Density of paths of iterated Levy transforms of Brownian motion.

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Density of paths of iterated Levy transforms of

Brownian motion.

Marc Malric

To cite this version:

Marc Malric.

Density of paths of iterated Levy transforms of Brownian motion..

2007.

�hal-00013282v5�

Brownian motion

Mar Malri

June 24, 2009

Abstra t : The Lévy transform of a Brownian motion B is the Brownian motion

B

′

t

=

Z

t

0

sgn (B

s

) dB

s

. Call

T

the orresponding transformation on

the Wiener spa e

W

. We establishthat a. s. the orbit of

w(∈ W )

under

T

is dense in

W

for the ompa t uniform onvergen e topology.

related tools.

Let

(Ω, F , P)

betheprobabilityspa ewhereallrandomelementsaredened, and

(W, W, π)

the Wiener spa e. Any measurable mapfrom

Ω

toany mea-sure spa e, dened

P

−

a.e., will be alled a random variable. If

X

is a r.v. withvaluesinsome measurablespa e

A

,theprobabilitymeasure

P

◦ X

−1

on

A

is alled the law of

X

, and denoted by

L(X)

. Forinstan e, a

W

−

valued r.v. with law

π

is aBrownian motion.

The Lévytransform

T

: W −→ W

isdened

π−

a.e. and preserves

π

. Given a Brownian motion

B

, we denote by

B

n

its

n−

Lévy iterate, that is, the Brownian motion

B

n

_{= T}

n

_{◦ B}

.

From nowon,

T > 0

,

ϕ ∈ C([0, T ], R)

and

ε > 0

arexed, and

B

aBrownian motion. Thegoalistoprovethattheevent

E = E

ε

_{= {∀n ≥ 0, kB}

n

_−ϕk

∞

>

ε}

isnegligible,where

kf k

∞

= kf

|[0,T ]

k

∞

. Itsu esinfa ttoshow

P

(E) < ε

, be ause

E

ε

1

_{⊂ E}

ε

2

when

ε

2

< ε

1

.

The idea is to onstru t from B another sto hasti pro ess

Γ : Ω −→ W

, whi hdepends on

T

,

ϕ

and

ε

, and has the following threeproperties :

(i) The law of the pro ess

Γ

, i.e., the probability

F −→ P(Γ

−1

_{F )}

on

(W, W)

, isabsolutely ontinuous w.r.t. the law

π

of

B

.

(ii) Forsomedeterministi

r ≥ 0

,onehas

Γ

r

_{= B}

r

,thatis,

T

r

_{◦Γ = T}

r

_◦B

. (iii)

P

(∀n ≥ 0kΓ

n

_{− ϕk}

∞

> ε) < ε

.

Property (i) impliesthat

T

◦ Γ

an be (almosteverywhere) dened, inspite of

T

not being everywhere dened. Indeed, if

T

′

_{: W −→ W}

is another version of

T

,that isif

T

′

_{= T}

a.e., the set

{T

′

_{6= T}}

is

π−

negligible;hen e, by (i),

Γ

−1

_{T

′

_{6= T}}

is

P

−

negligible,and

T

′

_{◦ Γ = T ◦ ΓP−}

a.s.. Similarly, one andenethe sto hasti pro esses

Γ

n

_{= T}

n

_{◦ Γ}

,whi hverify

Γ

0

_{= Γ}

and

T

_{◦ Γ}

n

_{= Γ}

n+1

.

Proposition 1. For xed

G ∈ W

and

ε > 0

, let us suppose that there exists a sto hasti pro ess

Γ : Ω −→ W

satisfying properties (i), (ii) and

(1)

P

(∀n ≥ 0, Γ

n

_{∈ G) < ε.}

Then :

Proof Take

G ∈ W

and put

F =

T

n≥0

T

−n

G

. Then, for

r ≥ 0

, (3)

T

−r

F =

\

n≥r

T

−n

_{G ⊃}

\

n≥0

T

−n

_{G = F.}

But these two sets,

F

and

T

−r

_F

, in luded in one another, have the same

π−

probabilityby

T

−

invarian e;soequality

F = T

−r

_F

holdsupto

π−

negligibility. As the laws of

Γ

and

B

are absolutely ontinuous w.r.t.

π

(this is where (i) is used), we have

Γ

−1

_{(F ) = Γ}

−1

_(T

−r

_{F )}

and

B

−1

_{F = B}

−1

_(T

−r

_{F )}

up to

P

₋

negligible events. In other words, almost surely, we have

{Γ ∈ F } =

{Γ

r

_{∈ F }}

and

{B ∈ F } = {B

r

_{∈ F }}

. Consequently, hoosing

r

given by (ii) and using

Γ

r

_{= B}

r

, we have

{Γ ∈ F } = {B ∈ F }

a.s.. That is tosay : (4)

P

(∀n ≥ 0, Γ

n

_{∈ G) = P (∀n ≥ 0, B}

n

_{∈ G)}



Spe ializing

G = {w ∈ W

;

kw − ϕk

∞

> ε}

, we obtain:

P

_{(E) < ε}

Proposition 1 redu es the proof of the approximation theorem to the on-stru tion of a pro ess

Γ

verifying (i), (ii), and (iii). We shall rst hoose

r

in asuitableway, then work ba kwards, in

r

steps, from

Γ

r

_{= B}

r

to

Γ = Γ

0

; ea h step ( alled a Lévy raise) will onstru t

Γ

n−1

from its Lévy transform

Γ

n

_{= T ◦ Γ}

n−1

. The sequen e

(Γ

r

_{, Γ}

r−1

_{, ..., Γ}

0

₎

is given aname : Denition 1. Given

r ∈ N

, a sequen e

(Γ

r

_{, Γ}

r−1

_{, ..., Γ}

0

₎

is alled a sequen e of

B−

raisedBrownianmotionsof index

r

ifea h

Γ

n

isa

W

−

valuedr.v. with law absolutely ontinuous w.r.t.

π

, if

Γ

r

_{= B}

r

, and ifwe have

Γ

n

_{= T ◦ Γ}

n−1

for

0 < n ≤ r

.

Infa t,for onvenien eof exposition,let usenlarge thelteredprobability spa e

Ω

, wesuppose it ontainsthe whole sequen e

(B

′n

₎

n∈N

of the Brownian iterates of

B

′

, B.M. independent from

B

. So we an assert :

Corollary 1. To prove the approximation theorem, it su es to exhibit a sequen e

(Γ

r

_{, Γ}

r−1

_{, ..., Γ}

0

₎

of

B−

raised Brownian motions of index

r

su h that

(5)

P

(kΓ

n

_{− ϕk}

of a sequen e of

B−

raised Brownian motions,and (iii)is implied by (1).



A Lévy raise starts with a given

W

−

valued r.v.

Γ

n

, and yields some r.v.

Γ

n−1

with Lévy transform

Γ

n

. Given a

W

−

valued r.v.

V

, how an one nd a r.v.

