COMPACT DIFFERENCE OF WEIGHTED COMPOSITION OPERATORS ON N p -SPACES IN THE BALL

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1 COMPACT DIFFERENCE OF WEIGHTED COMPOSITION OPERATORS ON N p -SPACES IN THE BALL HU BINGYANG and LE HAI KHOI Communicated by Mihai Putinar We obtain necessary and sucient conditions for the compactness of dierences of weighted composition operators acting from N p-spaces into the Beurling/Bergman-type spaces A q of holomorphic functions in the unit ball. AMS 00 Subject Classication: 3A36 47B33. Key words: N p-spaces weighted composition operators compact dierence.. INTRODUCTION.. Notation and denitions Let B be the unit ball in the complex vector space C n O(B) denotes the space of functions that are holomorphic in B with the compact-open topology and H (B) denotes the Banach space of bounded holomorphic functions on B with the norm f = f(z). If z = (z z... z n ) ζ = (ζ ζ... ζ n ) C n then z ζ = z ζ + + z n ζn and z = (z z + + z n z n ) /. Let p > 0 the Beurling-type space (sometime also called the Bergmantype space) A p (B) in the unit ball is dened as A p (B) := f(z) O(B) : f p = f(z) ( z ) p <. For a holomorphic self-mapping ϕ of B and a holomorphic function u : B C the linear operator W uϕ : O(B) O(B) W uϕ (f)(z) = u(z) (f ϕ(z)) f O(B) z B is called the weighted composition operator with symbols u and ϕ. Observe that W uϕ (f) = M u C ϕ (f) where M u (f) = uf is the multiplication operator with symbol u and C ϕ (f) = f ϕ is the composition operator with symbol ϕ. If u is identically then W uϕ = C ϕ and if ϕ is the identity then W uϕ = M u. REV. ROUMAINE MATH. PURES APPL. 60 (05) 06

2 0 Hu Bingyang and Le Hai Khoi Composition operators and weighted composition operators acting on spaces of holomorphic functions in the unit disk D of the complex plane have been studied quite well. We refer the readers to the monographs 9] for detailed information. Composition operators on A p (D) have also been intensively studied (see e.g. 5] and references therein)... N p -spaces in the unit ball Given a point a B we can associate to it the automorphism Φ a Aut(B) (see e.g. 8] Section.). In 7] we introduce the N p -spaces in B for p > 0 which is dened as follows: N p (B) := f O(B) : f p = a B ( f(z) ( Φ a (z) ) p dv (z) B ) / < where dv is the Lebesgue normalized volume measure on B (i.e. V (B) = ). Several important properties of the N p -spaces and of the weighted composition operators from N p -spaces to the spaces A q have been characterized in 7]. We cite here main results from 7] for the reader's convenience. Theorem.. The following statements hold: (a) For p > q > 0 we have H (B) N q (B) N p (B) A n+ (B) where the last embedding is given by f n+ p+n 3 p/ f p f N p (B). (b) For p > 0 if p > k k (0 n+ ] then A k (B) N p (B). In particular when p > n all N p (B) = A n+ (B). (c) N p (B) is a functional Banach space with the norm p and moreover its norm topology is stronger than the compact-open topology. (d) For 0 < p < B(B) N p (B) where B(B) is the Bloch space in B. Here the symbol X Y means the continuous embedding of X into Y. Theorem.. Let ϕ: B B be a holomorphic mapping u: B C a holomorphic mapping and p q > 0. The weighted composition operator W uϕ : N p (B) A q (B) () is bounded if and only if () is compact if and only if lim r ϕ(z) >r u(z) ( z ) q ( ϕ(z) ) n+ < ; u(z) ( z ) q = 0. ( ϕ(z) ) n+

