Bounded Toeplitz products on the Bergman space of the polydisk

J. Math. Anal. Appl. 278 2003) 25 35 www.elsevier.com/locate/jmaa Bounded Toeplitz products on the Bergman space of the polydisk Karel Stroethoff a, and echao Zheng b, a Mathematical Sciences, University of Montana, Missoula, MT 5982, USA b Mathematics epartment, Vanderbilt University, Nashville, TN 37240, USA Received 7 January 2002 Submitted by H. Gaussier Abstract We consider the question for which square integrable analytic functions f and g on the polydisk the densely defined products T f Tḡ are bounded on the Bergman space. We prove results analogous to those we obtained in the setting of the unit disk [K. Stroethoff,. Zheng, J. Funct. Anal. 69 999) 289 33]. 2003 Elsevier Science USA). All rights reserved.. Introduction Throughout let n be a fixed integer n 2. enote the unit disk in C by, andletν be Lebesgue volume measure on, normalized so that ν ) =. For λ,letϕ λ be the fractional linear transformation on given by ϕ λ z) = λ z)/ λz). Each ϕ λ is an automorphism on the disk, in fact, ϕλ = ϕ λ.forw = w,..., w n ) the mapping ϕ w on the polydisk given by ϕ w z) = ϕ w z ),...,ϕ wn z n )) is an automorphism on. The Bergman space L 2 a n ) is the space of analytic functions h on which are square-integrable with respect to Lebesgue volume measure on.the reproducing kernel in L 2 a n ) is given by * Corresponding author. E-mail addresses: ma_kms@selway.umt.edu K. Stroethoff), zheng@math.vanderbilt.edu. Zheng). The author was supported in part by the National Science Foundation. 0022-247X/03/$ see front matter 2003 Elsevier Science USA). All rights reserved. doi:0.06/s0022-247x02)00578-4

26 K. Stroethoff,. Zheng / J. Math. Anal. Appl. 278 2003) 25 35 K w z) = w j z j ) 2, for z, w.if, denotes the inner product in L 2 ),then h, K w =hw),forevery h L 2 a n ) and w. The orthogonal projection P of L 2 ) onto L 2 a n ) is given by Pg)w) = g,k w = gz) w j z j ) 2 dνz), for g L 2 ) and w.givenf L ), the Toeplitz operator T f is defined on L 2 a n ) by T f h = Pfh).Wehave T f h)w) = f z)hz) w j z j ) 2 dνz), for h L 2 a n ) and w. Note that the above formula makes sense, and defines a function analytic on,alsoiff L 2 ).So,ifg L 2 a n ) we define Tḡ by the formula Tḡh)w) = gz)hz) w j z j ) 2 dνz), for h L 2 a n ) and w.ifalsof L 2 a n ),thent f Tḡh is the analytic function ftḡh. We consider the following problem, which for n = was raised by Sarason in [2]. Problem of boundedness of Toeplitz products on L 2 a n ). For which f and g in L 2 a n ) is the operator T f Tḡ bounded on L 2 a n )? In this paper we extend our results for boundedness of these Toeplitz products on the Bergman space of the unit disk [4] to higher dimension. In the next section we will first give a necessary condition for boundedness of the Toeplitz product T f Tḡ on L 2 a n ). A recent counter-example of Nazarov [] for Toeplitz products on the Hardy space indicates that it may not be possible to prove that this necessary condition is also sufficient. In the final section of the paper we will show that this condition is, however, very close to being sufficient, as shown for Toeplitz products on the Hardy space of the unit circle in [5]. 2. Necessary condition for boundedness Suppose f and g are in L 2 ). Consider the operator f g on L 2 a n ) defined by f g)h = h, g f, for h L 2 a n ). It is easily proved that f g is bounded on L 2 a n ) with norm equal to f g = f g. We will obtain an expression for the operator f g, wheref,g L 2 a n ).Thisis most easily accomplished by using the Berezin transform: writing k w for the normalized

