BM O spaces related to Schrödinger operators with
potentials satisfying a reverse Hölder inequality ∗
J. Dziubański G. Garrigós T. Martı́nez J. Torrea J. Zienkiewicz
November 13, 2003
∗
Abstract
We identify the dual space of the Hardy-type space HL1 related to the time
independent Schrödinger operator L = −∆ + V , with V a potential satisfying
a reverse Hölder inequality, as a BM O-type space BM OL . We prove the
boundedness in this space of the versions of some classical operators associated
to L (Hardy-Littlewood, semigroup and Poisson maximal functions, square
function, fractional integral operator). We also get a characterization of BM OL
in terms of Carlesson measures.
1
Introduction
Let V be a fixed non-negative function on Rd , d ≥ 3, satisfying a reverse Hölder
inequality RHs (Rd ) for some s > d2 ; that is, there exists C = C(s, V ) > 0 such that
µZ
s
V (x) dx
¶ 1s
Z
≤ C
B
V (x) dx,
(1.1)
B
for every ball B ⊂ Rd . Consider the time independent Schrödinger operator with the
potential V :
L = −∆ + V,
∗
This research was partially supported by the European Commission, within the IHP Network
“HARP 2002-2006”, contract number HPRN-CT-2001-00273-HARP. Second author also supported
by Programa Ramón y Cajal and grant “BMF2001-0189”, MCyT (Spain). The first and the fifth
author supported by grants 5P03A05020 and 5P03A02821 from KBN (Poland) and by Foundation
of Polish Sciences, Subsidy 3/99. The third and the fourth author supported by grant MCyT
BMF2002-04013-C02-02.
∗
2000 Math Subject Classification: 35J10, 42B35, 42B30
1
2
and its associated semigroup:
Tt f (x) = e
−tL
f (x) =
Z
Rd
kt (x, y)f (y)dy,
f ∈ L2 (Rd ),
t > 0.
(1.2)
A Hardy-type space related to L is naturally defined by:
HL1 = {f ∈ L1 (Rd ) : T ∗ f (x) = sup |Tt f (x)| ∈ L1 (Rd )},
t>0
∗
with kf kHL1 := kT f kL1 (Rd ) .
(1.3)
For the above class of potentials, it was shown in [2] that HL1 admits a special atomic
characterization, where cancellation conditions are only required for atoms with small
supports. In this paper we shall be interested in properties of the dual space of HL1 ,
which we shall identify with a subclass of BM O functions, namely:
½
¾
Z
1
BM OL = f ∈ BM O :
|f | ≤ C, for all B = BR (x) : R > ρ(x) . (1.4)
|B| B
The precise definition of the norm in this space is given in Definition 3.5. The critical
radii above are determined by the function ρ(x; V ) = ρ(x) which takes the explicit
form
½
¾
Z
1
ρ(x) = sup r > 0 : d−2
V (y)dy ≤ 1 .
(1.5)
r
B(x,r)
Throughout the paper shall assume that V 6≡ 0, so that 0 < ρ(x) < ∞ (see [7]). This
BM OL space turns out to be the suitable extreme point for p = ∞ concerning the
boundedness of the classical operators associated to the operator L. We shall use the
following notations:
Z
1
M f (x) = sup
|f (y)| dy,
(1.6)
B3x |B| B
T ∗ f (x) = sup |e−tL f (x)|,
(1.7)
t>0
Z ∞ −u
¯
¯
√
e
¯
¯
∗
−t L
√ Tt2 /4u du,
P f (x) = sup¯Pt f (x)¯, where Pt = e
=
(1.8)
u
t>0
0
³Z ∞
´ 12
,
(1.9)
sQ f (x) =
|Qt f (x)|2 dtt
0
Z ∞
−α/2
Iα f (x) = L
f (x) =
e−tL f (x) tα/2−1 dt for 0 < α < d. (1.10)
0
These notations correspond, respectively, to the Hardy-Littlewood maximal function, the semigroup and Poisson-semigroup maximal functions, the L-square function
and the L-fractional integral operator. We observe that in the classical case (i.e.
V ≡ 0) these operators fail to be bounded in BM O, in fact may be identically infinity for functions with certain growth (see §5 below). However in our case it turns out
that they behave correctly in BM OL as the following results shows.
3
THEOREM 1.11 Let V 6≡ 0 be a non-negative potential in RHs (Rd ) for some s > d2 .
The operators M , T ∗ , P ∗ and sQ are well-defined and bounded in BM OL . For all
0 < α < d, the operator Iα is bounded from Ld/α (Rd ) into BM OL .
We also show a characterization of BM OL in terms of Carleson measures. A
positive measure µ on Rd+1
:= Rd × (0, ∞) is said to be a Carleson measure if
+
kµkC :=
µ(Br (x) × (0, r))
< ∞.
|Br (x)|
x∈Rd , r>0
sup
(1.12)
Our result characterizes the elements of BM OL by a Carleson measure condition
related to appropriate square function. To be more precise let
¯
¶
µ
dTs ¯¯
2
f (x), (x, t) ∈ Rd+1
(Qt f )(x) = t
(1.13)
+ ,
¯
ds s=t2
then the following theorems holds
THEOREM 1.14 Let V 6≡ 0 be a non-negative potential in RHs (Rd ) for some s >
and ρ(x) = ρ(x) be the weight defined in (1.5).
d
2
1. If f ∈ BM OL , then dµf (x, t) := |Qt f (x)|2 dxdt/t is a Carleson measure.
2. Conversely, if f ∈ L1 ((1 + |x|)−(d+1) dx) and dµf (x, t) is a Carleson measure,
then f ∈ BM OL .
Moreover, in either case, there exists C > 0 such that
1
kf k2BM OL ≤ kdµf kC ≤ C kf k2BM OL .
C
The outline of the paper is as follows. In Section 2 we gather the required estimates
on the kernel Qt (x, y), which complement those presented for kt (x, y) in [2, 3, 4]. In
Section 3 we recall the atomic decomposition for HL1 and establish the identification
(HL1 )∗ ≡ BM OL . In Section 5 we study the boundedness of classical operators in
BM OL . Some of the techniques needed in the proofs of Section 5 appear naturally
when proving Theorem 1.14, it is because of this reason that we present in Section 4
the proof of Theorem 1.14.
2
Estimates on the kernels
We begin by recalling some basic properties of the function ρ(x) under the assumption
(1.1) on V (see [7, Lemma 1.4]).
4
PROPOSITION 2.1 There exist c > 0 and k0 ≥ 1 so that, for all x, y ∈ Rd
³
c−1 ρ(x) 1 +
|x−y|
ρ(x)
´−k0
³
≤ ρ(y) ≤ c ρ(x) 1 +
|x−y|
ρ(x)
´ k k+1
0
0
.
(2.2)
In particular, ρ(x) ∼ ρ(y) when y ∈ Br (x) and r ≤ Cρ(x).
From the Feynman-Kac formula, it is well-known that the semigroup kernels
kt (x, y), associated with Tt = e−tL , satisfy the estimates
2
d
0 ≤ kt (x, y) ≤ ht (x − y) := (4πt)− 2 exp(− |x−y|
).
4t
(2.3)
These estimates can be improved in time when V 6≡ 0 satisfies the reverse Hölder
condition RHs for some s > d/2. The function ρ(x) arises naturally in this context.
PROPOSITION 2.4 (see [3], [6] ) For every N , there is a constant CN such that
³
√
√ ´−N
2
d
t
t
0 ≤ kt (x, y) ≤ CN t− 2 exp(− |x−y|
)
1
+
+
.
(2.5)
5t
ρ(x)
ρ(y)
Below, we shall also make use of the Lipschitz regularity of the kernels, which is
again a consequence of (1.1) (see [4, Proposition 4.11]).
PROPOSITION 2.6 If V ∈ RHs (Rd ), s > d/2, then there exists δ = δ(s) >√0 and
c > 0 such that for every N > 0 there is a constant CN so that, for all |h| ≤ t
³ ´δ d
³
√
√ ´−N
|h|
c|x−y|2
−2
t
√
|kt (x + h, y) − kt (x, y)| ≤ CN
t exp(− t ) 1 + ρ(x) + ρ(y)t
. (2.7)
t
We will also need estimates for the integral kernels of the operators Qt in (1.13):
¯
¯
∂k
(x,
y)
s
¯
Qt (x, y) = t2
(2.8)
∂s ¯s=t2
PROPOSITION 2.9 There exist constants c, δ > 0 such that for every N there is a
constant CN so that
³
´−N
2
t
t
(a)
|Qt (x, y)| ≤ CN t−d exp(− c|x−y|
)
1
+
+
;
t2
ρ(x)
ρ(y)
³
(b)
(c)
|Qt (x + h, y) − Qt (x, y)| ≤ CN
for all |h| ≤ t;
¯Z
¯
¯
¯
|h|
t
´δ
¯
¯
(t/ρ(x))δ
Qt (x, y)dy ¯¯ ≤ CN
.
