COMPACT DOUBLE DIFFERENCES OF COMPOSITION OPERATORS ON THE BERGMAN SPACES OVER THE BALL

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1 COMPACT DOUBLE DIFFERENCES OF COMPOSITION OPERATORS ON THE BERGMAN SPACES OVER THE BALL BOO RIM CHOE, HYUNGWOON KOO, AND JONGHO YANG ABSTRACT. Choe et. al. have recently characterized comact double differences formed by four comosition oerators acting on the standard weighted Bergman saces over the disk of the comlex lane. In this aer, we extend such a result to the ball setting. Our characterization is obtained under a suitable restriction on inducing mas, which is automatically satisfied in the case of the disk. We exhibit concrete examles, for the first time even for single comosition oerators, which shows that such a restriction is essential in the case of the ball. 1. INTRODUCTION The interlay between the oerator theoretic and the function theoretic roerties of comosition oerators, acting on classical holomorhic function saces such as the Hardy sace and the standard weighted Bergman saces over the disk or the ball, has been extensively studied over the ast several decades; we refer to standard monograhs by Cowen- MacCluer [5] and Shairo [11] for an overview of various asects on the theory of comosition oerators. As is well known, the several-variable theory of comosition oerators is much more subtle than one-variable theory. For examle, one-variable comosition oerators are always bounded by Littlewood s Subordination Princile on the aforementioned function saces, which is no longer guaranteed in the several-variable case; see [7]. As a consequence, while quite an extensive study on one-variable theory of comosition oerators has been established during the ast four decades, the several-variable theory has been relatively less known. Recently, study on differences, or more generally linear combinations, of comosition oerators has been a toic of growing interest; see [3] and references therein. In the setting of the weighted Bergman saces over the disk, Moorehouse [9] first characterized comact differences by means of a natural angular derivative cancellation roerty; note that boundedness is out of question. Moorhouse s characterization is then extended to several-variable settings; see [1] and [2]. In a more recent aer [3] Choe et. al. extended Moorehouse s characterization to the case of double differences. Our goal in this aer is to extend such a characterization for comact double differences to the ball. For a fixed ositive integer n, let B = B n be the unit ball of the comlex n-sace C n. For α > 1, let dv be the normalized volume measure on B and ut dv α (z) := c α (1 z 2 ) α dv(z) Date: July 6, Mathematics Subject Classification. Primary 47B33; Secondary 30H20. Key words and hrases. Comosition oerator; Comact oerator; Double difference; Weighted Bergman sace; Ball. B. R. Choe was suorted by NRF(2015R1D1A1A ) of Korea, H. Koo was suorted by NRF(2017R1A2B ) of Korea and J. Yang was suorted by NRF(2017R1A6A3A ) of Korea. 1

2 2 B. CHOE, H. KOO, AND J. YANG where the constant c α is chosen so that v α (B) = 1. For 0 < <, the α-weighted Bergman sace A α(b) is the sace of all holomorhic function f on B such that the norm { f A α := } 1/ f dv α B is finite. As is well known, the sace A α(b) equied with the norm above is a Banach sace for 1 <. On the other hand, it is a comlete metric sace for 0 < < 1 with resect to the translation-invariant metric (f, g) f g A α. Let S = S(B) be the class of all holomorhic self-mas of B. Each function ϕ S induces a comosition oerator C ϕ defined by C ϕ f := f ϕ for functions f holomorhic on B. Clearly, C ϕ takes the sace of all holomorhic functions into itself. As is mentioned earlier, C ϕ is always bounded on each A α(b) when n = 1, but not always when n > 1. We now introduce some notation to be used throughout the aer. We reserve four inducing mas ϕ 1, ϕ 2, ϕ 3, ϕ 3 S, not necessarily distinct, to form the double difference Setting note T := (C ϕ1 C ϕ2 ) (C ϕ3 C ϕ4 ). T ij := C ϕi C ϕj, i, j = 1, 2, 3, 4, T = T 12 T 34 = T 13 T 24. (1.1) Using the seudohyerbolic distance ρ (see Section 2.2 for the definition) on B, we also ut and M ij (z) := for each i, j = 1, 2, 3, 4. Finally, we ut ρ ij (z) := ρ ( ϕ i (z), ϕ j (z) ) (1.2) ( 1 z 1 ϕ i (z) + 1 z ) ρ ij (z) (1.3) 1 ϕ j (z) M := M 12 + M 34 and M := M13 + M 24 for short. The next theorem is our main result. Theorem 1.1. Let α > 1 and 0 < <. With the notation as above, consider the following two assertions: (a) T is comact on A α(b); (b) M M C 0 (B). Then the imlication (a) = (b) holds. The imlication (b) = (a) also holds, rovided that each C ϕj is bounded on A q β (B) for some β ( 1, α) and 0 < q <. We would like to emhasize that the inducing self-mas in Theorem 1.1 are comletely general. So, considering aroriate secial cases, one may recover several known results. For examle, the secial case ϕ 2 = ϕ 3 = ϕ 4 0 reduces to the comactness characterization for single comosition oerators as in Corollary 5.1. See Section 5 for more secial cases which might be of indeendent interest. In addition, we exhibit a concrete examle

3 COMPACT DOUBLE DIFFERENCES 3 showing that the additional boundedness assumtion for the imlication (b) = (a) is essential; see Examle 5.8. As far as we know, such an examle even for single comosition oerators has not aeared yet in the literature. In Section 2 we recall some basic facts to be used in later sections. In Section 3 we rove the sufficiency art of Theorem 1.1. In Section 4 we rove the necessity art of Theorem 1.1. In Section 5 we observe some consequences of Theorem 1.1 concerning some secial cases. We also exhibit concrete examles showing that the additional boundedness assumtion for sufficiency cannot be removed. Constants. In the rest of the aer we use the same letter C to denote various ositive constants which may change at each occurrence. Variables (other than n) indicating the deendency of constants C will be often secified in the arenthesis. We use the notation X Y for nonnegative quantities X and Y to mean X CY for some inessential constant C > 0. Similarly, we use the notation X Y if both X Y and Y X hold. 2. PRELIMINARIES In this section we recall some basic facts which will be used in later sections Comact Oerator. It seems better to clarify the notion of comact oerators, since when 0 < < 1 the saces under consideration are not Banach saces. Let X and Y be toological vector saces whose toologies are induced by comlete metrics. A continuous linear oerator L : X Y is said to be comact if the image of every bounded sequence in X has a convergent subsequence in Y. We have the following convenient comactness criterion for a linear combination of comosition oerators acting on the weighted Bergman saces. Lemma 2.1. Let α > 1 and 0 < <. Let L be a linear combination of comosition oerators and assume that L is bounded on A α(b). Then L is comact on A α(b) if and only if Lf k 0 in A α(b) for any bounded sequence {f k } in A α(b) such that f k 0 uniformly on comact subsets of B. A roof can be found in [5, Proosition 3.11] for a single comosition oerator over the disk and it can be easily modified for a linear combination over the ball Pseudohyerbolic Distance. The seudohyerbolic distance between z, w B is given by ρ(z, w) := σ z (w) where σ z is the involutive automorhism of B that exchange 0 and z. More exlicitly, we have 1 ρ 2 (z, w) = (1 z 2 )(1 w 2 ) 1 z, w 2 (2.1) where z, w denotes the Hermitian inner roduct of z, w C n. This yields an inequality 1 z 2 1 z, w 1 ρ(z, w) 2 for w z, (2.2) which is useful for our urose. As is well known, each holomorhic self-ma F of B is a ρ-contraction, i.e., see [10, Theorem 8.1.4]. ρ ( F (z), F (w) ) ρ(z, w), z, w B; (2.3)