K. Stroethoff,. Zheng / J. Math. Anal. Appl. 278 2003) 25 35 27 reproducing kernels, we define the Berezin transform of a bounded linear operator S on L 2 a n ) to be the function S defined on by Sw) = Sk w,k w, for w. The boundedness of S implies that the function S is bounded on.the Berezin transform is injective, for Sw) = 0, for all w, implies that S = 0, the zero operator on L 2 a n ) see [3] for a proof). Using the reproducing property of K w we have thus K w 2 = K w,k w =K w w) = k w z) = w j 2 ) 2, w j 2 w j z j ) 2, 2.) for z, w. It follows from 2.) that Sw) = wj 2) 2 SKw,K w, for w. It is easily seen that TḡKw = gw)k w. Thus T f TḡKw,K w = TḡKw,Tf K w = gw)k w, fw)k w =fw)gw) K w,k w, and we see that T f Tḡw) = fw)gw). 2.2) Since f g)k w,k w = K w,g f,k w = K w,g f,k w =fw)gw), wealsohave f gw) = wj 2) 2 fw)gw). 2.3) For a multi-index β = β,...,β n ), it follows from 2.2) that the Berezin transform of the Toeplitz product T z β T f TḡT z β = T z β f T is equal to the function w w β fw)gw)w gz β = β w 2β... w n 2β n fw)gw). Writing x j ) 2 = ) ) ) 2 2 2 ) k... x k k k...xk n n, 2 where k = k + k 2 + +k n and the sum is over all k,...,k n from {0,, 2}, wesee from 2.3) and the injectivity of the Berezin transform that f g = ) ) ) 2 2 2 ) k... T k k k 2 k n z...zn kn T f TḡT k z,...zn kn where k = k +k 2 + +k n and the sum is over all k,...,k n ranging over the set {0,, 2}. Using that T zj =, for each j, it follows that f g 4 n T f Tḡ, and thus f g 4 n T f Tḡ, 2.4) for f and g in L 2 a n ). k n

28 K. Stroethoff,. Zheng / J. Math. Anal. Appl. 278 2003) 25 35 We will next apply the invariance under composition of the symbols with the fractional transformations. We need some preliminaries to make this precise. The mapping ϕ w has real Jacobian n ϕ w j z j )ϕ w j z j ) = n ϕ w j z j ) 2, which by 2.) is equal to k w z) 2, so we have the following change-of-variable formula h ϕw z) ) k w z) 2 dνz) = hu) dνu), 2.5) for every h L ). It follows from 2.5) that the mapping U w h = h ϕ w )k w is an isometry on L 2 a n ): U w h 2 = h ϕw z) ) 2 kw z) 2 dνz) = hu) 2 dνu) = h 2, for all h L 2 a n ). It is easily verified that k w ϕw z) ) = k w z). Since ϕ w ϕ w = id, we see that Uw U w h) ) z) = U w h) ϕ w z) ) k w z) = hz)k w ϕw z) ) k w z) = hz), for all z and h L 2 a n ). Thus Uw = U w, and hence U w is unitary. Furthermore, T f ϕw U w = U w T f. 2.6) Proof. For h H and g L 2 a n ) we have U w T f h, U w g = T f h, g = fh,g = f u)hu)gu)dνz) n = ϕw z) ) h ϕ w z) ) g ϕ w z) ) kw z) 2 dνz) establishing 2.6). = f f ϕw z) ) h ϕ w z) ) k w z)g ϕ w z) ) k w z) dνz) = fu w h, U w g = T f ϕw U w h, U w g, It follows from 2.6), applied to f and ḡ,that T f ϕw Tḡ ϕw = T f ϕw U w )U w Tḡ ϕw U w )U w = U w T f )U w U w Tḡ)Uw = U w T f Tḡ)Uw. So if the Toeplitz product T f Tḡ is bounded on L 2 a n ), then so is the product T f ϕw Tḡ ϕw, and T f ϕw Tḡ ϕw = T f Tḡ. By 2.4) we have f ϕ w 2 g ϕ w 2 4 n T f ϕw Tḡ ϕw =4 n T f Tḡ,

K. Stroethoff,. Zheng / J. Math. Anal. Appl. 278 2003) 25 35 29 hence f 2 w) g 2 w) 4 n T f Tḡ 2, for all w.so,forf,g L 2 a n ), a necessary condition for the Toeplitz product T f Tḡ to be bounded on L 2 a n ) is sup f 2 w) g 2 w) <. 2.7) w In the next section we will show that this condition is very close to being sufficient for boundedness. 3. Sufficient condition In this section we will prove that a condition slightly stronger than 2.7) is sufficient for boundedness of the Toeplitz product T f Tḡ. In fact we have the following result. Theorem 3.. Let f and g be in L 2 a n ).Ifforε>0, sup f 2+ε w) g 2+ε w) <, w then the operator T f Tḡ is bounded on L 2 a n ). In the proof of Theorem 3. we will need estimates on Tf h and its derivatives, as well as an alternative way to write the inner product formula in L 2 a n ). 3.. Inner product formula in L 2 a n ) In this subsection we will establish a formula for the inner product in L 2 a n ) needed to prove our sufficiency condition for boundedness of Toeplitz products. Our point of departure is the following inner product formula proved in [4]: uz)vz)daz) = 3 uz)vz) z 2) 2 daz) + 6 u z)v z) z 2) 2 5 2 z 2 ) daz), for u, v L 2 a ).Letf,g L2 a 2 ).Forfixedz 2 we have fz,z 2 )gz,z 2 )daz ) = 3 fz,z 2 )gz,z 2 ) z 2) 2 daz ) + 6 f z z,z 2 ) g z z,z 2 ) z 2) 2 5 2 z 2) daz ).

