HARNACK INEQUALITIES IN INFINITE DIMENSIONS

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
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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