The balayage of a Carleson measure lies of course in the space of functions of bounded mean oscillation (BMO). We show that the converse statement is false. We also make a two-sided estimate of the Carleson norm of a positive measure in terms of &lt;i&gt;certain&lt;/i&gt; balayages

Pott, Sandra

Volberg, Alexander

English

arXiv

The balayage of a Carleson measure lies of course in the space of functions of bounded mean oscillation (BMO). We show that the converse statement is false. We also make a two-sided estimate of the Carleson norm of a positive measure in terms of certain balayages

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Copyedited by: Roda Vee Villariza
Pott, S., and A. Volberg. (2009) “Carleson Measure and Balayage,”
International Mathematics Research Notices, Article ID rnp222, 10 pages.
doi:10.1093/imrn/rnp222
Carleson Measure and Balayage 1
Sandra Pott 1 and Alexander Volberg 2 2
1Institut fu¨r Mathematik, Universita¨t Paderborn, Warburger Str. 100, 3
33098 Paderborn, Germany, and 2Department of Mathematics, 4
Michigan State University, East Lansing, MI 48824, USA 5
Correspondence to be sent to: sandrapo@math.uni-paderborn.de 6
The balayage of a Carleson measure lies of course in bounded mean oscillation (BMO). 7
We show that the converse statement is false. We also make a two-sided estimate of the 8
Carleson norm of a positive measure in terms of certain balayages. 9
1 Introduction and Notation 10
In this note, we consider a question that naturally appeared in the recent work of 11
Frazier–Nazarov–Verbitsky [3]. The question is: 12
How does the Carleson norm of a positive measure in the disk relate to the 13
bounded mean oscillation (BMO) norm of its balayage on the circle? 14
A related question is: 15
How can one describe measures on the disk (say, positive measures) whose bala- 16
yage is a BMO function? 17
The second author is grateful to Igor Verbitsky, who called our attention to these 18
questions. 19
We show that the seemingly answer: “These are exactly the Carleson measures” 20
is false. The Carleson property is indeed of course sufficient, but not at all necessary. 21
Received 7 August 2009; Revised 11 January 2009; Accepted 21 January 2009
Communicated by A. Editor
c© The Author 2009. Published by Oxford University Press. All rights reserved. For permissions,
please e-mail: journals.permissions@oxfordjournals.org.
2 S. Pott and A. Volberg
However, we can characterize the Carleson property in terms of the BMO norms of the 22
balayages of restrictions of the measure. 23
Throughout the paper, we will use the notation ,  for one-sided estimates up 24
to an absolute constant, and the notation ≈ for two-sided estimates up to an absolute 25
constant. 26
We will use the setting of the upper half plane R2+ rather than the unit disk. Given 27
a positive regular Borel measure μ on the upper half plane R2+ = {(t, y) ∈ R2 : y > 0}, its 28
balayage is defined as the function 29
Sμ(t) =
∫
R
2+
px,y(t)dμ(t, y),
where px,y(t) = 1π xy2+(t−x)2 is the Poisson kernel for R2+. We say that μ is a Carleson 30
measure if there exists a constant C > 0 such that for each interval I ⊂ R, the 31
inequality 32
μ(QI ) ≤ C |I | (1)
holds. Here, QI denotes the Carleson square {(x, y) : x ∈ I, 0 < y ≤ |I |} over I . It is easy 33
to see that it is sufficient to consider dyadic intervals in this definition. We denote the 34
infimum of all constants C > 0 such that (1) holds for all dyadic intervals by Carl(μ). 35
Recall that the space of functions of BMO (R) is defined as 36
{
b ∈ L2(R) : sup
I⊂R interval
1
|I |
∫
I
|b(t) − 〈b〉I |dt < ∞
}
,
with ‖b‖BMO = supI⊂R interval 1|I |
∫
I |b(t) − 〈b〉I |dt. By the John–Nirenberg inequality, the 37
L1 norm in the definition of BMO can be replaced by any ‖ · ‖p norm, 1 ≤ p < ∞. We thus 38
obtain a family of equivalent norms on BMO(R), with equivalent constants depending 39
on p. 40
The connection between the properties of a measure μ and its balayage Sμ have 41
long been studied. In particular, it is well known that the BMO norm of Sμ is controlled 42
by the Carleson constant of μ, 43
‖Sμ‖BMO  Carl(μ). (2)
For this and other basic facts on BMO functions, we refer the reader to [4]. 44
Carleson Measure and Balayage 3
A partial reverse of (2) was found in [2], [7], and in the dyadic case, [5]. Namely, 45
it was shown that for each b ∈ BMO, there exists an L∞(R) function φ and a Carleson 46
measure μ such that b = φ + Sμ, ‖φ‖∞ + Carl(μ)  ‖b‖BMO. If we allow μ to be a complex 47
measure, one even has the representation b = Sμ with Carl(μ)  ‖b‖BMO [6]. 48
The purpose of this note is to show that reverse inequality to (2) in the strict 49
sense does not hold, and to give a characterization of the Carleson property of a measure 50
μ in terms of the BMO norm of the balayage of restrictions of μ. 51
2 The Dyadic Balayage 52
We start by examining the dyadic case. We will use the standard Whitney-type decom- 53
position of the upper half plane, indexed by the set D of left-half open dyadic intervals 54
in R, 55
TI =
{
(x, y) : x ∈ I, |I |
2
< y ≤ |I |
}
for I ∈ D.
That means, TI is the “top half” of the Carleson square QI defined above. 56
For a positive regular Borel measure μ on R2+, we define the dyadic balayage by 57
Sdμ(t) =
∑
I∈D
χI (t)
|I | μ(TI ) (t ∈ R),
which is well defined as a function taking values in [0,∞]. By comparing box kernel and 58
Poisson kernel, one easily verifies the pointwise estimate Sdμ  Sμ. 59
We recall the definition of dyadic BMO, BMOd(R), as the class of L2(R) functions 60
for which 61
‖b‖2
BMOd
= sup
I∈D
1
|I |
∫
I
|b(t) − 〈b〉I |2dt = sup
I∈D
1
|I | ‖PI b‖
2 = sup
I∈D
1
|I |
∑
J∈D,J⊆I
|bI |2
is finite. Here, hJ denotes the L2-normalized Haar function, bJ := (b, hJ) denotes the cor- 62
responding Haar coefficient of function b, and PI denotes the orthogonal projection on to 63
span{hJ : J ⊆ I }. Again, by the John–Nirenberg inequality the L2 norm in the definition 64
can be replaced by any L p norm, 1 ≤ p < ∞, yielding an equivalent norm. 65
4 S. Pott and A. Volberg
We say that a sequence of nonnegative numbers (αI )I∈D is a Carleson sequence, 66
if there exists a constant C > 0 such that 67
1
|I |
∑
J∈D,J⊆I
aI ≤ C for each I ∈ D.
Again, we denote the infimum of such constants by Carl((aI )). With this notation, one 68
verifies immediately the following well-known lemma. 69
Lemma 2.1. Let b ∈ L2(R). Then the following are equivalent: 70
1. μ is a Carleson measure 71
2. (μ(TI ))I∈D is a Carleson sequence 72
3. bμ =∑I∈D hI μ(TI )1/2 ∈ BMOd(R). 73
In this case, Carl(μ) = Carl((μ(TI ))) = ‖bμ‖2BMOd.  74
Notice that with the above definition of bμ, 75
Sdμ =
∑
I∈D
χI
|I |μ(TI ) =
∑
I∈D
χI
|I | |(bμ)I |
2 = S[bμ],
where S denotes the square of the dyadic square function, S[ f] =∑I∈D χI|I | | fI |2 for f ∈ 76
L2(R). In this sense, we have identified the dyadic balayage of a positive regular Borel 77
measure μ with the square of a dyadic square function of bμ. Conversely, for any f ∈ 78
L2(R), S[ f] can be written as a dyadic balayage of a measure μ f , for example by letting 79
μ f =∑I∈D | fI |2δz(I ), z(I ) denoting the center of TI . 80
The well-known dyadic analog of (2) is therefore equivalent to the inequality 81
‖S[b]‖BMOd  ‖b‖2BMOd , (3)
which can be now be proved as a simple application of the John–Nirenberg inequality.
Notice that for any dyadic inverval I ∈ D, all summands in S[b] =∑J∈D χJ|J| |bJ |2 except
those corresponding to dyadic intervals J ⊂ I are constant on I . Thus
1
|I |
∫
I
|S[b](t) − 〈S[b]〉I |dt = 1|I |
∫
I
|S[PI b](t) − 〈S[PI b]〉I |dt
≤ 1|I |
∫
I
S[PI b](t)dt + 〈S[PI b]〉I = 2 1|I |
∫
I
∑
J⊆I
χJ(t)
|J| |bJ |
2dt = 2‖PI b‖22 ≤ 2‖b‖2BMOd ,
which proves (3). 82
Carleson Measure and Balayage 5
Here are the main results of this section, which concern the reverse inequality 83
to (3). The first says that the BMO norm of the dyadic balayage can be very much smaller 84
than the Carleson constant of a measure, even if one increases the BMO norm by the L2 85
norm. 86
Theorem 2.2. Let ε > 0. Then there exists a Carleson measure μ on R2+ with Carl(μ) = 1, 87
‖Sdμ‖BMO + ‖Sdμ‖2 < ε.  88
Proof. By Lemma 2.1 and the argument following it, we want to find a BMOd(R) func- 89
tion b of norm 1 such that both the BMOd norm and the L2 norm of S[b] are small. To this 90
end, let I0 = (0, 1], I−1 = (−2, 0], Ik = (2k − 1, 2k+1 − 1] for k > 0 and Ik = (−2−k,−2−k−1] 91
for k < 0. In particular, |Ik| = 2|k| for all k ∈ N. Let r1 denote the first Rademacher func- 92
tion on R, r1 =∑ j∈Z(−χ( j, j+ 12 ] + χ( j+ 12 , j+1]), and let rn = r1(2n−1·) be the nth Rademacher 93
function on R. Let N ∈ N, N to be determined later, and let 94
b =
∞∑
k=−∞
N−|k|∑
n=1
χIk(t)rn(t).
One verifies without difficulty that ‖b‖2
BMOd
= N. Clearly, 95
S[b] =
∞∑
k=−∞
N−|k|∑
n=1
χIk =
N∑
k=0
(N − k)χIk∪I−k.
This is a “dyadic log”, and it is not difficult to show that 96
‖S[b]‖BMO ≤ C ,
where C is an absolute constant independent of N. Notice that we have an estimate here 97
not only for the dyadic BMO norm, but for the full BMO norm. 98
Now choose N so large that CN <
ε
2 and replace b by
1
N1/2
b. This already guaran- 99
tees that ‖b‖2
BMOd
= 1, ‖S[b]‖BMO < ε2 . To deal with the desired L2 estimate, observe that 100
the estimates achieved so far do not change at all if b is dilated with an integer power of 101
2. By choosing a suitable power 2K of 2, K ∈ N, and replacing b by b(2K ·), we obtain the 102
desired estimate 103
‖b‖2
BMOd
= 1, ‖S[b]‖BMO + ‖S[b]‖2 < ε. 
6 S. Pott and A. Volberg
The next theorem says that we can retrieve the Carleson constant of a measure 104
up to an absolute constant from its dyadic balayage, if we restrict the measure to certain 105
sets. 106
Theorem 2.3. Let μ be Carleson measure μ on R2+. Then 107
Carl(μ) ≈ sup
E⊆R2+,E Borel set
‖SdμE ‖BMOd ≈ sup
I∈D
‖SdμQI ‖BMOd .
Here, μE stands for the restriction of μ to E , given by μE (A) = μ(E ∩ A).  108
Proof. Clearly, Carl(μE ) ≤ Carl(μ) for each Borel set E ⊆ R2+, so 109
sup
I∈D
‖SdμQI ‖BMOd ≤ sup
E⊆R2+,E Borel set
‖SdμE ‖BMOd  sup
E⊆R2+,E Borel set
Carl(μE ) ≤ Carl(μ).
To prove the reverse inequality, let I ∈ D. Observe that SdμQI is supported on the closure 110
of I . Therefore, with I ′ denoting the dyadic sibling of I , we have 111
‖SdμQI ‖BMOd ≥ |〈S
d
μQI
〉I − 〈SdμQI 〉I ′ | = 〈S
d
μQI
〉I = 1|I |
∫
I
∑
J∈D,J⊆I
χJ(t)
|J| μ(TJ)dt =
1
|I |μ(QI ).
Thus, Carl(μ)  supI∈D ‖SdμQI ‖BMOd.  112
3 The Algebra of Paraproducts 113
This section contains a short operator-theoretic motivation for the choice of the coun- 114
terexample, in particular the appearance of Rademacher functions, in the previous sec- 115
tion, in terms of paraproducts. Recall that for b ∈ L2(R), the standard dyadic paraprod- 116
uct πb is defined by 117
πb f =
∑
I∈D
hI bI 〈 f〉I for f ∈ L∞(R) ∩ L2(R).
It is well known, and indeed a reformulation of the classical Carleson Embedding Theo- 118
rem, that πb extends to a bounded linear operator on L2(R), if and only if b ∈ BMOd(R). 119
In this case, ‖πb‖ ≈ ‖b‖BMOd. 120
Carleson Measure and Balayage 7
Such dyadic paraproducts have the nice property that π∗bπb is essentially a dyadic 121
paraproduct again, with symbol S[b] (see [1]): 122
π∗bπb = πS[b] + (πS[b])∗ + Diag(b), (4)
where Diag(b) denotes the diagonal of π∗bπb with respect to the Haar basis, Diag(b)hI = 123
‖πbhI ‖2hI for I ∈ D. Moreover, 124
‖πS[b]‖ ≈ ‖πS[b] + (πS[b])∗‖ ≈ ‖S[b]‖BMOd . (5)
As pointed out in the previous section, the problem of finding a Carleson measure with 125
Carleson constant 1 and small BMOd norm of the dyadic balayage is equivalent to finding 126
b ∈ BMOd(R) of norm 1 such that S[b] has small BMOd norm. 127
In light of (4) and (5), this means finding b ∈ BMOd(R) such that π∗bπb is “almost 128
diagonal”, in the sense that 129
‖S[b]‖BMOd ≈ ‖πS[b] + (πS[b])∗‖ = ‖π∗bπb − Diagb‖  ‖π∗bπb‖ = ‖πb‖2 ≈ ‖b‖2BMOd .
Note the elementary identity 130
π∗bπbhI =
1
|I |1/2
⎛
⎝∑
J⊆I+
χJ
|J| |bJ |
2 −
∑
J⊆I−
χJ
|J| |bJ |
2
⎞
⎠ . (6)
The function
∑
J⊆I+
χJ|J| |bJ |2 +
∑
J⊆I−
χJ|J| |bJ |2 is constant on its support I for each I , if 131
b is a sum of Rademacher functions. In this case, the right-hand side
∑
J⊆I+
χJ|J| |bJ |2 − 132∑
J⊆I−
χJ|J| |bJ |2 of (6) is always a multiple of hI , and π∗bπb is diagonal in the Haar basis. 133
In our counterexample, we have to introduce cutoffs on the Rademacher functions in 134
order to control the L2 norm. This introduces nondiagonal terms, but these can then be 135
controlled by the logarithmic staggering of the cutoffs. 136
4 The Poisson Balayage 137
We are now going to construct a compactly supported positive measure μ on the 138
upper half plane such that its Carleson constant Carl(μ) is very large (say m), but 139
‖Sμ‖BMO + ‖Sμ‖L1 is bounded by absolute constant. From here, one can easily construct 140
finite positive measure μ which is not Carleson, but whose balayage is a nice BMO 141
function. 142
8 S. Pott and A. Volberg
Fix m∈ N. For 0 ≤ j ≤ m, let I j denote the interval [−2 j, 2 j] and I˜ j = I j \ I j−1. 143
Furthermore, let I˜0 = I0 and let I˜m+1 = R \ Im. 144
Let μ j denote one-dimensional Lebesgue measure on the segment I j × {2− j}, and 145
let μ =∑mj=0 mj. Clearly, Carl(μ) = m+ 1. 146
Here is the elementary technical lemma which will show the desired properties 147
of μ. 148
Lemma 4.1. There exists an absolute constant c > 0 (independent of m) such that 149
|Sμ j (t) − χI j (t)| ≤ c2−2 j for |t| ≤ 2 j−1 or |t| ≥ 2 j+1, j ∈ {0, . . . , m}. 
Proof. Observe that 150
Sμ j (t) =
1
π
∫ 2 j
−2 j
2− j
(x − t)2 + 2−2 j dx ≤ Sμ j (0) ≤ 1 for all t ∈ R, j ∈ {0, . . . , m}.
Now let |t| ≤ 2 j−1. Then 151
Sμ j (t) − 1 =
1
π
∫ 2 j
−2 j
2− j
(x − t)2 + 2−2 j dx −
1
π
∫ ∞
−∞
2− j
(x − t)2 + 2−2 j dx
= 1
π
∫ −2 j
−∞
2− j
(x − t)2 + 2−2 j dx +
1
π
∫ ∞
2 j
2− j
(x − t)2 + 2−2 j dx
≤ 2
π
∫ ∞
0
2− j
(x + 2 j−1)2 + 2−2 j dx
= 2
π
∫ ∞
22 j−1
1
x2 + 1dx ≤
∞∑
l= j
2
π
∫ 22l+1
22l−1
1
x2 + 1dx
≤ 6
π
∞∑
l= j
22l−1 1
(22l−1)2
= 8
π
2−2 j+1.
If |t| ≥ 2 j+1, then 152
Sμ j (t) =
1
π
∫ 2 j
−2 j
2− j
(x − t)2 + 2−2 j dx
≤ 1
π
∫ 2 j
−2 j
2− j
22 j + 2−2 j dx
≤ 1
π
2−2 j+1. 
Carleson Measure and Balayage 9
Writing Sμ =∑mj=0 Sμ j =∑mj=0 χI j +∑mj=0(Sμ j − χI j ), we see that the first term 153
is a dyadic log function, and therefore in BMO(R) with some absolute norm bound 154
independent of m. To estimate the second term, let t ∈ I˜k. By the previous lemma, 155
|Sμ j (t) − χI j (t)| ≤ c2− j for j /∈ {k − 1, k, k + 1}, therefore 156
m∑
j=0
|Sμ j (t) − χI j (t)| ≤
m∑
j=0
c2− j + 6 = 2c + 6.
Thus, the second term is in L∞(R), with L∞ norm bounded by 2c + 6. Altogether, we find 157
that there is an absolute constant c˜, independent of m, such that ‖Sμ‖BMO ≤ c˜. However, 158
an elementary calculation shows that 159
‖Sμ‖1 =
m∑
j=0
‖Sμ j‖1 =
m∑
j=0
2 j+1 = 2m+2 − 2,
and we would like to control the L1 norm of Sμ as well. But by scaling our con- 160
struction with a small h > 0, that is, replacing each μ j by μ˜ j, the one-dimensional 161
Lebesgue measure on [−h2 j, h2 j] × {h2− j} and letting μ˜ =∑mj=0 μ˜ j, we obtain a measure 162
μ˜ with Carl(μ˜) = Carl(μ) = m+ 1, Sμ˜(t) = Sμ( th). Thus, we have ‖Sμ‖1 = h(2m+2 − 2) and 163
‖Sμ˜‖BMO = ‖Sμ‖BMO ≤ c˜. 164
After choosing an appropriate h > 0 and dividing by an appropriate multiple of 165
m, we obtain 166
Theorem 4.2. Let ε > 0. Then there exists a Carleson measure μ on R2+ with Carl(μ) = 1, 167
‖Sμ‖BMO + ‖Sμ‖1 < ε.  168
We will now show a continuous analog to Theorem 2.3. 169
Theorem 4.3. Let μ be Carleson measure μ on R2+. Then 170
Carl(μ) ≈ sup
E⊆R2+,E Borel set
‖SdμE ‖BMOd ≈ sup
I⊂R interval
‖SμQI ‖BMO.

Proof. We only have to prove that supI⊂R interval ‖SμQI ‖BMO  Carl(μ). After translation 171
and dilation of μ, we can assume without loss of generality that μ(QJ) ≥ 14Carl(μ) for 172
10 S. Pott and A. Volberg
J = [1/4, 3/4]. Let I = [0, 1] and let I ′ denote the translated interval [2, 3]. Then 173
‖SμQI ‖BMO  |〈SμQI 〉I − 〈SμQI 〉I ′ |
=
∫ 1
0
1
π
∫
QI
y
(t − x)2 + y2 −
y
(t + 2 − x)2 + y2 dμ(x, y)dt
= 1
π
∫
QI
∫ 1−x
−x
y(4 + 4t)
(t2 + y2)((t + 2)2 + y2)dtdμ(x, y)
≥ 1
π
∫
[1/4,3/4]×[0,1]
∫ 1−x
−x
y(4 + 4t)
(t2 + y2)((t + 2)2 + y2)dtdμ(x, y)
≥ 1
π
∫
[1/4,3/4]×[0,1]
∫ 1/4
−1/4
y(4 + 4t)
(t2 + y2)((t + 2)2 + y2)dtdμ(x, y)
 1
π
∫
[1/4,3/4]×[0,1]
∫ 1/4
−1/4
y
t2 + y2 dt dμ(x, y)
≥ 1
π
∫
[1/4,3/4]×[0,1]
∫ 1/4
−1/4
1
t2 + 1dt dμ(x, y)  μ(QJ)  Carl(μ). 
References 174
[1] Blasco, O., and S. Pott. “Dyadic BMO on the bidisk.” Revista Matema´tica Iberoamericana 21, 175
no. 2 (2005): 483–510. 176
[2] Carleson, L. “Two remarks on H1 and BMO.” Advances in Mathematics 22, no. 3 (1976): 269– 177
77. 178
[3] Frazier, M., F. Nazarov, and I. Verbitsky. “Global estimates for kernels of Neumann series, 179
Green’s functions, and the conditional gauge.” (2009): preprint.Q1 180
[4] Garnett, J. B. Bounded Analytic Functions. Pure and Applied Mathematics 96. New York: 181
Academic Press, Inc., 1981. 182
[5] Garnett, J. B., and P. Jones. “BMO from dyadic BMO.” Pacific Journal of Mathematics 99, no. 2 183
(1982): 351–71. 184
[6] Smith, W. S. “BMO(ρ) and Carleson measures.” Transactions of the American Mathematical 185
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Mathematical Society 79, no. 1 (1980): 35–41. 188


Carleson measure and balayage

Enlighten

        Pott, S., and Volberg, A. (2010) Carleson measure and balayage. International Mathematics Research Notices, 2010(13), pp. 2427-2436.  Copyright © 2009 The Authors.  A copy can be downloaded for personal non-commercial research or study, without prior permission or charge   The content must not be changed in any way or reproduced in any format or medium without the formal permission of the copyright holder(s)   When referring to this work, full bibliographic details must be given    http://eprints.gla.ac.uk/47601/       Deposited on:  8 May 2015          Enlighten – Research publications by members of the University of Glasgow http://eprints.gla.ac.uk Copyedited by: Roda Vee VillarizaPott, S., and A. Volberg. (2009) “Carleson Measure and Balayage,”International Mathematics Research Notices, Article ID rnp222, 10 pages.doi:10.1093/imrn/rnp222Carleson Measure and Balayage 1Sandra Pott 1 and Alexander Volberg 2 21Institut fu¨r Mathematik, Universita¨t Paderborn, Warburger Str. 100, 333098 Paderborn, Germany, and 2Department of Mathematics, 4Michigan State University, East Lansing, MI 48824, USA 5Correspondence to be sent to: sandrapo@math.uni-paderborn.de 6The balayage of a Carleson measure lies of course in bounded mean oscillation (BMO). 7We show that the converse statement is false. We also make a two-sided estimate of the 8Carleson norm of a positive measure in terms of certain balayages. 91 Introduction and Notation 10In this note, we consider a question that naturally appeared in the recent work of 11Frazier–Nazarov–Verbitsky [3]. The question is: 12How does the Carleson norm of a positive measure in the disk relate to the 13bounded mean oscillation (BMO) norm of its balayage on the circle? 14A related question is: 15How can one describe measures on the disk (say, positive measures) whose bala- 16yage is a BMO function? 17The second author is grateful to Igor Verbitsky, who called our attention to these 18questions. 19We show that the seemingly answer: “These are exactly the Carleson measures” 20is false. The Carleson property is indeed of course sufficient, but not at all necessary. 21Received 7 August 2009; Revised 11 January 2009; Accepted 21 January 2009Communicated by A. Editorc© The Author 2009. Published by Oxford University Press. All rights reserved. For permissions,please e-mail: journals.permissions@oxfordjournals.org.2 S. Pott and A. VolbergHowever, we can characterize the Carleson property in terms of the BMO norms of the 22balayages of restrictions of the measure. 23Throughout the paper, we will use the notation ,  for one-sided estimates up 24to an absolute constant, and the notation ≈ for two-sided estimates up to an absolute 25constant. 26We will use the setting of the upper half plane R2+ rather than the unit disk. Given 27a positive regular Borel measure μ on the upper half plane R2+ = {(t, y) ∈ R2 : y > 0}, its 28balayage is defined as the function 29Sμ(t) =∫R2+px,y(t)dμ(t, y),where px,y(t) = 1π xy2+(t−x)2 is the Poisson kernel for R2+. We say that μ is a Carleson 30measure if there exists a constant C > 0 such that for each interval I ⊂ R, the 31inequality 32μ(QI ) ≤ C |I | (1)holds. Here, QI denotes the Carleson square {(x, y) : x ∈ I, 0 < y ≤ |I |} over I . It is easy 33to see that it is sufficient to consider dyadic intervals in this definition. We denote the 34infimum of all constants C > 0 such that (1) holds for all dyadic intervals by Carl(μ). 35Recall that the space of functions of BMO (R) is defined as 36{b ∈ L2(R) : supI⊂R interval1|I |∫I|b(t) − 〈b〉I |dt < ∞},with ‖b‖BMO = supI⊂R interval 1|I |∫I |b(t) − 〈b〉I |dt. By the John–Nirenberg inequality, the 37L1 norm in the definition of BMO can be replaced by any ‖ · ‖p norm, 1 ≤ p < ∞. We thus 38obtain a family of equivalent norms on BMO(R), with equivalent constants depending 39on p. 40The connection between the properties of a measure μ and its balayage Sμ have 41long been studied. In particular, it is well known that the BMO norm of Sμ is controlled 42by the Carleson constant of μ, 43‖Sμ‖BMO  Carl(μ). (2)For this and other basic facts on BMO functions, we refer the reader to [4]. 44Carleson Measure and Balayage 3A partial reverse of (2) was found in [2], [7], and in the dyadic case, [5]. Namely, 45it was shown that for each b ∈ BMO, there exists an L∞(R) function φ and a Carleson 46measure μ such that b = φ + Sμ, ‖φ‖∞ + Carl(μ)  ‖b‖BMO. If we allow μ to be a complex 47measure, one even has the representation b = Sμ with Carl(μ)  ‖b‖BMO [6]. 48The purpose of this note is to show that reverse inequality to (2) in the strict 49sense does not hold, and to give a characterization of the Carleson property of a measure 50μ in terms of the BMO norm of the balayage of restrictions of μ. 512 The Dyadic Balayage 52We start by examining the dyadic case. We will use the standard Whitney-type decom- 53position of the upper half plane, indexed by the set D of left-half open dyadic intervals 54in R, 55TI ={(x, y) : x ∈ I, |I |2< y ≤ |I |}for I ∈ D.That means, TI is the “top half” of the Carleson square QI defined above. 56For a positive regular Borel measure μ on R2+, we define the dyadic balayage by 57Sdμ(t) =∑I∈DχI (t)|I | μ(TI ) (t ∈ R),which is well defined as a function taking values in [0,∞]. By comparing box kernel and 58Poisson kernel, one easily verifies the pointwise estimate Sdμ  Sμ. 59We recall the definition of dyadic BMO, BMOd(R), as the class of L2(R) functions 60for which 61‖b‖2BMOd= supI∈D1|I |∫I|b(t) − 〈b〉I |2dt = supI∈D1|I | ‖PI b‖2 = supI∈D1|I |∑J∈D,J⊆I|bI |2is finite. Here, hJ denotes the L2-normalized Haar function, bJ := (b, hJ) denotes the cor- 62responding Haar coefficient of function b, and PI denotes the orthogonal projection on to 63span{hJ : J ⊆ I }. Again, by the John–Nirenberg inequality the L2 norm in the definition 64can be replaced by any L p norm, 1 ≤ p < ∞, yielding an equivalent norm. 654 S. Pott and A. VolbergWe say that a sequence of nonnegative numbers (αI )I∈D is a Carleson sequence, 66if there exists a constant C > 0 such that 671|I |∑J∈D,J⊆IaI ≤ C for each I ∈ D.Again, we denote the infimum of such constants by Carl((aI )). With this notation, one 68verifies immediately the following well-known lemma. 69Lemma 2.1. Let b ∈ L2(R). Then the following are equivalent: 701. μ is a Carleson measure 712. (μ(TI ))I∈D is a Carleson sequence 723. bμ =∑I∈D hI μ(TI )1/2 ∈ BMOd(R). 73In this case, Carl(μ) = Carl((μ(TI ))) = ‖bμ‖2BMOd.  74Notice that with the above definition of bμ, 75Sdμ =∑I∈DχI|I |μ(TI ) =∑I∈DχI|I | |(bμ)I |2 = S[bμ],where S denotes the square of the dyadic square function, S[ f] =∑I∈D χI|I | | fI |2 for f ∈ 76L2(R). In this sense, we have identified the dyadic balayage of a positive regular Borel 77measure μ with the square of a dyadic square function of bμ. Conversely, for any f ∈ 78L2(R), S[ f] can be written as a dyadic balayage of a measure μ f , for example by letting 79μ f =∑I∈D | fI |2δz(I ), z(I ) denoting the center of TI . 80The well-known dyadic analog of (2) is therefore equivalent to the inequality 81‖S[b]‖BMOd  ‖b‖2BMOd , (3)which can be now be proved as a simple application of the John–Nirenberg inequality.Notice that for any dyadic inverval I ∈ D, all summands in S[b] =∑J∈D χJ|J| |bJ |2 exceptthose corresponding to dyadic intervals J ⊂ I are constant on I . Thus1|I |∫I|S[b](t) − 〈S[b]〉I |dt = 1|I |∫I|S[PI b](t) − 〈S[PI b]〉I |dt≤ 1|I |∫IS[PI b](t)dt + 〈S[PI b]〉I = 2 1|I |∫I∑J⊆IχJ(t)|J| |bJ |2dt = 2‖PI b‖22 ≤ 2‖b‖2BMOd ,which proves (3). 82Carleson Measure and Balayage 5Here are the main results of this section, which concern the reverse inequality 83to (3). The first says that the BMO norm of the dyadic balayage can be very much smaller 84than the Carleson constant of a measure, even if one increases the BMO norm by the L2 85norm. 86Theorem 2.2. Let ε > 0. Then there exists a Carleson measure μ on R2+ with Carl(μ) = 1, 87‖Sdμ‖BMO + ‖Sdμ‖2 < ε.  88Proof. By Lemma 2.1 and the argument following it, we want to find a BMOd(R) func- 89tion b of norm 1 such that both the BMOd norm and the L2 norm of S[b] are small. To this 90end, let I0 = (0, 1], I−1 = (−2, 0], Ik = (2k − 1, 2k+1 − 1] for k > 0 and Ik = (−2−k,−2−k−1] 91for k < 0. In particular, |Ik| = 2|k| for all k ∈ N. Let r1 denote the first Rademacher func- 92tion on R, r1 =∑ j∈Z(−χ( j, j+ 12 ] + χ( j+ 12 , j+1]), and let rn = r1(2n−1·) be the nth Rademacher 93function on R. Let N ∈ N, N to be determined later, and let 94b =∞∑k=−∞N−|k|∑n=1χIk(t)rn(t).One verifies without difficulty that ‖b‖2BMOd= N. Clearly, 95S[b] =∞∑k=−∞N−|k|∑n=1χIk =N∑k=0(N − k)χIk∪I−k.This is a “dyadic log”, and it is not difficult to show that 96‖S[b]‖BMO ≤ C ,where C is an absolute constant independent of N. Notice that we have an estimate here 97not only for the dyadic BMO norm, but for the full BMO norm. 98Now choose N so large that CN <ε2 and replace b by1N1/2b. This already guaran- 99tees that ‖b‖2BMOd= 1, ‖S[b]‖BMO < ε2 . To deal with the desired L2 estimate, observe that 100the estimates achieved so far do not change at all if b is dilated with an integer power of 1012. By choosing a suitable power 2K of 2, K ∈ N, and replacing b by b(2K ·), we obtain the 102desired estimate 103‖b‖2BMOd= 1, ‖S[b]‖BMO + ‖S[b]‖2 < ε. 6 S. Pott and A. VolbergThe next theorem says that we can retrieve the Carleson constant of a measure 104up to an absolute constant from its dyadic balayage, if we restrict the measure to certain 105sets. 106Theorem 2.3. Let μ be Carleson measure μ on R2+. Then 107Carl(μ) ≈ supE⊆R2+,E Borel set‖SdμE ‖BMOd ≈ supI∈D‖SdμQI ‖BMOd .Here, μE stands for the restriction of μ to E , given by μE (A) = μ(E ∩ A).  108Proof. Clearly, Carl(μE ) ≤ Carl(μ) for each Borel set E ⊆ R2+, so 109supI∈D‖SdμQI ‖BMOd ≤ supE⊆R2+,E Borel set‖SdμE ‖BMOd  supE⊆R2+,E Borel setCarl(μE ) ≤ Carl(μ).To prove the reverse inequality, let I ∈ D. Observe that SdμQI is supported on the closure 110of I . Therefore, with I ′ denoting the dyadic sibling of I , we have 111‖SdμQI ‖BMOd ≥ |〈SdμQI〉I − 〈SdμQI 〉I ′ | = 〈SdμQI〉I = 1|I |∫I∑J∈D,J⊆IχJ(t)|J| μ(TJ)dt =1|I |μ(QI ).Thus, Carl(μ)  supI∈D ‖SdμQI ‖BMOd.  1123 The Algebra of Paraproducts 113This section contains a short operator-theoretic motivation for the choice of the coun- 114terexample, in particular the appearance of Rademacher functions, in the previous sec- 115tion, in terms of paraproducts. Recall that for b ∈ L2(R), the standard dyadic paraprod- 116uct πb is defined by 117πb f =∑I∈DhI bI 〈 f〉I for f ∈ L∞(R) ∩ L2(R).It is well known, and indeed a reformulation of the classical Carleson Embedding Theo- 118rem, that πb extends to a bounded linear operator on L2(R), if and only if b ∈ BMOd(R). 119In this case, ‖πb‖ ≈ ‖b‖BMOd. 120Carleson Measure and Balayage 7Such dyadic paraproducts have the nice property that π∗bπb is essentially a dyadic 121paraproduct again, with symbol S[b] (see [1]): 122π∗bπb = πS[b] + (πS[b])∗ + Diag(b), (4)where Diag(b) denotes the diagonal of π∗bπb with respect to the Haar basis, Diag(b)hI = 123‖πbhI ‖2hI for I ∈ D. Moreover, 124‖πS[b]‖ ≈ ‖πS[b] + (πS[b])∗‖ ≈ ‖S[b]‖BMOd . (5)As pointed out in the previous section, the problem of finding a Carleson measure with 125Carleson constant 1 and small BMOd norm of the dyadic balayage is equivalent to finding 126b ∈ BMOd(R) of norm 1 such that S[b] has small BMOd norm. 127In light of (4) and (5), this means finding b ∈ BMOd(R) such that π∗bπb is “almost 128diagonal”, in the sense that 129‖S[b]‖BMOd ≈ ‖πS[b] + (πS[b])∗‖ = ‖π∗bπb − Diagb‖  ‖π∗bπb‖ = ‖πb‖2 ≈ ‖b‖2BMOd .Note the elementary identity 130π∗bπbhI =1|I |1/2⎛⎝∑J⊆I+χJ|J| |bJ |2 −∑J⊆I−χJ|J| |bJ |2⎞⎠ . (6)The function∑J⊆I+χJ|J| |bJ |2 +∑J⊆I−χJ|J| |bJ |2 is constant on its support I for each I , if 131b is a sum of Rademacher functions. In this case, the right-hand side∑J⊆I+χJ|J| |bJ |2 − 132∑J⊆I−χJ|J| |bJ |2 of (6) is always a multiple of hI , and π∗bπb is diagonal in the Haar basis. 133In our counterexample, we have to introduce cutoffs on the Rademacher functions in 134order to control the L2 norm. This introduces nondiagonal terms, but these can then be 135controlled by the logarithmic staggering of the cutoffs. 1364 The Poisson Balayage 137We are now going to construct a compactly supported positive measure μ on the 138upper half plane such that its Carleson constant Carl(μ) is very large (say m), but 139‖Sμ‖BMO + ‖Sμ‖L1 is bounded by absolute constant. From here, one can easily construct 140finite positive measure μ which is not Carleson, but whose balayage is a nice BMO 141function. 1428 S. Pott and A. VolbergFix m∈ N. For 0 ≤ j ≤ m, let I j denote the interval [−2 j, 2 j] and I˜ j = I j \ I j−1. 143Furthermore, let I˜0 = I0 and let I˜m+1 = R \ Im. 144Let μ j denote one-dimensional Lebesgue measure on the segment I j × {2− j}, and 145let μ =∑mj=0 mj. Clearly, Carl(μ) = m+ 1. 146Here is the elementary technical lemma which will show the desired properties 147of μ. 148Lemma 4.1. There exists an absolute constant c > 0 (independent of m) such that 149|Sμ j (t) − χI j (t)| ≤ c2−2 j for |t| ≤ 2 j−1 or |t| ≥ 2 j+1, j ∈ {0, . . . , m}. Proof. Observe that 150Sμ j (t) =1π∫ 2 j−2 j2− j(x − t)2 + 2−2 j dx ≤ Sμ j (0) ≤ 1 for all t ∈ R, j ∈ {0, . . . , m}.Now let |t| ≤ 2 j−1. Then 151Sμ j (t) − 1 =1π∫ 2 j−2 j2− j(x − t)2 + 2−2 j dx −1π∫ ∞−∞2− j(x − t)2 + 2−2 j dx= 1π∫ −2 j−∞2− j(x − t)2 + 2−2 j dx +1π∫ ∞2 j2− j(x − t)2 + 2−2 j dx≤ 2π∫ ∞02− j(x + 2 j−1)2 + 2−2 j dx= 2π∫ ∞22 j−11x2 + 1dx ≤∞∑l= j2π∫ 22l+122l−11x2 + 1dx≤ 6π∞∑l= j22l−1 1(22l−1)2= 8π2−2 j+1.If |t| ≥ 2 j+1, then 152Sμ j (t) =1π∫ 2 j−2 j2− j(x − t)2 + 2−2 j dx≤ 1π∫ 2 j−2 j2− j22 j + 2−2 j dx≤ 1π2−2 j+1. Carleson Measure and Balayage 9Writing Sμ =∑mj=0 Sμ j =∑mj=0 χI j +∑mj=0(Sμ j − χI j ), we see that the first term 153is a dyadic log function, and therefore in BMO(R) with some absolute norm bound 154independent of m. To estimate the second term, let t ∈ I˜k. By the previous lemma, 155|Sμ j (t) − χI j (t)| ≤ c2− j for j /∈ {k − 1, k, k + 1}, therefore 156m∑j=0|Sμ j (t) − χI j (t)| ≤m∑j=0c2− j + 6 = 2c + 6.Thus, the second term is in L∞(R), with L∞ norm bounded by 2c + 6. Altogether, we find 157that there is an absolute constant c˜, independent of m, such that ‖Sμ‖BMO ≤ c˜. However, 158an elementary calculation shows that 159‖Sμ‖1 =m∑j=0‖Sμ j‖1 =m∑j=02 j+1 = 2m+2 − 2,and we would like to control the L1 norm of Sμ as well. But by scaling our con- 160struction with a small h > 0, that is, replacing each μ j by μ˜ j, the one-dimensional 161Lebesgue measure on [−h2 j, h2 j] × {h2− j} and letting μ˜ =∑mj=0 μ˜ j, we obtain a measure 162μ˜ with Carl(μ˜) = Carl(μ) = m+ 1, Sμ˜(t) = Sμ( th). Thus, we have ‖Sμ‖1 = h(2m+2 − 2) and 163‖Sμ˜‖BMO = ‖Sμ‖BMO ≤ c˜. 164After choosing an appropriate h > 0 and dividing by an appropriate multiple of 165m, we obtain 166Theorem 4.2. Let ε > 0. Then there exists a Carleson measure μ on R2+ with Carl(μ) = 1, 167‖Sμ‖BMO + ‖Sμ‖1 < ε.  168We will now show a continuous analog to Theorem 2.3. 169Theorem 4.3. Let μ be Carleson measure μ on R2+. Then 170Carl(μ) ≈ supE⊆R2+,E Borel set‖SdμE ‖BMOd ≈ supI⊂R interval‖SμQI ‖BMO.Proof. We only have to prove that supI⊂R interval ‖SμQI ‖BMO  Carl(μ). After translation 171and dilation of μ, we can assume without loss of generality that μ(QJ) ≥ 14Carl(μ) for 17210 S. Pott and A. VolbergJ = [1/4, 3/4]. Let I = [0, 1] and let I ′ denote the translated interval [2, 3]. Then 173‖SμQI ‖BMO  |〈SμQI 〉I − 〈SμQI 〉I ′ |=∫ 101π∫QIy(t − x)2 + y2 −y(t + 2 − x)2 + y2 dμ(x, y)dt= 1π∫QI∫ 1−x−xy(4 + 4t)(t2 + y2)((t + 2)2 + y2)dtdμ(x, y)≥ 1π∫[1/4,3/4]×[0,1]∫ 1−x−xy(4 + 4t)(t2 + y2)((t + 2)2 + y2)dtdμ(x, y)≥ 1π∫[1/4,3/4]×[0,1]∫ 1/4−1/4y(4 + 4t)(t2 + y2)((t + 2)2 + y2)dtdμ(x, y) 1π∫[1/4,3/4]×[0,1]∫ 1/4−1/4yt2 + y2 dt dμ(x, y)≥ 1π∫[1/4,3/4]×[0,1]∫ 1/4−1/41t2 + 1dt dμ(x, y)  μ(QJ)  Carl(μ). References 174[1] Blasco, O., and S. Pott. “Dyadic BMO on the bidisk.” Revista Matema´tica Iberoamericana 21, 175no. 2 (2005): 483–510. 176[2] Carleson, L. “Two remarks on H1 and BMO.” Advances in Mathematics 22, no. 3 (1976): 269– 17777. 178[3] Frazier, M., F. Nazarov, and I. Verbitsky. “Global estimates for kernels of Neumann series, 179Green’s functions, and the conditional gauge.” (2009): preprint.Q1 180[4] Garnett, J. B. Bounded Analytic Functions. Pure and Applied Mathematics 96. New York: 181Academic Press, Inc., 1981. 182[5] Garnett, J. B., and P. Jones. “BMO from dyadic BMO.” Pacific Journal of Mathematics 99, no. 2 183(1982): 351–71. 184[6] Smith, W. S. “BMO(ρ) and Carleson measures.” Transactions of the American Mathematical 185Society 287, no. 1 (1985): 107–26. 186[7] Uchiyama, A. “A remark on Carleson’s characterization of BMO.” Proceedings of the American 187Mathematical Society 79, no. 1 (1980): 35–41. 188

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