www.imstat.org/aihp 2011, Vol. 47, No. 3, 679–698
DOI:10.1214/10-AIHP380
© Association des Publications de l’Institut Henri Poincaré, 2011
Spectral gaps and exponential integrability of hitting times for linear diffusions
Oleg Loukianov
a, Dasha Loukianova
band Shiqi Song
baDépartement Informatique, IUT de Fontainebleau, Université Paris Est, route Hurtault, 77300 Fontainebleau, France.
E-mail:oleg.loukianov@u-pec.fr
bDépartement de Mathématiques, Université d’Evry-Val d’Essonne, Bd François Mitterrand, 91025 Evry, France.
E-mail:dasha.loukianova@univ-evry.fr;shiqi.song@univ-evry.fr Received 1 October 2009; revised 18 May 2010; accepted 28 June 2010
Abstract. LetX be a regular continuous positively recurrent Markov process with state spaceR, scale functionS and speed measurem. Fora∈Rdenote
Ba+=sup x≥a
m
]x,+∞[
S(x)−S(a) , Ba−=sup
x≤a m
]−∞;x[
S(a)−S(x) .
It is well known that the finiteness ofB±a is equivalent to the existence of spectral gaps of generators associated withX. We show how these quantities appear independently in the study of the exponential moments of hitting times ofX. Then we establish a very direct relation between exponential moments and spectral gaps, all by improving their classical bounds.
Résumé. SoitX un processus de Markov récurrent positif à trajectoires continues et à valeurs dansR.SoientS sa fonction d’échelle etmsa mesure de vitesse. Poura∈Rnotons
Ba+=sup x≥a
m
]x,+∞[
S(x)−S(a) , Ba−=sup
x≤a m
]−∞;x[
S(a)−S(x) .
Il est bien connu que la finitude deBa±est équivalente à l’existence d’un trou spectral du générateur associé àX.Nous montrons comment ces quantités apparaissent d’une manière indépendante dans l’étude des temps d’atteinte deX.Ensuite nous établissons une relation directe entre les moments exponentiels et le trou spectral, en améliorant en plus leurs encadrements classiques.
MSC:60J25; 60J35; 60J60
Keywords:Recurrence; Linear Markov process; Exponential moments; Hitting times; Poincaré inequality; Spectral gap; Dirichlet form
Introduction
Let(Xt; t≥0)be a regular linear continuous Markov process with the state space R. We assume throughout the paper thatX is positively recurrent and conservative (the killing time is identically +∞). Denote byS(x) a scale function ofXandm(dx)the speed measure associated withS (cf. [23], Chapter VII). Recall thatS is a continuous strictly increasing function andm(dx)is a symmetric measure forX, charging every not empty open set. Moreover, the positive recurrence ofXimplies limx→±∞S(x)= ±∞andm(R) <∞.
In this paper we study some relations between the exponential moments of hitting times ofX, the finiteness of the quantitiesBa+,Ba−given by (0.1), the Hardy and Poincaré inequalities for Dirichlet forms associated withX. As a consequence we give a chain of equalities linking all these objects.
Leta∈RandTa=inf{t≥0: Xt=a}be the hitting time ofabyX. The first question we are interested in is the estimation of exponential momentsEx[eλTa],x∈R,λ >0.
In some particular cases such moments have been well studied. We mention, for example, Ditlevsen [7] for Ornstein–Uhlenbeck process, Giorno et al. [12] for Bessel and Ornstein–Uhlenbeck processes, Deaconu and Wantz [6] for diffusion with strong drift, and the book of Borodin and Salminen [3] for an overview of known formulas.
But we were not able to find in the literature a simple general estimate of exponential moments in terms of the scale functionS(x)and the speed measurem(dx).
In the present paper this question receives a very satisfactory response with the quantities:
+a(x)=m
]x,+∞[
S(x)−S(a) , −a(x)=m
]−∞;x[
S(a)−S(x) .
