) = Ex eA(TD)Ifl(X(TD))} +Ex eAtdBt;

    (x) = p(v(x)).

    Then v is bounded on Rd, hence 0 is bounded on Rd. Furthermore, for any x, O(x)t (t) is increasing on [-v(x), v(x)] as a function of t.

    Define a measure on Rd as follows:


    Then Kd,~( is a negative Radon measure and. Then j7 E Kf,o is a negative Radon measure and









    85



    2 I Oi~(X,)ds

    is the continuous additive functional associated with jT.

    For any x E R d, define

    uo(x) = Ex (A (7D) f+ (X (TD)) + TD e AtdB+~

    wo(x) = -E' (eA(rD)-(X(T-D)) + JT e A' dBt). Then -v < wo uO < v. From Lemma 4.3.1 we can get that

    uo(x) = E (eA(rD)f+(X(TD)) + JTD eA d+ + (4'uo)(Xt)dt))

    wo(x) =-Ex (eA(TD)f-(X(TrD) + JD e At (dBT (Okwo)(Xt)dt)) where At = At + At.

    For any function h on R', we can define a new function H(h) on Rd as follows:

    Hl(h) (x) = V(x) h(x) (h (x)), VIx E Rd

    Now if we define

    u I(x) = Ex (CA(TD)f(X(TrD)) + jTD e At(dBt + H(uo)(Xt)dt)) w, (x) = Ex (C A(TD)f(X(TrD) + JTD CeAt(dBt + H(wo)(Xt)dt))

    then

    uo(x) ui(x) =E' (e A(TD)f-(X(,rD)) + JTD e A,(dBT + uo(td)

    > 0,

    WOWX Wi(x) = -Ex (CATDf+(X(TD)) + T C1 eAyB+ + (wo)(Xt)dt))

    < 0,

    ui(x) wl(x) = X Ex eT At(H(uo) H(wo))(Xt)dt

    > 0.








    86


    Therefore ul and w, satisfy the following relations: v >- UO U, W, > WO) > -V. For any x E Rd, define


    U2(X) = E' (eA(rD)f(X(rD)) +1JT e At(dBt + H(ul)(Xt)dt))

    W2 (X) =Ex (e A(TD) f(X (TD) + JTD e A(dBt + Hw)X~t

    then


    U2(X) U1(X) =Ex J0T e At(H(ul) H(uo))(Xt)dt < 0, w2(x) w1(x) =Ex fT e A(H(wi) H(wo))(Xt)dt > 0, U2(X) W2(X) =E' JT e At(H(ul) H(w1))(Xt)dt > 0. Thus U2 and W2 satisfy the following relations: V > UO -> U1 >- U2 W2 W1 > WO > -V. By induction, we can define two sequences {u,} and {w,,} of functions on R'd such that

    (a) for any n > 0,

    V >- Un > Un+I > Wn+i Wn > -V.

    (b) for any x E Rd


    u,,+1(x) =Ex (C A(rD f (X (7D)) + jTD e A t(dBt + H(Un)(Xt)dt))I Wn+1 (X) =Ex (CeA(TD)f(X(TrD) + J0 D e A ,(dBt + H(Wn)(Xt)dt))








    87


    Using the monotone convergence theorem we can get two functions u and w such that

    u(x) = E (eA'T) f(X(rD)) + jD e At(dBt + H(u)(X,)dt) ,

    w(x) = E' (e^'T)f(X(TD)+ jTD eA(dB, + H(w)(X)dt))

    Therefore by Theorem 4.2.2 we know that u and w are continuous solutions to the Dirichlet problem of (4.6) on D with the exterior function f.
    Q.E.D.
    Similar to Theorem 4.3.1, we can get the following two results. Theorem 4.3.2 If the gauge function g is bounded and if ( satisfies

    (i) (0) = 0;

    (ii) for any s E (0, oo), i(s) > 0, then for any f E De which is continuous on Dc, the Dirichlet problem for (4.6) on D with exterior function f has at least one continuous solution. Theorem 4.3.3 If the gauge function g is bounded and if ( satisfies

    (i) ((0) = 0;

    (ii) for any s E (-oo, 0), a(s) < 0, then for any f E De which is continuous on Dc, the Dirichlet problem for (4.6) on D with exterior function f has at least one continuous solution.
















