Integral operators on analytic Morrey spaces

Transcription

1 SCIENCE CHINA Mathematics ARTICLES September 4 Vol 57 No 9: doi: 7/s Integral operators on analytic Morrey spaces LI PengTao, LIU JunMing & LOU ZengJian epartment of Mathematics, Shantou University, Shantou 5563, China ptli@stueducn, 8jmliu@stueducn, zjlou@stueducn Received May 5, 3; accepted September 3, 3; published online April 4, 4 Abstract In this note, we characterize the boundedness of the Volterra type operator T g and its related integral operator I g on analytic Morrey spaces Furthermore, the norm and essential norm of those operators are given As a corollary, we get the compactness of those operators Keywords MSC() analytic Morrey space, Volterra type operator, essential norm 45P5, 4B35 Citation: Li P T, Liu J M, Lou Z J Integral operators on analytic Morrey spaces Sci China Math, 4, 57: , doi: 7/s Introduction Morrey spaces were initially introduced by Morrey [9] in 938 As useful tools, Morrey spaces play an important role in the study of harmonic analysis and partial differential equations See Taylor [5], Olsen [], Kukavica [4], Palagachev and Softova [], and the references therein In recent decades, in real and complex settings, Morrey spaces have been studied extensively For example, on Euclidean spaces R n, Adams and Xiao [, ] studied Morrey spaces by potential theory and Hausdorff capacity uong et al [] characterized Morrey spaces by the operators with heat kernel bounds The multipliers of Morrey spaces were studied by Gilles and Rieusset [6] In the unit disc, the analytic Morrey spaces, L,λ, were introduced and studied by Wu and Xie [7] Xiao and Xu [3] studied the composition operators of L,λ spaces Cascante et al [8] studied the corona theorem of L,λ For analytic functions g in, the Volterra type operator T g is defined as T g f(z) = z f(w)g (w)dw, on the space of analytic functions f in The operator T g was firstly investigated on Hardy spaces by Pommerenke [] Another integral operator related to T g (denoted by I g ) is defined by I g f(z) = z f (w)g(w)dw The boundedness and compactness of I g and T g between spaces of analytic functions were studied by many authors The T g on Hardy spaces and Bergman spaces were studied by Aleman and Cima [4], and Aleman and Siskakis [5, 6] Siskakis and Zhao [4] studied T g on the BMOA, the space of analytic Corresponding author c Science China Press and Springer-Verlag Berlin Heidelberg 4 mathscichinacom linkspringercom

2 96 Li P T et al Sci China Math September 4 Vol 57 No 9 functions with bounded mean oscillation Xiao [3] considered I g and T g on Q p spaces Constantin [9] studied the boundedness and the compactness of T g on Fock spaces Wu [6] considered T g from Hardy spaces to analytic Morrey spaces The closed ranges of I g and T g were studied by Anderson [7] See [3] and references therein for more information on those integral operators The essential norm can be seen as a useful tool to study the operators on function spaces By the essential norms, we can understand better the relation between bounded operators and compact operators The essential norms were first introduced by Shapiro [3] to study the composition operators Recently, the essential norms were applied to study the integral operators I g and T g, see Laitila et al [5] and Liu et al [8] We refer the reader to [5, 8, 3] for further information In this paper, we consider the operators I g,t g : L,λ () L,λ (), <λ<, I g,t g : H p () L, p (), <p The aim is to study the boundedness of I g and T g, and to estimate the norms and essential norms of I g and T g We will prove that the norms of I g and T g is equivalent to g and g BMOA, respectively By these equivalence, we could obtain the necessary and sufficient conditions of the boundedness of I g and T g See Theorems 33, 34, 37 and 38 On essential norms, we prove that the essential norm of I g is equivalent to g, and that the essential norm of T g is equivalent to the distance between g and the VMOA, the space of analytic functions with vanishing mean oscillation, see Theorems 4, 44, 47 and 49 As corollaries, we obtain necessary and sufficient conditions of the compactness for I g and T g, respectively We should point out that these necessary and sufficient conditions can be also obtained via other methods without using the essential norms Compared with such method, our results provide more information The rest of this paper is organized as follows In Section, we state some notation and preliminaries which will be used in the sequel Section 3 is devoted to the study of the boundedness of I g and T g and of the norm estimate of those integral operators The essential norms of I g and T g are given in Section 4 Notation For two functions F and G, if there is a constant C>dependent only on indexes p,λ, such that F CG,thenwesaythatF G Furthermore, denote that F G (F is comparable with G) whenever F G F Notation and Preliminaries Let and = {z : z =} denote respectively the open unit disc and the unit circle in the complex plane C LetH() be the space of all analytic functions on and da(z) = π dxdy the normalized area Lebesgue measure For <p<, the Hardy space H p () consists of all functions f H() with π f p H = sup f(re iθ ) p dθ < p <r< π For an arc I, let I = π I dζ be the normalized length of I, f I = I I f(ζ) dζ π, f H(), and be the Carleson box based on I with { = z : I z < and z } z I Clearly, if I =, then =