4 4 B. CHOE, H. KOO, AND J. YANG For 0 < r < 1 and z B, we denote by E r (z) := {w B : ρ(z, w) < r} the seudo-hyerbolic ball with center z and radius r. Clearly, E r (0) = rb. Using the well-known automorhism invariance of ρ, we also note E r (z) = σ z (rb). Given 0 < r < 1 and α > 1, we will use the well-known estimates and 1 z 1 w 1 z, w, w E r (z) (2.4) v α [E r (z)] (1 z ) n+1+α ; (2.5) the constants suressed in these estimates deend only on r and α. In most of the cases we will work with r = 1/2. So, we ut for brevity. E(z) := E 1/2 (z) 2.3. Test Function. Given α > 1, we have the submean value tye inequality f(a) C (1 z 2 ) n+1+α f, a B, (2.6) A α valid for all f A α(b) and 0 < <. Here the constant C deends only on α. So each oint evaluation in B is a bounded linear functional on the Hilbert sace A 2 α(b). Thus, to each a B corresonds a unique reroducing kernel whose exlicit formula is actually given by z (1 a, z ) (n+1+α). Suitable owers of those kernel functions will be our test functions in connection with Lemma 2.1. To this end we introduce functions τ a on B defined by τ a (z) := 1 1 z, a (2.7) for a B. When s > n α, we have the otimal norm estimates (see, for examle, [12, Theorem 1.12]) and thus τa s A 1, a B (2.8) α (1 a 2 ) s (n+1+α) as a 1. τ s a τ s a A α 0 uniformly on comact subsets of B (2.9) 2.4. Carleson Measure. Let α > 1 and µ be a ositive finite Borel measure on B. For 0 < r < 1 and 0 < <, it is well known (see [12, Section 2.4]) that the embedding A α(b) L µ[e r (z)] (dµ) is bounded su z B v α [E r (z)] <. We say that µ is an α-carleson measure if either side of the above holds. Note that the notion of α-carleson measures is indeendent of the size of or r. So, setting µ[e(z)] µ α = su z B v α [E(z)],

5 COMPACT DOUBLE DIFFERENCES 5 we see that µ is an α-carleson measure if and only of µ α <. Moreover, we have su f L (µ) µ α (2.10) f A 1 α for α-carleson measures µ on B; the constants suressed above deend only on α. The connection between comosition oerators and Carleson measures comes from the standard identity (see [6,. 163]) (h ϕ) dµ = h d(µ ϕ 1 ) (2.11) B valid for holomorhic self-mas ϕ of B and Borel functions h 0 on B. Here, µ ϕ 1 denotes the ullback measure defined by (µ ϕ 1 )(E) = µ[ϕ 1 (E)] for Borel sets E B. In articular, one can easily see from (2.11) that C ϕ : A α(b) L (µ) is bounded if and only if µ ϕ 1 α < Invariant Lalacian. For a function f on B, we define f(z) := (f σ z )(0), z B, where is the ordinary Lalacian. This oerator is called the invariant Lalacian because it is automorhism invariant in the sense that B (f σ) = ( f) σ for all automorhisms σ of B. Given α > 1 and 0 < <, we will use the norm equivalence ( f 2 ) /2 dv α f(z) f(0) dv α (z) (2.12) B B for function f holomorhic on B; the constants suressed here deends only on n and α. We refer to [12, Section 2.3] for more details. The next lemma is taken from [2, Lemma 4.4]. Lemma 2.2. Let α > 1, 0 < < and 0 < r 1 < r 2 < 1. Then there is a constant C = C(α,, r 1, r 2 ) > 0 such that f(a) f(b) ρ(a, b) C (1 a 2 ) n+1+α ( f 2 ) /2 dv α for all f H(B) and a, b B with a E r1 (b). E r2 (a) 2.6. Angular Derivative. The well-known notion of the angular derivatives on the disk has a natural extension to the ball. To state it we first recall some terminology. Let ζ B. A continuous function γ : [0, 1) B with lim t 1 γ(t) = ζ is said to be a restricted ζ-curve if γ(t) γ(t), ζ ζ 2 ζ γ(t), ζ ζ lim t 1 1 γ(t), ζ 2 = 0 and su <. 0 t<1 1 γ(t), ζ Note that a ζ-curve contained in the unit disk of the comlex line through ζ is restricted if and only if it aroaches to ζ nontangentially. We say that f : B C has restricted limit at ζ, denoted by f(ζ), if lim f(γ(t)) = f(ζ) t 1 for every restricted ζ-curve γ. In this case, we write R lim z ζ f(z) = f(ζ).

6 6 B. CHOE, H. KOO, AND J. YANG For a holomorhic self-ma ϕ of B, we say that ϕ has finite angular derivative at ζ, denoted by Dϕ(ζ), if there exists η B such that 1 ϕ(z), η Dϕ(ζ) = R lim. z ζ 1 z, ζ Note that the above forces ϕ to have restricted limit η at ζ with ϕ(ζ) = η. By the well-known Julia-Carathéodory Theorem we have ϕ has finite angular derivative at ζ d ϕ (ζ) := lim inf z ζ 1 ϕ(z) 1 z < where the liminf is taken as z ζ unrestrictedly in B. Moreover, if this is the case, then Dϕ(ζ) = d ϕ (ζ); (2.13) see [5, Theorem 2.81] or [10, Theorem 8.5.6] for details and further results. Thus, the angular derivative set A(ϕ) := {ζ B : d ϕ (ζ) < } consists of all boundary oints at which ϕ has finite angular derivatives. Moreover, it is known that lim λ 1 1 ϕ(λζ) 2 1 λ 2 = d ϕ (ζ), ζ A(ϕ) (2.14) where lim denotes the nontangential limit; see [2, Lemma 3.1]. Finally, we remark d ϕ (ζ) 1 ϕ(0) 1+ ϕ(0) > 0 by the Schwarz-Pick Lemma (see [10, Theorem 8.1.4]). 3. NECESSITY FOR COMPACTNESS In this section we rove the first art of Theorem 1.1. Our roof will be comleted by alying the comactness of the oerator under consideration to a suitably-chosen collection of test functions. For that urose we need several auxiliary lemmas. Before roceeding, we introduce some auxiliary functions. For a 1, a 2, a 3, a 4 B and a ositive integer m, ut and I m (z) = I m (z; a 1, a 2, a 3 ) : = (1 z 2 ) m τ m a1 (z) τ m a 2 (z) τ m a 3 (z) (3.1) J m (z) = J m (z; a 1, a 2, a 3, a 4 ) : = (1 z 2 ) m τ m a1 (z) τ m a 2 (z) τ m a 3 (z) + τ m a 4 (z) (3.2) for z B; recall that τ a denotes the function secified in (2.7). We will need certain lower estimates for these functions. We begin with the following inequality. Lemma 3.1. The inequality ( ) 1 a 2 m ( ) 1 a 2 m ( 1 a m2 m+2 ρ(a 1, a 2 ) 1 a 1, a 1 a 2, a 1 a a ) 1 a 2 holds for a, a 1, a 2 B and ositive integers m.

