Derivatives of Self-Intersection Local Times

Derivatives of self-intersection local times

Jay Rosen?

Department of Mathematics College of Staten Island, CUNY Staten Island, NY 10314 e-mail: [email protected]

Summary. We show that the renormalized self-intersection local time γt(x) for both the Brownian motion and symmetric stable process in R1 is diﬀerentiable in 0 the spatial variable and that γt(0) can be characterized as the continuous process of zero quadratic variation in the decomposition of a natural Dirichlet process. This Dirichlet process is the potential of a random Schwartz distribution. Analogous results for fractional derivatives of self-intersection local times in R1 and R2 are also discussed.

1 Introduction

In their study of the intrinsic Brownian local time sheet and stochastic area integrals for Brownian motion, [14, 15, 16], Rogers and Walsh were led to analyze the functional Z t A(t, Bt) = 1[0,∞)(Bt − Bs) ds (1) 0 where Bt is a 1-dimensional Brownian motion. They showed that A(t, Bt) is not a semimartingale, and in fact showed that Z t Bs A(t, Bt) − Ls dBs (2) 0 x has finite non-zero 4/3-variation. Here Ls is the local time at x, which is x R s formally Ls = 0 δ(Br − x) dr, where δ(x) is Dirac’s ‘δ-function’. A formal d d2 0 application of Ito’s lemma, using dx 1[0,∞)(x) = δ(x) and dx2 1[0,∞)(x) = δ (x), yields Z t 1 Z t Z s Bs 0 A(t, Bt) − Ls dBs = t + δ (Bs − Br) dr ds (3) 0 2 0 0 ? This research was supported, in part, by grants from the National Science Foun- dation and PSC-CUNY. 2 Jay Rosen which motivates the subject matter of this paper. We study the process which is formally defined as Z t Z s 0 0 γt = − δ (Xs − Xr) dr ds (4) 0 0 0 where δ is the derivative of the delta-function, and Xt is Brownian motion 1 0 or, more generally, a symmetric stable process in R . The process γt is related to the self-intersection local time process which is formally defined as Z t Z s αt = δ(Xs − Xr) dr ds. (5) 0 0 If we set Z t Z s αt(y) = δ(Xs − Xr − y) dr ds (6) 0 0 we obtain a ‘near intersection’ local time, and formally differentiating in y d suggests that (4) is the derivative dy αt(y) y=0. The process αt has not been studied much in one dimension since it can y 1 R y 2 be expressed in terms of the local time Lt of the process Xt: αt = 2 (Lt ) dy RR t x−y x y and αt(y) = 0 Ls dsLs dx. The fact that Lt is not differentiable in the spatial variable y indicates that the existence of (4) and its identification as a derivative requires some care. In two dimensions, even for Brownian motion, αt does not exist and must be ‘renormalized’ by subtracting off a counterterm. This was first done by Varadhan [22], and has been the subject of a large literature, see Dynkin [4], Le Gall [12], Bass and Khoshnevisan [1], Rosen [20, 21]. The resulting renormalized self-intersection local time turns out to be the right tool for the solution of certain “classical” problems such as the asymptotic expansion of the area of the Wiener and stable sausage in the plane and fluctuations of the range of stable random walks. (See Le Gall [11, 10], Le Gall–Rosen [13] and Rosen [19]). 0 1 The process γt in R , in a certain sense, is even more singular than self- intersection local time in R2, but as we shall see, due to the symmetry properties of δ0, there is no need for a counterterm. We begin with a precise definition of γ0, show that it exists, is the spatial derivative of the renormalized t self-intersection local time γt(x) = αt(x) − E αt(x) , and has zero quadratic variation. We then show how it can be characterized as the continuous process of zero quadratic variation in the decomposition of a natural Dirichlet process. This Dirichlet process is the potential of a random Schwartz distribution. Let t s def Z Z αt,(y) = f(Xs − Xr − y) dr ds (7) 0 0 and Z t Z s 0 def 0 αt,(y) = − f(Xs − Xr − y) dr ds (8) 0 0 Derivatives of self-intersection local times 3 where f is an approximate δ-function at zero, i.e. f(x) = f(x/)/ with f a positive, C1, even function of x supported in the unit interval with R f dx = 1. We then define

0 0 αt(y) = lim αt,(y), αt(y) = lim αt,(y) (9) →0 →0 γt(y) = lim αt,(y) − E αt,(y) (10) →0 and 0 0 0 γt(y) = lim αt,(y) − E αt,(y) (11) →0 0 0 whenever the limit exists. We set γt = γt(0). Let c(β) sgn(x)|x|β−2 if x 6= 0 h (x) = (12) β 0 if x = 0 where c(β) = −π−1Γ (2 − β) cos((1 − β)π/2) if 1 < β < 2 and c(2) = −1. Note that hβ(x) is not continuous at x = 0.

Theorem 1. Let Xt denote the symmetric stable process of order β > 3/2 in 1 0 p R . Then αt,(y) and αt,(y) converge a.s. and in all L spaces as → 0 for 1 any (t, y) ∈ R+ × R . The following hold almost surely: 1. For any continuous function g(y) we have

Z t Z s Z g(Xs − Xr) dr ds = g(y)αt(y) dy. (13) 0 0 and if g ∈ C1 Z t Z s Z 0 0 g (Xs − Xr) dr ds = − g(y)αt(y) dy. (14) 0 0 0 2. αt(y) − hβ(y)t is continuous in y. 1 0 1 3. {γt(y) , (t, y) ∈ R+ × R } and {γt(y) , (t, y) ∈ R+ × R } are continuous 0 d and γt(y) is diﬀerentiable in y with γt(y) = dy γt(y). 0 0 4. αt(0) = γt(0). 0 We see from (13) and (14) that αt(y) is the distributional derivative of 0 αt(y). However, we see from 2. that αt(y) is not continuous at y = 0 so that 0 these equations do not allow us to characterize αt(0). For any function gt and any sequence τ = {τn} of partitions τn = {0 = t0 < tn,1 < ··· < tn,n = T } of [0,T ], with mesh size |τn| = maxi |tn,i − tn,i−1| going to 0, we set

n X p Vp(g; τ) = lim |gt − gt | (15) n→∞ n,i n,i−1 i=1 4 Jay Rosen whenever it exists. 0 In [16], Rogers and Walsh show that for Brownian motion V4/3(γ ; τ) is a ﬁnite non-zero constant, independent of τ. Let β0 denote the usual conjugate 1 1 exponent to β, i.e. β + β0 = 1. Theorem 2. Let X be a symmetric stable process of order β > 3/2 in R1. 0 2β0 Then Vp(γ ; τ) = 0 for any τ and any p > 3 . 0 Note that for Brownian motion this shows that Vp(γ ; τ) = 0 for any p > 4/3. We conjecture that for the symmetric stable process of order β > 3/2 1 0 in R we have that V 2β0 (γ ; τ) is a ﬁnite non-zero constant, independent of τ. 3 0 We now obtain an intrinsic characterization for γt, which doesn’t involve 0 limits. The key idea is that γt has zero quadratic variation. In the case of renormalized intersection local time γt for Brownian motion in the plane this was observed by Bertoin, [2] and extended by us in [21]. Recall that a continuous adapted process Zt is said to have zero quadratic variation, if for each T > 0 and any sequence of partitions τn = {0 = t0 < t1 < ··· < tn = T } of [0,T ], with mesh size |τn| = maxi |ti − ti−1| going to 0 X 2 lim E (Zt − Zt ) = 0. (16) n→∞ i i−1 ti∈τn

Föllmer [7] has coined the term “Dirichlet process” to refer to any process which can be written as the sum of a martingale and a process of zero quadratic variation. It is important to note that such a decomposition is unique. The class of Dirichlet processes is much wider than the class of semimartingales. We use Yt to denote our stable process Xt killed at an independent expo- 0 nential time λ. In the following theorem γt will be defined for the process Yt in place of Xt. Let us begin with a special case of the Doob–Meyer decomposition for µ semimartingales. Let Lt denotes the continuous additive functional of Xt µ with Revuz measure µ. Using the additivity of Lt and the Markov property we have x µ µ 1 E (Lλ | Ft) = Lt∧λ + U µ(Yt) (17) 1 µ where Ft = σ(Ys , s ≤ t). Equivalently, U µ(Yt) = Mt − Lt∧λ where Mt = x µ E (Lλ | Ft) is a martingale. This is the Doob-Meyer decomposition for the 1 0 potential U µ(Yt). We will show that γt arises in a similar decomposition for the potential of a random Schwartz distribution. This new potential will no 0 longer be a semimartingale but a Dirichlet process, and γt will correspond to the process of zero quadratic variation in the decomposition of this Dirichlet process. Let Ft be the random additive distribution-valued process defined by Z t Z 0 Ft(g) = − lim g(y + Yr)f(y) dy dr (18) →0 0 Derivatives of self-intersection local times 5 whenever the limit exists. It follows by integration by parts that for all g ∈ ∞ 1 C0 (R ) Z t 0 Ft(g) = g (Yr) dr. (19) 0 With this definition it is natural to set Z t Z 1 1 0 U Ft(x) = − lim u x − (y + Yr) f(y) dy dr (20) →0 0 whenever the limit exists. Note that formally

Z t 1 1 du U Ft(x) = − (x − Ys) ds. (21) 0 dx We have the following analogue of the Doob–Meyer decomposition.

Theorem 3. Let Y be a symmetric stable process of order β > 3/2 in R1, 0 killed at an independent exponential time λ. Then γt is continuous a.s. with zero quadratic variation and

1 0 U Ft(Yt) = Mt − γt (22)

x 0 where Mt is the martingale E (γ∞ | Ft). In view of Theorem 3 we can characterize the renormalized intersec- 0 tion local time γt as the continuous process of zero quadratic variation 1 in the decomposition of the random potential U Ft(Yt) which is formally R t du1 − 0 dx (Yt − Ys) ds.

1.1 Fractional derivatives

All our results can be extended to fractional derivatives. There are in fact sev- eral natural candidates for the fractional derivative of order 0 < ρ < 1 in Rd: (ρ) −d R ipx g (x) = (2π) e w(p)gb(p) dp with w(p) positively homogeneous of index ρ, i.e. w(λp) = |λ|pw(p) for all λ > 0. Our results can be extended for any such w(p), but for simplicity we work with symmetric fractional derivatives: w(−p) = −w(p) which allows us to avoid introducing counterterms. In one dimension this determines w(p) up to a constant factor: w(p) = sgn(p)|p|ρ. Then we study Z t Z s (ρ) (ρ) γt = − δ (Xs − Xr) dr ds. (23) 0 0 More precisely, let

Z t Z s (ρ) def (ρ) αt, (y) = − f (Xs − Xr − y) dr ds, (24) 0 0 6 Jay Rosen

(ρ) (ρ) αt (y) = lim αt, (y) (25) →0 and (ρ) (ρ) (ρ) γt (y) = lim αt, (y) − E αt, (y) (26) →0 (ρ) (ρ) whenever the limit exists. We set γt = γt (0).

Theorem 4. Let Xt denote the symmetric stable process of order β > 1 ∨ 1 (ρ) p (ρ + 1/2) in R . Then αt, (y) converges a.s. and in all L spaces as → 0 1 for any (t, y) ∈ R+ × R . The following hold almost surely: 1. For any Cρ function g(x) we have

Z t Z s Z (ρ) (ρ) g (Xs − Xr) dr ds = − g(y)αt (y) dy. (27) 0 0

(ρ) 2. αt (y) − hβ+1−ρ(y)t is continuous in y. (ρ) 1 (ρ) 3. {γt (y) , (t, y) ∈ R+ ×R } is a.s. continuous and γt (y) is the derivative of order ρ in y of γt(y). (ρ) (ρ) 4. αt (0) = γt (0). (ρ) 2β 5. Vp(γ ; τ) = 0 for any τ and any p > 3β−1−2ρ .

1 Once again, we use Yt to denote our R valued Lévyprocess Xt killed at (ρ) an independent exponential time λ and in the following theorem γt will be defined for the process Yt in place of Xt. Let now Φt be the random additive distribution-valued process defined by Z t Z (ρ) Φt(g) = − lim g(y + Yr)f (y) dy dr (28) →0 0

∞ 1 whenever the limit exists. It is easy to check that for all g ∈ C0 (R ) Z t (ρ) Φt(g) = g (Yr) dr. (29) 0 With this deﬁnition it is natural to set Z t Z 1 1 (ρ) U Φt(x) = − lim u x − (y + Yr) f (y) dy dr (30) →0 0 whenever the limit exists. Note that formally Z t 1 1 (ρ) U Φt(x) = − (u ) (x − Yr) dr. (31) 0 We have the following analogue of the Doob–Meyer decomposition. Derivatives of self-intersection local times 7

Theorem 5. Let Y be a symmetric stable process of order β > ρ + 1/2 in 1 (ρ) R , killed at an independent exponential time λ. Then γt is continuous a.s. with zero quadratic variation and

1 (ρ) U Φt(Yt) = Mt − γt (32)

x (ρ) where Mt is the martingale E (γλ | Ft). We leave to the interested reader the task of formulating analogous results for fractional derivatives of renormalized self-intersection local times in the plane. This paper is organized as follows. In section 2 we prove Theorem 1 and in section 3 prove Theorem 2. Section 4 contains the short proof of Theorem 3. The proofs of Theorems 4 and 5 are similar and left to the reader. Acknowledgement. I would like to thank J. Walsh for some very helpful conversations.

0 2 Existence of αt(x)

1 2 Proof of Theorem 1. For any x ∈ R and bounded Borel set B ⊆ R+ let Z Z 0 0 α(x, B) = − f(Xs − Xr − x) dr ds. (33) B

0 α(x, B) is clearly continuous in all parameters as long as > 0. We use |B| 2 to denote the Lebesgue measure of B ⊆ R+. For any random variable Y we set {Y }0 = Y − E(Y ). The following Lemma will be proven at the end of this section.

Lemma 1. Let X be the symmetric stable process of index β > 3/2 in R1. Then for some ζ > 0

0 0 0 j 0 0 jζ E α (x, B) − α 0 (x ,B) ≤ c (ζ, j) |(, x) − ( , x )| (34) 0 0 and 0 j jζ E α(x, B) 0 ≤ c0(ζ, j) |B| (35) 0 0 1 1 for all j ∈ Z+, , ∈ (0, 1], x, x ∈ R and all Borel sets B ⊆ A1 =: [0, 1/2] × [1/2, 1].

Let

n −n −n −n −n Ak = [(2k − 2)2 , (2k − 1)2 ] × [(2k − 1)2 , (2k)2 ]. (36)

L 1/β 0 1 0 Using the scaling Xλt = λ Xt and fλ(x) = λ2 f(x/λ) we have 8 Jay Rosen

0 L −n(2−2/β) 0 n/β n α(x, B) = 2 α2n/β (2 x, 2 B) (37)

n+1 so that from (34) and (35) we have that for all Borel sets B ⊆ Ak

0 0 0 j −nj(2−2/β−ζ/β) 0 0 ζj E {α(x, B)−α0 (x ,B)}0 ≤ c0(ζ, j)2 |(, x)−( , x )| (38) and 0 j −nj(2−2/β−2ζ) ζj E {α(x, B)}0 ≤ c0(ζ, j)2 |B| . (39) For any B let 0 0 γ(x, B) = {α(x, B)}0. (40) Following Le Gall [12] we write

0 X 0 n γ(x, B) = γ(x, B ∩ Ak ) (41) n,k

for any B ⊆ {0 ≤ r ≤ s ≤ 1}. Using (38) and (39) together with independence we then have, (see Prop. 3.5.2 of [8]),

2n ! j X 0 n+1 0 0 n+1 E γ(x, B ∩ Ak ) − γ0 (x ,B ∩ Ak ) (42) k=1 nj/2 −nj(2−2/β−ζ/β) 0 0 ζj ≤ c0(ζ, j)2 2 |(, x) − ( , x )| ,

and

2n ! j X 0 n+1 nj/2 −nj(2−2/β−2ζ) ζj E γ(x, B ∩ Ak ) ≤ c0(ζ, j)2 2 |B| (43) k=1 so that 0 0 0 0 0 ζ kγ(x, B) − γ0 (x ,B)kj ≤ c |(, x) − ( , x )| (44) and 0 ζ kγ(x, B)kj ≤ c |B| (45) if we choose ζ > 0 so that 1/2 − 2 + 2/β + ζ(2 + 1/β) < 0. This is possible for β > 4/3 (and we are assuming that β > 3/2). 0 0 0 Let Bt = {0 ≤ r ≤ s ≤ t} and set γ,t(x) =: γ(x, Bt). If t, t ≤ M < ∞ 0 then |Bt − Bt0 | ≤ M|t − t | so that by (45) for some c < ∞

0 0 0 ζ kγ,t(x) − γ,t0 (x)kj ≤ c |t − t | (46)

and combined with (44) this shows that for some ζ > 0

0 0 0 0 0 0 ζ kγ,t(x) − γ0,t0 (x )kj ≤ c |(, x, t) − ( , x , t )| . (47)

Kolmogorov’s lemma then shows that locally Derivatives of self-intersection local times 9

0 0 0 0 0 0 ζ0 0 |γ,t(x) − γ0,t0 (x )| ≤ cω |(, x, t) − ( , x , t )| , , > 0 (48) for some ζ0 > 0, which assures us of a locally uniform and hence continuous limit 0 0 γt(x) = lim γ,t(x). (49) →0 The next Lemma is proven in section 5.

Lemma 2. For the symmetric stable process of index β > 1 we can ﬁnd a continuous function h(x) with h(0) = 0 such that for each x

β−2 0 c(β) sgn(x)|x| t if x 6= 0 lim E α,t(x) − h(x) = (50) →0 0 if x = 0 where c(β) = −2Γ (2 − β) cos((1 − β)π/2) if 1 < β < 2 and c(2) = −1. Furthermore, (50) converges locally uniformly in x away from 0 and locally in L1.