U

su h that

V = T ◦ U

? Knowing

V

is equivalent toknowing

|U|

, so todene

U

one onlyneedstode idewhi hsign isassignedtoea hex ursion of

|U|

away fromzero. To makethis rigorous, weneed aformaldenition of the ex ursions of a path and of their signs.

Notation 1. For

w ∈ W

and

q > 0

, denote by

Z(w) = {s ≥ 0/w(s) = 0}

the set of zeros of

w

, and dene

g

q

(w) = sup ([0, q] ∩ Z(w)) ≥ 0

(last zero before q) and

d

q

(w) = inf ([q, ∞] ∩ Z(w)) 6= ∞

(rst zero after q).

Fix a dense sequen e

(q

n

)

in

[0, ∞]

. To ea h

w ∈ W

, we an atta h the sequen e

(e

p

)

of disjoint,open intervalsobtained from the sequen e

(6)

((g

q

1

, d

q

1

), (g

q

2

, d

q

2

), ..., (g

q

n

, d

q

n

), . . . )

bydeletinganintervalwheneveritalreadyo ursearlierinthesequen e. The

e

p

are the ex ursionintervals of

w.π−

almost surely,there are innitely many of them, and they are the onne ted omponents of the open set

[0, ∞]\Z(w)

. The interval

e

p

(w)

will be alled the p-th ex ursion interval of

w

;

e

p

is an interval-valued measurable map, dened on

(W, W)

up to

π−

negligibility.

Sin e

w

does not vanish on

e

p

(w)

, its sign is onstant on this interval; this sign willbe denoted by

S

p

(w)

, and the sequen e

(S

p

)

will be alled

S

. If

B

isaBrownianmotion,the sequen e ofr.v.

S

◦ B = (S

p

◦ B)

isa oin-tossing; thismeans,itisani.i.d. sequen e,withea hr.v.

S

p

◦B

uniformlydistributed onthe set

{−1, +1}

. Moreover,

S

p

◦ B

and

|B|

are independent. (SeeChap. XII of [R,Y℄).

Lemma 1. Dene

I

: W −→ W

by

I

(w)(s) = inf

[0,s]

w

; that is,

|w| =

T

_{w − ITw}

for

π−

a.a.

w

.

Proof Fix

s ≥ 0

. On

[0, s]

,

B

1

_{= |B| − L ≥ −L}

s

= L

g

s

= B

1

g

s

. So

B

1

g

s

= inf

[0,s]

B

1

, and

|B

s

| = B

1

s

+ L

s

= B

s

1

− inf

[0,s]

B

1

.



Notation 2. If

f

is a ontinuous fun tion on

[u, v]

, we all

arg min

[u,v]

f

, resp.

arg max

[u,v]

f

, the largest

t ∈ [u, v]

su h that

f (t) = min

[

u, v]f

, resp.

f (t) = max

[

u, v]f

.

Lemma 2. Let

A

,

A

′

and

A

′′

be three measure spa es; let

µ

1

and

µ

2

be two measures on A,

f

a measurable map from

A

to

A

′

, and

ν

a measure on

A

′′

. If

µ

1

≪ µ

2

, then (i)

µ

1

◦ f

−1

_{≪ µ}

2

◦ f

−1

; (ii)

µ

1

⊗ ν ≪ µ

2

⊗ ν

. Proof (i)If

F ⊂ A

′

ismeasurableandif

(µ

2

◦ f

−1

_{)(F ) = 0}

,then

µ

2

(f

−1

_{F ) =}

0

, so

µ

1

◦ f

−1

_{(F ) = µ}

1

(f

−1

F ) = 0

. (ii)Ifameasurablesubset

F

of

A×A

′′

isnegligiblefor

µ

2

⊗ν

,then

ν−

almost all itsse tions

F

y

verify

µ

2

(F

y

) = 0

. Hen e they alsoverify

µ

1

(F

y

) = 0

, and onsequently

(µ

1

⊗ ν)(F ) =

R

µ

1

(F

y

)ν(dy) = 0

.



Lemma 3. Let

τ = (τ

p

)

be oin-tossing,

τ

′

_{= (τ}

′

p

)

a r.v. with values in

{−1, 1}

N

su h that

τ

′

p

= τ

p

for all but a.s. nitely many

p

, and

X

a r.v. independent of

τ

.

Then

L(X, τ

′

_{) ≪ L(X, τ )}

.

Thislemmasaysthat hangingnitelymanyvaluesof

τ

doesnotperturb too mu h the joint law of

X

and

τ

. For instan e, it implies that a pro ess obtained from a Brownian motion by hanging the signs of nitely many ex ursions has alawabsolutely ontinuous w.r.t.

π

. This is alled 'prin ipe de retournement des ex ursions' in [M℄.

Proof If

u = (u

1

, u

2

, ...)

is an innite sequen e, denote by

u

p]

the nite sequen e

(u

1

, ..., u

p

)

and by

u

[p+1

the innite sequen e

(u

p+1

, u

p+2

, ...)

. We have

(x, u) = f

p

(x, u

p]

, u

[p+1

)

forsome fun tion

f

p

.

We have to show that if

F

is measurable set su h that

P

[(X, τ ) ∈ F ] = 0

, then

P

[(X, τ

′

_{) ∈ F ] = 0}

. So assume

P

[(X, τ ) ∈ F ] = 0

. For

p ∈ N

, sin e

τ

p]

takesvalues in

{−1, 1}

p

, we an write (7)

X

σ∈{−1,1}

p

P



_f

_p

_{(X, σ, τ}

_[p+1

_{) ∈ F and τ}

_p]

_{= σ}



_{= P [(X, τ ) ∈ F ] = 0.}

Usingthe independen e of

τ

p]

and

(X, τ

[p+1

)

, this be omes :

(8)

X

σ∈{−1,1}

p

2

−p

P



_f

_p

_{(X, σ, τ}

_[p+1

_{) ∈ F}



_{= 0;}

Sofor ea h

p ∈ N

and ea h

σ ∈ {−1, 1}

p

, the event

{f

p

(X, σ, τ

[p+1

∈ F }

is negligible. Sin e

τ

′

p

(ω) = τ

p

(ω)

for all

p

larger than some

N(ω)

, one has

P

_{[(X, τ}

′

_{) ∈ F ] =}

_lim

p→∞

P



_{(X, τ}

′

_{) ∈ F and τ}

′

[p+1

= τ

[p+1



=

lim

p→∞

X

σ∈{−1,1}

p

P



_f

_p

_{(X, σ, τ}

_[p+1

_{) ∈ F and τ}

_p]

′

_{= σ and τ}

_[p+1

′

_{= τ}

_[p+1



_.

This is null be ause the event

f

p

(X, σ, τ

[p+1

) ∈ F

is negligible, as shown above.



Proposition 2. (me hanism of a Lévy raise) Suppose given the followingthree r.v. :

(i)

V

, a

W

−

valued r.v., su hthat

L(V ) ≪ π

; (ii)

τ = (τ

p

)

p∈N

, a oin-tossing independent of

V

; (iii)

τ

′

_{= (τ}

′

p

)

p∈N

, a r.v. valued in

{−1, 1}

N

, su h that the random set

p ∈ N

τ

p

′

(ω) 6= τ

p

(ω)

isa.s. nite.