3 Li P T et al Sci China Math September 4 Vol 57 No efinition For <p<, we say that a non-negative measure μ on is a p-carleson measure if μ() sup I I p < If p =,p-carleson measure is the classical Carleson measure The following theorem due to Carleson is a significant result on Carleson measure (see, for example, [, 35]) Theorem A Suppose that μ is a non-negative measure on Then μ is a Carleson measure if and only if the following inequality: π f(z) dμ(z) f H = f(e iθ ) dθ holds for all f in Hardy space H () Moreover, sup f H = f(z) dμ(z) sup I μ() I Recall that BMOA is the set of f in the Hardy space H () whose boundary value functions satisfy ( / f = sup f(e iθ ) f I dθ) < I I I The norm of functions f in BMOA can be expressed as f = f() + f From [3], we know that f can be comparable with the norm f BMOA = f() +sup f σ a f(a) H, a where a and σ a is the Möbius transformation with Furthermore, if lim I σ a (z) = a z az, z f(e iθ ) f I dθ =, I I we say that f VMOA, the space of analytic functions with vanishing mean oscillation, we also know that f VMOAif and only if lim a f σ a f(a) H = For more information on BMOA and VMOA, see [3] The following theorem (see [3, Theorem 65] or [9, Theorem 4]) is a Carleson measure characterization of BMOA Theorem B Let f H(), thenf BMOA if and only if the measure μ f with is a Carleson measure Moreover, dμ f (z) = f (z) ( z )da(z) ( f BMOA f() + sup I ) / μ f () I

4 964 Li P T et al Sci China Math September 4 Vol 57 No 9 efinition Let <λ The Morrey space L,λ () isthesetofallf belongs to the Hardy space H () such that ( sup I I λ f(ζ) f I dζ ) / < I π Clearly, L, () =BMOA The following lemma gives some equivalent conditions of L,λ () (see [3, Theorem 3] or [8, Theorem 3]) Lemma 3 Suppose that <λ< and f H() Then the following statements are equivalent: (i) f L,λ (); (ii) sup I I f (z) ( z )da(z) < ; λ (iii) sup a ( a ) λ f (z) ( σ a (z) )da(z) < ; (iv) sup a ( a ) λ f (z) log σ a(z) da(z) < From Lemma 3, the norm of functions f L,λ () can be defined as follows: ( / f L,λ = f() +sup I I λ f (z) ( z )da(z)) Remark 4 From the lemma above, it is easy to see that for f L,λ (), f L,λ f() +sup (( a ) λ a f() +sup (( a ) λ a f (z) ( σ a (z) )da(z) f (z) log ) / ) / σ a (z) da(z) Now we give a result about the growth rate of functions in L,λ () Lemma 5 Let <λ< Iff L,λ (), then f f(z) L,λ, z ( z ) λ Proof For any b, wehave f L,λ f() +sup (( a ) λ f (z) ( σ a (z) )da(z) a ) / (( b ) λ f (z) ( σ b (z) )da(z) = (( b ) λ f (σ b (w)) σ b (w) ( w )da(w) ( b ) 3 λ f (b), ) / ) / where we used [35, Lemma 4] in the last inequality Thus, f(b) f() = b f (bt)dt b Since the point b is arbitrary, we get f (bt) dt f L,λ b ( b t) 3 λ dt f L,λ ( b ) λ f(z) f L,λ ( z ) λ, z

