Bull. Malays. Math. Sci. Soc. DOI 10.1007/s40840-017-0538-0
A Note on Carleson Measures for Besov–Sobolev Spaces in the Unit Ball Cheng Yuan1 · Cezhong Tong2
Received: 10 November 2016 / Revised: 2 August 2017 © Malaysian Mathematical Sciences Society and Penerbit Universiti Sains Malaysia 2017
Abstract We show that a class of logarithmic Carleson measures is the Carleson measure for the Besov–Sobolev space in the unit ball of Cn . Keywords Carleson measures · Besov–Sobolev spaces Mathematics Subject Classification 32A37 · 30H25
1 Introduction Let Bn be the unit ball of Cn with boundary Sn and H (Bn ) the space of holomorphic functions on Bn . In particular, B1 is the unit disk D. Let dv denote the normalized Lebesgue measure of Bn , i.e., v(Bn ) = 1, and dσ denote the normalized rotation invariant Lebesgue measure of Sn satisfying σ (Sn ) = 1.
Communicated by Saminathan Ponnusamy. Cheng Yuan is supported by the National Natural Science Foundation of China (Grant No. 11501415); Cezhong Tong is supported by the National Natural Science Foundation of China (Grant No. 11301132) and Natural Science Foundation of Hebei Province (Grant No. A2013202265).
B
Cheng Yuan
[email protected] Cezhong Tong
[email protected]
1
Institute of Mathematics, School of Science, Tianjin University of Technology and Education, Tianjin 300222, China
2
Department of Mathematics, School of Science, Hebei University of Technology, Tianjin 300401, China
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If ζ ∈ Sn and r > 0, let B(ζ, r ) = {z ∈ Bn : |1 − z, ζ | < r }. For 0 ≤ t, s < ∞ with t +s > 0, we say that a nonnegative Borel measure μ on Bn is a (t, s)-logarithmic Carleson measure if 2 t μ(B(ζ, r )) : ζ ∈ S , r > 0 < ∞. μLCMt,s = sup log n r r ns We write LCMt,s for the class of all (t, s)-logarithmic Carleson measures. When t = 0, the (t, s)-logarithmic Carleson measure becomes the s-Carleson measure, and we will denote by CMs = LCM0,s . See [9] for more details. Logarithmic Carleson measures were first introduced in [7] and were used to characterize the Riemann– Stieltjes operator on the Q p space, see [4,6]. For f ∈ H (Bn ) with homogeneous expansion f (z) =
∞
f k (z),
k=0
the radial derivative of f is defined as R f (z) =
∞
k f k (z).
k=1
It is easy to see that R f ∈ H (Bn ) with
1
f (z) − f (0) = 0
R f (t z) dt. t
(1)
More generally, for any two real parameters α and λ with the property that neither n+α nor n+α+λ is a negative integer, we define a class of fractional radial derivatives on H (Bn ) as R α,λ f (z) =
∞ (n + 1 + α)(n + 1 + k + α + λ) k=0
(n + 1 + α + λ)(n + 1 + k + α)
f k (z).
For integer m > 0, and for 0 ≤ σ < ∞, 1 < p < ∞, m + σ > n/ p , where neither n + α nor n + α + m is a negative integer, the Besov–Sobolev space B σp (Bn ) is defined to consist of those f ∈ H (Bn ) on the ball such that p |R α,m f (z)| p (1 − |z|2 )mp+ pσ −n−1 dv(z) < ∞. (2) f B σ (Bn ) = p
Bn
A positive Borel measure μ on Bn is called a Carleson measure for B σp (Bn ) if there is a constant C > 0 such that p | f (z)| p dμ(z) ≤ C f B σ (Bn ) Bn
for all f ∈ B σp (Bn ).
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A Note on Carleson Measures for Besov–Sobolev Spaces in…
The following theorem is quoted from [5], which characterizes the relation between the Carleson measure for B σp (Bn ) and the s-Carleson measure. Theorem 1 (Theorem 2.1 in [5]) Suppose integer m > 0, 0 ≤ σ < ∞, 1 < p < ∞, m + σ > n/ p , neither n + α nor n + α + m is a negative integer. Let μ be a positive Borel measure in Bn . (1) If σ > 0, then μ is a Carleson measure for B σp (Bn ) implies μ is a pσ/n-Carleson measure, and on the other hand, for any ε > 0, suppose μ is a ( pσ + ε)/nCarleson measure, we can get μ is a Carleson measure for B σp (Bn ). (2) If σ = 0, then μ is a Carleson measure for B 0p (Bn ) implies
2 log r
sup
p−1
μ(B(ζ, r )) : ζ ∈ Sn , r > 0 < ∞,
and on the other hand, for any ε > 0, suppose sup
2 log r
p+ε−1
μ(B(ζ, r )) : ζ ∈ Sn , r > 0 < ∞,
we can get μ is a Carleson measure for B 0p (Bn ). In the above characterization, there is an ε difference between the sufficient and necessary conditions. Motivated by this result, we show that the above ε conditions can be substituted by the logarithmic conditions, which seems a little bit sharper than the ε conditions. Our results are stated as follows: Theorem 2 Suppose integer m > 0, 0 ≤ σ < ∞, 1 < p < ∞, t > p − 1, m + σ > n/ p , neither n + α nor n + α + m is a negative integer. Let μ be a positive Borel measure in Bn . (1) If σ > 0, suppose μ is a (t, pσ n )-logarithmic Carleson measure, then μ is a Carleson measure for B σp (Bn ). (2) If σ = 0, suppose sup
4 log log r
t 2 p−1 log μ(B(ζ, r )) : ζ ∈ Sn , r > 0 < ∞, r
then μ is a Carleson measure for B 0p (Bn ). The key point in [5] is that the duality theorem is adapted. In this manuscript we use a different strategy, which can also be used to prove Peng and Ouyang’s theorem in [5]. Some techniques in this paper are inspired in [3]. Notation Throughout this paper, we only write U V (or V U ) for U ≤ cV for a positive constant c, and moreover U ≈ V for both U V and V U . For 1 < p < ∞, we also write p = p/( p − 1) be the conjugate index of p.
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2 Preliminaries For each real parameter λ we define an operator R λ on H (Bn ) as λ
R f (z) =
∞
λ
k f k (z),
f (z) =
k=1
∞
f k (z).
(3)
k=0
The operator R λ is clearly invertible on H (Bn )/C, with its inverse given by Rλ f (z) = R
−λ
f (z) =
∞
k
−λ
f k (z),
f (z) =
k=1
∞
f k (z).
(4)
k=0
An application of Stirling’s formula yields that (n + 1 + α)(n + 1 + k + α + λ) ≈ kλ. (n + 1 + α + λ)(n + 1 + k + α) Thus, the semi-norm in definition of B σp (Bn ) is equivalent to
p
f B σ (Bn ),∗ = p
Bn
|R m f (z)| p (1 − |z|2 )mp+ pσ −n−1 dv(z) < ∞.
(5)
We will need the following lemma modified by Lemma 3.2 in [2]. Lemma 3 Let t, s, γ ∈ (0, ∞) and μ be a nonnegative Borel measure on Bn . Then (1) μ ∈ LCMt,s if and only if μLCMt,s ,γ
= sup log w∈Bn
2 1 − |w|2
t Bn
(1 − |w|2 )γ dμ(z) < ∞; |1 − z, w|γ +ns
(6)
(2) sup
4 t 2 s log log log μ(B(ζ, r )) : ζ ∈ Sn , r > 0 < ∞ r r
(7)
if and only if
t 4 log log 1−|w| 2 (1 − |w|2 )γ dμ(z) < ∞. sup
−s |1 − z, w|γ 2 Bn w∈Bn log 1−|w| 2
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(8)
A Note on Carleson Measures for Besov–Sobolev Spaces in…
Proof (1) is Lemma 3.2 in [2]. For (2), first suppose Eq. (8) holds, given B(ζ, r ) = {z ∈ Bn : |1 − z, ζ | < r }. Letting w = (1 − r )ζ , we have
