HARNACK INEQUALITIES IN INFINITE DIMENSIONS RICHARD F. BASS† AND MARIA GORDINA∗ Abstract. We consider the Harnack inequality for harmonic functions with respect to three types of inﬁnite-dimensional operators. For the inﬁnite dimensional Laplacian, we show no Harnack inequality is possible. We also show that the Harnack inequality fails for a large class of Ornstein-Uhlenbeck processes, although functions that are harmonic with respect to these processes do satisfy an a priori modulus of continuity. Many of these processes also have a coupling property. The third type of operator considered is the inﬁnite dimensional analog of operators in Hörmander’s form. In this case a Harnack inequality does hold. 1. Introduction The Harnack inequality is an important tool in analysis, partial diﬀerential equations, and probability theory. For over half a century there has been intense interest in extending the Harnack inequality to more general operators than the Laplacian, with seminal papers by Moser [24] and KrylovSafonov [21]. See [20] for a survey of some recent work. It is a natural question to ask whether the Harnack inequality holds for inﬁnite-dimensional operators. If L is an inﬁnite dimensional operator and h is a function that is non-negative and harmonic in a ball with respect to the operator L and B2 is a ball with the same center as B1 but of smaller radius, does there exist a constant c depending on B1 and B2 but not on h such that h(x) 6 ch(y) for all x, y ∈ B2 ? When one considers the inﬁnite-dimensional Laplacian, or alternatively the inﬁnitesimal generator of inﬁnite-dimensional Brownian motion, there is ﬁrst the question of what one means by a ball. In this case there are two diﬀerent norms present, one for a Banach space and one for a Hilbert space. We show that no matter what combination of deﬁnitions for B1 and B2 that are used, no Harnack inequality is possible. Our technique is to use Date: September 24, 2012. 1991 Mathematics Subject Classiﬁcation. Primary 60J45; Secondary 58J35, 47D07. Key words and phrases. Harnack inequality; abstract Wiener space; OrnsteinUhlenbeck operator; coupling; inﬁnite dimensional processes. † This research was supported in part by NSF Grant DMS-0901505. ∗This research was supported in part by NSF Grant DMS-1007496. 1 2 BASS AND GORDINA estimates for Green functions for ﬁnite dimensional Brownian motions and then to go from there to the inﬁnite dimensional Brownian motion. For more on the potential theory of inﬁnite-dimensional Brownian motion we refer to the classic work of L. Gross [19], as well as to [10,11,15,22,25,26]. V. Goodman [16, 17] has several interesting papers on harmonic functions for the inﬁnite-dimensional Laplacian. We next turn to the inﬁnite-dimensional Ornstein-Uhlenbeck process and its inﬁnitesimal generator. See [13,22,28] for the construction and properties of these processes. In this case, the question of the deﬁnitions of B1 and B2 is not an issue. We show that again, no Harnack inequality is possible. We again use estimates for the Green functions of ﬁnite dimensional approximations, but unlike in the Brownian motion case, here the estimates are quite delicate. We also establish two positive results for a large class of inﬁnite dimensional Ornstein-Uhlenbeck processes. First we show that functions that are harmonic in a ball are continuous and satisfy an a priori modulus of continuity. Secondly, it is commonly thought that there is a close connection between coupling and the Harnack inequality. See [4] for an example where this connection is explicit. By coupling, we mean that given B2 ⊂ B1 with the same center but diﬀerent radii and x, y ∈ B2 , it is possible to construct two Ornstein-Uhlenbeck processes X and Y started at x, y, resp. (by no means independent), such that the two processes meet (or couple) before either process exits B1 . Even though the Harnack inequality does not hold, we show that for a large class of Ornstein-Uhlenbeck processes it is possible to establish a coupling result. Finally we turn to the inﬁnite-dimensional analog of operators in Hörmander’s form. These are operators of the form Lf (x) = n ∑ ∇2Aj f (x), j=1 where ∇Aj is a smooth vector ﬁeld. For these operators we are able to establish a Harnack inequality. To deﬁne a ball in this context we use a distance intimately tied to the vector ﬁelds A1 , . . . , An . In addition, we connect this distance to another distance introduced in [9] for Dirichlet forms, and later used in connection with parabolic Harnack inequalities in diﬀerent settings in [27]. Our technique to prove the Harnack inequality for these operators in Hörmander’s form is to employ methods developed by Bakry, Émery, and Ledoux. For general reviews on their approach with applications to functional inequalities see [1, 23]. We prove a curvature–dimension inequality, derive a Li-Yau estimate from that, and then prove a parabolic Harnack HARNACK INEQUALITIES IN INFINITE DIMENSIONS 3 inequality, from which the usual Harnack inequality follows. For this approach on Riemannian manifolds with Ricci curvature bounded below we refer to [3]. We are not the ﬁrst to investigate Harnack inequalities for inﬁnite dimensional operators. In addition to the papers [9] and [8] mentioned above, they have been investigated by Bendikov and Saloﬀ-Coste [7], who studied the related potential theory as well. Their context is quite diﬀerent from ours, however, as they consider inﬁnite-dimensional spaces which are close to ﬁnite-dimensional spaces, such as inﬁnite products of tori. This allows them to modify some of the techniques used for ﬁnite dimensional spaces. We mention three open problems that we think are of interest: 1. Our positive result is for operators that are the inﬁnite-dimensional analog of Hörmander’s form, but we only have a ﬁnite number of vector ﬁelds. The corresponding processes need not live in any ﬁnite dimensional Euclidean space, but one would still like to allow the possibility of there being inﬁnitely many vector ﬁelds. 2. Are there any inﬁnite-dimensional processes of the form Laplacian plus drift for which a Harnack inequality holds? 3. Restricting attention to the inﬁnite-dimensional Ornstein-Uhlenbeck process, can one deﬁne B1 and B2 in terms of some alternate deﬁnition of distance such that the Harnack inequality holds? The outline of our paper is straightforward. Section 2 considers inﬁnitedimensional Brownian motion, Section 3 contains our results on inﬁnite Ornstein-Uhlenbeck processes, while our Harnack inequality for operators of Hörmander form appears in Section 4. We use the letter c with or without subscripts for ﬁnite positive constants whose exact value is unimportant and which may change from place to place. Acknowledgement. We are grateful to Leonard Gross and Laurent SaloﬀCoste for providing us with necessary background on the subject. Our thanks also go to Bruce Driver, Tai Melcher and Sasha Teplyaev for stimulating discussions. 2. Brownian Motion We ﬁrst prove a proposition that contains the key idea. Let B (n) (x, r) = )1/2 (∑ n 2 . {y ∈ Rn : |x − y| < r}, where |x − y| = i−1 |xi − yi | Proposition 2.1. Let K > 0. For all n suﬃciently large, there exists a function hn which is non-negative and harmonic on its domain B (n) (0, 1) and points xn , zn ∈ B (n) (0, 1/2) such that hn (zn ) > K. hn (xn ) 4 BASS AND GORDINA Proof. Let Gn (x, y) = |x − y|2−n , a constant multiple of the Newtonian potential density on Rn . Let e1 = (1, 0, . . . , 0). If we set hn (x) = Gn (x, e1 ), then it is well-known that hn is harmonic in Rn \ {0}. Let xn = 0 and zn = 14 e1 . Both are in B (n) (0, 1/2) and (3/4)2−n hn (zn ) = >K hn (xn ) 12−n if n is suﬃciently large. Next we embed the above ﬁnite-dimensional example into the framework of inﬁnite-dimensional Brownian motion. Let (W, H, µ) be an abstract Wiener space, where W is a separable Banach space, H is a Hilbert space, and µ is a Gaussian measure. For background about abstract Wiener spaces, see [10] or [22]. We use ∥ · ∥H and ∥ · ∥W for the norms on H and W , respectively. We denote the inner product on H by ⟨·, ·⟩H . The classical example of an abstract Wiener space has W equal to the continuous functions on [0, 1] that are 0 at 0 and has H equal to the functions in W that are absolutely continuous and whose derivatives are square integrable. Another example that perhaps better illustrates follows is ∑ what 2 < ∞ and let to let H be the set of sequences (x1 , x∑ x 2 , . . .) such that i i W be the set of such that i λ2i x2i < ∞, where {λi } is a ﬁxed ∑ sequences sequence with i λ2i < ∞. Let H∗ be the set of h ∈ H such that ⟨·, h⟩H ∈ H ∗ extends to a continuous linear functional on W. Here H ∗ is the dual space of H, and is, of course, isomorphic to H. (We will continue to denote the continuous extension of ⟨·, h⟩H to W by ⟨·, h⟩H .) Next suppose that P : H → H is a ﬁnite rank orthogonal projection such that P H ⊂ H∗ . Let {ej }nj=1 be an orthonormal basis for P H and ℓj = ⟨·, ej ⟩H ∈ W ∗ . Then we may extend P to a unique continuous operator from W → H (still denoted by P ) by letting (2.1) P w := n ∑ j=1 ⟨w, ej ⟩H ej = n ∑ ℓj (w) ej for all w ∈ W. j=1 For more details on these projections see [14]. Let Proj (W ) denote the collection of ﬁnite rank projections on W such that P W ⊂ H∗ and P |H : H → H is an orthogonal projection, i.e. P has the form given in (2.1). As usual a function f : W → R is a (smooth) cylinder function if it may be written as f = F ◦ P for some P ∈ Proj (W ) and some (smooth) function F : Rn → R, where n is the rank of P . For example, let {en }∞ n=1 be an orthonormal basis of H such that en ∈ H∗ , and Hn be the span of {e1 , . . . , en } identiﬁed with Rn . For each n, deﬁne Pn ∈ Proj (W ) by Pn : W → Hn ⊂ H∗ ⊂ H as in (2.1). HARNACK INEQUALITIES IN INFINITE DIMENSIONS 5 ( √) For t > 0 let µt be the rescaled measure µt (A) := µt A/ t with µ0 = δ0 . Then as was ﬁrst noted by Gross in [19, p. 135] there exists a stochastic process Bt , t > 0, with values in W which is continuous a.s. in t with respect to the norm topology on W , has independent increments, and for s < t has Bt −Bs distributed as µt−s , with B0 = 0 a.s. Bt is called standard Brownian motion on (W, µ). Let B(W ) be the Borel σ-algebra on W . If we set µt (x, A) := µt (x − A), for A ∈ B (W ) , then it is well known that {µt } forms a family of Markov transition kernels, and we may thus view (Bt , Px ) as a strong Markov process with state space W , where Px is the law of x+B. We do not need this fact in what follows, but want to point out that Bn (t) := Pn B (t) ∈ Pn H ⊂ H ⊂ W give a natural approximation to B (t) as is pointed out in [14, Proposition 4.6]. We denote the open ball in W of radius r centered at x ∈ W by B(x, r) and its boundary by Sr (x). The ﬁrst exit time of Bt from B(0, r) will be denoted by τr . By [19, Remark 3.3] the exit time τr is ﬁnite a.s. A set E is open in the ﬁne topology if for each x ∈ E there exists a Borel set Ex ⊂ E such that P(σEx > 0) = 1, where σEx is the ﬁrst exit from Ex . Let f be a locally bounded, Borel measurable, ﬁnely continuous, realvalued function f whose domain is an open set in W . Then f is harmonic if ∫ (2.2) f (x) = f (x + y) πr (dy) Sr (0) for any r such that the closure of B(x, r) is contained in the domain of f , where πr (dy) = P0 (Bτr ∈ dy). Let f be a real-valued function on W . We can consider F (h) = f (x + h) as a function on H. If F has the Fréchet derivative at 0, we say that f is H-diﬀerentiable. Similarly we can deﬁne the second H-derivative D2 , and ﬁnally ∆f (x) := tr D2 f (x) whenever D2 f (x) exists and of trace class. The following properties can be found in [16, Theorems 1, 2, 3] Theorem 2.2. Let (W, H, µ) be an abstract Wiener space. (1) A harmonic function on W is inﬁnitely H-diﬀerentiable. The second derivative of a harmonic function at each point of its domain is a HilbertSchmidt operator. (2) If a harmonic function on W satisﬁes a uniform Lipschitz condition in a neighborhood of a point x, then the Laplacian of u exists at x and (∆u) (x) = 0. Remark 2.3. So far the theory of harmonic functions in inﬁnite dimensions may not seem that diﬀerent from the ﬁnite dimensional case. There are, however, striking diﬀerences. For example, Goodman [16, Proposition 4] 6 BASS AND GORDINA shows there exists a harmonic function that is not continuous with respect to the topology of W . In view of the previous theorem, however, it is smooth with respect to the topology of H. Let (W, H, µ) be an abstract Wiener space. Denote by Gn (x, z) the function on Rn ×Rn deﬁned by Gn (x, z) = |x−z|2−n . Consider Pn ∈ Proj (W ) as deﬁned by (2.1), and deﬁne the cylinder function gn (w) := Gn (Pn w, Pn z) for any w ∈ W and z = e1 . Proposition 2.4. The function gn is harmonic on W away from the set {w ∈ W : Pn w = e1 } = {w ∈ W : e1 (w) = 1}. Proof. We need to check that gn is locally bounded, Borel measurable, ﬁnely continuous, and (2.2) holds with f replaced by gn for all r > 0 whenever the closure of Br (x) is contained in the domain of gn . One can show that gn is locally bounded, Borel measurable, and ﬁnely continuous similarly to [16, p. 455]. Now we check the last part. Suppose x ∈ / {w ∈ W : Pn w = e1 }. ∫ ∫ Gn ◦ Pn (x + y) πr (dy) gn (x + y) πr (dy) = Sr (0) Sr (0) x = E (Gn ◦ Pn (Bτr )) = Ex (Gn ◦ Pn (Pn Bτr )) . Note that Pn Bt is a martingale, and τr is a stopping time, and we would like to use the optional stopping time theorem. We need to point out here that e1 ∈ H∗ ⊂ H and therefore Pn e1 = e1 . So if we choose r < 1/2∥e1 ∥W ∗ , then e1 ∈ / Pn B(0, r). Indeed, if there is a w ∈ B(0, r) such that Pn w = e1 , then e1 (w) = ⟨w, e1 ⟩ = 1. But 1 |e1 (w) | 6 ∥e1 ∥W ∗ ∥w∥W < r∥e1 ∥W ∗ < 2 which is a contradiction. Thus Gn is harmonic in Pn B(0, r) ⊆ Pn H ∼ = Rn and therefore ∫ gn (x + y) πr (dy) = Gn (Pn x) = gn (x) . Sr (0) Our main theorem of this section is now simple. Theorem 2.5. For each n there exist functions gn that are non-negative and harmonic in the ball of radius 1 about 0 with respect to the norm of W and points x, z in the ball of radius 1/2 about 0 with respect to the norm of H such that gn (z) →∞ gn (x) as n → ∞. In particular, the Harnack inequality fails. HARNACK INEQUALITIES IN INFINITE DIMENSIONS 7 Proof. We let gn be as above and x = 0 and z = 14 e1 for all n. Our result follows by combining Propositions 2.1 and 2.4. 3. Ornstein-Uhlenbeck process Let H be a separable Hilbert space with inner product ⟨·, ·⟩ and corresponding norm | · |. Deﬁne ∥f ∥0 := sup |f (x) |. x∈H Recall (see [13]) that for an arbitrary positive trace class operator Q on H and a ∈ H there exists a unique measure Na,Q on B (H) such that ∫ 1 ei⟨h,x⟩ Na,Q (dx) = ei⟨a,h⟩− 2 ⟨Qh,h⟩ , h ∈ H. H We call such Na,Q (dx) a Gaussian measure with mean a and covariance Q. It is easy to check that ∫ xNa,Q (dx) = a, H ∫ |x − a|2 Na,Q (dx) = Tr Q, H ∫ ⟨x − a, y⟩⟨x − a, z⟩Na,Q (dx) = ⟨Qy, z⟩, and H 1 dNb,Q −1/2 (a−b)|2 +⟨Q−1/2 (y−a),Q−1/2 (b−a)⟩ (dy) = e− 2 |Q . dNa,Q We consider the Ornstein-Uhlenbeck process in a separable Hilbert space H. The process in question is a solution to the stochastic diﬀerential equation (3.1) dZt = −AZt dt + Q1/2 dWt , Z0 = x, where A is the generator of a strongly continuous semigroup e−At on H, W is a cylindrical Wiener process on H, and Q : H → H is a positive bounded operator. The solution to (3.1) is given by ∫ t x −At Zt = e x+ e−A(t−s) Q1/2 dWs . 0 The corresponding transition probability is deﬁned as usual by (Pt f ) (x) = Ef (Ztx ), f ∈ Bb (H) , where Bb (H) are the bounded Borel measurable functions on H. It is known that the law of Zt is a Gaussian measure centered at e−At x with covariance ∫ t ∗ Qt = e−A(t−s) Qe−A (t−s) ds, 0 which we called Ne−tA x,Qt (dy). Note that for the corresponding parabolic equation in H to be well-posed we need a basic assumption on Qt to be non-negative and trace-class for all t > 0 [13, p. 99]. 8 BASS AND GORDINA We assume the controllability condition e−At (H) ⊂ Qt 1/2 (3.2) (H) for all t > 0 holds. As is described in [13, p. 104], under the condition (3.2) the stochastic diﬀerential equation in question has a classical solution. We deﬁne −1/2 −tA (3.3) Λt := Qt −1/2 e , t > 0, 1/2 where Qt is the pseudo-inverse of Qt . By the closed graph theorem we see that Λt is a bounded operator in H for all t > 0. Suppose Q = I, the identity operator, and A is a self-adjoint invertible operator on H, then ∫ t ∫ t 1 −sA −sA∗ Qt = e e xds = e−2sA ds = A−1 (I − e−2tA ), t > 0. 2 0 0 If in addition we assume that A−1 is trace-class, then there is an orthonormal basis {en }∞ n=1 of H and the corresponding eigenvalues an such that Aen = an en , an > 0, an ↑ ∞, ∞ ∑ a−1 n < ∞. n=1 Then Qt is diagonal in the orthonormal basis {en }∞ n=1 : ( ) t e2tan − 1 Q t en = en . 2tan e2tan Then Qt is trace class with ( ) ∞ ∞ ∑ ∑ t e2tan − 1 1 Tr A−1 Tr Qt = 6 = < ∞. 2tan e2tan 2an 2 n=1 Now we see that n=1 √ 2tan √ Λ t en = en , 1/2 t e2tan − 1 √ and so |Λt x| 6 |x|/ t. This proves the following proposition. Proposition 3.1. Assume that Q = I and A−1 is trace-class. Then the √ operator Qt is a trace-class operator on H and ∥Λt ∥ 6 1/ t. Using the properties of Gaussian measures, we see that the the OrnsteinUhlenbeck semigroup can be described by the following Mehler formula ∫ ( ) (3.4) (Pt f ) (x) = f z + e−tA x N0,Qt (dz) . H HARNACK INEQUALITIES IN INFINITE DIMENSIONS 9 3.1. Modulus of continuity for harmonic functions. We establish an a priori modulus of continuity for harmonic functions. Lemma 3.2. Suppose (3.2) is satisﬁed. If f is a bounded Borel measurable function on H and t > 0, there exists a constant c(t) not depending on f such that (3.5) |Pt f (x) − Pt f (y)| 6 c∥f ∥0 |x − y|, x, y ∈ H. Moreover, for any u ∈ H ( )1/2 Du Pt f (x) 6 Pt f 2 (x) ∥Λt u∥2 . Proof. Consider N0,Qt (dz), a centered Gaussian measure with covariance Qt . By the Cameron-Martin theorem the transition probability Ptx (dz) = Ne−tA x,Qt (dz) has a density with respect to N0,Qt (dz) given by ( ) Ne−tA x,Qt (dz) 1 −1/2 2 = exp ⟨Λt x, Qt z⟩ − |Λt x| . (3.6) Jt (x, z) := N0,Qt (dz) 2 Thus ∫ (3.7) (Pt f ) (x) = Jt (x, z) f (z) N0,Qt (dz) . H Now we can use (3.7) to estimate the derivative Du Pt f for any u ∈ H by ∫ ) −1/2 ( z − e−At x ⟩f (z) Jt (x, z) N0,Qt (dz) Du Pt f (x) = ⟨Λt u, Qt ∫H ( ) −1/2 = ⟨Λt u, Qt z⟩f z + e−At x N0,Qt (dz) H ( 2 )1/2 6 Pt f (x) ( 2 (∫ H −1/2 |⟨Λt u, Qt z⟩|2 N0,Qt )1/2 (dz) )1/2 = Pt f (x) ∥Λt u∥2 . Note that Λt is bounded, therefore for bounded measurable functions f we see that Pt f is uniformly Lipschitz, and therefore strong Feller. Assumption 3.3. We now suppose Q = I and that A is diagonal in an orthonormal basis {en }∞ n=1 of H with eigenvalues an being a sequence of positive numbers. Moreover, we assume that an /np → ∞ for some p > 3. Note that under this assumption A−1 is trace-class for p > 3, and therefore by Proposition 3.1 the operator Qt is trace-class as well. We need the following lemma. Lemma 3.4. Suppose Xt is an Ornstein-Uhlenbeck process with Q and A satisfying Assumption 3.3. Let r > q > 0 and ε > 0. Then there exists t0 such that Px (sup |Xs | > r) 6 ε, x ∈ B(0, q). s6t0 10 BASS AND GORDINA Proof. We ﬁrst consider the nth component of Xs . Taking the stopping time τ identically equal to t0 , the main theorem of [18, Theorem 2.5] tells us that √ c log(1 + an t0 ) n E sup |Xs | 6 . √ an s6t0 Then by Chebyshev’s inequality, √ c log(1 + an t0 ) n (3.8) P(sup |Xs | > dn ) 6 √ dn an s6t0 for any positive real number dn . Choose δ > 0 small so that ∑ (p − 1)/2 > 1 + δ. Take dn = C(r − q)n−1/2−δ , ∞ 2 −1−2δ = 1. Then P(sup n where C is chosen so that C s6t0 |Xs | > dn ) n=1 n is summable in n, and if we choose n0 large enough, ∞ ∑ P(sup |Xsn | > dn ) < ε/2. n=n0 s6t0 By taking t0 smaller if necessary, we then have ∞ ∑ n=1 P(sup |Xsn | > dn ) < ε. s6t0 Suppose |x| 6 q and we start the process at x. By symmetry, we may assume each coordinate of x is non-negative. Since |Xs | 6 |Xs − x| + |x|, we observe that in order for the process to exit the ball B(0, r) before time t0 , for some coordinate n we must have |Xsn | increasing by at least dn . The probability of this happening is largest when xn = 0. But the probability that for some n we have |Xsn | increasing by at least dn in time t0 is bounded by ε. Theorem 3.5. Suppose Xt is an Ornstein-Uhlenbeck process with Q and A satisfying Assumption 3.3. If h is a bounded harmonic function in the ball B(0, 1), there is a constant c such that (3.9) |h (x) − h (y) | 6 c∥h∥0 |x − y|, x, y ∈ B(0, 1/2). Proof. Let ε > 0 and let τ be the exit time from B(0, 1). By Lemma 3.4 we can choose t0 such that Px (τ < t0 ) < ε, x ∈ B (0, 1/2) . If h is harmonic in B(0, 1) and x, y ∈ B(0, 1/2), h(x) = Ex h(Xτ ) = Ex [h(Xτ ); τ < t0 ] + Ex [h(Xτ ) : τ > t0 ]. The ﬁrst term is bounded by ∥h∥0 ε. By the Markov property the second term is equal to Ex [EXt0 h(Xτ ); τ > t0 ] = Ex [h(Xt0 ); τ > t0 ], HARNACK INEQUALITIES IN INFINITE DIMENSIONS 11 which diﬀers from Pt0 h(x) by at most ∥h∥0 ε. We have a similar estimate for h(y). Therefore by Lemma 3.2 |h(x) − h(y)| 6 |Pt0 h(x) − Pt0 h(y)| + 4∥h∥0 ε 6 c(t0 )|x − y| ∥h∥0 + 4∥h∥0 ε. This proves the uniform modulus of continuity. Remark 3.6. We remark that the constant c in the statement of Theorem 3.5 depends on r. Moreover, there does not exist a constant c independent of z0 such that (3.9) holds for x, y ∈ B(z0 , r/2) when h is harmonic in B(z0 , r). It is not hard to see that this is the case even for the two-dimensional Ornstein-Uhlenbeck process. 3.2. Counterexample to the Harnack inequality. As we have seen, the transition probabilities for the Ornstein-Uhlenbeck process Zt are Ptx (dz) := Ne−tA x,Qt (dz) . Suppose now that Q = I and A satisfy Assumption 3.3 with p = 1, but also that an is an increasing sequence with A−1 being a trace-class operator on H. As examples of such an , we can take an = np for p > 1. Denote by Pn the orthogonal projection on Hn := Span{e1 , ..., en }. Then PtPn x (dPn z) := pn (t, Pn x, Pn z) dz, where pn (t,Pn x, Pn z) ( )2 ) ( )1/2 n ( ∏ 2aj zj − e−aj t xj 2aj 1 = . exp − 2π 1 − e−2aj t 2 (1 − e−2aj t ) j=1 We would like to consider the Green function hn with pole at zn = 4en for Zt killed when Zt1 exceeds 6 in absolute value. We use a killed process to insure transience. We will show that hn (xn ) →∞ hn (0) as n → ∞, where xn = en . The key is to estimate the Green function ∫ ∞ hn (x, z) := p̃n (t, Pn x, Pn z) dt, 0 where p̃n is the density for the killed process. We will prove an upper estimate on hn (0, zn ) and a lower estimate on hn (xn , zn ). First we need the following lemma. Lemma 3.7. Let a > 0 and let Yt be a one-dimensional Ornstein-Uhlenbeck process that solves the stochastic diﬀerential equation dYt = dBt − aYt dt, where Bt is a one-dimensional Brownian motion and a > 0. Let Ye be Y killed on ﬁrst exiting [−6, 6], let q(t, x, y) be the transition densities for Y , 12 BASS AND GORDINA and let qe(t, x, y) be the transition densities for Ye . (1) There exist constants c and β such that qe(t, 0, 0) 6 ce−βt , t ≥ 1. (2) We have qe(t, 0, 0) →1 q(t, 0, 0) as t → 0. 2 Proof. The transition densities of Ye with respect to the measure e−x /2 dx are symmetric and by Mercer’s theorem can be written in the form ∞ ∑ e−βi t φi (x)φi (y) i=1 with 0 < β1 ≤ β2 6 β3 6 · · · . Here the βi are the eigenvalues and the φi are the corresponding eigenfunctions for the Sturm-Liouville problem { Lf (x) = 12 f ′′ (x) − af ′ (x) = −βf (x), f (−6) = f (6) = 0. See [6, Chapter IV, Section 5] for details. (1) is now immediate. Let U be the ﬁrst exit of Y from [−6, 6]. Using the strong Markov property at U , we have the well known formula ∫ t E0 [q(t − s, Ys , 0); U ∈ ds] . q(t, 0, 0) = qe(t, 0, 0) + 0 Using symmetry, this leads to (3.10) ∫ q(t, 0, 0) = qe(t, 0, 0) + t q(t − s, 6, 0)P0 (U ∈ ds). 0 Now by the explicit formula for q(r, x, y), we see that q(t−s, 6, 0) is bounded in s and t and so the second term on the right hand side of (3.10) is bounded by a constant times P0 (U 6 t), which tends to 0 as t → 0. On the other hand, q(t, 0, 0) ∼ (2πt)−1/2 → ∞ as t → 0. (2) now follows by dividing both sides of (3.10) by q(t, 0, 0). We now proceed to an upper estimate for the Green function. Proposition 3.8. There are constants K > 0 and c > 0 such that hn (0, z) 6 Kcn ann/2 e−16an . Proof. First for x = 0 and z = 4en we have pn (t, Pn 0, Pn z) = )1/2 ( ) n ( ∏ 2aj 1 16an exp − . 2π 1 − e−2aj t 1 − e−2an t j=1 HARNACK INEQUALITIES IN INFINITE DIMENSIONS 13 Step 1. Let t be in the interval 0 < t 6 2a1n < 1. Then )1/2 ( )n/2 n ( ∏ 2aj 1 1 6 , −2a t 2π 1 − e j tπ j=1 where we used the fact that an is an increasing sequence. For any t we have 8 16an > ; −2a t n 1−e t 1 therefore for 0 < t < 2an , ( )n/2 1 −8/t pn (t, 0, 4en ) 6 e . tπ 1 The right hand side has its maximum at 16 n which is larger than 2an for all large enough n by our assumptions on Q and A. Thus we can estimate the right hand side by its value at the endpoint 2a1n : ( ) 2an n/2 1 −16an pn (t, 0, 4en ) 6 e , 0<t6 . π 2an Step 2. Let t be in the interval 2a1n < t 6 1. Denote by n0 the index for which 2an1 +1 < t 6 2a1n . As before 0 0 ( )1/2 ( ) n ∏ 1 2aj 16an exp − 2π 1 − e−2aj t 1 − e−2an t j=1 ( )1/2 ( ) ( )n0 /2 ∏ n 2aj 1 16an 1 exp − 6 tπ 2π 1 − e−2aj t 1 − e−2an t j=n0 +1 ( )1/2 ( )n0 /2 ∏ n 2aj 1 1 6 e−16an . tπ 2π 1 − e−2aj t j=n0 +1 There is constant c independent of n such that 2aj 1 6 caj 6 can , j = n0 + 1, ..., n. 2π 1 − e−2aj t Since 1/t < 2an , there is a constant c such that )1/2 ) ( n ( ∏ 2aj 1 16an −16an exp − 6 cn an/2 . n e 2π 1 − e−2aj t 1 − e−2an t j=1 Step 3. For t > 1 the transition density of the killed process can be estimated by )1/2 ( ) n ( ∏ 2aj 16an 1 exp − e−βt 2π 1 − e−2aj t 1 − e−2an t j=2 for some β > 0, using Lemma 3.7(1). Similarly to Step 2, p̃ (t, 0, 4en ) 6 cn−1 an(n−1)/2 e−16an e−βt 1 14 BASS AND GORDINA for some constant c1 . Thus we have that there is a constant c > 0 such that { n/2 cn an e−16an , 0 < t < 1, p̃ (t, 0, 4en ) 6 n/2 −16an −βt n c an e e , 1 < t. Integrating over t from 0 to ∞ yields the result. We now obtain the lower bound for the Green function. Proposition 3.9. Let x = en . There are constants M > 0, c > 0 and ε > 0 such that eεan hn (x, z) > M cn e−16an an/2 . n an Proof. For x = en and z = 4en we have ( ( )2 ) )1/2 n ( ∏ an 4 − e−an t 2aj 1 pn (t, Pn x, Pn z) = exp − . 2π 1 − e−2aj t (1 − e−2an t ) j=1 Observe that )1/2 ( ) n ( ∏ 2aj 1 1 n/2 > . 2π 1 − e−2aj t 2πt j=1 Consider t in the interval [1/an , 2/an ]. When n is large, 2/an 6 1. Set v = e−an t , so that v ∈ [1/e2 , 1/e] when t ∈ [1/an , 2/an ]. Note that (4 − v)2 >0 1 − v2 for v ∈ [0, 8/17] ⊃ [1/e2 , 1/e], so there is a constant ε > 0 such that 16 − (4 − v)2 > ε, 1 − v2 16 − Thus v ∈ [1/e2 , 1/e]. ( ( )2 ) an 4 − e−an t exp − > e−16an +εan . (1 − e−2an t ) We now apply Lemma 3.7(2) and obtain ∫ 2/an hn (x, z) > p̃n (t, Pn x, Pn z) dt 1/an > e−16an +εan cn2 ∫ 2/an 1/an = e−16an +εan cn3 ann/2−1 t−n/2 dt ( 1 − 2− 2 +1 n 2 −1 Thus we have hn (x, z) > M cn e−16an an/2 n n eεan . an ) . HARNACK INEQUALITIES IN INFINITE DIMENSIONS 15 Theorem 3.10. Let K > 0. There exist functions hn harmonic and nonnegative on B(0, 4) and points xn in B(0, 2) such that hn (xn ) ≥K hn (0) for all n suﬃciently large. Thus the Harnack inequality does not hold for the Ornstein-Uhlenbeck process. Proof. The embedding of the ﬁnite dimensional functions hn into the Hilbert space framework is done similarly to the proof of Theorem 2.5, but is simpler here as there is no Banach space W to worry about. We leave the details to the reader. The theorem then follows by combining Propositions 3.8 and 3.9. 3.3. Coupling. It is commonly thought that coupling and the Harnack inequality have close connections. Therefore it is interesting that there are inﬁnite-dimensional Ornstein-Uhlenbeck processes that couple even though they do not satisfy a Harnack inequality. We now consider the inﬁnite-dimensional Ornstein-Uhlenbeck deﬁned as in the previous subsection, but with an = np and p = 6. We have the following theorem. Given a process X, let τX (r) = inf{t : |Xt | ≥ r}. Theorem 3.11. Let x0 , y0 ∈ B(0, 1). We can construct two inﬁnite-dimensional Ornstein-Uhlenbeck processes Xt and Yt such that X0 = x0 a.s., Y0 = y0 a.s., and if Px0 ,y0 is the joint law of the pair (X, Y ), then Px0 ,y0 (TC < τX (2) ∧ τY (2)) > 0, where TC = inf{t : Xt = Yt }. Proof. Let WjX (t), WjY (t), j = 1, 2, . . ., all be independent one-dimensional Brownian motions. Let dXtj = dWjX (t) − aj Xtj dt, X0j = xj0 , and the same for Ytj , where we replace dWjX by dWjY and x0 by y0 . Let TCj = inf{t : X j (t) = Y j (t)}. We deﬁne { Y j (t), t < TCj ; j Y (t) = X j (t), t ≥ TCj . Let Px be the law of X when starting at x and similarly for Py . Deﬁne to be the law of X j (t) started at xj and so on. Use Lemma 3.4 to choose t0 small such that j Px sup Px,y (τX (5/4) ∧ τY (5/4) 6 t0 ) 6 1/4. x,y∈B(0,1) Our ﬁrst step is to show (3.11) ∞ ∑ j=1 Px j ,y j (TCj > t0 ) < ∞. 16 BASS AND GORDINA j The law of Xtj0 /2 under Px is that of a normal random variable with mean j e−aj t0 /2 xj and variance (1−e−aj t0 /2 )/2aj . If AX j is the event where X (t0 /2) −1/4 −1/4 is not in [−aj , aj ], then standard estimates using the Gaussian density ∑ xj X show that j P (Aj ) is summable. The same holds if we replace X by Y . −1/4 Suppose |x′j |, |yj′ | 6 aj (3.12) j Z (t) = (x′j − . Let yj′ ) + ∫ (WjX (t) − WjY (t)) − aj t Zj (s) ds. 0 Now Z j is again a one-dimensional √ Ornstein-Uhlenbeck process, but with the Brownian motion replaced by 2 times a Brownian motion. Using (3.12) the probability that Zt does √ not hit 0 before time t0 /2 is less than or equal to the probability that 2 times a Brownian motion does not hit 0 before time t0 /2. This latter probability is less than or equal to √ √ −1/4 c|x′j − yj′ |/ t0 /2 6 2caj / t0 /2, which is summable in j. Let Bj be the event (TCj > t0 /2). We can therefore conclude that if ′ ′ −1/4 |x′j |, |yj′ | 6 aj , then Pxj ,yj (Bj ) is summable in j. Now use the Markov property at time t0 /2 on the event (AjX )c ∩ (AjY )c to obtain Pxj ,yj (TCj > t0 , (AjX )c ∩ (AjY )c ) [ ] = Exj ,yj PXj (t0 /2),Yj (t0 /2) (TCj > t0 /2); (AjX )c ∩ (AjY )c ) ( ′ ′ 6 sup Pxj ,yj (TCj > t0 /2) Pxj ,yj ((AjX )c ∩ (AjY )c ). −1/4 |x′j |,|yj′ |6aj Therefore Pxj ,yj (TCj > t0 , (AjX )c ∩ (AjY )c ) is summable in j. Since we already know that Pxj ,yj (AjX ) and Pxj ,yj (AjY ) are summable in j, we conclude that (3.11) holds. Now choose j0 such that ∞ ∑ j j Px ,y (TCj ≥ t0 ) < 1/4. j=j0 +1 Choose ε such that (1+ε)j0 6 5/4. We will show that there exists a constant c1 such that for each j 6 j0 we have (3.13) Px j ,y j (TCj < τX (1 + ε) ∧ τY (1 + ε)) ≥ c1 . We know that with probability at least 1/2, for each j > j0 each pair j (X j (t), Y (t)) couples before (X, Y ) exits B(0, 5/4). Once we have (3.13), j we know that with probability at least c1 , the pair (X j (t), Y (t)) couples before exiting [−1 − ε, 1 + ε] for j 6 j0 . Hence, using independence, with HARNACK INEQUALITIES IN INFINITE DIMENSIONS 17 j probability at least cj10 we have that for all j 6 j0 , each pair (X j (t), Y (t)) couples before either X j (t) or Y j (t) exits the interval [−1 − ε, 1 + ε]. Using the independence again, we have coupling with√probability at least cj10 /2 of X and Y before either exits the ball of radius 2(5/4) < 2. To show (3.13), on the interval [−1 − ε, 1 + ε], the drift term of the Ornstein-Uhlenbeck process is bounded, so by using the Girsanov theorem, it suﬃces to show with positive probability WjX hits WjY before either exits [−1 − ε, 1 + ε]. The pair (WjX (t), WjY (t)) is a two-dimensional Brownian motion started inside the square [−1, 1]2 and we want to show that it hits the diagonal {y = x} before exiting the square [−1 − ε, 1 + ε]2 with positive probability. This follows from the support theorem for Brownian motion. See, e.g., [5, Theorem I.6.6]. 4. Operators in Hörmander form We let Cb (H) denote the set of bounded continuous functions on H with the supremum norm and Cbn (H) the space of n times continuously Fréchet diﬀerentiable functions with all derivatives up to order n being bounded. Cb0,1 (H) will be the space of all Lipschitz continuous functions with ∥f ∥0,1 := sup |f (x)| + sup x x̸=y |f (x) − f (y)| . |x − y| Finally, Cb1,1 (H) will be the space of Fréchet diﬀerentiable functions f wih continuous and bounded derivatives such that Df is Lipschitz continuous; we use the norm ∥f ∥1,1 = ∥f ∥0,1 + sup x̸=y |Df (x) − Df (y)|H ∗ . |x − y| Suppose H is a separable Hilbert space, and {en }∞ n=1 is an orthonormal basis in H. We set (∂j f )(x) := (Dej f )(x). 4.1. Stochastic diﬀerential equation. Let m ≥ 1 and suppose A1 , . . . , Am are bounded maps from H to H. Let A := (A1 , . . . Am ). We assume that (4.1) aki (x) := ⟨Ak (x) , ei ⟩ > 0 for any x ∈ H, and that we have ai ∈ Cb1,1 (H) with (4.2) ∥Ak ∥21,1 := ∞ ∑ i=1 ∥aki ∥21,1 < ∞. 18 BASS AND GORDINA For any f ∈ Cb1 (H) we deﬁne ∞ ∑ (∇Ak f ) (x) := aki (x) (∂i f ) (x) , i=1 (∇A f ) (x) := ((∇A1 f ) (x) , ..., (∇Am f ) (x)) . Note that ( | (∇Ak f ) (x) |2 6 6 ∞ ∑ )( |aki (x) |2 ∞ ∑ ) | (∂i f ) (x) |2 i=1 i=1 k 2 2 ∥A |1,1 | (Df ) (x) | , so ∇Ak f and ∇A f are well-deﬁned for f ∈ Cb1 (H). Fix a probability space (Ω, F, P) with a ﬁltration Ft , t > 0, satisfying the usual conditions, that is, F0 contains all( null sets in )F, and Ft = Ft+ = ∩ 1 m s>t Fs for all t ∈ [0, T ]. Suppose Wt = ( 1Wt , ...,mW ) t is a Wiener process m on H with covariance operator Q = Q , ..., Q . We assume that each Qk , k = 1, ..., m is a non-negative trace-class operator on H such that ∞ ∑ Qk ei = λki ei , with λki > 0 and λki = 2, k = 1, ..., m. i=1 We consider a stochastic diﬀerential such that the inﬁnitesimal ∑m equation 2 (∇ generator of the solution is L = ) . k A k=1) ( Deﬁne B (x) := B 1 (x) , ..., B m (x) , x ∈ H as a linear operator from H to H m by ⟨B k (x) h, ei ⟩ := aki (x) , for any h ∈ H, k = 1, ..., m, and F : H → H m by ⟨F k (x) , ei ⟩ := ∞ ∑ akj (x) ∂j aki (x) , k = 1, ..., m. j=1 We can also re-write B and F as B (x) (h1 , ..., hm ) = A (x) , for any (h1 , ..., hm ) ∈ H m , ) (∞ ∞ ∑ ∑ 1 m F (x) = ∇A1 ai (x) ei , ..., ∇Am ai (x) ei . i=1 i=1 Theorem 4.1. (1) Suppose X0 is an H m -valued random variable. Then the stochastic diﬀerential equation ∫ t ∫ t T Xt = X0 + B (Xs ) dWs + F (Xs ) ds, 0 0 has a unique solution (up to a.s. equivalence) among the processes satisfying (∫ T ) P |Xt |2H m dt < ∞ = 1. 0 HARNACK INEQUALITIES IN INFINITE DIMENSIONS 19 (2) If in addition X0 ∈ L2 (Ω, F0 , P), then there is a constant CT > 0 such that E|Xt |2 6 CT E|X0 |2 . (3) Suppose f ∈ Cb2 (H). Then v (t, x) := E (f (Xtx )) = Pt f (x) is in Cb1,2 (H) and is the unique solution to the following parabolic equation t > 0, x ∈ H m , ∂t v (t, x) = Lv, v (0, x) = f (x) , where L is the operator (Lf ) (x) := = m ∑ (∇Ak ∇Ak f ) (x) k=1 ∞ m ∑ ∑ ( akj (x) ∂j ) aki (x) ∂i f (x) i=1 k=1 j=1 = ∞ ∑ ∞ m ∑ ∑ 2 aki akj ∂ij f (x) + k=1 i,j=1 m ∑ ∞ ∑ akj (x) ∂j aki (x) ∂i f (x) , x ∈ H. k=1 i,j=1 Proof. For simplicity of notation we take m = 1, and write A1 for A with corresponding functions aj . The proof for the general case is very similar. In this case B (x) , x ∈ H, is a linear operator on H deﬁned by ⟨B (x) h, ei ⟩ := ai (x) , for any h ∈ H, and F : H → H by ⟨F (x) , ei ⟩ := ∞ ∑ aj (x) ∂j ai (x) , j=1 ∑ or equivalently B (x) ej = A (x) , F (x) = ∞ i,j aj (x) ∂j ai (x) ei . According to [12, Theorem 7.4], for this stochastic diﬀerential equation to have a unique mild solution it is enough to check that (a) B (x) (·) is a measurable map from H to the space L02 of Hilbert-Schmidt operators from Q1/2 H to H; (b) ∥B (x) − B (y) ∥L02 6 C|x − y|, x, y ∈ H; ( ) (c) ∥B (x) ∥2L0 6 K 1 + |x|2 , x ∈ H; 2 ( ) (d) F is Lipschitz continuous on H and |F (x) | 6 L 1 + |x|2 , x ∈ H. 20 BASS AND GORDINA ∞ Let {ej }∞ j=1 be an orthonormal basis of H. Then {λj ej }j=1 is an orthonormal basis of Q1/2 H. First observe that since A is bounded we have 1/2 ∥B (x) ∥2L0 2 ∞ ∑ = 1/2 |⟨B (x) λj ej , ei ⟩|2 , i,j=1 |A (x) |2 ∞ ∑ λj = 2|A (x) |2 6 C, j=1 and similarly ∥B (x) − B (y) ∥L02 6 ∥A∥1,1 |x − y|. The last estimate implies ∥B (x) ∥L02 6 max{C, |B (0) |} (1 + |x|) which proves (a) and (c). We also have |F (x) − F (y) | = 2 ∞ ∑ i=1 = ∞ ∑ ⟨F (x) − F (y) , ei ⟩2 2 ∞ ∑ aj (x) ∂j ai (x) − aj (y) ∂j ai (y) i=1 62 j=1 ∞ ∑ i=1 +2 ∞ ∑ 2 (aj (x) − aj (y)) ∂j ai (x) j=1 ∞ ∑ 2 ∞ ∑ aj (y) (∂j ai (x) − ∂j ai (y)) i=1 j=1 ∞ ∞ ∑ ∑ 6 2 (aj (x) − aj (y))2 (∂j ai (x))2 j=1 + 2 ∞ ∑ j=1 (aj (y))2 i,j=1 ∞ ∑ i,j=1 (∂j ai (x) − ∂j ai (y))2 . HARNACK INEQUALITIES IN INFINITE DIMENSIONS 21 Now we can use our assumptions on A to see that ∞ ∑ 2 (aj (x) − aj (y)) 6 j=1 ∞ ∑ ∥ai ∥21,1 |x − y|2 = ∥A∥21,1 |x − y|2 , j=1 ∞ ∑ |aj (y) |2 6 ∥A∥21,1 , j=1 ∞ ∑ i,j=1 ∞ ∑ |∂j ai (x) | = 2 ∞ ∑ |Dai (x) |2 6 ∥A∥21,1 , and i=1 ∞ ∑ (∂j ai (x) − ∂j ai (y))2 = i,j=1 |Dai (x) − Dai (y) |2 i=1 6 ∞ ∑ ∥ai ∥21,1 |x − y|2 6 ∥A∥21,1 |x − y|2 , i=1 which gives Lipschitz continuity for F . Finally the estimate for |F (x) | follows from the Lipschitz continuity of F together with boundedness of A in a similar fashion to what we did for B. Assertion (2) follows directly from [12, Theorem 9.1]. Assertion (3) follows from [12, Theorem 9.16] which says that Pt f is the solution to the parabolic type equation with operator ( ) 1 tr vxx B (x) Q1/2 , B (x) Q1/2 + ⟨vx , F (x)⟩ 2 ∞ ∞ ( ) ⟨ ∑ ⟩ 1∑ 1/2 1/2 = vxx B (x) Q en , B (x) Q en + vx , aj (x) ∂j ai (x) ei 2 n=1 i,j ∞ ∞ ∞ ∞ ∑ ∑ ∑ 1∑ = λn vxx ai (x) ei , aj (x) ej + aj (x) ∂j ai (x) ⟨vx , ei ⟩ 2 Lv = n=1 = = ∞ ∑ i,j=1 ∞ ∑ i=1 j=1 ai (x) aj (x) vxx (ei , ej ) + i,j ∞ ∑ aj (x) ∂j ai (x) ⟨vx , ei ⟩ i,j 2 ai (x) aj (x) ∂ij v+ i,j=1 ∞ ∑ aj (x) ∂j ai (x) ∂i v. i,j Remark 4.2. Denote Lk f := ∇2Ak f = ∞ ∑ i,j=1 2 aki (x) akj (x) ∂ij f+ ∞ ∑ i,j akj (x) ∂j aki (x) ∂i f, 22 BASS AND GORDINA where k = 1, ..., m. Suppose f ∈ Cb2 (H). Then ∞ ∞ ( 2 ) ∑ ∑ k k k 2 2 L f (x) 6 |a a (x) | |∂ij f (x) |2 i j i,j=1 + i,j=1 ∞ ∑ ∞ 2 ∞ ∑ ∑ k (x) | ∂j ai (x) ∂i f (x) |akj 2 j=1 6 ∥Ak ∥41,1 ∥f ∥22 + j=1 i=1 ∞ ∑ k ∥Ak ∥21,1 ∂j ai i,j=1 ∞ 2 ∑ (x) |∂i f (x)|2 6 2∥Ak ∥41,1 ∥f ∥22 , and therefore Lk is well-deﬁned on Cb2 (H), and so is L = i=1 ∑m k=1 Lk . 4.2. Curvature-dimension inequality. We can write L= m ∑ Lk = k=1 For any f, g ∈ Cb2 (H) m ∑ ∇2Ak . k=1 we deﬁne 1 (L (f g) − f L (g) − gL (f )) , 2 1 Γ2 (f ) := L (Γ (f, f )) − Γ (f, Lf ) . 2 (4.3) Γ (f, g) := (4.4) Theorem 4.3. For any f, g ∈ Cb2 (H), (4.5) Γ (f, g) = Γ2 (f ) = (4.6) m ∑ k=1 m ∑ (∇Ak f ) (∇Ak g) , Γ(k) (∇Al f ) , k,l=1 where Γ(k) (f ) := (∇Ak f )2 . Proof. Note that for functions f, g ∈ Cb2 (H) ( (4.7) Lk (f g) = f Lk (g) + gLk (f ) + 2 ∑ ) ∑ aki ∂i f akj ∂j g i j = f Lk (g) + gLk (f ) + 2 (∇Ak f ) (∇Ak g) , and therefore (4.8) L (f g) = f L (g) + gL (f ) + 2 m ∑ k=1 (∇Ak f ) (∇Ak g) . HARNACK INEQUALITIES IN INFINITE DIMENSIONS 23 Hence ∑ 1 (L (f g) − f L (g) − gL (f )) = (∇Ak f ) (∇Ak g) , 2 k=1 ∑m and in particular Γ (f ) := Γ (f, f ) = k=1 (∇Ak f )2 . Before we ﬁnd Γ2 (f ) we need the following calculation. ) ∑( ∑ 3 2 akj akm ∂ijm f [Lk , ∂i ] := (Lk ∂i − ∂i Lk ) f = akj ∂j am ∂im f+ m Γ (f, g) = jm jm ∑ ∑ 2 akj ∂j akm ∂m f + − ∂i akj akm ∂jm f (4.9) jm jm ) ) ∑( ∑( 2 k 2 =− ∂i akj ∂j akm + akj ∂ij am ∂m f − 2 akm ∂i akj ∂jm f. jm jm Use (4.9) to see that ∑ ali ([Lk , ∂i ]f ) i =− ∑ ∑( m ) 2 l ali ∂i alj ∂j alm + ali alj ∂ij am ∂ m f ij ) ∑( 2 −2 ali alm ∂i alj ∂jm f ijm ) ) ∑( ∑( 2 =− Lk alm ∂m f − 2 ali alm ∂i alj ∂jm f m ijm m j m j ( )( ) ) ∑( ∑ ∑ ∑ l l l l 2 =− Lk am ∂m f − 2 ai ∂i aj am ∂mj f (4.10) m i ) ) ∑( ∑( =− Lk alm ∂m f − 2 ∇Al alj (∇Al ∂j f ) . Now we can deal with Γ2 (f ). We use (4.8) in the ﬁrst line. ) (m m m ∑ 1 1∑ 1∑ (∇Al f )2 L (Γ (f )) = Lk (Γ (f )) = Lk 2 2 2 = k=1 m ∑( k=1 l=1 ) (∇Al f ) (Lk ∇Al f ) + Γ(k) (∇Al f ) . k,l=1 The second term in Γ2 (f ) is Γ (f, Lf ) = m ∑ l=1 (∇Al f ) (∇Al Lf ) = m ∑ k,l=1 (∇Al f ) (∇Al Lk f ) . 24 BASS AND GORDINA Thus Γ2 (f ) = m ∑ m ∑ (∇Al f ) ([Lk , ∇Al ]f ) + k,l=1 Γk (∇Al f ) . k,l=1 By (4.7) we have ∞ ∞ ∑ ∑ l [Lk , ∇Al ]f = Lk aj ∂j f − alj ∂j Lk f j=1 = ∞ ∑ j=1 ( ) Lk alj ∂j f + j=1 ∞ ∑ j=1 − ∞ ∑ alj Lk ∂j f +2 ∞ ( ∑ ) ∇Ak alj (∇Ak ∂j f ) j=1 alj ∂j Lk f j=1 = ∞ ∑ ∞ ∞ ( ) ( ) ∑ ∑ l l Lk aj ∂j f + aj [Lk , ∂j ]f + 2 ∇Ak alj (∇Ak ∂j f ) . j=1 j=1 j=1 We can use (4.10) to see that [Lk , ∇Al ]f = 0 for k, l = 1, ..., m. Thus (4.6) holds. Corollary 4.4. L satisﬁes the curvature-dimension inequality CD (0, m) 1 (Lf )2 . (4.11) Γ2 (f ) > m Moreover, for m = 1 we have Γ2 (f ) = (Lf )2 . Proof. Note that by the Cauchy-Schwarz inequality (m )2 m m ∑ ∑ 1 1 ∑ 2 2 Γk (∇Al f ) = (∇Ak ∇Al f ) > ∇Ak f = (Lf )2 . m m k,l=1 k,l=1 Therefore Γ2 (f ) > m ∑ k=1 (∇Al f ) ([Lk , ∇Al ]f ) + k,l=1 1 (Lf )2 . m We need chain rules for the operators Γ and Γ2 . Proposition 4.5. Let Ψ be a C ∞ function on R and suppose f is in the domain of L. Then (4.12) LΨ (f ) = Ψ′ (f ) Lf + Ψ′′ (f ) Γ (f, f ) , (4.13) Γ (Ψ (f ) , g) = Ψ′ (f ) Γ (f, g) , (4.14) ( )2 ( )2 Γ2 (Ψ (f )) = Ψ′′ (f ) (Γ (f ))2 + Ψ′ (f ) Γ2 (f ) + Ψ′ (f ) Ψ′′ (f ) Γ (f, Γ (f )) . HARNACK INEQUALITIES IN INFINITE DIMENSIONS Proof. Suppose Ψ ∈ C ∞ (R). Recall that we can write L as Lf = ∑∞ k ∑m 2 k=1 ∇Ak f, where ∇Ak f := i=1 ai ∂i f. It is clear that ∑m k=1 Lk = ∇Ak (Ψ (f )) = Ψ′ (f ) ∇Ak f. (4.15) Then 25 ( ) ∇Ak ∇Ak (Ψ (f )) = ∇Ak Ψ′ (f ) ∇Ak f + Ψ′ (f ) ∇Ak (∇Ak f ) = Ψ′′ (f ) (∇Ak f )2 + Ψ′ (f ) ∇Ak (∇Ak f ) = Ψ′ (f ) Lk f + Ψ′′ (f ) Γk (f ) , which implies (4.12) by Theorem 4.3. Now we can easily show (4.13). Indeed, using (4.15) we have Γk (Ψ (f ) , g) = (∇Ak Ψ (f )) (∇Ak g) = Ψ′ (f ) (∇Ak f ) (∇Ak g) = Ψ′ (f ) Γk (f, g) . In particular, (4.13) implies ( )2 Γ (Ψ (f )) = Ψ′ (f ) Γ (f ) . Now we would like to prove (4.14). First, using (4.13) twice we see that ( )2 (4.16) Γ (Ψ (f )) = Ψ′ (f ) Γ (f ) . By (4.8) and (4.12) 1 LΓ (Ψ (f )) 2 (( (( ) )2 ) 1 ( ′ )2 )2 1 + Ψ (f ) LΓ (f ) + Γ Ψ′ (f ) , Γ (f ) = Γ (f ) L Ψ′ (f ) 2 2 ( ) ( )2 ′ ′′ = Ψ (f ) Ψ (f ) (Lf ) Γ (f ) + Ψ′′ (f ) + Ψ′ (f ) Ψ′′′ (f ) (Γ (f ))2 )2 1( ′ Ψ (f ) LΓ (f ) + 2Ψ′ (f ) Ψ′′ (f ) Γ (f, Γ (f )) . 2 Now use (4.8) and (4.14) repeatedly to obtain ( ) ( ) Γ (Ψ (f ) , LΨ (f )) = Γ Ψ (f ) , Ψ′ (f ) Lf + Γ Ψ (f ) , Ψ′′ (f ) Γ (f ) ( )2 = Ψ′ (f ) Γ (f, Lf ) + Ψ′ (f ) Ψ′′ (f ) (Lf ) Γ (f ) + + Ψ′ (f ) Ψ′′ (f ) Γ (f, Γ (f )) + Ψ′ (f ) Ψ′′′ (f ) (Γ (f ))2 . Note that we also used the fact that Γ (f, gh) = gΓ (f, h) + hΓ (f, h) . Combining these two calculations gives (4.14). Corollary 4.6. By (4.14) with Ψ (x) = log x, x > 0, and g > 0 we see that (4.17) (Γ (g))2 Γ (g, Γ (g)) Γ2 (g) Γ2 (log g) = − + . g4 g3 g2 26 BASS AND GORDINA 4.3. Li-Yau estimate. The following is the Li-Yau estimate in our context. In this proof we follow an argument in [2], which they used to prove a ﬁnitedimensional logarithmic Sobolev inequality for heat kernel measures. Theorem 4.7. L (log Pt f ) > − (4.18) 1 . 2t Proof. By (4.13) with Ψ (x) = log x, x > 0, f > 0, and 0 6 s 6 t, Γ (Pt−s f ) := Γ (Pt−s f, Pt−s f ) = (Pt−s f )2 Γ (log Pt−s f ) Deﬁne for f > 0 ( φ (s) := Ps (Pt−s f Γ (log Pt−s f )) = Ps Γ (Pt−s f ) Pt−s f ) . Then with g := Pt−s f and ∂s g = −Lg we see that by (4.12) and (4.13) ( ( Γ(g) )) φ′ (s) = ∂s Ps g ( ( Γ(g) ) 2Γ(g, Lg) Γ(g)Lg ) = Ps L − + g g g2 (( (1) ( 1 ) 2Γ(g, Lg) Γ(g)Lg ) = Ps LΓ(g)g + Γ(g)L + 2Γ Γ(g), − + g g g g2 ( ( 2Γ(g) Lg ) 2Γ(Γ(g), g) LΓ(g) − 2Γ(g, Lg) Γ(g)Lg ) = Ps Γ(g) − 2 − + + g3 g g2 g g2 ) ( (Γ(g))2 Γ(g, Γ(g) ( ) Γ2 (g) ) = 2Ps − + = 2P gΓ (log g) s 2 g3 g2 g by (4.17). We use the curvature-dimension inequality (4.11) to obtain ( ) 2 (4.19) φ′ (s) > Ps g (L log g)2 . m In particular, this means that φ is non-decreasing, and therefore φ (0) = Pt f Γ (log Pt f ) 6 Pt (f Γ (log f )) = φ (t) . Using the chain rule (4.13) we get Γ (Pt f ) 6 Pt Pt f Γ (log Pt f ) = Pt f ( Γ (f ) f ) This inequality together with (4.12) gives Γ (Pt f ) Pt f L (log Pt f ) = LPt f − > LPt f − Pt Pt f = Pt (f Γ (log f )) . ( Γ (f ) f Thus (4.20) Pt f L (log Pt f ) > Pt (f L (log f )) . ) = Pt (f L (log f )) . HARNACK INEQUALITIES IN INFINITE DIMENSIONS 27 We need more information about φ to complete the proof. Our expression for φ′ can be rewritten using the chain rule (4.12) as ( ( )2 ) ( ) 1 Γ (g) Lg − φ′ (s) = Ps g (L log g)2 = Ps . g g Note that since g > 0 we have ) ( ( ))) ( ( Γ (g) Γ (g) 1 √ = Ps Lg − Ps Lg − g √ g g g ( ( ( )2 ))1/2 1 Γ (g) 6 (Ps g)1/2 Ps Lg − , g g so ( ( ))2 ( ( Γ(g) )2 ) P Lg − s g 1 Γ (g) Lg − > Ps g g Ps g ( ) Since φ (s) = Ps Γ(g) , the last estimate becomes g φ′ (s) > 2 (Ps Lg − φ (s))2 . Ps g Now use the deﬁnition of g and the fact that L and Ps commute to see that Ps g = Pt f , so we have that for 0 6 s 6 t φ′ (s) > 2 (LPt f − φ (s))2 (φ (s) − LPt f )2 =2 . Pt f Pt f Thus for all s such that φ′ (s) > 0 we have ) ( 2 1 > > 0. −∂s φ (s) − LPt f Pt f By (4.19) we know that φ′ (s) > 0, and by integrating this estimate from 0 to t, we obtain 1 1 2t − > . φ (0) − LPt f φ (t) − LPt f Pt f That is, φ (t) − φ (0) 2t > > 0. (φ (0) − LPt f ) (φ (t) − LPt f ) Pt f Since φ is non-decreasing, the numerator on the left is non-negative. Since the right hand side of the estimate is positive, no matter what the sign of the denominator on the left, the following estimate holds: φ (t) − φ (0) > 2t (φ (0) − LPt f ) (φ (t) − LPt f ) . Pt f 28 BASS AND GORDINA Similarly to the proof of (4.20) Γ (Pt f ) − LPt f = −Pt f L (log Pt f ) , Pt f ( ) Γ (f ) φ (t) − LPt f = Pt − LPt f = −Pt (f L (log f )) . f φ (0) − LPt f = Finally we have (4.21) Pt f L (log Pt f ) > Pt (f L (log f )) (1 + 2tL (log Pt f )) . Now we are ready to prove (4.18). We only need to check (4.18) when L (log Pt f ) < 0. In this case, by (4.20) Pt (f L (log f )) < 0, and therefore (4.21) implies 1 + 2tL (log Pt f ) > 0. Corollary 4.8. For f > 0 −∂t (log Pt f ) < 1 − Γ (log Pt f ) . 2t Proof. By (4.12) and (4.16) LPt f Γ (Pt f ) − Pt f (Pt f )2 1 ∂t Pt f − Γ (log Pt f ) = ∂t (log Pt f ) − Γ (log Pt f ) > − . = Pt f 2t L (log Pt f ) = 4.4. Distances. For the purposes of the next subsection we need to introduce several distances related to the gradient ∇A . A natural distance as described in [1] is: d (x, y) := sup {f :Γ(f )61} (f (y) − f (x)) , x, y ∈ H. We will need another distance which is better suited for the proof of the parabolic Harnack inequality, and it will turn out that this distance is equal to the one we have just deﬁned. First we note that for any x ∈ H there is a smooth path γA : [0, ∞) → H m (possibly deﬁned only on a ﬁnite subinterval [0, T ] of R+ ) such that (4.22) γ˙A (t) = A (γA (t)) , γA (0) = x. This is equivalent to solving a system of ordinary diﬀerential equations, which gives γA implicitly as the solution to ∫ dγj xj + = t. aj (γ) HARNACK INEQUALITIES IN INFINITE DIMENSIONS 29 Using the assumption that aj > 0, we can determine γA as a function of t. An admissible component of x is deﬁned as VA (x) := {γA (s) , where s ∈ [0, T ], γ˙A (s) = A (γA (s)) , γA (0) = x} as described by (4.22). Example 4.9. Suppose aj (x) = cj . Then γ is a straight line, and so VA is a straight line through x in the direction of (c1 , c2 , ....). In particular, if H = R2 , and a1 (x) = 1 and a2 (x) = 0, then VA is a horizontal line through x. Deﬁnition 4.10. Let x ∈ H, and deﬁne y ∈ VA (x) ; Ty , darc (x, y) := +∞, y ∈ / VA (x) , where the path γA is described by (4.22) with γA (Ty ) = y. Remark 4.11. Note that our assumptions on A are essential for the deﬁnition of the distance function darc as we use the ordinary diﬀerential equations (4.22) to ﬁnd γA . Theorem 4.12. For any x, y ∈ H d (x, y) = darc (x, y) . Proof. Fix x ∈ H. We will consider the case when darc (x, y) = ∞ or d (x, y) = ∞ later, so for now we assume that both distances are ﬁnite. Let γ be any path connecting x and y with γ (s) = y. Note that since darc (x, y) < ∞, we have y ∈ VA (x). Then ∫s (4.23) d (x, y) = sup {f :Γ(f )61} (f (y) − f (x)) = Choosing fA such that ∇fA = A , |A|2 ⟨∇f |γ(t) , γ̇ (t)⟩ dt sup {f :Γ(f )61} 0 then Γ (fA ) = |∇A fA |2 = ⟨∇fA , A⟩2 = 1, and therefore for the function fA ∫Ty ∫Ty d (x, y) > fA (y) − fA (x) = ⟨∇fA , γ˙A (t)⟩ dt = 1 dt = Ty = darc (x, y) . 0 0 30 BASS AND GORDINA Again, by (4.23), d (x, y) = sup {f :Γ(f )61} (4.24) = 0 ∫Ty ⟨∇f |γA (t) , γA (γ (t))⟩ dt sup {f :Γ(f )61} = ∫Ty ⟨∇f |γA (t) , γ˙A (t)⟩ dt 0 ∫Ty ∇A f |γA (t) dt 6 sup {f :Γ(f )61} ∫Ty 0 1 dt = darc (x, y) . 0 Finally we want to show that both distances are inﬁnite for the same y. Deﬁne a function 0, z ∈ VA (x) ; fN (z) := N, z ∈ / VA (x) for some N . Note that Γ (fN ) = 0. Suppose darc (x, y) = ∞, so fN (y) = N . Then d (x, y) > fN (y) − fN (x) = N. By taking N → ∞ we see that d (x, y) = +∞. Next suppose that d (x, y) = ∞. Then there are functions fN with Γ (fN ) 6 1 such that fN (y) − fN (x) → +∞ as N → ∞. Similarly to (4.24) (if we assume that darc (x, y) < ∞ to ﬁnd γA ) we see that +∞ = lim fN (y) − fN (x) 6 Ty = darc (x, y) , N →∞ and therefore darc (x, y) = +∞. 4.5. The parabolic Harnack inequality. Theorem 4.13. Suppose u is a positive solution to the heat equation ∂t u = Lu, u(0, ·) = f. Then for any 0 6 t1 < t2 6 1 and x, y in the same admissible component, say, VA (x), we have log u (t1 , x) − log u (t2 , y) 6 Tx2 1 t2 + log , 4 (t2 − t1 ) 2 t1 where Tx is deﬁned in Deﬁnition 4.10. Proof. The proof is standard. Let u (t, x) := Pt f (x) for a positive function f ∈ Cb2 (H). Then by Theorem 4.1, u is the solution to the heat equation ∂t g = Lg, g(0, ·) = f. Denote g (t, x) := log u (t, x). Let t2 > t1 > 0, x, y ∈ H. Since y ∈ VA (x), we can ﬁnd a smooth path γA : [0, Ty ] → H m such that γ (0) = y, γ (Tx ) = x, and γ̇ (t) = A (γ (t)). Deﬁne σ : [0, Tx ] → [t1 , t2 ] × H m by HARNACK INEQUALITIES IN INFINITE DIMENSIONS ( σ (s) := t2 − Then t2 −t1 Tx s, γ 31 ) (s) . Note that σ (0) = (t2 , y) and σ (Tx ) = (t1 , x). g (t1 , x) − g (t2 , y) = g (σ (0)) − g (σ (Tx )) ∫ Tx d = g (σ (s)) ds ds 0 ) ) ( ∫ Tx ( t2 − t1 ∂t g (σ (s)) ds = ⟨∇g, γ˙A ⟩ − Tx 0 ∫ Tx ∫ Tx t2 − t1 6 Γ (g) ∇A f |γA (s) ds − Tx 0 0 ∫ 1 Tx (t2 − t1 ) + ds 2 0 Tx t2 − (t2 − t1 ) s by Corollary 4.8. Note that Γ (g) = |∇A g|2 , so ∇A f − t2 − t1 Tx Γ (g) 6 , Tx 4 (t2 − t1 ) where we used the elementary estimate ax − bx2 6 a2 /4b for b > 0 with x = ∇A g. Finally, we have g (t1 , x) − g (t2 , y) 6 1 t2 Tx2 + log . 4 (t2 − t1 ) 2 t1 References 1. Dominique Bakry, Functional inequalities for Markov semigroups, Probability measures on groups: recent directions and trends, Tata Inst. Fund. Res., Mumbai, 2006, pp. 91–147. 2. Dominique Bakry and Michel Ledoux, A logarithmic Sobolev form of the Li-Yau parabolic inequality, Rev. Mat. Iberoam. 22 (2006), no. 2, 683–702. 3. Dominique Bakry and Zhongmin M. Qian, Harnack inequalities on a manifold with positive or negative Ricci curvature, Rev. Mat. Iberoamericana 15 (1999), no. 1, 143– 179. 4. Martin T. Barlow and Richard F. Bass, Brownian motion and harmonic analysis on Sierpinski carpets, Canad. J. Math. 51 (1999), no. 4, 673–744. 5. Richard F. Bass, Probabilistic techniques in analysis, Probability and its Applications (New York), Springer-Verlag, New York, 1995. 6. , Diﬀusions and elliptic operators, Probability and its Applications (New York), Springer-Verlag, New York, 1998. 7. A. Bendikov and L. Saloﬀ-Coste, On- and oﬀ-diagonal heat kernel behaviors on certain inﬁnite dimensional local Dirichlet spaces, Amer. J. Math. 122 (2000), no. 6, 1205– 1263. 8. Christian Berg, Potential theory on the inﬁnite dimensional torus, Invent. Math. 32 (1976), no. 1, 49–100. 9. Marco Biroli and Umberto Mosco, Formes de Dirichlet et estimations structurelles dans les milieux discontinus, C. R. Acad. Sci. Paris Sér. I Math. 313 (1991), no. 9, 593–598. 32 BASS AND GORDINA 10. Vladimir I. Bogachev, Gaussian measures, Mathematical Surveys and Monographs, vol. 62, American Mathematical Society, Providence, RI, 1998. 11. René Carmona, Inﬁnite-dimensional Newtonian potentials, Probability theory on vector spaces, II (Proc. Second Internat. Conf., Blażejewko, 1979), Lecture Notes in Math., vol. 828, Springer, Berlin, 1980, pp. 30–43. 12. Giuseppe Da Prato and Jerzy Zabczyk, Stochastic equations in inﬁnite dimensions, Encyclopedia of Mathematics and its Applications, vol. 44, Cambridge University Press, Cambridge, 1992. 13. , Second order partial diﬀerential equations in Hilbert spaces, London Mathematical Society Lecture Note Series, vol. 293, Cambridge University Press, Cambridge, 2002. 14. Bruce K. Driver and Maria Gordina, Heat kernel analysis on inﬁnite-dimensional Heisenberg groups, J. Funct. Anal. 255 (2008), no. 9, 2395–2461. 15. Constance M. Elson, An extension of Weyl’s lemma to inﬁnite dimensions, Trans. Amer. Math. Soc. 194 (1974), 301–324. 16. Victor Goodman, Harmonic functions on Hilbert space, J. Functional Analysis 10 (1972), 451–470. 17. , A Liouville theorem for abstract Wiener spaces, Amer. J. Math. 95 (1973), 215–220. 18. S. E. Graversen and G. Peskir, Maximal inequalities for the Ornstein-Uhlenbeck process, Proc. Amer. Math. Soc. 128 (2000), no. 10, 3035–3041. 19. Leonard Gross, Potential theory on Hilbert space, J. Functional Analysis 1 (1967), 123–181. 20. Moritz Kassmann, Harnack inequalities: an introduction, Bound. Value Probl. Art. ID 81415 (2007), 21pp. 21. N. V. Krylov and M. V. Safonov, A property of the solutions of parabolic equations with measurable coeﬃcients, Izv. Akad. Nauk SSSR Ser. Mat. 44 (1980), no. 1, 161–175, 239. 22. Hui Hsiung Kuo, Gaussian measures in Banach spaces, Springer-Verlag, Berlin, 1975, Lecture Notes in Mathematics, Vol. 463. 23. Michel Ledoux, The geometry of Markov diﬀusion generators, Ann. Fac. Sci. Toulouse Math. (6) 9 (2000), no. 2, 305–366, Probability theory. 24. Jürgen Moser, On Harnack’s theorem for elliptic diﬀerential equations, Comm. Pure Appl. Math. 14 (1961), 577–591. 25. M. Ann Piech, Diﬀerential equations on abstract Wiener space, Paciﬁc J. Math. 43 (1972), 465–473. 26. , Regularity of the Green’s operator on abstract Wiener space, J. Diﬀerential Equations 12 (1972), 353–360. 27. L. Saloﬀ-Coste, Parabolic Harnack inequality for divergence-form second-order differential operators, Potential Anal. 4 (1995), no. 4, 429–467, Potential theory and degenerate partial diﬀerential operators (Parma). 28. Ichiro Shigekawa, Stochastic analysis, Translations of Mathematical Monographs, vol. 224, American Mathematical Society, Providence, RI, 2004, Translated from the 1998 Japanese original by the author, Iwanami Series in Modern Mathematics. Department of Mathematics, University of Connecticut, Storrs, CT 06269, U.S.A. E-mail address: r.bass@uconn.edu Department of Mathematics, University of Connecticut, Storrs, CT 06269, U.S.A. E-mail address: maria.gordina@uconn.edu

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