Zero products of Toeplitz operators with harmonic symbols

Journal of Functional Analysis 233 2006) 307 334 www.elsevier.com/locate/jfa Zero products of Toeplitz operators with harmonic symbols Boorim Choe, Hyungwoon Koo Department of Mathematics, Korea University, Seoul 36-70, Korea Received 29 April 2005; accepted 25 August 2005 Communicated by L. Gross Available online 29 September 2005 Abstract On the Bergman space of the unit ball in C n, we solve the zero-product problem for two Toeplitz operators with harmonic symbols that have continuous extensions to some part of) the boundary. In the case where symbols have Lipschitz continuous extensions to the boundary, we solve the zero-product problem for multiple products with the number of factors depending on the dimension n of the underlying space; the number of factors is n + 3. We also prove a local version of this result but with loss of a factor. 2005 Elsevier Inc. All rights reserved. MSC: Primary 47B35; secondary 32A36 Keywords: Zero product; Toeplitz operator; Harmonic symbol; Bergman space. Introduction Let B be the unit ball in the complex n-space C n. Let L p = L p B) denote the usual Lebesgue space with respect to the volume measure V on B normalized to have total mass. The Bergman space A 2 is then the space of all L 2 -holomorphic functions on B. Due to the mean value property of holomorphic functions, the space A 2 is a closed subspace of L 2, and thus is a Hilbert space. The Bergman projection P is defined to This research was supported by KOSEF R0-2003-000-0243-0). Corresponding author. E-mail addresses: cbr@korea.ac.kr B. Choe), koohw@korea.ac.kr H. Koo). 0022-236/$ - see front matter 2005 Elsevier Inc. All rights reserved. doi:0.06/j.jfa.2005.08.007

308 B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 be the Hilbert space orthogonal projection from L 2 onto A 2. For a function u L, the Toeplitz operator T u with symbol u is defined by T u f = Puf ) for f A 2. It is clear that T u : A 2 A 2 is a bounded linear operator. The books by Douglas [] and by Zhu [6] are standard sources for one variable theory of Toeplitz operators. In this paper we consider the so-called zero-product" problem of characterizing zero products of several Toeplitz operators: The zero-product problem. Suppose that T u T un = 0. Does it then follow that one of u j is identically zero? While we learned about this problem from a recent work of Ahern and Cu cković [2], its history goes back to as early as the 960s. In [7] Brown and Halmos studied Toeplitz operators on the Hardy space of the unit disk. We refer to [7] for the definition of Hardy space and their Toeplitz operators. They proved that T u T v = T τ on the Hardy space iff either u or v is holomorphic and τ = uv. From this they easily deduced that if T u T v = 0 then either u or v must be identically zero. This result initiated the zero-product problem on the Hardy space. In the study done so far, the number of factors depends in some essential way on the methods the authors used. So far, the problem has been answered yes for five factors by Guo [4] and for six factors by Gu [3] on the Hardy space of the unit disk. More recently, Ding [0] solved the problem for two factors on the Hardy space of the polydisk. The ball case seems to have not been studied yet. Returning to the Bergman space case, we notice that the study of the zero-product problem has begun only recently. Given that the work of Brown Halmos was published long time ago in 964, it is somewhat mysterious to us) that the zero-product problem on the Bergman space has not been studied in the literature until very recently Ahern and Cu cković [2] first studied it on the unit disk. In [2] Ahern and Cu cković solved the zero-product problem for two Toeplitz operators with harmonic symbols and the problem for arbitrary symbols still remains open. More recently, Ahern [] has given a more general approach that leads to the same zero-product theorem as in [2]. For study of higher dimensional analogues of such a one-variable result, the simplest substitutes for the disk might be the balls or polydisks and the simplest substitutes for harmonic symbols might be pluriharmonic ones. The polydisk case was studied by Choe et al. [9]. They solved the zero-product problem on the polydisks for two Toeplitz operators with pluriharmonic symbols by means of extending the methods of [2]. In the case of the balls it appears to be more subtle to extend the methods of [2] and, in fact, no progress has been made yet even for the simplest case of two factors with pluriharmonic symbols. In this paper we investigate the zero-product problem on the balls by devising a completely new approach to the problem. Roughly speaking, using careful analysis of the behavior of the operator on test functions which peak near boundary points, we show that one of the symbol functions must vanish on an open subset of the boundary

B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 309 and have radial derivatives which also vanishes and that such a symbol function must vanish identically. Our method has some aspects which we wish to emphasize. Our method allows us not only to cover up to harmonic symbols, but also to deal with multiple products on higher dimensional balls. The price we pay for extending symbols up to harmonic ones seems to be certain amount of boundary regularity. More explicitly, our harmonic symbols are restricted to those that have continuous extensions at least to some part of the boundary. Harmonicity hypothesis, together with such boundary regularity imposed on symbol functions, plays key roles in our arguments of the present paper. It allows us to use a uniqueness theorem Proposition 4.) for harmonic functions. In addition, it provides us quite explicit information on local behavior of symbol functions near a boundary vanishing point Lemma 4.2). We do not know whether either harmonicity or boundary regularity can be removed in the hypotheses of our results to be stated below. In what follows we let h denote the class of all bounded harmonic functions on B. Also, a boundary open set refers to a relatively open subset of B. Our first result is the next theorem. The disk case is contained in the result of Ahern and Cu cković mentioned earlier. Theorem.. Suppose that functions u,u 2 h are continuous on B W for some boundary open set W. If T u T u2 = 0, then either u = 0 or u 2 = 0. In the case where symbols have Lipschitz continuous extensions to the boundary, our method applies to multiple products. Given a subset X C n containing more than one point, recall that the Lipschitz class of order ε 0, ], denoted by Lip ε X), isthe class of all functions f on X such that fz) fw) =O z w ε ) for z, w X. For a given point ζ B, welet Lip ε ζ) = Lip ε U B), where the union is taken over all neighborhoods U of ζ. Our next result solves the zero-product problem for several Toeplitz operators with harmonic symbols that have Lipschitz continuous extensions to the whole boundary. Unfortunately, our method requires some restriction, depending on the dimension n, on the number of factors, as in the next theorem. Theorem.2. Let u,...,u n+3 Lip ε B) h for some ε > 0. If T u T un+3 = 0, then u j = 0 for some j. Our proof of Theorem.2 actually works with a local Lipschitz condition but with loss of a factor in the product. Theorem.3. Let u,...,u n+2 Lip ε ζ) h for some ε > 0 and ζ B. If T u T un+2 = 0, then u j = 0 for some j.

30 B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 Note that the identity operator is also a Toeplitz operator with symbol ). Thus, if the zero-product theorem holds for a certain number of factors, it also holds for any smaller number of factors. This paper is organized as follows. In Section 2 we recall and collect some basic facts to be used later. In Section 3 we prove some auxiliary integral identities and integral estimates. In Section 4 we prove Theorems.,.2 and.3. 2. Preliminaries Throughout the section we let a B denote an arbitrary point, unless otherwise specified. Since every point evaluation is a bounded linear functional on A 2, there exists a unique function K a A 2 which has following reproducing property: fa)= f, K a, f A 2, 2.) where the notation, denotes the inner product in L 2. The function K a well-known Bergman kernel and its explicit formula is given by is the K a z) =, z B. z a) n+ Here and elsewhere, we use the notation z a = n j= z j a j to denote the Hermitian inner product of points z = z,...,z n ) and a = a,...,a n ) in C n. We let k a denote the normalized kernel, namely, k a z) = a 2 ) n+)/2, z B. 2.2) z a) n+ By the reproducing property 2.), the Bergman projection P can be represented by for functions ψ L 2. For z, w B,z = 0, define P ψa) = B ψk a dv φ z w) = z z 2 w z)z z 2 [w z 2 w z)z] w z and φ 0 w) = w. The map φ a is an automorphism of B such that φ a φ a = id. Its real Jacobian is equal to k a z) 2, so we have the following change of variables

B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 3 formula: h φ a z) ) k a z) 2 dv z) = B B hw) dv w), 2.3) for every h L. We also have a very useful formula φ a z) φ a w) = a a) z w) z a) a w) 2.4) and thus, in particular, φ z w) 2 = z 2 ) w 2 ) z w 2. 2.5) We refer to Chapter 2 of [5] for details. We define a linear operator U a on A 2 by U a ψ = ψ φ a )k a for ψ A 2. It follows from 2.3) that each U a is an isometry on A 2. A straightforward calculation by means of 2.4) yields U a k a = k a φ a )k a =. It follows that U a U a = I and thus Ua = U a. Now, being an invertible linear isometry, U a is unitary. It is well known that U a T u U a = T u φa ; 2.6) see, for example, [4] or [5] where U a is defined with an extra factor ) on the disk and [8] on the ball. Recall that, given a bounded linear operator L on A 2, its Berezin transform is the function L defined by La) = Lk a,k a. We remark in passing that this Berezin transform plays the key role in characterizing the compactness of Toeplitz operators: see [5,2]. It is not hard to see that L is continuous on B. It turns out that L preserves the boundary continuity of symbols as in the next proposition. Proposition 2.. Suppose that functions u,...,u N L are continuous on B {ζ} for some ζ B. Let L = T un T u. Then L continuously extends to B {ζ} and Lζ) = u N u )ζ).