7 COMPACT DOUBLE DIFFERENCES 7 Proof. Fix a, a 1, a 2 B and a ositive integer m. Put λ := 1 a 1 a 1, a and η := 1 a 1 a 2, a for short. Since λ < 1 and η < 1, we have ( ) 1 a 2 m ( ) 1 a 2 m 2 m λ m η m 1 a 1, a 1 a 2, a m = 2 m λ η λ m j η j 1 m2 m λ η. Thus, in order to comlete the roof, it is sufficient to show ( 1 1 a 1, a 1 1 a 2, a 4ρ(a 1, a 2 ) j=1 1 1 a a 2 ). (3.3) To rove this inequality, ut a =: rζ where r = a and ζ B. Also, ut λ j = a j, ζ for j = 1, 2. Using the exlicit formula (see [10, Section 2.2] or [12, Section 1.2]) for the involutive automorhisms of B, we note ρ(λ 1 ζ, λ 2 ζ) = λ 1 λ 2 1 λ 1 λ. 2 We also note 1 ξ 2 1 rξ for comlex numbers ξ with ξ < 1. Accordingly, we have rλ 1 1 rλ 2 = r λ 1 λ 2 1 rλ 1 1 rλ 2 4 λ 1 λ 2 1 λ 1 1 λ 2 = 4ρ ( λ 1 ζ, λ 2 ζ ) 4ρ (a 1, a 2 ) 1 λ 1 λ 2 1 λ 1 1 λ 2 1 λ 1 λ 2 1 λ 1 1 λ 2 ; the last inequality holds by (2.3). Now, using the elementary inequality 1 λ 1 λ 2 1 λ λ 2, we conclude (3.3), as desired. The roof is comlete. We now rove the following lower estimate for the functions I m. Lemma 3.2. Given 0 < ɛ < 1, there is a collection of ositive integers {N k } 4 k=1 such that whenever a 1, a 2, a 3 B n satisfy max 1 k 4 1 j 3 I Nk (a j ) 1 2 (3.4) ɛ 1 a i 1 a j 1, i, j = 1, 2, 3. (3.5) ɛ In addition, the integers N k s can be chosen arbitrarily large.

8 8 B. CHOE, H. KOO, AND J. YANG Proof. Fix 0 < ɛ < 1 and suose that a 1, a 2, a 3 B satisfy (3.5). For simlicity we ut and r ij := ρ(a i, a j ), i, j = 1, 2, 3 d := min{r 12, r 13 }. Now we fix any ositive integer m. We consider the following three cases searately: (i) m2 m+4 d ɛ; (ii) m2 m+4 d > ɛ and a 1 max{ a 2, a 3 }; (iii) m2 m+4 d > ɛ and a 1 < max{ a 2, a 3 }. Case (i) : First, assume d = r 12. By Lemma 3.1, we have ( I m (a 3 ) 1 1 a3 2 ) m ( m 1 a3 2 1 a 1, a 3 1 a 2, a 3 ) ( 1 1 m2 m+2 a3 r 12 1 a a ) 3 1 a 2 1 ɛ 4 2 ɛ = 1 2. Similarly, we obtain I m (a 2 ) 1/2 if d = r 13. So, choosing N 1 = m, we obtain max I N 1 (a j ) 1 2 j 3 2. (3.6) Case (ii) : Since m2 m+4 d > ɛ and a 1 max{ a 2, a 3 }, we have by (2.2) 1 a a j, a 1 1 d 2 ɛ 1 2 m 2 4 m+4 for j = 2, 3. So, we have I ν (a 1 ) 1 3 ( 1 a1 2 j=2 1 a j, a 1 for any ν. Thus, choosing N 2 = N 2 (ɛ, m) m so that ( 1 we obtain ɛ 2 m 2 4 m+4 ) ν ( ɛ 2 ) ν/ m 2 4 m+4 ) N2 1 16, (3.7) I N2 (a 1 ) 1 2. (3.8) Case (iii) : Without loss of generality, we assume a 2 a 3. Since r 12 d > and a 2 > a 1, we have by (2.2) 1 a a 1, a 2 1 r ɛ 2 m 2 4 m+4. ɛ m2 m+4

9 COMPACT DOUBLE DIFFERENCES 9 For N 2 chosen in (3.7) and integers ν N 2, we obtain ( I ν (a 2 ) a2 2 ) ν ( 1 a2 2 1 a 3, a 2 1 a 1, a 2 ( a2 2 ) ν ( ɛ a 3, a 2 m 2 4 m+4 3 ( Re a2 2 ) ν. 1 a 3, a 2 ) ν ) ν/2 In conjunction with the estimate above, we note that for any comlex number λ at least one of λ, λ 2 and λ 3 has the nonnegative real art. Consequently, we have where N 3 = 2N 2 and N 4 = 3N 2. max I N k (a 2 ) 3 2 k 4 4, (3.9) Now, combining inequalities (3.6), (3.8) and (3.9), we conclude (3.4). In addition, recalling that m is an arbitrary ositive integer and N k m for each k, we see that the integers N k s can be chosen arbitrary large. The roof is comlete. For the functions J m, we rove the following lower estimate. Lemma 3.3. Given 0 < ɛ < 1, there is a collection of ositive integers {N k } 6 k=1 such that whenever a 1, a 2, a 3, a 4 B n satisfy max 1 k 6 1 j 4 J Nk (a j ) 1 2 (3.10) and min a j 1 1 j 3 2, ɛ min{ρ(a 1, a 2 ), ρ(a 1, a 3 )} (3.11) ɛ 1 a i 1 a j 1, i, j = 1, 2, 3. (3.12) ɛ In addition, the integers N k s can be chosen arbitrarily large. Proof. Fix 0 < ɛ < 1 and choose a ositive integer m = m(ɛ) such that (1 ɛ 2 ) m 1 ( and ɛ ) m ɛ 6. (3.13) Note that m can be chosen arbitrary large. For a 1, a 2, a 3, a 4 B satisfying (3.11) and (3.12), ut r ij := ρ(a i, a j ), i, j = 1, 2, 3, 4. for simlicity. In case a 1 max{ a 2, a 3 }, we obtain by (2.2) 1 a a j, a 1 1 r 2 1j 1 ɛ 2,

