# Fredholm Determinant of an Integral Operator Driven by a Diffusion Process

Authors: Adrian P. C. Lim
Journal: Journal of Applied Mathematics and Stochastic Analysis
 (2008)

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## Abstract

This article aims to give a formula for differentiating, with respect toT, an expression of the form λ(T,x):=𝔼x[f(XT)e−∫0TV(Xs)ds(det(I+KX,T))P], where p≥0 and X is a diffusion process starting from x, taking values in a manifold, and the expectation is taken  with respect to the law of this process. KX,T:L2([0,T)→ℝN)→L2([0,T)→ℝN) is a trace class operator defined by KX,Tf(s)=∫0TH(s⋀t)Γ(X(t))f(t)dt, where H, Γ are locally Lipschitz, positive N×N matrices.


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## Body

## 1. Introduction

Suppose we have a differentiable manifoldM of dimension d.
By the Whitney's embedding theorem, there exists an embedding i:M→ℝN′ such that i(M) is a closed subset of ℝN′.
It turns out that N′=2d+1 will do. We will identify M with the image i(M) and assume that M itself is a closed submanifold of ℝN′.
We will also assume that M does not have a boundary. Let M∪{∂M} be a one-point compactification of M.Definition 1.1.
AnM-valued path ω with explosion time e=e(ω)>0 is a continuous map ω:[0,∞)→M∪{∂M} such that ωt∈M for 0≤t<e and ωt=∂M for all t≥e if e<∞.
The space of M-valued paths with explosion time is called
the path space of M and is denoted by W(M).Let(Ω,ℱ∗,ℙ) be a filtered probability space and
let  L be a smooth
second-order elliptic operator on M.
Using the coordinates of the ambient space {x1,…,xN′},
and extending L smoothly to L˜ in the ambient space, we may write(1.1)L˜=12∑i,j=1N′Aij∂2∂xi∂xj+∑i=1N′bi∂∂xi,with σ:=A, A a positive matrix. Since A is smooth, its square root is locally
Lipschitz. Construct a time homogeneous Itô diffusion process X:Ω→W(M) which solves the following stochastic
differential equation:(1.2)dXs=b(Xs)ds+σ(Xs)dBs,s≥t;Xt=xin the ambient space ℝN′,
where Bs is N′-dimensional Euclidean-Brownian motion and b:ℝN′→ℝN′,σ:ℝN′→ℝN′×N′ such that(1.3)∥b(x)−b(y)∥+∥σ(x)−σ(y)∥≤D(R)|x−y|for some constant D(R) dependent on an open ball centered at x with radius R.
Till the explosion time e(X), Xs∈M for 0≤s<e(X).
On M, L˜=L.
Furthermore, μX:=ℙ∘X−1 is an L-diffusion measure on W(M).
As a result, we use μX to be the probability measure on W(M).
Refer to [1] for a
more detailed description.Fix some
positive integerN.
Let MN(ℝ) and 𝒮N(ℝ) be the spaces of N×N real-valued matrices and symmetric N×N real-valued matrices, respectively. Also let MN+(ℝ)⊆𝒮N(ℝ) be the space of nonnegative matrices. Suppose
that H:[0,∞)→MN+(ℝ) is locally Lipschitz with 0≤H(s)<H(t) if s<t, H(0)=0,
and Γ:M→MN+(ℝ) is locally Lipschitz. Also assume that supx∈M∥Γ(x)∥ is bounded, where ∥⋅∥ is the operator norm. Define MΓ(X) as the multiplication operator with Γ(X) and ϒT as the integral operator with kernel H(s⋀t),
that is, for any f∈L2([0,T]→ℝN),(1.4)(ϒTf)(s)=∫0TH(s⋀t)f(t)dt,where s⋀t is the minimum of s and t.
Note that under the assumptions on H, ϒT is a positive operator and is trace class
(see Proposition 3.1.).Consider the
following integral operator:KX,T:L2([0,T]→ℝN)→L2([0,T]→ℝN),(1.5)(KX,Tf)(s)=(ϒMΓ(X)f)(s)=∫0TH(s⋀t)Γ(X(t))f(t)dt.It is a fact that for any trace
class operator K,
if we let ∥K∥1 denote the trace of |K|,
then(1.6)∥KX,T∥1≤supx∈M∥Γ(x)∥∥ϒ∥1≤C∫0TtrH(s)ds,for some constant C.
Here, tr means taking the trace of a matrix. Thus KX,T is trace class. Therefore,(1.7)|det(I+KX,T)|≤exp(∥KX,T∥1)≤exp(C∫0TtrH(s)ds).Hence the Fredholm determinant
is bounded for each T.Letf,V be continuous bounded functions taking x∈M∪{∂M}↦ℝ.