5 Li P T et al Sci China Math September 4 Vol 57 No Boundedness and norm estimate of I g and T g In this section, we prove the boundedness and estimate the norms of I g and T g The following two lemmas will be used throughout this paper Lemma 3 (See [34, Lemma ]) Suppose that s> and r, t > Ift<s+<r, then we have ( z ) s bz r az t da(z) ( b ) r s ab t, where a, b Lemma 3 Let < λ < and b Then functions f b (z) = ( b ) λ (σ b (z) b) and F b (z) =( b )( bz) λ 3 belong to L,λ () Moreover, we have f b (z) L,λ and F b (z) L,λ, where the implicit constants are independent of z and b Proof For b, by Lemma 3, we have f b L,λ That is f a, F a L,λ () sup ( a ) λ f b (z) ( σ a (z) )da(z) a =sup( a ) λ ( b ) λ+ a ( a ) λ ( b ) λ sup a ab, F b L +sup( a ) λ ( b ),λ a ( a ) λ ( b ) λ +sup a ab We first consider the boundedness of I g on L,λ () ( z ) bz 4 az da(z) ( z ) bz 5 λ az da(z) Theorem 33 Let < λ < and g H() Then I g is bounded on L,λ () if and only if g belongs to H (), the space of bounded analytic functions on Moreover, the operator norm satisfies I g g, where g =sup z g(z) Proof Let g H () For any f L,λ (), we have I g f L,λ ( = sup I I λ / f (z) g(z) ( z )da(z)) f L,λ sup g(z) z This leads to the boundedness of I g and I g g On the other hand, if I g is bounded on L,λ () For any b, suppose that f b is defined as in Lemma 3, then ) / I g I g f b L,λ sup (( a ) λ f b (z) g(z) ( σ a (z) )da(z) a ( ) / σ b(z) g(z) ( σ b (z) )da(z) ( ) / = g(σ b (w)) ( w )da(w) g(b), where we used [35, Lemma 4] again in the last inequality Since b is arbitrary, we have I g g Theorem 33 isproved

6 966 Li P T et al Sci China Math September 4 Vol 57 No 9 As a main result of this section, we give the boundedness of T g on L,λ () in the next theorem Theorem 34 Suppose that <λ< and g H() ThenT g is bounded on L,λ () if and only if g BMOA Moreover, T g g BMOA Proof Suppose g BMOA For f L,λ () andanyarci, letζ be the center of arc I and b =( I )ζ Wehave I λ (T g f) (z) ( z )da(z) = I λ f(z) g (z) ( z )da(z) I λ I + I f(b) g (z) ( z )da(z)+ I λ f(z) f(b) g (z) ( z )da(z) We first estimate the term I By Lemma 5, we get It follows from Theorem B that I = I λ f(b) f L,λ ( b ) λ f L,λ I λ f(b) g (z) ( z )da(z) g BMOA f L,λ Now we estimate the term I Since σ b (z) z I for all z, we have I = I λ f(z) f(b) g (z) ( z )da(z) I λ f(z) f(b) g (z) ( σ b (z) )da(z) I λ f σ b (w) f(b) (g σ b ) (w) ( w )da(w) ( b ) λ f σ b (w) f(b) (g σ b ) (w) ( w )da(w) Since g BMOA, wehaveg σ b BMOA and (g σ b ) (w) ( w )da(w) is a Carleson measure by Theorem B Note that f L,λ () H (), then f σ b f(b) H () Combining this with Theorem A yields π I ( b ) λ g σ b BMOA f σ b (e iθ ) f(b) dθ ( b ) λ g BMOA f (z) log σ b (z) da(z) g BMOA f L,λ, where we used Littlewood-Paley identity in the second inequality (see, for example, [, 35]) Thus, T g f L I,λ + I g BMOA f L,λ That is T g g BMOA On the other hand, suppose that T g is bounded on L,λ () For any I, letb =( I )ζ, where ζ is the center of I Then (see [35]) ( b ) bz I, z

7 Li P T et al Sci China Math September 4 Vol 57 No Using the test function F b as in Lemma 3, we get g (z) ( z )da(z) I I λ = I λ F b (z) g (z) ( z )da(z) (T g F b ) (z) ( z )da(z) T g F b L,λ T g F b L,λ T g Since I is arbitrary, we obtain g BMOA and g BMOA T g The proof is completed Remark 35 Theorem 33 holdsforλ = (see [3]) But Theorem 34 isnottureforλ = In fact, Siskakis and Zhao [4] proved that T g is bounded on BMOA if and only if g belongs to logarithmic BMOA space For g H(), the multiplication operator M g is defined by M g f(z) =f(z)g(z) It is easy to see that M g is related with I g and T g by M g f(z) =f()g() + I g f(z)+t g f(z) The multiplication operator can be also seen as a Toeplitz operator Corollary 36 Let <λ< and g H() Then the multiplication operator M g is bounded on L,λ () if and only if g H () Proof If g H,thenI g and T g are both bounded on L,λ () bytheorems33 and34 So M g is bounded on L,λ () If M g is bounded on L,λ (), consider the function F b (z) in Lemma 3 Applying Lemma 5, we have b g(z) ( bz) 3 λ M gf b L,λ M ( z ) λ g ( z ) λ Taking z = b in the inequality above yields g(b) M g and then g H () by the arbitrariness of b We now characterize the boundedness of I g : H p () L, p () (<p ) Theorem 37 Let <p and g H() Then I g : H p () L, p () is bounded if and only if g H () Moreover, I g g Proof Suppose I g : H p () L, p () is bounded Consider the test function f(z) = ( b ) p b( bz), b <, z Applying the well-known inequality, π ze iθ +t dθ ( z, ) t z, t >, we have f H p () with f H p So, ( ) / I g I g f, sup ( a ) p f (z) g(z) ( σ a (z) )da(z) a ( ) / σ b(z) g(z) ( σ b (z) )da(z) ( ) / = g(σ b (w)) ( w )da(w) g(b), where we used [35, Lemma 4] again in the last inequality Since b <, we have g = sup g(z) I g z <

8 968 Li P T et al Sci China Math September 4 Vol 57 No 9 On the other hand, suppose g H () Applying Littlewood-Paley identity implies whereweusedthefactthat ( I g f, sup ( a ) p a ( sup ( a ) p f (z) g(z) log a g sup( a ) p f σa f(a) H a g sup( a ) p f σa H a g sup( a ) p f σa H p a g f H p, f σ a H p ) / f (z) g(z) ( σ a (z) )da(z) ( ) /p + a f H p a (see [, Theorem 36]) Therefore, I g g The proof is completed ) / σ a (z) da(z) In [6, Theorem 9], Wu proved that, for g H(), T g : H p () L, p () (<p ) is bounded if and if g BMOA WenextestimatethenormofT g Theorem 38 Let < p and g H() If T g : H p () L, p () is bounded, then T g g BMOA Proof The proof is similar to that of Theorem 34, for the completeness of the paper, we give the sketch of the proof below For any I, letζ be the center of I and b =( I )ζ Consider the function h b (z) = b, z ( bz) + p It is easy to see that h b H p () with h b H p So, g (z) ( z )da(z) I I p = I p T g h b L, p h b (z) g (z) ( z )da(z) (T g h b ) (z) ( z )da(z) T g It follows that g BMOA with g BMOA T g On the other hand, if p =, then, for f H (), (T g f) (z) ( z )da(z) f I g BMOA, which implies T g g BMOA For f H p (), write I p I p (T g f) (z) ( z )da(z) + I p E + E f(b) g (z) ( z )da(z) f(z) f(b) g (z) ( z )da(z)

9 Li P T et al Sci China Math September 4 Vol 57 No Note that (see [35, Theorem 9]), f(b) f H p ( b ) p f H p, b I p We have E g BMOA f H p Since σ b (z) z I for all z, we have E I p f(z) f(b) g (z) ( σ b (z) )da(z) ( b ) p f σ b (w) f(b) (g σ b ) (z) ( w )da(w) Note that (g σ b ) (z) ( w )da(w) is a Carleson measure and f H p () (p>) implies that f σ b H () Using Theorem A again yields E ( b ) p g σb BMOA f σ b f(b) H ( b ) p g BMOA f σ b H ( b ) p g BMOA f σ b H p g BMOA f H p Hence, T g g BMOA The theorem is proved Remark 39 If p =,thenl, p () =H () From the proof of Theorem 37, we know that Theorem 37 holdsforp = Theorem38 forp = is also true (see, for example [4, ]) Corollary 3 Let <p and g H() Then the multiplication operator M g : H p () L, p () is bounded if and only if g H () Proof If M g : H p () L, p () is bounded, consider the function f(z) = ( b ) p ( bz) It is easy to see that f H p () and f H p Applying Lemma 5 again, we get ( b ) p g(z) ( bz) M gf, M ( z ) g p ( z ) p Taking z = b in the inequality above yields g(b) M g Wehaveg H () If g H (), from [6, Theorem 9] and Theorem 37, we know that I g and T g are both bounded from H p () tol, p () So Mg : H p () L, p () is bounded 4 Essential norm of I g and T g Let X be a Banach space and T a bounded linear operator on X The essential norm of T (denoted by T e )isdefinedasfollows, T e =inf{ T K : K are compact operators on X} Since that T is compact if and only if T e =, then the estimation of T e indicates the condition for T to be compact In this section, we estimate the essential norm of I g,t g