(1 + t/ρ(x))N
Rd
−d
t
³
2
exp(− c|x−y|
)
t2
1+
t
ρ(x)
+
t
ρ(y)
´−N
,
5
PROOF: Corollary 6.4 of [4] asserts that the integral kernels kζ (x, y) of the extension
of {Tt }t>0 to the holomorphic semigroup {Tζ }ζ∈∆π/4 satisfy
³
|kζ (x, y)| ≤ CN (<ζ)−d/2 1 +
<ζ
ρ(x)2
+
<ζ
ρ(y)2
´−N
¡
¢
exp −c|x − y|2 /<ζ .
The Cauchy integral formula combined with (2.10) gives
¯d
¯
¯1 Z
kζ (x, y) ¯¯
¯
¯
¯
dζ ¯
¯ kt (x, y)¯ = ¯
dt
2π |ζ−t|=t/10 (ζ − t)2
´−N
¡
¢
CN −d/2 ³
t
t
≤
t
1 + ρ(x)
exp −c0 |x − y|2 /t
2 + ρ(y)2
t
(2.10)
(2.11)
which, by (2.8), implies (a).
We now turn to prove (b). By the semigroup property, Proposition 2.6, and the
part (a) already proved, we obtain
¯ Z ³
¯
´
¯
¯
¯
√
¯Qt (x + h, y) − Qt (x, y)¯ = ¯¯2
kt2 /2 (x + h, z) − kt2 /2 (x, z) Qt/ 2 (z, y) dz ¯
³
≤ CN
|h|
t
³
. CN
´δ Z ³
|h|
t
´δ
t−d
1+
t
ρ(x)
´−N
−d −c|x−z|2 /t2 −d −c|z−y|2 /t2
t e
³
0
2
exp(− c |x−y|
)
1+
t2
t e
t
ρ(x)
+
t
ρ(y)
³
1+
t
ρ(y)
´−N
dz
´−N
which establishes (b).
It follows from [7] Lemmas 1.2 and 1.8 that there is a constant C0 such that for a
nonnegative Schwartz class function ϕ there exists a constant C such that

³ √ ´δ

Z
t
 Ct−1
for t ≤ ρ(x)2
ρ(x)
³ √ ´C0 +2−d
ϕt (x − y)V (y) dy ≤
(2.12)

 C ρ(x)t
for t > ρ(x)2 ,
√
where ϕt (x) = t−d/2 ϕ(x/ t). Therefore
¯ Z
¯ ¯
¯Z d
¯
¯ ¯
¯
kt (x, y) dy ¯ = ¯Tt L1(x)¯ = kt (x, y)V (y) dy
¯
dt
and (c) follows by first using (2.5) with N sufficiently large and then (2.12).
(2.13)
2
Finally, we recall some results about covering Rd by critical balls. These depend
entirely on the estimates in Proposition 2.1, and can be found in [2]. Throughout the
paper, given a ball B we denote by B ∗ the ball with same center and twice radius.
6
PROPOSITION 2.14 (see [2, Lemma 2.3]). There exists a sequence of points
d
∞
{xk }∞
k=1 in R , so that the family of “critical balls” Q = {Qk }k=1 , defined by Qk :=
{|x − xk | < ρ(xk )}, satisfy
S
d
(a)
k Qk = R .
∗∗
(b) There exists N = N (ρ) so that, for every k ≥ 1, card {j : Q∗∗
j ∩Qk 6= ∅} ≤ N .
Combining the previous with (2.2) one easily gets
COROLLARY 2.15 There is a constant c = c(ρ) so that, for every ball BR (x) with
R > ρ(x), we have
X
|BR (x)| ≤
|Qk | ≤ c |BR (x)|.
Qk ∩BR (x)6=∅
COROLLARY 2.16 There exists aP
family of C ∞ functions ϕk such that supp ϕk ⊂
Q∗k , 0 ≤ ϕk ≤ 1, |∇ϕk | ≤ C/ρ(xk ), k ϕk = 1.
3
The dual space of HL1
We shall assume that V 6≡ 0 is a potential in the reverse Hölder class as stated in the
introduction. For such potentials it was shown in [2] the following characterization
of HL1 .
A function a : Rd → C is an HL1 -atom associated with a ball Br (x0 ) when
supp a ⊂ Br (x0 ),
and in addition,
kak∞ ≤ 1/|Br (x0 )|,
(3.1)
Z
a(x) dx = 0,
whenever 0 < r < ρ(x0 ).
(3.2)
1
THEOREM 3.3PAn integrable function f in Rd belongs
Pto HL if and only if it can be
1
written as f = j λj aj , where aj are HL -atoms and j |λj | < ∞. Moreover, there
exists a constant c > 0 such that
P
P
c−1 kf kHL1 ≤ inf { j |λj | : f = j λj aj } ≤ c kf kHL1 .
REMARK 3.4 We note that in the above atomic decomposition we may restrict to
atoms supported by balls Br (x) with r ≤ ρ(x). Indeed, if we are given an HL1 -atom
a associated with a ball Br (y0 ) with r > ρ(y0P
) then it easily follows
from Proposition
P
2.14 that the atom a can be written as a = j λj aj , where j |λj | . 1, and aj are
HL1 -atoms supported by critical balls.
7
DEFINITION 3.5 We shall say that a locally integrable function f belongs to BM OL
whenever there is a constant C ≥ 0 so that
Z
Z
1
1
|f − fBs | ≤ C and
|f | ≤ C,
(3.6)
|Bs | Bs
|Br | Br
for all balls Bs = Bs (x), Br = Br (x) such that s ≤ ρ(x) ≤ r. We let
R kf kBM OL denote
−1
f (x) dx.
the smallest C in (3.6) above. Here and subsequently, fB = |B|
B
We observe that kf kBM OL is actually a norm (and not only a seminorm) making
BM OL a Banach space. Moreover, kf kBM O ≤ 2kf kBM OL . We also observe that, by
Proposition 2.14 and its corollary, it is enough to consider the condition on the right
hand side of (3.6) only for balls from the family Q.
From the previous atomic decomposition, and the definition of f ∈ BM OL , it is
clear that
Z
f (x)a(x) dx, acting on HL1 -atoms a,
Φf (a) :=
Rd
defines a linear functional on HL1 with kΦf k ≤ ckf kBM OL . Our first result shows that
all linear functionals on HL1 do actually arise in this way.
THEOREM 3.7 The correspondence
BM OL 3 f 7−→ Φf ∈ (HL1 )∗
is a linear isomorphism of Banach spaces.
We shall need a lemma about the size of HL1 -functions.
LEMMA 3.8 The space L2c (Rd ), of square-integrable functions with compact support
is contained in HL1 . Moreover, there is a constant C = C(L) > 0 so that, for large
balls B = BR (x0 ) with R ≥ ρ(x0 ), we have
1
kgkHL1 ≤ C |B| 2 kgkL2 (B) ,
∀ g ∈ L2 (B).
(3.9)
PROOF: Assume
g ∈ L2 (BR (x0 )), R ≥
P that P
Pρ(x0 ). Then, using Corollary 2.16, we
can write g = k ϕk g = k gk and kgk2L2 ∼ k kgk k2L2 . By Schwarz’s inequality and
Corollary 2.15 it suffices to prove (3.9) for g = gk . Obviously,
kT ∗ gk kL1 (Q∗∗
≤ C |Qk |1/2 kT ∗ gk kL2 (Q∗∗
≤ C |Qk |1/2 kgk kL2 .
k )
k )
(3.10)
/ Q∗∗
If x ∈
k , then applying Propositions 2.4 and 2.1, we get
√ ´
Z ³
t −N −d/2 −c|x−y|2 /t
t
e
|gk (y)| dy
|Tt gk (x)| ≤ CN
1+
ρ(y)
ρ(xk )N
≤ CN
|Qk |1/2 kgk kL2 .
|x − xk |d+N
(3.11)
Now (3.9) for g = gk follows from (3.11) and (3.10).
8
2
PROOF of Theorem 3.7:
Let Φ ∈ (HL1 )∗ , and denote BN = BN (0), N > ρ(0).
1
By Lemma 3.8, there exists a unique fN ∈ L2 (BN ) with kfN kL2 (BN ) ≤ C |BN | 2 kΦk,
and so that
Z
Φ(g) =
fN g, ∀ g ∈ L2 (BN ).
BN
Iterating in N , and noticing that fN +1 |BN = fN , one defines a unique locally squareintegrable function f in Rd so that
Z
f (x) g(x) dx, for all g ∈ L2c (Rd ).
Φ(g) =
Rd
Thus, Φ = Φf , and only remains to show that kf kBM OL ≤ CkΦk. If we test first on
atoms supported by large balls B = BR (x), R > ρ(x), and use again Lemma 3.8, we
1
see that kf kL2 (B) ≤ C |B| 2 kΦk. Thus, from Hölder’s inequality we conclude
Z
Z
³
´1
2 2
1
1
|f
|
≤
|f
|
≤ C kΦk.