10 10 B. CHOE, H. KOO, AND J. YANG for j = 2, 3. It follows that ( J ν (a 1 ) a1 2 ) ν 3 ( 1 a1 2 1 a 4, a 1 1 a j=2 j, a 1 ( a1 2 ) ν 2(1 ɛ 2 ) ν/2 1 a 4, a Re ( 1 a1 2 1 a 4, a 1 for all integers ν m. Thus, by the same argument as in the roof of Case (iii) of Lemma 3.2, we obtain ) ν max 1 k 3 J N k (a 1 ) 1 2 ) ν (3.14) where N 1 = m, N 2 = 2m and N 3 = 3m. We now consider the case when max{ a 2, a 3 } > a 1 for the rest of the roof. By symmetry we may further assume a 2 a 3 so that a 2 max{ a 1, a 3 }. Note from (2.2) and (3.11) 1 a a 1, a 2 1 r ɛ 2. (3.15) Fixing an integer N 4 = N 4 (ɛ, m) m such that ( 1 we claim ɛ 4 m 2 4 m+5 max 1 k 6 1 j 4 ) N4 1 16, (3.16) J Nk (a j ) 1 2 (3.17) where N 5 := 2N 4 and N 6 := 3N 4. This estimate, together with (3.14), comletes the roof. In order to rove (3.17), we consider the following three cases searately: (i) 1 a 2 1 a 4 > 1 ɛ or 1 a 2 1 a 4 < ɛ; (ii) min{r 24, r 34 } > ɛ 2 (m2 m+5 ) 1 ; (iii) ɛ 1 a 2 1 a 4 1 ɛ and min{r 24, r 34 } ɛ 2 (m2 m+5 ) 1. Case (i): First, assume 1 a2 1 a 4 > 1 ɛ. Since a 2 max{ a 1, a 3 }, we have 1 a j, a 4 1 a 4 1 a j a 4 1 a 4 1 a 2 a 4 1 a 4 for j = 1, 2, 3. Meanwhile, since a by (3.11), we have 1 a 4 < ɛ(1 a 2 ) < 1 2 and thus a 4 > 1 2. So, we obtain 1 a 2 a 4 1 a 4 2 = 1 + a ( ) 4 1 a2 1 + a 4 1 a 4 1 > ɛ (3.18) 3ɛ

11 COMPACT DOUBLE DIFFERENCES 11 and thus J N1 (a 4 ) = J m (a 4 ) 3 ( 1 a a j=1 j, a 4 ( ɛ ) m 3ɛ ) m 1 2 ; the last inequality holds by (3.13). Now, we consider the case 1 a2 1 a < ɛ. In this case, exchanging the roles of a 4 2 and a 4 in (3.18), we have 1 a 4 a 2 1 a 2 2 > ɛ 3ɛ. This, together with (3.13) and (3.15), yields ( J ν (a 2 ) a2 2 ) ν ( 1 a2 2 ) ν ( 1 a2 2 1 a 3, a 2 1 a 1, a 2 1 a 4, a 2 ( a2 2 ) ν ( (1 ɛ 2 ) ν/ ɛ ) ν 1 a 3, a 2 3ɛ Re ( 1 a2 2 1 a 3, a 2 for all integers ν m. We thus conclude ) ν max 1 k 3 J N k (a 2 ) 1 2 as in the roof of (3.14). This comletes the roof of (3.17) for Case (i). Case (ii): We first consider the case of a 2 > a 4. Since m2 m+5 r 24 > ɛ 2, we have by (2.2) 1 a a 4, a 2 1 r ɛ 4 m 2 4 m+5. Thus we have by (3.15) and (3.16) ( J ν (a 2 ) a2 2 ) ν ( 1 a2 2 ) ν ( 1 a2 2 1 a 3, a 2 1 a 1, a 2 1 a 4, a 2 ( a2 2 ) ν ( (1 ɛ 2 ) ν/2 ɛ 4 ) ν/2 1 1 a 3, a 2 m 2 4 m+5 1 ( Re a2 2 ) ν, 1 a 3, a 2 for all ν N 4. We thus conclude ) ν ) ν as in the roof of (3.14). max 4 k 6 J N k (a 2 ) 1 2

12 12 B. CHOE, H. KOO, AND J. YANG We now consider the case a 2 a 4. This time we have by (2.2) 1 a a j, a 4 1 rj4 1 2 ɛ 4 m 2 4 m+5 for j = 2, 3. Thus, reeating a similar argument, we have J ν (a 4 ) 1 ( Re a4 2 ) ν, ν N 4 1 a 1, a 4 and thus conclude max J N k (a 4 ) 1 4 k 6 2. This comletes the roof of (3.17) for Case (ii). Case (iii): First, we consider the case r 24 r 34. Since 1 a2 1 a < 1 4 ɛ and m2m+2 r 24 ɛ2 8, we have by Lemma 3.1 and (3.12) ( 1 a1 2 ) m ( 1 a1 2 ) m ( 1 m2 m+2 a1 r 24 1 a 2, a 1 1 a 4, a 1 1 a a ) 1 1 a 4 ( ɛ2 1 a1 8 1 a a 1 1 a 2 1 a ) 2 1 a 4 ( < ɛ2 1 8 ɛ + 1 ) ɛ Note that the same estimate holds with a 3 in lace of a 1. Also, recall a 2 max{ a 1, a 3 }. Accordingly, we may assume a 1 a 3 so that 1 a a 3, a 1 1 r ɛ 2 (3.19) by (2.2) and (3.11). It follows that J N1 (a 1 ) = J m (a 1 ) ( 1 a1 2 ) m ( 1 1 a1 2 ) m ( m 1 a1 2 1 a 3, a 1 1 a 2, a 1 1 a 4, a 1 ) 3 4 (1 ɛ2 ) m/2 1 2 ; the last inequality holds by (3.13). Next, we consider the case r 34 r 24. In this case a similar argument using Lemma 3.1 yields ( 1 a2 2 ) m ( 1 a2 2 ) m 1 1 a 3, a 2 1 a 4, a 2 4. Thus, using (3.15) in lace of (3.19), we obtain J N1 (a 2 ) = J m (a 2 ) 1 2. This comletes the roof of (3.17) for Case (iii). The roof is comlete. Having Lemmas 3.2 and 3.3, we now roceed to the roof of the first art of in Theorem 1.1, which can be restated as follows.

13 COMPACT DOUBLE DIFFERENCES 13 Proosition 3.4. With the notation as in Theorem 1.1, assume that T is comact on A α(b). Then M M C 0 (B). Proof. We assume M M / C 0 (B) and comlete the roof by deriving a contradiction to the comactness of T. Since M M / C 0 (B), there is a sequence {z k } B with z k 1 and inf k M(z k) M(z k ) > 0 (3.20) as k. For simlicity we introduce some temorary notation associated with this sequence. Put a jk := ϕ j (z k ) and Q jk := 1 z k 1 a jk for each j and k. By the Schwarz-Pick Lemma, we have su Q jk 1 + ϕ j(0) k 1 ϕ j (0) (3.21) for each j. Thus, after assing to a subsequence of {z k } if necessary, we may assume that the sequence {Q jk } k converges for each j. So, we note V := {j : lim k Q jk > 0} by (3.20). Thus, for each j V, we have 1 z k 1 a jk for all k and thus a jk 1 as k. Finally, ut r ijk := ρ ij (z k ) = ρ(a ik, a jk ) for each i, j and k. With these notations, we have M ij (z k ) = (Q ik + Q jk )r ijk for each i, j and k. Here, ρ ij and M ij are the functions secified in (1.2) and (1.3). Now, we introduce our test function. Let Λ be the set of all ositive integers greater. Given j V and m Λ, let than n+1+α f j,m k (z) := (1 a jk 2 ) m n+1+α (1 z, a jk ) m, z B for ositive integers k. For fixed j and m, note from (2.8) and (2.9) that {f j,m k } k is bounded in A α(b) and that f j,m k 0 uniformly on comact subsets of B as k. So, we have j,m lim T fk A k α = 0 by Lemma 2.1. This, together with (2.6), yields T f j,m k A α T f j,m k (z k ) (1 z k 2 ) n+1+α T f j,m k (z k ) (1 a jk 2 ) n+1+α = J m (a jk ) where J m = J m ( ; a 1k, a 2k, a 3k, a 4k ) is the function introduced in (3.2). Accordingly, we have lim J m(a jk ) = 0. k

