# Bilateral Harnack Inequalities for Stochastic Differential Equation with Multiplicative Noise

Authors: Zihao An; Gaofeng Zong
Journal: Journal of Function Spaces
 (2022)

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## Abstract

By constructing a coupling with unbounded time-dependent drift, a lower bound estimate of dimension-free Harnack inequality with power is obtained for a large class of stochastic differential equation with multiplicative noise. The key is an application of the inverse Hölder inequality. Combining this with the well-known upper bound, bilateral dimension-free Harnack inequality with power is established. As a dual inequality, the bilateral shift-Harnack inequalities with power are also investigated for stochastic differential equation with additive noise. Applications to the study of heat kernel inequalities are provided to illustrate the obtained inequalities.


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## Body

## 1. Introduction

We consider the following stochastic differential equation (SDE for brief) onℝd:
(1)dXt=bt,Xtdt+σt,XtdBt,whereBtt≥0 is the d-dimensional Brownian motion on a filtered complete probability space Ω,F,Ftt≥0,ℙ satisfying the usual hypotheses, and
(2)b:Ω×0,∞×ℝd⟶ℝd,σ:Ω×0,∞×ℝd⟶ℝd⊗ℝd,are progressively measurable and locally bounded in the second variable and are continuous in the third variable. We will assume throughout this paper that, for any initial valueX0∈ℝd, the SDE above has a unique strong solution which is nonexplosive and continuous in time t. To establish the bilateral Harnack inequalities, we will introduce the following assumptions:Assumption 1.
Forb,σ, there exists an increasing function Kon0,∞ so that for all t≥0, all y′,y∈ℝd, and
(3)2bt,y′−bt,y,y′−y+σt,y′−σt,yHS2≤Kty′−y2,a.s.Assumption 2.
There exists a decreasing functionλ:0,∞⟶0,∞ s.t. for all x∈ℝd,t≥0,(4)σt,x∗σt,x≥λt2I,a.s.Assumption 3.
Forσ, there exists an increasing positive function δon0,∞ s.t. for all t≥0, all y′,y∈ℝd,δt<λt, and
(5)σt,y′−σt,yy′−y≤δty′−y,a.s.Assumption 4.
Forn≥1, there exists a constant cn>0 s.t. for any y′,y,t≤n,(6)σt,y′−σt,yHS+bt,y′−bt,y≤cny′−y,a.s.It follows from the above assumptions that the existence and uniqueness of the strong solution of SDE (1) are ensured (see Protter [1]). For more results about existence and uniqueness of the solution of differential equations, we can see [2–4].LetXtx be the solution to the SDE (1) for X0=x. In this paper, we will establish the bilateral Harnack inequalities for the operator Pt; for any f∈Bb+ℝd, here Bb+ℝd is the class of all bounded nonnegative measurable functions:
(7)Ptfx≔EfXtx,x∈ℝd,t≥0,whereE is the expectation with respect to the probability measure ℙ. In Wang [5], under the above assumptions (Assumptions 1–4), the author has successfully established the dimension-free Harnack inequalities for a large class of SDEs with multiplicative noise by constructing a coupling.The dimension-free Harnack inequality was firstly established by Wang [6] for diffusion semigroups on Riemannian manifolds under a curvature condition. As a useful tool in the study of diffusion semigroups, in particular, for the uniform integrability, contractivity properties, and estimates on heat kernels, Wang’s remarkable work has cause a lot of studies in the last two decades (see Aida and Zhang [7], Arnaudon et al. [8], and Da Prato [9]).Wang’s Harnack inequality has been extended in a large number of papers. For instance, by using coupling method, the upper bounded of dimension-free Harnack inequality has been established in [10] for a delay SDE with additive noise, in Wang and Yuan [11] for a delay SDE with multiplicative noise, in Wang and Zhang [12] for SDE with non-Lipschitz drift and driven by additive anisotropic cylindrical α-stable process, and in Wang et al. [13] for stochastic Burgers equation. Recently, using coupling by change of measures, the dimension-free Harnack inequality is investigated in Wang [14] for a distribution-dependent stochastic differential equation with regular coefficients and in Huang and Wang [15] for a distribution-dependent SDE with singular coefficients. For more details, we can refer to the book of Wang [16] for a deep analysis about dimension-free Harnack inequalities. All existing literatures focus on the upper bound of the dimension-free Harnack inequality. So far, we have not seen the lower bound estimate of dimension-free Harnack inequality with power in any previous literature. The aim of this paper is to establish the lower bound of the dimension-free Harnack inequality with power for SDE with multiplicative noise.Theorem 1 in Wang [5] gives an upper bound of the dimension-free Harnack inequality with power for SDE with multiplicative noise. Hence, combining Theorem 1 in Wang [5] with our lower bound of the dimension-free Harnack inequality with power for SDE with multiplicative noise, actually, we have the following bilateral dimension-free Harnack inequality with power for SDE with multiplicative noise.Theorem 5.