Fix some number p≥0.
Define a function λ:[0,∞)×M→ℝ by(1.8)λ(T,x):=𝔼x[f(XT)e−∫0TV(Xs)ds(det(I+KX,T))p],where the expectation is taken
with respect to μX and the paths start from x.
Note that λ is finite for any x,T from the above discussion. Let h:[0,∞)×M×MN(ℝ)→MN(ℝ) such that(1.9)h(t,x,v)=(v−H(t))Γ(x)(v−H(t)).The main result is
as follows.Theorem 1.2.
Let(t,x,v)∈[0,∞)×M×𝒮N(ℝ).
Then(1.10)λ(T,x)=(eTH^f)(0,x,0),where(1.11)H^=L+∂∂t+∑i,j=1Nh(t,x,v)ij∂∂vij−V(x)+ptr(H(t)Γ(x)−vΓ(x)).Here, h(t,x,v)ij is the i,j component of the matrix h.Clearly,λ is not in the Feynman-Kac formula form using
the process X.
The idea is to construct a diffusion process Ws=(s,Xs,Zs)∈[0,∞)×M×𝒮N(ℝ) such that(1.12)det(I+KX,T)=exp(∫0TG(Ws)ds),and G is given by(1.13)G(t,x,v)=tr(H(t)Γ(x)−vΓ(x)).If we can achieve this, then our
result follows from a simple application of the Feynman-Kac formula. Proving
(1.12) requires the following 2 steps.First, we have
to prove an essential formula for the derivative oflogdet(I+zKX,T) with respect to T,
given by(1.14)ddTlogdet(I+zKX,T)=tr(zH(T)Γ(XT)−Zz,TΓ(XT)),where z is a complex number and Zz,T is some adapted process. For a precise
definition of Zz,T,
see (4.1), with KX,s replaced by zKX,s.
When z=1, Z1,T=ZT.
The goal is to show that the formula holds for z=1 by analytic continuation.Fix a timeT.
By making |z| small such that ∥zKX,T∥<1,
we can use the perturbation formula and apply it
to the determinant; see (2.3). Differentiating this equation with respect to T will give us (1.14). By analytic continuation,
we can extend the formula in some domain O⊆ℂ containing the origin, provided we avoid the
poles of the resolvent of zKX,s for all s≤T.
If 1∈O,
then (1.14) holds with z=1.
By integrating both sides and raising to the exponent, we will get (1.12). Note
that if ∥KX,T∥<1 for some time T,
(1.14) and hence (1.12) hold. The details are given in Sections 2 and 3.Now assume that
(1.14) holds with z=1.
The second step consists of constructing a diffusion process W from X by using a stochastic differential equation.
To do this, we differentiate ZT with respect to T and show that it satisfies the differential
equation(1.15)dZT=h(T,XT,ZT)=(ZT−H(T))Γ(XT)(ZT−H(T))dT,and hence WT=(T,XT,ZT) satisfies the following stochastic
differential equation:(1.16)dWT=(1,b(XT),h(WT))dT+(0,σ(XT),0)dBT,with explosion time e(W).
From this stochastic differential equation, it is clear that W is a diffusion process and by replacing the
Fredholm determinant by the formula in (1.12), λ(T,x) can be written as a Feynman-Kac form using
this process W.
However, if e(W)<T<e(X),
then (1.12) may fail to hold.The positivity
ofH and Γ are used to show that Z1,T=ZT exists for all time T and hence (1.14) holds at z=1.
This will also imply that e(W)=e(X).
In particular, only the positivity of H is required to show that KX,T is a trace class operator. To avoid e(W)<e(X),
we can restrict ourselves to small time T such that (2.24) holds true.We can weaken our assumptions onH and Γ by not insisting that they are symmetric
matrices. If we only assume that KX,T is trace class, then we can replace 𝒮N(ℝ) with MN(ℝ).
Under these weaker assumptions, we have the following result.Theorem 1.3.
Suppose
that, for a given locally LipschitzH and Γ, KX,T is trace class. Assume that there exists some
constant C such that(1.17)supx∈M∥Γ(x)∥≤C<∞,∥H(s)∥≤Cs,s≥0. Let (t,x,v)∈[0,∞)×M×MN(ℝ).
Then for all T<1/C2,(1.18)λ(T,x)=(eTH^f)(0,x,0),where(1.19)H^=L+∂∂t+∑i,j=1Nh(t,x,v)ij∂∂vij−V(x)+ptr(H(t)Γ(x)−vΓ(x)).From Lemma2.4, using the assumptions on H and Γ, ∥KX,T∥≤C2T.
If T<1/C2,
then the norm is less than 1. Hence (1.14) holds and thus (1.12) holds true.