14 14 B. CHOE, H. KOO, AND J. YANG Note that this holds for each j V and m Λ. As a consequence we obtain lim J m (a jk ) = 0 (3.22) k m Λ F j V for any finite set Λ F Λ. On the other hand, we will rove below that this is not ossible by (3.20), which is a contradiction. Note from (3.21) that both M and M are bounded above on B. We also note from (3.20) that M and M both bounded below along the sequence {z k } by some ositive number, say 2c. So, we have max{m 12 (z k ), M 34 (z k )} c and max{m 13 (z k ), M 24 (z k )} c for each k. Thus, for each k, at least two of the following four cases hold: (a) min{m 12 (z k ), M 13 (z k )} c; (b) min{m 12 (z k ), M 24 (z k )} c; (c) min{m 34 (z k ), M 13 (z k )} c; (d) min{m 34 (z k ), M 24 (z k )} c. First, consider Case (a). Restating Case (a) more exlicitly, we have (Q 1k + Q 2k )r 12k c and (Q 1k + Q 3k )r 13k c (3.23) for each k. This, together with (3.21), yields δ > 0 such that min{r 12k, r 13k } δ (3.24) for each k. Note that the sequence {Q jk } k is bounded below by some ositive number for each j V. Thus we may further assume for each k. This, together with (3.21), yields so that min Q jk δ j V δ 1 z k 1 + ϕ i (0) s := max 1 a jk 1 i 4 1 ϕ i (0) ɛ 1 a ik 1 a jk 1 ɛ for i, j V and for each k. We also have by (3.24) for each k. where ɛ := δ s (3.25) ɛ min{r 12k, r 13k } (3.26) We now slit the roof according to the number V of the elements of the set V : Subcase (a1): Assume V = 4 so that V = {1, 2, 3, 4}. For j V, note a jk 1 2 for k sufficiently large, because a jk 1 as k. Having (3.25) and (3.26), we may aly Lemma 3.3 to find a finite set Λ 1 Λ, indeendent of a jk s, such that for k sufficiently large. max J m (a jk ) 1 2 m Λ 1 j V

15 COMPACT DOUBLE DIFFERENCES 15 Subcase (a2): Assume V = 3. We rovide details for V = {1, 2, 3}; other cases can be treated in a similar way. Note as k. This yields 1 a jk 2 1 a 4k, a jk 2 1 z k δ 1 a 4k, a jk 2 δ Q 4k 0, j V J m (a jk ) I m (a jk ) ( 2Q4k δ ) m I m (a jk) 1 4, j V, m Λ for k sufficiently large. Accordingly, having (3.25), we may aly Lemma 3.2 to find a finite set Λ 2 Λ, indeendent of a jk s, such that max I m (a jk ) 1 2 m Λ 2 j V where I m = I m ( ; a 1k, a 2k, a 3k ) is the function introduced in (3.1). This in turn yields for k sufficiently large. max J m (a jk ) 1 4 m Λ 2 j V Subcase (a3): Assume V = 2. In this case we claim that there is some m 1 Λ such that max J m (a jk ) 1 4 m Λ 3 j V where Λ 3 := {m 1, 2m 1, 3m 1 } (3.27) for k sufficiently large. We rovide below details for the cases V = {1, 2} and V = {1, 4}; other cases can be treated in a similar way. In conjunction with this we note from (3.23) that V {2, 4} and V {3, 4}. First, consider the case V = {1, 2}. In this case we may assume (after assing to a subsequence if necessary) a 1k a 2k by symmetry so that 1 a 1k 2 1 a 1k, a 2k 1 r12k 2 1 δ 2 by (2.2) and (3.24). Now, since Q 3k + Q 4k 0 as k, we have as in the roof of Subcase (a2) ( 1 a1k 2 ) m J m (a 1k ) a 1k, a 2k 4, m Λ 3 4 (1 δ2 ) m/2 for k sufficiently large. Accordingly, choosing m 1 Λ sufficiently large so that (1 δ 2 ) m1 1 4, we conclude (3.27). We now consider the case V = {1, 4}. Since Q 2k + Q 3k 0 as k, we obtain as in the roof of Subcase (a2) ( J m (a 1k ) a1k 2 ) m 1 1 a 4k, a 1k 4 3 ( Re a1k 2 ) m, m Λ 1 a 4k, a 1k for k sufficiently large. We thus conclude (3.27) as in the roof of (3.14).

16 16 B. CHOE, H. KOO, AND J. YANG Subcase (a4): Assume V = 1. We rovide details for V = {1}; other cases can be treated in a similar way. Since Q 2k + Q 3k + Q 4k 0 as k, we have as in the roof of Subcase (a2) J m (a 1k ) 1 4, m Λ for k sufficiently large. Thus, fixing any m 3 Λ, we conclude for k sufficiently large. Now, utting max J m (a jk ) 1 4 m Λ 4 j V Λ a := where Λ 4 := {m 3 } where Λ l s are the finite sets obtained in Subcases (a1)-(a4), we conclude for Case (a) 4 l=1 Λ l max J m (a jk ) 1 4 m Λ a j V for k sufficiently large. So far we have constructed a finite set Λ a Λ that corresonds to Case (a). One may reeat similar arguments to find finite sets Λ b, Λ c and Λ d that corresond to Case (b), Case (c) and Case (d), resectively. Finally, setting we obtain Λ F0 := Λ a Λ b Λ c Λ c, max J m (a jk ) 1 m Λ F0 4 j V for k sufficiently large, which is a contradiction to (3.22). The roof is comlete. 4. SUFFICIENCY FOR COMPACTNESS In this section we rove the second art of Theorem 1.1. We first recall a sufficient condition for boundedness of a comosition oerator on A α(b). The next lemma is taken from [4, Theorem 3.3]. Lemma 4.1. Let β > 1 and 0 < q <. Let ϕ S. If C ϕ is bounded on A q β (B), then C ϕ is bounded on A α(b) for any α β and 0 < <. Given α > 1 and a bounded nonnegative Borel function W on B, ut dw α := W dv α. Associated with this measure is the weighted ullback measure d(w α ϕ 1 ) defined by (W α ϕ 1 )(E) := W α [ϕ 1 (E)] for Borel subsets E of B. In the setting of the disk the next lemma is imlicit in the roof of [9, Lemma 1]. The roof below is included for comleteness.