(3.12)
|B|
|B|
B
B
1
On the other hand, since classical H -atoms are particular cases of HL1 -atoms, it
follows that Φ|H 1 ∈ (H 1 )∗ , and thus Φ|H 1 = Φh for a unique (modulo constants) h ∈
BM O(Rd ). Now, testing on H 1 -atoms over a fixed ball B we see that f = h + cB , for
some constant cB , and therefore f ∈ BM O(Rd ) with kf kBM O ≤ 2 khkBM O ≤ C 0 kΦk.
This establishes the theorem.
2
Observe that we have shown, for large balls, the local square integrability estimate
(3.12). The exponent 2 can actually be replaced by any 1 ≤ p < ∞, by just using in
the proof above the corresponding p0 -version of Lemma 3.8. This, together with the
John-Nirenberg inequality gives the following corollary.
COROLLARY 3.13 For every p ∈ [1, ∞) there exists c = c(p, ρ) > 0 such that, for
all f ∈ BM OL we have
Z
´1
³
p p
1
|f − fB |
≤ c kf kBM OL , for all balls B,
|B|
³
B
Z
1
|B|
|f |p
B
´ p1
≤ c kf kBM OL ,
for B = Br (x) : r ≥ ρ(x).
We conclude with the following lemma which will be used often below. Its proof
is elementary and left to the reader.
LEMMA 3.14 There exists c > 0 so that, for all f ∈ BM OL and B = Br (x) with
r < ρ(x), then
|fB ∗ | ≤ c (1 + log ρ(x)
) kf kBM OL .
r
9
4
Proof of Theorem 1.14
LEMMA 4.1 For all f ∈ L2 (Rd ) we have ksQ f k2 =
Z N
dt
f (x) = 8
lim
Q2t f (x) ,
ε→0, N →∞ ε
t
√1
8
kf k2 . Moreover,
in L2 (Rd ).
(4.2)
PROOF: The proof is standard and follows by using spectral techniques, as in
[8,
3]. For completeness we provide some details. Since Tt = e−tL =
R ∞ Chapter
−tλ
e dE(λ), we have
0
Z ∞
d Tt
t
= −tL Tt = −
tλ e−tλ dE(λ).
dt
0
Thus, for all f ∈ L2 (Rd ), using the self-adjointness of Qt , we get
Z Z ∞
dt
2
ksQ f k2 =
|Qt f (x)|2 dx
t
d
ZR∞ 0
¡
¢
dt
2
h t4 ddsTs |s=t2 f, f i
=
t
Z0 ∞ hZ ∞
i
dt
2
=
t4 λ2 e−2t λ
dEf,f (λ) = 18 kf k22 .
t
0
0
For the second part, it suffices to show that, for every pair of sequences nk % ∞,
εk & 0
Z nk+m
Z εk
dt
2 dt
lim
Qt f
= lim
Q2t f
= 0, ∀ m ≥ 1.
(4.3)
k→∞ n
k→∞ ε
t
t
k
k+m
Rn
Indeed, when this is case we can find h ∈ L2 (Rd ) so that limk→∞ εkk Q2t f dtt = h, and
therefore, by using also a polarized version of the first part
Z nk
dt
hQt f, Qt gi
hh, gi = lim
k→∞ ε
t
Z ∞ k
dt
=
hQt f, Qt gi
= 81 hf, gi, ∀ g ∈ L2 (Rd ),
(4.4)
t
0
which implies h = 18 f . To prove (4.3) we use again functional calculus, so that
° Z nk+m
°2 Z ∞ ¯Z nk+m
¯2
°
°
dt
¯
2
4 2 −2t2 λ dt ¯
°
°
≤
Qt f
tλ e
dEf,f (λ).
¯
t ¯
°
t °
0
nk
nk
Computing the integral inside one is led to estimate
Z ∞
2
(1 + 2λn2k )e−2λnk dEf,f (λ), as nk → ∞,
0
which by dominated convergence tends to 0. Observe that the last step makes use of
the fact that 0 is not an eigenvalue of L because, V (x) > 0 for almost every x, and
hLf, f i ≥ hV f, f i > 0 unless f ≡ 0). One proceeds similarly when εk → 0.
2
10
4.1
Proof of part (1) of Theorem 1.14
Our first observation is that
Qt f (x) =
Z
Rd
Qt (x, y)f (y) dy
is a well-defined absolutely convergent integral for all (x, t) ∈ Rd+1
+ , as it follows from
the kernel decay in Proposition 2.9 and the integrability of (1 + |y|)−d−1 |f (y)| (see [9,
page 141]). Let us fix a ball B = Br (x0 ). We wish to show that
Z rZ
1
dxdt
|Qt f (x)|2
≤ C kf k2BM OL .
(4.5)
|B| 0 B
t
To do this, we split the function f into local, global, and constant parts as follows
f = (f − fB ∗ ) χB ∗ + (f − fB ∗ ) χ(B ∗ )c + fB ∗
= f1 + f2 + fB ∗ ,
As we shall see, the novelty of the BM OL condition appears in the control of the
constant term. For the other two terms the proof follows from standard arguments
(cf. e.g. [9]) and estimates for the kernel Qt (x, y). Indeed, in the local case, a simple
use of Lemma 4.1 gives us
Z rZ
Z
1
C
2 dxdt
|Qt f1 (x)|
≤
|sQ f1 (x)|2 dx
|B| 0 B
t
|B| B
Z
C
1
2
≤
kf1 k2 =
|f − fB ∗ |2 ,
|B|
|B| B ∗
which is smaller than a multiple of kf k2BM O by Corollary 3.13.
To estimate the global term, we only need a very mild decay of |Qt (x, y)|: if
x ∈ B = Br (x0 ) and t < r then
Z
t−d
|Qt f2 (x)| .
dy
|f2 (y)|
(1 + |x−y|
)d+1
Rd
t
Z
t
dy
.
|f (y) − fB ∗ |
|x0 − y|d+1
(B ∗ )c
·Z
¸
∞
X
t
k d
|f (y) − fB2k r | dy + (2 r) |fB2k r − fB ∗ |
.
k r)d+1
(2
kr
|y−x
|∼2
0
k=1
.
∞
t X −k
t
2 [ kf kBM O + k kf kBM O ] . kf kBM O .
r k=1
r
Thus, integrating over B × (0, r) we obtain
Z rZ
Z r 2
t dt
1
2 dxdt
|Qt f2 (x)|
.
kf k2BM O =
2 t
|B| 0 B
t
r
0
1
2
kf k2BM O . kf k2BM OL .
11
It remains to estimate the constant term, for which we shall make use of part (c) of
Proposition 2.9. Assuming first that r < ρ(x0 ), and using ρ(x) ∼ ρ(x0 ) for x ∈ B (cf.
Proposition 2.1), we have
Z rZ
Z r Z ¯Z
¯2
¯
¯
|fB ∗ |2
2 dxdt
1
|Qt (fB ∗ 1)(x)| t
= |B|
(4.6)
¯ Qt (x, y)dy ¯ dxdt
|B|
t
0
B
0
B
Z rZ
¡
¢2δ dxdt
|fB ∗ |2
. |B|
t/ρ(x)
t
0
B
¡
¢2δ
2
. |fB ∗ | r/ρ(x0 )
¡
¢2 ¡
¢2δ
. kf k2BM OL 1 + log ρ(xr 0 )
r/ρ(x0 )
. kf k2BM OL ,
where the last line follows from Lemma 3.14.
Suppose finally that r ≥ ρ(x0 ), and select
from Corollary 2.15 a finite family of
P
critical balls {Q` } so that B ⊂ ∪Q` and
|Q` | . |B|. Then, using again part (c) of
∗
Proposition 2.9 and |fB | ≤ kf kBM OL , we can bound the left hand side of (4.6) by
ÃZ
!
Z ∞ Z
ρ(x` ) Z
kf k2BM O X
dx
dt
L
(t/ρ(x` ))2δ dxdt
+
|B|
t
2N −2δ t
(1
+
t/ρ(x
))
`
0
Q`
ρ(x` ) Q`
`
X
|Q` | . kf k2BM OL ,
. Ckf k2BM OL |B|−1
`
which, by Corollary 2.15, establishes the first part of Theorem 1.14.
REMARK 4.7 It is worthwhile to notice that, from the previous proof, it actually
follows that
sup kQt f k∞ = sup |Qt f (x)| . kf kBM OL .
t>0
Rd
t>0,x∈
That is, the solution to the evolution equation
½
ut (t, x) = −Lu(t, x),
u(0, ·) = f
(t, x) ∈ Rd+1
+
(4.8)
with initial data f ∈ BM OL , satisfies the regularity estimate
kLu(t, ·)k∞ . t−1 kf kBM OL .