Using this Lemma and the locally uniform convergence (49) we see that

0 0 αt(x) = lim α,t(x) (51) →0

0 0 exists for all x, t, and αt(0) = γt(0). Furthermore the convergence is locally uniform in x away from 0 and locally in L1. A similar and simpler analysis shows that locally

0 0 0 0 ζ0 0 |γ,t(x) − γ0,t0 (x )| ≤ cω |(, x, t) − ( , x , t )| , , > 0 (52) for some ζ0 > 0, which assures us of a locally uniform and hence continuous limit γt(x) = lim γ,t(x). (53) →0 1/β R t R s Using the bound ps(y) ≤ c/s we can check that 0 0 ps−r(y) dr ds is bounded and continuous in y and that uniformly in x Z t Z sZ lim E α,t(x) = lim f(y − x)ps−r(y) dy dr ds (54) →0 →0 0 0 Z Z t Z s = lim f(y − x) ps−r(y) dr ds dy →0 0 0 Z t Z s = ps−r(x) dr ds 0 0 Together with the locally uniform convergence of (53) we see that

αt(x) = lim α,t(x). (55) →0 locally uniformly. 10 Jay Rosen

Now let g(x) be a C1 function with compact support. The locally L1 convergence in (51) and the fact that {Xs ; 0 ≤ s ≤ t} is bounded a.s. shows that Z Z 0 0 g(x)αt(x) dx = lim g(x)α,t(x) dx (56) →0 Z Z t Z s 0 = − lim g(x) f(Xs − Xr − x) dr ds dx →0 0 0 Z t Z sZ 0 = − lim g (x)f(Xs − Xr − x) dx dr ds →0 0 0 Z t Z s 0 = − lim f ∗ g (Xs − Xr) dr ds →0 0 0 Z t Z s 0 = − g (Xs − Xr) dr ds. 0 0

0 Since the path {Xs ; 0 ≤ s ≤ t} is bounded a.s. we have that αt(x) has compact support a.s. so that (56) holds for all C1 functions g(x). Similarly, using the locally uniform convergence (55) we see that for any continuous function h(x) we have

Z Z t Z s h(x)αt(x) dx = h(Xs − Xr) dr ds. (57) 0 0 Therefore Z Z 0 0 g(x)αt(x) dx = − g (x)αt(x) dx (58) holds for all C1 functions g(x). d 0 It is clear that dx γ,t(x) = γ,t(x) for any > 0 and hence Z x 0 γ,t(x) = γ,t(y) + γ,t(z) dz. (59) y The locally uniform convergence shown above then implies that Z x 0 γt(x) = γt(y) + γt(z) dz (60) y

d 0 and therefore dx γt(x) = γt(x). This completes the proof of Theorem 1. Proof of Lemma 1. We begin by showing how to ﬁnd a bound 0 j j E α(x, B) ≤ c (61)

1 1 uniform in ∈ (0, 1] , x ∈ R and B ⊆ A1. We use Derivatives of self-intersection local times 11 i Z f 0(x) = eipxpfb(p) dp (62) 2π and independence to write that

0 j E α(x, B) (63) j j ! 1 ZZ X X = j exp −ix pk E exp ipk(Xsk − Xrk ) (2πi) j B k=1 k=1 j Y pkfb(pk) dsk drk dpk k=1 j j ! 1 ZZ X X = j exp −ix pk E exp ipk(X1/2 − Xrk ) (2πi) j B k=1 k=1 j ! j X Y E exp ipk(Xsk − X1/2) pkfb(pk) dsk drk dpk. k=1 k=1 We write j j X X pk(X1/2 − Xrk ) = vk(Xtk+1 − Xtk ) (64) k=1 k=1 def where the t1, . . . , tj are the ri’s relabeled so that t1 ≤ t2 ≤ · · · ≤ tj ≤ tj+1 = 1/2 and v = P p so that the v ’s span Rj. Similarly we rewrite i l:rl≤ti l i

j j X X 0 pk(Xs − X ) = v (X 0 − X 0 ) (65) k 1/2 k tk tk−1 k=1 k=1

0 def with t0 = 1/2. Then using (63) and independence we have

0 j E α(x, B) (66) ZZ j j 1 X X β 0 0 = j exp −ix pk exp − |vk| (tk+1 − tk) (2πi) j B k=1 k=1 j j X 0 β Y exp − |vk| (tk − tk−1) pkfb(pk) dsk drk dpk. k=1 k=1 Using this and the simple bound

Z 1 −t|v|β c e dt ≤ β (67) 0 1 + |v| we have the uniform bound 12 Jay Rosen

j Z 1 1 0 j j Y Y Y E α(x, B) ≤ c β 0 β |pk| dpk (68) 1 + |vk| 1 + |v | k k=1 1/2 1/2 j Y |pk| Y |pk| ≤ c β 0 β . 1 + |vk| 2 1 + |vk| 2 Since v = P p , we see that each p can be represented as the diﬀerence i l:rl≤ti l k pk = vi −vi−1 for some i, and each vi appears in the representation of at most two pk’s. Thus

1/2 2 Z Y |pk| Y |pk| β = 2β dpk (69) 1 + |vk| 2 1 + |vk| Z 2 j Y 1 + |vk| + |vk| ≤ c 2β dvk 1 + |vk| which is bounded if β > 3/2. We can now establish (34). To handle the variation in x we replace (66) by

0 0 0 j E α(x, B) − α(x ,B) (70) ZZ j ! 1 Y 0 = j {exp(−ipkx) − exp(−ipkx )} (2πi) j B k=1 j j X β 0 0 X 0 β exp − |vk| (tk+1 − tk) exp − |vk| (tk − tk−1) k=1 k=1 j Y pkfb(pk) dsk drk dpk k=1 and use the bound

0 ζ 0 ζ | exp(−ipkx) − exp(−ipkx )| ≤ C|pk| |x − x | (71) for any 0 ≤ ζ ≤ 1. Similarly to handle the variation in we replace (66) by

0 0 j E α(x, B) − α0 (x, B) (72) ZZ j j 1 X X β 0 0 = j exp −ix pk exp − |vk| (tk+1 − tk) (2πi) j B k=1 k=1 j j X 0 β Y 0 exp − |vk| (tk − tk−1) pk fb(pk) − fb( pk) dsk drk dpk k=1 k=1 and use the bound Derivatives of self-intersection local times 13

0 ζ 0 ζ fb(pk) − fb( pk) ≤ C|pk| | − | (73) for any 0 ≤ ζ ≤ 1. Since (70)-(72) hold also when j = 1, we obtain (34). To prove (35) we ﬁrst apply Holder’s inequality to (66):

0 j E α(x, B) (74) ZZ j j 1 X X β 0 0 = j exp −ix pk exp − |vk| (tk+1 − tk) (2πi) j B k=1 k=1 j j X 0 β Y exp − |vk| (tk − tk−1) pkfb(pk) dsk drk dpk k=1 k=1 Z Z j j j/a 0 X β 0 0 ≤ c |B| exp −a |vk| (tk+1 − tk) j B k=1 0 j j !1/a j 0 X 0 β Y Y exp −a |vk| (tk − tk−1) dsk drk |pk| dpk k=1 k=1 k=1 for any 1/a + 1/a0 = 1. The last integral can be bounded as before if a0 is chosen close to 1. As in the proof of (34), this completes the proof of (35) and therefore of Lemma 1.

0 3 p-variation of γt

0 0 2 Proof of Theorem 2. Since we know that αt, → γt in L , we have

0 0 2 E (γt − γt0 ) (75) 0 0 2 = lim E (αt, − αt0,) →0 Z t Z s 2! 0 = lim E f(Xs − Xr) dr ds →0 t0 0 ZZ = lim E exp −ip(Xs2 − Xs1 ) − i(p + q)(Xs3 − Xs2 ) →0 n 0≤s1≤s2≤s3≤s4 o 0 0 t ≤s3≤t ; t ≤s4≤t 4 Y exp −iq(Xs4 − Xs3 ) dsk pqfb(p)fb(q) dp dq k=1 ZZ + lim E exp −ip(Xs2 − Xs1 ) − i(p + q)(Xs3 − Xs2 ) →0 n 0≤s1≤s2≤s3≤s4 o 0 0 t ≤s3≤t ; t ≤s4≤t 4 Y exp −ip(Xs4 − Xs3 ) dsk pqfb(p)fb(q) dp dq k=1

ZZ β β β = lim e−(s2−s1)|p| −(s3−s2)|p+q| −(s4−s3)|q| →0 n 0≤s1≤s2≤s3≤s4 o 0 0 t ≤s3≤t ; t ≤s4≤t 14 Jay Rosen

4 Y dsk pqfb(p)fb(q) dp dq k=1

ZZ β β β + lim e−(s2−s1)|p| −(s3−s2)|p+q| −(s4−s3)|p| →0 n 0≤s1≤s2≤s3≤s4 o 0 0 t ≤s3≤t ; t ≤s4≤t 4 Y dsk pqfb(p)fb(q) dp dq k=1 Consider the ﬁrst summand on the right hand side of (75). It is bounded by

4 ZZ β β β −r2|p| −r3|p+q| −r4|q| Y 0 e drk|p||q| dp dq (76) n t ≤r1+r2+r3≤t o 0 t ≤r1+r2+r3+r4≤t k=1 Z 1 1 1 ≤ C(t − t0)1+a |p||q| dp dq 1 + |p|β 1 + |p + q|β 1 + |q|(1−a)β where we ﬁrst integrated with respect to dr4 using H¨older’s inequality with 0 ≤ a ≤ 1

t−(r1+r2+r3) t (1−a) Z β Z β −r4|q| 0 a −r4|q| /(1−a) e dr4 ≤ (t − t ) e dr4 , (77) 0 t −(r1+r2+r3) 0

R t−(r2+r3) 0 then with respect to dr1 using 0 dr1 ≤ t − t , and ﬁnally the dr2, t −(r2+r3) dr3 integrals are bounded using (67). It is easily seen that (76) is bounded as long as β − 1 + (1 − a)β − 1 > 1, i.e. a < 2 − 3/β. Now consider the second summand on the right hand side of (75). An attempt to use a bound similar to (76) where we bound pq by |pq| would β be fatal. Rather, we ﬁrst observe that R e−(s3−s2)|q| q dq = 0 and use this to rewrite the second summand on the right hand side of (75) as

ZZ β β lim e−(s2−s1)|p| −(s4−s3)|p| (78) →0 n 0≤s1≤s2≤s3≤s4 o 0 0 t ≤s3≤t ; t ≤s4≤t 4 n β β β o −(s3−s2)|p+q| −(s3−s2)|p| −(s3−s2)|q| Y e − e e dsk pqfb(p)fb(q) dp dq. k=1 Now we bound this as in (76) by

ZZ β β −r2|p| −r4|p| 0 e (79) n t ≤r1+r2+r3≤t o 0 t ≤r1+r2+r3+r4≤t 4 β β β Y −r3|p+q| −r3|p| −r3|q| e − e e drk|p||q| dp dq k=1 Z 0 1+a 1 1 1 ≤ C(t − t ) − |p||q| dp dq. 1 + |p|(2−a)β 1 + |p + q|β 1 + |p|β + |q|β Derivatives of self-intersection local times 15

Here we proceeded as in (76) except that for the dr3 integral we used Z −r −r|p+q|β −r(|p|β +|q|β ) 1 1 e e − e dr ≤ − (80) 1 + |p + q|β 1 + |p|β + |q|β by arguing seperately depending on whether or not |p|β + |q|β > |p + q|β. We claim that once again the integral on the right hand side of (79) is ﬁnite whenever β − 1 + (1 − a)β − 1 > 1, i.e. a < 2 − 3/β. This is clear in the region 1 1 c where |q| ≤ 2|p| since we can use the bound 1+|p|(2−a)β ≤ 1+|p|β 1+|q|(1−a)β . If, however, |q| > 2|p|, we can use the bound

1 1 l|p + q|β − |q|β − |p|β − = (81) β β β β β β 1 + |p + q| 1 + |p| + |q| (1 + |p + q| )(1 + |p| + |q| ) c|p||q|β−1 + |p|β ≤ (1 + |p + q|β)(1 + |p|β + |q|β) c|p||q|β−1 ≤ . (1 + |q|β)2

This allows us to bound the resulting integral from the right hand side of (79) by Z |p|2 1 c dp dq (82) 1 + |p|(2−a)β (1 + |q|β) which leads to the same result as before. Finally, for h ≤ 2 we have

0 0 h 0 0 2 h/2 0 (1+a)h/2 E (γt − γt0 ) ≤ E (γt − γt0 ) ≤ C(t − t ) . (83) We will have h-variation 0 when (1 + a)h/2 > 1 for some a < 2 − 3/β, i.e. (3 − 3/β)h/2 > 1. Thus we will have h-variation 0 when h > (2/3)β0. This completes the proof of Theorem 2.

4 The Doob–Meyer decomposition

Proof of Theorem 3. It is easy to check that the proofs of Theorems 1 and 2 go through with X replaced by Y and in that case

0 0 0 0 γ∞ = lim γ,∞ and γt = lim γ,t (84) →0 →0 converge a.s. and in all Lp spaces. Let Z t Z 1 1 0 U F,t(y) = − u x − (y − Yr) f(x) dx dr. (85) 0 The proof of Theorem 3 then follows from (84) and the next Lemma. 16 Jay Rosen

1 Lemma 3. Let {Ys ; s ∈ R+} be the exponentially killed symmetric stable process of index β > 3/2 in R1. Then for any > 0

1 x 0 0 U F,t(Yt) = E (γ,∞ | Ft) − γ,t (86)

Proof of Lemma 3. We have Z t Z s 0 0 γ,t = − f(Ys − Yr) dr ds (87) 0 0 so that Z ∞ Z s 0 0 γ,∞ = − f(Ys − Yr) dr ds (88) 0 0 Z ∞ Z s 0 0 = γ,t − f(Ys − Yr) dr ds t t Z ∞ Z t 0 − f(Ys − Yr) dr ds . t 0

1 Now, using the fact that {Ys ; s ∈ R+} has independent increments Z ∞ Z s 0 E f(Ys − Yr) dr ds Ft (89) t t Z ∞ Z s Z ∞ Z s 0 0 = E f(Ys − Yr) dr ds Ft − E f(Ys − Yr) dr ds = 0 t t t t and Z ∞ Z t 0 E f(Ys − Yr) dr ds Ft (90) t 0 Z ∞ Z t Z ∞ Z t 0 0 = E f(Ys − Yr) dr ds Ft − E f(Ys − Yr) dr ds . t 0 t 0 We have Z ∞ Z t 0 E f(Ys − Yr) dr ds Ft (91) t 0 Z ∞ Z t 0 = E f (Ys − Yt) + (Yt − Yr) dr ds Ft t 0 Z ∞ Z t ¯ 0 ¯ = E f Ys + (Yt − Yr) dr ds 0 0