    CHAPTER 5
    CONCLUSIONS

    Let X = (Xt, Px) be the symmetric stable process of index a, 0 < a < 2 on Rd (d > 2), let yt and v be Radon measures on Rd belonging to the Kato class Kd,a (see Definition 3.1.1) and let F be a Borel function on Rd x Rd belonging to Ad,a (see Definition 3.1.3). Suppose that A and Bt are the continuous additive functionals with Revuz measures it and v, respectively, and that


    At= A"+ E F(XS-,X).
    O The topic of this thesis is to study the perturbation of the symmetric stable process Xt by the multiplicative functional: eAt

    and the results can be divided into two groups. The first group consists of the results given in Chapter 3, while the second group consists of the results in Chapter 4.

    The results of the first group are related to the semigroup defined by

    Tf(x) = EZ{eA f(Xt)}.

    First we studied the properties of the semigroup (Tt)t>0 in great detail. More precisely, we proved the following:

    (1) If 1 < p p' < oc, then for any t > 0, Tt is a bounded operator from LP(Rd) into LP'(Rd);

    (2) For any p E [1, co), (Tt)t>0 is a strongly continuous semigroup on LP(Rd); 88








    89


    (3) For any t > 0, T maps L-(Rd) into bC(Rd);

    (4) For any t > 0 and any pE [1, oo), Tt maps LP(Rd) into Co(Rd);

    (5) (Tt)t>o is a strongly continuous semigroup on Co(Rd). Then we identified the bilinear form corresponding to (T)t>o0.
    Let D be a regular domain in Rd. The results of the second group are related to the gauge function defined by g(x) = Ex{eA(ID)}.

    First, we proved a criterion, called the gauge theorem, about the boundedness of g. Then we applied the gauge theorem to prove that for a suitable exterior function f, u(x) = Ex ( A(rD)f(X( D))) + Ex eD e A tdB is the unique continuous solution to the Dirichlet problem of S-(-) +,) u(x) + I G(x, y)u(y)dy + v 0,
    -YT ady + v = 0,


    on D, where G = eF 1. Finally, we proved the existence of a solution to the Dirichlet problem for the semilinear equation + P) u(x) + J G(, y)u~)d + ((u) + ii 0 on D, where is a continuous differentiable function on R. = 0 on D, where is a continuous differentiable function on R'.
















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    BIOGRAPHICAL SKETCH

    Renaming Song was born in Linzhang, Hebei Province, China, on April 6, 1963. He entered the Mathematics Department of Hebei University in September 1979, received his B. S. degree in July 1983 and his M. S. degree in July 1986. Upon receiving his master's degree, he joined the faculty of the Mathematics Department of Hebei University and he was promoted to lecturer one year later.

    In August 1988, he came to the United States and enrolled at the Mathematics Department of the University of Florida as a graduate student to pursue his doctor's degree. Since then he has been working as a teaching assistant at the Mathematics Department of the University of Florida.

    He is married to Junge Guo and they have a daughter, Linda Song.




























    92











    I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in -scope and quality, as a dissertation for the degree of Doctor of Philosophy.






    Jps h Gl ver, Chairman
    Professor of Mathematics





    I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy.






    Murali Rao
    Professor of Mathematics





    I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy.






    Zoran R. Pop-Stoj~novi6
    Professor of Mathematics











    I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy.




    I~~ I C e' 1

    Nicolae Dinculeanu
    Professor of Mathematics




    I certify that I have read this study and that in my opinion it conforms to acceptable standards of scholarly presentation and is fully adequate, in scope and quality, as a dissertation for the degree of Doctor of Philosophy.




    aS6
    Malay Ghosh
    Professor of Statistics





    This dissertation was submitted to the Graduate Faculty of the Department of Mathematics in the College of Liberal Arts and Sciences and to the Graduate School and was accepted as partial fulfillment of the requirements for the degree of Doctor of Philosophy.




    August 1993

    Dean, Graduate School