4.2
Proof of part (2) of Theorem 1.14
Now, let us fix f ∈ L1 ((1 + |x|)−(d+1) dx) so that µf (x, t) := |Qt f (x)|2 dxdt/t is a
Carleson measure. We wish to show that such f must belong to BM OL . By Theorem
3.7 it suffices to show that the linear functional
Z
1
f (x)g(x) dx,
HL 3 g 7−→ Φf [g] :=
Rd
12
defined at least over finite linear combinations of HL1 -atoms, satisfies the estimate
1
¯
¯
¯ Φf [g] ¯ ≤ c kµf k 2 kgkH 1 .
C
L
(4.9)
To do this, we shall proceed in three steps. First, we shall write Φf in terms of
the extended functions
(x, t) ∈ Rd+1
+ .
F (x, t) := Qt f (x) and G(x, t) := Qt g(x),
More precisely, we shall show the following identity.
LEMMA 4.10 Let f ∈ L1 ((1 + |x|)−(d+1) dx) and g be an HL1 -atom. Then
Z
Z
dxdt
1
F (x, t) G(x, t)
f (x) g(x) dx =
.
8
t
Rd
Rd+1
+
(4.11)
In our second step we shall bound the right hand side of (4.11) using a general
result about tent spaces.
LEMMA 4.12 : (see [9, p. 162]). Let F (x, t), G(x, t) be measurable functions on
Rd+1
satisfying
+
Ã
I(F )(x) := sup
x∈B
1
|B|
Z
r(B)
0
Z
|F (y, t)|2
B
µZ Z
dydt
|G(y, t)| d+1
t
Γ(x)
2
G(G)(x) :=
dydt
t
¶ 21
! 12
∈ L∞ (Rd ),
∈ L1 (Rd ),
where r(B) denotes the radius of B and Γ(x) = {(y, t) ∈ Rd+1
: |y − x| < t}. Then,
+
there is a universal c > 0 so that
Z Z
Z
dydt
|F (y, t)G(y, t)|
≤ c
I(F )(x) G(G)(x) dx
t
Rd+1
Rd
+
≤ c kI(F )kL∞ kG(G)kL1 .
Observe that kµf kC = kI(F )k2L∞ . Thus, in order to establish (4.9) we just need
to show that kG(G)kL1 ≤ C kgkHL1 . Note that G(G)(x) = SQ g(x), where SQ is the
following area integral operator,
µZ
∞
SQ g(x) :=
0
Z
dydt
|Qt g(y)| d+1
t
|x−y|<t
2
¶ 21
,
x ∈ Rd .
Then, the following result about the boundedness of the area integral operator gives
us the desired inequality.
13
LEMMA 4.13 There exists c > 0 so that kSQ gkL1 ≤ c kgkHL1 for every g being a
finite linear combination of HL1 -atoms.
Thus, we have reduced the proof of the theorem to show Lemmas 4.10 and 4.13.
Let us start with the proof of the second one.
PROOF of Lemma 4.13:
By Theorem 1.5 in [2], it is enough to consider sums
of atoms associated to balls Br (x0 ) with r . ρ(x0 ). Let us fix an HL1 -atom, g(x),
associated with a ball B = Br (x0 ). Now, observe that
#
Z "Z Z
dydt
2
2
kSQ gkL2 (Rd ) =
|Qt g(y)| χΓ(x) (y, t) d+1 dx
t
Rd
Rd+1
+
·Z
¸
Z Z
dydt
=
|Qt g(y)|2
dx d+1
t
Rd+1
|x−y|<t
Z +Z
dydt
|Qt g(y)|2
= cd
t
Rd+1
+
cd
= cd ksQ gk2L2 (Rd ) =
kgk2L2 (Rd ) ,
8
after using Lemma 4.1 in the last step. Therefore,
µZ
Z
SQ g(x) dx ≤ |B
∗∗∗
|
1
2
2
SQ g(x) dx
B ∗∗∗
B ∗∗∗
¶ 21
1
. |B| 2 kgkL2 . 1.
To complete the proof of Lemma 4.13, we must find a uniform bound for
Z
I=
SQ g(x) dx.
(B ∗∗∗ )c
We consider first the case when r < ρ(x0 ). Then, by the moment condition on g,
·Z
´2 dydt ¸ 12
SQ g(x) =
(Qt (y, x ) − Qt (y, x0 ))g(x ) dx
td+1
0
|x−y|<t
Rd
# 21
" Z |x−x0 | Z
³Z
0 ´2
2
dx
dydt
|Qt (y, x0 ) − Qt (y, x0 )|
≤
|B| td+1
0
|x−y|<t
B
"Z
#1
Z
³ ´2 dydt 2
∞
+
···
= I1 (x) + I2 (x).
(4.14)
|x−x0 |
td+1
|x−y|<t
2
∞
Z
³Z
0
0
0
We now use the smoothness of Qt (x, y) = Qt (y, x) established in Proposition 2.9. In
the first range of integration we have |y − x0 | ∼ |y − x0 | ∼ |x − x0 | and |x0 − x0 | <
14
|y − x0 |/4, so that
"Z
|x−x0 |
2
I1 (x) .
Z
0
"Z
|x−y|<t
|x−x0 |
2
≤
|x0 −x0 |
´δ
t
B
t−d (1 +
|y−x0 | −(d+1)
)
t
Z
0
"Z
³Z ³
|x−y|<t
( rt )2δ t−2d (1 +
|y−x0 | −2(d+1)
)
t
dydt
td+1
´2 dydt
|B|
td+1
# 21
dx0
# 21
#1
´2(d+1) dt 2
t
.
( rt )2δ t−2d |x−x
0|
t
0
1
" Z |x−x0 |
#2
2
rδ
dt
rδ
2−2δ
'
t
'
.
|x − x0 |d+1
t
|x − x0 |d+δ
0
³
|x−x0 |
2
Thus, integrating over (B ∗∗∗ )c ,
Z
Z
I1 (x) dx .
(B ∗∗∗ )c
|x−x0 |>8r
(4.15)
rδ
dx . 1.
|x − x0 |d+δ
For I2 (x) we have |x0 − x0 | ≤ r < |x − x0 |/2 ≤ t, so that Proposition 2.9 gives
|Qt (y, x0 ) − Qt (y, x0 )| .
³
|x0 −x0 |
t
´δ
t−d .
Thus, for x ∈ (B ∗∗∗ )c a similar argument to that presented above leads to
"Z
I2 (x) .
"Z
≤
Z
∞
|x−x0 |
2
|x−y|<t
B
|x0 −x0 |δ
tδ
Z
∞
|x−x0 |
2
"Z
' r
³Z
|x−y|<t
∞
δ
|x−x0 |
2
r2δ
t2δ
t−2d
# 12
dt
t2d+2δ+1
t−d
dydt
td+1
´2 dydt
dx0
|B|
td+1
# 21
# 21
rδ
'
.
|x − x0 |d+δ
Integrating over (B ∗∗∗ )c one gets theR required bound.
We now turn to the estimate of (B ∗∗∗ )c SQ g(x) dx when r is comparable to ρ(x0 ).
As before, we shall estimate pointwise SQ g(x), for each x ∈ (B ∗∗∗ )c . Proceeding as
in (4.14), we break up the integral in t > 0 defining SQ g(x) into three parts
Z
r
2
2
SQ g(x) =
0
Z
|x−x0 |
4
··· +
r
2
Z
··· +
∞
|x−x0 |
4
· · · = I10 (x) + I20 (x) + I30 (x).
15
In the first integrand we have |x0 − y| ∼ |x − x0 |, so applying the estimate (a) of
Proposition 2.9, we get
I10 (x) .
r2
,
|x − x0 |2(d+1)
x ∈ (B ∗∗∗ )c .
For the second term we use the extra decay in time, together with |x0 − y| ∼ |x − x0 |
and ρ(x0 ) ∼ ρ(x0 ) ∼ r (see Propositions 2.1 and 2.9):
Z
I20 (x)
|x−x0 |
4
.
Z
r
2
Z
³Z
|x−y|<t
|x−x0 |
.
t−2d
³
r
2
B
t
|x−x0 |
r2(M +1)
'
|x − x0 |2(d+M +1)
t−d (1+|x−x0 |/t)−(d+M +1) dx0
(1+t/ρ(x0 ))M
|B|
´2(d+M +1) ³
Z
2|x−x0 |
r
1
t2
ρ(x0 )
t
´2M dt
t
dt
r2M
'
.
t
|x − x0 |2(d+M )
Finally, for the last term the extra decay just gives
Z ∞ Z
³Z
0
I3 (x) .
t−d (1 + t/ρ(x0 ))−M
Z
.
|x−x0 |
4
∞
|x−x0 |
4
|x−y|<t
t−2d
³
ρ(x0 )
t
B
´2 dydt
td+1
dx0
|B|
´2 dydt
td+1
´2M dt
t
ρ(x0 )2M
r2M
'
.
|x − x0 |2(d+M )
|x − x0 |2(d+M )
R
So, in the three cases we obtain bounds that lead to (B ∗∗∗ )c SQ g(x) dx . 1. This
completes the proof of Lemma 4.13.
2
'
To conclude with the proof of Theorem 1.14, it only remains to justify the identity
in Lemma 4.10. We observe that such identity is clearly valid when f, g ∈ L2 (Rd )
(see (4.4)), while we must justify the convergence of the integrals in the case when
f ∈ L1 ((1 + |x|)−(d+1) dx) and g is an HL1 -atom. As in the classical case (when V ≡ 0)
this requires further estimates of the kernels (cf. [9]). A sketch of the modifications
needed in this situation is the following.
PROOF of Lemma 4.10:
First one observes that, by Lemmas 4.12 and 4.13, and
the dominated convergence theorem, the following integral is absolutely convergent
and satisfies
Z
Z NZ
dxdt
dxdt
I=
F (x, t) G(x, t)
= lim
Qt f (x) Qt g(x)
.
ε→0
t
t
Rd+1
ε
Rd
N →∞
+
16
Next, a formal use of Fubini’s theorem allows us to write, for each t > 0,
Z
Z Z
Z
Qt f (x) Qt g(x) dx =
Qt (x, y)f (y) Qt g(x) dydx =
f (y) Q2t g(y) dy,
Rd
Rd Rd
and, consequently, again formally
Z
I = lim
ε→0
N →∞
=
Z
lim
ε→0
N →∞
Rd
N
Rd
ε
Rd
¸
·Z
f (y) Q2t g(y) dy
·Z
N
f (y)
ε
Q2t g(y)
dt
t
¸
dt
dy.
t
(4.16)
The absolute integrability justifying these steps is a simple exercise, which combines
the hypothesis f ∈ L1 ((1 + |x|)−(d+1) dx), the kernel decay |Qt (x, y)| . t−d (1 + |x −
y|/t)−N , and the following general estimate on HL1 -atoms:
LEMMA 4.17 Let qt (x, y) be a function satisfying
³
|qt (x, y)| ≤ cN 1 +
t
ρ(x)
+
t
ρ(y)
´−N
t−d (1 + |x − y|/t)−N .
Then, for every HL1 -atom g supported by Br (y0 ), there is Cy0 ,r > 0 such that
¯Z
¯
¯
¯
¯
Mq g(x) = sup ¯
qt (x, y)g(y) dy ¯¯ ≤ Cy0 ,r (1 + |x|)−(d+1) , x ∈ Rd .
t>0
PROOF:
Rd
(4.18)
(4.19)
There is no loss of generality in assuming that r < 2ρ(y0 ). Obviously,
¯Z
¯
¯
¯
(4.20)
¯ qt (x, y)g(y) dy ¯ ≤ CkgkL∞ .
If x ∈
/ B2r (y0 ) then for y ∈ Br (y0 ) we have |x − y| ∼ |x − y0 |, ρ(y0 ) ∼ ρ(y). Hence,
applying (4.18), we get
¯Z
¯
³
t ´−N −d ³
|x − y0 | ´−N
¯
¯
t
1+
¯ qt (x, y)g(y) dy ¯ ≤ CN kgkL1 1 +
ρ(y0 )
t
≤ CN |x − y0 |−d−N ρ(y0 )N
Now (4.19) easily follows from (4.20) and (4.21).
(4.21)
2
Finally, in order to complete proof of the lemma we also need to justify the estimate
¯Z N
¯
¯
¯
dt
2
Qt g(y) ¯¯ ≤ Cy0 ,r (1 + |y|)−(d+1) , y ∈ Rd .
sup ¯¯
(4.22)
t
ε,N >0
ε
17
Indeed, such bound allows passing the limit inside the integral in (4.16), and conclude
from (4.2) that
Z
1
I=
f (y) g(y) dy.
8 Rd
For the justification
one defines a new kernel Hε (x, y) as the one associated
R ∞ of2 (4.22)
dt
to the operator ε Qt g(y) t and observes that
¯Z
¯
¯
N
ε
Q2t g(y) dtt
¯
¯Z
¯
¯
¯ ≤ sup¯
ε>0
Rd
¯
¯Z
¯
¯
Hε (x, y)g(y) dy ¯ + sup ¯
N >0
Rd
¯
¯
HN (x, y)g(y) dy ¯.
So, it suffices to verify that Hε (x, y) satisfies the assumption of Lemma 4.17. But
this is an immediate consequence of the same properties for the kernels Tt2 (x, y) and
Qt (x, y), due to the identity
¡
¢
Hε (x, y) = 18 T2ε2 (x, y) − Q√2 ε (x, y) .
The verification of this last identity is left to the reader, which may use spectral
techniques as in the proof of (4.2). This establishes Lemma 4.10, and completes the
proof of Theorem 1.14.
2
5
5.1
Operators acting on BM OL
The Hardy-Littlewood maximal operator
Bennet, DeVore and Sharpley [1] proved that for a function f ∈ BM O, the (uncentered) Hardy-Littlewood maximal function M f (x) (see (1.6)) is either identically
infinity or belongs to BM O, with norm kM f kBM O ≤ Ckf kBM O . The first situation
occurs when f grows at infinity, e.g. with f (x) = log+ |x|.
Below we show that M f (x) is always well-defined when f ∈ BM OL , and moreover,
that M preserves this space. This extends some known properties about the action
of M over the local bmo space of Goldberg.
THEOREM 5.1 The operator M maps BM OL into BM OL . Moreover, there exists
C > 0 such that kM f kBM OL ≤ Ckf kBM OL for all f ∈ BM OL .
PROOF: We first show that, for every f ∈ BM OL the maximal function is finite
a.e. This is a consequence of the following lemma.
LEMMA 5.2 Given f ∈ BM OL , x0 ∈ Rd and any C0 ≥ 1, then M f (x) < ∞ at
almost every x ∈ B0 = B(x0 , C0 ρ(x0 )).
18
PROOF: Let us split f = f1 + f2 , with f1 (x) = f (x)χB ∗ . Since every function
0
in BM OL is locally integrable, we have M f1 (x) < ∞, for a.e. x ∈ Rd . To bound
the second term, we use that supp f2 ⊂ (B0∗ )c . Then, we may compute M f2 (x) at
x ∈ B0 by just considering the integrals over balls B 3 x for which B ∩ (B0∗ )c 6= ∅,
and therefore with diameter 2r ≥ C0 ρ(x0 ). Thus, we have B ⊂ B4r (x0 ) = B̃, and
consequently, by the definition of BM OL (observe that r(B̃) = 4r > ρ(x0 ))
Z
Z
1
4d
|f2 (y)| dy ≤
|f (y)| dy ≤ c kf kBM OL .
(5.3)
|B| B
|B4r (x0 )| B4r (x0 )
2
We turn to the boundedness of M in BM OL . By invoking the results in [1] and
the definition of BM OL , it suffices to prove that for every B = Br (x0 ) with r ≥ ρ(x0 )
Z
1
|M f (y)| dy ≤ Ckf kBM OL .
(5.4)
|B| B
To show this, we split f = f1 + f2 , with f1 (x) = f (x)χB ∗ . From (5.3) it follows that
M f2 (x) ≤ ckf kBM OL for all x ∈ B, so (5.4) follows with f replaced by f2 . For the
other term, we may use the boundedness of M in L2 to obtain:
Z
³ 1 Z
´1/2
1
|M f1 (y)| dy ≤
|M f1 (y)|2 dy
|B| B
|B| B
³ 1 Z
´1/2
≤ C
|f (y)|2 dy
. kf kBM OL ,
|B| B ∗
where in the last inequality we have used Corollary 3.13.
5.2
2
Semigroup maximal functions
In our setting, we are also interested in maximal functions arising from the semigroup
Tt . These are more naturally related with the definition of our spaces (recall the
definition of HL1 in §1), and give us information about the solution to the evolution
equation in (4.8).
More precisely, we shall consider T ∗ and P ∗ , that is, the “heat” and the “Poisson”
maximal functions related to L (see (1.7) and (1.8)). We observe that, in general,
these maximal operators are not bounded in the classical BM O(Rd ). E.g., when
V (x) = |x|2 (Hermite operator), then T ∗ 1(x) is not a constant function, and so T ∗
cannot be bounded in BM O (see [11]). Our next result shows that the natural space
for boundedness of T ∗ and P ∗ is BM OL .
THEOREM 5.5 Let f ∈ BM OL . Then T ∗ f and P ∗ f belong to BM OL , and moreover, there exists C > 0 such that
kT ∗ f kBM OL + kP ∗ f kBM OL ≤ C kf kBM OL .
19
For a fixed ball B ⊂ Rd we say that f ∈ BM O(B) if
Z
1
kf kBM O(B) =
sup
|f (y) − fBr (x) | dy
Br (x)⊂B |Br (x)| Br (x)
Z
1
|f (y) − c| dy.
∼
sup inf
Br (x)⊂B c∈C |Br (x)| Br (x)
is finite.
PROOF of Theorem 5.5:
lemma.
(5.6)
We will use systematically the following elementary
LEMMA 5.7 Let h ∈ BM O(Q∗k ) and g1 and g2 be functions in L∞ . If f is any
measurable function satisfying
h − g1 ≤ f ≤ h + g2 ,
a.e.
then f ∈ BM O(Q∗k ) and kf kBM O(Q∗k ) ≤ khkBM O(Q∗k ) + max{kg1 kL∞ , kg2 kL∞ }.
We shall consider first the operator T ∗ . By the definition of BM OL and Proposition 2.1 it suffices to prove the following: there exists a constant C > 0 such that for
every fixed Qk ∈ Q (see Proposition 2.14), we have
Z
1
(i)
|T ∗ f (x)| dx ≤ C kf kBM OL ;
|Qk | Qk
(ii) kT ∗ f kBM O(Q∗k ) ≤ C kf kBM OL .
Observe that (i) implies the almost everywhere finiteness of the operator. This
part is immediate from Lemma 5.2 and (5.4), since T ∗ f (x) ≤ supt>0 |f | ∗ ht (x) .
M |f |(x), and therefore
Z
Z
1
1
∗
|T f (x)| dx .
M |f |(x) dx ≤ C kf kBM OL .
|Qk | Qk
|Qk | Qk
We just need to show (ii), which we shall split into three different estimates:
°
°
°
°
(5.8)
° sup |Tt f (x)|° ∞ ∗ ≤ C kf kBM OL ,
L (Qk )
t≥ρ(xk )2
°
°
°
°
(5.9)
° sup |(Tt − T̃t )f (x)|° ∞ ∗ ≤ C kf kBM OL ,
L (Qk )
t≤ρ(xk )2
°
°
°
°
sup
≤ C kf kBM OL ,
(5.10)
|
T̃
f
(x)|
°
°
t
∗
t≤ρ(xk )2
BM O(Qk )
where T̃t f = f ∗ ht . It is not difficult to verify that (ii) follows from these three
estimates by using Lemma 5.7.
20
The proof of each of them requires a different use of the decay and the smoothness
of the kernels. For instance, (5.8) follows easily from (2.5). More precisely,
Z
√
|Tt f (x)| .
|f (y)| t−d/2 (1 + |x − y|/ t)−M dy
Rd
Z
∞
X
1 1
. d/2
|f (y)| dy +
|f (y)| dy.
√
t
2jN td/2 |x−y|∼2j √t
|x−y|≤ t
j=1
√
Now, for j ≥ 0, we have 2j t ≥ ρ(xk ) ∼ ρ(x) for every x ∈ Q∗k , thus
Z
Z
1
2jd
|f (y)| dy .
|f (y)| dy ≤ 2jd kf kBM OL .
td/2 |x−y|∼2j √t
|B2j √t (x)| B2j √t (x)
1
Z
Therefore,
sup |Tt f (x)| .
t≥ρ(xk )2
∞
X
j=0
1
2j(N −d)
kf kBM OL = Ckf kBM OL .
(5.11)
To prove (5.9) we shall need the following estimates on the difference kt − ht ,
which can be found in [3].
LEMMA 5.12 (See Proposition 2.16 in [3].) There exists a nonnegative Schwartz
class function w in Rd so that
 µ √ ¶δ
√


 ρ(x)t wt (x − y), for t ≤ ρ(x)


µ √ ¶δ
√
|ht (x − y) − kt (x, y)| ≤
(5.13)
t

wt (x − y), for t ≤ ρ(y)

ρ(y)



wt (x − y), elsewhere,
√
where wt (x − y) = t−d/2 w((x − y)/ t).
√
Going back to (5.9), since ρ(x) ∼ ρ(xk ) for all x ∈ Q∗k , and in this case t ≤ ρ(xk ),
we can use Lemma 5.12 and proceed as in the previous estimate to obtain:
Z µ √ ¶δ
t
|(Tt − T̃t )f (x)| ≤
wt (x − y)|f (y)| dy
ρ(x)
Rd
µ
.
µ
=
√
t
ρ(xk )
¶δ X
∞
Z
−j(N −d)
2
|B2j √t (x)|
j=0
√
t
ρ(xk )
¶δ µ
X
1≤2j ≤
+
1
Z
−j(N −d)
2
ρ(x )
2j > √k
t
1
|B2j √t (x)|
ρ(xk )
√
t
X
|f (y)| dy
B2j √t (x)
|f (y)| dy
B2j √t (x)
¶
−j(N −d)
2
kf kBM OL .
21
Now, by Lemma 3.14, for j such that 1 ≤ 2j ≤
ρ(xk )
√ ,
t
we have
Z
1
|B2j √t (x)|
B2j √t (x)
ρ(xk )
√ .
√
|f (y)| dy . 1 + log 2ρ(x)
j t . 1 + log
t
Therefore,
µ
|(Tt − T̃t )f (x)| .
√
t
ρ(xk )
¶δ µ
¶
1 + log
ρ(xk )
√
t
kf kBM OL
∞
X
2−j(N −d) . kf kBM OL .
j=0
Finally, let us sketch the proof of (5.10). This seems to be a classical result,
following from a vector-valued singular integral theory. Consider B = Br (x0 ) ⊂ Q∗k
and split f as
£
¤
f = (f − fB )χB ∗ + (f − fB )χ(B ∗ )c + fB = f1 + f2 .
(5.14)
Denote kT̃t f (x)k`∞ (t) = supt≤ρ(xk )2 |T̃t f (x)| and choose cB = kT̃t f2 (x0 )k`∞ (t) , which is
a finite real number by (5.4). Then,
Z
Z
¯
¯
1
¯ sup |T̃t f (x)| − cB ¯ dx ≤ 1
kT̃t f (x) − T̃t f2 (x0 )k`∞ (t) dx
|B| B t≤ρ(xk )2
|B| B
Z
Z
1
1
≤
kT̃t f1 (x)k`∞ (t) dx +
kT̃t f2 (x) − T̃t f2 (x0 )k`∞ (t) dx = I + II.
|B| B
|B| B
For the first integral, observe that by the L2 boundedness of supt≤ρ(xk )2 |T̃t f (x)|, we
have
µ
¶1/2
Z
1
2
I≤
|f (x) − fB | dx
≤ Ckf kBM O ≤ Ckf kBM OL .
|B| B ∗
For the second integral, the standard arguments of singular integrals and the smoothness of the kernel will give:
Z
1
kT̃t f2 (x) − T̃t f2 (x0 )k`∞ (t) dx
|B| B
°
Z °Z
°
°
1
°
°
(h
(x,
y)
−
h
(x
,
y))(f
(y)
−
f
)
dy
≤
t
t
0
B
° ∞ dx ≤ Ckf kBM O .
|B| B ° (B ∗ )c
` (t)
It should be observed that in this last step the constant C is independent of Qk .
We now turn to the Poisson maximal function P ∗ f , for which we indicate the main
differences with respect to the previous proof. As before, it is enough to prove (i)–(ii)
with T ∗ replaced by P ∗ . By subordination we have P ∗ f (x) . T ∗ f (x). Hnence there is
22
no problem in verifying (i). It thus suffices to prove that kP ∗ f kBM O(Q∗k ) ≤ Ckf kBM OL
for any fixed Qk ∈ Q. Set ρ = ρ(xk ). Observe that
P1∗ f (x) − P2∗ f (x) ≤ P ∗ f (x) ≤ P1∗ f (x) + P2∗ f (x),
where
¯Z ∞
¯
¯
¯
e−u
¯
√ Tt2 /4u f (x) du¯¯,
= sup ¯
u
t≥0
t2 /4ρ2
¯
¯ Z t2 /4ρ2 −u
¯
¯
e
∗
√ Tt2 /4u f (x) du¯¯.
P2 f (x) = sup ¯¯
u
t≥0
0
P1∗ f (x)
Moreover, |P2∗ f (x)| ≤ sups≥ρ2 |Ts f (x)|, which by (5.8), belongs to L∞ (Q∗k ). An application of Lemma 5.7 reduces matters to show that P1∗ f is in BM O(Q∗k ). We shall
repeat this argument two more times. Indeed
∗
∗
∗
∗
P11
f (x) − P12
f (x) ≤ P1∗ f (x) ≤ P11
f (x) + P12
f (x),
where
¯Z ∞
¯
= sup ¯¯
t≥0
t2 /4ρ2
¯Z ∞
¯
∗
P12
f (x) = sup ¯¯
∗
P11
f (x)
t≥0
t2 /4ρ2
¯
¯
e−u
√ T̃t2 /4u f (x) du¯¯,
u
¯
−u
¯
e
√ (Tt2 /4u − T̃t2 /4u )f (x) du¯¯.
u
∗
Now P12
f (x) ≤ C sups≤ρ(xk )2 |(Ts − T̃s )f (x)|, which belongs to L∞ by (5.9). Next for
∗
the remaining term P11
f (x) we have
∗
∗
∗
∗
∗
P111
f (x) − P112
f (x) ≤ P11
f (x) ≤ P111
f (x) + P112
f (x),
where
¯ Z ∞ −u
¯
¯
¯
e
√ T̃t2 /4u f (x) dx¯¯,
= sup ¯¯
u
t≥0
0
¯
¯ Z t2 /4ρ2 −u
¯
¯
e
∗
√ T̃t2 /4u f (x) dx¯¯.