t s (1 − |w|2 )γ dμ(z) 4 2 log log log 1 − |w|2 1 − |w|2 |1 − z, w|γ Bn t s 2 (1 − |w|2 )γ 4 log ≥ log log μ(B(ζ, r )) inf z∈B(ζ,r ) |1 − z, w|γ r r t s 2 4 log log log μ(B(ζ, r )). r r
This means that 2 s 4 t log log log μ(B(ζ, r )) : ζ ∈ Sn , r > 0 r r
t 4 log log 1−|w| 2 (1 − |w|2 )γ dμ(z) sup
. −s |1 − z, w|γ 2 Bn w∈Bn log 1−|w| 2
sup
On the other hand, if sup
2 s 4 t log μ(B(ζ, r )) : ζ ∈ Sn , r > 0 < ∞. log log r r
When |w| < 3/4, it is easy to check that log log
4 1 − |w|2
t log
2 1 − |w|2
s Bn
(1 − |w|2 )γ dμ(z) < ∞. |1 − z, w|γ
When 3/4 ≤ |w| < 1, let ξ = w/|w| and Jw be the integer part of | log2 (1−|w|2 )|−1. For k = 0, 1, . . . , Jw , consider the sets E 0 = ∅, E k = {z ∈ Bn : |1 − z, ξ | < 2k (1 − |w|2 )}, k ≥ 1. Moreover, let E Jw +1 be the unit ball. It follows that log log
4 1 − |w|2
t log
2 1 − |w|2
s Bn
(1 − |w|2 )γ dμ(z) |1 − z, w|γ
t
4 J w +1 log log 1−|w| 2 (1 − |w|2 )γ
μ(E k \ E k−1 ) −s 2γ (k−1) (1 − |w|2 )γ 2 log 1−|w| k=1 2
t 4 J w +1 log log 1−|w| 2 μ(E k )
−s 2γ (k−1) 2 log 1−|w| k=1 2
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C. Yuan, C. Tong
t
−s 4 2 −t J w +1 log log log 1−|w| 2 k 2 4 2 (1−|w| )
. log log −s 2k (1 − |w|2 ) 2γ (k−1) 2 log 1−|w| k=1 2 Similar to the proof of Lemma 3.2 in [2], it can be checked that log
2 (1 − |w|2 )
log
2 k 2 (1 − |w|2 )
−1
k + 1,
and log log
4 (1 − |w|2 )
log log
4 2k (1 − |w|2 )
−1
1 + log(k + 1).
Then we can easily get that
t
−s 4 2 −t J w +1 log log log 1−|w| 2 k 2 4 2 (1−|w| ) log log
−s 2k (1 − |w|2 ) 2γ (k−1) 2 log 1−|w| k=1 2
∞ (k + 1)s (1 + log(k + 1))t
2γ (k−1)
k=1
< ∞.
This completes the proof.
It is easy to check that if Eqs. (6) and (8) hold for some γ > 0, they hold for all γ > 0 in the above lemma. We also need the following standard result from [9]. Lemma 4 Suppose s > −1, b > 0 and c < 0. Then (1)
Bn
(2)
(1 − |w|2 )s dv(w) 1; |1 − z, w|n+1+s+c
(9)
dσ (ζ ) 1 ≈ |1 − z, ζ |n+b (1 − |z|2 )b
(10)
Sn
for all z ∈ Bn . Straightforward computation can verify the following lemma. Lemma 5 Suppose t > 1 and s > −1. Then (1)
Bn
123
log
2 1 − |w|2
−t
(1 − |w|2 )s dv(w) 1; |1 − z, w|n+1+s
(11)
A Note on Carleson Measures for Besov–Sobolev Spaces in…
(2) Bn
log log
4 1 − |w|2
−t log
2 1 − |w|2
−1
(1 − |w|2 )s dv(w) 1 |1 − z, w|n+1+s (12)
for all z ∈ Bn . Proof Inequality (11) can be seen as a special case of Lemma 3.4 in [1]. Indeed, let ζ = z and s = t0 + t1 in Lemma 3.4 of [1], (11) follows. Here we focus on the verification of (12). (11) can also be checked in the same way. It follows from the polar coordinate integral that −t −1 4 2 (1 − |w|2 )s log dv(w) 1 − |w|2 1 − |w|2 |1 − z, w|n+1+s Bn −t −1 1 4 2 (1 − r 2 )s log log log = 2n r 2n−1 dr dσ (ζ ) 2 2 1−r 1−r |1 − z, r ζ |n+1+s Sn 0 −t −1 1 4 2 dσ (ζ ) 2 )s dr log = 2n r 2n−1 log log (1 − r 2 2 n+1+s 1−r 1−r Sn |1 − r z, ζ | 0 −t −1 1 4 2 (1 − r 2 )s log ≈ 2n r 2n−1 log log dr 2 2 1−r 1−r (1 − r 2 |z|2 )1+s 0 −t −1 1 4 2 (1 − r 2 )s log r 2n−1 log log dr ≤ 2n 2 2 1−r 1−r (1 − r 2 )1+s 0 −t −1 1 4 2 1 log r 2n−1 log log dr = 2n 2 2 1−r 1−r 1 − r2 0 −t −1 1 4 2 1 du. log u n−1 log log =n 1 − u 1 − u 1 − u 0
log log
Obviously, the above “≈” line follows from (10) since 1 + s > 0. Notice that n ≥ 1. So last integral is less than or equal to 1
I =
log log 0
Let x =
4 1−u , ∞
I =
then u = 1 −
(log log x)
−t
4
∞
I ≤2 4
−t log
and du =
4 x2
x 2
≥
1 2
−1
1 du. 1−u
∞
4
x −1 1 dx. (log log x)−t log 2 x
log x when x ≥ 4. Recall that t > 1, we have
dx =2 x log x(log log x)t
This completes the proof.
2 1−u
dx. Thus,
x −1 x 4 · log dx = 2 4 x2
It is easy to check that log
4 x
4 1−u
∞ log 4
dy < ∞. y(log y)t
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C. Yuan, C. Tong
3 Proof of Theorem 2 Proof For f ∈ B σp (Bn ), R m f ∈ H (Bn ), we can use the reproducing formula to get that R m f (w)dvβ (w) , R m f (z) = n+1+β Bn (1 − z, w) where dvβ (z) =
(n + 1 + β) (1 − |z|2 )β dv(z), n!(β + 1)
and β > −1. Without loss of generality, we can suppose β > m + σ − n+1 p . m Since R f (0) = 0, we have 1 m m R f (w) − 1 dvβ (w) R f (z) = (1 − z, w)n+1+β Bn for all z ∈ Bn . According to (1), 1 1 dt1 dtm f (z) − f (0) = ··· R m f (t1 · · · tm z) ··· t tm 1 0 0 = R m f (w)L m (z, w)dvβ (w), Bn
where the kernel L m (z, w) =
1
1
···
0
0
1 dt1 dtm − 1 ··· . n+1+β (1 − t1 · · · tm z, w) t1 tm
Similar to [9, page 74, Exercise 2.24], we have |L m (z, w)|
1 . |1 − z, w|n+1+β−m
Indeed, if we let h w (z) =
1 − 1, (1 − z, w)n+1+β
then h w (0) = 0. We can deduce from (1) and (4) that
1
1
dt1 dtm ··· t1 tm 0 0 1 1 dt1 dtm = ··· R R −1 h w (t1 · · · tm z) ··· t1 tm 0 0 −m = R h w (z) = Rm h w (z).
L m (z, w) =
123
···
h w (t1 · · · tm z)
A Note on Carleson Measures for Besov–Sobolev Spaces in…
Recall that 1 −1 (1 − z, w)n+1+β ∞ (n + 1 + β + k) z, wk ; = k!(n + 1 + β)
h w (z) =
k=1
thus,
Rm h w (z) =
∞ (n + 1 + β + k) z, wk . k!(n + 1 + β)k m k=1
It can be checked from Stirling’s formula that (n + 1 + β + k) (n + 1 + β + k − m) . ≈ m (n + 1 + β)k (n + 1 + β − m) This gives that |L m (z, w)| = |Rm h w (z)|
1 . |1 − z, w|n+1+β−m
It is not loss of generality to suppose f (0) = 0, and then we have the estimate | f (z)| p
Bn
p |R m f (w)|(1 − |w|2 )β dv(w) . |1 − z, w|n+1+β−m
(1) When σ > 0, notice that t/( p − 1) > 1, and it follows from the Hölder’s inequality and Lemma 5 that t 2 |R m f (w)| p (1 − |w|2 ) pβ | f (z)| log dv(w) n+1+ pβ− pm 1 − |w|2 Bn |1 − z, w|
p−1 t 1− p 2 dv(w) log × 1 − |w|2 |1 − z, w|n+1 Bn t 2 |R m f (w)| p (1 − |w|2 ) pβ log dv(w). n+1+ pβ− pm 1 − |w|2 Bn |1 − z, w|
p
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C. Yuan, C. Tong
Recall that β > m + σ − n+1 p , which implies n + 1 − pm − pσ + pβ > 0, and then we can use Fubini’s theorem and Lemma 3 to get that | f (z)| p dμ(z) Bn
|R f (w)| log
t 2 (1 − |w|2 ) pβ dμ(z) dv(w) n+1+ pβ− pm 1 − |w|2 Bn |1 − z, w| 2 t |R m f (w)| p (log 1−|w| 2) (1 − |w|2 )n+1− pm− pσ + pβ dμ(z) dv(w) (1 − |w|2 )n+1− pm− pσ Bn |1 − z, w|n+1+ pβ− pm m
Bn
Bn
Bn
p
|R m f (w)| p (1 − |w|2 ) pm+ pσ −n−1 dv(w) p
= f B σ (Bn ),∗ , p
as desired. (2) When σ = 0, similarly to the σ > 0 case, we have
| f (z)| p
4 t |R m f (w)| p (1 − |w|2 ) pβ (log log 1−|w|2 ) dv(w) 2 n+1+ pβ− pm 1− p (log 1−|w| Bn |1 − z, w| 2) ⎛ ⎞ p−1 (log log 4 ) 1−t p 2 dv(w) 1−|w| ⎠ ×⎝ |1 − z, w|n+1 (log 2 2 ) Bn 1−|w|
Thus,
Bn
4 t |R m f (w)| p (1 − |w|2 ) pβ (log log 1−|w|2 ) dv(w). |1 − z, w|n+1+ pβ− pm (log 2 2 )1− p 1−|w|
Bn
| f (z)| p dμ(z)
Bn
|R m f (w)| p
4 t (log log 1−|w| 2)
(log
2 )1− p 1−|w|2
Bn
(1 − |w|2 ) pβ dμ(z) dv(w) |1 − z, w|n+1+ pβ− pm
|R m f (w)| p 2 n+1− pm Bn (1 − |w| ) 4 t (log log 1−|w| 2) (1 − |w|2 )n+1− pm+ pβ dμ(z) × dv(w) 2 (log 1−|w|2 )1− p Bn |1 − z, w|n+1+ pβ− pm |R m f (w)| p (1 − |w|2 ) pm−n−1 dv(w)
Bn
p
= f B 0 (B p
n ),∗
.
This completes the proof.
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A Note on Carleson Measures for Besov–Sobolev Spaces in…
Remark 6 In [5], Peng and Ouyang proved the sufficient part of Theorem 1 by the duality theorem. Indeed, it can also be proved by the method of this paper. (1) When σ > 0, if ε > 0 and μ is a sup
w∈Bn
Bn
σ p+ε n -Carleson
measure, then
(1 − |w|2 )γ dμ(z) < ∞ |1 − z, w|γ +σ p+ε
for γ > 0. See Theorem 50 of [8] or Lemma 2.1 in [5]. We can use Lemma 4 and Hölder’s inequality to estimate that | f (z)|
Bn
Bn
|R m f (w)| p (1 − |w|2 ) pβ dv(w). |1 − z, w|n+1+ pβ− pm+ε
|R m f (w)| p (1 − |w|2 ) pβ dv(w) |1 − z, w|n+1+ pβ− pm+ε
p
Bn
p−1
dv(w) |1 − z, w|
ε n+1− p−1
Then Bn
| f (z)| p dμ(z)
Bn
Bn
Bn
(1 − |w|2 ) pβ dμ(z) dv(w) n+1+ pβ− pm+ε Bn |1 − z, w| |R m f (w)| p (1 − |w|2 )n+1− pm− pσ + pβ dμ(z) dv(w) 2 n+1− pm− pσ (1 − |w| ) |1 − z, w|n+1+ pβ− pm+ε Bn
|R m f (w)| p
|R m f (w)| p (1 − |w|2 ) pm+ pσ −n−1 dv(w) p
= f B σ (Bn ),∗ . p
(2) When σ = 0, if ε > 0 and μ is a nonnegative Borel measure with
2 log r
sup
p+ε−1
μ(B(ζ, r )) : ζ ∈ Sn , r > 0 < ∞,
then sup
w∈Bn
log
2 1 − |w|2
p−1+ε Bn
(1 − |w|2 )γ dμ(z) < ∞ |1 − z, w|γ
(13)
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C. Yuan, C. Tong
for γ > 0. See Lemma 2.2 in [5]. Then | f (z)| p
Bn
|R m f (w)| p (1 − |w|2 ) pβ |1 − z, w|n+1+ pβ− pm
⎛ ×⎝
Bn
Bn
log
2 1 − |w|2
log
2 1 − |w|2
p−1+ε dv(w)
⎞ p−1 dv(w) ⎠ |1 − z, w|n+1 p−1+ε 2 log dv(w). 1 − |w|2
− p−1+ε p−1
|R m f (w)| p (1 − |w|2 ) pβ |1 − z, w|n+1+ pβ− pm
It follows from (13) that | f (z)| p dμ(z) Bn
Bn
|R m f (w)| p log |R m
2 ) p−1+ε 1−|w|2 (1 − |w|2 )n+1− pm
Bn
|R f (w)| (1 − |w| ) m
Bn
p
p
= f B 0 (B p
p−1+ε
f (w)| p (log
2 1 − |w|2
n ),∗
2 pm−n−1
Bn
Bn
(1 − |w|2 ) pβ dμ(z) dv(w) |1 − z, w|n+1+ pβ− pm
(1 − |w|2 )n+1− pm+ pβ dμ(z) dv(w) |1 − z, w|n+1+ pβ− pm
dv(w)
.
Acknowledgements The authors would like to thank the editor and the referee(s) for their helpful comments and valuable suggestions for improving this manuscript.
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