## 2. Functional Analytic Tools

Notation 2.1.
Suppose thatK is an integral operator, acting on L2([0,T]→ℝN)↦L2([0,T]→ℝN).
We will write Kf to mean (2.1)(Kf)(s)=∫0TK(s,t)f(t)dt,where T<∞.
To distinguish the operator K from its kernel, we will write K(s,t) to refer to its kernel. This may be confusing,
but it is used to avoid too many symbols. In this
article, our integral operator is always trace class and the kernel is a
continuous N×N matrix-valued function. By abuse of notation, Kn(s,t) refers to the kernel of the integral operator Kn.Notation 2.2.
We will usetr to denote the trace of a matrix and Tr to denote taking the trace of a trace class
operator. ∥⋅∥ will denote the operator norm.It is well known that for a trace class operatorA and z∈ℂ, Trlog(I+zA) is a meromorphic function and has
singularities at points z such that −z−1∈σ(A).
Define the determinant det(I+zA),
given by(2.2)det(I+zA)=eTrlog(I+zA).However, this determinant, also
known as the Fredholm determinant of A,
is analytic in z because the singularities z such that −z−1∈σ(A) are removable; see [2, Lemma 16].We want to differentiate the functionlogdet(I+KT) with respect to T,
where we write KT to denote the dependence on the domain [0,T].
If the kernel of KT is given by K(s,t),
then for small z such that ∥zKT∥<1,
using the perturbation formula,(2.3)det(I+zKT)=exp(∑n=1∞(−1)n+1nTr(zKT)n).If we let r=∥KT∥<1,
then(2.4)∑n=1∞|(−1)n+1nTrKTn|≤∑n=1∞rn−1n∥KT∥1≤∥KT∥11−r.Thus the series in the exponent
converges absolutely.We will define
the resolventRT by KT(I+KT)−1.
Since we can write KT(I+KT)−1=KT−KT(I+KT)−1KT,
the kernel of RT can be written as(2.5)RT(s,t)=K(s,t)−KT(I+KT)−1KT(s,t).When we write KT(I+KT)−1KT(s,t),
we mean(2.6)KT(I+KT)−1KT(s,t)=∫0TK(s,u)((I+KT)−1K(⋅,t))(u)du.We will also write the resolvent
of the operator zKT, z∈ℂ as RT,z.
One more point to note is that we assume that K(⋅,0)=K(0,⋅)=0.Now the
operatorKs is an operator defined on different Hilbert
spaces L2[0,s].
Therefore, we will now think of our operator Ks as acting on L2[0,T],
defined as(2.7)(Ksf)(u)=∫0TK(u,v)χ[0,s](v)f(v)dv,where χ is the characteristic function. Hence now our
operator Ks has a kernel K(u,v)χ[0,s](v) dependent on the parameter s.
Note that our Ks is continuous in the u variable but is discontinuous at v=s.
Thus when we write Ks(u,s),
we mean(2.8)Ks(u,s):=limv↑sKs(u,v).Definition 2.3.
LetK(⋅,⋅) be a continuous matrix-valued function and
let  ∥K(⋅,⋅)∥ be the matrix norm of K(⋅,⋅).
Define CT to be the maximum value of ∥K(⋅,⋅)∥ on [0,T]×[0,T].The next lemma allows us to control the operator norm
of the operator by controlling the sup norm of the kernel.Lemma 2.4.
For0≤s<s′≤T,(2.9)∥Ks′−Ks∥≤CTT(s′−s).Proof.
Forf,g∈L2 and any s∈[0,T],(2.10)|〈(Ks−Ks′)f,g〉|≤∬0T∥g(u)Tχ[s,s′](v)K(u,v)f(v)∥dvdu≤∬0Tχ[s,s′](v)CT|g(u)||f(v)|dvdu≤CT(∫0T|g(u)|du)(∫0Tχ[s,s′](u)|f(u)|du)≤CT(∫0T|g(u)|2du⋅∫0T1du)1/2(∫0T|f(u)|2du⋅∫0Tχ[s,s′](u)du)1/2≤CTT(s′−s)∥f∥2∥g∥2.Hence,(2.11)∥Ks′−Ks∥≤CTT(s′−s)for all 0≤s<s′≤T.Lemma 2.5.