Namely, letλ+a be the supremum ofλ >0 such thatExeλTa <∞for somex > a (hence for allx > a, see the
“all-or-none” property, Proposition1.2). Respectively, letλ−a be the supremum ofλsuch thatExeλTa <∞for some (all)x < a. Put
Ba+=sup
x≥a
+a(x), Ba−=sup
x≤a
−a(x) (0.1)
and
Ca±=lim inf
x→±∞±a(x).
Our first result (see Section1, Theorem1.1) asserts that 1
4Ba±
≤λ±a ≤ 1 4Ca±
∧ 1
Ba±, (0.2)
whereBa+orBa−can eventually be infinite. The positivity ofλ±a is thereby equivalent to the finiteness ofBa±. More- over, ifλ±a >0 for somea∈R, it is so for alla∈R.
As it turns out, the importance of the quantitiesBa±is well-known in some different context. Using Krein’s method, Kac and Krein [15] and Kotani and Watanabe [17] have shown that the spectral gapsγa+ (respectivelyγa−) of the generator ofXkilled when it exits]a,+∞[(resp.]−∞, a[) satisfy
1 4Ba±
≤γa±≤ 1
Ba±. (0.3)
We see therefore that the positivity ofλ±a is equivalent to this ofγa±, a fact that can actually be derived e.g. from the works of Down et al. [8] and Bakry [1] or Rökner and Wang [24]. But comparing (0.2) and (0.3) leads to the more explicit conjectureλ±a =γa±. This identity would not be surprising, since for exit times from a bounded domainDit is well known from the works of Khasminskii [16] and Friedman [10]. Namely, ifτDis the exit time fromDandXD is a process killed atτD, then the equality holds between the widthγD of the spectral gap of the generator ofXD and the supremumλD ofλsuch thatExeλτD<∞for allx∈D. However, the PDE methods of [16] requireExτ to be bounded, which is not the case in general, and in particular forD= ]a,∞[.
What was surprising, is that we have not found in the literature an analogue of Khasminskii identity for unbounded domains. So in the second section of this article we firstly show (Theorem2.1) in the setting ofm-symmetric Hunt processes that
λD=γD ifλD=sup
λ: ExeλτD∈L1(mID)
>0.
The proof is based on the spectral calculus, available thanks to the symmetry of the generator of XD. Since this symmetry is automatically fulfilled in dimension one, we get for anya∈R,
γa+=λ+a and γa−=λ−a. (0.4)
Notice that we cannot establish such kind of equality directly onR, because our process is conservative and the exit time fromRis identically infinite. But it is well known that the spectral gapγ of am-symmetric operator can be characterized in terms of Poincaré inequality for its Dirichlet form. We recall in the sequel of Section2that
γa±= 1
A±a and γ= 1 cP,
whereA±a andcP are the best possible constants satisfying ∞
a
F (x)−F (a)2
m(dx)≤A+a ∞
a
dF dS
2
(t)dS(t), a
−∞
F (x)−F (a)2
m(dx)≤A−a a
−∞
dF dS
2
(t)dS(t) and
+∞
−∞
F (x)−m(F ) m(R)
2
m(dx)≤cP +∞
−∞
dF dS
2
(x)dS(x) for allF in an appropriate functional spaceF.
A classical Mukenhoupt result [22] yields Ba+≤A+a ≤4Ba+ and Ba−≤A−a ≤4Ba−.
Looking at (0.2) and (0.4), we see that the lower bounds above can actually be replaced by 4Ca±∨Ba±, which are sometimes more precise. They also allow to recover (in our settings) the case of equalityA±a =4Ba±studied in Miclo [21].
Further,cP can be easily related toA±a by sup
a
A+a ∧A−a
≤cP ≤inf
a
A+a ∨A−a
which yields sup
a
Ba+∧Ba−
≤cP ≤4 inf
a
Ba+∨Ba− .