17 COMPACT DOUBLE DIFFERENCES 17 Lemma 4.2. Let α > β > 1 and 0 <, q <. Put γ := min{α β, 1}. Let ϕ S and assume that C ϕ is bounded on A q β (B). Let ɛ > 0 and W : B [0, 1] be a Borel function. If [ ] 1 z su W (z) ɛ, (4.1) z B 1 ϕ(z) then there is a constant C = C(α, β) > 0 such that for functions f A α(b). B f ϕ W dv α Cɛ γ f A α Proof. Assume (4.1). Since β 1 := α γ β, we note v β1 ϕ 1 β1 < by Lemma 4.1 and (2.10). Also, note W 1 γ 1, because γ 1. We thus have for z B (W α ϕ 1 )[E(z)] = W (w) dv α (w) = ɛ γ ϕ 1 [E(z)] ϕ 1 [E(z)] ϕ 1 [E(z)] W (w) 1 γ [ W (w)(1 w 2 ) ] γ dvβ1 (w) (1 ϕ(w) ) γ dv β1 (w) ɛ γ (1 z ) γ v β1 [ϕ 1 (E(z))]; the last estimate holds by (2.4). This, together with (2.5), yields W α ϕ 1 [E(z)] v α [E(z)] ɛ γ (v β 1 ϕ 1 )[E(z)] ; v β1 [E(z)] the constant suressed here deends only on n, α and β. It follows that W α ϕ 1 α Cɛ γ v β1 ϕ 1 β1 for some constant C = C(α, β) > 0. We therefore conclude the lemma by (2.10). The roof is comlete. The following lemma is a key ste in the roof of sufficiency art. Lemma 4.3. Let α > β > 1 and 0 <, q <. Let ϕ, ψ S and assume that C ϕ, C ψ are bounded on A q β (B). Let K B be a Borel set. If [( 1 z 1 ϕ(z) + 1 z ) ρ ( ϕ(z), ψ(z) )] ɛ, (4.2) 1 ψ(z) su z K then there exists a constant h(ɛ) = h(ɛ,, α, β, ϕ, ψ) > 0 such that and for functions f A α(b). K lim h(ɛ) = 0 ɛ 0 f ϕ f ψ dv α h(ɛ) f A α

18 18 B. CHOE, H. KOO, AND J. YANG Proof. Assume (4.2) and let f A α(b). For a number δ = δ(ɛ) (0, 1 3 ) to be fixed later, we decomose the integral under consideration into two arts as f ϕ f ψ dv α = (4.3) where K K δ + K δ := {z K : ρ(ϕ(z), ψ(z)) < δ} and K δ := K \ K δ. First, we estimate the first term in the right-hand side of (4.3). Alying Lemma 2.2 (with r 1 = 1 3 and r 2 = 1 2 ) and then Fubini s Theorem, we obtain [ δ ( f 2 K δ B (1 ϕ(z) 2 ) n+1+α (w) ) ] /2 dvα (w) dv α (z) E(ϕ(z)) = δ ( f 2 (w) ) [ ] /2 dv α (z) (1 ϕ(z) 2 ) n+1+α dv α (w); B ϕ 1 [E(w)] recall E( ) = E 1/2 ( ). Meanwhile, noting that ρ ( ϕ(z), w ) < 1 2 for z ϕ 1 [E(w)], we obtain by (2.4) and (2.5) dv α (z) (1 ϕ(z) 2 ) n+1+α (v α ϕ 1 )[E(w)] v α ϕ 1 α v α [E(w)] ϕ 1 [E(w)] for all w B. In addition, since C ϕ is bounded on A α(b) by Lemma 4.1, we note v α ϕ 1 α < by (2.10). Combining these observations and then using (2.12), we obtain δ f(w) f(0) dv α (w) δ f A. (4.4) α K δ B Next, we estimate the second term in the right-hand side of (4.3). Note δχ K δ ρ(ϕ, ψ) by (4.2) where χ K δ denotes the characteristic function of K δ. Accordingly, we have ( 1 z χ K δ (z) 1 ϕ(z) + 1 z ) ɛ 1 ψ(z) δ for all z B. It follows from Lemma 4.2 that ( ɛ ) γ ( f ϕ + f ψ )χ K δ dv α f δ A α K K δ B where γ := min{α β, 1}. Now, we deduce from (4.5) and (4.4) [( ɛ ) ] γ f ϕ f ψ dv α + δ f δ A, α K δ f A α(b). (4.5) One may kee track of the constant suressed above to see that it is indeendent of f, ɛ and δ. Consequently, a suitable choice of δ, say δ =, comletes the roof. ɛ 1+3 ɛ Now, we are ready to rove the second art of Theorem 1.1. Proosition 4.4. With the notation as in Theorem 1.1, assume M M C 0 (B). Then T is comact on A α(b), rovided that each C ϕj is bounded on A q β (B) for some β ( 1, α) and 0 < q <.

19 COMPACT DOUBLE DIFFERENCES 19 Proof. Assume that each C ϕj is bounded on A q β (B) for some β ( 1, α) and 0 < q <. Consider an arbitrary sequence {f k } in A α(b) such that f k A α 1 and f k 0 uniformly on comact subsets of B. We claim T f k 0 in A α(b). (4.6) Note that, with this claim granted, the asserted comactness of T follows from Lemma 2.1. We now roceed to the roof of (4.6). Let ɛ > 0. Since M M C 0 (B) by assumtion, there is some r (0, 1) such that for z with z r. Thus, setting we note M(z) M(z) ɛ 2 U ɛ := {z B : M(z) ɛ} and Ũ ɛ := {z B : M(z) ɛ}, According to this observation, we obtain T f k dv α + B B \ rb U ɛ Ũɛ. rb U ɛ + Ũ ɛ =: I 1k + I 2k + Ĩ2k (4.7) for each k. Note f k 0 uniformly on the set 4 j=1 ϕ j(rb) which is relatively comact in B. So, for the integral over rb in (4.7), we conclude as k. Meanwhile, note I 1k 0 (4.8) M(z) = M 12 (z) + M 34 (z) ɛ for z U ɛ. So, for the integral over U ɛ in (4.7), we see by Lemma 4.3 that there is a constant h(ɛ) > 0 satisfying h(ɛ) 0 as ɛ 0 and I 2k T 12 f k dv α + U ɛ T 34 f k dv α h(ɛ) U ɛ for all k. It follows that lim su I 2k h(ɛ). (4.9) k Recalling T = T 13 T 24 from (1.1), one may reeat the same argument to conclude lim su Ĩ 2k h(ɛ) (4.10) k for some constant h(ɛ) > 0 such that h(ɛ) 0 as ɛ 0. One may check that the constants suressed in (4.9) and (4.10) are indeendent of k and ɛ. Thus, combining the observations in (4.8), (4.9) and (4.10), we obtain lim su T f k C[h(ɛ) + h(ɛ)] k B for some constant C > 0 indeendent of ɛ. Finally, taking the limit ɛ 0, we conclude (4.6), as required. The roof is comlete.

20 20 B. CHOE, H. KOO, AND J. YANG 5. REMARKS Note that Theorem 1.1 can be alied even when some of the oerators C ϕj s coincide or already comact. So, in this section we consider three secial cases to recover or derive some consequences which might be of indeendent interest. When ϕ 2 = ϕ 3 = ϕ 4 0, note M M = M In conjunction with this, we also note ρ 12 = ρ ( ϕ 1 (z), 0 ) = ϕ 1 (z) so that [ ] 1 z M 12 (z) = + (1 z ) ϕ 1 (z) 1 ϕ 1 (z) = 1 z 1 ϕ 1 (z) (1 z )(1 ϕ 1(z) ). We thus recover the following characterization due to Zhu [13, Theorem 11]. (5.1) Corollary 5.1. Let α > 1 and 0 < <. Let ϕ S and assume that C ϕ is bounded on A q β (B) for some β ( 1, α) and 0 < q <. Then C ϕ is comact on A α(b) if and only if lim z 1 1 z 1 ϕ(z) = 0. Meanwhile, when ϕ 1 = ϕ 4 and ϕ 2 = ϕ 3, note M M = 4M Thus we recover the following result (see [2, Theorem 4.7]), which is a ball version of Moorhouse s characterization for comact differences over the disk. Corollary 5.2. Let α > 1 and 0 < <. Let ϕ, ψ S and assume that C ϕ and C ψ are both bounded on A q β (B) for some β ( 1, α) and 0 < q <. Then C ϕ C ψ is comact on A α(b) if and only if lim z 1 ( 1 z 1 ϕ(z) + 1 z 1 ψ(z) Also, consider the case ϕ 1 = ϕ 4. In this case we have M M = (M 12 + M 13 ) 2. ) ρ ( ϕ(z), ψ(z) ) = 0. Thus, as a consequence of Corollary 5.2, we obtain following result. Corollary 5.3. Let α > 1 and 0 < <. For j = 1, 2, 3 let ϕ j S and assume that C ϕj is bounded on A q β (B) for some β ( 1, α) and 0 < q <. Then the following two assertions are equivalent: (a) 2C ϕ1 C ϕ2 C ϕ3 is comact on A α(b); (b) C ϕ1 C ϕ2 and C ϕ1 C ϕ3 are both comact on A α(b). Finally, we consider the case when one of the oerators is already comact. Recall that A(ϕ) denotes the angular derivative set of ϕ S. Corollary 5.4. Let α > 1 and 0 < <. For j = 1, 2, 3 let ϕ j S and assume that C ϕj is bounded on A q β (B) for some β ( 1, α) and 0 < q <. Then the following three assertions are equivalent: (a) C ϕ1 C ϕ2 C ϕ3 is comact on A α(b);

21 COMPACT DOUBLE DIFFERENCES 21 [ (b) lim M 12 (z) + 1 z ] [ M 13 (z) + 1 z ] = 0; z 1 1 ϕ 3 (z) 1 ϕ 2 (z) (c) A(ϕ 1 ) = A(ϕ 2 ) A(ϕ 3 ) and A(ϕ 2 ) A(ϕ 3 ) =. Moreover, for j = 2, 3. Proof. As in (5.1), we have M j4 (z) = lim M 1j(z) = 0, ζ A(ϕ j ) (5.2) z ζ 1 z 1 ϕ j (z) (1 z )(1 ϕ j(z) ) for j = 2, 3. Thus the condition M M C 0 (B) reduces to Assertion (b). So, (a) and (b) are equivalent by Theorem 1.1. While the equivalence of (a) and (c) is contained in [2, Theorem 5.4], we include below another roof by establishing the equivalence of (b) and (c). Assume that (c) holds. Since A(ϕ 1 ) is the disjoint union of A(ϕ 2 ) and A(ϕ 3 ), we have either lim z ζ 1 z 1 ϕ 2 (z) 0 or 1 z 1 ϕ 3 (z) 0 as z ζ for each ζ A(ϕ 1 ). Thus we have by (5.2) [ M 12 (z) + for any ζ A(ϕ 1 ). Note 1 z 1 ϕ 3 (z) ] [ M 13 (z) + 1 z ] = 0 (5.3) 1 ϕ 2 (z) 1 z 1 ϕ 2 (z) 0 and 1 z 1 ϕ 3 (z) 0 as z ζ for ζ B \ A(ϕ 1 ). Thus, (5.3) remains valid for ζ B \ A(ϕ 1 ) by (5.2). So, (b) holds, as asserted. Conversely, assume that (b) holds. Since (b) imlies lim z ζ 1 z 1 ϕ 2 (z) 1 z = 0, ζ B, 1 ϕ 3 (z) we note A(ϕ 2 ) A(ϕ 3 ) = by the Julia-Carathéodory Theorem. Next, assume ζ A(ϕ 2 ) but ζ / A(ϕ 1 ). We then have by (2.1) 1 ρ 2 12(λζ) 4 for λ B 1. So we have by (2.13) and (2.14) and thus 1 ϕ 2(λζ) 1 ϕ 1 (λζ) = 1 ϕ 2(λζ) 1 λζ 1 λζ 1 ϕ 1 (λζ) lim λ 1 ρ 12 (λζ) = 1 ( ) 1 λ lim M 12 (λζ) = 1 λ 1 1 ϕ 2 (λζ) d 2 ϕ 2 (ζ) > 0, which contradicts to (b). We thus conclude A(ϕ 2 ) \ A(ϕ 1 ) =, i.e., A(ϕ 2 ) A(ϕ 1 ). Similarly, we have A(ϕ 3 ) A(ϕ 1 ). On the other hand, if ζ A(ϕ 1 ) but ζ / A(ϕ 2 ) A(ϕ 3 ), then a similar argument yields lim M 12 (λζ)m 13 (λζ) = 1 λ 1 d 2 ϕ 1 (ζ) > 0, which again contradicts to (b). Accordingly, we conclude A(ϕ 1 ) = A(ϕ 2 ) A(ϕ 3 ).

22 22 B. CHOE, H. KOO, AND J. YANG We now show (5.2). By symmetry it is enough to consider the case j = 2 only. In order to derive a contradiction, suose that (5.2) fails for some ζ A(ϕ 2 ). We then have inf k M 12(z k ) > 0 (5.4) for some sequence {z k } B converging to ζ. This imlies Since by (b), we also have by (5.4) and inf k ρ 12(z k ) > 0. (5.5) ( lim M 12(z k ) M 13 (z k ) + 1 z ) k = 0 k 1 ϕ 2 (z k ) lim M 13(z k ) = 0 (5.6) k lim k It follows from (5.4), (5.5) and (5.7) that inf k which, together with (5.6), in turn yields 1 z k = 0. (5.7) 1 ϕ 2 (z k ) 1 z k > 0, (5.8) 1 ϕ 1 (z k ) lim ρ 13(z k ) = 0. k In articular, by (2.4), we have 1 ϕ 1 (z k ) 1 ϕ 3 (z k ) for all k and thus obtain by (5.8) lim su k 1 z k 1 ϕ 3 (z k ) > 0. Consequently, we conclude ζ A(ϕ 3 ), which contradicts to the fact that A(ϕ 2 ) and A(ϕ 3 ) are disjoint. The roof is comlete. We now turn to the construction of exlicit examles showing that the additional boundedness assumtion in Theorem 1.1(b) cannot be removed. For simlicity we take n = 2 for the rest of the aer. We introduce some notation. For the rest of the aer we use the notation h(z) := z 1 + z2 2 2 for z = (z 1, z 2 ) B 2. Given 0 < ɛ < 1, we ut ( ( 1 h(z) ψ ɛ (z) := 1 2 ) ɛ, 0). Since h(b 2 ) B 1, we have ψ ɛ S(B 2 ). Put e := (1, 0). Since Arg(1 ψ ɛ (z), e ) ɛπ 2, we also note that ψ ɛ(z), e is contained in a nontangential region with vertex at 1. So, there is a constant c = c(ɛ) > 1 such that for all z B 2. Finally, we ut 1 ψ ɛ (z), e < c(1 ψ ɛ (z), e ) (5.9) S δ (ζ) := {z B 2 : 1 z, ζ < δ}

