HOLOMORPHIC MEAN LIPSCHITZ SPACES AND HARDY SOBOLEV
HOLOMORPHIC MEAN LIPSCHITZ SPACES AND HARDY SOBOLEV SPACES ON THE UNIT BALL HONG RAE CHO∗ AND KEHE ZHU A BSTRACT. We study two classes of holomorphic functions in the unit ball Bn of Cn : mean Lipschitz spaces and Hardy Sobolev spaces. Main results include new characterizations in terms of fractional radial differential operators and various comparisons between these two classes.
1. I NTRODUCTION For points z = (z1 , · · · , zn ) and w = (w1 , · · · , wn ) in Cn we write p hz, wi = z1 w1 + · · · + zn wn , |z| = |z1 |2 + · · · + |zn |2 . Let Bn = {z ∈ Cn : |z| < 1} denote the open unit ball and let Sn = {ζ ∈ Cn : |ζ| = 1} denote the unit sphere in Cn . The normalized Lebesgue measures on Bn and Sn will be denoted by dv and dσ, respectively. Let H(Bn ) denote the space of all holomorphic functions in Bn . Given 0 < r < 1, 0 < p < ∞, and f ∈ H(Bn ), we define Z 1/p p Mp (r, f ) = |f (rζ)| dσ(ζ) Sn p
and call it the L -mean of f over the sphere |z| = r. When p = ∞, we write M∞ (r, f ) = sup{|f (rζ)| : ζ ∈ Sn }. For 0 < p ≤ ∞ the Hardy space H p consists of all functions f ∈ H(Bn ) such that kf kH p = sup Mp (r, f ) < ∞. 0 0, an application of the Stirling’s formula shows that Γ(n + 1 + s)Γ(n + 1 + k + s + t) ∼ kt Γ(n + 1 + s + t)Γ(n + 1 + k + s)
as k → ∞.
Therefore, Rs,t is indeed a fractional radial differential operator of order t. It is obvious that Rs,t is the inverse of Rs,t and so will be called a fractional integral operator of order t. One of the main advantages of the operators Rs,t and Rs,t is that they interact well with Bergman type and Cauchy type kernels on the unit ball. This is made precise by the following result. Lemma 2.1 ([20]). Suppose neither n + s nor n + s + t is a negative integer. Then Rs,t
1 1 = , (1 − hz, wi)n+1+s (1 − hz, wi)n+1+s+t
Rs,t
1 1 = . (1 − hz, wi)n+1+s+t (1 − hz, wi)n+1+s
and
For s > −1 the weighted Lebesgue measure dvs is defined by dvs (z) = cs (1 − |z|2 )s dv(z), where Γ(n + s + 1) n! Γ(s + 1) is a normalizing constant so that dvs is a probability measure on Bn . The following lemma is a consequence of the above formulas. cs =
Lemma 2.2. Suppose f is holomorphic in Bn . If s > −1 and n + s + t is not a negative integer, then Z f (rw)dvs (w) s,t . R f (z) = lim− (1 − hz, wi)n+1+s+t r→1 n B
HONG RAE CHO∗ AND KEHE ZHU
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If s = −1 and n + t − 1 is not a negative integer, then Z f (rζ)dσ(ζ) −1,t R f (z) = lim− . (1 − hz, ζi)n+t r→1 n S Moreover, the limits above always exist. Proof. The case s > −1 is proved in [20] and the proof is based on the reproducing property of the Bergman kernel. The case s = −1 can be proved in a similar fashion using the Cauchy integral formula for the unit ball. By the definition of the two-parameter fractional radial differential operator Rs,t , we have the following elementary relations. Lemma 2.3. Suppose that neither n + s nor n + s + t is a negative integer. Then we have Rs,t = Rs+t,−t . Lemma 2.4. Suppose that none of n + λ, n + λ + t, and n + λ + s + t is a negative integer. Then we have Rλ,t Rλ+t,s = Rλ+t,s Rλ,t = Rλ,s+t . Lemma 2.5. Suppose f ∈ H(Bn ) and n + s is not a negative integer. Then Rs,1 f = f +
Rf . n+s+1
Proof. 1 1 = Rs,1 , (1 − hz, wi)n+2+s (1 − hz, wi)n+1+s and 1 (1 − hz, wi)n+2+s
= =
1 hz, wi + n+1+s (1 − hz, wi) (1 − hz, wi)n+2+s 1 1 1 + R . (1 − hz, wi)n+1+s n + 1 + s (1 − hz, wi)n+1+s
Differentiating with respect to w ¯ at w = 0 then leads to the result.
Lemma 2.6. Let m be a positive integer. If n + s is not a negative integer, then there exist constants {c0 , c1 , · · · , cm } such that Rs,m =
m X
cj Rj ,
j=0
where c0 = 1 and cm 6= 0. Proof. Since Rs+j,m−j = Rs+j,1 Rs+j+1,m−j−1 ,
j = 0, 1, . . . , m − 2,
we have Rs,m = Rs,1 Rs+1,1 · · · Rs+m−1,1 . By Lemma 2.5, we have X m R R R Rs,m = 1 + 1+ ··· 1 + = cj Rj . n+s+1 n+s+2 n+s+m j=0
MEAN LIPSCHITZ SPACES AND HARDY SOBOLEV SPACES
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3. H OLOMORPHIC MEAN L IPSCHITZ SPACES Note that our definition of the holomorphic mean Lipschitz spaces Λp,q α makes perfect sense when α ≤ 0. The restriction α > 0 ensures that the functions we consider will have boundary values. p Lemma 3.1. Suppose 0 < α < ∞, 1 ≤ p < ∞, and 1 ≤ q ≤ ∞. Then Λp,q α ⊂H .
Proof. Let 1/2 < r < 1 and σα = α − m, where m = [α] is the integer part of α. By the fundamental theorem of calculus and Minkowski’s inequality, we have Z r Mp (t, Rf )dt. Mp (r, f ) . sup |f (z)| + |z|
1. I NTRODUCTION For points z = (z1 , · · · , zn ) and w = (w1 , · · · , wn ) in Cn we write p hz, wi = z1 w1 + · · · + zn wn , |z| = |z1 |2 + · · · + |zn |2 . Let Bn = {z ∈ Cn : |z| < 1} denote the open unit ball and let Sn = {ζ ∈ Cn : |ζ| = 1} denote the unit sphere in Cn . The normalized Lebesgue measures on Bn and Sn will be denoted by dv and dσ, respectively. Let H(Bn ) denote the space of all holomorphic functions in Bn . Given 0 < r < 1, 0 < p < ∞, and f ∈ H(Bn ), we define Z 1/p p Mp (r, f ) = |f (rζ)| dσ(ζ) Sn p
and call it the L -mean of f over the sphere |z| = r. When p = ∞, we write M∞ (r, f ) = sup{|f (rζ)| : ζ ∈ Sn }. For 0 < p ≤ ∞ the Hardy space H p consists of all functions f ∈ H(Bn ) such that kf kH p = sup Mp (r, f ) < ∞. 0 0, an application of the Stirling’s formula shows that Γ(n + 1 + s)Γ(n + 1 + k + s + t) ∼ kt Γ(n + 1 + s + t)Γ(n + 1 + k + s)
as k → ∞.
Therefore, Rs,t is indeed a fractional radial differential operator of order t. It is obvious that Rs,t is the inverse of Rs,t and so will be called a fractional integral operator of order t. One of the main advantages of the operators Rs,t and Rs,t is that they interact well with Bergman type and Cauchy type kernels on the unit ball. This is made precise by the following result. Lemma 2.1 ([20]). Suppose neither n + s nor n + s + t is a negative integer. Then Rs,t
1 1 = , (1 − hz, wi)n+1+s (1 − hz, wi)n+1+s+t
Rs,t
1 1 = . (1 − hz, wi)n+1+s+t (1 − hz, wi)n+1+s
and
For s > −1 the weighted Lebesgue measure dvs is defined by dvs (z) = cs (1 − |z|2 )s dv(z), where Γ(n + s + 1) n! Γ(s + 1) is a normalizing constant so that dvs is a probability measure on Bn . The following lemma is a consequence of the above formulas. cs =
Lemma 2.2. Suppose f is holomorphic in Bn . If s > −1 and n + s + t is not a negative integer, then Z f (rw)dvs (w) s,t . R f (z) = lim− (1 − hz, wi)n+1+s+t r→1 n B
HONG RAE CHO∗ AND KEHE ZHU
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If s = −1 and n + t − 1 is not a negative integer, then Z f (rζ)dσ(ζ) −1,t R f (z) = lim− . (1 − hz, ζi)n+t r→1 n S Moreover, the limits above always exist. Proof. The case s > −1 is proved in [20] and the proof is based on the reproducing property of the Bergman kernel. The case s = −1 can be proved in a similar fashion using the Cauchy integral formula for the unit ball. By the definition of the two-parameter fractional radial differential operator Rs,t , we have the following elementary relations. Lemma 2.3. Suppose that neither n + s nor n + s + t is a negative integer. Then we have Rs,t = Rs+t,−t . Lemma 2.4. Suppose that none of n + λ, n + λ + t, and n + λ + s + t is a negative integer. Then we have Rλ,t Rλ+t,s = Rλ+t,s Rλ,t = Rλ,s+t . Lemma 2.5. Suppose f ∈ H(Bn ) and n + s is not a negative integer. Then Rs,1 f = f +
Rf . n+s+1
Proof. 1 1 = Rs,1 , (1 − hz, wi)n+2+s (1 − hz, wi)n+1+s and 1 (1 − hz, wi)n+2+s
= =
1 hz, wi + n+1+s (1 − hz, wi) (1 − hz, wi)n+2+s 1 1 1 + R . (1 − hz, wi)n+1+s n + 1 + s (1 − hz, wi)n+1+s
Differentiating with respect to w ¯ at w = 0 then leads to the result.
Lemma 2.6. Let m be a positive integer. If n + s is not a negative integer, then there exist constants {c0 , c1 , · · · , cm } such that Rs,m =
m X
cj Rj ,
j=0
where c0 = 1 and cm 6= 0. Proof. Since Rs+j,m−j = Rs+j,1 Rs+j+1,m−j−1 ,
j = 0, 1, . . . , m − 2,
we have Rs,m = Rs,1 Rs+1,1 · · · Rs+m−1,1 . By Lemma 2.5, we have X m R R R Rs,m = 1 + 1+ ··· 1 + = cj Rj . n+s+1 n+s+2 n+s+m j=0
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3. H OLOMORPHIC MEAN L IPSCHITZ SPACES Note that our definition of the holomorphic mean Lipschitz spaces Λp,q α makes perfect sense when α ≤ 0. The restriction α > 0 ensures that the functions we consider will have boundary values. p Lemma 3.1. Suppose 0 < α < ∞, 1 ≤ p < ∞, and 1 ≤ q ≤ ∞. Then Λp,q α ⊂H .
Proof. Let 1/2 < r < 1 and σα = α − m, where m = [α] is the integer part of α. By the fundamental theorem of calculus and Minkowski’s inequality, we have Z r Mp (t, Rf )dt. Mp (r, f ) . sup |f (z)| + |z|