In the works of Bobkov and Götze [2] and Malrieu and Roberto [20] it was shown that these inequalities can be replaced by
1
2Bme+ ∨Bme− ≤cP ≤4
Bme+ ∨Bme− ,
wheremeis a median ofm(see [20], Theorem 6.6.2). At the end of Section2we give another refinement of the bounds oncP. After proving thatA±a areS-Hölder ina, we show that there exists a pointc∈Rsuch thatA+c =A−c =cP, which yields
Bc+∨Bc−∨4Cc+∨4Cc−≤cP ≤4
Bc+∧Bc− .
The pointcis, however, unknown (except for obvious symmetric cases) and worths further investigations.
Finally, the last section contains some examples to illustrate the above-mentioned results.
1. Exponential integrability of hitting times
In this section we study the exponential moments of hitting timesTa. Fora∈R, denote λ+a =sup
λ≥0:∀x > a,ExeλTa <∞
; and
λ−a =sup
λ≥0:∀x < a,ExeλTa <∞ .
As we will see (Proposition1.2), an important “all-or-none” property holds for anyλ >0:
∃x > a, ExeλTa<∞ ⇐⇒ ∀x > a, ExeλTa <∞, the same being true forx < a.
Recall the definitions Ba+=sup
x≥a
m
]x,+∞[
S(x)−S(a) , Ba−=sup
x≤a
m
]−∞;x[
S(a)−S(x) and
Ca+=lim inf
x→∞ m
]x,+∞[
S(x)−S(a) , Ca−=lim inf
x→−∞m
]−∞;x[
S(a)−S(x) . The main result of this section is
Theorem 1.1. For alla∈R, 1
4Ba+
≤λ+a ≤ 1 4Ca+
∧ 1
Ba+ and 1 4Ba−
≤λ−a ≤ 1 4Ca−
∧ 1 Ba− with the convention1/∞ =0.
In the sequel we often prove only assertions concerningBa+andλ+a, since the proofs of their “left” counterparts are completely similar.
1.1. Kac formula
The Kac formula, first derived in [13,14] for linear Brownian motion, then generalized in [5] and in [9], permits to calculate the moments ofAv= 0Tv(Xt)dt for a functionvof a Markov process(X)and a suitable random timeT. In our proof we need a particular case of this formula, wherev=1 andT is an exit time from an interval or a hitting time.
Fora < x < bconsider Ta,b=inf
t≥0;Xt∈ ]/ a, b[ .
The Green potential kernel on[a, b]is given by (see e.g. [23], Chapter VII) G(a, b, x, y)=(S(b)−S(x∨y))(S(x∧y)−S(a))
S(b)−S(a) .
This kernel defines the Green operator Gf (x)=
b
a
G(a, b, x, y)f (y)dm(y).
Notice that sinceG(a, b, x, a)=G(a, b, x, b)=0, the integration interval may or may not includeaandb.
With the help of this operator we can calculate the moments ofTa,busing Kac formula ExTa,bn =n
b
a
G(a, b, x, y)EyTa,bn−1dm(y)=n!Gn1(x). (1.1) To obtain an analogous formula for the moments of hitting times, recall that limt→±∞S(t)= ±∞, and consider the limits ofG(a, b, x, y)whena→ −∞(resp.b→ ∞):
G(−∞, b, x, ξ )= S(b)−S(ξ )
, x≤ξ≤b, S(b)−S(x)
, −∞< ξ≤x,
G(a,+∞, x, ξ )= S(ξ )−S(a)
, a≤ξ≤x, S(x)−S(a)
, x≤ξ <∞.
Taking monotone limits in (1.1), we get a formula for thenth moment of hitting times (see also [18]):
ExTbn=n b
−∞G(−∞, b, x, ξ )EξTbn−1dm(ξ ) ifx < b,
(1.2) ExTan=n
+∞
a
G(a,+∞, x, ξ )EξTan−1dm(ξ ) ifx > a.