30 K. Stroethoff,. Zheng / J. Math. Anal. Appl. 278 2003) 25 35 Integrating both sides of the above equation with respect to z 2 results in fz,z 2 )gz,z 2 )daz )daz 2 ) = 9 fz,z 2 )gz,z 2 ) z 2 2) 2 z 2) 2 daz )daz 2 ) + 2 + 2 + 36 f z g z z 2) 2 z 2 2) 2 5 2 z 2) daz )daz 2 ) f g z 2) 2 z 2 2) 2 5 2 z 2 2) daz )daz 2 ) z 2 z 2 2 f z z 2 2 g z z 2 z 2) 2 z2 2) 2 5 2 z 2) 5 2 z 2 2) dνz,z 2 ). To formulate a formula for the inner product in L 2 a n ) we first introduce some notation. For a nonempty subset α ={α,...,α m } of {,...,n} with α < <α m let µ α be the measure on defined by dµ α z) = 3n m z 6 m 2) 2... zn 2) 2 5 2 zj 2) daz )...daz n ), j α for z = z,...,z n ),wherem is the cardinality of α, andlet α h = α... αm h, where j hz) = h/ z j.efine h = h. Note that and dµ z) = 3 n z 2) 2... zn 2) 2 daz )...daz n ) dµ α z) 3 n z 2) 2... zn 2) 2 daz )...daz n ), for all subsets α of {,...,n}. Then the following formula for the inner product in L 2 a n ) is proved by repeating the above procedure: fz)gz)dνz) = α fz) α gz)dµ α z), 3.2) α where α runs over all subsets of {,...,n}.

K. Stroethoff,. Zheng / J. Math. Anal. Appl. 278 2003) 25 35 3 3.2. Estimates In the proof of our sufficiency condition for boundedness of Toeplitz products we will also need the estimates contained in the following lemmas. Lemma 3.3. For f L 2 a n ) and h H ) we have T f h)w) f w j 2 2 w) /2 h, for all w. Proof. By the inequality of Cauchy Schwarz, T f h)w) 2 fz) hz) w j z j 2 dνz) = fz) 2 w j z j 4 dνz) and the stated inequality follows. w j 2 ) 2 f 2 w) h 2, ) 2 hz) 2 dνz) Lemma 3.4. Let f L 2 a n ), h H ) and ε>0. Ifα ={α,...,α m } is a subset of {,...,n} with α < <α m,then α Tf h)w) 2 2n w j 2 f 2+ε w) /2+ε) hz) δ for all w,whereδ = 2 + ε)/ + ε). ) /δ w j z j 2 dνz), Proof. We will first prove the estimate for α ={,...,n}. Forf L 2 a n ) and h H ) we have Tf h)w) = f z)hz) w j z j ) 2 dνz), thus n Tf w... w h)w) = 2n f z)hz) n z j w j z j ) 3 dνz).

32 K. Stroethoff,. Zheng / J. Math. Anal. Appl. 278 2003) 25 35 Let ε>0. Applying Hölder s inequality we get n Tf w... w h)w) n 2 n fz) 2 n fz) 2+ε = 2 n hz) δ hz) w j z j 4/2+ε) ) /2+ε) w j z j 4 dνz) dνz) w j z j 2+3ε)/+ε) w j 2 ) 2 f 2+ε w) hz) δ ) /2+ε) dνz) w j z j 2+3ε)/2+ε) ) /δ ) /δ dνz). w j z j 2+3ε)/+ε) For each j, w j z j 2+3ε)/+ε) 2 w j z j 2 w j 2 ) ε/+ε), so n Tf w... w h)w) n 2 2n w j 2 f 2+ε w) /2+ε) hz) δ ) /δ w j z j 2 dνz), proving the estimate for α ={,...,n}. Now suppose that α ={α,...,α m },whereα < <α m.forf L 2 a n ) and h H ) we have α Tf h)w) = 2 l α z l w l z l f z)hz) Noting that w l z l = w z... w n z n l α 2 n m w z... w n z n, w j z j ) 2 dνz). j {,...,n}\α w j z j