Let(8)κ1p≔1−pKT4pδp,Tδp,T−λT1−p1−e−KTT,
whereδp,T≔maxδT,λT1−p/2, and
(9)κ2p≔p−1KT4pδp,TλTp−1−δp,T1−e−KTT,
whereδp,T≔maxδT,λTp−1/2.
ForP1≔0,1−δT/λT2 and P2≔1+δT/λT2,+∞, the bilateral dimension-free Harnack inequality with power
(10)supp∈P1PTfpx1/pexpκ1px−y2≤PTfy≤infp∈P2PTfpx1/pexpκ2px−y2,
holds for allT>0,x,y∈ℝd and f∈Bb+ℝd.In this paper, we will give a direct proof about the lower bound estimate of dimension-free Harnack inequality with power for a large class of SDE with multiplicative noise, rather than the reciprocal transformation based on the upper bound of dimension-free Harnack inequality with power. Of course, based on the reciprocal transformation of our lower bound estimate, we can also get the upper bound of dimension-free Harnack inequality with power. Unfortunately, our method does not adapt to establish the lower bound estimate of dimension-free log-Harnack inequality. At the same time, there is no evidence that the lower bound estimate of dimension-free log-Harnack inequality can be easily obtained by reciprocal transformation. This also leaves an open problem of how to establish the lower bound estimate of dimension-free log-Harnack inequality for a large class of SDE with additive noise or multiplicative noise.Letptx,y be the density of the operator Pt with respect to a Radon measure μ. It follows from Theorem 5 that the following corollary on bilateral heat kernel inequalities is a direct consequence.Corollary 6.
Assume that Assumptions1–4 hold and the operator Pt have a strictly positive density Ptx,y with respect to μ. Then, for p>1+δT/λT2, the upper bound of heat kernel inequality
(11)∫ℝdpTx,zptx,zpty,z1/p−1,μdz≤exppκ2pp−1x−y2,
holds, and for0<p<1−δT/λT2, the lower bound of heat kernel inequality, and for all T>0,
(12)intMpTx,zpTx,zpTy,z1/pμdz≥expκ1px−y2,∀x,y∈ℝd,
holds.The main aim of this paper is to establish lower bound of dimension-free Harnack inequality with power for SDE with multiplicative noise. As a dual inequality, the shift-Harnack inequality has been developed and applied in Wang [17]. Obviously, it is relatively easy to establish lower bound of shift-Harnack inequality by the method used in this paper.Next, we establish the bilateral dimension-free Wang’s shift-Harnack inequalities with power forPt introduced in Wang [17]. We now only consider the additive noise for which the SDE (1) reduces to
(13)dXt=bt,Xtdt+σtdBt,X0=x.Theorem 7.
Letσ from 0,∞ to ℝd⊗ℝd and b from 0,∞×ℝd to ℝd be measurable and satisfy the assumptions in Section 1. Furthermore, the function σt is invertible and satisfies the assumption.Assumption 8.
There exists a decreasing functionλ from 0,∞ to 0,∞ s.t. for all t≥0,
(14)σt∗σt≥λt2I,a.s.
Then,(1)
for anyp∈0,1,T>0,x,v∈ℝd, and for any f∈Bb+ℝd,(15)PTfx≥PTfpv+⋅1/pxexp12p−1λT21T+KT+KT2T3(2)
for anyp>1,T>0,x,v∈ℝd, and for any f∈Bb+ℝd,(16)PTfx≤PTfpv+⋅1/pxexp12p−1λT21T+KT+KT2T3It is quite remarkable that the lower bound of dimension-free Harnack inequality with power in Theorem5 is essentially equivalent to the upper bounded of dimension-free Harnack inequality with power through appropriate transformation. The same remark as Theorem 5 is also true for bilateral dimension-free shift-Harnack inequalities with power in Theorem 7. In the future, we want to establish bilateral Harnack inequalities for various models, for example, functional SDEs with additive/multiplication noise, distribution-dependent SDEs, distribution-dependent SDEs with singular coefficients, and SDEs driven by cylindrical α-stable processes. However, the method we use here is not suitable for the lower bounded dimension-free log-Harnack inequality. How to establish the lower bounded dimension-free log-Harnack inequality is a very interesting and meaningful question.In Section2, we construct a coupling and prove several auxiliary results which will be needed for the proof of theorem. The dimension-free bilateral Harnack inequality with power and dimension-free bilateral shift-Harnack inequality with power are given in Sections 3–4, respectively. As applications, bilateral heat kernel estimates are derived in Section 5.


## 2. Auxiliary Results

Letx,y∈ℝd,T>0, and
(17)0<p<1−δTλT2,be fixed such thatx≠y. Due to δT<λT, we have
(18)θT≔2δTλT1−p∈0,2.For anyθ∈0,2, write
(19)κt=2−θKT1−eKTt−T,t∈0,T.Then, functionκ is smooth and strictly positive on 0,T such that
(20)2−KTκt+κt′=θ,t∈0,T.We now construct the following coupling:(21)dXt=σt,XtdBt+bt,Xtdt,X0=x,dYt=σt,YtdBt+bt,Ytdt+1κtσt,Ytσt,Xt−1Xt−Ytdt,Y0=y.We setinf∅=T by convention. Let
(22)ζn≔inft∈0,T:Yt≥n,andζ=limn⟶∞ζn is the explosion time of Yt, and then, the coupling Xt,Yt is a well-defined continuous process for t<T∧ζ. For t<T∧ζ, let
(23)dB~t=dBt+1κtXt−Ytσt,Xt−1dt.Ifζ=T and
(24)Rs≔exp−∫0sκt−1Xt−Ytσt,Xt−1,dBt−12∫0sκt−2σt,Xt−1Xt−Yt2dt,is a uniformly integrable martingale fors∈0,T, then by the martingale convergence theorem, RT≔limt↑TRt exists and Rtt∈0,T is a martingale. In this case, by the Girsanov theorem, B~tt∈0,T is a d-dimensional Brownian motion under the probability RTℙ. We can rewrite (21) as
(25)dXt=σt,XtdB~t+bt,Xtdt−Xt−Ytκtdt,X0=x,dYt=σt,YtdB~t+bt,Ytdt,Y0=y.Let(26)τn≔inft∈0,Tt:Xt+Yt≥n.We haveτn↑ζ as n⟶∞ because of the nonexplosibility of Xt.By the Lemma 2.1 in Wang [5], we know that for anys∈0,T, the following limits
(27)Rs∧ζ≔limn⟶∞Rs∧τn∧T−1/n,RT∧ζ≔lims↑TRs∧ζ,exist such thatRs∧ζs∈0,T with respect to ℙ is a uniformly integrable martingale, and
(28)ℚζ=T=1,so thatℚ=RTℙ.Lemma 9.
Assume Assumptions1–4. Then, for s∈0,T,
(29)Es,ζℚ∫0s∧ζXt−Ytκt22dt≤1θκ0x−y2,where the expectation Es,ζℚ is under the probability measure ℚ=Rs∧ζℙ.Proof.
Fixeds∈0,T. Using the Itô’s formula for Xt−Yt2 and Assumption 1, we have, for t≤s∧τn,
(30)dXt−Yt2=−2κtXt−Yt2dt+2bt,Xt−bt,Yt,Xt−Ytdt+σt,Xt−σt,YtHS2dt+2Xt−Ytσt,Xt−σt,Yt,dB~t≤−2κtXt−Yt2dt+KTXt−Yt2dt+2Xt−Ytσt,Xt−σt,Yt,dB~t.
Combining this with the fact2−KTκt+κ′t=θ, we obtain
(31)dXt−Yt2κt≤−2−KTκt+κt′κtXt−Yt2dt+2κtXt−Ytσt,Xt−σt,Yt,dB~t=−θκt2Xt−Yt2dt+2κtXt−Ytσt,Xt−σt,Yt,dB~t,
fort≤s∧τn. Multiplying this inequality by 1/θ and then integrating in 0,s∧τn, we get
(32)∫0s∧τnXt−Yt2κtdt≤−Xs∧τn−Ys∧τn2θκs∧τn+1θκ0x−y2+∫0s∧τn2θκtXt−Ytσt,Xt−σt,Yt,dB~t.
By the Girsanov theorem,B~tt≤s∧τn ia the d-dimensional Brownian motion under the probability measure Rs∧τnℙ. Taking expectation Es,nℚ with respect to Rs∧τnℙ, hence we obtain, for any s∈0,T,
(33)Es,nℚ∫0s∧τnXt−Yt2κt2dt≤1θκ0y−x2,n≥1.
By the dominated convergence theorem, we have(34)Es,ζℚ∫0s∧ζXt−Yt2κt2dt≤1θκ0x−y2,
where the expectationEs,ζℚ is under the probability measure ℚ=Rs∧ζℙ; hence, we obtain the desired result.Lemma 10.
Assume Assumptions1–4. Let
(35)rT=λT2θT24δT2−4θTλTδT<−1.
Then,(36)sups∈0,TERs∧ζ1+rT≤expθTKT2δT−λTθTx−y28δT22−θTδT−λTθT1−e−KTT.Proof.
By the definition ofRs, we have
(37)ERs∧τn1+rT=Es,nℚRs∧τnrT=Es,nℚexp−rT∫0s∧τn1κtσt,Xt−1Xt−Yt,dB~t+rT2∫0s∧τnσt,Xt−1Xt−Yt2κt2dt.
Observing that for any exponential integrable martingaleΛt with respect to Rs∧τnℙ, one has
(38)Es,nℚexprTΛt+rTΛt2=Es,nℚexprTΛt−rT2βΛt2+rTβrT+1Λt2≤Es,nℚexprTβΛt−rT2β2Λt21/β×Es,nℚexprTβrTβ+12β−1Λtβ−1/β=Es,nℚexprTβrTβ+12β−1Λtβ−1/β,
whereβ>1. It follows that
(39)ERs∧τn1+rT≤Es,nℚexprTβrTβ+12β−1∫0s∧τnσt,Xt−1Xt−Yt2κt2dtβ−1/β≤Es,nℚexprTβrTβ+12β−1λT2∫0s∧τnXt−Yt2κt2dtβ−1/β.
Noting thatrT<−1, take β=1+1+rT−1, i.e.,
which minimizesrTβrTβ+1/2β−1forβ∈1,∞ (see Figure 1), such that
(40)rTβrTβ+12β−1λT2=rT−rT1+rTrT+1−rT1+rT−2/rTrT1+rTλT2=rT−rT1+rT22λT2=θT28δT2.
Hence, noting thatβ−1/β∈0,1, we can write (39) as
(41)ERs∧τn1+rT≤Es,nℚexprTβrTβ+12β−1λT2∫0s∧τnXt−Yt2κt2dtβ−1/β≤Es,nℚexpθT28δT2∫0s∧τnXt−Yt2κt2dtβ−1/β.
Furthermore, we see from (40) that
(42)rT−rT1+rT=−λTθT2δT.
Noting thatrTβ=rT−rT1+rT, it follows from the above relation (42) that
(43)rT1+rT=rT+λTθT2θT=λT2θT24δT2−4λTθTδT+λTθT2δT=λT2θT22δT2δT−2λTθT+λTθT2δT=λT2θT22δT2δT−2λTθT+λTθT2δT−2λTθT2δT2δT−2λTθT=λT2θT2+λTθT2δT−2λTθT2δT2δT−2λTθT=2λTθTδT−λT2θT22δT2δT−2λTθT.
Due to the choice ofβ and the definition of rT (35) and (43), we can compute
(44)β−1β=−1/rTrT1+rT1−1/rTrT1+rT=rT1+rTrT1+rT−rT=2λTθTδT−λT2θT2/2δT2δT−2λTθTλTθT/2δT=2λTθTδT−λT2θT2λTθT/2δT×2δT2δT−2λTθT=2λTθTδT−λT2θT2λTθT2δT−2λTθT=2δT−λTθT2δT−2λTθT.
For anyα>0, it follows from (32) that
(45)Es,nℚexpα∫0s∧τnXt−Yt2κt2dt≤expαθTκ0x−y2⋅Es,nℚexp2αθT∫0s∧τn1κtXt−Ytσt,Xt−σt,Yt,dB~t≤expαKTθT2−θT1−e−KTTx−y2⋅Es,nℚexp8α2δT2θT2∫0s∧τnXt−Yt2κt2dt1/2
Takingα=θT2/8δT2, we have
(46)Es,nℚexpθT28δT2∫0s∧τnκt−2Xt−Yt2dt≤expθTKTx−y24δT22−θT1−e−KTT.
Now, returning to (41), and noticing that β−1/β∈0,1, then we obtain
(47)ERs∧τn1+rT≤Es,nℚexprTβrTβ+12β−1∫0s∧τnσt,Xt−1Xt−Yt2κt2dtβ−1/β≤Es,nℚexpθT28δT2∫0s∧τnXt−Yt2κt2dtβ−1/β≤expθKT4δT22−θT1−e−KTTy−x2β−1/β=expθKT2δT−λTθT8δT22−θTδT−λTθT1−e−KTTy−x2.
Therefore, by lettingn⟶∞, the Fatou lemma implies that
(48)sups∈0,TERs∧ζ1+rT≤expθTKT2δT−λTθT8δT22−θTδT−λTθT1−e−KTTx−y2.
This proves the assertion.Figure 1
MinimizerTβrTβ+1/β−1 when rT<−1.


## 3. Proof of Lower Bounded of Harnack Inequality

We will only give the proof of the lower bounded of the dimension-free Harnack inequality with power, since the upper bounded of the dimension-free Harnack inequality with power has been established in Wang [5]. We will first introduce an inverse Hölder inequality, which is the key building stone for this paper to obtain successfully lower bound of the dimension-free Harnack inequality with power. In the literature, we can find the earlier version of inverse Hölder inequality with integral form in Hardy et al. [18], P23-24, and more inverse Hölder inequalities can be found in Nehari [19] and Lin and Bai [20] for the following type and the references there cited.Lemma 11.
For anyp∈0,1, q≔p/p−1<0, then
(49)1p+1q=1,
and for any positive random variablesX,Y, the inverse Hölder inequality holds:
(50)EXY≥EXp1/pEYq1/q.For our purpose to establish the dimension-free Harnack inequality with power and for simplicity, we only considerp∈0,1, although the inverse Hölder inequality also holds for p<0. With the inverse Hölder inequality in hand, we now can start to prove the lower bounded of dimension-free Harnack inequality with power.Theorem 12.
Let(51)κ1p≔1−pKT4pδp,Tδp,T−λT1−p1−e−KTT,
whereδp,T≔maxδT,λT1−p/2. If Assumptions 1–4 hold, then for 0<p<1−δT/λT2, the dimension-free Harnack inequality with power
(52)PTfy≥PTfpx1/pexpκ1px−y2,
holds for allT>0,x,y∈ℝd and f∈Bb+ℝd.Proof.
SinceRs∧ζs∈0,T with respect to ℙ is a uniformly integrable martingale and B~tt∈0,T is the d-dimensional Brownian motion under new probability measure ℚ, thus Yt can be solved up to time T. Let
(53)τ≔inft∈0,T:Xt=Yt,
and setinf∅=∞ by convention. We claim τ≤T and thus, XT=YT,ℚ-a.s. Indeed, if for some ω∈Ω such that τω>T, by the continuity of the processes, we have
(54)inft∈0,TXtω−Ytω2>0.
So, on the setτ>T, we have
(55)∫0TXt−Yt2κt2dt=∞.
But according to Lemma9, we obtain
(56)Eℚ∫0TXt−Yt2κt2dt<∞.
Hence, we conclude thatℚτ>T=0. Therefore, XT=YT,ℚ-a.s.
Now, sinceXT=YT and B~tt∈0,T is the d-dimensional Brownian motion under new probability measure ℚ, we have
(57)PTfyp=EℚfYTp=ERT∧ζfXTp.
Using the inverse Hölder inequality, for0<p<1, hence −1<p−1<0, we have
(58)ERT∧ςfXTp≥PTfpxERT∧ζp/p−1p−1.
Hence, we arrive at(59)PTfyp≥PTfpxERT∧ζp/p−1p−1=PTfpxERT∧ς1+rTp−1.
Here,rT=1/p−1<−1.
Letθ=θT. Noting the choice of the θ (18), we have
(60)rT=1p−1=λT2θ24δT2−4θλTδT,
that is,(61)p−1λT2θ2=4δT2−4θλTδT.
In fact, in (18), we choose
(62)θ=2δTλT1−p,
which solves the equationp−1λT2θ2=4δT2−4θλTδT.
Observing that the functionxp−1 is a decreasing function with respect to x in 0,∞ if 0<p<1, it follows from Lemma 10 that
(63)PTfyp≥PTfpxERT∧ς1+rTp−1≥PTfpxexpp−1θKT2δT−λTθ8δT22−θδT−λTθ1−e−KTTx−y2.
Replacingθ with 2δT/λT1−p in (63), we can rewrite (63) and then obtain
(64)PTfyp≥PTfpxexp2δTp1−pKT8δT2δTp+1+λTp−11−e−KTTx−y2=PTfpxexpp1−pKT4δTδT−λT1−p1−e−KTTx−y2.
This completes the proof of this theorem.Remark 13.
In this theorem, we give a direct proof about the lower bound estimate of dimension-free Harnack inequality with power for a large class of stochastic differential equation with multiplicative noise, rather than the reciprocal transformation based on the upper bound of dimension-free Harnack inequality with power. Of course, based on the reciprocal transformation of our lower bound estimate, we can also get the upper bound of dimension-free Harnack inequality with power. Unfortunately, our method does not adapt to establish the lower bound estimate of dimension-free log-Harnack inequality. At the same time, there is no evidence that the lower bound estimate of dimension-free log-Harnack inequality can be easily obtained by reciprocal transformation. This also leaves an open problem of how to establish the lower bound estimate of dimension-free log-Harnack inequality for a large class of stochastic differential equation with additive noise or multiplicative noise.


## 4. Bilateral Shift-Harnack Inequalities

In this section, we establish the bilateral dimension-free Wang’s shift-Harnack inequalities with power forPt, and the upper bounded of this inequality has been introduced in Wang (2014). As seen in the previous section, the study for the multiplicative noise case is very complicated; we hence only consider the additive noise for which the SDE (1) reduces to
(65)dXt=bt,Xtdt+σtdBt,X0=x.Proposition 14.
Let functionσ from 0,∞toℝd×ℝd and b from 0,∞×ℝd to ℝd be two progressively measurable process and satisfy assumptions in Section 1. Furthermore, the function σt is invertible and satisfies the following assumption.Assumption 15.
There exists a decreasing functionλ from 0,∞ to 0,∞ s.t. for all x∈ℝd,t≥0,
(66)σt∗σt≥λt2I,a.s.
Then,(1)
for anyp∈0,1, T>0,x,v∈ℝd, and any f∈Bb+ℝd,(67)PTfx≥PTfpv+⋅1/pxexpv22p−1λT21T+KT+KT2T3(2)
for anyp>1, T>0,x,v∈ℝd, and any f∈Bb+ℝd,(68)PTfx≤PTfpv+⋅1/pxexpv22p−1λT21T+KT+KT2T3Proof.
LetYt=Xt+t/Tv,t∈0,T. Then, we have
(69)dYt=bt,Ytdt+σtdB~t,Y0=x,t∈0,T,
where(70)B~t≔Bt+∫0tηsds,ηt≔σ−1tvT+bt,Xt−bt,Xt+tv/T.
Let(71)RT=e−∫0Tηt,dBt−1/2∫0Tηs2ds.
By the Girsanov theorem, we have(72)PTfx=ERTfYT=ERTfXT+v.
Similar to (58), by inverse Hölder inequality, we obtain
(73)PTfx≥PTfpv+⋅1/pxERTp/p−1p−1/p.
Due to (71), we have
(74)ERTp/p−1≤EMtexpp2p−12∫0Tηs2ds=EMTexpp2p−12∫0Tσ−2tvT+bt,Xt−bt,Xt+tvT2ds≤EMTexpp2p−12λT2∫0TvT+bt,Xt−bt,Xt+tvT2ds.
Here, the last inequality holds according to Assumption15, and
(75)Mt≔exp−pp−1∫0tηsdBs−p22p−12∫0tηs2ds.
Now, noticing from Assumption1 that
(76)bt,x−bt,y≤KTx−y,
we see that for allx∈ℝ,
(77)vT+bt,x−bt,x+tv/T≤1+KTtTv.
Thus, in view of the above assertion (77) and EMT≤1, estimate (74) becomes
(78)ERTp/p−1≤exppv22p−12λT21T+KT+KT2T3.
Using again the fact that the functionxp−1 is a decreasing function with respect to x in 0,∞ if −1<p−1<0, it follows from (73) that
(79)PTfx≥PTfpv+⋅1/pxERTp/p−1p−1/p≥PTfpv+⋅1/pxexpv22p−1λT21T+KT+KT2T3.
This completes the proof of the first part of this theorem. Noticing that (78) holds for all p>0, the second part of this theorem can be obtained.In fact, we can get a better conclusion than Proposition14.Theorem 16.
Let functionσ from 0,∞ to ℝd×ℝd and function b from 0,∞×ℝdtoℝd be measurable and satisfy assumptions in Section 1. Furthermore, the function σt is invertible and satisfies Assumption 15. Then,
(1)
for anyp∈0,1, T>0,x,v∈ℝd, and f∈Bb+ℝd,(80)PTfx≥PTfpv+⋅1/pxexppv221+pp−1λT21T+KT+KT2T3(2)
for anyp>1, T>0,x,v∈ℝd, and f∈Bb+ℝd,(81)PTfx≤PTfpv+⋅1/pxexppv221+pp−1λT21T+KT+KT2T3Proof.
Similar to (39), we have
(82)ERTp/p−1≤Eℚexpp2p−12∫0Tηs2dsp/p+1=Eℚexpp2p−12∫0Tσ−2tvT+bt,Xt−bt,Xt+tvT2dsp/p+1≤Eℚexpp2p−12λT2∫0TvT+bt,Xt−bt,Xt+tvT2dsp/p+1,
Due to (77), we have
(83)ERTp/p−1≤expppv221+pp−12λT21T+KT+KT2T3.(1)
Forp that satisfies 0<p<1, using the fact that the function xp−1 is a decreasing function with respect to x in 0,∞, it follows from (73) based on the inverse Hölder inequality that(84)PTfx≥PTfpv+·1/pxERTp/p−1p−1/p≥PTfpv+·1/pxexppv221+pp−1λT21T+KT+KT2T3.This completes the proof of the lower bounded
(2)
Forp>1, by Hölder inequality, we have(85)PTfx≤PTfpv+·1/pxERTp/p−1p−1/p
Notice that (83) holds for all p>0; hence, we get
(86)PTfx≤PTfpv+·1/pxexppv221+pp−1λT21T+KT+KT2T3.
This completes the proof of this theorem.Remark 17.
From this simplified proof of the bilateral dimension-free shift-Harnack inequality with power for additive case, we can easily see that the moment estimation (83) is the key to establishing the bilateral dimension-free shift-Harnack inequality with power. We need to specifically point out that the moment estimation (83) holds for all p>0. To establish the upper bound, we need to use the moment estimation (83) and Hölder inequality, but to establish the lower bound, we need to use the moment estimation (83) and the inverse Hölder inequality.


## 5. Proof of the Corollary6

Letptx,y be the density of Pt with respect to μ which is a Radon measure. Then, it follows from Corollary 1.2 in Wang [5] that for p>1+δT/λT2, the Harnack inequality with power in Wang [5] is equivalent to the following heat kernel inequalities:
(87)∫ℝpTx,zptx,zpty,z1/p−1μdz≤expKTpx−y24δp,Tp+1p−1λT−δp,T1−e−KTT.For allx,y∈ℝd,T>0, hence, (11) holds. Due to this, then we will complete the proof of the Corollary 6, if we prove the following proposition:Proposition 18.
Assume that Assumptions1–4 hold. Let ptx,y be a strictly positive density of the operator Pt with respect to a Radon measure μ. Then, for 0<p<1−δT/λT2, the following heat kernel inequality
(88)∫MpTx,zpTx,zpTy,z1/pμdz≥exp1−pKT4pδTδT−λT1−p1−e−KTTx−y2,
holds for allx,y∈ℝd.Proof.
By Theorem12, we have
(89)PTfyp≥PTfpxexpκ~px−y2.
Here,(90)κ~p≔p1−pKT4δTδT−λT1−p1−e−KTT.
LetPTx,dy be a transition probability such that
(91)PTfx=∫EfyPTx,dy,x∈E.
Obviously, the above inequality implies thatPTx,· and PTy,· are equivalent to each other. Indeed, if PTx,A=0, then applying the above inequality to f=IA, we conclude that PTy,A=0.
Let(92)pT,x,yz≔PTx,dzPTy,dz,
be the Radon-Nikodym derivative ofPTx,· with respect to PTy,·.
Applying (89) to fz≔pT,x,yz1/p, we have
(93)PTfxp≥PTfpyexpκ~px−y2=expκ~px−y2∫EpT,x,yzp/pPy,dz=expκ~px−y2∫EpT,x,yzPy,dz=expκ~px−y2∫EPx,dz=expκ~px−y2.
So we obtain(94)PTfxp≥expκ~px−y2,
that is,(95)PTfx≥expκ~ppx−y2.
Thus,(96)∫MpTx,zpTx,zpTy,z1/pμdz≥expκ~ppx−y2.
Therefore, the desired result in (88) holds.



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*Source: 5464688-2022-07-30.xml*