The summation overnyields a formula for exponential moments:
Exexp(λTb)=1+λ b
−∞G(−∞, b, x, ξ )Eξexp(λTb)dm(ξ ), x < b,
(1.3) Exexp(λTa)=1+λ
+∞
a
G(a,+∞, x, ξ )Eξexp(λTa)dm(ξ ), x > a.
Remark. The expressions(1.2)–(1.3)are always defined,since all functions therein are positive.
The following proposition will be referred to as “all-or-none” property in the sequel:
Proposition 1.2 (All-or-none). Leta∈Randλ >0.The following properties are equivalent:
• for somex > a,Exexp(λTa) <∞,
• for allx > a,Exexp(λTa) <∞,
• a+∞Eξexp(λTa)dm(ξ ) <∞.
The same holds forx < a.
Proof. Observe thatG(a,+∞, x, ξ )≡const>0 forξ > x, and thatE•exp(λTa)is increasing on]a,∞[. Using the exponential Kac formula we then see that forx > a,Exexp(λTa) <∞if and only if Eξexp(λTa)ism-integrable on ]a,∞[. In this case, since mcharges every interval ofR, Eξexp(λTa) <∞ for allξ > a by monotonicity of
Eξexp(λTa).
1.2. Exit time from an interval
It is known for a while (Khasminskii condition, see [9]) that the exit timeTa,bfrom a bounded interval[a, b]admits an exponential moment for someλ >0. Let
λa,b=sup
λ: ∀x∈ ]a, b[,Exexp(λTa,b) <∞ .
In this subsection we establish some upper and lower bounds forλa,b. The upper bound will be particularly important in the proof of the Theorem1.1.
Lemma 1.3. λa,b≥1/Ca,b,where Ca,b= 1
S(b)−S(a) b
a
S(b)−S(y)
S(y)−S(a) m(dy).
Proof. Observe that
G(a, b, x, y)≤(S(b)−S(y))(S(y)−S(a)) S(b)−S(a) . Iff is positive and bounded, then
Gf (x)≤ 1 S(b)−S(a)
b
a
S(b)−S(y)
S(y)−S(a)
f (y)m(dy)≤Ca,bf∞.
It follows thatExTa,bn /n! =Gn1(x)≤Ca,bn , whenceExeλTa,b<∞forλ <1/Ca,b. Makingb→ ∞, we getExeλTa<∞for allx > aif
1
λ > Ca,∞= ∞
a
S(y)−S(a) m(dy),
but the last integral is generally infinite, so this inequality does not provide a satisfactory lower bound forλ+a. Observe, however, that whenCa,∞<∞, we getExeλTa ≤(1−λCa,∞)−1 for allx > a andλ <1/Ca,∞(we postpone an example for the last section).
To find an upper bound forλa,b, fix some interval[a, b] ⊂ ]a, b[.Defineκ1>0 andκ2>0 by S
a
−S(a)=κ1 S
b
−S a
, S(b)−S b
=κ2 S
b
−S a
(1.4) and denote
c= κ1κ2 1+κ1+κ2
S b
−S a
m a, b
.
Lemma 1.4. Ifλ≥1/cthen for allx∈ [a, b],ExeλTa,b= ∞.In particular,λa,b≤1/cfor any choice of[a, b].
Proof. Observe that for allx, yin[a, b],
G(a, b, x, y)≥(S(b)−S(b))(S(a)−S(a))
S(b)−S(a) = κ1κ2 1+κ1+κ2
S b
−S a
. It follows that for allx∈ [a, b]
ExTa,b≥ b
a
G(a, b, x, y)dm(y)≥ κ1κ2 1+κ1+κ2
S b
−S a
m a, b
=c.
