doi:10.1155/2008/130940 Research Article Fredholm Determinant of an Integral Operator Driven by a Diffusion Process

This article aims to give a formula for differentiating, with respect to 𝑇, an expression of the form 𝜆(𝑇,𝑥)∶=𝔼𝑥[𝑓(𝑋𝑇)𝑒−∫𝑇0𝑉(𝑋𝑠)𝑑𝑠(det(𝐼


Introduction
Suppose we have a differentiable manifold M of dimension d.By the Whitney's embedding theorem, there exists an embedding i : M → R N such that i M is a closed subset of R N .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 R N .We will also assume that M does not have a boundary.Let M ∪ {∂ M } be a one-point compactification of M. Definition 1.1.An M-valued path ω with explosion time e e ω > 0 is a continuous map ω : 0, ∞ → M ∪ {∂ M } such that ω t ∈ M for 0 ≤ t < e and ω t ∂ 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 {x 1 , . . ., x N }, and extending L smoothly to L in the ambient space, we may write with σ : √ A, A a positive matrix.Since A is smooth, its square root is locally Lipschitz.Construct a time homogeneous It ô diffusion process X : Ω → W M which solves the following stochastic differential equation: dX s b X s ds σ X s dB s , s ≥ t; X t x 1.2 in the ambient space R N , where B s is N -dimensional Euclidean-Brownian motion and b : for some constant D R dependent on an open ball centered at x with radius R. Till the explosion time e X , X s ∈ M for 0 ≤ s < e X .On M, L L. Furthermore, μ X : P • X −1 is an L-diffusion measure on W M .As a result, we use μ X to be the probability measure on W M .Refer to 1 for a more detailed description.Fix some positive integer N. Let M N R and S N R be the spaces of N ×N real-valued matrices and symmetric N × N real-valued matrices, respectively.Also let M N R ⊆ S N R be the space of nonnegative matrices.Suppose that H : 0, ∞ → M N R is locally Lipschitz with 0 ≤ H s < H t if s < t, H 0 0, and Γ : M → M N R is locally Lipschitz.Also assume that sup x∈M Γ x is bounded, where • is the operator norm.Define M Γ X as the multiplication operator with Γ X and Υ T as the integral operator with kernel where s ∧ t is the minimum of s and t.Note that under the assumptions on H, Υ T is a positive operator and is trace class see Proposition 3.1. .Consider the following integral operator: K X,T : It is a fact that for any trace class operator K, if we let K 1 denote the trace of |K|, then for some constant C. Here, tr means taking the trace of a matrix.Thus K X,T is trace class.Therefore, Hence the Fredholm determinant is bounded for each T .
Let f, V be continuous bounded functions taking where the expectation is taken with respect to μ X and the paths start from x.Note that λ is finite for any x, T from the above discussion.
The main result is as follows. where

1.11
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 12 and G is given by 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 of log det I zK X,T with respect to T , given by where z is a complex number and Z z,T is some adapted process.For a precise definition of Z z,T , see 4.1 , with K X,s replaced by zK X,s .When z 1, Z 1,T Z T .The goal is to show that the formula holds for z 1 by analytic continuation.
Fix a time T .By making |z| small such that zK X,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 zK X,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 K X,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 Z T with respect to T and show that it satisfies the differential equation and hence W T T, X T , Z T satisfies the following stochastic differential equation: 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 of H and Γ are used to show that Z 1,T Z T 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 K X,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 on H and Γ by not insisting that they are symmetric matrices.If we only assume that K X,T is trace class, then we can replace S N R with M N R .Under these weaker assumptions, we have the following result.
Theorem 1.3.Suppose that, for a given locally Lipschitz H and Γ, K X,T is trace class.Assume that there exists some constant C such that where From Lemma 2.4, using the assumptions on H and Γ, K X,T ≤ C 2 T .If T < 1/C 2 , then the norm is less than 1.Hence 1.14 holds and thus 1.12 holds true.

Functional analytic tools
Notation 2.1.Suppose that K is an integral operator, acting on 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, K n s, t refers to the kernel of the integral operator K n .
Notation 2.2.We will use tr 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 operator A and z ∈ C, Tr log 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 det I zA e Tr log I zA .

2.2
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 function log det I K T with respect to T , where we write K T to denote the dependence on the domain 0, T .If the kernel of K T is given by K s, t , then for small z such that zK T < 1, using the perturbation formula, Thus the series in the exponent converges absolutely.We will define the resolvent R T by

6 Journal of Applied Mathematics and Stochastic Analysis
When we write K T I K T −1 K T s, t , we mean We will also write the resolvent of the operator zK T , z ∈ C as R T,z .One more point to note is that we assume that K •, 0 K 0, • 0. Now the operator K s is an operator defined on different Hilbert spaces L 2 0, s .Therefore, we will now think of our operator K s as acting on L 2 0, T , defined as where χ is the characteristic function.Hence now our operator K s has a kernel K u, v χ 0,s v dependent on the parameter s.Note that our K s is continuous in the u variable but is discontinuous at v s.Thus when we write K s u, s , we mean The next lemma allows us to control the operator norm of the operator by controlling the sup norm of the kernel.Lemma 2.4.For 0 ≤ s < s ≤ T , 2.9 Proof.For f, g ∈ L 2 and any s ∈ 0, T ,

2.10
Hence, where s n 1 s 1 .Differentiate with respect to T and using the fundamental theorem of calculus, we get T T, T .

2.15
Let C T T α.Thus,

2.16
Thus tr R T T, T .
Proof.Fix a z and write zK s as K s .By assumption, I K s is invertible for all s.By Lemma 2.4, K s → K s 0 as s → s 0 .Note that and if we let G s K s − K s 0 , then

2.19
By the open mapping theorem, because I K s o is a surjective continuous map, it is an open map.Therefore, its inverse is a bounded operator.Since G s K s − K s 0 → 0, thus

2.24
Proof.The corollary follows from differentiating 2.21 .By Lemma 2.6, tr R s s, s is continuous and hence the fundamental theorem of calculus applies.

Fredholm determinant
The kernel we are interested in is K X,T H s ∧ t Γ X t , for some process X.More generally, the kernel we are interested in is of the form K T s, t H s ∧ t Λ t for some continuous matrix-valued Λ.The Hilbert space is L 2 0, T → R N for some positive number T .Without any ambiguity, we will in future write this space as L 2 .We will also use • 2 to denote the L 2 norm.Proposition 3.1.If H is continuous, 0 ≤ H s ≤ H t for any s ≤ t and Λ continuous, then Υ T 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.Let μ 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, for any x 1 , . . ., x J ∈ X, z 1 , . . ., z J ∈ C J and for any J.Then K x, x ≥ 0 for all x.Suppose that, in addition, K x, x dμ x < ∞.

Journal of Applied Mathematics and Stochastic Analysis
Then there exists a unique trace class integral operator A such that Af x K x, y f y dμ y , A 1 K x, x dμ.

3.3
Proof of Proposition 3.1.Let X 0, T and μ be Lebesgue measure.Using Theorem 3.2, it suffices to show that H s ∧ t is Hermitian positive.Let z 1 , z 2 , . . ., z J be any complex column vectors.Note that there are N entries in each column and the entries are complex valued.Let s 1 , . . ., s J ∈ 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 s 1 ≤ s k , k 2, . . ., J. Hence s 1 ∧ s k s 1 for any k.Let •, • be the usual dot product.Then

3.5
Since s i ∧ s j ≥ s 1 , i 2, . . ., J and hence H s i ∧ s j ≥ H s 1 by assumption on H. Thus by induction hypothesis, replace H by H s i ∧ s j z j , z i ≥ 0. 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 that H 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 This follows using a Riemann lower sum approximation on a double integral and that for any complex vectors z 1 , . . ., z J , under the strict positivity assumptions.
The next proposition is a crucial statement.For the time being, we will assume that I K T −1 exists for any time T without any further justification.Later on, we will prove that for our operator K X,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.Let K T be an integral operator with kernel K T s, t H s ∧ t Λ t for some continuous matrix-valued Λ:

3.11
Taking trace completes the proof.

3.12
To ease the notation, we will now write L : I K T −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 Z T .Define a distributional kernel where δ is the Dirac delta function and R is the resolvent.
For any operator depending smoothly on some parameter T , we have the differentiation formula

3.14
For the integral operator K , its kernel is given, by the fundamental theorem of calculus, by K s, t K s, T δ t − T ; 3.15 and hence combining with 3.14 , we have R s, T ρ T, t .

3.16
Notation 3.7.Let K be an integral operator with kernel K s, t .We define the adjoint K * by Here, f is an N × N matrix-valued function.We will also write Λ T Λ T and H T H T .
The following lemma defines the relationship between R and L.

3.22
Hence differentiating with respect to T , using the fundamental theorem and 3.22 , gives

3.26
This completes the proof.

Integral operator driven by a diffusion process
Now back to the integral operator K X,T defined in Section 1. Define a process Z s : where Γ X : Γ X see Notation 3.3 for the definition of the angle brackets .From the definition, it is clear that Z s is F * adapted.In fact, Z s is a symmetric matrix under the usual assumptions on H and Γ.
Proposition 4.1.If H and Γ are symmetric matrices, then Z is symmetric as a matrix.
Proof.Since s 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, Proof.Now by the definition of X s , 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 W s .

Long-time existence of W s
We had addressed the existence and uniqueness of the solution to 4.5 , given by W s s, X s , Z s , with e W ≤ e X .We will now give sufficient conditions for e W e X .
Proposition 5.1.Suppose that the integral operator Υ T 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 zK X,T −1 exists for all T < e X .
Proof.When z 0 is trivial.So assume z / 0. Fix an ω ∈ Ω and any T < e X ω .Since K X K X,T is a compact operator, it suffices to show that the kernel of I zK X is 0. Write Γ X ω Γ and K X ω K. Let 0 / v ∈ L 2 such that v, v > 0 and Γv 0. Then I zK X v v is nonzero.Hence we can assume that Γv, Γv > 0. Note that K ΥM Γ and M Γ M Γ ΥM Γ is a symmetric operator see Section 1 for definitions of Υ and M Γ .Therefore, Γv, v z ΥΓv, Γv .

5.1
Since Γ is a nonnegative matrix, Γv, v ≥ 0, and because Υ is a strictly positive operator, ΥΓv, Γv > 0. If Re z > 0, Otherwise, Im z / 0 and hence we have Either way, if Re z ≥ 0, is nonzero, and therefore I zΥM Γ v is nonzero.Thus for any nonzero L 2 0, T function v, I zΥM Γ v is never zero and since ω is arbitrary, hence I zK X is invertible for any ω ∈ Ω.
Proposition 5.2.Suppose that the usual assumptions on Γ and H hold. Then I K X,s −1 exists for all 0 ≤ s < e X .Furthermore, 3.10 holds for all 0 ≤ s < e X .
Proof.Under the assumptions on Γ and H, Proposition is an open-connected set containing 0, and I zK X,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.

Proof of main result
The proof of Theorem

3 . 1 and
Remark 3.4 will imply Proposition 5.1.Fix an ω ∈ Ω, a T < e X ω and let C T be as defined in Lemma 2.4.Then on U {z | |z| < 1/C T }, I zK X,s is invertible for all s ∈ 0, T .Hence O U ∪ {z | Re z > 0} Since z is fixed, we will replace zK T by K T and hence assume that |C T T | < 1. Lemma 2.4 tells us that K T < 1 and thus 2.3 holds true.Taking log on both sides of 2.3 , we have 11for all 0 ≤ s < s ≤ T .Lemma 2.5.Fix a z ∈ C. For any T such that |zC T T | < 1 and if K s, t is continuous, then K s ≤ TC T for all s ∈ 0, T .Thus if we choose U : {z | |z| < 1/ TC T }, then U is an open set containing 0 and for z ∈ U, zK s < 1 for all s ∈ 0, T .From Lemma 2.5, for z ∈ U,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.Fix T > 0 and C T sup s,t∈ 0,T K s, t .There exists an open set U T : {z | |z| < 1/ TC T } such that 2.21 holds for all z ∈ U T .Let O be an open-connected set as in Lemma 2.7 such that 1 ∈ O. Then for s ∈ 0, T , s .2.20 Since K s, s , K s and I K s −1 are continuous in s, hence R s s, s is continuous in s.Lemma 2.7.Let K •, • be continuous.If there exists an open-connected set O containing 0 such that for z ∈ O, I zK s −1 is analytic for all s ∈ 0, T , then 1.2 now follows from Theorem 4.2 and Proposition 5.2.Integrating 3.10 , we have for T < e X , By the Feynman-Kac formula, Ψ satisfies the following partial differential equation: