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TRANSACTIONS OF THEAMERICAN MATHEMATICAL SOCIETYVolume 351, Number 3, March 1999, Pages 1151–1169S 0002-9947(99)02090-5

RIESZ TRANSFORMS FOR 1 p 2

THIERRY COULHON AND XUAN THINH DUONG

Abstract. It has been asked (see R. Strichartz, Analysis of the Laplacian. . . ,J. Funct. Anal. 52 (1983), 48–79) whether one could extend to a reasonableclass of non-compact Riemannian manifolds the Lp boundedness of the Riesztransforms that holds in Rn. Several partial answers have been given since.In the present paper, we give positive results for 1 p 2 under very weakassumptions, namely the doubling volume property and an optimal on-diagonalheat kernel estimate. In particular, we do not make any hypothesis on thespace derivatives of the heat kernel. We also prove that the result cannot holdfor p > 2 under the same assumptions. Finally, we prove a similar result forthe Riesz transforms on arbitrary domains of Rn.

1. Introduction

Let M be a complete Riemannian manifold, d be the geodesic distance on M ,and dµ be the Riemannian measure. Denote by B(x, r) the geodesic ball of centerx 2 M and radius r > 0 and by V (x, r) its Riemannian volume µ(B(x, r)).

One says that M satisfies the doubling volume property if there exists C suchthat

V (x, 2r) C V (x, r), 8x 2 M, r > 0.

Let � be the Laplace-Beltrami operator on M , e

�t� be the associated heatsemigroup, and p

t

(x, y) be the heat kernel on M , i.e. the kernel of e

�t�. Let r bethe Riemannian gradient.

For f 2 C

10 (M), denote by k f k

p

the L

p norm of f with respect to dµ, and byk f k1,1 the quantity sup

�>0 �µ({x; |f(x)| > �}).Our main result is

Theorem 1.1. Let M be a complete Riemannian manifold satisfying the doubling

volume property and such that

p

t

(x, x) C

V (x,

p

t),

for all x 2 M , t > 0 and some C > 0. Then the Riesz transform T = r��1/2is

weak (1, 1) and bounded on L

p

, 1 < p 2. That is, there exists C

p

, 1 p 2,such that, 8 f 2 C

10 (M),

k |rf | k

p

C

p

��1/2f

��p

, 1 < p 2,

Received by the editors October 1, 1996 and, in revised form, March 20, 1997.1991 Mathematics Subject Classification. Primary 42B20, 58G11.

c�1999 American Mathematical Society

1151

License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use

1152 THIERRY COULHON AND XUAN THINH DUONG

and

k |rf | k1,1 C1

��1/2f

��1.

Remarks. -Integration by parts shows that

k |rf | k2 =��1/2

f

��2, 8 f 2 C

10 (M),

therefore T is obviously bounded from L

2 into itself, and the real issue is to provethat T is weak (1, 1); the L

p boundedness, 1 < p < 2, then follows by interpolation.-It is a standard fact (see [3], p. 172 or [5], p. 36) that for a fixed p 2]1, +1[

the inequality

k |rf | k

p

C

��1/2f

��p

, 8 f 2 C

10 (M),

implies by duality��1/2

f

��q

C k |rf | k

q

, 8 f 2 C

10 (M),

where q is the conjugate exponent of p. The converse is not clear.

Let us compare Theorem 1.1 with the existing results about Riesz transforms.If M has non-negative Ricci curvature (in which case the assumptions of Theorem1.1 are satisfied), Bakry proves in [4], using the Littlewood-Paley theory, that theRiesz transforms on M are L

p bounded, 1 < p < +1. Note that he does not gothrough weak (1, 1) type.

Another approach consists in showing that the Riesz transforms are Calderon-Zygmund operators, using estimates of the heat kernel (see [31] for the classicalEuclidean setting, and [7], [22] for the Riemannian setting). At first sight, thisrequires a pointwise estimate of two spatial derivatives of the heat kernel that ishighly nontrivial. Already the estimate of the first derivative uses the gradientestimates of Li and Yau ([23]).

This approach has also been followed in the Lie group setting (which is alsocovered by our method, see the remark below): for 1 p < 2, Salo↵-Coste, in [28],uses a trick that is specific to the group structure to get the estimate on the firstderivative and then the estimate on the second one for free. Alexopoulos gets thecomplete result for 1 p < +1 by more intricate methods ([1]).

Finally, in a general situation, [14] shows that again for 1 p < 2, it is enoughto get an pointwise estimate on the first derivative. Such an estimate follows fromsome strong form of the parabolic Harnack principle, but there are many naturalsituations where it can be false, for example on manifolds that are quasi-isometric toa manifold with non-negative Ricci curvature, or where it is not clear, for exampleon covering manifolds.

However, [19] shows that at least a weighted L

2 estimate of the first spatialderivative of the heat kernel can be derived from the upper estimate of the heatkernel itself without any further assumption. Our contribution here is to point outthat this L

2 estimate is enough to apply the method of [14].