K. Stroethoff,. Zheng / J. Math. Anal. Appl. 278 2003) 25 35 33 we get α Tf h)w) 2 n m+ fz) hz) w j z j 3 dνz), and the stated inequality follows from the proof of the first part of the lemma. 3.3. Sufficient condition for boundedness We are now in a position to prove our sufficiency condition for boundedness of Toeplitz products. Proof of Theorem 3.. Let h and k be bounded analytic functions on. It follows from Lemma 3.3 that Tf h)w)t ḡk)w) f w j 2 ) 2 2 w) /2 g 2 w) /2 h k, thus T f h)z)t ḡ)z) dµ z) 3 n h k sup f 2 w) /2 g 2 w) /2. w UsingLemma3.4wehave α Tf hw) α Tḡkw) f w j 2 ) 2 2+ε w) /2+ε) g 2+ε w) /2+ε) Q [ h δ] w) /δ Q [ k δ] w) /δ, where Q is the integral operator defined by Q[u]w) = uz) w j z j 2 dνz), for u L,dν) and w.if f 2+ε w) /2+ε) g 2+ε w) /2+ε) M, for all w, then the above inequality implies α Tf hw) α Tḡkw) dµ α z) 3 n M Q [ h δ ] z) /δ Q [ k δ] z) /δ dνz). That the operator Q is L p bounded, for each <p<, is easily shown as in the onedimensional case see, for example, Chapter 4 in [6]). Since δ<2, p = 2/δ >, so there exists a constant C such that Q[u]z) p dνz) C p uz) p dνz).

34 K. Stroethoff,. Zheng / J. Math. Anal. Appl. 278 2003) 25 35 In particular, [ h δ ] z) p dνz) C p h 2, Q and a similar inequality for function k. By the Cauchy Schwarz inequality, [ h δ ] z) /δ Q [ k δ] z) /δ dνz) Thus Q Q [ h δ ] z) 2/δ dνz) ) /2 Q C p h 2) /2 C p k 2) /2 = C 2/δ h k. [ k δ ] z) 2/δ dνz) α T f hw) α Tḡkw) dµ α z) 3 n MC 2/δ h k, ) /2 for every subset α of {,...,n}. Using 3.2) we conclude that there is a finite constant C such that Tf Tḡk,h C h k, for all bounded analytic functions h and k on. Hence the operator T f Tḡ is bounded on L 2 a n ). 3.4. Compact Toeplitz products The following theorem states that the Toeplitz product T f Tḡ is only compact in the trivial case that it is the zero operator. Theorem 3.5. Let f and g be in L 2 a n ).ThenT f Tḡ is compact if and only if f 0 or g 0. Proof. If T f Tḡ is compact on L 2 a n ), then its Berezin transform vanishes near the boundary of : T f Tḡw) 0 as w in approaches ). We have seen that T f Tḡw) = fw)gw), so f w)gw) = T f Tḡw) 0 as w in approaches ), and it follows from the maximum modulus principle that fg 0.

K. Stroethoff,. Zheng / J. Math. Anal. Appl. 278 2003) 25 35 35 References [] F. Nazarov, A counter-example to Sarason s conjecture, preprint; available at http://www.math.msu.edu/~ fedja/prepr.html. [2]. Sarason, Products of Toeplitz operators, in: V.P. Khavin, N.K. Nikol skiĭ Eds.), Linear and Complex Analysis Problem Book 3, Part I, in: Lecture Notes in Math., Vol. 573, Springer, Berlin, 994, pp. 38 39. [3] K. Stroethoff, The Berezin transform and operators on spaces of analytic functions, in: J. Zemánek Ed.), Linear Operators, in: Banach Center Publications, Vol. 38, Polish Academy of Sciences, Warsaw, 997, pp. 36 380. [4] K. Stroethoff,. Zheng, Products of Hankel and Toeplitz operators on the Bergman space, J. Funct. Anal. 69 999) 289 33. [5]. Zheng, The distribution function inequality and products of Toeplitz operators and Hankel operators, J. Funct. Anal. 38 996) 477 50. [6] K. Zhu, Operator Theory in Function Spaces, Marcel ekker, New York, 990.