3 3 Compact dierence 03 The aim of this paper is to obtain necessary and sucient conditions for the compactness of dierences of weighted composition operators acting from N p (B)-spaces into the spaces A q (B). It should be noted that the problem of compact dierence of composition operators and weighted composition operators on spaces of holomorphic functions in several variables (either polydisk or ball) has been treated in several papers in which the methods of proof in general follow the standard lines using the characterization of compact operators in the corresponding function spaces (see e.g ]). In the present paper for N p -spaces we focus on dierent technical issues beginning with giving an estimate the quantity f(z) f(w) where f N p (B) and z w B; and then inspiring by the pseudohyperbolic metric in B we use the function ρ ϕφ (z) = Φ ϕ(z) (φ(z)) to characterize the compact dierence between N p (B) and A q (B). Although our approach is standard the results have their own interest due to the novelty of N p (B)-spaces. Especially the criterion of compactness of dierences depends on n the dimension of the complex space C n while it is independent of p > 0.. COMPACT DIFFERENCE Let ϕ ϕ be two holomorphic self-mappings on B u u : B C two holomorphic mappings and p q > 0. Let further W u ϕ and W u ϕ be two weighted composition operators acting from N p (B) into A q (B). We need some auxiliary results. Lemma.. The operator W u ϕ W u ϕ : N p (B) A q (B) is compact if and only if (W u ϕ W u ϕ )(f m ) q 0 as m for any bounded sequence f m in N p (B) such that f m converges to zero uniformly on every compact subset of B. The proof of Lemma. follows by standard arguments and hence we omit the details. Now recall that the pseudohyperbolic metric in the ball is dened as ρ(z w) = Φ w (z) z w B. It is a true metric (see e.g. 3]). Also it is easy to verify in particular that ρ(0 w) = w ρ(φ w (z) w) = z. The following lemmas play an important role in the proof of our main result. They also have their own interest. Lemma.. For z w B if ρ(z w) then 6 z w 6.

4 04 Hu Bingyang and Le Hai Khoi 4 Proof. Let z w B. For simplicity denote r = ρ(z w). By (3] Theorem (c)) we have ρ(φ w (z) 0) ρ(0 w) ρ(φ w (z) 0)ρ(0 w) ρ(φ w(z) w) ρ(φ w(z) 0) + ρ(0 w) + ρ(φ w (z) 0)ρ(0 w) or equivalently r w r + w z r w + r w. Actually for the left inequality we need only a weaker version. Namely (.) w r w + r z r w + r w. Furthermore since r (0 ] from the left inequality of (.) it follows that z w w while the right inequality of (.) gives We also have Therefore we get w r r w w +r +r w z w w = + r r w 3 = r + r w 3. + w + z + w + w. 6 z w = z w + z + w 6. Lemma.3. For f N p (B) and z w B we have f(z) f(w) c f p max ( z ) n+ ( w ) n+ Here c = 6 n+ p+n+ (3+ 3) n 3 p/. Proof. We consider two cases: ρ(z w). Case : ρ(z w) 4. Since f(z) f(w) f(z) + f(w) by Theorem.(a) we have min ( z ) n+ ( w ) n+ f(z) f(w) ( z ) n+ f(z) + ( w ) n+ f(w)

5 5 Compact dierence 05 f n+ which implies that f(z) f(w) p+n+3 3 p/ f p max p+n+ 3 p/ f p p+n+3 3 p/ f p ρ(z w) ( z ) n+ ( w ) n+ ρ(z w). Case : ρ(z w) < 4. Take and x w B. From ρ(φ w (z) w) = z it follows that if z B / then ρ(φ w (z) w). In this case by Theorem.(a) and Lemma. we have f(φ w (z)) = f n+ ( Φ w (z) ) n+ p+n f p 3 p/ ( w ) n+ p+n f p 3 p/ ( Φ w (z) ) n+ w Φ w (z) ] n+ 6 n+ p+n f p. 3 p/ ( w ) n+ Now we follow the standard scheme to estimate a quantity f(z) f(w). Set g w = f Φ w then f(z) f(w) = f(φ w (Φ w (z)) f(φ w (0)) = g w (Φ w (z)) g w (0). For each z B with ρ(z w) = Φ w (z) < 4 we have f(z) f(w) = g w (Φ w (z)) g w (0) g w (t) Φ w (z) = g w (t) ρ(z w) where t = (t t... t n ) is some point in B with t Φ w (z) 4. Furthermore g w (t) ρ(z w) nρ(z w) max g w kn (t) z k g w (t t... ξ k... t n ) nρ(z w) max kn πi ξ k = 3 (ξ 4 k t k ) dξ k nρ(z w) max g w (t t... ξ k... t n ) π kn ξ k = 3 (ξ 4 k t k ) dξ k. Note that for (t t... ξ k... t n ) with t 4 and ξ k = 3 4 we have ρ (Φ w (t t... ξ k... t n ) w) = ρ((t t... ξ k... t n ) 0) = (t t... ξ k... t n ) t + ξ k ( ) ( ) 3 + = 4 4 and so g w (t t... ξ j... t n ) = f ( Φ w (t t... ξ k... t n ) ) 6 n+ p+n f p. 3 p/ ( w ) n+