Examples. - The assumptions of Theorem 1.1 are satisfied on manifolds where aparabolic Harnack principle holds (see [29]), for instance manifolds that are quasi-isometric to a manifold with non-negative Ricci curvature or cocompact covering


RIESZ TRANSFORMS FOR 1 p 2 1153

manifolds whose deck transformation group has polynomial growth. The L

p bound-edness of the Riesz transforms is new in these two situations. One can expect it tohold also for p > 2.

- It is easy to construct manifolds that satisfy the assumptions of Theorem 1.1 butwhere the parabolic Harnack principle is false. A typical example is the following,considered in [30], see also [9], §4: take two copies of R2

\ B(0, 1), and glue themsmoothly along the unit circles. For more information, see §4.

- A natural question is whether Theorem 1.1 also holds for p > 2. We will usethe above example to answer negatively in §4. This answer could be expected froman unpublished counterexample of Kenig [21] (using an idea of Meyers) of a self-adjoint second order elliptic operator (with discontinuous coe�cients) on R2 suchthat the associated Riesz transforms are unbounded for p > 2 + ". For a precisestatement, see [2], §5, Lemma 4. For another negative result, see [11].

Remark. The Riemannian structure is simply a convenient setting for this paper,but our method covers more general situations, e.g. manifolds endowed with asecond-order subelliptic operator. Even more generally, we could consider abstractdi↵usion semigroups in the sense of Bakry ([3]). It is proved in [27] that the Bakry-Emery assumption �2 � 0 is enough to get Li-Yau’s gradient estimate, therefore theheat kernel hypothesis of Theorem 1.1. Note that strangely enough, the inequalitygiven by Theorem 1.1 for 1 < p < 2 is the one out of the four cases (p smalleror bigger than 2, domination of the “carre du champ” by the square root of thegenerator or the converse) that Bakry does not get in [3]. Note also that it followsfrom [31] that the conclusion of Theorem 1.1 cannot hold for Markov semigroupswithout the di↵usion assumption. This was observed by Silverstein (see [26]).

The above theorem admits a local version:

Theorem 1.2. Let M be a complete Riemannian manifold satisfying the local dou-

bling volume property

8 r0 > 0, 9C

r0 such that V (x, 2r) C

r0 V (x, r), 8x 2 M, r 2]0, r0[,

and whose volume growth at infinity is at most exponential in the sense that

V (x, ✓r) Ce

c✓

V (x, r), 8x 2 M, ✓ > 1, r 1.

Suppose that

p

t

(x, x) C

0

V (x,

p

t),

for all x 2 M and t 2]0, 1]. Then there exists C

p

, 1 p 2, such that, 8 f 2

C

10 (M),

k |rf | k

p

C

p

✓��1/2f

��p

+ k f k

p

◆, 1 < p 2,

and

k |rf | k1,1 C1

⇣��1/2f

��1

+ k f k1

⌘.

Again, Theorem 1.2 is a generalisation of the result of Bakry ([4], thm. 4.1)that treats the case of Ricci curvature bounded below. As a corollary, one getsthe following generalisation of the result of Lohoue ([24]) about Cartan-Hadamardmanifolds. Let us recall that in their specific situations, [4] and [24] are able totreat also the case p > 2.



Theorem 1.3. Let M be a complete Riemannian manifold such that

V (x, 2r) C V (x, r), 8x 2 M, r 2]0, 1[,

V (x, ✓r) Ce

c✓

V (x, r), 8x 2 M, ✓ > 1, r 1,

and

p

t

(x, x) C

0

V (x,

p

t), 8x 2 M, t 2]0, 1].

Assume further that M has a spectral gap � > 0:

� k f k2 k�f k2 , 8 f 2 C

10 (M).

Then there exists C

p

, 1 < p 2, such that

k |rf | k

p

C

p

��1/2f

��p

, 8 f 2 C

10 (M).

Let us explain here why Theorem 1.3 follows from Theorem 1.2. The spectralgap assumption means that

��e

�t��

2!2 e

��t

.

For 1 p 2, by interpolation��

e

�t��

p!p

e

�2(1� 1p

)�t

,

hence

��1/2 =p

⇡

2

Z +1

0e

�t� dt

p

t

is bounded on L

p if 1 < p 2; therefore��1/2

f

��p

+ k f k

p

C

��1/2f

��p

.

Note that our method to prove Theorem 1.3 also works for non-amenable Liegroups endowed with a family of Hormander vector fields and therefore gives an-other proof of the corresponding result in [25].

2. The tools

2.1. Calderon-Zygmund decomposition. Let (M, d, µ) be a metric measuredspace, B(x, r) be the ball of center x 2 M and radius r, and V (x, r) be its volume.Suppose that M satisfies the doubling volume property, i.e.

V (x, 2r) C V (x, r), 8x 2 M, r > 0.

Then there exists c = c(M) such that, given f 2 L

1(M)\L

2(M) and � > 0, onecan decompose f as

f = g + b = g +X

i

b

i

,

so that(a) |g(x)| c� for almost all x 2 M ;



(b) there exists a sequence of balls B

i

= B(xi

, r

i

) so that the support of each b

i

is contained in B

i

:Z

|b

i

(x)|dµ(x) c�µ(Bi

) andZ

b

i

(x)dµ(x) = 0;

(c)P

i

µ(Bi

) c

�

R|f(x)|dµ(x);

(d) there exists k 2 N⇤ such that each point of M is contained in at most k ballsB

i

.Note that conditions (b) and (c) imply that k b k1

Pi

k b

i

k1 c k f k1. Hencek g k1 (1 + c) k f k1.

For a proof of the existence of the Calderon-Zygmund decomposition in thissetting, see for example [8].

2.2. Upper estimates of the heat kernel and its time derivative. Our mainassumption on M , apart from the doubling volume property, is the heat kernelestimate

p

t

(x, x) C

V (x,

p

t).

A necessary and su�cient geometric condition, in isoperimetric terms, for thisestimate to hold is given in [16], Prop. 5.2.

From this on-diagonal estimate, the corresponding o↵-diagonal estimate auto-matically follows ([19], thm. 1.1):

p

t

(x, y) C

↵qV (x,

p

t)V (y,

p

t)exp

✓�↵

d

2(x, y)t

◆, 8x, y 2 M, t > 0

for any ↵ 2]0, 1/4[.With the doubling volume property, this implies

p

t

(x, y) C

0↵

V (y,

p

t)exp

✓�↵

d

2(x, y)t

◆, 8x, y 2 M, t > 0

for any ↵ 2]0, 1/4[. Indeed B(y,

p

t) ⇢ B(x,

p

t + d(x, y)). Now an obvious con-sequence of the doubling volume property is that there exists D > 0 such thatV (x, ✓r) C✓

D

V (x, r), if ✓ > 1. Therefore

V (y,

p

t) V (x,

p

t + d(x, y)) C(1 +d(x, y)p

t

)D

V (x,

p

t),

and the estimate follows.A similar estimate for the time derivative of the heat kernel also follows from

[13], thm.4 (see also [19], cor. 3.3):

|

@p

t

@t

(x, y)| C

00↵

tV (y,

p

t)exp

✓�↵

d

2(x, y)t

◆, 8x, y 2 M, t > 0.

From now on ↵ 2]0, 1/4[ is fixed. Note that since p

t

(x, y) = p

t

(y, x) one canexchange x and y in the right hand sides of the above estimates.



2.3. Weighted estimates of the space derivative of the heat kernel.

Lemma 2.1. For all � > 0,Z

d(x,y)�t

1/2e

�2�

d

2(x,y)s

dµ(x) C

�

V (y,

p

s)e��t/s

, 8 y 2 M, s, t > 0.

Proof. Note first thatZ

d(x,y)�t

1/2e

�2�

d

2(x,y)s

dµ(x) e

��

t

s

Z

M

e

��

d

2(x,y)s

dµ(x) = e

��

t

s

I.

Now split the region of integration into annular regions i s

1/2 d(x, y) < (i+1) s

1/2,i 2 N, to estimate I. One gets

I

1X

i=0

V (y, (i + 1)s1/2)e��i

2 V (y, s

1/2)1X

i=0

(i + 1)D

e

��i

2,

which proves the lemma.

Applying the upper bound of p

t

(x, y) and Lemma 2.1 with t = 0, one gets

Lemma 2.2. For all � 2]0, 2↵[,Z

M

|p

s

(x, y)|2e�

d

2(x,y)s

dµ(x) C

�

V (y,

p

s), 8 y 2 M, s > 0.

From Lemma 2.2, one deduces, using [18], an estimate of a weighted L

2 norm of|r

x

p

s

(., y)|. Here we need a slightly less sophisticated version of the estimate thanin [18]; it admits a simpler proof, that goes over to a more general setting.

Lemma 2.3. For all � 2]0, 2↵[,Z

M

|r

x

p

s

(x, y)|2e�

d

2(x,y)s

dµ(x) C

�

V (y,

p

s)s

�1, 8 y 2 M, s > 0.

Proof. By integration by parts,

I(s, y) =Z

M

|r

x

p

s

(x, y)|2e�

d

2(x,y)s

dµ(x)

=Z

M

p

s

(x, y)�x

p

s

(x, y)e�

d

2(x,y)s

dµ(x)

�

Z

M

p

s

(x, y)rx

p

s

(x, y)rx

(e�

d

2(x,y)s ) dµ(x)

= I1(s, y) + I2(s, y).

Now

I1(s, y) = �

Z

M

p

s

(x, y)@p

s

@s

(x, y)e�

d

2(x,y)s

dµ(x).

Using the estimate

|

@p

s

@s

(x, y)| C

sV (y,

p

s)exp

✓�↵

d

2(x, y)s

◆,

(see §2), one gets as in Lemma 2.2

|I1(s, y)| C

�

sV (y,

p

s).