Fix az∈ℂ.
For any T such that |zCTT|<1 and if K(s,t) is continuous, then(2.12)ddT(logdet(I+zKT))=trRT,z(T,T).Proof.
Sincez is fixed, we will replace zKT by KT and hence assume that |CTT|<1.
Lemma 2.4 tells us that ∥KT∥<1 and thus (2.3) holds true. Taking log on both sides of (2.3), we have(2.13)logdet(I+KT)=∑n=1∞(−1)n+1nTrKTn.Now(2.14)TrKTn=∫0T⋯∫0Ttr∏i=1nK(si,si+1)ds1⋯dsn,where sn+1=s1.
Differentiate with respect to T and using the fundamental theorem of calculus,
we get(2.15)ddT∫0T⋯∫0Ttr∏i=1nK(si,si+1)ds1⋯dsn=ntr∫0T⋯∫0TK(T,s2)(∏i=2n−1K(si,si+1))K(sn,T)ds2⋯dsn=ntrKTn(T,T).Let CTT=α.
Thus,(2.16)∑n=1∞|trKTn(T,T)|≤N∑n=1∞∥KTn(T,T)∥≤N∑n=1∞CTnTn−1≤NCT∑n=0∞αn=NCT1−α<∞.Thus(2.17)ddT(logdet(I+KT))=∑n=1∞ddT((−1)n+1ntr∫0T⋯∫0T∏i=1nK(si,si+1)ds1⋯dsn)=tr∑n=1∞(−1)n+1KTn(T,T)=trKT(I+KT)−1(T,T)=trRT(T,T).Lemma 2.6.
LetK(⋅,⋅) be continuous. For all z∈ℂ such that −z−1∉σ(Ks) for all s∈[0,T],
then trR⋅,z(⋅,⋅) is continuous.Proof.
Fix az and write zKs as Ks.
By assumption, I+Ks is invertible for all s.
By Lemma 2.4, Ks→Ks0 as s→s0.
Note that(2.18)I+Ks=I+Ks0+Ks−Ks0=(I+Ks0)(I+(I+Ks0)−1(Ks−Ks0)),and if we let Gs=Ks−Ks0,
then(2.19)(I+Ks)−1=(I+(I+Kso)−1Gs)−1(I+Kso)−1.By the open mapping theorem,
because I+Kso is a surjective continuous map, it is an open
map. Therefore, its inverse is a bounded operator. Since Gs=Ks−Ks0→0,
thus (I+(I+Kso)−1Gs)−1→I and hence (I+Ks)−1 converges to (I+Ks0)−1 as s→s0.
This shows that (I+Ks)−1 is continuous in s.
Note that(2.20)Rs(s,s)=K(s,s)−Ks(I+Ks)−1K(s,s).Since K(s,s), Ks and (I+Ks)−1 are continuous in s,
hence Rs(s,s) is continuous in s.Lemma 2.7.
LetK(⋅,⋅) be continuous. If there exists an
open-connected set O containing 0 such that for z∈O, (I+zKs)−1 is analytic for all s∈[0,T],
then(2.21)logdet(I+zKT)=∫0TtrRs,z(s,s)dsfor all z∈O.Proof.
For allz∈O, I+zKs is invertible for all s∈[0,T] and hence trRs,z(s,s) is analytic in O.
Therefore, it follows that ∫0TtrRs,z(s,s)ds is analytic in O,
because(2.22)ddz∫0TtrRs,z(s,s)ds=∫0TddztrRs,z(s,s)ds.By Lemma 2.4, ∥Ks∥≤TCT for all s∈[0,T].
Thus if we choose U:={z||z|<1/(TCT)},
then U is an open set containing 0 and for z∈U, ∥zKs∥<1 for all s∈[0,T].
From Lemma 2.5, for z∈U,(2.23)logdet(I+zKT)=∫0TtrRs,z(s,s)ds.Since logdet(I+zKT) is also analytic in O and agrees with ∫0TtrRs,z(s,s)ds in U,
it follows that both functions are equal for all z∈O.The proof in the previous theorem gives us the existence
of a small neighbourhood containing 0 such that (2.21) holds. Hence we have the
following corollary.Corollary 2.8.
FixT>0 and CT=sups,t∈[0,T]∥K(s,t)∥.
There exists an open set UT:={z||z|<1/(TCT)} such that (2.21) holds for all z∈UT.Corollary 2.9.
LetO be an open-connected set as in Lemma 2.7 such
that 1∈O.
Then for s∈(0,T),(2.24)ddslogdet(I+Ks)=trRs(s,s).Proof.
The
corollary follows from differentiating (2.21). By Lemma 2.6, trRs(s,s) is continuous and hence the fundamental
theorem of calculus applies.