6 06 Hu Bingyang and Le Hai Khoi 6 Also max kn max kn ξ k = 3 4 ξ k = 3 4 Consequently nρ(z w) f(z) f(w) π dξ k ξ k t k max kn ( dξ k 3 4 π 4 ) ξ k = n+ p+n f p 3 p/ ( w ) n+ = 6 n+ p+n+ (3 + 3) n 3 p/ ( 3 dξ k 4 t k ) ( ) 4 = 4π(3 + 3). 3 4π(3 + 3) ( w ) n+ Combining the results of the two cases yields f(z) f(w) c f p max ( z ) n+ ( w ) n+ f p ρ(z w). ρ(z w) where c = 6 n+ p+n+ (3+ 3) n 3 p/. Inspiring by the pseudohyperbolic metric in the unit ball for two holomorphic mappings ϕ ψ : B B we dene ρ ϕψ (z) = Φϕ(z) (ψ(z)) z B. Evidently ρ ϕψ = ρ ψϕ. For each w B set ( w (.) k w (z) = ( z w ) ) n+ z B. By (7] Lemma 3.) we have k w N p (B) and k w p. Also note w B ( ) n+ that k w (w) = w w B. Now we are ready to formulate our main result of this paper. Theorem.4. Let ϕ ϕ : B B be two holomorphic mappings u u : B C two holomorphic mappings and p q > 0. Let further W u ϕ and W u ϕ be two weighted composition operators acting from N p (B) into A q (B). Then W u ϕ W u ϕ is compact if and only if the following conditions are satised: (i) lim r ϕ k (z) >r u k (z) ( z ) q ( ϕ k (z) ) n+ ρ ϕ ϕ (z) = 0 (k = );

7 7 Compact dierence 07 (ii) lim r min ϕ (z) ϕ (z) >r u (z) u (z) min ( z ) q ( ϕ (z) ) n+ ( z ) q ( ϕ (z) ) n+ Proof. Necessity. Suppose W u ϕ W u ϕ is a compact operator. Prove (i). It suces to prove for k =. Since W u ϕ is bounded by Theorem. and the fact that ρ ϕ ϕ (z) z B we have u (z) ( z ) q u (z) ( z ) q ρ ( ϕ (z) ) n+ ϕ ϕ (z) <. ( ϕ (z) ) n+ Set G(r) = ϕ (z)) >r u (z) ( z ) q ( ϕ (z) ) n+ ρ ϕ ϕ (z) r (0 ). It is clear that G is bounded and decreasing on (0 ) and hence lim G(r) r does always exist. Assume that (i) is not true. Then there exists an L > 0 such that lim G(r) > L. r By the standard diagonal process as sketched in (7] Theorem 3.4) we can choose a sequence z m B such that ϕ (z m ) as m and (.3) u (z m ) ( z m ) q ( ϕ (z m ) ) n+ Consider the functions ρ ϕ ϕ (z m ) > L (m =...). 4 g m (z) = Φ ϕ (z m)(z) k wm (z) z B where w m = ϕ (z m ) and k wm is dened by (.) m =.... Obviously g m O(B). Moreover since Φ ϕ (z m)(z) z B we have g m p k wm p which shows that g m (z) N p (B) for all m N and that the sequence g m is bounded in N p (B). Furthermore by the fact that k wm m N converges to zero uniformly on every compact subset of B and g m (z) k wm (z) z B we have g m m N also converges to zero uniformly on every compact subset of B. By Lemma. (.4) (W u ϕ W u ϕ )(g m ) q 0 as m. Note that for each m N Φ ϕ (z m)(ϕ (z m )) = 0 which implies g m (ϕ (z m )) = 0. Then ] =0.