Thenthere exists a unique

w

−

valued r.v.

U

su h that (9)

|U| = V − I ◦ V and S

p

◦ U = τ

′

p

f or each p.

It is measurable w.r.t. the

σ−

eld

σ(V, τ

′

₎

and we have

L(U) ≪ π

and

T

_{◦U = V}

. Forany

n ≥ 0

,wehave

U = B

n

ontheevent

{V = B

n+1

_{and τ}

′

₌

Proof We start from

L(V ) ≪ π = L(B) = L(B

1

₎

. Using Lemma 2 (i) we write

L(V, V −I◦V ) ≪ L(B

1

_{, B}

1

_−I◦B

1

₎

. ByLemma2(ii),the oin-tossing

τ

(resp.

S

◦ B

)whi hisindependentof

V

(resp.

B

1

) anbeaddedontheleft (resp. right), and we obtain

L(V, V − I ◦ V, τ ) ≪ L(B

1

_{, B}

1

_{− I ◦ B}

1

_{, S ◦ B)}

; by Lemma1,the right-handside is

L(B

1

_{, |B|, S ◦ B)}

. Lemma 3allows us to repla e

τ

by

τ

′

in the left-hand side, sowe nallyhave

(10)

L(V, V − I ◦ V, τ ) ≪ L(B

1

_{, |B|, S ◦ B)}

Now,we all

W

+

thesetofnon-negativepathsand

f : W

+

_{×{−1, 1}}

N

−→

W

the measurable fun tion su h that

w = f (|w|, S(w))

. We remark that

B = f (|B|, S ◦ B)

andwe dene

U = f (V − I ◦ V, τ

′

₎

;this isthe unique r.v.

U

su h that

|U| = V − I ◦ V

and

S

◦ U = τ

′

. To verify that

L(U) ≪ π

and

T

_{◦U = V}

,weapplyLemma2(i)to(2)withthefun tions

g(x, y, σ) = f (y, σ)

and

h(x, yσ) = (f (y, σ), x)

. With

g

weobtain

L(U) ≪ L(B)

,the rst laim. With

h

we obtain

L(U, V ) ≪ L(B, B

1

₎

; this implies

T

◦ U = V

sin e the jointlaw

L(B), (B

1

₎

is arried by the graphof

T

. Last, onthe event

{V = B

n+1

_andτ

′

_{= s ◦ B}

n

_}

,using the denition of

U

and Lemma 1 wehave

(11)

U = f (V −I◦V, τ

′

) = f (B

n+1

−I◦B

n+1

, S◦B

n

) = f (|B

n

|, S◦B

n

) = B

n

.



Proposition 3. Denote by

P

f

(N)

the set of all nite subsets of

N

. Fix

r

in

N

. For ea h

n ≤ r

, let be given

N

n

, a r.v. with values in

P

f

(N)

, and

Σ

n

_{= (Σ}

n

p

, p ∈ N

n

)

, a r.v. taking its values in

S

M∈P

f

(N)

{−1, 1}

M

, su h that

Σ

n

_{(ω) ∈ {−1, 1}}

N

n

_(ω)

. Starting with

Γ

r

_{= B}

r

, we an dene a sequen e

(Γ

n

₎

n≤r

su h that

Γ

n−1

is the

W

−

valued r.v.

U

obtained in Proposition2 from

V = Γ

n

_,

_{τ = S ◦ B}

′n−1

_,

_τ

′

p

=



Σ

n

p

(ω)

if

p ∈ N

n

_(ω)

τ

p

(ω)

else

Then the sequen e

(Γ

n

₎

n≤r

is a

B

-raised Brownian motions of index

r

.

Proof First, we verify that the

Γ

n

an be onstru ted stepwise. Assuming

Γ

n

Proposition 2appliesto

V = Γ

n−1

and

τ = S ◦ B

′n−1

(they are independent. The r.v.

Γ

n−1

_{= U}

yielded by Proposition 2 alsosatises

L(Γ

n−1

_{) ≪ π}

, and is measurable in

σ(F , τ

′

₎

.

The rest ofthe proof willexhibit asequen e

(Γ

n

₎

n≥r

of B-raised motions su hthat

Γ

J−n

_{= C}

n

. Startingwith

Γ

r

_{= B}

r

,theother

Γ

m

willbeindu tively dened : if

m < r

, suppose

Γ

m+1

has been dened, is

σ(B

m+1

₎₋

measurable, and veries

L(Γ

m+1

_{≪ π}

;dene

Γ

m

as the r.v.

U

obtained in Proposition2 from (12)

V = Γ

m+1

,

τ = S ◦ B

m

,

τ

_p

′

=



Σ

J−m−1

_(ω)

if

p ∈ N

J−m−1

_(ω)

τ

p

(ω)

else

This is possible sin e

V

and

τ

are independent and

N

J−m−1

is a.s. nte; the result

Γ

m

veries

L(Γ

n

_{) ≪ π}

and

T

◦ Γ

m

_{= Γ}

m+1

. To show that

Γ

m

is

σ(B

m

₎₋

measurable, it su es to show that so is

τ

′

; this may be done separatelyonea hoftheevents

{J ≤ m}

,

{J = m + 1}

,...,

{J = r}

,be ause they forma

σ(B

m

₎₋

partitionof

Ω

. On

{J ≤ m}

, wehave

τ

′

_{= S ◦ B}

m

; this is

σ(B

m

₎₋

measurable. To see what happens for other values of

j

, introdu e

ϕ

n

and

ψ

n

su h that

N

n

_{= ϕ}

n

_(B

J−n

₎

and

Σ

n

_{= ψ}

n

_(B

J−n

₎

for

0 ≤ n < k

. For

j ∈ {m + 1, ..., m + k}

, wehave on

{J = j}

(13)

τ

′

_{= ψ}

n

_(B

J−n

)1

ϕ

n

_(B

J −n

₎

+ τ 1

_Ω

− 1

_ψ

n

_(B

J −n

₎

Thisis

σ(B

m

₎₋

measurable too. We have established that

Γ

r

_{, ..., Γ}

0

exist and formasequen e of

B−

raisedmotions;itremainstosee that

Γ

J−n

_{= C}

n

. This is done in two steps. Firstly, by indu tion on

m

, we have

Γ

m

_{= B}

m

on

{J ≤ m}

: this holds for

m = r

, and if it holds for

m + 1

, it holds for

m

too, owing to the last statement in Proposition 1. Consequently,

Γ

m

₌

B

m

_{on{J = m}}

,thatis

Γ

J

_{= B}

J

_{= C}

0

. Se ondly,topro eedbyindu tion on

n

,wewillassumethat

Γ

J−n

_{= C}

n

forsome

n ≥ 0

,and show

Γ

J−n−1

_{= C}

n+1

. It su es to show this equality on the event

{J = j}

; on this event, using the denitionof

Γ

m

with

m = j − n − 1

andtheinequality

m = j − n − 1 < j

, the r.v.

