COMPOSITION OPERATORS ON HARDY-SOBOLEV SPACES

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1 Indian J. Pure Appl. Math., 46(3): 55-67, June 015 c Indian National Science Academy DOI: /s x COMPOSITION OPERATORS ON HARDY-SOBOLEV SPACES Li He, Guang Fu Cao 1 and Zhong Hua He Department of Mathematics, Guangzhou University, Guangzhou, , P.R. China Department of Mathematics, Guangdong University of Finance Department of Applied Mathematics, Guangzhou, 51051, P.R. China s: helichangsha1986@163.com; guangfucao@163.com; gfcao@gzhu.edu.cn; zhonghuahe010@163.com (Received 14 May 013; accepted 1 July 014) In this paper, we obtain some growth estimates for Hardy-Sobolev functions of the unit ball. We also give the representation of the spectra of composition operators C ϕ on H (B n) and describe the Fredholmness of C ϕ equivalently. Key words : Hardy-Sobolev space; composition operator; spectrum; Fredholm operator. 1. INTRODUCTION For any points z = (z 1, z,..., z n ) and w = (w 1, w,..., w n ) in C n, we denote z, w = z 1 w 1 + z w z n w n and z = z, z = z 1 + z z n. Let B n be the open unit ball in C n, and = B n its boundary. The normalized Lebesgue measure on B n and will be denoted as dv and dσ respectively. Let H(B n ) be the space of all holomorphic functions in B n. For any R, we consider the special fractional radial differential operator R defined on H(B n ) as R f(z) = (1 + k) f k (z), k=0 1 Corresponding author supported by National Natural Science Foundation of China (No ) and Doctoral Fund of Ministry of Education of Guangdong Province (No ).

2 56 LI HE et al. where f(z) = k=0 f k(z) is the homogeneous expansion of f. Given 0 < r < 1 and 0 < p < +, we define and the norm of f in H p (B n) is H p (B n) = {f H(B n ) R f H p } { f H p = R f H p = sup R f(rζ) p p dσ(ζ), 0<r<1 then (H p (B n), H p) is a Banach space called as Hardy-Sobolev space. For p =, we define the Hardy-Sobolev space H as H = {f R f H }. Let ϕ be a self-map of B n, the composition operator C ϕ for f C ϕ f = f ϕ. H p (B n) is defined by For any point z B n, α N n, by direct calculation, we have z α H = sup 0<r<1 = sup 0<r<1 { R (z α )(rζ) dσ(ζ) { (1 + α ) (rζ) α dσ(ζ) = (1 + α ) { ζ α dσ(ζ) = (1 + α ) M 1 α where M α = ζ α dσ(ζ) = (n 1)!α! (n 1 + α )!. Set e α (z) = zα z α H, then {e α } α N n is the orthogonal base of H (B n). For any points z B n and w B n, it is easy for us to get the reproducing kernel of H (B n) is and K z (w) = R K z (w) = R [ α N n e α (z)e α (w) = α N n α N n zαwα (1 + α ) ] = M α zαwα (1 + α ), M α α N n zα wα (1 + α ) M α.

3 COMPOSITION OPERATORS ON HARDY-SOBOLEV SPACES 57 Therefore, K z H = sup 0<r<1 = sup 0<r<1 { = = = = { { R K z (rw) dσ(w) { α N n { α N n zαrwα (1 + α ) dσ(w) M α zαwα (1 + α ) dσ(w) M α zα wα (1 + α ) dσ(w) M α Nn α zα (1 + α ) M α Nn α { zα (1 + α ) M α Nn α w α dσ(w). For any point a B n, we define the Möbius transform on B n as z, a = 0; ϕ a (z) = a P a (z) S a Q a (z), 1 z, a a 0, where S a = 1 a z, a, P a (z) = a a and Q a = I P a. As we known, ϕ a is an automorphism of B n and has the following properties: ϕ a (0) = a, ϕ a (a) = 0, ϕ a ϕ a (z) = z, 1 ϕ a (z) = (1 a )(1 z ) 1 z, a. As general analytic function spaces, the Hardy-Sobolev spaces have very complex constructions and the classical methods on classical spaces such as Hardy spaces or Bergman spaces can not be applied to them directly. Naturally, to characterize the operators on these spaces are even more difficult. In recent years, there are a series of papers investigate these spaces. For instance, Tchoundja [1] gives a characterization of Carleson measures for some Hardy-Sobolev spaces by means of Green s formula. Cho and Zhu [] show the relation and difference between Hardy-Sobolev spaces and holomorphic mean Lipschitz spaces in terms of fractional radial differential operators and integral operators. Ahern and Cohn [3] equivalently characterize the exceptional sets for Hardy-Sobolev spaces by