8 08 Hu Bingyang and Le Hai Khoi 8 (W u ϕ W u ϕ )(g m ) q = u (z)g m (ϕ (z)) u (z)g m (ϕ (z)) ( z ) q u (z m )g m (ϕ (z m )) u (z m )g m (ϕ (z m )) ( z m ) q = u (z m )g m (ϕ (z m )) ( z m ) q = u (z m ) ( z m ) q Φ ( ϕ (z m ) ) n+ ϕ (z m)(ϕ (z m )) = u (z m ) ( z m ) q ρ ( ϕ (z m ) ) n+ ϕ ϕ (z m ) > L 4 by (.3) which contradicts (.4). Thus we must have u (z) ( z ) q lim ρ r ϕ (z) >r ( ϕ (z) ) n+ ϕ ϕ (z) = 0 and (i) is proved. Prove (ii). Since both W u ϕ and W u ϕ are bounded by Theorem. we have ] ( z ) q ( z ) q + u (z) u (z) min u (z) min u (z) min u (z) ( z ) q ( ϕ (z) ) n+ For each r (0 ) set H(r) = min ϕ (z) ϕ (z) >r ( ϕ (z) ) n+ ( z ) q ( ϕ (z) ) n+ ( z ) q ( ϕ (z) ) n+ + u (z) ( z ) q ( ϕ (z) ) n+ u (z) u (z) min ( ϕ (z) ) n+ ] ( z ) q ( ϕ (z) ) n+ ( z ) q ( ϕ (z) ) n+ ( z ) q <. ( ϕ (z) ) n+ ] ( z ) q ( ϕ (z) ) n+ This function H(r) is clearly bounded and decreasing on (0 ) and hence lim H(r) exists. We prove that this limit must be zero. r We follow the same scheme of proving (i) but it requires more delicate arguments. Assume that lim H(r) = L > 0. Again by the standard diagonal r process we can choose a sequence z m B such that min ϕ (z m ) ϕ (z m ) ].

9 9 Compact dierence 09 as m and that for each m N ( z m ) q u (z m ) u (z m ) min ( ϕ (z m ) ) n+ ( z m ) q ( ϕ (z m ) ) n+ > L 4. For each m N we have either ϕ (z m ) ϕ (z m ) or ϕ (z m ) ϕ (z m ). Choose a subsequence z mk of z m such that for each k N ϕ (z mk ) ϕ (z mk ). Otherwise there are only nitely many indexes m such that ϕ (z m ) ϕ (z m ) and in this case we choose a subsequence z mk of z m such that for each k N ϕ (z mk ) ϕ (z mk ). We only consider the rst case omitting the second case (which can be proved by interchanging the role of ϕ and ϕ ) and without loss of generality for the purpose of convenience we write z mk as z m. Since ϕ (z m ) ϕ (z m ) for each m N we have ( z m ) q ( z m ) q u (z m ) u (z m ) min = u (z m ) u (z m ) ( ϕ (z m ) ) n+ ( z m ) q ( ϕ (z m ) ) n+ > L 4. ( ϕ (z m ) ) n+ Since the sequence ρ ϕ ϕ (z m ) is bounded it contains a convergent subsequence. Without loss of generality we can assume that lim ρ ϕ m ϕ (z m ) = l 0. There are two cases for l: Case : l > 0. In this case there exists an m 0 N such that ρ ϕ ϕ (z m ) > l m > m 0. In this case we have ( z m ) q u (z m ) u (z m ) ( ϕ (z m ) ) n+ ( z m ) q u (z m ) ( ϕ (z m ) ) n+ l ρ ϕ ϕ (z m ) l ρ ϕ ϕ (z m ) + ( z m ) q u (z m ) ( ϕ (z m ) ) n+ ( z m ) q u (z m ) ( ϕ (z m ) ) n+ ( z m ) q u (z m ) ( ϕ (z m ) ) n+ + ( z m ) q u (z m ) ( ϕ (z m ) ) n+ + ( z m ) q u (z m ) ( ϕ (z m ) ) n+ ] ]