Γ

J−n−1

satisesboth

T

(Γ

j−n−1

_{) = Γ}

j−n

_{= C}

n

_{= T(C}

n+1

)and (14)

S

◦ Γ

j−n−1

₌



Σ

n+1

on

N

n

S

_{◦ B}

j−n−1

else

= S ◦ C

n+1

.

These twoequalities entail

Γ

J−n−1

_{= C}

n+1

a.s. on

{J = j}

.

Lemma 4. Let

(j, k) ∈ N

2

and

Q

and

R

be two r.v. su hthat

k ≤ j − 1

and

0 ≤ Q ≤ R

. On the event

{∀n ∈ {k, ..., j − 1}Z ◦ B

n

_{∩ (Q, R) = ∅}}

that the rst iterates of

B

do not vanish between

Q

and

R

, there exists a (random) isometry

i : R −→ R

su h that

B

j

_{= i ◦ B}

k

on the interval

(Q, R)

.

Proof By indu tion, it su es toshow that if

B

j−1

doesnot vanish on the interval

(Q, R)

, then

B

j

_{= i ◦ B}

j−1

on

(Q, R)

, for some random isometry

i

. This is just Lemma 5 with

j = k + 1

and

B

j−1

instead of

B

k

, so we may suppose that

j = 1

.

On theevent

{Q = R}

,theresultistrivial. On

{Q < R} ∩ {Z ◦ B ∩ (Q, R) =

∅}

, the lo al time

L

is onstant on

[Q, R]

be ause its support is

Z ◦ B

, and the signof

B

is onstanton

(Q, R)

;so

B

1

_{= |B| − L = i(B)}

on

(Q, R)

,where

i

is the randomisometry

x 7→ xsgn B

(Q+R)/2

− L

(Q+R)/2

.



Notation 3. For

w ∈ W

, the p-th ex ursion interval

e

p

(w)

was dened earlier; the number

h

p

(w) = max

s∈e

p

(w)

|w(s)|

will be alled the height of the

orresponding ex ursion.

Lemma 5. Let

X

be a pro ess whose law is absolutely ontinuous w.r.t. Wiener measure. Almost surely,

• lim

p→∞

h

_p

_(X)1

_{e

p

(X)⊂[0,t]}

= 0

;

•

P

p∈N

h

_p

_(X)1

_{e

p

(X)⊂[0,t]}

= ∞

;

•

the set

(

P

p∈M

h

_p

_(X)1

_{e

p

(X)⊂[0,t]}

, M ∈ P

f

(N)

)

is dense in [0, ∞)

;

•

betweenanytwo dierentex ursionsof

X

, thereexistsathirdone,with height smaller than any given random variable

η > 0

.

Proof By a hange of probability, we may suppose that

X

is a Brownian motion. It is known (see Exer ise (VI.1.19) of [RY℄) that when

η → 0

+

, the number

P

p

1

{e

p

(X)⊂[0,t]}

1

{h

p

(X)>η}

of down rossings of the interval

[0, η]

by

|X|

before

t

is a.s. equivalent to

η

−1

_L

at

0

. This easily implies(i) and (ii),wherefrom (iii)follows.

Last, between any two ex ursions of

X

there are innitely many other ones (be ause

X

hasnoisolatedzeroes)and,by(i),onlynitelymanywithheights above

η

, when e (iv).



Notation 4. An ex ursion whose interval is in luded in

[0, t]

will be alled a

t−

ex ursion.

Itremains todes ribe the

N

n

and

Σ

n

,i.e., to hoose the signs of nitely manyex ursions whenLévy-raisingfrom

Γ

n

to

Γ

n−1

. Thiswillbedone soon; we rst need some notation and alemma.

Notation 5. If

e

′

and

e

′′

are two ex ursions of a path (or of a pro ess),

e

′

_{≺ e}

′′

means that

e

′

is anterior to

e

′′

:

s

′

_{< s}

′′

for all

s

′

_{∈ e}

′

and

s

′′

_{∈ e}

′′

. For an ex ursion

e

of

w

, we denote by

i

w

e := inf{w

s

; s ∈ [0, d

e

]}

.

Denition 2. An ex ursion

e

of a path

w ∈ W

is said to be tall if it is positive (this implies that the pro ess

I

w

remains onstant during

e

); and if for any ex ursion

e

′

of

w

su hthat

i

w

e

′

_{= i}

w

e

and higherthan

e

, then

e

′

_{= e}

. Formally,

e

istall if it is positive and if

(15)

max (w(s); s ≥ 0, (Iw)(s) = i

w

e) = max (w(s); s ∈ e) .

Lemma 6. Let

η

be a positive number,

m ≥ 1

be an integer and

w ∈ W

a path. Let

e

1

, . . . , e

m+1

be

m + 1

dierent

t−

ex ursions of

w

, numbered in hronologi al order :

e

1

≺ · · · ≺ e

m+1

; all

h

1

, . . . , h

m+1

their respe tive heights. Let

f

1

, . . . , f

p

denote all ex ursions of

w

whi h are anterior to

e

m+1

and whose heights are

≥ min(η, h

1

, . . . , h

m+1

)

, numbered in reverse hrono-logi al order : let

g

1

, . . . , g

p

be

p

ex ursions of

w

verifying

f

p

≺ g

p

≺ · · · ≺

f

1

≺ g

1

≺ e

m+1

. Suppose that

•

the ex ursion

e

m+1

is negative, and all

t−

ex ursions higher than

e

m+1

are positive;

•

theex ursions

g

1

, . . . , g

p

arenegative; andeverynegative ex ursion an-terior to

g

p

is smaller than

g

q

.

We all the

g

i

's the plug-ex ursions, and the

e

j

's the prote ted ex ursions. Then

e

1

, . . . , e

m

are tall, and

|i

w

e

1

| < |i

w

e

2

| < · · · < |i

w

e

m

| < η

.

Proof Firstly,

|i

w

f

1

| < η

be ause

f

1

≺ e

m+1

and any ex ursion anterior to

e

m+1

and havingheight

≥ η

is one of the

f

q

, hen e positive.

Se ondly, for

1 ≤ q ≤ p

, the ex ursion

g

q

is negative and higher than any negative ex ursion, anterior to it; so

I

w

is not onstant during

g

q

, and on-sequently we have

(16)

|i

w

f

p

| < |i

w

f

p−1

| < · · · < |i

w

f

1

| < height of g

1

,

where ea h

<

sign is due to

I

w

varying on the orresponding

g

q

.

thirdly, ombining (20) with

|i

w

f

1

| < η

(rst step), and noti ing that, by denition of the

f

q

,

(e

1

, . . . , e

m

)

is asub-sequen e of

(f

p

, . . . , f

1

)

, weobtain (17)

|i

w

e

1

| < · · · < |i

w

e

m

| < η.