## 3. Fredholm Determinant

The kernel we are interested in isKX,T=H(s⋀t)Γ(X(t)),
for some process X.
More generally, the kernel we are interested in is of the form KT(s,t)=H(s⋀t)Λ(t) for some continuous matrix-valued Λ.
The Hilbert space is L2([0,T]→ℝN) for some positive number T.
Without any ambiguity, we will in future write this space as L2.
We will also use ∥⋅∥2 to denote the L2 norm.Proposition 3.1.
IfH is continuous, 0≤H(s)≤H(t) for any s≤t and Λ continuous, then ϒT as defined in Section 1 is a trace class
operator.To prove this result, we need the following theorem,
which is [3, Theorem 2.12].Theorem 3.2.
Letμ be a Baire measure on a locally compact space X.
Let K be a function on X×X which is continuous and Hermitian positive,
that is,(3.1)∑i,j=1Jzi¯zjK(xi,xj)≥0for any x1,…,xJ∈X, z1,…,zJ∈ℂJ and for any J.
Then K(x,x)≥0 for all x.
Suppose that, in addition,(3.2)∫K(x,x)dμ(x)<∞.Then there exists a unique trace
class integral operator A such that(3.3)(Af)(x)=∫K(x,y)f(y)dμ(y),∥A∥1=∫K(x,x)dμ.Proof of Proposition3.1.
LetX=[0,T] and μ be Lebesgue measure. Using Theorem 3.2, it
suffices to show that H(s⋀t) is Hermitian positive. Let z1,z2,…,zJ be any complex column vectors. Note that there
are N entries in each column and the entries are
complex valued. Let s1,…,sJ∈[0,T].
The proof is obtained using induction. Clearly,
when J=1,
it is trivial. Suppose it is true for all values from k=1,2,…,J−1.
By relabelling, we can assume that s1≤sk, k=2,…,J.
Hence s1⋀sk=s1 for any k.
Let 〈⋅,⋅〉 be the usual dot product. Then(3.4)∑j=1J〈H(s1⋀sj)zj,z1¯〉+∑j=1J〈H(sj⋀s1)z1,zj¯〉=∑j=1J〈H(s1)zj,z1¯〉+∑j=1J〈H(s1)z1,zj¯〉=〈∑j=1JH(s1)zj,∑j=1Jzj¯〉−∑i,j=2J〈H(s1)zj,zi¯〉.Therefore,(3.5)∑i,j=1J〈H(si⋀sj)zj,zi¯〉=〈∑j=1JH(s1)zj,∑j=1Jzj¯〉−∑i,j=2J〈H(s1)zj,zi¯〉+∑i,j=2J〈H(si⋀sj)zj,zi¯〉=〈∑j=1JH(s1)zj,∑j=1Jzj¯〉+∑i,j=2J〈(H(si⋀sj)−H(s1))zj,zi¯〉.Since si⋀sj≥s1,i=2,…,J and hence H(si⋀sj)≥H(s1) by assumption on H.
Thus by induction hypothesis, (replace H by H(⋅)−H(s1)),(3.6)∑i,j=2J〈(H(si⋀sj)−H(s1))zj,zi¯〉≥0and hence(3.7)∑i,j=1J〈H(si⋀sj)zj,zi¯〉≥0.Notation 3.3.
By
abuse of notation,〈⋅,⋅〉T will denote integration over [0,T],
(3.8)〈f,g〉T:=∫0Tf(u)⋅g(u)du, where ⋅ should be interpreted as matrix multiplication
or inner product, depending on the context. To ease the notation, we will write
in future K(T):=K(T,T).Remark 3.4.
If we assume thatH(s)−H(t) is strictly positive if s>t,
which is the case we are interested in this article, then the proof of
Proposition 3.1 shows that the operator ϒT with kernel H(s⋀t) is strictly positive, that is, 〈ϒTf,f〉T>0 if 〈f,f〉T>0.
This follows using a Riemann lower sum approximation on a double integral and
that for any complex vectors z1,…,zJ,(3.9)∑i,j=1J〈H(si⋀sj)zj,zi¯〉T2J2>0under the strict positivity
assumptions.The next proposition is a crucial statement. For the
time being, we will assume that(I+KT)−1 exists for any time T without any further justification. Later on,
we will prove that for our operator KX,T,
this is true; (see Proposition 5.2.). Writing in our new notation, we obtain
the next proposition from Corollary 2.9.Proposition 3.5.
LetKT be an integral operator with kernel KT(s,t)=H(s⋀t)Λ(t) for some continuous matrix-valued Λ:(3.10)ddT(logdet(I+KT))=tr[K(T)−〈HΛ,(I+KT)−1H〉TΛ(T)].Proof.
We will
use (2.5). So(3.11)RT(T,T)=K(T,T)−〈K(T,⋅),(I+KT)−1K(⋅,T)〉T=K(T,T)−〈H(⋅)Λ(⋅),(I+KT)−1H(⋅)Λ(T)〉T=K(T)−〈HΛ,(I+KT)−1H〉TΛ(T).Taking trace completes the
proof.Definition 3.6.
Let(3.12)ZT=〈HΛ,(I+KT)−1H〉T.To ease the notation, we will now writeL:=(I+KT)−1 and L(s,t) to be the kernel of L.
Note that in future we will drop the subscript T from the operator K and it should be understood that K is dependent on T.
Operators with a prime will denote its derivative with respect to T.
Our task now is to differentiate ZT.Define a
distributional kernel(3.13)ρ(s,t)=R(s,t)−δ(s−t),where δ is the Dirac delta function and R is the resolvent.For any
operator depending smoothly on some parameterT,
we have the differentiation formula(3.14)LT′=ddT(I+KT)−1=−(I+KT)−1KT′(I+KT)−1.For the integral operator K′,
its kernel is given, by the fundamental theorem of calculus, by(3.15)K′(s,t)=K(s,T)δ(t−T);and hence combining with (3.14),
we have(3.16)L′(s,t)=−LK′L(s,t)=−LK′(s,t)−〈LK(s,T)δ(⋅−T),−KL(⋅,t)〉T=−LK(s,T)δ(t−T)+LK(s,T)〈δ(⋅−T),KL(⋅,t)〉T=−R(s,T)δ(t−T)+R(s,T)R(T,t)=R(s,T)ρ(T,t).Notation 3.7.
LetK be an integral operator with kernel K(s,t).
We define the adjoint K∗ by (3.17)(K∗f)(t)=∫0Tf(s)K(s,t)dt.Here, f is an N×N matrix-valued function. We will also write ΛT=Λ(T) and HT=H(T).The following lemma defines the relationship betweenR and L.Lemma 3.8.
It holds that(3.18)R(s,T)=(LH)(s)ΛT,R(T,t)=(L∗HΛ)(t).Proof.
We write the identity operatorI as If(s)=f(s)=∫0Tδ(s−t)f(t)dt,
where δ is Dirac delta function. Then(3.19)R(s,T)=〈L(s,⋅),K(⋅,T)〉T=〈L(s,⋅),H(⋅⋀T)ΛT〉T=〈L(s,⋅),H(⋅)ΛT〉T=(LH)(s)ΛT,R(T,t)=〈K(T,⋅),L(⋅,t)〉T=〈H(T⋀⋅)Λ(⋅),L(⋅,t)〉T=(L∗HΛ)(t).Theorem 3.9.
ZT satisfies the following differential equation:(3.20)ZT′=(ZT−HT)ΛT(ZT−HT).Proof.
Now by definition ofZT,(3.21)ZT=〈HΛ,(I+KT)−1H〉T=〈HΛ,LH〉T=〈L∗HΛ,H〉T.Using (3.16) and from Lemma 3.8,
(3.22)(L′H)(s)=〈R(s,T)ρ(T,⋅),H(⋅)〉T=〈R(s,T)R(T,⋅),H(⋅)〉T−R(s,T)HT=(LH)(s)ΛT〈L∗HΛ,H〉−(LH)(s)ΛTHT=(LH)(s)ΛTZT−(LH)(s)ΛTHT.Hence differentiating with
respect to T,
using the fundamental theorem and (3.22), gives(3.23)ddTZT=(HΛLH)(T)+〈HΛ,L′H〉T=(HΛLH)(T)+〈(HΛ)(⋅),(LH)(⋅)ΛTZT−(LH)(⋅)ΛTHT〉T=(HΛLH)(T)−ZTΛTHT+ZTΛTZT.But(3.24)(KLH)(T)=(K(I+K)−1H)(T)=〈H(T⋀⋅)Λ(⋅),(LH)(⋅)〉T=〈HΛ,LH〉T=ZT.Hence(3.25)(HΛLH)(T)=(HΛH)(T)−(HΛKLH)(T)=(HΛH)(T)−(HΛ)(T)ZT.Therefore(3.26)ddTZT=(HΛH)(T)−(HΛ)(T)ZT−ZTΛTHT+ZTΛTZT=(ZT−HT)ΛT(ZT−HT).This completes the proof.


## 4. Integral Operator Driven by a Diffusion Process

Now back to the
integral operatorKX,T defined in Section 1. Define a process Zs:Ω→MN(ℝ),(4.1)Zs=〈HΓX,((I+KX,s)−1H)〉s,where ΓX:=Γ(X) (see Notation 3.3 for the definition of the
angle brackets). From the definition, it is clear that Zs is ℱ∗ adapted. In fact, Zs is a symmetric matrix under the usual
assumptions on H and Γ.Proposition 4.1.
IfH and Γ are symmetric matrices, then Z is symmetric as a matrix.Proof.
Sinces is fixed, we will drop the subscript s.
Also fix an ω∈Ω,
so we will also drop the subscript X.
Let K∗ be the adjoint of K with kernel Γ(s)H(s⋀t).
By assumption of symmetry and by definition,(4.2)ZT=〈H((I+K∗)−1),ΓH〉=〈HΓ,H〉−〈HK∗((I+K∗)−1),ΓH〉=〈HΓ,H〉−〈HΓK((I+K)−1),H〉=〈HΓ((I+K)−1),H〉=〈HΓ,((I+K)−1)H〉=Z.Theorem 4.2.
LetXs be an L-diffusion process satisfying (1.2) and H:[0,∞)→MN+(ℝ) and let Γ:M→MN+(ℝ) be continuous. Further assume that H(s)≥H(t)≥0 for s≥t.
Let KX,s be an integral operator defined by (1.5) and(4.3)Zs=〈HΓX,((I+KX,s)−1H)〉s.Let e(Z) be the explosion time of Z.
Then for s<e(Z), Ws=(s,Xs,Zs):Ω→[0,∞)×M×𝒮N(ℝ) satisfies the following stochastic
differential equation:(4.4)dWs=(1,b(Xs),h(Ws))ds+(0,σ(Xs),0)dBs,s≥t;Wt=(t,x,v),where(4.5)h(t,x,v)=(v−H(t))Γ(x)(v−H(t)).Proof.
In the ambient spaceℝN′, (s,Xs) is a diffusion satisfying the stochastic
differential equation of the form(4.6)d(s,Xs)=(1,b(Xs))ds+(0,σ(Xs))dBs.Now by Theorem 3.9, Zs satisfies the following differential equation:(4.7)dZs=h(s,Xs,Zs)ds,where h is defined by (4.5) and by Proposition 4.1, Zs∈𝒮N(ℝ).
Thus we can write(4.8)d(s,Xs,Zs)=(1,b(Xs),h(s,Xs,Zs))ds+(0,σ(Xs),0)dBs,which is (4.4). The existence of Zs for small time is guaranteed by Lemma 2.4.Lemma 4.3.
IfΓ is locally Lipschitz, then Ws is the unique solution (path-wise) to (4.4).Proof.
Now by the definition ofXs, b and σ are locally Lipschitz. However, since Γ is locally Lipschitz on the manifold M with bounded operator norm, it follows that h is locally Lipschitz. Therefore (4.4) has a
unique solution and is given by Ws.


## 5. Long-Time Existence ofWs

We had addressed the existence and uniqueness of the
solution to (4.5), given by Ws=(s,Xs,Zs),
with e(W)≤e(X).
We will now give sufficient conditions for e(W)=e(X).Proposition 5.1.
Suppose that the integral operatorϒT with kernel H(s⋀t) is a strictly positive operator and Γ is a symmetric nonnegative matrix. Then for z∈ℂ such that Re(z)≥0, (I+zKX,T)−1 exists for all T<e(X).Proof.
Whenz=0 is trivial. So assume z≠0.
Fix an ω∈Ω and any T<e(X(ω)).
Since KX=KX,T is a compact operator, it suffices to show
that the kernel of I+zKX is 0. Write Γ(X(ω))=Γ and KX(ω)=K.
Let 0≠v∈L2 such that 〈v,v〉>0 and Γv=0.
Then (I+zKX)v=v is nonzero. Hence we can assume that 〈Γv,Γv〉>0.
Note that K=ϒMΓ and (MΓ+MΓϒMΓ) is a symmetric operator (see Section 1 for
definitions of ϒ and MΓ.)
Therefore,(5.1)〈(MΓ+zMΓϒMΓ)v,v〉=〈MΓv,v〉+z〈MΓϒMΓv,v〉=〈Γv,v〉+z〈ϒΓv,Γv〉.Since Γ is a nonnegative matrix, 〈Γv,v〉≥0,
and because ϒ is a strictly positive operator, 〈ϒΓv,Γv〉>0.
If Re(z)>0,(5.2)Re(〈MΓv,v〉+z〈ϒMΓv,MΓv〉)>0.Otherwise, Im(z)≠0 and hence we have(5.3)Im(z)〈ϒMΓv,MΓv〉≠0.Either way, if Re(z)≥0,(5.4)(MΓ+zMΓϒMΓ)v=MΓ(I+zϒMΓ)vis nonzero, and therefore (I+zϒMΓ)v is nonzero. Thus for any nonzero L2[0,T] function v, (I+zϒMΓ)v is never zero and since ω is arbitrary, hence I+zKX is invertible for any ω∈Ω.Proposition 5.2.
Suppose that the usual assumptions onΓ and H hold. Then (I+KX,s)−1 exists for all 0≤s<e(X).
Furthermore, (3.10) holds for all 0≤s<e(X).Proof.
Under the assumptions onΓ and H,
Proposition 3.1 and Remark 3.4 will imply Proposition 5.1. Fix an ω∈Ω,
a T<e(X(ω)) and let CT be as defined in Lemma 2.4. Then on U={z∣|z|<1/CT}, (I+zKX,s) is invertible for all s∈[0,T].
Hence O=U∪{z∣Re(z)>0} is an open-connected set containing 0, and (I+zKX,s)−1 exists for all s∈[0,T].
In particular, at z=1.
Then the assumptions in Corollary 2.9 are met, and hence (3.10) holds.


## 6. Proof of Main Result

The proof of Theorem1.2 now follows from Theorem 4.2
and Proposition 5.2. Integrating (3.10), we have for T<e(X),(6.1)logdet(I+KX,T)=∫0Ttr(H(s)Γ(Xs)ds−ZsΓ(Xs))ds.For (t,x,v)∈[0,∞)×M×𝒮N(ℝ),
define(6.2)Ψ(T,t,x,v)=𝔼(t,x,v)[f(XT)exp(−∫0TV(Xs)−ptr(H(s)Γ(Xs)−ZsΓ(Xs))ds)]and observe that(6.3)Ψ(T,0,x,0)=𝔼x[f(XT)e−∫0TV(Xs)ds(det(I+KX,T))p]=λ(T,x).By the Feynman-Kac formula, Ψ satisfies the following partial differential
equation:(6.4)∂∂TΨ(T,t,x,v)=H^Ψ(T,s,x,v).Thus(6.5)Ψ(T,t,x,v)=(eTH^f)(t,x,v),and therefore from
(6.3),(6.6)λ(T,x)=eTH^f(0,x,0).



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*Source: 130940-2008-11-02.xml*