Approximation via regularization of the of and Blandine Berard Bergery, Pierre Vallois

To cite this version:

Blandine Berard Bergery, Pierre Vallois. Approximation via regularization of the local time of semi- martingales and Brownian motion. Stochastic Processes and their Applications, Elsevier, 2008, 118, pp.2058-2070. ￿10.1016/j.spa.2007.12.002￿. ￿hal-00169518￿

HAL Id: hal-00169518 https://hal.archives-ouvertes.fr/hal-00169518 Submitted on 4 Sep 2007

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t x g(Xs)ds= g(x)Lt (X)dx. (1.1) R Z0 Z Moreover, t Lx(X) is continuous and non-decreasing, for any x R. → t ∈ Besides, an extension of Itˆo’s formula in the case of convex functions f may be obtained thanks to local time processes. Namely,

t 1 x f(Xt)= f(X0)+ f ′ (Xs)dXs + Lt f ′′(dx), R Z0 − 2 Z with f ′ the left derivative of f and f ′′ the second derivative of f in the distribution− sense.

2. The density occupation formula (1.1) gives a relation between the quadratic variation of a semimartingale and its local time process. We would like to show that the existence of the quadratic variation implies the weak existence of the local time process at a fixed level. For our purpose, it is convenient to use the definition of the quadratic variation which has been given in [23], in the setting of stochastic integration by regularization [24]: for any continuous process X, its quadratic variation [X]t is the process:

t 1 2 [X]t = lim(ucp) (Xs+ǫ Xs) ds, (1.2) ǫ 0 ǫ 0 → Z − provided that this limit exists in the (ucp) sense. We denote by (ucp) the convergence in probability, uniformly on the compact sets (c.f. Section II.4 of [18]). In the case of a continuous semimartingale X, the quadratic variation defined by (1.2) coincides with the usual quadratic variation .

Let ǫ> 0 and X be a continuous process. Let us introduce a family (Jǫ(t, y),y R, t > 0) of processes, which will play a central role in our study: ∈ 1 t Jǫ(t, y)= 1I y

As a consequence, if we suppose that X is a semimartingale, the occupation

2 times formula implies:

y lim(ucp) f(y)Jǫ(t, y)dy = f(y)Lt (X)dy. ǫ 0 R R → Z Z y Thus, in a certain sense, the measures (Jǫ(t, y)dy) converge to (Lt (X)dy) as ǫ 0. As a result, it seems natural to study the convergence of Jǫ(t, y) to Ly→(X) when ǫ 0 and y is a fixed real number. t →

Let us remark that, if Jǫ(t, y) converges in (ucp) sense, then its limit is equal to the covariation [X, 1I y

3. Our first approximation result concerns (Jǫ(t, y)) when X belongs to a class of diffusions stable under time reversal. This kind of diffusions has been studied in [17] and [15]. We consider the generalization made in Section 5 of [25]. Let X be a diffusion which satisfies

t t Xt = X0 + σ(s,Xs)dBs + b(s,Xs)ds. (1.5) Z0 Z0 It is moreover assumed that the following conditions hold:

t [0, T ],Xt has a density p(t, x) with respect to Lebesgue measure, ∀ ∈  σ, b are jointly continuous,    2 2,1 1,1 2,  σ (s, .) W (R), b(s, .) W (R), xp(s, .) W ∞(R),  loc loc loc  ∈ ∈ ∈  2 1,1 R pσ (s, .) Wloc ( ) hold for almost every s [0, T ], ∈ ∈  ∂pσ2 ∂2xp  , L1([0, t] R), t ]0, T ].  ∂x ∂x2  ∈ × ∈  (1.6)  To a fixed T > 0, we associate the process

Xu = XT u, u [0, T ]. (1.7) − ∈ f According to Theorem 5.1 of [25], (Xu)u [0,T ] is a diffusion which verifies the ∈ following equation: f u u Xu = XT + σ(T s, Xs)dβs + ˜b(T s, Xs)ds, (1.8) Z0 − Z0 − f f f where β is a Brownian motion on a possibly enlarged space and ˜b is explicitly known.

Theorem 1.1 Let X be a diffusion which satisfies (1.5)-(1.6). Then

x R lim(ucp) Jǫ(t, x)= Lt (X), x . ǫ 0 → ∀ ∈

3 Our limit in Theorem 1.1 is valid when x is fixed. We have not been able to prove that the convergence is uniform with respect to x varying in a compact set.

4. For simplicity of notation, we take x = 0 and we note Jǫ(t) instead of Jǫ(t, 0). The proof of Theorem 1.1 is based on a decomposition of Jǫ(t) as a sum of two terms. It is actually possible to prove (see Section 3) that each term has a limit. However, theses limits cannot be expressed only through 0 Lt (X). Modifying the factor 1I 0

3 4 Jǫ(t)= Iǫ (t)+ Iǫ (t)+ Rǫ(t), (1.9) where

t t 3 1 + 1 Iǫ (t)= X(u+ǫ) t1I Xu60 du + X(−u+ǫ) t1I Xu>0 du, (1.10) ǫ 0 ∧ { } ǫ 0 ∧ { } Z t Z t 4 1 1 + Iǫ (t)= Xu−1I X(u+ǫ)∧t>0 du + Xu 1I X(u+ǫ)∧t60 du, (1.11) ǫ Z0 { } ǫ Z0 { }

and (Rǫ(t))t> is a process which goes to 0 a.s. as ǫ 0, uniformly on compact 0 → sets in time. Note that in (1.10) and (1.11), we have systemically introduced 3 4 X(u+ǫ) t instead of Xu+ǫ, in order to guarantee that Iǫ (t), Iǫ (t) are adapted processes.∧ This will play an important role in the proof of our results, in particular to obtain the convergence in the (ucp) sense.

The decomposition (1.9) seems at first complicated. However, it is interesting since it may be proved (see Theorem 1.2 below) that, under suitable assump- tions, I3(t) and I4(t) converge to a fraction of L0(X), as ǫ 0. ǫ ǫ t → Theorem 1.2 i) If X is a continuous semimartingale, then

3 1 0 lim (ucp) Iǫ (t)= Lt (X). ǫ 0 → 2 ii) If X is a diffusion which satisfies (1.5) and (1.6), then

4 1 0 lim (ucp) Iǫ (t)= Lt (X). ǫ 0 → 2 Remark 1.3 (1) We would like to emphasize that the choice of strict or large inequalities in (1.10) and (1.11) is important. Indeed, the Lebesgue mea- sure of u; Xu =0 may not vanish. Note that, in Tanaka’s formula (c.f { + } (3.4) below), Xt is associated with 1I Xt>0 . This heuristically explains { } our choice. (2) When X is a standard Brownian motion, we will prove (see Theorem 1.4 3 4 below) that the two terms of the sum in Iǫ (t) and Iǫ (t) converge separately 1 0 to 4 Lt (X).

4 (3) Applying Theorem 1.2 to the process (Xt x)t> gives − 0

3 1 x lim (ucp) Iǫ (t, x)= Lt (X), where ǫ 0 → 2

t t 3 1 + 1 Iǫ (t, x)= (X(u+ǫ) t x) 1I Xu6x du+ (X(u+ǫ) t x)−1I Xu>x du. ǫ Z0 ∧ − { } ǫ Z0 ∧ − { } 3 4 There exists obviously a similar result where Iǫ (t, x) is replaced by Iǫ (t, x).

5. In the Brownian case, we will give complements to Theorem 1.1 and The- 3 4 orem 1.2. The first one concerns Theorem 1.2. Obviously, Iǫ (t) and Iǫ (t) can be decomposed as:

3 3,1 3,2 3 4 4,1 4,2 4 Iǫ (t)= Iǫ (t)+ Iǫ (t)+ rǫ (t), Iǫ (t)= Iǫ (t)+ Iǫ (t)+ rǫ (t), (1.12) where

t t 3,1 1 3,2 1 + Iǫ (t)= X(−u+ǫ) t1I Xu>0 du, Iǫ (t)= X(u+ǫ) t1I Xu<0 du. (1.13) ǫ 0 ∧ { } ǫ 0 ∧ { } Z t Z 3 1 + rǫ (t)= X(u+ǫ) t1I Xu=0 du, (1.14) ǫ 0 ∧ { } Z t t 4,1 1 4,2 1 + Iǫ (t)= Xu−1I X(u+ǫ)∧t>0 du, Iǫ (t)= Xu 1I X(u+ǫ)∧t<0 du, (1.15) ǫ 0 { } ǫ 0 { } Z t Z 4 1 + rǫ (t)= Xu 1I X(u+ǫ)∧t=0 du, (1.16) ǫ Z0 { } and r3(t),r4(t) converge a.s. to 0 as ǫ 0 (c.f point 1. of Section 5). ǫ ǫ → Theorem 1.4 Let X be a standard Brownian motion. Then,

(1) 3,1 3,2 1 0 lim (ucp) Iǫ (t) = lim (ucp) Iǫ (t)= Lt (X), ǫ 0 ǫ 0 → → 4 (2) 4,1 4,2 1 0 lim (ucp) Iǫ (t) = lim (ucp) Iǫ (t)= Lt (X). ǫ 0 ǫ 0 → → 4 Moreover, for all T > 0, δ ]0, 1 [, there exists a constant C such as ∈ 2

4,i 1 0 δ sup Iǫ (t) Lt (X) 6 Cǫ 2 , i =1 or 2. t [0,T ] − 4 2 ∈ L (Ω)

4,1 4,2 Remark 1.5 The ucp convergence of Iǫ (t), Iǫ (t) in Theorem 1.4 is still t true (see [3]) if Xt = 0 σ(s)dBs, where B is the standard Brownian motion and σ : R R a function which is H¨older continuous of order γ > 1 and + → R 4 such as σ(s) > a> 0. | |

5 Then, we will give below a complement related to Theorem 1.1. In Theorem x 2 1.6, we determine the rate of convergence of Jǫ(t, x) to Lt (X) in L (Ω), as ǫ 0. → Theorem 1.6 Let X be the standard Brownian motion. For all T > 0, x R, δ ]0, 1 [, there exists a constant C such as: ∈ ∈ 2

0 δ ǫ ]0, 1], sup Jǫ(t) Lt (X) 6 Cǫ 2 . ∀ ∈ t [0,T ] − ∈ L2(Ω)

In the setting of stochastic integration by regularization (see for instance [22], [20], [21], [23] and [24]), the ucp convergence is mainly used. So, Theorem 1.1, 1.2, 1.4 are of this type. It seems interesting to investigate the almost sure convergence. There exists few results using this type of convergence in [9]. Our version of Theorem 1.1 formulated in terms of a.s. convergence is the following.

Proposition 1.7 Let X be the standard Brownian motion and (ǫn)n N be a ∈ decreasing sequence of non-negative real numbers which satisfies i∞=1 √ǫi < . Then, for any x R, almost surely, ∞ ∈ P

x lim sup Jǫ (t, x) L (X) =0, n n t →∞ t [0,T ] | − | ∈ 1 lim sup I4,i(t, x) Lx(X) =0, i =1, 2. n ǫn t →∞ t [0,T ] − 4 ∈

6. Let (Xt)t>0 be a continuous process with values in R. Let us briefly enu- x R merate processes which admit local time processes (Lt (X), x , t > 0). The most popular ones are semimartingales (see for instance Sec∈tion VI of [19]). x R x Moreover, there exists a version of (Lt (X), x , t > 0) such that x Lt is right-continuous for any t. When X is a Markov∈ process, the local→ time x process (Lt )t>0 at level x is defined as a specific additive functional (see [6]), and through excursions in Chapter IV of [5]. A construction of the local time process related to one dimensional diffusions has been given in Chapter VI of [11]. A large class of L´evy processes admits local time process (see Chapter V of [5]). Beyond semimartingales and Markov processes, local time process associated with Gaussian processes may exist (c.f. [8]) as occupation densi- ties. A particular example of which has local time process is the fractional Brownian motion. Tanaka formula and Itˆo-Tanaka formula have been given in [7].

x R The definition of (Lt (X), x , t > 0) may not directly refer to the paths ∈ x of (Xt)t>0. For instance, if X is a continuous , (Lt (X), t >

6 0) may be defined as the unique adapted process vanishing at 0 such that x Xt x Lt (X) is a local martingale. Therefore, it may be interesting to | − | − x R perform approximation schemes of (Lt (X), x , t > 0) which involve more directly the paths of X. In the case of semimartingales,∈ we deduce easily x from the occupation times formula and right continuity of t Lt (X) that 1 t x → 0 1I x6Xs6x+ǫ ds tends to Lt (X) as ǫ 0. ǫ { } → R The L´evy excursion theory (c.f. [27]) gives other kind of approximation by the count of the downcrossings number or of the excursion number before a given time. In the case of diffusion, the convergence of normalized sums to local time have been studied in [1] and [13]. Finaly, we may refer to [16] and [2] for approximations of the local time for L´evy’s processes.

7. Let us briefly detail the organization of the paper. Section 2 contains few preliminary lemmas and complements related to some results stated in the Introduction. The proof of Theorem 1.1 is given in Section 3. Section 4 contains the proof of Theorem 1.2. The Brownian case is studied in Section 5 (proof of Theorem 1.4) and in Section 6 (proof of Theorem 1.6).

Some results of this paper were announced without any proof in [4]. Moreover, the setting in [4] was the Brownian one.

Let us adopt two conventions, which will be used in the sequel of the paper:

[0, T ] will denote a given compact interval of time, • in the calculations, C will denote a generic constant. •

2 Decomposition of Jǫ(t) and preliminary lemmas

This section has five independent parts. In the two first ones, we recall some known results related to Fubini’s stochastic theorem (c.f. Lemma 2.1 below) and H¨older continuity properties (c.f. Lemma 2.2 below). Proofs of develop- ments given in the Introduction may be found in points 3. and 4. We end this Section by a technical lemma.

1. Let us start with a modification of Fubini’s theorem. This result may be found in Section IV.5 of [19] and is crucial in most of our proofs. It permits to express some Lebesgue integrals as stochastic integrals with respect to martingales. This allows to obtain (ucp) convergence via Doob’s inequality (see for instance the proof of Proposition 3.1).

Lemma 2.1 Let (H(u,s),s R,u > 0) be a collection of predictable processes which are measurable with respect∈ to (u,s,ω). If one of the following hypotheses holds:

7 t t 2 i. (Xs)s> is the standard Brownian motion and E(H(u,s) )dsdu < , 0 0 0 ∞ ii. (Xs)s>0 is a continuous semimartingale and R(HR (u,s))s,u>0 is uniformly bounded,

then, almost surely,

t t t t H(u,s)dXs du = H(u,s)du dXs, t > 0. Z0 Z0  Z0 Z0  ∀

2. Let us recall some H¨older continuity properties of the Brownian motion and its local time process.

Lemma 2.2 Let δ ]0, 1 [,T > 0 and X be the standard Brownian motion. ∈ 2 2 i) Then, there exists a positive random constant Cδ L (Ω) such as a.s. ∈ δ Xy Xy′ 6 Cδ y y′ , y,y′ [0, T ]. | − | | − | ∀ ∈ ii) The Brownian local time is H¨older continuous in space: there exists a posi- 2 tive random constant Kδ L (Ω) such as a.s. ∈ a a′ δ L (X) L (X) 6 Kδ a a′ , t [0, T ], a, a′ R. | t − t | | − | ∀ ∈ ∈ 1 In the further proofs, as soon as δ ]0, 2 [ and T are given, the constants Kδ,Cδ may be considered as fixed. ∈

3. Let Y and Z be continuous processes. In [21], the covariation of Y and Z has been defined as: 1 t [Y,Z]t = lim(ucp) (Ys+ǫ Ys)(Zs+ǫ Zs) ds, (2.1) ǫ 0 ǫ 0 → Z − − if the limit exists.

Let X be a continuous process with finite quadratic variation [X] = [X,X] (c.f.(1.2)). According to Proposition 2.1 of [21]:

t 1 [f(X),g(X)]t = f ′(Xs)g′(Xs)d[X]s, f,g . (2.2) Z0 ∀ ∈ C The aim of this section is to explain how (2.2) combined with (2.1) leads to (1.4). Let f be a continuous function with compact support and let F be its primitive which vanishes at 0. Taking Yt = F (Xt) and Zt = Xt in (2.1) and using (2.2) comes to:

1 t lim (ucp) (F (Xs+ǫ) F (Xs))(Xs+ǫ Xs) ds = [F (X),X]t, ǫ 0 → ǫ 0 − − Z t = f(Xs)d[X]s. (2.3) Z0

8 We now express the integral on the left hand-side of (2.3) through f instead of

F . Since F (x)= R 1I y

4. We would like to prove (1.9). First, let us introduce Rǫ(t):

1 t Rǫ(t)= 1I 0 0. − ǫ (t ǫ)+ { } − { } − Z −   Then, 1 t Jǫ(t) Rǫ(t)= (1I 0

1I X >0 1I Xu>0 = 1I X >0,Xu60 1I X 60,Xu>0 , (2.5) { (u+ǫ)∧t } − { } { (u+ǫ)∧t } − { (u+ǫ)∧t } and developing the product of the integrand in (2.4) lead to

1 t Jǫ(t) Rǫ(t)= X(u+ǫ) t1I X(u+ǫ)∧t>0,Xu60 X(u+ǫ) t1I X(u+ǫ)∧t60,Xu>0 − ǫ 0 ∧ { } − ∧ { } Z h Xu1I X >0,Xu60 + Xu1I X 60,Xu>0 du. − { (u+ǫ)∧t } { (u+ǫ)∧t } i + Since X.1I X.>0 = X. and X.1I X.60 = X.−, (1.9) follows. { } − { }

5. In (1.9), Rǫ(t) may be viewed as a remainder term. The lemma below ensures the convergence to 0, in the a.s. sense, of this kind of terms. A proof of Lemma 2.3 may be found in [3].

Lemma 2.3 Let X be a continuous process and let a,b,c,d : [0, 1] [0, T + 1] R, be Borel functions such that × → 0 6 b(ǫ, t) a(ǫ, t) 6 ǫ, c(ǫ, s) d(ǫ, s) 6 ǫ, t, s [0, T + 1], ǫ [0, 1]. − | − | ∈ ∈ Let us consider 1 b(ǫ,t) Rǫ(t)= (Xc(ǫ,s) Xd(ǫ,s))1IAs ds, 0 6 t 6 T, ǫ Za(ǫ,t) − b where (As)s [0,T +1] is a collection of measurable events. ∈ Then Rǫ(t) tends a.s. to 0 as ǫ 0, uniformly on [0, T ]. Furthermore, if X → b 9 1 is the standard Brownian motion and δ ]0, 2 [, there exists a positive random 2 ∈ constant Cδ L (Ω) such that, almost surely, ∈ δ sup Rǫ(t) 6 Cδǫ . t [0,T ] | | ∈ b

3 Proof of Theorem 1.1 and associated results

1. Let X be a continuous process. We have the following decomposition of Jǫ(t): 1 2 Jǫ(t)= I (t)+ I (t), (3.1) − ǫ ǫ with

t 1 Xs+ǫ Xs Iǫ (t)= − 1I 0

Proposition 3.1 Let X = M + V be a continuous semimartingale, with M a continuous local martingale and V an adapted continuous process with bounded variation. It is moreover supposed that both dV and d are absolutely continuous with respect to the Lebesgue measure. Then

t 1 lim(ucp) Iǫ (t)= 1I 0

Proposition 3.3 If X is a diffusion which satisfies (1.5) and (1.6), then

2 + + 1 0 lim(ucp) Iǫ (t)= Xt X + Lt (X). ǫ 0 0 → − 2

10 The proof of this result is postponed in point 3. below.

Remark 3.4 (1) When X is a diffusion which satisfies (1.5) and (1.6), then 1 2 Iǫ (t) and Iǫ (t) converge as ǫ 0. Theorem 1.1 is a direct consequence of Tanaka’s formula: →

t + + 1 0 Xt = X0 + 1I 0

2. Proof of Proposition 3.1. Let X be a continuous semimartingale with 1 canonical decomposition X = M + V . The key of our proof is to write Iǫ (t) as the sum of a semimartingale plus a remainder term. From a technical point of 1 view, the semimartingale will be obtained by expressing Iǫ (t) through X(s+ǫ) t ∧ instead of Xs+ǫ. Namely,

t 1 1 1 Iǫ (t) 1I 0

t t 1 1 Iǫ (t)= (M(s+ǫ) t Ms)1I 0

By Lemma 2.3, ∆1(t, ǫ) tends a.s. to 0, uniformly on the compact sets. We 1 1 will prove the convergence of Iǫ (t) (resp. Iǫ (t)) in step a) (resp. b)) below.

b1 e a) First, we will show that Iǫ (t) tends to 0 almost surely. The key of our 1 approach is to write Iǫ (t) as an integral with respect to dV , through Fubini’s theorem: b b

t (s+ǫ) t t 1 1 ∧ Iǫ (t)= 1I 0

b

11 We observe that 1 u lim 1I 0

t u 1 1 Iǫ (t)= 1I 0

e Let us suppose that T is bounded. It is clear that

t u 2 1 1 < Iǫ >t = 1I 0u, 0 ǫ (u ǫ)+ { } − { }! Z Z − e T 6 4du=4 T Z0

1 2 Consequently, (Iǫ )t [0,T ] is a martingale which is bounded in L (Ω). This allows ∈ to apply Doob’s inequality: e

1 2 1 2 E sup (Iǫ (t)) 6 4E Iǫ (T ) , (3.10) 06t6T !    e eT 1 u 2 6 4E 1I 0u .  0 ǫ (u ǫ)+ { } − { }!  Z Z −   By using (3.9) and by repeating the arguments developed in item a) above, we may conclude that E sup (I1(t))2 goes to 0 when ǫ 0. This implies 06t6T ǫ → 1   that limǫ 0 (ucp) Iǫ (t)=0. → e e Since T is not necessarly bounded, let us introduce the following se-

12 quence of stopping times:

Tn = inf t > 0,t> n , { } with the convention inf =+ . Since M T n is a local martingale so that T n ∅ ∞ 1 < M >t=t Tn 6 n, then, for any n > 0, Iǫ (t Tn) goes to 0 ∧ ∧ as ǫ 0, in the (ucp) sense. Using the fact that Tn is a non decreasing sequence→ of stopping time converging to + as n e+ , it follows that 1 2 ∞ → ∞ limǫ 0 (ucp) Iǫ (t)=0. → 3. Proof of Propositione 3.3. Let us now suppose that X is a diffusion which satisfies (1.5) and (1.6) . Since (1I 00 is not a ( s)-adapted process, { } F we are not allowed to use Fubini’s theorem as in point 2. above. We will use time reversal. Roughly speaking, time reversal allows to reduce the study of 2 1 Iǫ (t) to the one of Iǫ (t). First, we make the change of variable u = T s ǫ 2 − − and we write Iǫ (t) as a sum of two terms:

2 2 Iǫ (t)= Iǫ (t) + ∆2(ǫ, t), (3.11) with e

T ǫ 2 1 − Iǫ (t) = 1I ǫ6t (XT u XT u ǫ)1I 0

By Lemma 2.3, ∆2(t, ǫ) tends a.s. to 0, uniformly on the compact sets. Thus, 2 2 the convergence of Iǫ (t) implies the one of Iǫ (t) to the same limit. Let us recall that X has been defined by (1.7). By Theorem 5.1 of [25], e (Xu)u [ 0,T ] is a diffusion which satisfies (1.8). Let us note that < X >t= ∈ t 2 f σ (T s, Xs)ds. Then, 0f − f R T ǫ f 2 1 − I (t)= 1I 6 (Xu ǫ Xu)1I du. ǫ ǫ t + 0

T lim(ucp) I2(t)= 1I dX . ǫ 0

f f 2 + + 1 0 0 lim(ucp) Iǫ (t)= Xt X + (LT (X) LT t(X)). ǫ 0 0 → − 2 − − e f f 13 0 0 x In a last step, we express (Lt (X)t [0,T ] via (Lt (X))t [0,T ]. Since x Lt (X) is ∈ ∈ right-continuous, we obtain: → f f

T 0 0 1 L (X) L (X) = lim 1I d< X >u, T T t 0s= Lt (X). (3.12) α 0 α 0 { } → Z 2 As a result, I2(t) tends in the ucp sense to X+ X+ + 1 L0(X) when ǫ 0. 2 ǫ t − 0 2 t → e

4 Proof of Theorem 1.2

We have already observed in the proof of Theorem 1.1 (see Section 3) that 2 time reversal property allows to reduce the convergence of Iǫ (t) to the one 1 4 of Iǫ (t). The same property permits again to obtain the convergence of Iǫ (t) 3 3 via the one of Iǫ (t). We begin with the study of Iǫ (t) in point 1, and we will 4 deduce the convergence of Iǫ (t) in point 2.

1. Proof of point i). In this part, X will be a continuous semimartingale with canonical decomposition X = M + V . Reasoning by stopping (see point 2.b of the proof of Proposition 3.1) allows to suppose that M, and 3 the total variation of V are bounded processes. Let us recall that Iǫ (t) has 3 1 0 been defined by (1.10). Our aim is to prove that Iǫ (t) goes to 2 Lt as ǫ 0. Our approach is based mainly on Tanaka’s formula and Fubini’s theorem.→

1.a. By using Tanaka’s formula, we get:

(u+ǫ) t + + ∧ 1 0 0 X(u+ǫ) t = Xu + 1I Xs>0 dXs + (L(u+ǫ) t(X) Lu(X)). ∧ Zu { } 2 ∧ − + Since Xu 1I Xu60 du = 0, integrating the previous relation over [0, T ] gives { }

t t (u+ǫ) t 1 + 1 ∧ X u ǫ t1I Xu60 du = 1I Xu60 1I Xs>0 dXs ( + ) { } { } { } ǫ Z0 ∧ ǫ Z0 " Zu 1 0 0 + (L(u+ǫ) t(X) Lu(X))1I Xu60 du. 2 ∧ − { }

The process (1I Xu60 )u>0 is adapted, thus { }

14 t t (u+ǫ) t 1 + 1 ∧ X(u+ǫ) t1I Xu60 du = 1I Xs>0 1I Xu60 dXs du ǫ Z0 ∧ { } ǫ Z0 Zu { } { } ! t 1 0 0 + (L(u+ǫ) t(X) Lu(X))1I Xu60 du. (4.1) { } 2ǫ Z0 ∧ − + The same method applies to X− instead of X . Combining the two expressions comes to:

t (u+ǫ) t 3 1 ∧ Iǫ (t)= 1I Xs>0,Xu60 dXs du ǫ Z0 Zu { } ! t (u+ǫ) t 1 ∧ 1I Xs60,Xu>0 dXs du − ǫ Z0 Zu { } ! t 1 0 0 + (L(u+ǫ) t(X) Lu(X))du. 2ǫ Z0 ∧ −

1 t 0 0 1.b. Study of 2ǫ 0 (L(u+ǫ) t(X) Lu(X))du. 0 ∧ 0− First, we write (LR(u+ǫ) t(X) Lu(X)) as a Stieltjes integral with respect to 0 ∧ − dL. (X):

t t (u+ǫ) t 1 0 0 1 ∧ 0 (L(u+ǫ) t(X) Lu(X))du = dLs(X) du. 2ǫ Z0 ∧ − 2ǫ Z0 Zu ! Next, Fubini’s theorem gives

t t s 1 0 0 1 1 0 (L(u+ǫ) t(X) Lu(X))du = du dLs(X), 2ǫ 0 ∧ − 2 0 ǫ (s ǫ)+ ! Z Z Z − t 1 s ǫ 0 = ∧ dLs(X). 2 Z0 ǫ 0 t 0 Since Lt (X)= 0 dLs(X), we obtain

t R T 0 0 0 s ǫ 0 (L(u+ǫ) t(X) Lu(X))du Lt (X) 6 ∧ 1 dLs(X), 0 6 t 6 T. Z0 ∧ − − Z0 ǫ −

s ǫ Observe that ∧ 1 is bounded by 2 and tends to 0 for all s ]0, T ] as ǫ 0. ǫ − ∈ → Consequently, Lebesgue’s theorem implies

t 1 0 0 1 0 lim (L(u+ǫ) t(X) Lu(X))du Lt (X)=0, a.s., ǫ 0 ǫ 0 ∧ → 2 Z − − 2 uniformly on t [0, T ]. ∈ 1 t (u+ǫ) t 1.c. Study of 0 u ∧ 1I Xs>0,Xu60 dXs du. ǫ { } Obvioulsy, R R 

15 t (u+ǫ) t t (u+ǫ) t 1 ∧ 1 ∧ 1I Xs>0,Xu60 dXs du = 1I Xs>0,Xu60 dVs du ǫ Z0 Zu { } ! ǫ Z0 Zu { } ! t (u+ǫ) t 1 ∧ + 1I Xs>0,Xu60 dMs du. { } ǫ Z0 Zu ! Since V has bounded variation, we can proceed as in step 1.b. above.

t (u+ǫ) t t s 1 ∧ 1 1I Xs>0,Xu60 dVs du = 1I Xs>0 1I Xu60 du dVs , ǫ 0 u { } ! 0 { } ǫ (s ǫ)+ { } ! Z Z Z Z −

T 1 s 6 1I Xs>0 1I Xu60 du d V s. 0 { } ǫ (s ǫ)+ { } | | Z Z −

1 s Let s [0, T ] so that Xs > 0. Since t Xt is continuous, ǫ (s ǫ)+ 1I Xu60 du = − { } ∈ → 1 s 0 as soon as ǫ is small enough. Observing that ǫ (s ǫ)+ 1I XRu60 du is bounded − { } by 1 allows to apply Lebesgue’s convergence theorem: R

t (u+ǫ) t 1 ∧ lim 1I Xs>0,Xu60 dVs du =0, ǫ 0 ǫ 0 u { } → Z Z ! a.s, uniformly with respect to t [0, T ]. ∈ Next we claim that

t (u+ǫ) t 1 ∧ lim (ucp) 1I Xs>0,Xu60 dMs du =0. (4.2) ǫ 0 ǫ 0 u { } → Z Z !

We may mimic the former analysis which involves dVs, since we have at our disposal a stochastic version of Fubini’s theorem (c.f. Lemma 2.1). More pre- cisely, t (u+ǫ) t t 1 ∧ 1I Xs>0,Xu60 dMs du = g(ǫ, s)dMs, { } ǫ Z0 Zu ! Z0 with 1 s g(ǫ, s) = 1I Xs>0 1I Xu60 du. { } ǫ (s ǫ)+ { } Z − Doob’s inequality comes to:

t 2 T 2 E sup g(ǫ, s)dMs 6 4E g(ǫ, s) ds . t [0,T ] 0 ! 0 | | ! ∈ Z  Z It is now clear that (4.2) holds.

1 t (u+ǫ) t 1.d. Study of 0 u ∧ 1I Xs60,Xu>0 dXs du ǫ { } 1 t (u+ǫ) t  Although 0 u R ∧R 1I Xs60,Xu>0 dXs du is quite similar to ǫ { } R R 

16 1 t (u+ǫ) t 0 u ∧ 1I Xs>0,Xu60 dXs du, the difference between large and strict in- ǫ { } equalityR R forces us to study it independently. Similary to point 1.c, we have:

t (u+ǫ) t t (u+ǫ) t 1 ∧ 1 ∧ 1I Xs60,Xu>0 dXs du = 1I Xs60,Xu>0 dVs du ǫ Z0 Zu { } ! ǫ Z0 Zu { } ! t (u+ǫ) t 1 ∧ + 1I Xs60,Xu>0 dMs du. ǫ Z0 Zu { } ! For the first term, we have

t (u+ǫ) t T (u+ǫ) t 1 ∧ 1 ∧ 1I Xs60,Xu>0 dVs du 6 1I Xu>0 1I Xs60 dVs du, ǫ 0 u { } ! 0 { } ǫ u { } Z Z Z Z

for 0 6 t 6 T . Let u [0, T ] so that Xu > 0. Using the continuity of X , it 1 (u+ǫ) t ∈ is clear that u ∧ 1I Xs60 dVs = 0 as soon as ǫ is small enough. Moreover, ǫ { } this term is boundedR by the total variation of V . Thus, Lebesgue’s convergence theorem gives

T (u+ǫ) t 1 ∧ lim 1I Xs60,Xu>0 dVs du =0, ǫ 0 ǫ 0 u { } → Z Z ! a.s, uniformly with respect to t [0, T ]. ∈ For the second term, we proceed as in point 1.c. Fubini’s stochastic theorem gives

t (u+ǫ) t t 1 ∧ 1I Xs60,Xu>0 dMs du = 1I Xs60 g˜(ǫ, s)dMs, ǫ Z0 Zu { } ! Z0 { } with 1 s g˜(ǫ, s)= 1I Xu>0 du. ǫ (s ǫ)+ { } Z − Doob’s inequality comes to:

t 2 T 2 E sup 1I Xs60 g˜(ǫ, s)dMs 6 4E 1I Xs60 g˜(ǫ, s) ds . t [0,T ] 0 { } ! 0 { }| | ! ∈ Z  Z

T 2 We claim that E 0 1I Xs60 g˜(ǫ, s) ds tends to 0 as ǫ 0. Indeed { }| | → we decompose thisR term as a sum of two terms:

T T 2 2 E g˜(ǫ, s) ds = E 1I Xs<0 g˜(ǫ, s) ds Z0 | | ! Z0 { }| | ! T 2 +E 1I Xs=0 g˜(ǫ, s) ds . Z0 { }| | !

17 Since Xs < 0 in the first term,g ˜(ǫ, s) = 0 as soon as ǫ is small enough and, by Lebesgue’s convergence theorem, this term converges to 0 as ǫ 0. Since T → g˜(ǫ, s) 6 1, the second term may be bounded by E 0 1I Xs=0 ds . By { } the occupation times formula, R  T x 1I Xs=0 ds= 1I x=0 LT dx =0. R Z0 { } Z { } 1 t (u+ǫ) t 2 Thus, supt [0,T ] ǫ 0 u ∧ 1I Xs60,Xu>0 dMs du tends to 0 in L (Ω) as ǫ 2 ∈ { } → 0. R R 

3 Remark 4.1 By definition, Iǫ (t) is a sum of two terms. We have not been able to prove that each term converges. This explains that they have been gathered. Let us explain where comes the difficulty. Let us start with (4.1) and let use Fubini’s theorem in the integral which contains the local time. We get

t t s 0 0 1 0 (L(u+ǫ) t(X) Lu(X))1I Xu60 du = 1I Xu60 du dLs(X). 0 ∧ − { } 0 ǫ (s ǫ)+ { } ! Z Z Z −

1 s We recall that ǫ (s ǫ)+ 1I Xu60 du tends to 1I Xs60 almost surely with respect − { } { }   1 s to Lebesgue measure.R This result cannot guarantee that (s ǫ)+ 1I Xu60 du ǫ { } 0 0− converges dLs(X)-everywhere, since the random measure dLR s(X) is singular with respect to Lebesgue measure.

2. Proof of point ii). Let us now suppose that X is a diffusion which satisfies (1.5) and (1.6) . We will use time reversal. 4 First, we decompose Iǫ (t) as

t ǫ t ǫ 4 1 − 1 − + 4,1 Iǫ (t)= Xu−1I Xu+ǫ>0 du + Xu 1I Xu+ǫ60 du + Rǫ (t), (4.3) ǫ Z0 { } ǫ Z0 { } with t t 4,1 1 1 + Rǫ (t)= Xu−1I Xt>0 du + Xu 1I Xt60 du. ǫ t ǫ { } ǫ t ǫ { } Z − Z − 1 t 1 t Since ǫ t ǫ Xu−1I Xt>0 du = ǫ t ǫ(Xu− Xt−)1I Xt>0 du, Lemma 2.3 applies and − { 4,1 } − − { } the firstR term of Rǫ (t) convergesR a.s. to 0. The second term in the definition 4,1 4,1 Rǫ (t) above converges to 0 by the same way. Thus, Rǫ (t) converges a.s. to 0 as ǫ 0, uniformly on [0, T ]. → Next, we make the change of variable s = T u ǫ in the two integrals in − − (4.3):

T ǫ T ǫ 4 1 − 1 − + 4,1 Iǫ (t)= XT− s ǫ1I XT −s>0 ds + XT s ǫ1I XT −s60 ds + Rǫ (t). ǫ T t − − { } ǫ T t − − { } Z − Z −

We recall that X is defined by (1.7). By Theorem 5.1 of [25], (Xu)u [0,T ] is a ∈ semimartingale. We obtain f f

18 T ǫ T ǫ 4 1 − 1 − + 4,1 I (t)= X− 1I ds + X 1I ds + R (t), ǫ s+ǫ Xs>0 s+ǫ Xs60 ǫ ǫ T t { } ǫ T t { } Z − Z − T T 1 f 1 f + = X− e1I ds + X e 1I ds (4.4) (s+ǫ) T Xs>0 (s+ǫ) T Xs60 " ǫ T t ∧ { } ǫ T t ∧ { } # Z − Z − 4,2 f f +Rǫ (t), e e with

T T 4,2 4,1 1 1 + R (t)= R (t) X−1I ds + X 1I ds. ǫ ǫ T Xs>0 T Xs60 − ǫ T ǫ { } ǫ T ǫ { } Z − Z − f f It is clear that R4,2(t) converges a.s to 0e as ǫ 0, uniformly one [0, T ]. ǫ → Since X is a semimartingale, point i) of Theorem 1.2 may be applied and the term in bracket in (4.4) converges. Moreover, using (3.12), we get: f

4 1 0 0 1 0 lim (ucp) Iǫ (t)= (LT (X) LT t(X)) = Lt (X). ǫ 0 → 2 − − 2 2 f f

5 Proof of Theorem 1.4

In this section, X will be a standard Brownian motion, and we fix δ ]0, 1 [ ∈ 2 (c.f. Lemma 2.2). Since ( X) is a standard Brownian motion, then − 3,1 3,2 4,1 4,2 lim Iǫ (t) = lim Iǫ (t), lim Iǫ (t) = lim Iǫ (t). ǫ 0 ǫ 0 ǫ 0 ǫ 0 → → → → Consequently, we have to prove three distinct properties:

4,2 1 0 (1) the (ucp) convergence of Iǫ (t) to 4 Lt (X) (see point 2.), 3,2 1 0 (2) the (ucp) convergence of Iǫ (t) to 4 Lt (X) (see point 3.), (3) the rate of decay of I4,i(t) 1 L0(X) as ǫ 0, i =1, 2 (see point 2.). ǫ − 4 t → 3 4 1. First, we briefly show that rǫ (t) and rǫ (t) (defined by (1.14) and (1.16)) tend to 0 as ǫ 0, uniformly on [0, T ]. By the occupation times formula, → t x 1I Xu=0 du = 1I x=0 Lt (X)dx =0. { } R { } Z0 Z 3 4 Consequently, rǫ (t) vanishes. Next, we decompose rǫ (t) and we make the change of variable v = u + ǫ:

t t 4 1 + 1 + rǫ (t)= Xv ǫ1I Xv=0 dv + Xu 1I Xt=0 du. ǫ ǫ − { } ǫ t ǫ { } Z Z −

19 4 As previously, the first integral in rǫ (t) vanishes. By Lemma 2.3, the second term tends a.s. to 0 as ǫ 0, uniformly on [0, T ], and we have → 4 δ r (t) 6 Cδǫ . (5.1) | ǫ |

4,2 2. Next, we study the convergence of Iǫ (t) (defined by (1.15)). Since 1I X <0 is not ( u)-measurable, we ”approximate” it by { (u+ǫ)∧t } F E(1I X <0 u). Let us remark that this term is a function of Xu. Indeed, { (u+ǫ)∧t }|F

E(1I X <0 u)= E(1I X Xu< Xu Xu), { (u+ǫ)∧t }|F { (u+ǫ)∧t− − }| X = Φ u , − ǫ (t u) ∧ −  q  where Φ is the distribution function of the (0, 1)-law. N 4,2 We introduce it in Iǫ (t). This leads us to consider a new decomposition of 4,2 Iǫ (t): t 4,2 1 2 1 + Xu Iǫ (t)= Aǫ (t)+ Aǫ (t)+ Xu Φ du, (5.2) ǫ Z0 −√ǫ ! where

t 1 1 + Xu Aǫ (t)= Xu 1I X <0 Φ du, (5.3)  { (u+ǫ)∧t }   ǫ Z0 − − ǫ (t u) ∧ − t   q  2 1 + Xu Xu Aǫ (t)= Xu Φ du Φ . (5.4) ǫ (t ǫ)+ " −√t u! − −√ǫ !# Z − −

1 t + Xu 1 0 The main term is ǫ 0 Xu Φ √ǫ since it converges to 4 Lt (X) (see point − 1 2 2.a. below). In pointR2.b, we will show that Aǫ (t)+ Aǫ (t) tends to 0.

2.a. Study of 1 t X+Φ Xu du. By the occupation times formula, ǫ 0 u − √ǫ R   t 1 + Xu 1 + x x Xu Φ du = x Φ Lt (X)dx. R ǫ Z0 −√ǫ ! ǫ Z −√ǫ! We make the change of variable y√ǫ = x and we get

t 1 + Xu ∞ y√ǫ Xu Φ du = yΦ( y)Lt (X)dy. ǫ Z0 −√ǫ ! Z0 − 1 Since ∞ yΦ( y)dy = , we have 0 − 4 Rt 1 + Xu 1 0 ∞ √ǫy 0 Xu Φ du Lt (X)= yΦ( y) Lt (X) Lt (X) dy. ǫ 0 −√ǫ ! − 4 0 − − Z Z  

20 By the H¨older property of the Brownian local time (c.f. Lemma 2.2),

t 1 + Xu 1 0 ∞ δ+1 δ Xu Φ Lt (X) 6 Kδ y Φ( y)dy ǫ 2 . (5.5) ǫ 0 −√ǫ ! − 4 0 − Z  Z 

Therefore 1 t X+Φ Xu converge to 1 L0(X) a.s, uniformly on [0, T ]. ǫ 0 u − √ǫ 4 t R   1 2 1 2.b. Study of Aǫ (t)+ Aǫ (t). First, we modify Aǫ (t) as follows:

(t ǫ)+ 1 1 − + Xu Aǫ (t)= Xu 1I Xu+ǫ<0 Φ du ǫ Z0 " { } − −√ǫ !# t 1 + Xu + Xu 1I Xt<0 Φ du. ǫ (t ǫ)+ " { } − −√t u!# Z − −

Xu Secondly, we decompose 1I Xt<0 Φ as: { } − − √t u  − 

Xt Xu Xu 1I Xt<0 Φ − + Φ − Φ − { } − √u + ǫ t! " √ǫ ! − √t u!# − − Xt Xu + Φ − Φ − . " √u + ǫ t! − √ǫ !# − + + Multiplying by Xu and integrating over [(t ǫ) , t] give rise to four integrals, but we remark that the third one is equal to− A2(t). Then, − ǫ 1 2 1 2 3 4 Aǫ (t)+ Aǫ (t)= Rǫ (t)+ Dǫ (t)+ Dǫ (t)+ Rǫ (t), (5.6) where

(t ǫ)+ 1 1 − + Xu Rǫ (t)= Xu 1I Xu+ǫ<0 Φ du, { } ǫ Z0 " − −√ǫ !# t 2 1 + Dǫ (t) = 1I Xt<0 Xu du, { } ǫ (t ǫ)+ Z − t 3 1 + Xt Dǫ (t)= Xu Φ − du, − ǫ (t ǫ)+ √u + ǫ t! Z − − t 4 1 + Xt Xu Rǫ (t)= Xu Φ − Φ − du. ǫ (t ǫ)+ " √u + ǫ t! − √ǫ !# Z − − For u fixed, Itˆo’s formula applied to φ(x, s) = Φ x , − √u+ǫ s (φ(x, s) 2(]0, + [, [u,u + ǫ[), comes to:  −  ∈ C ∞ X2 v s exp 2(u+ǫ s) − − φ(Xv, v) φ(Xu,u)=  dXs, v [u,u + ǫ[. − − Zu 2π(u + ǫ s) ∀ ∈ − q

21 Since the Lebesgue measure of u; Xu+ǫ =0 vanishes, this formula, used for v = u + ǫ if u [0, (t ǫ)+], and{ for v = t if}u ](t ǫ)+, t], leads to ∈ − ∈ −

X2 u+ǫ s Xu exp 2(u+ǫ s) − − dX , 1I Xu+ǫ<0 Φ =   s (5.7) { } − −√ǫ ! − Zu 2π(u + ǫ s) − q X2 t s Xt Xu exp 2(u+ǫ s) − − Φ Φ =  dXs. (5.8) −√u + ǫ t! − −√ǫ ! − Zu 2π(u + ǫ s) − − q 1 2 Using (5.7) and (5.8) permits to obtain a new expression of Aǫ (t)+ Aǫ (t):

1 2 1 2 3 Aǫ (t)+ Aǫ (t)= Dǫ (t)+ Dǫ (t)+ Dǫ (t), (5.9) where X2 t (u+ǫ) t exp s 1 1 + ∧ 2(u+ǫ s) D (t)= X − − dX du. ǫ u    s − ǫ Z0 Zu 2π(u + ǫ s) −  q i In the rest of the paragraph, we will prove that each term Dǫ(t) tends to 0, 2 3 uniformly on the compact set. Let us observe that, in Dǫ (t) and Dǫ (t), the lenghts of the intervals of integration are smaller than ǫ. It will be shown that the convergence of these terms holds almost surely and is a consequence of the 1 continuity of X. As for the first term Dǫ (t), we observe that it is a martingale (via Fubini’s theorem) and we use Doob’s inequality to get the convergence in the L2 sense.

2 + Convergence of Dǫ (t) to 0. Since 1I Xt60 Xt = 0, we have { } t 2 1 + + Dǫ (t)= (Xu Xt )1I Xt<0 du . ǫ (t ǫ)+ − { } ! Z − 2 By Lemma 2.3, Dǫ (t) tends a.s. to 0, uniformly on [0, T ] and we have

2 δ sup Dǫ (t) 6 Cδ ǫ . (5.10) t [0,T ] | | ∈

3 + 3 Convergence of Dǫ (t) to 0. Let us introduce Xt in Dǫ (t), we get

3 1 t + + Xt Dǫ (t)= ǫ (t ǫ)+ (Xt Xu )Φ √u−+ǫ t du − − − − 1 R t + Xt   + ǫ (t ǫ)+ Xt Φ √u−+ǫ t du. − − R   In the first term, by using the fact that Φ is bounded by 1 and Lemma 2.2 (H¨older property of t Xt), we get → t t 1 + + Xt Cδ δ δ (Xt Xu )Φ − du 6 | | t u du 6 Cδ ǫ . ǫ (t ǫ)+ − √u + ǫ t! ǫ (t ǫ)+ | − | | | Z − Z − −

22 2 3 −x Let us consider the second term in Dǫ (t). Since Φ(x) 6 Ce 4 for any x 6 0, we obtain:

t t X2 1 Xt C t + + 4(u+ǫ−t) Xt Φ − du 6 Xt e− du. ǫ (t ǫ)+ √u + ǫ t ! ǫ (t ǫ)+ Z − Z − −

x2 + The function defined on R by x xe− 4(u+ǫ−t) is non-negative and bounded 1 → by 2(u + ǫ t)e− 2 . Consequently, − q t t 1 + Xt 1 Xt Φ − du 6 C (u + ǫ t)du 6 C√ǫ. ǫ (t ǫ)+ √u + ǫ t! ǫ (t ǫ)+ − Z − Z − q −

3 Thus, Dǫ (t) converge a.s. to 0 uniformly on [0, T ] and

3 δ sup Dǫ (t) 6 Cδ ǫ + C√ǫ. (5.11) t [0,T ] | | | | ∈

1 Convergence of Dǫ (t) to 0. In order to use Fubini’s theorem (i.e. Theorem 2.1), first we prove that

t (u+ǫ) t ∧ 2 E Hǫ (u,s) ds du < , 0 u ! ∞ Z Z h i where + 2 Xu Xs Hǫ(u,s)= exp . ǫ 2π(u + ǫ s) −2(u + ǫ s)! − − By Cauchy-Schwarz inequality,q we have

1 2X2 E H2(u,s) 6 E[(X+)4]E exp s . ǫ 2 v u 2π(u + ǫ s)ǫ u " −(u + ǫ s)!# h i − u − t

Since Xr is a centered Gaussian random variable with variance r, we can calculate explicitly the two expectations which appear in the right-hand side of the previous inequality:

2 + 4 3 2 2Xs u + ǫ s E[(Xu ) ]= u , E exp = − . 2 " −(u + ǫ s)!# s3s + u + ǫ − Since s > u 3s + u + ǫ> 4u, reporting in the inequality gives ⇒

3 C u 4 E H2(u,s) 6 , u [0, t],s ]u,u + ǫ[. ǫ ǫ2 u + ǫ s ∀ ∈ ∈ h i  − 

23 3 t (u+ǫ) t u 4 By an easy calculation, it may be proved that 0 u ∧ u+ǫ s ds du <  −  . Consequently, Fubini’s stochastic theorem mayR R be applied , we have: ∞ t s + 2 1 1 Xu Xs Dǫ (t)= exp du dXs.  +  − Z0 Z(s ǫ) ǫ 2π(u + ǫ s) −2(u + ǫ s)! − − −  q  1 Thus, (Dǫ (t), 0 6 t 6 T ) is a square integrable martingale, and by Doob’s inequality, we get:

1 2 1 2 E sup (Dǫ (t)) 6 4E (Dǫ (T )) , t [0,T ] ! ∈ h i and

X2 2 T s X+ exp s 1 2 u − 2(u+ǫ s) E (Dǫ (T )) = E  − du ds . 0  (s ǫ)+    h i Z Z − ǫ 2π(u + ǫ s)  −    q   The Cauchy-Schwarz inequality applied to the term in brackets gives:

T s 2 1 Xs 1 2 + 2 u+ǫ−s E (Dǫ (T )) 6 E (Xu ) e− du ds. 0 " (s ǫ)+ 2πǫ(u + ǫ s) # h i Z Z − −   2 Xs + 2 u+ǫ−s Let s>u> 0, we introduce A := E (Xu ) e− . Since Xs Xu is   − independent from Xu, we have

2 (y+x)2 y2 1 ∞ 2 x A = x e− 2u e− u+ǫ−s − 2(s−u) dydx. R 2π (s u)u Z0 Z − q Since (y + x)2 y2 s u + ǫ 2(s u)x x2 + = − y + − + , u + ǫ s 2(s u) 2(s u)(u + ǫ s) s u + ǫ ! s u + ǫ − − − − − − we obtain

1 u + ǫ s (s+u+ǫ)x2 u(s u + ǫ)√u + ǫ s ∞ 2 2u(s−u+ǫ) A = − x e− dx = − 3 − . √2π su(s u + ǫ) 0 2(s + u + ǫ) 2 − Z > 6 1 6 1 Since u s ǫ s u + ǫ 2ǫ and s+u+ǫ u , reporting in the integral gives: − ⇒ −

T u+ǫ 1 2 1 du E (Dǫ (T )) 6 C ds 6 C√T ǫ. 0 u √u + ǫ s ! √u h i Z Z − Hence, 1 2 E sup Dǫ (t) 6 C√T ǫ, (5.12) "t [0,T ] # ∈  

24 1 2 and Dǫ (t) converge to 0 in L (Ω), uniformly on [0, T ].

2.c. We are now able to prove point 2. of Theorem 1.4. It is clear that (5.2) and (5.9) imply:

t 4,2 1 0 1 2 3 1 + Xu 1 0 Iǫ (t) Lt (X)= Dǫ (t)+ Dǫ (t)+ Dǫ (t)+ Xu Φ Lt (X) . − 4 ǫ Z0 −√ǫ ! − 4 ! (5.13) 4,2 1 0 The (ucp) convergence of Iǫ (t) to 4 Lt (X) and

4,2 1 0 δ sup Iǫ (t) Lt (X) 6 Cǫ 2 , t [0,T ] − 4 2 ∈ L (Ω)

is a direct consequence of inequalities (5.10), (5.11), (5.12) and (5.5).

3,2 3. Finally, we study the convergence of Iǫ (t) (defined by (1.13)).

+ We follow the same method as in point 2. Since Xu+ǫ is not ( u)-measurable, + F we ”replace” it by E(Xu+ǫ u). This leads us to consider the following decom- 3,2 |F position of Iǫ (t):

3,2 3,2 3,2 3,2 Iǫ (t)= Iǫ (t)+ Iǫ (t) + ∆ǫ (t), where e

t 3,2 1 + Iǫ (t)= E(Xu+ǫ u)1I Xu<0 du, (5.14) { } ǫ Z0 |F (t ǫ)+ e3,2 1 − + + Iǫ (t)= Xu+ǫ E(Xu+ǫ u) 1I Xu<0 du (5.15) ǫ 0 − |F { } Z t   1 + + + E(Xu+ǫ t) E(Xu+ǫ u) 1I Xu<0 du, (5.16) ǫ (t ǫ)+ |F − |F { } Z − t   3,2 1 + + ∆ǫ (t)= Xt E(Xu+ǫ t) 1I Xu<0 du. (5.17) ǫ (t ǫ)+ − |F { } Z −   3,2 1 0 The main term is Iǫ (t) since it converges to 4 Lt (X) (see point 3.a. below). 3,2 3,2 + In point 3.b , we will show that Iǫ (t) tends to 0. As for ∆ǫ (t), since Xt + e + + + − E(Xu+ǫ t) = Yt Yu+ǫ, with Ys = E(Xs t),s [(t ǫ) , t], Lemma 2.3 applies|F and ∆3,2(t−) converges a.s. to 0 as ǫ |F0, uniformly∈ − on [0, T ]. ǫ → 3,2 + 3.a. Study of I (t). First, let us remark that E(X u) is a function of ǫ u+ǫ|F Xu. Indeed, e + Xu E(Xu+ǫ u)= √ǫg , |F √ǫ ! where g(x)= E((G+x)+) and G is a Gaussian random variable with (0, 1)- N law.

25 By the occupation times formula, we have:

0 3,2 1 x x Iǫ (t)= g Lt (X)dx. √ǫ √ǫ ! Z−∞ e The change of variable y√ǫ = x gives

0 3,2 √ǫy Iǫ (t)= g(y)Lt (X)dy. Z−∞ e Since

0 0 + 0 g(y)dy = E (G + y) dy = E 1I G>0 G(G + y)dy , −∞ −∞ { } − R 1 R + 2 1   R  = 2 E((G ) )= 4 , we have 1 0 I3,2(t) L0(X)= g(y)(L√ǫy(X) L0(X))dy. ǫ − 4 t t − t Z−∞ Consequently,e the H¨older property of the Brownian local time (c.f. Lemma 2.2), implies that:

0 3,2 1 0 δ + δ δ Iǫ (t) Lt (X) 6 ǫ 2 KδE (G + y) y dy 6 Cǫ 2 . − 4 Z  −∞ e 3,2 1 0 Therefore Iǫ (t) converge to 4 Lt (X) a.s, uniformly on [0, T ].

e 3,2 3,2 3.b Study of Iǫ (t). Our goal is to show that Iǫ (t) is a martingale, in order + + to apply Doob’s inequality. To begin with, we write X E(X u) and u+ǫ − u+ǫ|F + +   E(X t) E(X u) as stochastic integrals using Ito’s formula. u+ǫ|F − u+ǫ|F   + Let u, ǫ be fixed. We define Ms = E(Xu+ǫ s),s [u,u + ǫ[. (Ms)s [u,u+ǫ[ is a ∈ martingale and |F ∈

+ Ms = E((Xu ǫ Xs + Xs) s)= f(s,Xs), + − |F where

+ + ∞ f(s,y)= (x + y) pu+ǫ s(x)dx = zpu+ǫ s(z y)dz, R Z − Z0 − −

2 − x e 2(u+ǫ−s) and pu+ǫ s(x)= . − √2π(u+ǫ s) −

2 Since 1 ∂ f (s,y) = ∂f (s,y), applying Itˆo’s formula to f 1,2([u,u + ǫ[, R) 2 ∂y2 − ∂s ∈ C comes to:

26 u+ǫ ∂f + f(u + ǫ, Xu+ǫ) f(u,Xu)= (s,Xs)dXs, u [0, (t ǫ) ],(5.18) − Zu ∂y ∈ − t ∂f + f(t, Xt) f(u,Xu)= (s,Xs)dXs, u [(t ǫ) , t]. (5.19) − Zu ∂y ∈ −

∂f + Let us evaluate ∂y (s,y). Writing f(s,y) = y∞(x + y)pu+ǫ s(x)dx allows to − − calculate the y-derivative of f: R

+ ∂f ∞ y (s,y)= pu+ǫ s(x)dx =1 Φ − , ∂y − y − − √u + ǫ s! Z− − where Φ is the distribution function of the standard Gaussian distribution.

Let us observe that

+ + f(u + ǫ, Xu ǫ) f(u,Xu)= Mu ǫ Mu = X E(X u), + − + − u+ǫ − u+ǫ|F and

+ + f(t, Xt) f(u,Xu)= Mt Mu = E(X t) E(X u). − − u+ǫ|F − u+ǫ|F 3,2 As a result, reporting (5.18) and (5.19) in Iǫ (t) gives:

t (u+ǫ) t 3,2 1 ∧ Xs Iǫ (t)= (1 Φ − )dXs 1I Xu<0 du. ǫ 0 " u − √u + ǫ s! # { } Z Z − Since x 1 Φ(x) is uniformly bounded by 1, Fubini’s theorem (i.e. Theorem → − 2.1) may be applied, we have

t s 3,2 1 Xs Iǫ (t)= (1 Φ − )1I Xu<0 du dXs. 0 " ǫ (s ǫ)+ − √u + ǫ s! { } # Z Z − − 3,2 Thus, (Iǫ (t), 0 6 t 6 T ) is a square integrable martingale. By Doob’s in- equality, we get:

T s 2 3,2 2 1 Xs E sup Iǫ (t) 6 4E (1 Φ − )1I Xu<0 du ds . t [0,T ] | | !  0 " ǫ (s ǫ)+ − √u + ǫ s! { } #  ∈ Z Z − −   Finally, we will show that the right-hand side of the inequality above converges T to 0. Recall that 0 1I Xs=0 ds = 0, let us introduce two cases: Xs < 0 and { } Xs > 0. R We have: • s 1 Xs 1I Xs>0 (1 Φ − )1I Xu<0 du { } ǫ (s ǫ)+ − √u + ǫ s! { } Z − −

27 1 s 6 1I Xs>0 1I Xu<0 du . { } ǫ (s ǫ)+ { } ! Z − We have already proven that the right-hand side of the prior inequality goes to 0, as ǫ 0, a.s. and in L1(Ω). → α2 if Xs < 0, by using 1 Φ(α) 6 Ce− 4 for any α> 0, we get • | − |

s s X2 1 Xs C s 4(u+ǫ−s) (1 Φ − )1I Xu<0 du 6 e− du. ǫ (s ǫ)+ − √u + ǫ s! { } ǫ (s ǫ)+ Z − Z − −

The change of variable v = u + ǫ s gives: − s ǫ 2 1 Xs C Xs 4v (1 Φ − )1I Xu<0 du 6 e− dv. ǫ (s ǫ)+ − √u + ǫ s! { } ǫ (ǫ s)+ Z − Z − − 2 ǫ Xs 1 4v 1 Since Xs = 0, ǫ (ǫ s)+ e− dv converges to 0 a.s and in L (Ω), as ǫ 0. 6 − → R 2 T 1 s Xs Then, by Lebesgue Theorem, E 0 ǫ (s ǫ)+ (1 Φ √u−+ǫ s )1I Xu<0 du ds  − − − { }  converges to 0. 2 R h R   i

6 Proofs of Theorem 1.6 and Proposition 1.7

1. In this Section, X is supposed to be the standard Brownian motion. It is convenient to adopt the convention that Xs = 0 if s < 0. The proof of Theorem 1.6 is based on the identity

t 0 1 2 + 1 0 Jǫ(t) Lt (X)= Iǫ (t) 1I 0

1 2. Study of the convergence of Iǫ (t). We will use (3.5). Let us remark 1 that Iǫ (t) = 0 since X is a martingale. Therefore (3.5) reduces to:

b t 1 1 Iǫ (t) 1I 0

28 1 Let us now deal with Iǫ (t). Recall that u= u, the inequality (3.10) becomes: e T u 2 1 2 1 E sup (Iǫ (t)) 6 4E 1I 0

′ 2 1I s

81I ′ 1 2 6 T {s

E (1I 0

T 1I s

1 2 E sup (Iǫ (t)) 6 C√T √ǫ. (6.4) 06t6T ! e

2 2 + 3. Study of the convergence of Iǫ (t). First, we decompose Iǫ (t) Xt 1 0 − − 2 Lt (X) as:

2 + 1 0 2 + 1 0 Iǫ (t) Xt Lt (X)= Iǫ (t) Xt Lt (X) + ∆2(t, ǫ), (6.5) − − 2  − − 2  e where

29 t 2 1 Iǫ (t)= (X(s+ǫ) t Xs)1I 0

t t+ǫ

X(s+ǫ) t1I 0

= ǫXt1I 0

Xu1I 0

t ǫ 2 + 1 1 + Iǫ (t)= Xt + Xu 1I 0

ǫ 2 + 4,1 4,2 1 + 4 Iǫ (t)= Xt + Iǫ (t)+ Iǫ (t) Xu du + rǫ (t), − ǫ Z0 e 4,1 4,2 4 where it is recalled that Iǫ (t), Iǫ (t),rǫ (t) are defined by (1.15)-(1.16). Hence, we get

2 + 1 0 4,1 1 0 4,2 1 0 Iǫ (t) Xt Lt (X)= Iǫ (t) Lt (X) + Iǫ (t) Lt (X) − − 2  − 4   − 4  ǫ e 1 + 4 Xu du + rǫ (t). (6.7) − ǫ Z0 According to Lemma 2.2, we have:

ǫ 1 + Cδ δ sup Xu du 6 ǫ . (6.8) t [0,T ] ǫ Z0 δ +1 ∈

30 Then, using (5.1) and Theorem 1.4 comes to

2 + 1 0 δ sup Iǫ (t) Xt Lt (X) 6 Cǫ 2 . (6.9) t [0,T ] − − 2 2 ∈ L (Ω) e

It is clear that Theorem 1.6 is a direct consequence of (5.1),(6.1), (6.2), (6.3), (6.4), (6.6), (6.8) and (6.9). 2

4. We now demonstrate Proposition 1.7. Let us consider a sequence (ǫn)n N ∈ of positive real numbers decreasing to 0, such that ∞ √ǫi < . i=1 ∞ P I4,2 t 1 L0 X n First, we show the almost sure convergence of ǫn ( ) to 4 t ( ), as . Note that Theorems 1.4 and 1.6 do not permit to obtain the a.s. convergence→ ∞ results stated in Proposition 1.7, via the Borel Cantelli lemma, since we cannot 1 take δ = 2 . Recall the identity (5.13):

t 4,2 1 0 1 2 3 1 + Xu 1 0 Iǫ (t) Lt (X)= Dǫ (t)+ Dǫ (t)+ Dǫ (t)+ Xu Φ Lt (X) . − 4 ǫ Z0 −√ǫ ! − 4 ! According to (5.5), (5.10) and (5.11) , the quantities 1 t X+Φ Xu 1 L0(X) , D2(t) and D3(t) goes to 0, as ǫ 0, with a ǫ 0 u − √ǫ − 4 t ǫ ǫ →    1 rateR of decay of order δ < 2 . But it does not matter since the convergence holds in the almost sure sense.

1 1 2 We now focus on Dǫ (t). According to (5.12), we have E supt [0,T ] (Dǫ (t)) 6 ∈ C√ǫ. Then, the Borel Cantelli Lemma implies that, forh all t > 0, D1 (t) i ǫn n N converge almost surely uniformly on [0, T ]. Hence, we get the a.s. convergence ∈ of I4,2(t) to 1 L0(X), as n . ǫn 4 t →∞ I4,1 t 1 L0 X The convergence of ǫn ( ) to 4 t ( ) may be obtained by using the symmetry of Brownian motion.

Let us now investigate the convergence of Jǫn (t). It is clear that (6.1), (3.5), 0 (6.5) and (6.7) imply that Jǫ(t) L (X) is equal to: − t I1(t) + ∆1(t)+ I4,1(t) 1 L0(X) + I4,2(t) 1 L0(X) ǫ ǫ ǫ − 4 t ǫ − 4 t  2 4 1 ǫ +  e +∆ (t)+ r (t) X du. ǫ ǫ − ǫ 0 u R 4,2 We proceed as in the convergence of Iǫn (t). From inequalities (6.3), (6.6), (5.1) 1 2 4 1 ǫ + and (6.8), it may be deduced that ∆ǫ (t) + ∆ǫ (t)+ rǫ (t) ǫ 0 Xu du tends − 4,i 1 0 a.s to 0 as ǫ 0. Note that we have already shown that I (Rt) L (X) , → ǫn − 4 t i = 1, 2, converges a.s. as n . Finally, the a.s. convergence of I1 (t), as → ∞ ǫn n , may be obtained through (6.4). 2 →∞ e

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