Last, it remains to establish that

e

l

is tall for

1 ≤ l ≤ m

. Let

e

′

denote a positiveex ursion of

w

with height

h

′

_{≥ h}

l

and su h that

i

w

e

′

_{= i}

w

e

l

. From (13), we have

|i

w

e

′

_{| = |i}

w

e

l

| < height of g

1

; so

e

′

is anterior to

g

1

and a fortiori anterior to

e

m+1

. As

h

′

_{≥ h}

l

,

e

′

must be one of the

f

q

(see their denition). But

e

l

isalsoone ofthe

f

q

and, dueto(13), all

i

w

f

q

aredierent; so

e

′

_{= e}

l

. This means that

e

l

is tall.



In the proof ofLemma6, thenegativeex ursions

g

q

areused toseparate the

f

q

fromea hother. Yet, inthe end,we are not interested inthe behavior of all

f

q

but onlyin the

e

l

. It is possible to repla e this lemma with a variant, where

2m

ex ursions (instead of

p

ones, the

g

q

) are made negative, ea h

e

l

being anked by two of them.

Lemma 7. Let

X

be a pro ess with law absolutely ontinuous w.r.t.

π

, and

E

a tall ex ursion of

T

◦ X

with height

H

. There exists an ex ursion of

X

, with interval

{s; (I ◦ T ◦ X)(s) = i

T◦X

E}

, and with height

H + |i

T◦X

E|

.

Proof First,re all a.s., Brownianmotion

B

doesnot rea h its urrent min-imum

I

◦ B

in the interior of a time-interval where

I

◦ B

is onstant. (This

is a onsequen e of (

I

◦ B)(s) < 0

for

s > 0

and of the Markov property at the rst time that

B = I ◦ B

after some rational).

Put

Y = T ◦ X

and all

F

the interval

{s ≥ 0; (I ◦ Y )(s) = i

X

E}

;

Y

rea hes its urrent minimum

I

◦ Y

at both endpoints of

F

but not in the interior of

F

(see above). Sin e

|X| = Y − I ◦ Y

by Lemma 1, we have that

F

is the supportof some ex ursion of

X

. The height of that ex ursionis

max(|X

s

|; s ≥ 0, and (I ◦ Y )(s) = i

Y

E)

=

max (Y

s

− (I ◦ Y )(s); s ≥ 0, (I ◦ Y )(s) = i

Y

E)

=

max (Y

s

; s ≥ 0, (I ◦ Y )(s) = i

Y

E) − i

Y

E

=

max (Y

s

; s ∈ E) − i

Y

E

because E is tall

=

H + |i

Y

E|.



Lemma 8. Let

(0,

−

→

i ,

−

→

j )

be an orthonormal basis of the plan in whi h we represent paths. Let

τ

a+

b

be the verti al translation of ve tor

(b − a)

−

→

j

and

τ

_b

a−

the ree tion along the horizontal axis of equation :

y =

a + b

2

. Consider

(t, k, p) ∈ R

+

∗

× N

2

su h that

w

k

t

= a

and

w

k+p

t

= b

and denote

γ

t

the rst time posterior to

t

when at least one of the iterated Lévy transforms

w

s

,

k ≤ s ≤ k + p − 1

, vanishes. Then we have :

w

k+p

|

[t,γ

t

]

=







τ

_b

a+

o w

k

|

[t,γ

t

]

if

k+p−1

Π

i=k

w

i

t

> 0

τ

_b

a−

o w

−k

|

[t,γ

t

]

else We will denote

τ

k

k+p

(w)

the plan transformation, whi h transforms

w

k

|

[t,γ

t

]

in

w

k+p

|

[t,γ

t

]

.

Proof It is animmediate onsequen e of Tanaka'sLemma, when

p = 1

. In general ase, we break up the displa ement

τ

whi h transforms

w

k

|

[t,γ

t

]

in

w

k+p

|

[t,γ

t

]

under the form

τ = τ

p

◦ τ

p−1

◦ ... ◦ τ

1

where

τ

i

transforms

w

k+i−1

|

[t,γ

t

]

in

w

k+i

|

[t,γ

t

]

. From the pre eding remark, ea h

τ

i

is a verti al translation or a ree tion alongan horizontalaxis, a ordingto thesign of

w

k+i−1

t

. Thenwe dedu e the laim.



To onstru t the desired pro ess

Γ

, we will pro eed by indu tion on dis- retized time, andsowe willperform,fromthe level

r

,twotypesofraisings. In a rst type, the so- alled horizontal raisings, at a level when the path vanishes on

[t

d

, t

d+1

]

, we prote t the material furnished by the indu tion hy-pothesis on

[0, t

d

]

, namely

Γ

e

. And we prepare the path on

[t

d

, t

d+1

]

to give it the formit ought to have at this level for being near to

ϕ

on the interval when

Γ

is near to

ϕ

on

[0, t

d

]

.

So we make positive the signi ative ex ursions of the path alled here the prote ted ex ursions, and insert between them small ex ursions alled the plug-ex ursions (see lemma 6). And on

[t

d

, t

d+1

]

, we prepare ex ursions , the building ones, whi h we prote t, and they will a t, one by one, during a su ession of horizontalraisings, to give the path the previewed form (see Lemma 13 and Proposition 14), at the ondition the path, after that, will not vanishedon

[t

d

, t

d+1

]

.

Inase ondtype,theso alledverti alraisings,wegiveanewtotheprote ted ex ursions the signs they have before the horizontal raisings. Then we get up while the path doesn't vanish on

[t

d

, t

d+1

]

. In fa t we must distinguish the last raising of a su ession of horizontal raisings, the so alled terminal horizontal raising, when we leave to prote t the prote ted ex ursions and givethem the goodsigns.

It is important to know the real level of the path, ie. the level without the horizontal raisings. Pre isely,wedene the r.v. indu tively:

RL

r

(w) = r

and f or all integer n ≤ r

RL

n−1

(w) =

(

RL

n

(w)

if thestep

n − 1 → n

orresponds to an horizontal raising

RL

n

(w) − 1

if it orresponds to a verti al raising.

Forourneeds, wewill allmap-ex ursion,orsimplyex ursion,ea hmap

e : R

+

_{→ R}

whosesupportisanot empty segmentand whi hdoesn't vanish at any point of the interior of the support. In parti ular, for

w ∈ W

and

t > 0

, we will all ex ursion straddling

t

, and denote itby :

e

t

(w)

, the map so dened :

e

t

(w) : R

+

→ R, ∀u ∈ R

+

, e

t

(w)(u) =



0

if

u ∈ [0, g

t

(w)] ∪ [d

t

(w), +∞[

w

u

else

Wewill introdu e the map

de

t

(w) : R

+

_{→ R}

∀u ∈ R

+

_{, de}

t

(w)(u) =







0

if

u ∈ [0, t] ∪ [R

t

(w), +∞[

where

R

t

(w) = sup{u > t, ∀s ∈]t, u], w

s

6= w

t

}

w

u

− w

t

else

and we allitadierentialex ursionof

w

wheneveritssupportisnotempty, ie.

∃ε > 0, ∀u ∈ (t, t + ε), w

u

6= w

ε

We denote them

e

t

and

de

t

when there is no ambiguity, and

e

s

t

and

de

s

t

in the ase of ex ursions of

w

s

.

Lemma 9. Let

w ∈ W

and

e

a negative

m−

ex ursionof

T

w

, lowerthanall pre eding it. Let

γ

be its beginningand

δ

its end. We set :

γ

1

= arg min Tw

[0,γ]

, γ

2

= inf{t ∈ supp(e); e(t) = Tw

γ

1

}, γ

3

= arg min e

[γ,δ]

Then :







de

γ

1

oin ides with an ex ursion of

|w|,

and its support is

[γ

1

, γ

2

]

de

γ

3

oin ides with an ex ursion of

|w|,

whi h begins at

γ

3

and whose support ontains

[γ

3

, δ]

Furthermore,

∀u ∈ [γ

2

, γ

3

]

,

de

u

oin ides with an ex ursion of

|w|

, if, and only if :



de

u

is a positive ex ursion

e(u) = inf{e(t), t ∈ [γ, u]}

Itisthe ase inparti ular when

de

u

istherst positiveex ursionof theform

de

v

,

v ∈ [γ

2

, γ

3

]

to overow a given value.

Proof

From Tanaka's formula:

|w

t

| = Tw

t

+ sup{−Tw

u

, u ∈ [0, t]}

Therefore,

|w

γ

1

| = Tw

γ

1

− Tw

γ

1

= 0,

while, for all

t > γ

1

, su iently small:

T

_w

_t

_{> Tw}

_γ

1

.

So,

de

γ

of

|w|

beginning at

γ

3

whose support ontains

[γ

3

, δ]

. Let

e

′

be an ex ursion of

w

with support in luded in

[γ

2

, γ

3

]

. Its beginning

u

, and its end

v

verify:

u = arg

[0,v[

min Tw

and

v = arg

[0,v]

min Tw.

So wededu e :

de

u

= |e

′

_|

.

Re ipro ally,let

u ∈ [γ

2

, γ

3

]

su h that

de

u

is a positive ex ursionand

u = arg

[0,u]

min Tw

.

Then,

u = arg

[0,v[

min Tw

,where

v

is the end of

de

u

, be ause

de

u

is positive.

Thus,

w

u

= w

v

= 0

, and for all

t ∈]u, v[

,

w

t

6= 0

. Consequently,

de

u

isan ex ursionof

|w|

.

Let

h > 0

be su h that there exists

u ∈ [γ

2

, γ

3

]

verifying

de

u

is the rst positive ex ursion of the form

de

v

,

v ∈ [γ

2

, γ

3

]

, whose height overows

h

. Then, for all

v < u

, the support of

de

v

an't ontain this of

de

u

without denying the minimality of

u

.



As an immediate onsequen e,we observe :

Corollary 2. The ex ursions of

w

oin ide with the positive dierential ex- ursions of

T

w

, beginning at

arg min

[0,t]

(Tw)

for all

t ∈ [0, +∞[

.

In thisparagraph,wewanttheraisedpathtoapproa hthemap

ϕ

uniformly on

[0, T ]

. Pre isely :

Whatever

ε

stri tlypositive, and

ϕ ∈ W

|

[0,T ]

, thereexists

Γ

a

B−

raised Brow-nian motion su h that :

P



_kΓ

_|

[0,T ]

− ϕ

|

[0,T ]

k

∞

< ε



> 1 − ε

We onsider a modulus of uniform ontinuity

α

0

asso iatedto

 ε

4

, ϕ, [0, T ]



and a real number

α

1

su h that

P (A

0 ε

) > 1 −

ε

2

where

A

0 ε

=

h

sup{|B

t

− B

u

|, (t, u) ∈ [0, T ]

2

and

|t − u| < α

1

} <

ε

2

i

,

then weset

α := min(α

0

, α

1

)

,

d

0

:=



_T

α



+ 1

,and forall

d ∈ N

,

t

d

= (dα) ∧ T

. Weset again, for allinteger

d ∈ [1, ..., d

0

]

,

A

d

_ε

:=

h

sup{|B

t

− B

u

|, (t, u) ∈ [t

d

, T ]

2

, |t − u| < α} <

ε

2

i

.

Our aimis to show, by indu tion on

d

,the following property

P

d

: "For all

ε > 0

,thereexists aninteger

r

d

and

Γ

a

B−

raisedBrownianmotionof index

r

d

su h that :

P

h

kΓ

|

_[0,td]

− ϕ

|

_[0,td]

k

∞

< ε

i

∩ [|Γ

t

d

− ϕ(t

d

)| < ε

1

] ∩ A

d

ε



> 1 − ε



1 +

d

d

0



”

Noti e that

P

0

immediately yields from the hoi e of

ϕ

whi h vanishes at

0

. We suppose now

P

d

true. We are going to apply this hypothesis to the Brownianmotion

B

s

0

, foraninteger

s

0

whi h,asthe realnumber

ε

1

,willbe later spe ied. As

A

d

ε

⊂



sup{|B

t

− B

u

|, (t, u) ∈ [t

d

, t

d+1

]

2

} <

₂

ε



∩ A

d+1

ε

,and fromthe inde-penden eofthein rementsofBrownianmotion,we andedu etheexisten e of adisjointedsum

Γ

e

of

B

s

0

−

raised Brownian motionsof index

r

d

su h that :

P

_(A

ε

0

) > 1 − ε



1 +

d

d

0



,

where

A

ε

₀

:=

h

ke

Γ

|

_[0,td]

− ϕ

|

_[0,td]

k

∞

<

ε

2

i

∩

h

|e

Γ

t

d

− ϕ(t

d

)| < ε

1

i

...

...

∩

h

sup{|B

s

0

+r

_{− B}

s

+r

_{|, (t, u) ∈ [t}

_{, t}

_]

2

_{} <}

ε

i

_{∩ A}

d+1

(It su es toapply

P

d

with

ε

2

instead of

ε

).

We willdenote :

∀i ∈ N, e

Γ

i

_{= e}

_w

i

. By denition,

∀i > r

d

, ˜

w

i

_{= w}

s

0

+i

. From the theorem of density of zeroes ([M℄), there exists a.s. an integer

ℓ

su h that

w

e

ℓ

vanishes atleast one time on

[t

d

, t

d+1

]

.

Let

L(w)

be the smallest of these integers

ℓ

.

L

is a r.v. almost surely nite. Then there exists aninteger

ℓ

0

whi hwe will hoose

> r

d

su hthat :

P

_(A

ε

1

) > 1 − ε



1 +

d

d

0



−

ε

2d

0

où

A

ε

1

:= A

ε

0

∩ [L ≤ ℓ

0

] .

Weset :

r = s

0

+ max(r

d

, l

0

)

.

Denition 3. Theprote ting tree of

e

Γ

. Let

N

be a set of ex ursionsof

w

|[0,d

t

(w)]

,

w ∈ W

, and

ε > 0

. We denote by

T

t,ε

_w

_{(N )}

the set of ex ursionsof

T

w

|[0,d

t

(Tw)]

so dened :

e

′

_{∈ T}

t,ε

w

(N )

i

e

′

is an ex ursion of

T

w

|[0,d

t

(Tw)]

whi h appears in the dif-ferential ex ursions

dE

of

T

w

orresponding to an ex ursion

E ∈ N

, and

h(e

′

) >

ε

₄

(we all them ex ursions of the rst type) or the support of

e

′

ontains

arg max

of the dierential ex ursion

dE

, we all them ex ursions of the se ond type, and all the ex ursions

e

′′

of

T

w

of height belonging to

[h(e

′

), h(e

′

−

)]

,

e

′

−

being the pre eding ex ursion of

T

w

of height

>

ε

4

, with

e

′

−

< e

′′

< e

′

, we allthese ex ursions

e

′′

the third type ex ursions.

Lemma 10. When

N

is a nite set, so is

T

t,ε

w

(N )

.

Proof We remark rstly that the number of distin t dierential ex ur-sionsof

T

w

orrespondingtotheelementsof

N

isnite,equaltothe ardinal of

N

. Thenumberofex ursionsofthersttypeisnitebe ausetheirheight isgreaterthan

ε

4

. Thenumberofex ursionsofthese ondtypeisnite follow-ingthe rst remark. And forea h ex ursionof the se ond type,the number of ex ursions of the third typeis nite too.



Thenwe all prote tingtree of

e

Γ

,the tree onstituted by :

•

at the

0

-generation : the elements of

N

e

0

, the set of

d

t

( e

w

0

₎

-ex ursions of height

> ε

.

•

and, for all

n ∈ N

,if we denoteby

N

e

n

, the set of prote tedex ursions of

w

e

n

whose elements, given in hronologi al order, onstitute the

n

generation of the tree, then at the

(n + 1)

th

generation, the elements of

N

e

n+1

_{:= T}

t

d

,ε/4

n

e

w

n



e

N

n



. As we have ordered ea h

N

e

n

, whi h is now a nite sequen e of ex ursions of

w

e

n

, we an onsider

Σ

e

n

the nite sequen e of their signs. And our next task is to denite

N

′n

the sequen e of prote ted

t

d

-ex ursions of

Γ

n

and

Σ

′n

the asso iated sequen e of their signs. We have dened the

N

e

n

's by getting down in the iterations. Wewilldenethe

N

′n

's by gettingup fromlevel

r

. We put

N

′r

_{(ω) = e}

_N

r−s

0

_(ω)

, and

Σ

′r

_{(ω) = e}

_Σ

r−s

0

_(ω)

. If

Γ

n+1

_{−→ Γ}

n

is an horizontal raising, then

N

0,n+1

is onstituted by the familyofthe ex ursions of

Γ

n

whoseabsolutevalueshavethe same

arg max

asthe ex ursions of

N

0,n+2

. Then we pro eed as in lemma6, onsidering the elements of

N

0,n+1

asthe

e

i

,

1 ≤ i ≤ m

. We havenow todenethe r.v.

η

: let

η

i

betheminimaldistan e between theheights of distin t prote ted ex ursions of

w

e

i

for

i ∈ {0, . . . , r − s

0

}

and

η

m

=

r−s

V

0

i=0

η

i

, e

ε

i

= ε ∧ η

m

₄

i+1

1

, for

i ∈ {0, . . . , r − s

0

}

, and

ε

i

= e

ε

RL(i)−s

0

1

4

i′ −i

, where

i

′

_{= sup{j > i/RL(j) = RL(i)}}

. Then we hoose for

η

the r.v.

ε

RL(n)−s

0

. And we ( an) dene the set

N

′n

as the set of elements the

f

j

's and the

g

j

's. And

Σ

′n

give sign

−1

to the

g

j

's and

+1

tothe

f

j

's if

Γ

n+1

_{−→ Γ}

n

is not a terminalhorizontalraising. In the ase of aterminalhorizontalraising,weput simply

N

′n+1

_{= N}

0,n+1

. If

Γ

n+1

_{−→ Γ}

n

isa verti al raising,

N

n+1

is onstituted by the family of the ex ursions of

Γ

n

orrespondingtothe positivedierentialex ursions of

Γ

n+1

whose support en ounters at least the supportof an element of

N

0,n+2

and beginning at an

arg min

of

Γ

n+1

. In these two ases, to spe ify

Σ

′n+1

, we need the following lemma. Let us all argext of an ex ursion the

arg max

(resp.

arg min

) of this ex ursion if itis positive (resp. negative).

Lemma 11. If for all

i

from level

r

to

n

(with

r ≥ n

) the elements of

N

0,i

and

N

e

RL(i)−s

0

have the same argext, whi h implies

|N

0,i

_{| =}





 e

N

RL(i)−s

0







, and

the same signs, then the elements of

N

0,n−1

and

N

e

RL(n−1)−s

0

have the same argext, hen e these sets have the same ardinality, and the same hierar hy, ie. the order of the heights between the respe tive ex ursions of ea h set is the same. So we put

Σ

0,n−1

_{= e}

_Σ

RL(n−1)−s

0

.

Proof The distin tion between

Γ

and

Γ

e

is due to the introdu tion of horizontal raises. If

Γ

n+1

_{−→ Γ}

n

is an horizontal raising, ea h prote ted ex ursion gains in height the height of a plug ex ursion, and its support enlarges. So, duringsu haraising,the"error"between

Γ

and

Γ

e

in reasedof

on the other side,

Γ

n+1

_{−→ Γ}

n

is a verti al raising, a prote ted ex ursion of

Γ

n

is onstituted by atmost two prote ted ex ursions of

Γ

n+1

. Soduring su h a raise, the error between

Γ

and

Γ

e

doubles. And it is easy, by means of our hoi eof the heightsof the plug ex ursions, toshowthat this erroris always majorized by

ε ∧ η

. Hen e the lemmafollows immediately.



Now we an onsider (see Proposition 2) that

Γ

|[0,t

d

]

is orre tly dened : for on e

RL

w

(n) = s

0

, we end the raises, if ne essary, by horizontal ones : prote ting allthe material a quainted. But is

Γ

near from

e

Γ

? To answer this question, we have to onsider the error between

Γ

and

e

Γ

now, as being

kΓ

|[0,t

d

]

− e

Γ

|[0,t

d

]

k

∞

.

We have just built

Γ

|[0,t

d

]

. We have now to ontrol :

kΓ

N

|[0,t

d

]

− e

Γ

0

|[0,t

d

]

k

∞

, where

N = sup {n ∈ {0, . . . , r}/RL(n) − s

0

= 0}

(with

N = −∞

if the set is empty).

N

is ar.v. Lemma 12.

kΓ

N

|[0,t

d

]

− e

Γ

0

|[0,t

d

]

k

∞

≤ 2ε

, on the event

[N ≥ 0]

.

Proof Letusrst remark,fromthe pre edinglemma,that, onthe event

[N ≥ 0]

,wehave prote tedex ursions with thesame argext, thesame signs, and heights near at

ε

. The supports of the prote ted ex ursions of

Γ

N

on-taining the supports of the orresponding prote ted ex ursions of

e

Γ

0

. F ur-thermore, onthe dieren eof their supports,

Γ

N

and

Γ

e

0

dierfrom atmost

2ε

, and likelyoutside the union of their supports.



The purpose of the following lemma is to prepare, at level

s

, when the iteratedBrownianmotionvanishes on

]t

d

, t

d+1

[

,the ex ursions whi hwill al-lowthe orre tly raised path toapproa h

ϕ

at level

0

on

]t

d

, t

d+1

[

.

Lemma 13. Full planing. Let

w

belongto

W

, and

t, t

′

_{, ε}

′

_{∈ R}

+

∗

be su hthat

t < t

′

. We suppose there is no interval in whi h

w

is onstant, and

w

vanishes in

(t, t

′

₎

. The following r.v. are fun tionalsof

|w|

:

• t

0

(w) = g

inf{s>d

t

;|w

s

|≥ε

′

}

∧ t

′

, with

inf ∅ = +∞

.

• ∀n ∈ N

, while

t

n

< t

′

, we set :

t

n+1

:=















inf{u ∈ [t

n

; arg max |de

t

n

|[, h(de

u

) > ε

′

and sgn

(de

u

) = −

sgn

(de

t

n

)}

if this set is not empty,

else :

arg max |de

t

n

| ∧ t

The sequen e

(t

n

)

n∈N

is stri tly in reasing and nite. Let

1 + K(w)

be its ardinality.

Proof By onstru tion, the sequen e

(t

n

)

is stri tly in reasing and lower than

t

′

. Suppose the number of its terms is innite. In this ase, it would admit a limit

t

∗

≤ t

′

, and the os illation of

w

at

t

∗

would be innite, so ontradi ting the ontinuity of

w

. Then

(t

n

)

n

isnite.

The measurability and the niteness of

K

are immediate.



Let us remark that this Lemma gives us the possibility of planing the path after

d

t

in

K

raises.

For, duringthe rst raise, we put negative the ex ursion beginningat

t

0

(w)

and positive all the other ex ursions in

(t

0

(w), t

′

₎

of height greater than

ε

′

. Then during the se ond raise, we put negative the ex ursion whose support ontains this of

de

t

1

, and so on. At the end of su h

K

raises, the path on

[t

0

, t

′

]

has an absolute value whi h doesn't ex eed

ε

′

_{+ Kε}

′′

, and

ε

′

on the ex ursion straddling

t

′

.

Sowe are going now to analyze its behavior on

(t

d

, t

d+1

)

. Letus denote

γ

t

d

the rst time after

t

d

at whi h one of the

Γ

σ

,

0 ≤ σ ≤ s

0

+ l

0

, vanishes on

(t

d

, t

d+1

)

, and

σ

0

the orresponding level.

Proposition 4. There exists a

σ(Γ

σ

′

₀

+1

₎₋

measurable,

N

−

valued r.v.,

K

′

σ

0

su h that there exists

K

′

σ

0

− 1

r.v.

P

1

, . . . , P

K

′

σ0

−1

themselves with values in

N

and

σ(Γ

σ

0

+1

₎₋

measurable, su h that : (i) the

K

′

σ

0

− 1

ex ursion intervals

e

P

1

(Γ

σ

0

_{), . . . , e}

P

K

′

_{σ0 −1}

(Γ

σ

0

₎

are disjoint and in luded in

(d

t

d

(Γ

σ

0

_{), t}

0

(Γ

σ

0

))

(ii) the heights

H

1

, . . . , H

K

′

−1

of these

K

′

_{− 1}

ex ursions of

Γ

σ

0

satisfy on

[σ

0

≥ 0]

:





τ

RL

0

_σ0

−s

0

)

( e

w)(ϕ(t

d+1

))





−ε

′

< H

1

+· · ·+H

K

′

σ0

−1

<





τ

RL

0

_σ0

−s

0

)

( e

w)(ϕ(t

d+1

))





+ε

′

.

Proof Noti ing that the pro ess :

X =

X

n∈N

Γ

n

1

[σ

0

=n]

onse-

Ofteninthesequel,wewilldenote

K

′

σ

0

simplybyK',ifthereisnoambiguity. Set :

σ

₀

′

= σ

0

− K(Γ

σ−1

) + K

σ

′

0

.

Proposition 5. For all

(n, ω)

su h that

σ

0

(ω) ≤ n < σ

0

(ω) − K

σ

0

(ω) −

K

_σ

′

₀

(ω)

,

N

n

_(ω)

and

Σ

n

_(ω)

an be hosen so that : (i)

0 ≤ Γ

σ

0

−K

t

d+1

< ε

′

and

ke

Γ

RL(σ

0

−1)−s

0

|[0,t

d

]

− Γ

σ

0

|[0,t

d

]

k

∞

< Kε

′′

. (ii)

−(I ◦ Γ

σ

0

_)(t

d+1

) = H

K

′

σ0

−1

(iii)

Γ

σ

0

has

K

′

_{− 2}

tall ex ursions in luded in

(t

d

, t

d+1

)

:

E

1

< E

2

<

· · · < E

K

′

₋₂

with respe tive heights

H

1

, . . . , H

K

′

−2

verifying

|i

Γ

σ0

E

1

| <

|i

Γ

σ0

E

₂

| < · · · < |i

_Γ

σ0

E

_K

′

₋₂

|

. (iv)

H

n+1

+ · · · + H

K

′

σ0

−1

≤ Γ

σ

0

−K−n

t

d+1

< H

n+1

+ · · · + H

K

σ0

′

+ ε

′

_{+ nK}

′

_ε

′′

. (v)

H

K

′

σ0

−n

< −(I ◦ Γ

σ

0

−K−n

_)(t

d+1

) < H

K

′

σ0

−n

+ K

′

_ε

′′

. (vi)

Γ

σ

0

−K

σ0

−n

has

K

′

σ

0

− n − 2

tall ex ursions in luded in

(t

d

, t

d+1

)E

n

1

<

· · · < E

n

K

′

σ0

−n−2

su hthat:

|i

Γ

σ0−Kσ0 −n

E

1

| < · · · <







i

Γ

σ0−Kσ0 −n

E

K′σ0−n−2







 <

ε

′′

, and whose heights

H

n

1

, . . . , H

K

n

′

J0

−n−2

satisfy :

H

l

≤ H

l

n

< H

l

+ nε

′′

f or n + 1 ≤ l < K

σ

′

0

− 1.

In our pursuitof the pro edure, we an state : Proposition6. Itispossibleto hoose

N

n

_(ω)

and

Σ

n

_(ω)

, forall

(n, ω)

su h that

n = σ

′

0

, in order to have :





Γ

σ

′

0

t

d+1

− τ

0

RL(σ

′

0

)−s

0

(ϕ(t

d+1

))





 < ε

′

+

1

2

(K

′

σ

0

− 1)(K

′

σ

0

− 2)ε

′′

Proof Wenoti ethatthebuildingex ursionswhi happearinProposition5, the

e

P

0

(Γ

σ

0

+K

_σ0

₎

's, are su essively prote ted. On eprote ted, ea hof them re eivesasmallex ursionofheightlowerthan

ε

′′

atea hraise. Sowededu e the result.