P112 f (x) = sup ¯¯
u
t≥0
∗
P111
f (x)
0
∗
f (x)| ≤ sups≥ρ2 |T̃s f (x)| which is bounded by the same reasoning as
Again, |P112
∗
f , which is the classical
in (5.8). Thus, we have reduced matters to show that P111
∗
Poisson maximal function, belongs to BM O(Qk ). This will again follow from classical
arguments of vector-valued singular integrals. Skipping these details, the proof of the
theorem is now complete.
2
23
REMARK 5.18 It is interesting to observe that an analog of the previous theorem
holds as well for the non-tangential maximal operator
T ∗∗ f (x) = sup |Tt f (y)|,
x ∈ Rd .
|x−y|<t
In fact, the reader can easily check that the same proof above goes along in this more
general situation.
5.3
Fractional integrals
In this section we shall be interested in the behavior of the fractional integral operator
Iα = L−α/2 (see (1.10)).
Recall that in the classical setting Iα = (−∆)−α/2 maps Lp (Rd ) into Lq (Rd ) for
0 < α < d and 0 < p < q < ∞ with 1/p = 1/q − α/d. Moreover, there is a
dichotomy in the limiting case q = ∞: for every f ∈ Ld/α (Rd ), either Iα f ≡ ∞
or Iα f ∈ BM O(Rd ) with kIα f kBM O ≤ C kf kLd/α (Rd ) (see page 221 in [10]). Simple
1
examples like f (x) = |x|α log
show that Iα f may be identically ∞.
|x|
Our next result shows that such pathological behaviors cannot happen to the
operator Iα , when the potential V 6≡ 0. Moreover, the natural target space is now
BM OL .
THEOREM 5.19 For all 0 < α < d, the operator Iα is bounded from Ld/α (Rd ) into
BM OL , that is, there is a constant C > 0 so that
for every f ∈ Ld/α (Rd ).
kIα f kBM OL ≤ C kf kLd/α ,
PROOF: The proof follows the same scheme as in the previous section. We shall
try to establish the analogs of (i)-(ii) with T ∗ replaced by Iα . To see (i), let us split
Z
Z
ρ(x)2
Iα f (x) =
Tt f (x)t
α/2−1
∞
dt +
Tt f (x)tα/2−1 dt = I1 f (x) + I2 f (x). (5.20)
ρ(x)2
0
For the first integral, we use the trivial estimate
Z
ρ(x)2
|I1 f (x)| ≤
T ∗ f (x)tα/2−1 dt . ρ(x)α M |f |(x).
(5.21)
0
For the second integral, the extra decay in time of kt (x, y) gives
¶N
Z ∞ Z µ
ρ(x)
√
hct (x − y)|f (y)| dy tα/2−1 dt
|I2 f (x)| ≤ CN
2
d
t
ρ(x)
R Z
∞
tα/2−N/2−1 dt . ρ(x)α M |f |(x).
. M |f |(x) ρ(x)N
ρ(x)2
(5.22)
24
Thus, combining these two estimates, and using ρ(x) ∼ ρ(xk ) if x ∈ Qk , we obtain
Z
Z
1
1
|Iα f (x)| dx .
|M f (x)| dx ≤ kM f kLd/α . kf kLd/α ,
|Qk | Qk
|Qk |1−α/d Qk
where in the last two steps we used Hölder’s inequality and the boundedness of M in
Ld/α .
We pass now to prove the analog of assertion (ii), that is, kIα f kBM O(Q∗k ) ≤
C kf kBM OL . As before, the strategy is based on an iterative use of Lemma 5.7.
Splitting Iα f = I1 f (x) + I2 f (x) as in (5.20), we shall show that I1 f ∈ BM O(Q∗k )
and I2 f ∈ L∞ (Q∗k ), with norms controlled by kf kLd/α . For the second term we must
slightly refine the argument in (5.22): if x ∈ Q∗k then
Z ∞ Z
|x−y|2
1
N
− 4ct
|I2 f (x)| . ρ(xk )
e
|f (y)| dy t−N/2−1 dt.
(d−α)/2
t
ρ(xk )2 Rd
Now, observe that
Z
|x−y|2
t−(d−α)/2 e− 4ct |f (y)| dy .
0 2j Z
∞
X
2j(d−α) e−c 2
√
|f (y)| dy
j t)d−α
|x−y|
(2
√ ≤2j
j=0
Z t
1
|f (y)| dy =: Mα f (x),
. sup
1−α/d
r>0 |Br (x)|
Br (x)
Rd
where Mα denotes the fractional maximal operator, which trivially maps Ld/α (Rd )
into L∞ (Rd ). Thus,
Z ∞
N
|I2 f (x)| . ρ(xk ) Mα f (x)
t−N/2−1 dt . kf kLd/α (Rd ) .
(5.23)
2
ρ(xk )
To deal with I1 f we make some further splittings:
Z ρ(xk )2
Z ρ(xk )2
α/2−1
I1 f =
T̃t f t
dt +
(Tt − T̃t )f tα/2−1 dt = I11 f + I12 f.
0
0
A new use of Lemma 5.12, Proposition 2.1, and reasoning with a similar estimate to
that presented above gives
Z ρ(xk )2 Z µ √ ¶δ
t
|I12 f (x)| .
wt (x − y)|f (y)| dy tα/2−1 dt
0
Rd ρ(x)
Z ρ(xk )2 δ/2−1
t
.
Mα f (x) dt . kf kLd/α (Rd )
ρ(x)δ
0
for all x ∈ Q∗k .
25
It remains to control the term I11 f , for which we must show that, given any ball
B = Br (x0 ) ⊂ Q∗k there is a constant cB so that
Z
1
| I11 f (x) − cB | dx ≤ C kf kLd/α (Rd ) .
(5.24)
|B|
B
This is elementary to verify with cB = 0 when the radius of the ball B is comparable
to ρ(xk ). Indeed, in this case Hölder’s and Minkowski’s inequalities give
µ
Z
1
|B|
| I11 f (x)| dx ≤
B
¶α/d
Z
1
|B|
| I11 f (x)|
dx
B
Z
1
|B|α/d
≤
d/α
ρ(xk )2
0
kT̃t f kLd/α tα/2−1 dt . kf kLd/α .
(5.25)
Suppose instead we are given a ball with r ¿ ρ(xk ). In this case we must further
split the integral defining I11 f (x) into two new pieces:
Z
Z
r2
I11 f (x) =
T̃t f (x)t
α/2−1
ρ(xk )2
dt +
T̃t f (x)tα/2−1 dt = I111 f (x) + I112 f (x).
r2
0
R
1
For the first piece we can repeat the previous calculation to obtain |B|
|I f (x)| dx .
B 111
kf kLd/α . For the second piece we write f = f1 + f2 with f1 (x) = f (x)χB ∗ (x), and
choose cB = I112 f2 (x0 ) (which is a finite number for a.e. x0 ). Then,
Z
Z
Z
1
1
1
|I112 f (x) − cB | dx ≤ |B|
|I112 f1 (x)| dx + |B|
|I112 f2 (x) − I112 f2 (x0 )| dx.
|B|
B
B
B
When x ∈ B, the first integrand is uniformly bounded by
Z
ρ(xk )2
Z
r2
ρ(xk )2
Z
t−d/2 e−
|I112 f1 (x)| .
Z
|f (y)| dy tα/2−1 dt
B∗
|f (y)| dy t−d/2+α/2−1 dt
.
r2
|x−y|2
4t
|x−y|<3r
Z ∞
d−α
−d/2+α/2−1
. Mα f (x) r
t
r2
dt ≤ kf kLd/α .
For the second integrand we use smoothness
¯
¯
¯ Z ρ(xk )2 Z
¯
|x0 −y|2 ¯
1 ¯¯ − |x−y|2
¯
¯
−
e 4t − e 4t ¯¯|f (y)| dy tα/2−1 dt
¯I112 f (x) − I112 f (x0 )¯ .
¯
d/2
t
r2
(B ∗ )c
¯
¯
Z ρ(xk )2 X
Z
∞
¯ − |x−y|2
|x0 −y|2 ¯
1
−
¯e 4t − e 4t ¯|f (y)| dy tα/2−1 dt.
.
¯
¯
d/2
t
j
2
r
j=1
|x0 −y|∼2 r
26
√
By the mean value theorem (and the fact |x − x0 | < r < t) we have:
Z
Z
¯ |x−y|2
¯
|x −y|2
¯ − 4t
|x−x0 | −c(2j r)2 /t
− 04t ¯
√
−e
e
¯e
¯|f (y)| dy .
t
|x0 −y|∼2j r
|f (y)| dy
B2j r (x0 )
|x−x0 |
√
t
.
(2j r)d−α Mα f (x0 )
√
,
(1 + 2j r/ t)N
for a sufficiently large integer N . Hence,
¯
¯
¯I112 f (x) − I112 f (x0 )¯ .
Z
ρ(xk )2
−d/2+α/2−1
t
∞
hX
r2
(2j r)d−α
√
(1+2j r/ t)N
i
|x−x0 |
√
t
dt Mα f (x0 ) .
j=1
Breaking
√ the inner sum into two parts it is easy to see that it is bounded by a constant
times ( t)d−α . Therefore
Z ∞
¯
¯
¯I112 f (x) − I112 f (x0 )¯ . Mα f (x0 )|x − x0 |
t−1/2−1 dt . Mα f (x0 ) . kf kLd/α .
r2
From here it is immediate that
the assertion (ii).
1
|B|
¯
R ¯
¯I112 f (x)−I112 f (x0 )¯ dx . kf kLd/α , establishing
B
2
5.4
Square functions
Consider the following square function associated with L:
³Z ∞
´1/2
2 dt
t
s(f )(x) =
|t dT
f
(x)|
.
dt
t
(5.26)
0
This is just a multiple of the square function sQ f (x) defined in (1.9).
THEOREM 5.27 There exists a constant C > 0 such that
ks(f )kBM OL ≤ Ckf kBM OL .
(5.28)
PROOF: Fix f ∈ BM OL and Qk ∈ Q with center xk . It is not difficult to prove
using (2.11) that
Z
1
|s(f )(x)| dx . kf kBM OL .
(5.29)
|Qk | Qk
In fact, if we split
Z
2
2
Z
ρ(xk )2
2
|s(f )(x)| = |s1 f (x)| + |s2 f (x)| :=
0
t
|t dT
f (x)|2 dtt
dt
∞
+
ρ(xk
t
|t dT
f (x)|2
dt
)2
dt
,
t
27
then (5.29) for s1 (f ) was shown in (4.5). The second term s2 (f )(x) has a uniform
bound when x ∈ Qk . In fact, by using (a) in Proposition 2.9 and the same reasoning
as in (5.11), we get
Z ∞ hZ
i2
¡ ρ(x) ¢N −d/2 − c|x−y|2
2
dt
t
√
|s2 f (x)| .
t
e
|f
(y)|
dy
t
t
h
.
ρ(xk )2
Rd
i2
sup |f | ∗ hs (x) . kf k2BM OL .
s≥ρ(xk )2
To complete the proof, it suffices to show that
ks(f )(x)kBM O(Q∗k ) ≤ Ckf kBM OL
(5.30)
with some constant C independent Qk . Decompose f = f1 + f2 as in (5.14) and set
³R
´1/2
ρ(xk )2 dTet
2 dt
|t dt f2 (x0 )| t
, which is a finite number. Then
cB = 0
³Z
∞
|s(f )(x) − cB | ≤
³Z
ρ(xk )2
t
|t dT
f (x)|2 dtt
dt
ρ(xk )2
+
0
´1/2
³Z
ρ(xk )2
+
0
e
t
|t dT
f (x) − t ddtTt f (x)|2 dtt
dt
e
e
|t ddtTt f (x) − t ddtTt f2 (x0 )|2 dtt
´1/2
´1/2
= S1 (x) + S2 (x) + S3 (x).
One can easily check using standard arguments combined with (2.11) that
Z
¡
¢
1
S1 (x) + S2 (x) dx . kf kBM OL .
|B| B
(5.31)
The proof of (5.30) will be finished if we show that
S3 (x) . kf kBM OL
for x ∈ B.
The perturbation formula implies
Z t/2
Z t
dTet dTt
d e
dTs
e
−
= Tt/2 V Tt/2 +
( Tt−s )V Ts ds +
T̃t−s V
ds.
dt
dt
dt
ds
0
t/2
(5.32)
(5.33)
Thus (5.32) will be established if we verify that
Z ρ(xk )2
Z ρ(xk )2 ¯ Z t/2
¯2
d
¯
¯
2 dt
|tT̃t/2 V Tt/2 f (x)| t +
t( T̃t−s )V Ts f (x) ds¯ dtt
(5.34)
¯
dt
0
0
0
Z ρ(xk )2 ¯ Z t
¯2
¯
¯
dTs
+
tT̃t−s V ds f (x) ds¯ dtt = S30 (x) + S300 (x) + S3000 (x) . kf k2BM OL .
¯
0
t/2
The following lemma is a direct consequence of the definition of BM OL and Lemma
3.14
28
LEMMA 5.35 Let qt (x, y) be a function satisfying (4.18). Then, for every N there
is a constant CN > 0 such that
¯Z
¯
³
´³
t ´−N
¯
¯
ρ(x)
1+
.
(5.36)
¯ qt (x, y)f (y) dy ¯ ≤ CN kf kBM OL 1 + log+ t
ρ(x)
Hence, from Propositions 2.4 and 2.1 we conclude
Z ρ(xk )2 ¯ Z
´ ¯2 dt
³
¯
¯
ρ(z)
0
√
S3 (x) .
kf k2BM OL
¯ tht/2 (x, z)V (z) 1 + log+ t dz ¯
t
0
Z ρ(xk )2 ¯ Z
¯
(5.37)
.
¯ tht/2 (x, z)V (z)
0
¯2 dt
³
√ ¢ k0 ¤´
£
¡
¯
|x−z|
t 1+k0
√
√
× 1 + log+ ρ(x)
1
+
dz
kf k2BM OL .
¯
t
t ρ(x)
t
By using (2.12) and the fact that ρ(x) ∼ ρ(xk ) for x ∈ B, we have
Z ρ(xk )2 ¯³ √ ´δ ³
´¯2 dt
¯
¯
0
√
S3 (x) .
1 + log+ ρ(x)
kf k2BM Oρ . kf k2BM Oρ .
¯ ρ(x)t
¯
t
t
0
The estimates for S300 (x) and S3000 (x) stated in (5.34) could be proved by applying the
d
ks (x, y) established in Section 2.
same arguments and the bounds for ks (x, y) and ds
2
References
[1] C. Bennet, R.A. DeVore and R. Sharpley. “Weak-L∞ and BMO”, Ann.
of Math. 113 (1981), 601–611.
“Hardy space H 1 associated to
[2] J. Dziubański and J. Zienkiewicz,
Schrödinger operator with potential satisfying reverse Hölder inequality”. Rev.
Mat. Iberoam. 15 (2), 1999, 279–296.
[3] J. Dziubański and J. Zienkiewicz, “H p spaces for Schrödinger operators,”
Fourier Analysis and Related Topics, Banach Center Publications, Vol. 56 (2002),
45-53.
[4] J. Dziubański and J. Zienkiewicz, “H p spaces associated with Schrödinger
operators with potentials from reverse Hölder classes”, Colloq. Math. to appear.
[5] D. Goldberg, “A local version of real Hardy spaces” Duke Math. J. 46 (1979),
27-42.
[6] K. Kurata, “An estimate on the heat kernel of magnetic Schrödinger operators
and uniformly elliptic operators with non-negative potentials”, J. London Math.
Soc. (2) 62 (2000), 885–903.
29
[7] Z. Shen, “Lp estimates for Schrödinger operators with certain potentials”. Ann.
Inst. Fourier 45 (2) 1995, 513-546.
[8] E. Stein, Topics in Harmonic Analysis Related to the Littlewood-Paley Theory,
Princeton University Press, Princeton NJ, 1970.
[9] E. Stein, Harmonic Analysis: Real Variable Methods, Orthogonality, and Oscillatory Integrals,. Princeton University Press, 1993.
[10] A. Torchinsky, Real-variable methods in harmonic analysis, Pure and Applied
Mathematics, 123. Academic Press Inc., Orlando, 1986.
[11] K. Stempak, J.L. Torrea, “Endpoint results for Poisson inetgrals and Riesz
transforms for Hermite function expansions”, preprint.
Dziubański and Zienkiewicz
Garrigós, Martı́nez and Torrea
Institute of Mathematics
University of WrocÃlaw
Pl. Grunwaldzki 2/4
50-384 WrocÃlaw, Poland
Departamento de Matemáticas
Universidad Autónoma de Madrid
Ciudad Universitaria Cantoblanco
28049 Madrid, Spain
E-mail address:
[email protected]
[email protected]
[email protected]
[email protected]
[email protected]