Thus far we have a sequence of processes {Xn, Pxn} with state space Vnand we intend to take a weak limit as n → ∞ to get a process on V . We use the following conven-tions, identify points on ∂aV0 as ∆ and f (∆) = 0 for all functions f . For x ∈ V0− Vn let Pxn be the probability measure for the process that is standard Brownian motion until the time t where it hits Vn and then behave like Xtn after that time. The proofs in this section are from [5], and require no modification.
If τn= inf{t > 0 : Xt ∈ Vn}, Qx is standard Weiner measure on paths in R2, A ∈ Fτn, then define
Pxn(A ∩ (B ◦ θτn)) = ExQ(PXτn(B); A). (2.6.1)
In this way we have defined Pxn on F∞. Now extend Un to all of V0.
Theorem 2.6.1. Suppose f : V0 → R is bounded. Then there exists ω(δ) that tends to 0 as δ → 0 such that
sup
n
sup
x,y∈V0
|x−y|<δ
|Unf (x) − Unf (y)| ≤ ω(δ) kf k∞ (2.6.2)
Proof
This is Theorem 6.1 from [5]. If , η > 0 and B is a 4N -gon of side r and τB = inf{t : Xt6∈ B}, then there exists δ > 0(independent of r) such that
ExQτB < and Qx(|XτB − x| > η) < (2.6.3)
whenever x ∈ B and d(x, Bc) < δ by the arguments of [87] in chapter 2 section 3.
Let > 0. We wish to show that for each x0 there exists a δ(x0) such that
|Unf (x) − Unf (x0)| ≤ 4 kf k∞ (2.6.4)
whenever |x − x0| < δ. Then the result follows by the compactness of V0.
Suppose x0 = ∆. Choose η small enough that Exτ < if x ∈ Vn and d(x, ∆) < 2η.
This choice can be made independently of n by Theorem 2.5.4. Now choose δ ∈ (0, η) small enough that 2.6.3 is satisfied. If d(x, ∆) < δ and x ∈ Vn, then Exnτ < . On the other hand if x 6∈ Vn then
Exnτ ≤ ExQτn+ ExQEXntnτ (2.6.5)
+ sup
x∈Vn
Exnτ + sup
x∈Vn
d(x,∆)<η+δ
Exnτ ≤ 3. (2.6.6)
Then using 2.5.3 and the definition of ∆, we have
|Unf (x) − Unf (x0)| = |Unf (x)| ≤ kf k∞Exnτ ≤ 3 kf k∞. (2.6.7)
Suppose x0 6∈ V , then there exists m and η such that Bη(x0)∩Vn= ∅ when n ≥ m.
Write ρ = inf{t : X6∈ Bη(x0)}. Choose η to be small enough that supx∈Bη(x0)ExQρ < .
Let g be a bounded function, then ExQg(Xρ) is harmonic in x in the interior of Bη(x0).
Furthermore there exists δ < η/2 such that
|ExQg(Xρ) − ExQ0g(Xρ) ≤ kgk∞ (2.6.8)
if |x − x0| < δ. For x ∈ Bη(x0) we have
Unf (x) = ExQ
Z ρ 0
f (Xs)ds + ExQU nf (Xρ) (2.6.9)
From this we see that
|Unf (x) − Unf (x0)| ≤ 2 kf k∞ sup
x∈Bη(x0)ExQρ + |ExQUnf (Xρ) − ExQ0Unf (Xp)| (2.6.10)
≤ 2 kf k∞+ kUnf k∞ (2.6.11)
≤ 3 kf k∞. (2.6.12)
Now, suppose x0 ∈ V . If we pick η so that if |x − x0| < 2η, x ∈ Vn, then
|Unf (x) − Unf (x0)| < kf k∞. (2.6.13)
By theorem 2.5.5 this choice is independent of n. Now pick δ < η to satisfy 2.6.3. If x ∈ Vn then 2.6.4 is true, and if x 6∈ Vn and |x − x0| < δ, then
|Unf (x) − Unf (x0)| ≤ |ExQ
Z τn
0
f (Xs)ds| + ExQUnf (Xτn) − Unf (x0)| (2.6.14)
≤ kf k∞ExQτn+ 2 kUnf k∞PxQ(|Xτn− x| ≥ η) + sup
y∈Vn
|y−x0|<2η
|Unf (y) − Unf (x)|
(2.6.15)
≤ 4 kf k∞. (2.6.16)
The tightness estimate for Pxn when x 6∈ Vn is routine and thus omitted.
Definition 2.6.2. We define the λ resolvent of U for bounded f and by
Using [21] V. 5.10, we have the identity
Unλf =
This argument is due to Barlow and Bass in [5] Theorem 6.2. Assume that λ ≤ 1/2.
By theorem 2.5.5,
Let > 0. Choose i0 so that P∞
Combining 2.6.19 and 2.6.22 with β = 0 gives
sup
Theorem 2.6.4. There is a subsequence nj for which Unλ
jf converges uniformly for
This is Proposition 6.3 from [5]. Let {fm}m be a sequence of continuous function on V0 such that kfmk ≤ 1 and the closure of the linear span of {fm} consists of all continuous functions on V0. Let λi be a countable dense subset of [0, ∞). Fix i and m. By the Arzela-Ascoli theorem there exists a subsequence of Unλifm converges uniformly. By using a diagonalization argument we can choose a subsequence nj so that Unλi
jfm converges uniformly for each i and m, and we call this limit Uλif . Note that each Unλi satisfies the inequality
Unλi
∞ ≤ λ−1 and thus Uλi ≤ λ−1. We may then conclude that Unλif converges uniformly and we call the limit Uλif for f
continuous on V0. Each Unλ satisfies the resolvent equation
Unλ− Unβ = (β − λ)UnλUnβ. (2.6.25)
A limiting argument shows that Uλ also does if λ = λi for some i. Further, assuming that β, λ ∈ {λi} then by 2.6.25
This next result comes from Barlow and Bass in [5], see Lemma 6.4.
Theorem 2.6.6. Suppose xj → x so that {xj}, {x} ∈ V0. Then {Pxnjj} converges weakly and we call the limit Px.
Proof
By Theorem 2.5.2 and the remark following 2.6.1 it suffices to show that any two limit points agree. Let f be a continuous function on V0. If a subsequence of Pxnjj
converges weakly to a limit which we call P0 then
Unλ
However, by the equicontinuity of Unλjf , we have Unλjf (xj) → Uλf (x). If P00 is limit point of a subsequence of Pxnjj we have
E0 Z ∞
0
eλsf (Xs)ds = Uλf (X) = E00 Z ∞
0
e−λsf (Xs)ds. (2.6.28)
By the uniqueness of the Laplace transform we have
E0f (Xs) = E00f (Xs) (2.6.29)
for almost every s. Since f is continuous and Xs is continuous a.s. under both P0 and P00, we have equality for all s. A standard limiting argument gives 2.6.21. for bounded f , and it follows that the one-dimensional distributions of Xt are equal under both P0 and P00.
Write E0f (Xs) by Psf (x). We have Pnsjf (xj) → Psf (x) for xj → x. By 2.6.5 Psf is continuous on V0 thus the convergence is uniform. Let s < t be times, and f and g be continuous functions. Using the Markov property of Pxnj
Exnjjg(Xs)f (Xt) = Exnjj((Pnt−sj f )g)(Xs) (2.6.30)
= Exnjj((Pt−sf )g)(Xs) + Exnjj((Pt−sf − Pnt−sj f )g)(Xs). (2.6.31)
The first term converges to E0((Pt−sf )g)(Xs) by argument above. The second term tends to 0 by by the uniform convergence of Pnt−sj f . Repeating the argument shows that the finite dimensional distributions under Pxnjj converge. Tightness gives the result.
Corollary 2.6.7. If f is continuous, Ptf is continuous.
Theorem 2.6.8. {Px} is a strong Markov family of probability measures.
Proof
This proof is due to Barlow and Bass in [5] Proposition 6.7. For each n, we have PntPns = Pnt+s since {Pxn} is a Markov process. If f is continuous then by 2.6.7 Pnsjf → Psf uniformly and P0f is continuous, Pnt+sj f → Pt+sf , so we have
PntjPnsjf − PtPsf
∞ ≤ kPnsjf − Psf k∞+
PntjPsf − PtPsf
∞ → 0 (2.6.32)
as j → ∞. This gives the result for functions, and a limiting argument gives the result for all bounded and measurable functions f .
Because Pxnj are tight and Pxis the weak limit, then Px( The Paths of Xt are continuous) = 1 and and Px(X0 = x) = 1. Thus Ptf (x) = Exf (Xt) → f (x) as t → 0 if f is
continu-ous. By if f is continuous then by 2.6.7 Ptf is continuous. By the proof of [21] Th.
I. 9.4 {Xt, Px} is a strong Markov Process.
Theorem 2.6.9. The process Xt is nowhere degenerate or more precisely
Px(Xt= x for all t) 6= 1. (2.6.33)
Proof
The ideas for this proof come from [5]. Let τ = inf{t : Xt 6∈ [0, 1 − ]2}. Write
Xs(i), i = 1, 2 for the x(i) coordinates of Xs. Then for any x and all t and we have
This next result is Proposition 6.9 in [5] and we include it as it is an important step toward the study of the Heat Kernel associated to the process that we have
created.
Theorem 2.6.11. If f is bounded, then Uλf is continuous on V . Proof
This result, also due to Barlow and Bass is Proposition 6.9 of [5]. Suppose f is continuous. Then
Since the bound ω(δ) kf k∞ does not depend on the modulus of continuity of f a limiting argument shows that we have 2.6.44.