10 0 Hu Bingyang and Le Hai Khoi 0 which gives ρ ϕ ϕ (z m ) ( z m ) q u (z m ) ( ϕ (z m ) ) n+ ] + ( z m ) q u (z m ) Ll ( ϕ (z m ) ) n+ 8 m > m 0. However since ϕ (z m ) ϕ (z m ) m N from it follows that and hence Hence by (i) ρ ϕ ϕ (z m ) which is impossible. lim min ϕ (z m ) ϕ (z m ) = m lim ϕ (z m ) = lim ϕ (z m ) =. m m lim ρ ϕ m ϕ (z m ) ( z m ) q u k (z m ) = 0 k = ( ϕ k (z m ) ) n+ ( ( z m ) q u (z m ) ( ϕ (z m ) ) n+ ) + ( z m ) q u (z m ) 0 as m ( ϕ (z m ) ) n+ Case : l = 0. We claim for the probe functions k wm where w m = ϕ (z m ) that (.5) ( ϕ (z m ) ) n+ kwm (ϕ (z m )) 0 as m. Indeed by Lemma.3 we have ( ϕ (z m ) ) n+ kwm (ϕ (z m )) = ( ϕ (z m ) ) n+ k wm (ϕ (z m )) k wm (ϕ (z m )) ( ϕ (z m ) ) n+ c k wm p ρ ϕ ϕ (z m ) max ( ϕ (z m ) ) n+ ( ϕ (z m ) ) n+ ( ϕ (z m ) ) n+ cρ ϕ ϕ (z m ) = cρ ( ϕ (z m ) ) n+ ϕ ϕ (z m ). The last expression converges to l = 0 as m and (.5) follows. Here c is the constant dened in Lemma.3. Furthermore since lim ρ ϕ m ϕ (z m ) = 0 there exists m N such that ρ ϕ ϕ (z m ) < m m ϕ (z m ). Then by Lemma. 6. Also ϕ (z m ) since W u ϕ is bounded from N p (B) into A q (B) by Theorem. there

11 Compact dierence ( z ) q u (z) exists some positive number K > 0 such that < K. ( ϕ (z) ) n+ Again by lim ρ ϕ m ϕ (z m ) = 0 there exists m N such that ρ ϕ ϕ (z m ) < L m > m. 8 6 n+ ck and Consequently for m > maxm m (W u ϕ W u ϕ )(k wm ) q = ( z ) q u (z)k wm (ϕ (z)) u (z)k wm (ϕ (z)) ( z m ) q u (z m )k wm (ϕ (z m )) u (z m )k wm (ϕ (z m )) = ( z m ) q u (z m ) u ( ϕ (z m ) ) n+ (z m )k wm (ϕ (z m )) ( z m ) q u (z m ) u (z m ) ( ϕ (z m ) ) n+ ( ϕ (z m ) ) n+ ( z m ) q u (z m ) u ( ϕ (z m ) ) n+ (z m )k wm (ϕ (z m )) = ( z m ) q u (z m ) u (z m ) ( ϕ (z m ) ) n+ ( z m ) q u (z m ) ( ϕ (z m ) ) n+ Moreover we also have ( z m ) q u (z m ) u (z m ) ( ϕ (z m ) ) n+ ( z m ) q u (z m ) ( ϕ (z m ) ) n+ ( ϕ (z m ) ) n+ ( ϕ (z m ) ) n+ Therefore we arrive at K 6 n+. ( ϕ (z m ) ) n+ ( z m ) q u (z m ) u (z m ) ( ϕ (z m ) ) n+ kwm (ϕ (z m )). > L 4 = ( z m ) q u (z m ) ( ϕ (z m ) ) n+ (.6) (W u ϕ W u ϕ )(k wm ) q L 4 6 n+ Kcρϕ ϕ (z m ) L 4 L 8 = L 8. However since W u ϕ W u ϕ is compact and k wm converges to zero uniformly on every compact subset of B we must have (.7) lim m (W u ϕ W u ϕ )(k wm ) q = 0

12 Hu Bingyang and Le Hai Khoi which contradicts (.6). Suciency. Let conditions (i)(ii) hold. Take an arbitrary bounded sequence f m in N p (B) that converges to zero uniformly on every compact subset of B. By Lemma. we show that (W u ϕ W u ϕ )(f m ) q 0 as m. Again we use a method of contradiction. Assume that there is an ε 0 > 0 and a subsequence f mk of f m such that (.8) (W u ϕ W u ϕ )(f mk ) q ε 0 k N. We may assume for the sake of simplicity that f mk is f m itself. That is m N (W u ϕ W u ϕ )(f m ) q = ( z ) q u (z)f m (ϕ (z)) u (z)f m (ϕ (z)) ε 0. From this it follows that there exists a sequence z m B such that m N (.9) H m = ( z m ) q u (z m )f m (ϕ (z m )) u (z m )f m (ϕ (z m )) ε 0. Here z m 's are not necessarily distinct. We may also without loss of generality assume that both sequences ϕ (z m ) and ϕ (z m ) converge (as otherwise we can consider their convergent subsequences instead). Note that since both W u ϕ and W u ϕ are bounded by Theorem. there exists K > 0 such that (.0) u k (z) ( z ) q u k (z) ( z ) q ( ϕ k (z) ) n+ K k =. Now we consider the sequence max ϕ (z m ) ϕ (z m ). It is clear that there exists lim m max ϕ (z m ) ϕ (z m ) = q. Claim : q =. Proof. Assume (.) max ϕ (z m ) ϕ (z m ) q < as m. Then by (.9) ε 0 H m ( z m ) q( ) u (z m )f m (ϕ (z m )) + u (z m )f m (ϕ (z m )

13 3 Compact dierence 3 ( ) K f m (ϕ (z m )) + f m (ϕ (z m )) m N. Furthermore by (.) there exists m 3 N such that max ϕ (z m ) ϕ (z m ) + q m > m 3. In particular for all m > m 3 ϕ k (z m ) z : z +q k =. Since is a compact set of B the sequence f m (z) converges uniformly z : z +q to zero on this set and hence both sequences f m (ϕ k (z m )) k = converge to zero as m which shows that H m K ( f m (ϕ (z m )) + f m (ϕ (z m )) ) 0 as m. But this contradicts the fact that H m ε 0 m N. Thus we have (.) max ϕ (z m ) ϕ (z m ) as m. Then at least one of the limits lim m ϕ k(z m ) (k = ) must be. So we may assume that (.3) lim ϕ (z m ) = P with P = m lim ϕ (z m ) = Q with Q. m Furthermore we may also assume that there exists the limit lim ρ ϕ m ϕ (z m ) = l 0 (otherwise we consider its convergent subsequence). Claim : l = 0. Proof. Assume in contrary that l > 0. Consider two cases of Q. Case : Q =. In this case from (i) and (.3) it follows that u k (z m ) ( z m ) q (.4) lim m ( ϕ k (z m ) ) n+ Then by Theorem. and (.9) we have = 0 (k = ). H m ( z m ) q u (z m ) ( ϕ ( ϕ (z m ) ) n+ (z m ) ) n+ fm (ϕ (z m )) + ( z m ) q u (z m ) ( ϕ ( ϕ (z m ) ) n+ (z m ) ) n+ fm (ϕ (z m ))

14 4 Hu Bingyang and Le Hai Khoi 4 ( z m ) q u (z m ) ( ϕ (z m ) ) n+ p+n 3 p/ ( z m ) q u (z m ) ( ϕ (z m ) ) n+ ] + ( z m ) q u (z m ) f ( ϕ (z m ) ) n+ m n+ ] + ( z m ) q u (z m ) ( ϕ (z m ) ) n+ f m p. In the last inequality letting m by (.4) as well as boundedness of f m in N p (B) we get lim H m = 0 m which is impossible because it contradicts (.9). Case : Q <. In this case the second limit in (.3) gives ϕ (z m ) z : z + Q for all m large enough say m > m 4. Then by (.0) and Theorem. for all m > m 4 we have H m ( z m ) q u (z m ) f m (ϕ (z m )) +( z m ) q u (z m ) f m (ϕ (z m )) = ( z m ) q u (z m ) ( ϕ ( ϕ (z m ) ) n+ (z m ) ) n+ fm (ϕ (z m )) +( z m ) q u (z m ) f m (ϕ (z m )) ( z m ) q u (z m ) f ( ϕ (z m ) ) n+ m n+ + K f m (ϕ (z m )) p+n 3 p/ ( z m ) q u (z m ) f ( ϕ (z m ) ) n+ m p + K f m (ϕ (z m )). Letting m from (.4) and the fact that f m (ϕ (z m )) converges to zero uniformly on the compact set z : z + Q it follows that right-hand side of the last inequality tends to 0 as m which again contradicts (.9). Thus we have (.5) lim m ρ ϕ ϕ (z m ) = 0. Claim 3: Q = that is lim m ϕ (z m ) =. Proof. Indeed for each m N we have ρ ϕ ϕ (z m ) = Φ ϕ (z m)(ϕ (z m ) = ( ϕ (z m ) )( ϕ (z m ) ) ϕ (z m ) ϕ (z m ). Since lim ϕ (z m ) = if lim ϕ (z m ) = Q < then we would have m m for all m large enough ϕ (z m ) ϕ (z m ) ϕ (z m ) ϕ (z m )

15 5 Compact dierence 5 ϕ (z m ) ϕ (z m ) Q > 0. But this implies that lim m ρ ϕ ϕ (z m ) = which contradicts (.5). Now by the same reason as in the necessity part we may assume that (.6) ϕ (z m ) ϕ (z m ) m N. Then from (.9)(.0) and Lemma.3 it follows that for each m N H m = ( z m ) q u (z m )f m (ϕ (z m )) u (z m )f m (ϕ (z m )) ( z m ) q u (z m )f m (ϕ (z m )) u (z m )f m (ϕ (z m )) +( z m ) q u (z m )f m (ϕ (z m )) u (z m )f m (ϕ (z m )) = ( z m ) q u (z m ) ( ϕ (z m ) ) n+ + ( z m ) q f m (ϕ (z m )) ( ϕ (z m ) ) n+ ck f m p max + p+n ( ϕ (z m ) ) n+ fm (ϕ (z m )) f m (ϕ (z m )) ( ϕ (z m ) ) n+ ( ϕ (z m ) ) n+ 3 p/ f m p ( z m ) q u (z m ) u (z m ). ( ϕ (z m ) ) n+ However by (.6) ( ϕ (z m ) ) n+ max ( ϕ (z m ) ) n+ and hence = max ( ϕ (z m ) ) n+ ( ϕ (z m ) ) n+ u (z m ) u (z m ) ( ϕ (z m ) ) n+ H m ck f m p ρ ϕ ϕ (z m ) ( ϕ (z m ) ) n+ ρ ( ϕ (z m ) ) n+ ϕ ϕ (z m ) ( ϕ (z m ) ) n+ ( ϕ (z m ) ) n+ = + p+n 3 p/ f m p ( z m ) q u (z m ) u (z m ) ( ϕ (z m ) ) n+ = ck f m p ρ ϕ ϕ (z m ) + f m p p+n 3 p/ u (z m ) u (z m ) ( z m ) q ( z m ) q min ( ϕ (z m ) ) n+ ( ϕ (z m ) ) n+.

16 6 Hu Bingyang and Le Hai Khoi 6 In the inequality above since f m p is bounded by (.5) lim f m p ρ ϕ ϕ m (z m ) = 0. Furthermore by (ii) and lim ϕ (z m ) = lim ϕ (z m ) = we get m m lim f ( z m ) q ( z m ) q m p u (z m ) u (z m ) min =0. m ( ϕ (z m ) ) n+ ( ϕ (z m ) ) n+ These equalities imply that lim m H m = 0 but this contradicts (.9). The theorem is proved completely. Acknowledgments. Supported in part by MOE's AcRF Tier grant M (RG4/3). REFERENCES ] B. Choe H. Koo and I. Park Compact dierences of composition operators over polydisks. Integral Equations Operator Theory 73 (03) 579. ] C. Cowen and B. MacCluer Composition Operators on Spaces of Analytic Functions. CRC Press Boca Raton ] P. Duren and R. Weir The pseudohyperbolic metrc and Bergman spaces in the ball. Trans. Amer. Math. Soc. 359 (007) ] L. Jiang and C. Ouyang Compact dierences of composition operators on holomorphic function spaces in the unit ball. Acta Math. Sci. Ser B Engl. Ed. 3 (0) ] H. Hedenmalm B. Korenblum and K. Zhu Theory of Bergman Spaces. Springer-Verlag ] K. Heller B.D. MacCluer and R.J. Weir Compact dierences of composition operators in several variables. Integral Equations Operator Theory 69 (0) ] B. Hu and L.H. Khoi Weighted-composition operators on N p-spaces in the ball. C.R. Math. Acad. Sci. Paris. Ser.I 35 (03) ] W. Rudin Function Theory in the Unit Ball of C n. Springer-Verlag New York ] J.H. Shapiro Composition Operators and Classical Function Theory. Springer-Verlag New York ] C. Tong Compact dierences of weighted composition operators on H (B N ). J. Comput. Anal. Appl. 4 (0) 34. Received 4 February 04 Nanyang Technological University (NTU) Division of Mathematical Sciences School of Physical and Mathematical Sciences Singapore bhu@e.ntu.edu.sg lhkhoi@ntu.edu.sg