23 COMPACT DOUBLE DIFFERENCES 23 for 0 < δ < 1 and ζ B 2. We need some reliminary lemmas. First, we observe that each function ψ ɛ does not have any finite angular derivatives. Lemma 5.5. Let 0 < ɛ < 1. Then lim z 1 1 z 1 ψ ɛ (z) = 0. Proof. By the Julia-Carathéodory Theorem it suffice to rove that ψ ɛ does not have any finite angular derivatives. Noting ψɛ 1 ( B 2 ) = {e}, we consider a restricted e-curve γ in B 2 given by ( γ(t) = t, 1 t ), 0 t < 1. 2 Note that ψ ɛ (e) = e and 1 ψ ɛ (γ(t)), e = 1 [ ] ɛ 1 h(γ(t)) 1 γ(t), e 1 t 2 = 1 ( ) ɛ ( ) ɛ 1 t 3 + t 1 t 2 4 as t 1. This shows that ψ ɛ does not have angular derivative at e. So, we see that ψ ɛ does not have any finite angular derivatives, as asserted. The roof is comlete. Next, we investigate the relation of weight arameters α and β for which an oerator of the form N L := a j C ϕj where ϕ j := λ j ψ ɛ (5.10) j=1 is bounded from A α(b 2 ) into A β (B 2). Here, N is a ositive integer, λ 1,..., λ N are distinct unimodular comlex numbers and a 1,..., a N are nonzero comlex numbers. To this end we recall the following otimal estimate which is imlicit in the roof of [8, Proosition 4.4]. Lemma 5.6. Let α > 1. Then there is a constant C = C(α) > 0 such that C 1 (v α h 1 ) [ Sδ (1) ] C δ 3+α 1/4 for 0 < δ < 1 where S δ (1) = {λ B 1 : 1 λ < δ}. Lemma 5.7. Let α, β > 1 and 0 < <. Let 0 < ɛ < 1 and L be an oerator as in (5.10). Then L : A α(b 2 ) A β (B 2) is bounded if and only if β + 3 ɛ(α + 3) Proof. We first establish the otimal estimate (v β ψɛ 1 )[S δ (ζ)] su δ (3+β 1/4)/ɛ 3 α (5.11) ζ B 2 v α [S δ (ζ)] for 0 < δ < 1. In conjunction with this estimate, we recall the well-known estimate v α [S δ (ζ)] δ 3+α, 0 < δ < 1 (5.12) uniformly in ζ B 2 ; see, for examle, [5, Exercise 2.2.8].

24 24 B. CHOE, H. KOO, AND J. YANG Let ζ B 2. To avoid triviality assume ψɛ 1 [S δ (ζ)] and let z ψɛ 1 [S δ (ζ)]. Since ψ ɛ (z), ζ ψ ɛ (z) = ψ ɛ (z), e, we note from (5.9) 1 ψ ɛ (z), e c 1 ψ ɛ (z), ζ < cδ for all z B 2. In other words, we have ψ 1 ɛ [S δ (ζ)] ψ 1 [S cδ (e)], which allows us to focus on the case ζ = e. Note and thus S δ (e) = {z B 2 : 1 z 1 < δ} ψ 1 ɛ [S δ (e)] = {z B 2 : 1 h(z) < 2δ 1/ɛ } = h 1[ S2δ 1/ɛ(1) ]. We thus have by Lemma 5.6 (v β ψɛ 1 )[S δ (e)] δ (3+β 1/4)/ɛ for 0 < δ < 1. By this and (5.12) we conclude (5.11), as required. We now roceed to the roof of the lemma. Note by (5.11) and the well-known Carleson Measure Criteria (see, for examle, [7, Proosition 3.1]) that ɛ C ϕj : A α(b 2 ) A β (B 2) is bounded β + 3 ɛ(α + 3) (5.13) for each j. So, the lemma holds for N = 1. Also, this imlies the sufficiency art of the lemma. We now rove the necessity art of the lemma for the case N 2. So, suose that L : A α(b 2 ) A β (B 2) is bounded. We may assume λ 1 = 1 so that ϕ 1 = ψ ɛ. We emloy test functions f δ given by f δ (z) := δ 1/ [1 (1 δ)z 1 ] (α+4)/, z = (z 1, z 2 ) B 2 for 0 < δ < 1. Note f δ A α 1 for all δ by (2.8). It follows that 1 Lf δ A β N a 1 f δ ϕ 1 dv β a j f δ ϕ j dv β B 2 B 2 =: I II j=2 To estimate the first integral of the above, we note 1 (1 δ)z 1 2δ, z S δ (e)

25 COMPACT DOUBLE DIFFERENCES 25 and thus f δ δ (α+3) on S δ (e). Accordingly, we obtain I = a 1 f δ d(v β ψɛ 1 ) B 2 a 1 f δ d(v β ψɛ 1 ) S δ (e) (v β ψɛ 1 )[S δ (e)] δ α+3 δ (3+β 1/4)/ɛ 3 α ; the last estimate holds by (5.11). Meanwhile, since for each j 1, we have (f δ ϕ j )(e) = f δ (λ j e) = O(δ 1/ ) II = O(δ) for 0 < δ < 1. Combining these observations, we obtain δ (3+β 1/4)/ɛ 3 α = O(1), which yields β + 3 ɛ(α + 3) + 1 4, as required. This comletes the roof. We now close the aer with the following examle in connection with Theorem 1.1 and its corollaries in this section. Examle 5.8. Let α > 1 and 1 1 4(α+3) Then lim z 1 for each j, but L is not bounded on A α(b) for any 0 < <. < ɛ < 1. Let L be an oerator as in (5.10). 1 z 1 ϕ j (z) = 0 (5.14) Proof. Clearly, (5.14) holds by Lemma 5.5. Meanwhile, since α + 3 < ɛ(α + 3) + 1 4, we see from Lemma 5.7 that L is not bounded on A α(b) for any 0 < <. REFERENCES [1] B. R. Choe, H. Koo and I. Park, Comact differences of comosition oerators over olydisks, Integral Equations Oerator Theory 73(1)(2012), [2] B. R. Choe, H. Koo and I. Park, Comact differences of comosition oerators on the Bergman saces over the ball, Potential Analysis 40(2014), [3] B. R. Choe, H. Koo and M. Wang, Comact double differences of comosition oerators on the Bergman saces, J. Funct. Anal. 272(2017), [4] D. Clahane, Comact comosition oerators on weighted Bergman saces of the unit ball, J. Oer. Theory 45(2001), [5] C. C. Cowen and B. D. MacCluer, Comosition oerators on saces of analytic fuctions, Studies in Advanced Mathematics, CRC Press, Boca Raton, FL, [6] P. Halmos, Measure Theory, Sringer-Verlag, New York, [7] H. Koo and W. Smith, Comosition oerators induced by smooth self-mas of the unit ball in C N, J. Math. Anal. Al. 329(2007), [8] H. Koo and M. Wang, Joint Carleson measure and the difference of comosition oerators on A α(b n), J. Math. Anal. Al. 419(2014), [9] J. Moorhouse, Comact differences of comosition oerators, J. Funct. Anal. 219(2005), [10] W. Rudin, Function theory in the unit ball of C n, Sringer, New York (1980). [11] J. H. Shairo, Comosition oerators and classical function theory, Sringer, New York, [12] K. Zhu, Saces of holomorhic functions in the unit ball, Sringer-Verlag, New York, 2005.

26 26 B. CHOE, H. KOO, AND J. YANG [13] K. Zhu, Comact comosition oerators on Bergman saces of the unit ball, Houston J. Math. 33, (2007). DEPARTMENT OF MATHEMATICS, KOREA UNIVERSITY, SEOUL 02841, KOREA address: DEPARTMENT OF MATHEMATICS, KOREA UNIVERSITY, SEOUL 02841, KOREA address: DEPARTMENT OF MATHEMATICS, KOREA UNIVERSITY, SEOUL 02841, KOREA address: