Weighted Hardy Spaces of Quasiconformal Mappings
aa r X i v : . [ m a t h . C V ] A p r Weighted Hardy Spaces of Quasiconformal Mappings
Sita Benedict, Pekka Koskela and Xining Li
Abstract
We establish a weighted version of the H p -theory of quasiconformal mappings. Let f : B n → R n be a quasiconformal mapping, see Section 2.2 for the definition. Analogously tothe setting of analytic functions defined in the unit disk, we say that f belongs to Hardy space H p , < p < ∞ , provided that ( ∗ ) sup 1. Then f belongs to H p if andonly if ˆ M ( r, f ) p dr < ∞ . In [10], the weighted Hardy space for − < α < ∞ and 0 < p < ∞ was defined as the class of allunivalent functions for which ( ∗∗ ) ˆ M ( r, f ) p (1 − r ) α < ∞ . Notice that sup 0. Consequently, one cannot give a simple definition for weighted Hardy spaces based ona variant of ( ∗ ) or on weighted radial limits. Several equaivalent characterizations for membershipin weighted Hardy spaces were given in [6] and [10].1iven 0 < p < ∞ , − < α < ∞ and a quasiconformal mapping f : B → R n we write f ∈ H pα whenever ˆ (1 − r ) n − α M p ( r, f ) dr < ∞ . (1.1)We establish the following characterization of membership in H pα . Theorem 1.1. Let f : B n → R n be a quasiconformal mapping, < p < ∞ , and − < α < ∞ .Then the following are equivalent: || f || pH pα := ˆ (1 − r ) n − α M p ( r, f ) dr < ∞ . (1.2) ˆ (1 − r ) n − α ˆ B (0 ,r ) | f ( x ) | p − n | Df ( x ) | n dxdr < ∞ (1.3) ˆ (1 − r ) n − α ˆ B (0 ,r ) | Df ( x ) | n dx ! p/n dr < ∞ (1.4) If α ≥ or p ≥ n , the above conditions are further equivalent to ˆ B n a pf ( x )(1 − | x | ) p − α dx < ∞ . (1.5)Actually, we prove a bit more than what is stated in Theorem 1.1. Namely, for α ≥ ˆ S n − (cid:18) sup Lemma 2.1. Let f : B n → Ω be a K -quasiconformal mapping. There exists a constant C = C ( n, K ) such that for all x ∈ B n , we havediam ( f ( B x )) /C ≤ d ( f ( x ) , ∂ Ω) ≤ C diam ( f ( B x )) ≤ C d ( f ( B x ) , ∂ Ω) , and d ( f ( x ) , ∂ Ω)) /C ≤ | f ( y ) − f ( x ) | ≤ Cd ( f ( x ) , ∂ Ω)) for every y ∈ ∂B x . A quasiconformal mapping is only almost everywhere differentiable and hence we will employthe the concept of averaged derivative a f ( x ) = exp[ ˆ B x log J f ( y ) dy/ ( n | B x | )] . If f is a conformal mapping, then | Df | n = J f and especially a f = | Df | , see [2] for details and [1]for the origins of the definition.The following lemmas is from [2]. Lemma 2.2. Let f : B n → Ω be a K -quasiconformal mapping. There exists a constant C ( n, K ) ,with d ( f ( x ) , ∂ Ω) /C ≤ a f ( x )(1 − | x | ) ≤ Cd ( f ( x ) , ∂ Ω) and C ˆ B x | Df ( y ) | n dy ≤ Ca f ( x ) ≤ ˆ B x | Df ( y ) | n dy for all x ∈ B n . The following result is [3, Lemma 2.5]. Lemma 2.3. Let f : B n → Ω be a K -quasiconformal mapping, and suppose that u > satisfies u ( y ) /C ≤ u ( x ) ≤ Cu ( y ) for all x ∈ B n and y ∈ B x . Let < q ≤ n and p ≥ q . Then ˆ B n a pf u dx ≈ ˆ B n a p − qf | Df ( x ) | q u dx. with constants only depending on p, q, n, C, K . We continue with a useful estimate. Lemma 2.4. Let f : B n → Ω be a K -quasiconformal mapping. Let < p < ∞ and α ∈ R . Then ˆ S n − sup x ∈ Γ( ω ) d ( f ( x ) , ∂ Ω) p (1 −| x | ) α dσ ≤ C ˆ S n − sup x ∈ Γ( ω ) a pf ( x )(1 −| x | ) p + α dσ ≤ C ˆ B n a pf ( x )(1 −| x | ) p + α dx with constants C , C that only depend on n, K, p, α. roof. By Lemma 2.2 and Lemma 2.1 applied to x ∈ Γ( ω ) we have the estimate d ( f ( x ) , ∂ Ω)(1 − | x | ) α/p ≤ C a f ( x )(1 − | x | ) α/p ≤ C ( ˆ B x a pf ( y ) dy ) /p (1 − | x | ) − n/p + α/p ≤ C ( ˆ Γ( ω ) a f ( x ) p (1 − | x | ) p − n + α dx ) /p with constants that only depend on n, K, p, α. For any integrable function h on B n , by the Fubini theorem ˆ B n | h ( x ) | dx ≈ ˆ S n − ˆ Γ( ω ) | h ( y ) | (1 − | y | ) − n dydσ. Especially this holds for h ( x ) = a pf ( x )(1 − | x | ) p − α and the claim follows.A measure µ on B n is called a Carleson measure if there is a constant C µ such that µ ( B n ∩ B ( ω, r )) ≤ C µ r n − for all ω ∈ S n − , r > 0. The following lemma, see [7] and [3, Lemma 5.6], gives us a family ofCarleson measures. Lemma 2.5. If f is quasiconformal in B n , < p < n , and f ( x ) = 0 for all x ∈ B n , then themeasure µ defined by dµ = | Df ( x ) | p | f ( x ) | − p (1 − | x | ) p − dx is a Carleson measure on B n . Given a collection of locally rectifiable curves Γ in R n , the modulus of Mod(Γ) is defined as:Mod(Γ) = inf ˆ R n ρ n dx, where the infimum is taken for all nonnegative Borel functions ρ such that ´ γ ρds ≥ γ ∈ Γ.Given a K -quasiconformal mapping f : Ω → R n , one has Mod(Γ) /C ≤ Mod( f Γ) ≤ C Mod(Γ),where C = C ( K, n ). See e.g. [11] for a proof.We recall two useful estimates, see [11]. Given E ⊂ S n − , < r < , and the family Γ of radialsegments joining rE := { rx/ | x | : x ∈ E } and E , we haveMod(Γ) = σ ( E )(log(1 /r )) − n , where σ ( E ) is the surface area of E . For an upper bound, we always haveMod(Γ) ≤ ω n − log( R/r ) n − , if each γ ∈ Γ joins S n − ( x, r ) to S n − ( x, R ), 0 < r < R .The following modulus estimate that can be found in [3] and [12], is one of our key tools. Lemma 2.6. [3, Lemma 4.2, Remark 4.3] There exists a constant C = C ( n, K ) such that if f isK-quasiconformal in B n , x ∈ B n , M > , and α ≥ , then σ ( { ω ∈ S x : d ( f ( w ) , f ( x ))(1 − | x | ) α > M d ( f ( x ) , ∂f ( B n ))(1 − | x | ) α } ) ≤ Cσ ( S ( x ))(log M ) n − . .4 Nontangential and Radial Maximal Functions Given p > , α ≥ , we define the weighted radial maximal and nontangential maximal functionsby setting M f p,α ( ω ) = sup Lemma 2.7. Let f : B n → R n be a K -quasiconformal mapping and let < p < ∞ and α ≥ .There exists a constant C = C ( n, K, p, α ) such that ˆ S n − ( M ∗ p,α ( ω )) p dσ ( ω ) ≤ C ( n, K ) ˆ S n − ( M p,α f ( ω )) p dσ ( ω ) . (2.1) Proof. Given ω ∈ S n − and x ∈ Γ( ω ), there exists 0 < r < x ∈ B r ω . By thedefinition of B r ω , we have (1 − r ) ≤ − | x | ≤ − r ) and | r ω − x | ≤ (1 − r ) . Set r = (1 + r ) / . Then r ω ∈ ∂B r ω . Hence Lemma 2.1 gives that | f ( r ω ) − f ( x ) | ≤ C ( n, K ) | f ( r ω ) − f ( r ω ) | . (2.2)By the triangle inequality | f ( x ) | ≤ | f ( r ω ) | + | f ( r ω ) − ∂f ( x ) | (2.3)and | f ( r ω ) − f ( r ω ) | ≤ | f ( r ω ) | + | f ( r ω ) | . (2.4)By combining (2.2),(2.3),(2.4) we obtain | f ( x ) | ≤ C ( n, K ) ( | f ( r ω ) | + | f ( r ω ) | ) . Since 1 − r = (1 − r ) / , we conclude that( M ∗ p,α f ( ω )) p ≤ C ( n, K, α )( M p,α f ( ω )) p , and (2.1) follows. We begin with the following lemma. We will employ it in the proof of Lemma 3.3 to fix the problemthat M ( r, f )(1 − r ) α/p for α > M ( r, f ) is. Lemma 3.1. Let M : [0 , → [0 , ∞ ) be increasing and continuous with M (0) = 0 . Let p > , α ≥ and define N ( r ) = sup ≤ t ≤ r M ( t )(1 − t ) α/p . Then ˆ (1 − r ) n − α M p ( r ) dr < ∞ (3.1) if and only if ˆ (1 − r ) n − N p ( r ) dr < ∞ . (3.2)5 roof. Since M p ( r )(1 − r ) n − α = M p ( r )(1 − r ) α (1 − r ) n − ≤ (cid:18) sup ≤ t ≤ r M p ( t )(1 − t ) α (cid:19) (1 − r ) n − = N p ( r )(1 − r ) n − , we have that (3.2) implies (3.1) for any p .For the other direction, we may assume that N ( r ) is unbounded. Moreover, if the desiredconclusion is true for the case p = 1 and all M as in our formulation, then by applying it to c M ( r ) := M p ( r ) we obtain our claim for all p > 0. So it suffices prove that (3.1) implies (3.2) for p = 1.We define a sequence of points r k ∈ [0 , 1) as follows. Let r = 0 and set r k = inf { r : N ( r ) = 2 k − } . Then the continuity and monotonicity of N ( r ) gives that 2 k − = N ( r k ) = M ( r k )(1 − r k ) α . Hence ˆ N ( r )(1 − r ) n − dr ≤ ∞ X k =0 N ( r k +1 ) ˆ r k +1 r k (1 − r ) n − dr = ∞ X k =0 N ( r k +1 ) n − (cid:2) (1 − r k ) n − − (1 − r k +1 ) n − (cid:3) = ∞ X k =0 M ( r k +1 )(1 − r k +1 ) α n − (cid:2) (1 − r k ) n − − (1 − r k +1 ) n − (cid:3) = 2 n − ∞ X k =0 M ( r k )(1 − r k ) α (cid:2) (1 − r k ) n − − (1 − r k +1 ) n − (cid:3) = 2 n − ∞ X k =0 M ( r k ) (cid:2) (1 − r k ) n − α − (1 − r k ) α (1 − r k +1 ) n − (cid:3) ≤ n − ∞ X k =0 M ( r k ) (cid:2) (1 − r k ) n − α − (1 − r k +1 ) n − α (cid:3) We also have that ˆ (1 − r ) n − α M ( r ) dr ≥ ∞ X k =0 M ( r k ) ˆ r k +1 r k (1 − r ) n − α dr = ∞ X k =0 M ( r k ) n − α (cid:2) (1 − r k ) n − α − (1 − r k +1 ) n − α (cid:3) . The desired implication follows.We continue with a result on Carleson measures. Lemma 3.2. Let f : B n → R n be a quasiconformal mapping, < p < ∞ , α ≥ and let µ be aCarleson measure on B n . Then there is a constant C = C ( n, K, C µ ) such that ˆ B n | f ( x ) | p (1 − | x | ) α dµ ≤ C ˆ S n − ( M f p,α ) p dσ ( ω ) . Proof. By Lemma 2.7 it suffices to show that there exists a constant C ( n, K ) such that ˆ B n | f ( x ) | p (1 − | x | ) α dµ ≤ C ˆ S n − ( M ∗ f p,α ) p dσ ( ω ) . λ > 0, set E λ = { x ∈ B n : | f ( x ) | (1 − | x | ) α/p > λ } and U λ = ( ω ∈ S n − : sup x ∈ Γ( ω ) | f ( x ) | (1 − | x | ) α/p > λ ) . Recall the definion of the shadows S x from Subsection 2.1. They are spherical caps. We candecompose U λ into a Whitney-type decomposition of spherical caps. That is, we can write, U ( λ ) = ∞ [ k =1 S x k , where any ω ∈ U λ belongs to no more than N ( n ) spherical caps S x k and d ( S x k , ∂U ( λ )) ≈ diam( S x k ) ≈ (1 − | x k | ) , with universal constants. If x ∈ E λ and x = 0, then M ∗ f p,α ( ω ) > λ whenever x ∈ Γ ω . More-over, x | x | ∈ S x k for some k. Thus by the definition of S k and the properties of the Whitney-type decomposition there exists a universal constant C such that 1 − | x | ≤ C (1 − | x k | ) . Hence E λ ⊂ ∪ ∞ k =1 B ( x k / | x k | , C (1 − | x k | )). Therefore µ ( E λ ) ≤ ∞ X k =1 µ ( B ( x k / | x k | , C (1 − | x k | )) ∩ B n ) ≤ C ( n, C µ ) ∞ X k =1 (1 − | x k | ) n − ≤ C ( n, C µ ) ∞ X k =1 σ ( S x k ) ≤ C ( n, C µ ) σ ( U λ ) . This together with the Cavalieri formula gives ˆ B n | f ( x ) | p (1 − | x | ) α dµ = ˆ ∞ pλ p − µ ( E λ ) dλ ≤ C ( n, C µ ) ˆ ∞ pλ p − σ ( U λ ) dλ = C ( n, C µ ) ˆ S n − sup x ∈ Γ( ω ) | f ( x ) | p (1 − | x | ) α dσ ( ω )= C ˆ S n − ( M ∗ f p,α ( ω )) p dσ ( ω ) . We are now ready to prove a maximal characterization for H pα . By Lemma 2.7 we could alsoreplace the radial maximal function by the nontangential one. Lemma 3.3. Let f : B n → R n be a quasiconformal mapping, < p < ∞ and α ≥ . Then ˆ S n − sup M f p,α ( ω ) = sup 1) such that N ( r, f ) = λ , since N ( r, f ) is continuous. Our function N ( r, f ) isalso nondecreasing and we let r λ = max { r : N ( r, f ) = λ/ } . (3.6)We may assume that λ is so large that 1 / < r λ < . Let Γ E be the family of radial line segmentsconnecting B (0 , r λ ) and E ⊂ S n − . ThenMod(Γ E ) = σ ( E )(ln(1 /r λ )) − n ≥ σ ( E )2 − n (1 − r λ ) − n . By the definitions of E and r λ , for any γ ∈ Γ E , the image curve f ( γ ) connects B (0 , ( λ/ − r λ ) − α/p )to R n \ B (0 , λ (1 − r λ ) − α/p ), and therefore the modulus of the image family f Γ E satisfiesMod( f Γ E ) ≤ σ ( S n − )(ln 2) − n . By combining the above two estimates and using the quasi-invariance of the modulus, we arriveat the upper bound σ ( E ) ≤ C ( n, K )(1 − r λ ) n − . In order to prove (3.3) we may assume that M f p,α is unbounded on S n − . Define a measure ν on [0 , 1] by setting dν = (1 − r ) n − dr and recall the definition of r λ from (3.6). Now ν ( { r : N ( r, f ) > λ/ } ) = ˆ r λ (1 − r ) n − dr = (1 − r λ ) n − n − . Thus ˆ ∞ pλ p − σ ( { ω ∈ S n − : M f p,α ( ω ) > λ } ) dλ ≤ σ ( S n − )2 p N p (1 / , f ) + ˆ ∞ N (1 / ,f ) pλ p − σ ( { ω ∈ S n − : M f p,α ( ω ) > λ } ) dλ ≤ σ ( S n − )2 p N p (1 / , f ) + C ( n, K, p ) ˆ ∞ λ p − (1 − r λ ) n − dλ ≤ σ ( S n − )2 p N p (1 / , f ) + C ( n, K, p ) ˆ ∞ λ p − ˆ { r : N ( r,f ) >λ/ } (1 − r ) n − drdλ ≤ σ ( S n − )2 p N p (1 / , f ) + C ( n, K, p ) ˆ (1 − r ) n − N p ( r, f ) dr < ∞ , f (0) = 0 , we consider the quasiconformal mapping g defined by setting g ( x ) = f ( x ) − f (0) . Then (3.4) also holds with f replaced by g, and by the first part of our proof (3.3)follows with f replaced by g. We conclude with (3.3) via the triangle inequality.For the other direction, suppose that (3.3) holds, set r k := 1 − − k and choose x k ∈ B n so that | x k | = r k and | f ( x k ) | = M ( r k , f ). Then ˆ (1 − r ) n − α M p ( r, f ) dr ≤ n ∞ X k =1 (2 − k ) n − α M p ( r k , f ) = 2 n ˆ B n | f ( x ) | p (1 − | x | ) α dµ, where dµ = P ∞ k =1 (1 − | x | ) n − δ x k . Notice that µ is a Carleson measure. Hence Lemma 3.2 gives usthat ˆ (1 − r ) n − α M p ( r, f ) dr ≤ C ( n, K, C µ ) ˆ S n − sup Let f : B n → R n be a K -quasiconformal mapping, < p < ∞ , and α ≥ . Let v ( ω ) = sup x ∈ Γ( ω ) d ( f ( x ) , ∂f ( B n ))(1 − | x | ) α/p ∈ L p ( S n − ) . There exists C = C ( n, K, p, α ) such that ˆ S n − M f pp,α ( ω ) dσ ( ω ) ≤ C ˆ S n − v p ( ω ) dσ ( ω ) . Proof. Recall that M ∗ f p,α ( ω ) = sup x ∈ Γ( ω ) | f ( x ) | (1 − | x | ) α/p and M f p,α ( ω ) = sup L > . By the Cavalieri formula ˆ S n − M f pp,α ( ω ) dσ ( ω ) = L p ˆ ∞ pλ p − σ ( { ω ∈ S n − : M f p,α ( ω ) > Lλ } ) dλ. (3.7)Set Σ λ = σ ( { ω ∈ S n − : M f p,α ( ω ) > Lλ } ) . Then, for any γ > 0, we haveΣ λ ≤ σ ( { ω ∈ S n − : M f p,α ( ω ) > Lλ, v ( ω ) ≤ γ } ) + σ ( { ω ∈ S n − : v ( ω ) > γ } ) . If γ is a fixed multiple of λ, then the latter term is what we want but we need to obtain a suitableestimate for the first term.Towards this end, set E Lλ,γ = { ω ∈ S n − : M f p,α ( ω ) > Lλ, v ( ω ) ≤ γ } and define U ( λ ) = { ω ∈ S n − : M ∗ f p,α ( ω ) > λ } . L > ≥ , clearly E Lλ,γ ⊂ U ( λ ) . We utilize a generalized Whitney decomposition of theopen set U ( λ ) as in the proof of Lemma 3.2: U ( λ ) = ∪ S x k . where the caps S x k have uniformly bounded overlaps and d ( S x k , ∂U ( λ )) ≈ diam( S x k ) ≈ (1 − | x k | ) . (3.8)Suppose ω ∈ S x k is such that v ( ω ) ≤ γ and M f p,α ( ω ) > Lλ . According to (3.8), we can choose¯ ω ∈ ∂U ( λ ) with d ( ω, ¯ ω ) ≤ C diam( S ( x k )) . (3.9)Let ¯ x k ∈ Γ(¯ ω ) satisfy | ¯ x k | = | x k | . By (3.8), we conclude that d ( x k , ¯ x k ) ≤ C (1 − | x k | ). Hence Lemma2.1 allows us to conclude that d ( f ( x k ) , f (¯ x k ))(1 − | x k | ) α/p ≤ Cd ( f ( x k ) , ∂ Ω)(1 − | x k | ) α/p ≤ Cv ( ω ) ≤ Cγ. (3.10)Since ¯ ω / ∈ U ( λ ) , we may deduce from (3.10) that | f ( x k ) | (1 − | x k | ) α/p ≤ ( | f (¯ x k ) | + d ( f ( x k , f (¯ x k )))(1 − | x k | ) α/p ≤ λ + Cγ. (3.11)Next, the assumption that M f p,α ( ω ) > Lλ, allows us to choose choose r ω ∈ (0 , 1) such that | f ( r ω ω ) | (1 − r ω ) α/p ≥ M f p,α ( ω ) ≥ Lλ. (3.12)We proceed to show that 1 − r ω ≤ C (1 − | x k | ) (3.13)for an absolute constant C . Suppose not. Then 1 − | x k | ≤ C (1 − r ω ), which implies by (3.9) that d ( w, ¯ w ) ≤ C diam( S ( x k )) ≤ C (1 − | x k | ) ≤ CC (1 − r ω ) . This shows that r ω ω ∈ Γ ¯ ω when C > C. Since L > , we conclude that M ∗ f p,α (¯ ω ) > λ, which contradicts the assumption that ¯ ω / ∈ U ( λ ) . We may assume that C ≥ . By (3.12) together with (3.13) we obtain Lλ ≤ | f ( r ω ω ) | (1 − r ) α/p ≤ C α/p | f ( r ω ω ) | (1 − | x k | ) α/p . (3.14)Let us fix the value of L by choosing L = 4 C α/p . Then (3.14) yields2 λ ≤ | f ( r ω ω ) | (1 − | x j | ) α/p . (3.15)We proceed to estimate σ ( S x k ∩ E Lλ,γ ) . Let ω ∈ S x k ∩ E Lλ,γ . Then there is r ω ∈ (0 , 1) so thatboth (3.13) and (3.15) hold. Consider the collection of all the corresponding caps S r ω ω . By theBesicovitch covering theorem we find a countable subcollection of these caps, say S r ω , S r ω , ..., so that S x k ∩ E Lλ,γ ⊂ [ j S r j ω j (3.16)10nd P j χ S rjωj ( w ) ≤ C n for all ω ∈ S n − . By (3.13) we further have X j σ ( S r j ω j ) ≤ C σ ( S x k ) (3.17)for an absolute constant C . Fix one of the caps S r j ω j =: S j and let A ≥ . Write E j ( A ) = { w ∈ S j ∩ S x k ∩ E Lλ,γ : | f ( w ) − f ( r j ω j ) | ≥ Ad ( f ( r j ω j ) , ∂ Ω) } and E j ( A ) = { w ∈ S j ∩ S x k ∩ E Lλ,γ : | f ( w ) − f ( x k ) | ≥ Ad ( f ( x k ) , ∂ Ω) } . We claim that we can find a constant C only depending on C , p, α so that the choice λ = C Aγ guarantees that S j ∩ S x k ∩ E Lλ,γ = E j ( A ) ∪ E j ( A ) . (3.18)Let ω ∈ S j ∩ S x k ∩ E Lλ,γ . Suppose first that | f ( ω ) − f ( r j ω j ) | (1 − r j ) α/p ≥ Aγ. (3.19)Since ω ∈ E Lλ,γ ∩ S j , we have γ ≥ d ( f ( r j ω j ) , ∂ Ω)(1 − r j ) α/p , and we deduce from (3.19) that ω ∈ E j ( A ) . We are left to consider the case | f ( ω ) − f ( r j ω j ) | (1 − r j ) α/p < Aγ. (3.20)Under this condition, the triangle inequality together with (3.13), (3.14) and (3.11) give | f ( ω ) − f ( x k ) | (1 − | x k | ) α/p ≥ | f ( ω ) | (1 − | x k | ) α/p − | f ( x k ) | (1 − | x k | ) α/p ≥ ( | f ( r j ω j ) | − | f ( ω ) − f ( r j ω j ) | ) (1 − r j ) α/p C α/p − | f ( x k ) | (1 − x k ) α/p ≥ Lλ C α/p − AγC α/p − ( λ + Cγ ) ≥ λ − ( λ + Cγ ) − AγC α/p . (3.21)We now fix the relation between λ and γ by setting λ = ( C + AC α/p + 1) γ. Then (3.21) reduces to | f ( ω ) − f ( x k ) | (1 − | x k | ) α/p ≥ Aγ ≥ Ad ( f ( x k ) , ∂ Ω)(1 − | x k | ) α/p and we conclude that ω ∈ E j ( A ) . According to Lemma 2.6, σ ( E j ( A )) ≤ C σ ( S j )(log A ) n − , (3.22)where C depends only on K, n. Thus (3.22) together with (3.17) gives X j σ ( E j ( A )) ≤ C C σ ( S x k )(log A ) n − . (3.23)11e also deduce via Lemma 2.6 that σ ( ∪ j E j ( A )) ≤ σ ( { ω ∈ S x k : | f ( w ) , f ( x k ) | ≥ Ad ( f ( x ) , ∂ Ω) } ≤ C σ ( S x k )(log A ) n − . (3.24)Now (3.18) together with (3.23) and (3.24) gives σ ( S x k ∩ E Lλ,γ ) ≤ C σ ( S x k )(log A ) n − , (3.25)where C depends only on K, n. By the choice of the caps S x k , the definition of E Lλ,γ and (3.25) give via summing over k theestimate Σ λ ≤ σ ( E Lλ,γ ) + σ ( { ω ∈ S n − : v ( ω > γ ) } ) ≤ C σ ( U ( λ ))(log A ) n − + σ ( { ω ∈ S n − : v ( ω > γ ) } ) . (3.26)We insert (3.26) into (3.7) and conclude that ˆ S n − M f pp,α ( ω ) dσ ( ω ) = L p ˆ ∞ pλ p − Σ λ dλ ≤ L p ˆ ∞ pλ p − C σ ( U ( λ ))(log A ) n − dλ + L p ˆ ∞ pλ p − σ ( { ω ∈ S n − : v ( ω > γ ) } ) dλ ≤ C L p (log A ) n − ˆ S n − M ∗ p,α f p ( ω ) dω + L p ˆ ∞ pλ p − σ ( { ω ∈ S n − : v ( ω > γ ) } ) dλ. (3.27)Suppose that the integral on the left-hand-side of (3.27) is finite. Then Lemma 2.7 allows usto choose A only depending on K, n, p, α, L, C so that the integral of M ∗ p,α f p can be embeddedinto the left-hand-side. In this case our claim follows via the Cavalieri formula, recalling that λ = ( C + AC α/p + 1) γ. We are left with the case where the integral on the left-hand-side of (3.27)is infinite. In this case, we replace f by the K -quasiconformal map f j defined by setting f j ( x ) = f ((1 − /j ) x ) . Since the corresponding integral is now finite, we obtain a uniform estimate for theintegral of M f jp,α in terms of the integral of v j , defined analogously. The desired estimate followsvia the Fatou lemma by letting j tend to infinity since it easily follows that v j ( ω ) ≤ v ( ω ) for all ω and that M f jp,α ( ω ) → M f p,α ( ω ) for a.e. ω. Lemma 3.5. Let f : B n → R n be a quasiconformal mapping, < p < ∞ and α ≥ . Then thefollowing are equivalent:1. ´ S n − M f pp,α dσ < ∞ ´ B n a pf ( x )(1 − | x | ) p − α dx < ∞ ´ S n − sup x ∈ Γ( ω ) a pf ( x )(1 − | x | ) p + α dσ < ∞ Proof. ( = ⇒ ) Suppose first that 0 < p ≤ . We may assume that f = 0 in B n . Then themeasure given by dµ = | Df | p | f | − p (1 − | x | ) p − dx is a Carleson measure by Lemma 2.5 and hence12emma 2.3 and Lemma 3.2 give ˆ B n a pf ( x )(1 − | x | ) p − α dx ≤ C ˆ B n | Df | p (1 − | x | ) p − α dx ≤ C ˆ B | f ( x ) | p (1 − | x | ) α dµ ( x ) ≤ C ˆ S n − M f pp,α ( ω ) dσ. Suppose finally that p > y ∈ ∂f ( B n ). By Lemma 2.3 and Lemma 2.2, we have ˆ B n a pf ( x )(1 − | x | ) p − α dx ≤ C ˆ B n | Df | a p − f ( x )(1 − | x | ) p − α dx ≤ C ˆ B n | Df | d ( f ( x ) , ∂f ( B n )) p − (1 − | x | ) α dx ≤ C ˆ B n | Df || f ( x ) − y | p − (1 − | x | ) α dx. Since f ( x ) − y = 0 in B n , the measure given by dµ = | Df ( x ) || f ( x ) − y | − dx is a Carleson measureLemma 2.5. Hence we can apply Lemma 3.2 to conclude that ´ B n a pf ( x )(1 − | x | ) p − α dx < ∞ . ( = ⇒ ) This follows from Lemma 2.4.( = ⇒ ) By Lemma 2.2 we have that d ( f ( x ) , ∂ Ω) ≤ Ca f ( x )(1 − | x | ) . Hence ( ) follows from( ) by Lemma 3.4. Lemma 3.6. Let f : B n → R n be a quasiconformal mapping with f (0) = 0 . Let p ≥ n and − < α .Then ˆ (1 − r ) n − α M p ( r, f ) dr ≤ C ˆ B n a f ( x ) p (1 − | x | ) p − α dx. Proof. Define v ( ω ) = sup x ∈ Γ( ω ) d ( f ( x ) , ∂f ( B n ))(1 − | x | ) α/p . By Lemma 2.4 we only need to showthat ˆ (1 − r ) n − α M ( r, f ) p dr ≤ C ˆ S n − v p dσ. (3.28)For each i ≥ , let r i = 1 − − i and pick x i ∈ S n − ( r i ) with | f ( x i ) | = M ( r i , f ). Then ˆ (1 − r ) n − α M ( r, f ) p dr = ∞ X i =1 ˆ r i r i − (1 − r ) n − α M ( r, f ) p dr ≤ C ∞ X i =1 | f ( x i ) | p (1 − | x i | ) n − α . (3.29)Let ˜ C be a constant, to be determined later, and let G ( f ) = { i ∈ N : | f ( x i ) | ≤ ˜ Cd ( f ( x i ) , ∂f ( B n )) } and B ( f ) = N \ G ( f ) . For i ∈ G ( f ) and ω ∈ S ( x i ), we have | f ( x i ) | p (1 − | x i | ) n − α ≤ ˜ C p d ( f ( x i ) , ∂f ( B n )) p (1 − | x i | ) n − α ≤ ˜ C p v f ( ω ) p (1 − | x i | ) n − . (3.30)Letting δ = 1 / ˜ C , for i ∈ B ( f ) , we have d ( f ( x i ) , ∂f ( B n )) ≤ δ | f ( x i ) | . Set ω i = x i | x i | , and let y i − = r i − ω i . Then we have x i ∈ B y i − . Hence Lemma 2.1 gives | f ( x i ) − f ( y i − ) | ≤ diam f ( B y i − ) ≤ Cd ( f ( B y i − ) , ∂ Ω) ≤ C d ( f ( x i ) , ∂ Ω) . Therefore, by the choice of x i − , we obtain | f ( x i ) | ≤ | f ( y i − ) | + C δ | f ( x i ) | ≤ | f ( x i − ) | + C δ | f ( x i ) | . 13f ˜ C is sufficiently large, then C δ < | f ( x i ) | ≤ λ | f ( x i − ) | , (3.31)where λ = 1 / (1 − C δ ) . By multiplying both sides of (3.31) to (1 − | x i | ) ( n − α ) /p and raising topower p , we conclude that | f ( x i ) | p (1 − | x i | ) n − α ≤ λ p | f ( x i − ) | p (1 − | x i | ) n − α = λ p | f ( x i − ) | p − ( n − α ) (1 − | x i − | ) n − α . (3.32)Now, notice that n − α > α > − 1. By recalling that δ = 1 / ˜ C and λ = 1 / (1 − C δ ), wefind ˜ C = ˜ C ( p, C ) big enough such that λ p − ( n − α ) < C ≥ C . Then there exists c < , suchthat | f ( x i ) | p (1 − | x i | ) n − α ≤ c | f ( x i − ) | p (1 − | x i − | ) n − α . Since x = 0 and f (0) = 0 , we have 0 ∈ G ( f ) . If i − ∈ B ( f ) , we repeat the above argument with i replaced by i − | f ( x i ) | p (1 − | x i | ) n − α ≤ c | f ( x i − ) | p (1 − | x i − | ) n − α . We repeat inductively until i − k ∈ G ( f ) : for each i ∈ B ( f ), there exists k such that l ∈ B ( f ), forall i − k < l ≤ i, i − k ∈ G ( f ) ande | f ( x i ) | p (1 − | x i | ) n − α ≤ c k | f ( x i − k ) | p (1 − | x i − k | ) n − α . Therefore, ∞ X i =0 | f ( x i ) | p (1 − | x i | ) n − α ≤ C X i ∈ G ( f ) | f ( x i ) | p (1 − | x i | ) n − α ≤ C ˜ C X i ∈ G ( f d ( f ( x i ) , ∂f ( B n )) p (1 − | x i | ) α (1 − | x i | ) n − . (3.33)Set S i = S x i for i ∈ G ( f ). Since | x i | = 1 − − i , the definition of the shadows S i , gives that X j>i σ ( S j ) ≤ σ ( S i ) . Thus, we also have σ ( { ω ∈ S i : X j>i χ S j ( ω ) ≥ } ) ≤ σ ( S i ) . Hence there is ˆ S i ⊂ S i with σ ( ˆ S i ) ≥ σ ( S i ) = C n (1 − | x i | ) n − and so that no point of ˆ S i belongsto more than one S j . Then X j ∈ G ( f ) χ ˆ S i ( ω ) ≤ . By combining (3.30) and (3.33), we finally obtain that ∞ X i =0 | f ( x i ) | p (1 − | x i | ) n − α ≤ C ˜ CC n X i ∈ S ( f ) ˆ ˆ S i v p ( ω ) dσ ≤ C ˜ CC n ˆ S n − v p ( ω ) dσ. This together with (3.29) gives (3.28) and hence our claim follows.14 roof of Theorem 1.1. We will first prove the equivalence of (1.2), (1.3) and (1.4).We assume f (0) = 0 and handle the cases 0 < p ≤ n and p > n separately. Case 1 First suppose 0 < p ≤ n . Then ˆ B (0 ,r ) | f | p − n | Df | n dx ≤ K ˆ B (0 ,r ) | f | p − n J f ( x ) dx = K ˆ f ( B (0 ,r )) | y | p − n dy ( ∗ ) ≤ K ˆ B (0 , n √ | f ( B (0 ,r )) | ) | y | p − n dy = K ˆ S n − ˆ n √ ( | f ( B (0 ,r ))0 t p − dtdσ = C ( K, n, p ) | f ( B (0 , r )) | p/n = C ˆ B (0 ,r ) J f ( x ) dx ! p/n ≤ C ˆ B (0 ,r ) | Df ( x ) | n dx ! p/n , where ( ∗ ) holds since the weight function | y | p − n is radially decreasing when 0 < p ≤ n. We haveproved that (1.4) ⇒ (1.3).Now let g = | f | ( p − n ) /n f. Since quasiconformal mappings are differentiable almost everywhere,we can calculate that Dg = | f | ( p − n ) /n (cid:18) I + p − nn f T f | f | (cid:19) Df (for a.e. x ∈ B n ) , and so | Dg | . | f | ( p − n ) /n | Df | . Then Fubini’s theorem and Lemma 2.3 give ˆ (1 − r ) n − α ˆ B (0 ,r ) | f | p − n | Df | n & ˆ (1 − r ) n − α ˆ B (0 ,r ) | Dg | n dxdr ≈ ˆ B n | Dg | n (1 − | x | ) n − α dx ≈ ˆ B n a ng ( x )(1 − | x | ) n − α dx. So by assuming (1.3) in the statement of the theorem holds, we can apply Lemma 3.6 to thequasiconformal mapping g , which gives ˆ (1 − r ) n − α M p ( r, f ) dr ≈ ˆ (1 − r ) n − α M n ( r, g ) dr . ˆ B n a g ( x ) n (1 − | x | ) n − α dx . ˆ (1 − r ) n − α ˆ B (0 ,r ) | f | p − n | Df | n dxdr. (3.34)We have shown that (1.3) = ⇒ (1.2) . The implication that (1.2) = ⇒ (1.4) is a straightforward application of H¨older’s inequalityand the definition of M ( r, f ). Indeed, ˆ (1 − r ) n − α ˆ B (0 ,r ) | Df ( x ) | n dx ! p/n dr ≤ ˆ (1 − r ) n − α ˆ B (0 ,r ) | Df ( x ) | n | f ( x ) | p − n M ( r, f ) n − p dx ! p/n dr = ˆ (1 − r ) ( n − α )( n − p ) /n M ( r, f ) ( n − p ) p/n (1 − r ) n − α ˆ B (0 ,r ) | Df ( x ) | n | f ( x ) | p − n dx ! p/n dr ≤ (cid:18) ˆ (1 − r ) n − α M ( r, f ) p dr (cid:19) ( n − p ) /n ˆ (1 − r ) n − α ˆ B (0 ,r ) | Df | n | f | p − n dxdr ! p/n . (3.35)15ince ˆ B (0 ,r ) | Df | n | f | p − n dxdr ≤ K ˆ f ( B (0 ,r )) | y | p − n dy ≤ K ˆ B (0 ,M ( r,f )) | y | p − n dy ≤ C ( n, K, p ) M ( r, f ) p , the desired result follows. Case 2 We assume that p > n. Now ˆ B (0 ,r ) | Df | n dx ≤ K ˆ B (0 ,r ) J f ( x ) dx ≤ K | f ( B (0 , r )) | ≤ KM ( r, f ) n Therefore, (1.2) implies (1.4).Set g = | f | ( p − n ) /n f. Then | g | n = | f | p , and M ( r, f ) p = M ( r, g ) n . Analogously to (3.34), byLemma 3.6. We have ˆ (1 − r ) n − α M p ( r, f ) dr ≤ ˆ (1 − r ) n − α M n ( r, g ) dr . ˆ (1 − r ) n − α ˆ B (0 ,r ) | f | p − n | Df | n dxdr. (3.36)Hence (1.3) = ⇒ (1.2) . We need to show that (1.4) = ⇒ (1.3) . First of all, ˆ (1 − r ) n − α ˆ B (0 ,r ) | f | p − n | Df | n dxdr ≤ ˆ (1 − r ) n − α M ( r, f ) p − n ˆ B (0 ,r ) | Df | n dxdr H¨older ≤ ( ˆ (cid:0) (1 − r ) n − α M ( r, f )) p dr (cid:1) ( p − n ) /p ˆ (1 − r ) n − α ˆ B (0 ,r ) | Df | n dx ! p/n dr n/p . (3.37)If the later term on the right-hand-side is finite, then we obtain via (3.36) that ˆ (1 − r ) n − α ˆ B n | f | p − n | Df | n dxdr ≤ C ˆ (1 − r ) n − α (cid:18) ˆ B n | Df | n dx (cid:19) p/n dr. Since the constant in this inequality only depends on n, K, p, α the general case easily follows byapplying this estimate with f replaced by f j , defined by setting f j ( x ) = f ((1 − /j ) x ) , and bypassing to the limit.We have shown the equivalence of (1.2),(1.3) and (1.4) under the additional assumption that f (0) = 0 . The general case follows since each of them holds for a f if and only if it holds for g, defined by setting g ( x ) = f ( x ) − f (0) . We are left with the equivalence of (1.2)–(1.5), when α ≥ − < α < p ≥ n .For α ≥ , by Lemma 3.3 and Lemma 3.5, we know that (1.2) is equivalent to (1.5).For − < α < , p ≥ n , by Lemma 3.6 we know that (1.5) implies (1.1), so we only need toshow that (1.3) implies (1.5) Fix y ∈ ∂f ( B ). Then, Lemma 2.2 and 2.3 give us that ˆ B n a f ( x ) p (1 − | x | ) p − α dx ≈ ˆ B n a f ( x ) p − n (1 − | x | ) p − n | Df ( x ) | n (1 − | x | ) n − α dx ≤ C ˆ B n d ( f ( x ) , ∂ Ω) p − n | Df ( x ) | n (1 − | x | ) n − α dx ≤ C ˆ B n | f ( x ) − y | p − n | Df ( x ) | n (1 − | x | ) n − α dx ≈ ˆ (1 − r ) n − α ˆ B (0 ,r ) | f ( x ) − y | p − n | Df ( x ) | n dxdr. (3.38)16otice that | f ( x ) − y | ≤ | f ( x ) | + | y | and | f ( x ) − y | p − n ≤ C ( p, n )( | f ( x ) | p − n + | y | p − n ). By (3.38), weneed to show that ˆ (1 − r ) n − α ˆ B (0 ,r ) ( | f ( x ) | p − n + | y | p − n ) | Df ( x ) | n dxdr == ˆ (1 − r ) n − α ˆ B (0 ,r ) | f ( x ) | p − n | Df ( x ) | n dxdr + | y | p − n ˆ (1 − r ) n − α ˆ B (0 ,r ) | Df ( x ) | n dxdr =( I ) + ( II ) < ∞ . From the equivalenc of (1.1) and (1.3), we know that ( I ) < ∞ . On the other hand, we have ´ (1 − r ) n − α ´ B (0 ,r ) | Df ( x ) | n dxdr ≤ (cid:18) ´ (1 − r ) n − α (cid:16) ´ B (0 ,r ) | Df ( x ) | n (cid:17) p/n dxdr (cid:19) n/p (cid:16) ´ (1 − r ) n − α dr (cid:17) ( p − n ) /p . Apply the equivalence of (1.1) and (1.4) and ´ (1 − r ) n − α dr < ∞ for − < α < . Then we haveshown that ( II ) < ∞ . Therefore, we have finished the proof of Theorem 1.1. References [1] K. Astala, F. Gehring, Injectivity, the BMO-norm and the universal Teichm¨uller space , J.Analyse Math. (1986), 16-57.[2] K. Astala, F. Gehring, Quasiconformal analogues of theorems of Koebe and Hardy-Littlewood, Mich. Math. J. (1985), 99-107.[3] K. Astala, and P. Koskela, H p theory for quasiconformal mappings , Pure Appl. Math. Q. (2011), no. 1, 19-50.[4] A. Baernstein, D. Girela and J. Pel´aez, Univalent functions, Hardy spaces and spaces of Dirich-let type , Illinois J. Math. (2004), no. 3, 837-859.[5] S. Benedict, Hardy-Orlicz Spaces of Conformal Densities , Conform. Geom. Dyn. (2015),146-158.[6] D. Girela, M. Pavlovi´c and J. A. Pel´aez, Spaces of analytic functions of Hardy-Bloch type , J.Anal. Math. (2006), 53-31.[7] P. Jones, Extension Theorems for BMO , Indiana Math. J. (1979), 41- 66.[8] P. Koskela, An inverse Sobolev Lemma, Rev. Mat. Iberoamericana (1994),123-141.[9] P. Koskela, and S. Benedict, Intrinsic Hardy-Orlicz spaces of conformal mappings. Bull. Lond.Math. Soc (2015), no.1, 75-84.[10] F. P´erez-Gonz´alez and J. R¨atty¨a, Univalent Functions in Hardy, Bergman, Bloch and RelatedSpaces. J. Anal. Math. (2008), 125-148.[11] J. V¨ais¨al¨a, Lectures on n-dimensional quasiconformal mappings, Lecture Notes Mathematics,Springer Verlage, vol. , 1971. 1712] M. Zinsmeister, A distortion theorem for quasiconformal mappings , Bull. Soc. Math France (1986), 123-133. Addresses:Sita Benedict: Department of Matematics, University of Hawaii, Honolulu, HI 96822, USAE-mail: [email protected] Pekka Koskela: Department of Mathematics, University of Jyv¨askyl¨a, P.O. Box 35, FIN-40351 Jyv¨askyl¨a,FinlandE-mail: [email protected] Xining Li: School of Mathematics, Sun Yat-Sen University, Guangzhou, China 510275E-mail: [email protected]@mail.sysu.edu.cn