By inductionExTa,bn ≥n!cn, as seen from ExTa,bn ≥n
b
a
G(a, b, x, y)EyTa,bn−1dm(y)≥n(n−1)!cn−1 b
a
G(a, b, x, y)dm(y)=n!cn.
HenceExeλTa,b = ∞pourλ≥1/candx∈ [a, b].
At this stage let us mention a theorem of Carmona–Klein [4]
“spectral gap”⇒ExeλTU <∞,
whereUis a set of positive invariant measure. The formulation of this theorem is somewhat confusing, since it does not precise thatλdepends onU. In fact, the following corollary shows that the propertyExeλTa <∞can not hold simultaneously for all(x, a)with a commonλ >0.
Corollary 1.5. ∀λ >0,∀x∈R,there exista < xandb > xsuch thatExeλTa,b=ExeλTa =ExeλTb = ∞.
Proof. Fixλ >0 and x∈R. Put, for example, κ1=κ2=1 and chose [a, b] and[a, b]in such a way that x∈ [a, b] ⊂ ]a, b[and the equalities (1.4) hold. Then
1
c= 3
(S(b)−S(a))m([a, b])< λ
as soon as (S(b)−S(a))m([a, b]) >3λ, which can always be achieved takinga orblarge enough. Hence, ac- cording to Lemma1.4,ExeλTa,b= ∞and therebyExeλTa=ExeλTb = ∞for suchaandb.
1.3. Hitting time moments
In this subsection we will prove the Theorem1.1:
1 4Ba+
≤λ+a ≤ 1 4Ca+
∧ 1
Ba+ and 1 4Ba−
≤λ−a ≤ 1 4Ca−
∧ 1 Ba−.
The proofs of two parts being completely similar, we only give one forλ+a. It will be split in a number of propositions.
Proposition 1.6. ∀a∈R, λ+a ≤B1+
a ,where1/∞ =0.
Proof. Fix somea∈R. From Lemma1.4we deduce that fora < a< x < b< b, for allx∈ [a, b],ExeλTa = ∞if λ≥1/c, where
1
c= 1+κ1+κ2
κ1κ2(S(b)−S(a))m([a, b]).
Now fixaandband makek2→ ∞(sob→ ∞), then 1
c→ 1
κ1(S(b)−S(a))m([a, b])= 1
(S(a)−S(a))m([a, b]). We conclude thatExeλTa= ∞forx∈ [a, b]and
λ > 1
(S(a)−S(a))m([a, b]).
It follows by the “all-or-none” Proposition1.2thatExeλTa = ∞for anyx > a, alla,bandλas above. Observing that
sup
a,b
S a
−S(a) m
a, b
=sup
a>a
S a
−S(a) m
a,∞
=Ba+
by the continuity ofS, we getExeλTa = ∞for any λ > inf
a,b
1
(S(a)−S(a))m([a, b])= 1 Ba+.
The inequalityλ+a ≤1/Ba+is thereby proved.
The bounds(4Ca+)−1≥λ+a ≥(4Ba+)−1require more work. To simplify the notations we putBa+=B. Letb≥aand define forx > aandf ≥0 two positive linear operators,JbandKb:
Kbf (x)= x
a
S(y)−S(a)
f (y)1b≤ym(dy), Jbf (x)=
S(x)−S(a) 1b≤x
∞
x
f (y)m(dy), where xyis understood as ]x,y]. Notice that
Gf =Jaf+Kaf≥Jbf +Kbf=G(f1[b,∞[).
Put
C(b)=inf
y>bJb1(y), B(b)=sup
y>b
Jb1(y), soB=B(a).
Proposition 1.7. We have n
l=0
an,lCl(b)Kbn−l1(x)≤(Jb+Kb)n1(x)≤ n
l=0
an,lBl(b)Kbn−l1(x) wherean,l≥0satisfy
an,l=0 ifl <0orl > n, a0,0=1, an+1,l=
i≤l
an,i.
Proof. For anyf we have JbKbf (x)=
S(x)−S(a) 1b≤x
∞
x
dm(y) y
a
S(u)−S(a)
f (u)1b≤udm(u)
=
S(x)−S(a) 1b≤x
∞
x
dm(y) x
a
S(u)−S(a)
f (u)1b≤udm(u) +
S(x)−S(a) 1b≤x
∞
x
dm(y) y
x
S(u)−S(a)
f (u)1b≤udm(u)
=Jb1(x)Kbf (x)+
S(x)−S(a) 1b≤x
∞
x
f (u)dm(u)
S(u)−S(a) 1b≤u
∞
u
dm(y)
=Jb1(x)Kbf (x)+Jb(f Jb1)(x)
whence
C(b)(Kbf +Jbf )(x)≤JbKbf (x)≤B(b)(Kbf+Jbf )(x),
the inequalities being trivially true forx≤b. By induction, we easily see that the following inequalities hold for all n∈N:
C(b)
Cn(b)+Cn−1(b)Kb1(x)+ · · · +Kbn1(x)
≤JbKbn1(x)≤B(b)
Bn(b)+Bn−1(b)Kb1(x)+ · · · +Kbn1(x) .
Now notice that forn=0,(Jb+Kb)n1(x)=1=a0,0. By induction again, for anyx > a, (Jb+Kb)n+11(x)≤(Jb+Kb)
l≥0
an,lBl(b)Kbn−l1(x)
=Jb
l≥0
an,lBl(b)Kbn−l1(x)+
l≥0
an,lBl(b)Kbn−l+11(x)
≤B(b)
l≥0
an,lBl(b)
n−l
i=0
Bn−l−i(b)Kbi1(x)+
l≥0
an,lBl(b)Kbn−l+11(x)
=
i≥0
Bn+1−i(b)Kbi1(x)
l≤n−i
an,l+
i≥0
an,n+1−iBn+1−i(b)Kbi1(x)
=
i≥0
Bn+1−i(b)Kbi1(x)
l≤n+1−i
an,l=
j≥0
Bj(b)Kbn+1−j1(x)
l≤j
an,l
=
j≥0
an+1,jBj(b)Kbn+1−j1(x).
The lower bound is proved in the same way.
An explicit formula foran,l will now be derived.
Lemma 1.8. For0≤l≤n,an,l=Cnl+l−Cnl−+1l,withCn−1=0.This implies 4n
n+2
√ 8
π(n+2) n k=0
an,k=an+1,n≤4n, whence
nlim→∞
√nan+1,n=4.
Proof. The expression ofan,lfollows by induction from the recursive definition ofan,land the equality C0n+Cn1+1+ · · · +Cnl+l=Cnl+l+1.
The second assertion follows then easily by Stirling’s formula.
Theorem 1.9. Recall that B=Ba+=sup
x>a
S(x)−S(a) m
]x,∞]
and
Ca+=lim inf
x→∞
S(x)−S(a) m
]x,∞]
.
• If0≤λ < (4Ba+)−1then for allx > a,ExeλTadm(x) <∞.
• Ifλ > (4Ca+)−1then for allx > a,ExeλTadm(x)= ∞.
Proof. Putfb= b∞f (z)m(dz)forb≥a. We have forf ≥0 Kbfb=
∞
b
m(dz) z
b
S(u)−S(a)
f (u)m(du)= ∞
b
f (u)m(du)
S(u)−S(a)
∞ u
m(dz), whence
C(b)fb≤ Kbfb≤B(b)fb. (1.5)
The first point of the theorem being obviously true forB= ∞, we can suppose thatB=B(a) <∞. Combining the inequality (1.5) forb=awith the Proposition1.7and Lemma1.8, we can write
1 n!
+∞
a
ExTandm(x)=Gn1
a≤ n l=0
an,lBlKan−l1
a
≤ n
l=0
an,lBlBn−l1a=an+1,nBnm ]a,∞[
≤4nBnm ]a,∞[
,
which implies the first assertion.
In the same way, for anyb≥a, Gn1
b≥ n
l=0
an,lC(b)lC(b)n−l1b=an+1,nC(b)nm ]b,∞[
.
Lemma1.8now implies that b∞ExeλTam(dx)= ∞for anyλ > (4C(b))−1. The second point follows by the “all-or-
none” property, since limb→∞C(b)=Ca+.
Finally, the Propositions1.6and1.9jointly imply the assertion of Theorem1.1, namely 1
4Ba+
≤λ+a ≤ 1 4Ca+
∧ 1
Ba+ and 1 4Ba−
≤λ−a ≤ 1 4C−a
∧ 1 Ba−, the inequalities concerningλ−a being proved in the same way.
Remark. It is easy to see thatBa±<∞implies∀a∈R, Ba±<∞.So the Theorem1.1yields yet another “all-or- none” property:
∃a∈R, λ±a >0 ⇐⇒ ∀a∈R, λ±a >0.
The Corollary1.5implies,however,that
alim→∞λ−a = lim
a→−∞λ+a =0.
2. Spectral gap
It turns out that the quantitiesBa±have already appeared in the context of the operators theory (see, e.g., [15,17]). In this section we discuss some relations between exponential moments of a diffusion and spectral gaps of associated operators.
We begin with a general remark. Consider a Hunt process X on a Polish space Ein the sense of Fukushima et al. [11]. Letmbe a Radon measure onE. Suppose thatmis bounded andXis am-symmetric process. Denote by (Pt)t≥0the transition semigroup ofX. Denote byPx, x∈E, the law of the processXissued fromx∈E.
For an open setG⊆E, set τG=inf{t >0:Xt ∈/G} the exit time ofXfromG. Introduce
PtG[A](x)=Px[Xt∈A;t < τG] for measurable subsetAofE, and set
XGt =
Xt, 0≤t < τG, , t≥τG.
Then, according to [11],XGis a Hunt process on the state spaceG, symmetric with respect to the measureIG·m(dx) with the transition semigroup(PtG). IfAGdenotes the infinitesimal generator of(PtG)inL2(IG·m(dx)),AGis a self- adjoint negative operator. Let us denote by(·,·)the scalar product inL2(IG·m(dx))and by(Eξ, ξ≥0)the spectral family of−AG.
Recall now the usual properties of the spectral decomposition. For any bounded and continuousf on[0,∞[one can definef (−AG)by
f
−AG u=
[0;∞[f (ξ )dEξu, u∈L2
IG·m(dx) , with
f
−AG u, g
−AG v
=
[0;∞[f (ξ )g(ξ )d(Eξu, v).
In particular, PtG=exp
t AG
=
[0;∞[e−ξ tdEξ.
Denote by EG the Dirichlet form associated with −AG on L2(m). Let Hξ be the image space ofEξ (which is a projection operator). The elements ofH0are those who satisfyPtGu=ufor allt >0. We know that−AGhas a spectral gap at 0 of width at leastγ >0 if and only if the following inequality
γu−E0u2=γ
]0,∞[d(Eξu, u)≤
]0,∞[ξd(Eξu, u)=EG(u, u) (2.1)
holds for alluin the domain ofEG. 2.1. Khasminskii identity
The main theorem of this section concerns the caseτG<∞.
For any bounded non negative functionu(x)∈L2(IG·m(dx)), for allλ >0, 0< N <∞, we have the formulas N
0
eλtPtG1(x)dt=1 λEx
eλτG∧N−1 , N
0
eλtPtGu(x)dt≤ u∞1 λEx
eλτG∧N−1 . The spectral calculus yields
1 λEx
eλτG∧N−1
=
[0,∞[
e(λ−ξ )N−1 λ−ξ dEξ1.
Hypothesis (λ0). λ0>0and for anyλ < λ0,Ex[eλτG]is an element ofL1(IG·m(dx)).
Theorem 2.1. Hypothesis(λ0)is equivalent toE(λ0−)=0,i.e.−AGhas a spectral gap of width at least equal toλ0. Remark. This equivalence for bounded domains G⊂Rn is well-known since the works of Khasminskii [16] and Friedman[10].However,the proof of[16],Theorem2,makes use of the boundedness ofExτGinG,which may not be the case in our general setting.
The proof is divided in two parts.
Lemma 2.2. Hypothesis(λ0)implies [0,λ
0[dEξ=0,i.e.E(λ0−)=0.
Proof. Let 0< λ < λ0. For any bounded non negative functionf (x)∈L2(IG·m(dx)), for allλ >0, 0< N <∞, we can write
f2∞ λ
E•
eλτG∧N−1 ,1
≥ N
0
eλtPtGfdt, f
=
[0,∞[d(Eξf, f ) N
0
e(λ−ξ )tdt
≥
[0,λ[d(Eξf, f ) N
0
e(λ−ξ )tdt
=
[0,λ[
e(λ−ξ )N−1
λ−ξ d(Eξf, f ).
Taking the limit whenN↑ ∞, the preceding computation gives (E(λ−)f, f )=
[0,λ[d(Eξf, f )=0.
The bounded non negative functions being dense inL2(IG·m(dx)), we conclude thatE(λ−)=0. Since this holds for
any 0< λ < λ0, the lemma is proved.
Lemma 2.3. Let0< λ < λ0.Suppose thatE(λ0−)=0,i.e. [0,λ
0[dEξ=0.Then,E•[eλ]is an element ofL1(IG· m(dx))and therefore Hypothesis(λ0)is true.
Proof. For 0< λ < λ0we look at the formula 1
λE•
eλτG∧N−1 ,1
=
[λ0,∞[
1−e(λ−ξ )N ξ−λ
d(Eξ1,1).
LetN↑ ∞. The dominated convergence theorem yields 1
λ E•
eλτG−1 ,1
=
[λ0,∞[
1
(ξ−λ)d(Eξ1,1)
≤ 1
(λ0−λ)(1,1)
= 1
(λ0−λ)m(G) <∞ whence
EmeλτG≤ m(G) 1−λ/λ0
. (2.2)
Now, forG= ]a,∞[andG= ]−∞, a[, denote byγa+andγa−the spectral gaps of the corresponding operators. In virtue of the “all-or-none” Proposition1.2, the Hypotheses(λ±a)are verified, and we obtain
Theorem 2.4. For anya∈R,γa±=λ±a,whence 1
4Ba±
≤γa±≤ 1 4Ca±
∧ 1 Ba±. 2.2. Hardy and Poincaré inequalities
The Theorem2.1and the above (in)equalities only make sense under the conditionτG<∞. In this subsection we would like to estimate the spectral gap of a (not-killed) diffusionXonG=E=R. To do so, we use the well-known relations between spectral gaps and Poincaré inequalities.
Recall thatS andmare a scale function and the corresponding speed measure ofX. Denote by dS the measure induced byS(x). LetF (x)be a real function onR. We shall write dF dS, if there exists a functionf (x)inL1(dS) such that
b
a
f (x)dS(x)=F (b)−F (a) ∀a < b.
The functionf (x)will be denoted dFdS(x). Introduce then the function spaces F=
F∈L2(m): dFdS,dF
dS ∈L2(dS)
, (2.3)
F]a,∞[=
F∈F: F (x)=0, x≤a , F]−∞,a[=
F∈F: F (x)=0, x≥a .
Theorem 2.5. The diffusionXism-symmetric.The Dirichlet space associated withXis the function spaceFgiven by(2.3),and the Dirichlet form has the expression
E(F, F )= ∞
−∞
dF dS
2
(x)dS(x), F ∈F.