10 97 Li P T et al Sci China Math September 4 Vol 57 No 9 Let X and Y be two Banach spaces with X Y Iff Y, then the distance from functions f to the Banach space X is defined as dist Y (f,x) = inf g X f g Y In the following lemma, Laitila et al [5, Lemma 3] characterized the distance from functions f BMOA to VMOAspace Here and afterward, we denote g r (z) =g(rz) with<r< Lemma 4 Let g BMOA Then dist(g, V MOA) lim sup r g g r BMOA lim sup g σ a g(a) H a Theorem 4 Suppose <λ<and g H() If I g is a bounded operator on L,λ (), then I g e g Proof For compact operators K, it follows from Theorem 33 that I g e =inf K I g K I g g On the other hand, choose the sequence {b n } such that b n asn Consider the sequence of functions f n (z) =( b n ) λ (σbn (z) b n ) It follows from Lemma 3 that f n L,λ Note that f n can be written as z f n (z) = ( b n ) λ+ dw ( b n w) and f n converges to zero uniformly on compact subsets of Then Kf n L,λ asfor any compact operator K on L,λ () Since I g K lim sup (I g K)f n L,λ lim sup I g f n L,λ, lim sup( I g f n L,λ Kf n L,λ) I g f n L,λ sup (( a ) λ f n(z) g(z) ( σ a (z) )da(z) a ( ) / σ b n (z) g(z) ( σ bn (z) )da(z) ( ) / = g(σ bn (w)) ( w )da(w) g(b n ), we get I g e lim sup g(b n ) The arbitrary choice of the sequence {b n } implies I g e g The proof of Theorem 4 is completed For the proof of the next theorem, we need the following technical lemma Lemma 43 Suppose <λ< Ifg BMOA, thent gr : L,λ () L,λ () is compact Proof Let {f n } be such that f n L,λ andf n uniformly on compact subsets of as n We need only to show that lim T g r f n L,λ = Note that g r BMOA g BMOA (see [33, Lemma ]) and g r(z) ( r z ) g (rz) r g BMOA r, z ) /

11 Li P T et al Sci China Math September 4 Vol 57 No 9 97 (see [7, Lemma ]) We get T gr f n L,λ sup a (( a ) λ g BMOA r sup g BMOA r a ( ( f n (z) g r(z) ( σ a (z) )da(z) f n (z) ( z )( a ) λ az da(z) / f n (z) ( z ) da(z)) λ Note that f n L,λ and f n (z) ( z ) λ by Lemma 5 The desired result follows from the dominated convergence theorem In the following theorem, as a main result of this section, we give the essential norm of T g on L,λ () Theorem 44 ) / ) / Suppose <λ< and g BMOA ThenT g : L,λ () L,λ () satisfies T g e dist(g, V MOA) lim sup g σ a g(a) H a Proof Let {I n } be the subarc sequence of such that I n asenoteb n =( I n )ζ n, whereζ n is the center of arc I n,n=,, We know that (see [35]) ( b n ) b n z I n, z S(I n ) Choose the function F n (z) =( b n )( b n z) λ 3,z Then F n uniformly on the compact subsets of as n and F n L,λ by Lemma 3 Thus, for any compact operator K on L,λ () Therefore, lim KF n L,λ T g K lim sup( T g F n L,λ KF n L,λ) = lim sup T g F n L,λ ( ) / lim sup I n λ F n (z) g (z) ( z )da(z) S(I n) ( / lim sup g (z) ( z )da(z)) I n S(I n) Since the sequence {I n } is arbitrary, we obtain ( T g e lim sup I I g (z) ( z )da(z)) / It follows from the proof of [4, Lemma 34] that, for g BMOA, ( / lim sup g σ a g(a) H lim sup g (z) ( z )da(z)) a I I Hence, T g e lim sup g σ a g(a) H a On the other hand, T gr : L,λ () L,λ () is a compact operator Combining this with Theorem 34 implies T g e T g T gr = T g gr g g r BMOA, whereweusedthelinearityoft g respect to g Hence, T g e lim sup r by Lemma 4 The theorem is proved g g r BMOA lim sup g σ a g(a) H a