4 58 LI HE et al. the sets of Hausdorff capacity zero and the sets of Bessel capacity zero. The purpose of this paper is to discuss more unknown properties about Hardy-Sobolev spaces and the composition operators on them. Definitely speaking, we mainly answer the following three questions: 1. How fast does a H p -function grow near the boundary?. Is there any way to calculate the spectra of composition operators on H? 3. When does C ϕ is Fredbolm? In section, we solve the question 1 and obtain some growth estimates for H p (B n)-functions. In section 3, we give a definite answer for the question, that is, we will show the representation of the spectra of composition operators C ϕ on H (B n). In section 4, we describe the Fredholmness of C ϕ equivalently.. SOME ESTIMATES ABOUT H p -FUNCTIONS, (0 < p < ) To obtain a growth estimate for H p -functions, we need the following two lemmas. Lemma.1 ([1]) Suppose and t be two positive real numbers such that t > 0, let 1 f(z) =. Then the following conclusion holds: (1 z, w ) t (1) There exists a positive constant C such that and for all w B n. R f(z) R f(z) C 1 z, w t+ C 1 z, w t Lemma. ([4]) Suppose 0 < p <, f H p (B n ), then for any z B n. f(z) n/p f p (1 z ) n/p For convenience, we simply denote f g if there exists a positive constant C which doesn t depend on the essential parameters such that f Cg. The following theorem tells us how fast an arbitrary function from H p grows near the boundary. Theorem.3 For 0 < p < and 0 < < n/p, then for any f H p, we have R f(z) f H p (1 z ) n/p

5 COMPOSITION OPERATORS ON HARDY-SOBOLEV SPACES 59 and f(z) f H p (1 z ) n/p. PROOF : For any f H p, we know that R f H p and R f p = f H p, then R f(z) R f p f (1 z ) n/p = H p (1 z ) n/p because of the conclusion in Lemma.. Consequently, it follows f(z) = R R 1 f f(z) f H p R (1 z ) n/p H p (1 z ) n/p from the fact given in Lemma.1 as 0 < < n/p. The proof is completed. 3. SPECTRA OF COMPOSITION OPERATORS ON H (B n) From the definition introduced in section 1, we know that H (B n) = {f H(B n ) R f H }. Obviously, H (B n) is a reflexive Hilbert space and the inner product in it is induced by f, g H = R fr gdσ, f H (B n), g H (B n). From now on, the inner product in H (B n) is simply denoted as,. In this section, we are to compute the spectra of composition operators on H (B n). At the beginning, we give an equivalent characterization of the composition operator on H (B n). Theorem 3.1 Suppose T is a bounded linear operator on H (B n). Then T is a composition operator on H (B n) if and only if T K z is a reproducing kernel for any z B n, where K z is the reproducing kernel of H (B n). PROOF : If T is a composition operator on H (B n), then there exists a holomorphic map ϕ mapping B n into itself such that T = C ϕ. For any f H (B n), z B n, we have f, CϕK z = C ϕ f, K z = C ϕ f(z) = f ϕ(z) = f, K ϕ(z), which implies that C ϕk z = K ϕ(z).

6 60 LI HE et al. Conversely, if T K z is a reproducing kernel, then for any f H (B n), we have T f(z) = T f, K z = f, T K z. Since T K z is a reproducing kernel we know that there is a map φ mapping B n into itself such that T K z = K φ(z), then T f(z) = f, K φ(z) = f(φ(z)). Let f j (z) = z j, then T f j (z) = f j, T K φ(z) = f j (φ(z)) = φ j (z), therefore, φ j is a holomorphic function and T = C φ. This completes the proof. Now, we turn to calculate the spectra of C ϕ. Firstly, we decompose the H (B n). Suppose e α = z α z α H, then {e α } α N n dictionary order. Set is the orthogonal base of H (B n). We arrange the {e α } α =k in H k = α N n, α =k {z α }, k = 0, 1,,..., then {e α } α =k is the orthogonal base of H k whose dimension is and H (B n) = k=0 H k. dim H k = C k n+k 1 Take T is a bounded linear operator on H (B n) and let (n + k 1)(n + k )... n =, k! S k (T ) = ( T e α, e γ ) α = γ =k. If T = C ϕ, then we denote S k (ϕ) = S k (C ϕ ). Secondly, we still need the following proposition. Proposition 3. Suppose ϕ is a holomorphic map mapping from B n into itself which satisfies ϕ(0) = 0. Then according to the above decomposition of H (B n) = k=0 H k, C ϕ has the following form C ϕ = S 1 (ϕ) S (ϕ) 0... S 3 (ϕ)