Then

I2(s, y) = �

Z

M

p

s

(x, y)rx

p

s

(x, y)2�d(x, y)

s

r

x

d(x, y)e�

d

2(x,y)s

dµ(x);

hence, since |r

x

d(x, y)| 1,

|I2(s, y)|

Z

M

p

s

(x, y)|rx

p

s

(x, y)|2�d(x, y)

s

e

�

d

2(x,y)s

dµ(x)

C

p

s

Z

M

p

s

(x, y)|rx

p

s

(x, y)|e�

0 d

2(x,y)s

dµ(x)

1p

s

✓Z

M

|p

s

(x, y)|2e�

00 d

2(x,y)s

dµ(x)◆1/2

⇥

✓Z

M

|r

x

p

s

(x, y)|2e�

d

2(x,y)s

dµ(x)◆1/2

.

Note that since � < 2↵, �

0 and �

00 may be chosen smaller than ↵. Therefore,according to Lemma 2.2,

I2(s, y) C

✓1

s V (y,

p

s)

◆1/2pI(s, y),

so that

I(s, y) C

�

sV (y,

p

s)+ C

s1

s V (y,

p

s)p

I(s, y).

The lemma follows.

Finally one can state

Lemma 2.4. There exists � > 0 such that

Z

d(x,y)�t

1/2|r

x

p

s

(x, y)| dµ(x) Ce

��t/s

s

�1/2, 8 y 2 M, s, t > 0.

Proof. Choose � < ↵, writeZ

d(x,y)�t

1/2|r

x

p

s

(x, y)| dµ(x)

✓Z

M

|r

x

p

s

(x, y)|2e2�

d

2(x,y)s

dx

◆1/2 Z

d(x,y)�t

1/2e

�2�

d

2(x,y)s

dx

!1/2

and apply Lemmas 2.1 and 2.3.

For more information about the di�culties that arise when one tries to getpointwise estimates on the gradient of the heat kernel, see [18].

3. The proof of the main result

The first part of the argument is taken from [14]. We reproduce it for the sakeof completeness.

Let T = r��1/2. We want to prove that

µ({x; |Tf(x)| > �}) C

k f k1

�

,

for all � > 0, f 2 L

1(M).



Fix f 2 L

1(M) \ L

2(M) and � > 0, and write the C-Z decomposition of f atthe level �. One has

µ({x; |Tf(x)| > �}) µ({x; |Tg(x)| > �/2}) + µ({x; |Tb(x)| > �/2}).

Using the facts that T is bounded on L

2(M) and that |g(x)| c�, we obtain

µ({x; |Tg(x)| > �/2}) C�

�2kTg k

22 C

0�

�2k g k

22

C

00�

�1k g k1 C

000�

�1k f k1 .

Let us estimate now Tb =P

i

Tb

i

. Write

Tb

i

= Te

�t

i

�b

i

+ T (I � e

�t

i

�)bi

,

where t

i

= r

2i

and r

i

is the radius of B

i

.One checks first that

��X

i

e

�t

i

�b

i

��

2

2

C� k f k1 .

The L

2 boundedness of T will then imply as above

µ({x; |TX

i

e

�t

i

�b

i

| > �/2}) C

0

�

k f k1 .

The heat kernel upper bound, the fact that b

i

is supported in B(xi

,

p

t

i

) and prop-erty (c) of the C-Z decomposition yield

|e

�t

i

�b

i

(x)|

Z

M

e

�↵

d

2(x,y)t

i

V (x,

p

t

i

)|b

i

(y)| dµ(y)

C

e

�↵

0 d

2(x,x

i

)t

i

V (x,

p

t

i

)

Z

M

|b

i

(y)| dµ(y)

C

0 e�↵

0 d

2(x,x

i

)t

i

V (x,

p

t

i

)�µ(B

i

)

C

00�

Z

M

e

�↵

0 d

2(x,y)t

i

V (x,

p

t

i

)1

B

i

(y) dµ(y).

It is therefore enough to show��

X

i

Z

M

e

�↵

0 d

2(.,y)t

i

V (.,p

t

i

)1

B

i

(y) dµ(y)

��2

C

��X

i

1B

i

��2

,

since

�

2

��X

i

1B

i

��

2

2

C�

2X

i

µ(Bi

) C

0� k f k1



because of properties (c) and (d) of the C-Z decomposition. Now

��

X

i

Z

M

e

�↵

0 d

2(.,y)t

i

V (.,p

t

i

)1

B

i

(y) dµ(y)

��2

= supk u k2=1

��

Z

M

0

@X

i

Z

M

e

�↵

0 d

2(x,y)t

i

V (x,

p

t

i

)1

B

i

(y) dµ(y)

1

Au(x)dµ(x)

��

supk u k2=1

Z

M

X

i

0

@Z

M

e

�↵

0 d

2(x,y)t

i

V (x,

p

t

i

)|u(x)|dµ(x)

1

A 1B

i

(y) dµ(y).

Since

V (y,

p

t

i

) (1 +d(x, y)p

t

i

)D

V (x,

p

t

i

)

(see §2.2), one can estimate

Z

M

e

�↵

0 d

2(x,y)t

i

V (x,

p

t

i

)|u(x)|dµ(x)

by

1V (y,

p

t

i

)

Z

M

e

�↵

00 d

2(x,y)t

i

|u(x)|dµ(x)

for ↵

00< ↵

0< ↵. Next

1V (y,

p

t

i

)

Z

M

e

�↵

00 d

2(x,y)t

i

|u(x)|dµ(x)

=1

V (y,

p

t

i

)

Z

d(x,y)p

t

i

e

�↵

00 d

2(x,y)t

i

|u(x)|dµ(x)

+X

k2N

Z

2k

pt

i

d(x,y)2k+1pt

i

e

�↵

00 d

2(x,y)t

i

|u(x)|dµ(x)

!

1V (y,

p

t

i

)

Z

B(y,

pt

i

)|u(x)|dµ(x) +

X

k2Ne

�↵

0022k

Z

B(y,2k+1pt

i

)|u(x)|dµ(x)

!

=

1 +

X

k2N

V (y, 2k+1p

t

i

)V (y,

p

t

i

)e

�↵

0022k

!Mu(y)

1 +

X

k2N2(k+1)D

e

�↵

0022k

!Mu(y),



where Mu(y) = supr>0

1V (y,r)

RB(y,r) |u(x)| dµ(x) is the Hardy-Littlewood maximal

function. Therefore��

X

i

Z

M

e

�↵

0 d

2(.,y)t

i

V (.,p

t

i

)1

B

i

(y) dµ(y)

��2

C supku k2=1

Z

M

Mu(y)X

i

1B

i

(y) dµ(y)

C

0

��X

i

1B

i

��2

,

since the sublinear operator M is bounded on L

2. This ends the estimate of theterm T

Pi

e

�t

i

�b

i

.Consider now the term T

Pi

(I � e

�t

i

�)bi

. Write

µ({x; |TX

i

(I � e

�t

i

�)bi

| > �})

X

i

µ(2B

i

) + µ({x 2 M \

[

i

2B

i

; |TX

i

(I � e

�t

i

�)bi

| > �}).

The first term is controlled by a constant times k f k1/� thanks to the doublingvolume property and property (c) in the C-Z decomposition. The second term isdominated by1�

Z

M\S

i

2B

i

|T

X

i

(I � e

�t

i

�)bi

(x)| dµ(x) 1�

X

i

Z

M\2B

i

|T (I � e

�t

i

�)bi

(x)| dµ(x).

Hence it is enough to showZ

M\2B

i

|T (I � e

�t

i

�)bi

(x)| dµ(x) C k b

i

k1 .

From now on drop the subscripts i. Let k

t

(x, y) be the kernel of the operatorT (I � e

�t�). Since b is supported in B, one hasZ

M\2B

|T (I � e

�t�)b(x)| dµ(x)

Z

M\2B

✓Z

B

|k

t

(x, y)||b(y)| dµ(y)◆

dµ(x)

Z

M

Z

d(x,y)�t

1/2|k

t

(x, y)| dµ(x)

!|b(y)| dµ(y).

It is therefore enough to proveZ

d(x,y)�t

1/2|k

t

(x, y)| dµ(x) C, 8y 2 M, 8t > 0.

Let us now compute the kernel k

t

(x, y). Since

��1/2 =Z +1

0e

�s� ds

p

s

(we forget a multiplicative constant which plays no role), one can write

��1/2(I � e

�t�) =Z +1

0e

�s� ds

p

s

�

Z +1

0e

�(s+t)� ds

p

s

=Z +1

0

✓1p

s

�

1{s>t}p

s� t

◆e

�s�ds,



and

T (I � e

�t�) = r��1/2(I � e

�t�) =Z +1

0

✓1p

s

�

1{s>t}p

s� t

◆re

�s�ds.

Therefore

k

t

(x, y) =Z +1

0

✓1p

s

�

1{s>t}p

s� t

◆r

x

p

s

(x, y) ds =Z +1

0g

t

(s)rx

p

s

(x, y) ds.

Now, according to Lemma 2.3,Z

d(x,y)�t

1/2|k

t

(x, y)| dµ(x)

Z +1

0|g

t

(s)|

Z

d(x,y)�t

1/2|rp

s

(x, y)| dµ(x)

!ds

C

Z +1

0|g

t

(s)|e��t/s

s

�1/2ds.

Finally

Z +1

0|g

t

(s)|e��t/s

s

�1/2ds =

Zt

0

e

��t/s

s

ds +Z +1

t

✓1

p

s� t

�

1p

s

◆e

��t/s

s

�1/2ds

= I1 + I2,

where

I1 =Z 1

0

e

��/u

u

du

is finite and independent of t, and

I2 I

02 =

Z +1

t

✓1

p

s� t

�

1p

s

◆s

�1/2ds

=Z +1

0

✓1p

u

�

1p

u + t

◆1

p

u + t

du.

Again, setting u = vt,

I

02 =

Z +1

0

1p

v(v + 1)�

1v + 1

!dv

is finite and independent of t. This ends the proof of Theorem 1.1.

4. Localisation

In this section, we shall explain the modifications that are necessary to proveTheorem 1.2. It is enough to prove that the operator

T = r(� + a)�1/2 =Z +1

0e

�as

re

�s� ds

p

s

is weak (1, 1), for some a > 0. To this end we shall use a version of the localisationtechnique of [15].

Let (xj

)j2J

be a maximal 1-separated subset of M : the collection of balls B

j =B(x

j

, 1), j 2 J , covers M , whereas the balls B(xj

, 1/2) are pairwise disjoint. Itfollows from the local doubling volume property that there exists N 2 N⇤ suchthat every x 2 M is contained in at most N balls 2B

j = B(xj

, 2). Consider a C

1

partition of unity '

j

, j 2 J such that '

j

is supported in B

j .



For f 2 C

10 (M), write

T f =X

j

12B

j

T (f'

j

) +X

j

(1 � 12B

j)T (f'

j

).

One has, for � > 0,

µ({x; |T f(x)| > �}) µ({x;X

j

12B

j

|T (f'

j

)(x)| > �/2})

+ µ({x;X

j

(1� 12B

j)|T (f'

j

)(x)| > �/2}).

Since at most N terms of the sumP

j

12B

j

|T (f'

j

)(x)| are non-zero, this yields

µ({x; |T f(x)| > �}) X

j

µ({x; 12B

j |T (f'

j

)(x)| > �/2N})

+2�

X

j

�� (1 � 12B

j )|T (f'

j

)|��

1.

The desired estimate of µ({x; |T f(x)| > �}) will follow if we prove that

µ({x; 12B

j

|T f(x)| > �}) C

�

k f k1 , 8f 2 C

10 (Bj),

and that Z

M\2B

j

|T f(x)| dµ(x) C k f k1 , 8f 2 C

10 (Bj),

where the constants are independent of j 2 J . Indeed one can add up theseestimates since

X

j

k f'

j

k1 C

0k f k1 .

To prove the first estimate, one performs the C-Z decomposition f = g+b in B

j ,which is a space with the doubling volume property. Note that the constants in theC-Z decomposition only depend on the constant in the doubling volume propertyand therefore are independent of j. One treats the good part as before. As for thebad part, write

T b =X

i

T13B

j

e

�t

i

�b

i

+X

i

T (1 � 13B

j)e�t

i

�b

i

+X

i

T (I � e

�t

i

�)bi

,

where t

i

= r

2i

and r

i

is the radius of B

i

. Note that since B

i

⇢ B

j , one has t

i

1.Therefore one can use the small time heat kernel and small radii volume estimatesto estimate

��X

i

13B

j

e

�t

i

�b

i

��

2

2

.

Indeed, the corresponding argument in Section 3 applies with easy modifications;one is able to use the L

2 boundedness of the Hardy-Littlewood maximal functionrestricted to 3B

j because the doubling volume property holds there.To control

Pi

T (I � e

�t

i

�)bi

, it is enough to prove thatZ

M\2B

i

|T (I � e

�t

i

�)bi

(x)| dµ(x) C k b

i

k1 .



To controlP

i

T (1 � 13B

j)e�t

i

�b

i

, it is enough to prove thatZ

2B

j

|T f(x)| dµ(x) C k f k1 , 8f 2 C

10 (M \ 3B

j).

Then it remains to proveZ

M\2B

j

|T f(x)| dµ(x) C k f k1 , 8f 2 C

10 (Bj).

These three estimates are similar; they follow from kernel estimates. Specifically,the first estimate follows fromZ

d(x,y)�t

1/2|k

t

(x, y)| dµ(x) C, 8y 2 M, t 1,(1)

where

k

t

=Z +1

0

e

�as

p

s

�

e

�a(s�t)1{s>t}p

s� t

!r

x

p

s

(x, y) ds

is the kernel of the operator T (I � e

�t�), the second and third estimates followfrom Z

d(x,y)�1|k(x, y)| dµ(x) C, 8y 2 M,(2)

where

k(x, y) =Z +1

0e

�as

r

x

p

s

(x, y)ds

p

s

is the kernel of the operator T .Let us gather the estimates on the heat kernel and its derivatives that are still

true under our weaker assumptions. Our volume growth assumptions are enoughfor Lemma 2.1 to hold for s 1. Lemmas 2.2, 2.3 and 2.4 follow, for all t > 0but only for s 1. For s � 1, one uses the integral maximum principle ([17], [18])according to which the quantity

Z

M

|p

s

(x, y)|2e�

d

2(x,y)s

dµ(x)

is non-increasing in s > 0. It follows thatZ

M

|p

s

(x, y)|2e�

d

2(x,y)s

dµ(x) C

�

V (y, 1), 8 y 2 M, s � 1.

Since

|

@p

s

@s

(x, y)| C

sV (y, 1)exp

✓�↵

d

2(x, y)s

◆

([18], [13]), the proof of Lemma 2.3 givesZ

M

|r

x

p

s

(x, y)|2e�

d

2(x,y)s

dµ(x) C

�

V (y, 1)s

�1, 8 y 2 M, s � 1;

therefore by HolderZ

d(x,y)�t

1/2|r

x

p

s

(x, y)| dµ(x) ✓

C

sV (y, 1)

◆1/2 Z

d(x,y)�t

1/2e

�2�

d

2(x,y)s

dx

!1/2

.



Proceeding as in Lemma 2.1 and using the fact that the volume growth is at mostexponential, one gets

Z

d(x,y)�t

1/2|r

x

p

s

(x, y)| dµ(x) Ce

��t/s

s

�1/2e

cs

, 8 y 2 M, s � 1, t > 0.

To prove (2), it is therefore enough to check thatZ 1

0e

�as

e

��t/s

ds

s

+Z +1

1e

�as

e

��t/s

e

cs

ds

s

is finite, which is obvious, provided a is chosen large enough.To prove (1), one checks that

Z 1

0|

e

�as

p

s

�

e

�a(s�t)1{s>t}p

s� t

|e

��t/s

ds

p

s

+Z +1

1|

e

�as

p

s

�

e

�a(s�t)1{s>t}p

s� t

|e

��t/s

e

cs

ds

p

s

=Z

t

0

e

�as

p

s

e

��t/s

ds

p

s

+Z 1

t

✓e

�a(s�t)

p

s� t

�

e

�as

p

s

◆e

��t/s

ds

p

s

+Z +1

1

✓e

�a(s�t)

p

s� t

�

e

�as

p

s

◆e

��t/s

e

cs

ds

p

s

is uniformly bounded for t 1.Indeed, the treatment of the first and third terms is obvious. As for the second

one, writeZ 1

t

✓e

�a(s�t)

p

s� t

�

e

�as

p

s

◆e

��t/s

ds

p

s

Z 1

t

✓e

at

p

s� t

�

1p

s

◆ds

p

s

= e

at

Z 1

t

✓1

p

s� t

�

1p

s

◆ds

p

s

+ (eat

� 1)Z 1

t

ds

s

.

The first term was treated in Section 3, and the second one is (eat

� 1)(� log t)which is bounded for t 2]0, 1].

5. A counterexample for p > 2

For n � 2, let M

n

be a manifold consisting in two copies of Rn

\ B(0, 1), withthe Euclidean metric, glued smoothly along the unit circles. One checks easily thaton this manifold the volume growth is uniformly Euclidean:

C

�1r

n

V (x, r) Cr

n

.

Moreover, the usual Sobolev inequality (or Nash inequality if n = 2) holds on M

n

:

k f k 2n

n�2 C k |rf | k2 , 8 f 2 C

10 (M

n

),

and, for p > n,

|f(x)� f(y)| Cd(x, y)1�n

p

k |rf | k

p

, 8 f 2 C

10 (M

n

), x, y 2 M

n

.

Fix p > n, and suppose that

k |rf | k

p

C

��1/2f

��p

.



Fix z 2 M

n

and choose f = p

t

(., z). One gets

|p

t

(x, z)� p

t

(y, z)| Cd(x, y)1�n

p

��1/2x

p

t

(., z)��

p

.

Now

�1/2x

p

t

(., z) = �1/2x

e

� t

2�x

p

t/2(., z);

therefore ��1/2x

p

t

(., z)��

p

Ct

�1/2��

p

t/2(., z)��

p

since the heat semigroup is analytic on L

p. By the Holder inequality

k p

t

(., z) kp

k p

t

(., z) k2/p

2 k p

t

(., z) k1�2p

1 .

Because the Euclidean L

2 Sobolev inequality holds on M

n

, one has

k p

t

(., z) k22 = p2t

(z, z) Ct

�n/2

and

k p

t

(., z) k1 Ct

�n/2

(see [33]). Thus

k p

t

(., z) kp

Ct

�n

2 (1� 1p

).

This yields

|p

t

(x, z)� p

t

(y, z)| Cd(x, y)1�n

p

t

� 12 (1�n

p

)�n

2.

The uniform lower bound

p

t

(z, z) � ct

�n/2

holds on M

n

(see [10], thm. 7.2); therefore

|p

t

(x, z)� p

t

(y, z)| C

✓d(x, y)p

t

◆1�n

p

p

t

(z, z),

in particular

|p

t

(z, z)� p

t

(y, z)| C

✓d(y, z)p

t

◆1�n

p

p

t

(z, z),

and for a small enough

|p

t

(z, z)� p

t

(y, z)| 12p

t

(z, z)

as soon as d(y, z) a

p

t. In other words

p

t

(y, z) �12p

t

(z, z) � ct

�n/2

for y 2 B(z, a

p

t). A standard iteration argument shows that this implies theGaussian lower bound

p

t

(y, z) � ct

�n/2 exp✓�C

d

2(y, z)t

◆, 8x, y 2 M

n

.

This estimate is false because the probability of going from y in the first copy ofRn

\B(0, 1) to z in the other one is at most the probability of reaching the centralhole from y times the probability of reaching z from the central hole, therefore



substantially smaller than the corresponding probability in Rn. This idea can easilybe made precise for n > 2 (see [6]), and also, in a more subtle way, for n = 2. Weowe the latter observation to Laurent Salo↵-Coste.

Without computations, one can also say that the upper and lower Gaussianestimates of the heat kernel would imply the parabolic Harnack principle, which isfalse on M

n

because the family of L

2 Poincare inequalities considered in [29] clearlycannot hold.

Therefore the estimate

k |rf | k

p

C

��1/2f

��p

is false for p > n on M

n

, n � 2.

6. The case of bad domains in Rn

In this section, let ⌦ be an open subset of the Euclidean space Rn. Define thequadratic form on C

1c

(⌦) by

Q(f) =Z

⌦

X

i

@f

@x

i

@f

@x

i

dx.

Then Q is closable and the self-adjoint operator � associated with the obtainedclosed form is called the (minus) Laplacian on ⌦ with Dirichlet boundary conditions.It is well known that � generates a holomorphic semigroup and its heat kernelp

t

(x, y) satisfies the Gaussian upper bound

p

t

(x, y) C

t

n/2exp

✓�↵

|x� y|

2

t

◆, 8x, y 2 ⌦, t > 0

for some ↵ > 0. See for example [12].A similar estimate is also true for the time derivative of the heat kernel, by the

method of [13]:

|

@p

t

@t

(x, y)| C

↵

t

n

2 +1exp

✓�↵

|x� y|

2

t

◆, 8x, y 2 ⌦, t > 0.

If the domain ⌦ has Lipschitz boundary, then ⌦ satisfies the doubling volumeproperty and the method of Theorem 1.1 is applicable, hence the Riesz transformis bounded on L

p(⌦) for all 1 < p 2 and is of weak type (1, 1). The boundednesson L

p in this case is also one of the results of [20]. However, a general open setneed not satisfy the doubling volume property, and the weak (1, 1) estimate of theRiesz transform on such a domain is an open question. The di�culty in this case isthat the usual Calderon-Zygmund theory is not applicable on a domain which doesnot satisfy the doubling volume property.

Section 3 of [14] gives us the right tools to prove a positive answer for thisquestion. We first note that the term t

n/2 in the Gaussian upper bounds of theheat kernel and its time derivative is the volume of the ball of radius

p

t in Rn,not the volume of the ball in ⌦. This observation plays an important role in theestimates of our next theorem.

Theorem 6.1. The Riesz transform T = r��1/2is weak (1, 1) and bounded on

L

p(⌦), 1 < p 2. That is, there exists C

p

, 1 p 2, such that, 8 f 2 C

10 (⌦),

k |rf | k

p

C

p

��1/2f

��p

, 1 < p 2,



and

k |rf | k1,1 C1

��1/2f

��1.

Proof. The boundedness of T = r��1/2 on L

2(⌦) is obvious from integration byparts.

Using the argument of Section 3 of [14], we define an associated operator T onL

2(Rn) by putting T (f)(x) = T (1⌦f)(x) for x 2 ⌦ and T (f)(x) = 0 for x 2 Rn

\⌦.It is straightforward to check that T is of weak type (1, 1) on ⌦ if and only if Tis of weak type (1, 1) on Rn (and that T is bounded on L

p(⌦) if and only if T isbounded on L

p(Rn)). Thus we transform the question of weak type (1, 1) estimateof T on the bad domain ⌦ to that of T on Rn which is a space with the doublingvolume property. We also observe that if k(x, y) is the associated kernel of T , thenT has an associated kernel K(x, y) given by K(x, y) = k(x, y) for x, y 2 ⌦ andK(x, y) = 0 otherwise. We now extend the semigroup e

�t� on L

2(⌦) to L

2(Rn) inthe same way. Note that all the weighted estimates in subsection 2.3 are still truefor e

�t� hence are also true for its extension defined as above. We then apply thesame method as in Theorem 1.1 to show that T is of weak type (1, 1) on Rn. See[14] for more details.

Remark. We say that a region ⌦ of Rn has the extension property if there existsa bounded linear map E from the Sobolev spaces W

1,p(⌦) into the Sobolev spacesW

1,p(Rn) such that Ef is an extension of f from ⌦ to Rn for all f 2 W

1,p(⌦) andall 1 p 1. Consider the Laplacian � with Neumann boundary conditions ona bounded domain ⌦ with the extension property. Then the Gaussian heat kernelbound is well known, see Chapter 3 of [12]. Using a limiting argument and thedefinition of the quadratic form, one can see that the estimate using integration byparts in the proof of Lemma 2.3 is still valid. Therefore, Theorem 6.1 is still truefor the Laplacian with Neumann boundary conditions.

Acknowledgements

The authors thank Laurent Salo↵-Coste for interesting comments, Damien Lam-berton for providing and explaining reference [26], and Pascal Auscher for informingthem about [21]. This work was started during the first author’s stay at MacquarieUniversity and the second author is supported by a Macquarie University ResearchFellowship. Both of them thank Alan McIntosh for creating a stimulating environ-ment and for helpful discussions. Finally, thanks are due to Emmanuel Russ andEl-Maati Ouhabaz who read carefully the manuscript and suggested some improve-ments.

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D

´

epartment de Math

´

ematiques, Universit

´

e de Cergy-Pontoise, 95302 Cergy Pontoise,

France

E-mail address: [email protected]

Department of Mathematics, Macquarie University, North Ryde NSW 2113, Aus-

tralia

E-mail address: [email protected]


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