¯ 1 1 ¯ where {Ys ; s ∈ R+} is an independent copy of {Ys ; s ∈ R+} and E denotes ¯ 1 expectation with respect to {Ys ; s ∈ R+}. Hence Derivatives of self-intersection local times 17 Z ∞ Z t Z t Z 0 0 1 E f(Ys − Yr) dr ds Ft = f(x + Yt − Yr)u (x) dx dr (92) t 0 0 Z t Z 1 0 = u x − (Yt − Yr) f(x) dx dr. 0 The same sort of argument shows that Z ∞ Z t Z t Z 0 0 1 E f(Ys − Yr) dr ds = f(x + y)u (x)pr(y) dx dy dr (93) t 0 0 which is zero by symmetry. This concludes the proof of Lemma 3.

5 Proof of Lemma 2

Proof of Lemma 2. We have Z t Z sZ 0 0 E α,t(x) = − f(y − x)ps−r(y) dy dr ds (94) 0 0 0 We ﬁrst consider the case 1 < β < 2. Using the fact that f is an odd fuction, Plancherel and then Fubini Z t Z sZ 0 − f(y − x)ps−r(y) dy dr ds (95) 0 0 t s Z Z Z β = i(2π)−1 eipxpfb(p)e−(s−r)|p| dp dr ds 0 0 t s Z Z Z β = i(2π)−1 eipxpfb(p) e−(s−r)|p| dr ds dp 0 0 and t s Z Z β e−(s−r)|p| dr ds (96) 0 0 t t t t−r Z Z β Z Z β = e−(s−r)|p| ds dr = e−s|p| ds dr 0 r 0 0 t r ∞ t ∞ Z Z β Z β Z Z β = e−s|p| ds dr = t e−s|p| ds − e−s|p| ds dr 0 0 0 0 r t Z β β = t|p|−β − e−r|p| |p|−β dr = t|p|−β − (1 − e−t|p| )|p|−2β. 0 Hence Z 0 −1 ipx 1−β E α,t(x) = i(2π) t e sgn(p)|p| fb(p) dp (97)

Z β −i(2π)−1 eipx sgn(p)(1 − e−t|p| )|p|1−2βfb(p) dp. 18 Jay Rosen

β It is easily checked that (1 − e−t|p| )|p|1−2β ∈ L1(R1) so that the last term converges uniformly as → 0 to a continuous limit. On the other hand, it follows from [9, formula (13), page 173] that Z i(2π)−1 eipx sgn(p)|p|1−βfb(p) dp (98) Z −1 β−2 = −π Γ (2 − β) cos (1 − β)π/2 sgn(y)|y| f(y − x) dy.

Since sgn(y)|y|β−2 is locally in L1 and continuous away from 0, this last term converges locally uniformly away from 0 and locally in L1 to c(β) sgn(x)|x|β−2. On the other hand, when x = 0, (94) is 0 by symmetry. This completes the proof of Lemma 2 for β 6= 2. For Brownian motion, we proceed diﬀerently. We ﬁrst write

Z t Z sZ 0 − f(y − x)ps−r(y) dy dr ds (99) 0 0 Z t Z sZ 0 = f(y − x)ps−r(y) dy dr ds. 0 0

0 −1 −y2/2s 0 We have ps(y) = (2π)1/2s3/2 ye so that, for y 6= 0, {|ps(y)| ; s ≥ 0} is the density of T|y|, the ﬁrst hitting time of |y| for Brownian motion. Hence Z ∞ 0 |ps(y)| ds = 1, y 6= 0. (100) 0 This justiﬁes the use of Fubini

Z t Z sZ 0 f(y − x)ps−r(y) dy dr ds (101) 0 0 Z Z t Z s 0 = f(y − x) ps−r(y) dr ds dy 0 0 and then, just as in (96)

Z t Z s Z ∞ Z t Z ∞ 0 0 0 ps−r(y) dr ds = t ps(y) ds − ps(y) ds dr, y 6= 0. (102) 0 0 0 0 r Just as in (100) we see that the ﬁrst integral is − sgn(y) for y 6= 0. Now 0 1 |ps(y)| ≤ (2π)1/2s3/2 |y|, so that by the dominated convergence theorem the second integral is continuous in y. As before, when y = 0, the left hand side of (99) is 0 by symmetry. Lemma 2 now follows. Derivatives of self-intersection local times 19 References

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