7 COMPOSITION OPERATORS ON HARDY-SOBOLEV SPACES 61 relative to the base {e α } α=(α1,α,...,α n) N n, where S k(ϕ) is a Cn+k 1 k Ck n+k 1-matrix whose entries satisfying C ϕ e α, e γ = α=(α j i ), n j=1 αj i =γ i, n i=1 αj i =α j n ( ϕ j ) αj i. i,j=1 PROOF : Set ϕ = ϕ 1 ϕ...ϕ n. Since ϕ(0) = 0, then ϕ j can be written as n ϕ j ϕ j (z) = (0)z i + ψ j (z) (j = 1,,..., n), i=1 where each term of the Fourier series of ψ j (z) has at least degree two. Therefore, it is not difficult to find that C ϕ z α, z γ = ϕ α, z γ = 0 for any α = (α 1, α,..., α n ) N n and γ = (γ 1, γ,..., γ n ) N n with α > γ. If α = γ, then we have C ϕ z α, z γ = ϕ α, z γ = ϕ 1 ϕ...ϕ n, z γ = n ϕ 1 n [ (0)z i + ψ 1 (z)] α 1 ϕ n...[ (0)z i + ψ n (z)] αn, z γ z i=1 i z i=1 i n n ϕ j = [ (0)z i + ψ j (z)] α j, z γ j=1 i=1 α j n n = Cα k j [ j=1 k=0 n n = [ = = = j=1 i=1 n { j=1 i=1 ϕ j (0)z i ] α j k [ψ j (z)] k, z γ ϕ j (0)z i ] α j, z γ n i=1 αj i =α j [ ϕ j (0)z i ] αj i }, z γ α=(α j i )n i,j=1, n j=1 αj i =γ i, n i=1 αj i =α j α=(α j i )n i,j=1, n j=1 αj i =γ i, n i=1 αj i =α j { n [ ϕ j (0)] αj i }z γ, z γ i,j=1 n [ ϕ j (0)] αj i z γ z H. i i,j=1 Thus C ϕ e α, e γ = α=(α j i ), n j=1 αj i =γ i, n i=1 αj i =α j n ( ϕ j ) αj i, i,j=1

8 6 LI HE et al. which implies that C ϕ = S 1 (ϕ) S (ϕ) 0... S 3 (ϕ) under the base {e α } α N n and this finishes the proof. Now, we give the main results of this section. Theorem 3.3 Suppose ϕ is a holomorphic map mapping from B n into itself which satisfies ϕ(0) = 0. Then σ p (C ϕ ) = σ(s k (ϕ)). k=0 PROOF : It is easy to see that k=0 σ(s k(ϕ)) σ p (C ϕ ) from the Proposition 3.. To complete the proof, we just need to show that σ p (C ϕ ) k=0 σ(s k(ϕ)). Suppose λ σ p (C ϕ ), then there is a non-zero function f H (B n) such that (C ϕ λ)f = 0. According to the decomposition of H (B n), we can find f j H j (j = 0, 1,...) such that f = j=0 f j. Let l = min{j f j 0}, then f = j=l f j satisfying f l 0. Consequently, by Proposition 3., we have 1 λ S 1 (ϕ) λ (C ϕ λ)f = S (ϕ) λ 0... S 3 (ϕ) λ f lk... = 0, and (S lk (ϕ) λ)f lk = 0, i.e. λ k=0 σ(s k(ϕ)). Therefore, σ p (C ϕ ) k=0 σ(s k(ϕ)) and the proof of this theorem is completed. Theorem 3.4 Suppose ϕ is a holomorphic map mapping from B n into itself which satisfies ϕ(0) = 0. If C ϕ is a compact operator on H (B n), then σ(c ϕ ) = σ(s k (ϕ)) {0, 1}. k=0 PROOF : It is easy to get that k=0 σ(s k(ϕ)) {0, 1} σ(c ϕ ) from the conclusion in Theorem 3.3. On the other hand, if λ σ(c ϕ ), without loss of generality, we assume that λ 0, 1. Since C ϕ

9 COMPOSITION OPERATORS ON HARDY-SOBOLEV SPACES 63 is compact, then λ σ p (C ϕ ). By using the Theorem 3.3 again, we obtain that λ k=0 σ(s k(ϕ)). The proof is completed. Corollary 3.5 Suppose ϕ is a holomorphic map mapping from B n into itself and z 0 B n is a fixed point of ϕ. If C ϕ is a compact operator on H (B n), then σ(c ϕ ) = σ(s k (ϕ z0 ϕ ϕ z0 )) {0, 1}, k=0 where ϕ z0 is the Möbius transform of B n defined in section 1. PROOF : Since z 0 is a fixed point of ϕ, we have ϕ(z 0 ) = z 0, then ϕ z0 ϕ ϕ z0 (0) = ϕ z0 ϕ(z 0 ) = ϕ z0 (z 0 ) = 0. By Theorem 3.4, we know that σ(c ϕz0 ϕ ϕ z0 ) = σ(s k (ϕ z0 ϕ ϕ z0 )) {0, 1}. k=0 Further, ϕ z0 corollary. is the Möbius transform of B n, then σ(c ϕ ) = σ(c ϕz0 ϕ ϕ z0 ) and this proves the Example : If C ϕ is a compact operator on H (B n) satisfying ϕ(0) = 0 and ϕ j (0) = 0 (i j), then according to the Proposition 3., we know that S k (ϕ) is a diagonal matrix which has diagonal entries { n ( ϕ j (0)) α j } α z j =k. i i,j=1 In this case, by Theorem 3.4, it is not difficult for us to obtain that σ(c ϕ ) = {( ϕ 1 z 1 (0)) α 1 ( ϕ z (0)) α...( ϕ n z n (0)) αn α j Z +, j = 1,,..., n} {0, 1}. In fact, if the Jacobi matrix of ϕ at 0 J C ϕ(0) is lower triangular, we can see easily that all S k (ϕ) are lower triangular and C ϕ has a lower triangular form, moreover, the diagonal entries of S k (ϕ) consist of all the possible k-times products of the diagonal entries of J C ϕ(0) in this case. Thus σ(s k (ϕ)) = {λ α λ = (λ 1 λ...λ n ), α N n, α j = k}, where λ i is an eigenvalue of J C ϕ(0) for i = 1,,..., n. This implies that σ p (C ϕ ) consist of all the possible finite products of the eigenvalues of J C ϕ(0). For general cases, if J C ϕ(0) is not lower

10 64 LI HE et al. triangular, then we can transform it to be lower triangular by some unitary transformation, that is, we can find a unitary matrix U such that UJ C ϕ(0)u 1 is a lower triangular matrix and set ψ = U ϕ U 1, we can obtain a similar conclusion holds in this case. Further, if there exists a point z 0 B n such that z 0 is a fixed point of ϕ, then we can see that ψ = ϕ z0 ϕ ϕ z0 and σ(c ψ ) = σ(c ϕ ). Thus we just need to discuss the spectrum of C ψ which satisfies ψ(0) = 0. Obviously, a similar conclusion holds, too. At last, we will show an important property of the compact composition operators on H (B n), which plays a significant role in calculating the spectra of compact composition operators. Theorem 3.6 Suppose ϕ is a holomorphic map mapping from B n into itself. If C ϕ is a compact operator on H (B n), then ϕ has a fixed point in B n. PROOF : Suppose {λ k } is an increasing positive sequence and λ k 1 as k. For any w B n, write ϕ k (w) = ϕ(λ k w), then {ϕ k } be elements of the ball algebra on B n. By the Brower fixed point theorem, ϕ k has fixed points in B n. We may assume that w k is a relative fixed point of ϕ k (k = 1,,...), that is ϕ k (w k ) = w k B n. Without loss of generality, we suppose that w k a as k. Therefore, ϕ(a) = a. To complete the proof, we only need to show that a < 1. Otherwise, if a = 1, then λ k w k a B n. Since K λk w k K λk w k H w 0 in H (B n) and C ϕ is compact on H (B n), we have C ϕk λk w k H K λk w k H 0. and Note we get K ϕ(λk w k ) H K λk w k H = { C ϕk λk w k H = K ϕ(λk w k ) H α N n [ϕ(λ k w k )] α (1+ α ) M α { α N n (λ k w k ) α (1+ α ) M α α N n w α k = { (1+ α ) M α α N n (λ k w k ) α (1+ α ) M α 0 n α N wk α α N (1+ α ) M α 1 } (λ k w k ) α, (1+ α ) M α as k, which implies that w k < λ k w k as k. However, 0 < λ k contradiction. Therefore, a is the fixed point of ϕ in B n. < 1, which is a

11 COMPOSITION OPERATORS ON HARDY-SOBOLEV SPACES 65 Combining this conclusion with corollary 3.5, we can compute the spectrum of any compact composition operator on H (B n). 4. FREDHOLMNESS OF COMPOSITION OPERATORS ON H (D) In this section, we are going to investigate some other properties of composition operators on H (B n) as n = 1. Let D be the open unit disk in C, and T = D its boundary. From section 1, we know that the reproducing kernel on H (D) is and which has the following properties: K z (w) = K z H = { m=0 m=0 z m w m (1 + m) z m (1 + m) Proposition 4.1 For any z D, the following proposition are true: 1. K z H 1;. K z H is an increasing function for z ; 3. If 1, then for any point z D, we have K z H is uniformly convergent on any compact subset of D and lim z 1 K z H =. If > 1, then for any point z D, we have K z H <. The following conclusion gives some equivalent descriptions about the Fredholmness of composition operators on H (D). Theorem 4. Suppose 1 and ϕ is a holomorphic map mapping from D into itself. If C ϕ is the composition operator on H (D). Then the following are equivalent: (1) ϕ is a disk automorphism; () C ϕ is invertible; (3) C ϕ is a Fredholm operator. PROOF : Clearly, (1) () (3), we just need to prove that (3) (1). Suppose that (3) holds, we notice that ϕ is not a constant function. Otherwise, the co-dimension of C ϕ must be infinite from the fact that the closure of the range of C ϕ consist of just the constant functions, which is contradicted with the Fredholmness of C ϕ. Now, we claim that ϕ is a bijection on D.

12 66 LI HE et al. Firstly, ϕ is one to one. Assume that ϕ is not one to one on D, then there exist points ζ and ζ in D with ζ ζ such that ϕ(ζ) = ϕ(ζ ). Let V and V be open neighborhoods of ζ and ζ in D satisfies that V V =, then the set W = ϕ(v ) ϕ(v ) is a non-empty open set since ϕ is analytic and not constant. And we can find a sequence {z k } in W with z k z l if k l. Suppose {ζ k } is the sequence in V and {ζ k } is the sequence in V such that ζ k = ϕ 1 (z k ) V and ζ k = ϕ 1 (z k ) V respectively. Let K ζk and K ζ k be the reproducing kernel at ζ k and ζ k. Then for any f H (D), we have K ζk K ζ k, C ϕ f = C ϕ(k ζk K ζ k ), f = K ϕ(ζk ) K ϕ(ζ k ), f = 0, this shows that K ζk K ζ k Ran(C ϕ ). But since {K ζk K ζ k } is independent, we have that dim[ker(c ϕ ) ] = dim[ran(c ϕ )] =, which contradicts that C ϕ is Fredholm. Thus, ϕ is one to one. Secondly, ϕ is onto as 1. In fact, if ϕ(d) D, then D [ϕ(d)] should not be empty. Suppose a D [ϕ(d)], that is we can find a sequence {ζ k } D such that ζ k 1 and ϕ(ζ k ) a as k. Since a D, we know that C ϕk ζk K a H = K ϕ(ζk ) K a H 0 as k. Set g k = k. Therefore, K ζ k K ζk H, then g k H = 1, C ϕk ζk H is bounded and 1 K ζk H 0 as C ϕg k H = C ϕk ζk H K ζk H 0 as k. Since (H ) = H and C ϕ is a Fredholm composition operator, there is a bounded operator T and a compact operator K on H (D) such that T C ϕ K = I. Consequently, T C ϕ(g k ) H 0 from the fact that C ϕg k H 0, which induces that Kg k + g k H 0 (4.1). Since K is compact and {g k } is a bounded sequence on H, we know that {Kg k} has a convergent subsequence, without loss of generality, we may assume that {Kg k } converges to g in H, then {g k} converges to g by (4.1). However, it is not difficult to check that {g k } w 0 in H, we get g = 0 and g k H g H = 0 as k, which is contradicted with the fact that g k H = 1 for any k Z +. Hence, ϕ is onto when 1. The proof is completed. Question : How about the case as > 1?

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