12 97 Li P T et al Sci China Math September 4 Vol 57 No 9 Remark 45 Theorem 4 holdsforλ = (see [8, Theorem ]) But Theorem 44 isnottruefor λ = (see [5, Theorem ]) Corollary 46 Let <λ< and g H() Then () I g is compact on L,λ () if and only if g =; () T g is compact on L,λ () if and only if g VMOA Now we give the essential norm of I g : H p () L, p () (<p ) in the following theorem Theorem 47 Let < p and g H() If I g : H p () L, p () is bounded, then I g e g Proof The proof is similar to that of Theorem 37 Choose the sequence {b n } with b n / such that b n asn Consider the sequence of functions f n (z) = ( b n ) p b n ( b n z) It is easy to see that f n H p andf n converges to zero uniformly on compact subsets of Then Kf n, asfor any compact operator K : Hp () L, p () Since we have I g K lim sup (I g K)f n, lim sup( I g f n, Kf L n, p ) lim sup I g f n,, ( ) / I g f n, sup ( a ) p f n (z) g(z) ( σ a (z) )da(z) a ( ) / σ b n (z) g(z) ( σ bn (z) )da(z) ( ) / = g(σ bn (w)) ( w )da(w) g(b n ), I g e lim sup g(b n ) The arbitrary choice of the sequence {b n } implies I g e g On the other hand, for compact operators K, it follows from Theorem 37 that The theorem is proved I g e =inf K I g K I g g As another main result of this section, we next estimate essential norm of T g : H p () L, p () (< p ), for its proof, we need the following lemma Lemma 48 Let <p and g BMOA ThenT gr : H p () L, p () is compact Proof The proof is similar to that of Lemma 43 Let {f n } H p () be such that f n H p and f n uniformly on compact subsets of as n Wehave T gr f n L, p ( sup ( a ) p a ( g BMOA r sup g BMOA r a ( ) / f n (z) g r (z) ( σ a )da(z) f n (z) ( z )( a ) + p az da(z) ) / f n (z) ( z ) p da(z) ) /

13 Li P T et al Sci China Math September 4 Vol 57 No and f n (z) ( z ) p by [35, Theorem 9] Using the Lebesgue dominated convergence theorem again implies lim T gr f n, = Theorem 49 Suppose <p and g BMOA ThenT g : H p () L, p () satisfies T g e dist(g, V MOA) lim sup g σ a g(a) H a Proof As the proof of Theorem 44 Let {I n } be the subarc sequence of such that I n as n, b n =( I n ζ n ) and ζ n be the center of arc I n Consider the function h n (z) = b n ( b n z) + p It is easy to check that h n uniformly on compact subsets of as n and h n H p For any compact operator K : H p () L, p (), we have So, lim Kh n L, p T g K lim sup( T g h n, Kh L n, p ) = lim sup T g h n, ( lim sup lim sup lim sup The arbitrary choice of {I n } yields, ( T g e lim sup I I I n p ( I n p ( I n S(I n) S(I n) S(I n) (T g h n ) (z) ( z )da(z) ) / ) / h n (z) g (z) ( z )da(z) g (z) ( z )da(z)) / / g (z) ( z )da(z)) lim sup g σ a g(a) H a On the other hand, it follows from Lemma 48 thatt gr Applying Theorem 38gives : H p () L, p () is a compact operator So T g e T g T gr = T g gr g g r BMOA T g e lim sup r by Lemma 4 The proof of Theorem 49 is finished g g r BMOA lim sup g σ a g(a) H a Remark 4 The proof of Theorem 47 shows that Theorem 47 istrueforp = Theorem49 holds also for p = (see [5, Theorem ]) Corollary 4 (See [6, Theorem 9]) Let p and g H() Then () I g : H p () L, p () is compact if and only if g =, () T g : H p () L, p () is compact if and only if g VMOA Acknowledgements This work was supported by National Natural Science Foundation of China (Grant Nos 73 and 8), Research Fund for the octoral Program of Higher Education of China (Grant No 443) and National Science Foundation of Guangdong Province (Grant Nos 5535 and S45)

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