32 B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 Proof. Let a B. Note that k a = U a. Also, note that U a LU a = U a T un U a ) U a T u U a ) = T un φ a T u φ a, because U a U a = I. Since U a = U a = U a, it follows that La) = U a LU a, = T un φ a T u φ a,. 2.7) Fix ζ B. We claim that, as a ζ, wehave T un φ a T u φ a u N u )ζ) in L 2 2.8) which, together with 2.7), implies the proposition. Now, we prove the claim. Let a ζ. Then, for a given function u continuous on B {ζ}, we observe that φ a ζ pointwise in fact, uniformly on compact sets) and thus u φ a uζ) in L 2 by the dominated convergence theorem. In particular, we have Pu φ a ) uζ) in L 2. So, we see that the claim holds for N =. We now proceed by induction on N. Assume that 2.8) holds for some N and consider the case of N +. Having 2.8) as induction hypothesis and writing p for the L p -norm, we have T un+ φ a T un φ a T u φ a u N+ u N u )ζ) 2 T un+ φ a [T un φ a T u φ a u N u )ζ)] 2 + u N u )ζ) T un+ φ a u N+ ζ) 2 u N+ T un φ a T u φ a u N u )ζ)] 2 0 + u N u )ζ) T un+ φ a u N+ ζ) 2 so that 2.8) also holds for N +. This completes the induction and the proof of the proposition. 3. Auxiliary identities and estimates The idea of our proofs is to decompose each factor into major and error parts and to employ suitable test functions whose sources are, of course, the Bergman kernel functions. In utilizing such decompositions and test functions, we need substantial amount of estimates for major parts and error parts, respectively. For the major parts we need to know precise information on how certain Toeplitz operators act on Bergman kernel functions. Also, for the error parts, we need to know the

B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 33 integral behavior of Bergman kernel with logarithmic weights. Such integral identities and estimates are very technical and thus collected in this section. Recall the elementary binomial expansion for positive integers k: x) k = C k,j x j, x <, 3.) j=0 where C k,j = k+j )! k )!j!. Associated with these coefficients are the coefficients a k,p,j = k p )!k j)! k )!k p j)! defined for positive integers k, p, j with k>pand j k p. The collection of these coefficients will be the source for coefficients of various identities below. The precise values of first two coefficients a k,p, and a k,p,2 are needed in our estimates. We list them here for easier reference later: a k,p, = k and a k,p,2 = k p ). 3.2) k 2 The main significance of the coefficients a k,p,j lies in the following elementary relation. Lemma 3.. Let k, p be positive integers with k>p. Then i=0 Proof. Fix i 0. Note that k p C k,i p + i xi = j= a k,p,j, x <. x) k j C p+,i p + i = C p,i p. In general, we have the recursion relation C q,i p + i = q + i Cq,i q p + i = + q p ) C q,i q p + i for integers q>p. Thus, iterating this recursion relation as many times as needed, we have k p C k,i p + i = a k,p,j C k j,i. j=

34 B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 Now, multiplying both sides by x i and then summing up the resulting equalities, we conclude the lemma by 3.). The proof is complete. In what follows we let λ t z) = tz, z B for each 0 t <. The powers of these functions will be our test functions later. Lemma 3.2. Let k>nbe an integer. Then T w λ k t = tλ k t k n + j= ka k+,n+,j+ λ j t for 0 t <. In the proof below we will use the explicit formula for L 2 -norm of monomials given by w j wj n n 2 dv w) = n!j! j n!, 3.3) n + J)! B where j k s are nonnegative integers and J = j + +j n. See Proposition.4.9 of [5] for details. Proof. Let 0 t < and z B. By 3.) we have T w λ k t z) = w B tw ) k dv w) z w) n+ = C k,i C n+,j w tw ) i z w) j dv w) = i=0 j=0 B C k,j+ C n+,j t j+ j=0 j=0 B w j+ w z w ) j dv w) = C k,j+ C n+,j t j+ z j w 2j+) dv w), B

B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 35 where second to the last equality holds by orthogonality of monomials. It follows from 3.3) that T w λ k t z) = C k,j+ C n+,j t j+ z j n!j + )! n + j + )! = t j=0 = tk j=0 j=0 C k,j+ j + n + j + tz ) j C k+,j n + j + tz ) j, where the last equality holds, because j + )C k,j+ = kc k+,j. Now, the lemma follows from Lemma 3. and 3.2). The proof is complete. Lemma 3.3. Let k>n+ be an integer. Then for 0 t <. T wl 2λk t = k n j= a k,n+,j λ k j t, l = 2,...,n Proof. Let 0 t <, z B and let l = 2 without loss of generality. As in the proof of Lemma 3.2, we get by 3.3) and orthogonality of monomials that T w2 2λk t = w 2 2 B tw ) k dv w) z w) n+ = C k,i C n+,i t i z i w 2i w 2 2 dv w) = = i=0 n!i!! C k,i C n+,i n + i + )! tz ) i i=0 i=0 C k,i n + + i tz ) i. B Now, the lemma follows from Lemma 3.. The proof is complete. We now turn to certain integral estimates. Those estimates will take care of error terms in repeated Toeplitz integrals which arise in the course of our proofs. To this end

36 B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 we need to know the volume) integral behavior of the Bergman kernel with logarithmic weights. Constants. In the rest of the paper we use the same letter C, often depending on the dimension n, to denote various constants which may change at each occurrence. For nonnegative quantities X and Y, we often write X Y or Y X if X is dominated by Y times some inessential positive constant. Also, we write X Y if X Y X. We begin with a lemma. Lemma 3.4. Let s 0 and c real be given. Then the following estimates hold: ε r +c log ) s ε c log ε s if c>0, dr log ε s+ if c = 0, 3.4) r if c<0, ε r +c log ) s dr ε c log ε s if c<0, 3.5) r 0 for 0 < ε < 2. The constants suppressed above are independent of ε. Proof. The case s = 0 is easily treated. So, assume s>0. Let 0 < ε < 2. We first consider the integral in 3.4), which we denote by I c for simplicity. The estimate for I c with c<0 is clear, because in that case log r s r c is integrable near r = 0. Meanwhile, for c = 0, we easily deduce that I 0 = s + log ε s+. Next, consider the case c>0. Note that we have r +c log ) { s + s log ) } dr = r c r cr c log ) s + C 3.6) r by elementary calculus. Note that log r s is integrable near. Thus, it follows from the above that cε c ) log ε s = and thus we have the estimate for c>0. ε r +c log ) { s + s log ) } dr r c r /2 + ε r +c log ) s dr r I c

B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 37 Now, let II c denote the integral in 3.5). Note that 3.6) also holds for c<0. Thus, since c<0, it follows from 3.6) that ε c ε c ) log ε s = r +c log ) { s + s log ) } dr II c r c r 0 and thus we have the estimate for II c. The proof is complete. The surface integral behavior of the Bergman kernel is well known as follows: B dsζ) z ζ n+c δ c z if c>0, + log δ z if c = 0, if c<0, 3.7) for z B; see Proposition.4.0 of [5] for details. Here, and in what follows, we use the notation δ z = z, z B for simplicity. Also, ds denotes the surface area measure on B normalized to have total mass. The volume integral behavior, similar to 3.7), of the Bergman kernel with ordinary weights is well known, as is also given in Proposition.4.0 of [5]. However, such estimates are not enough and more delicate estimates with logarithmic weights are needed in our arguments later. More explicitly, we need to estimate two types of integrals J c,s and I c,s defined for s 0 and c real as follows: J c,s z) = B log δ ξ s dvξ) z ξ n++c and I c,s z, w) = B Φ c,s ξ,w) dvξ) z ξ n+ for z, w B, where Φ c,s z, w) = + logδ zδ w ) s z w n++c. In our application the parameter c will be restricted to c n for the integrals I c,s.

38 B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 The estimates for the integrals J c,s unweighted case s = 0 is well known. are as follows. As is mentioned above, the Proposition 3.5. Let s 0 and c real be given. Then δ c z + log δ z s ) if c>0, J c,s z) + log δ z s+ if c = 0, if c<0, for z B. The constants suppressed above are independent of z. Proof. Let z be an arbitrary point of B. We may assume δ z < 2. By integration in polar coordinates we have J c,s z) = 2n 0 B dsζ) rz ζ n++c log ) s r 2n dr r and thus the estimate for c is clear. So, assume c>. By 3.7) we have J c,s z) 0 z r z ) +c + z 0 := I c + II c. log r z ) +c ) s dr r log ) s dr r We first consider I c. Note that z r z z 2 for z <r<. It follows that δz I c δ c z log ) s dr δ c z 0 r log δ z s, where the second equivalence holds by Lemma 3.4. Meanwhile, since r r z r 2 for 0 <r< z, it follows that II c z 0 r) +c log ) s dr r δ z r +c log ) s dr. r Now, combining the estimates for I c and II c above, we conclude the lemma by Lemma 3.4. The proof is complete. In order to estimate the integrals I c,s, we need not only estimates given by Proposition 3.5, but also a localized version for c<0. We need some notation. For 0 < ε <,

B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 39 we let Q ε z) ={ζ B : z ζ < ε}, z B and R ε z) ={ξ B : z ξ < ε}, z B. Note that Q ε z) = R ε z) = for δ z ε. Also, note that Q ε z) Q ε+δz η) for z = z η where η B. So, we have S[Q ε z)] ε n 3.8) for z B and 0 < ε < by Proposition 5..4 of [5]. We begin with a more precise localized version of the estimate 3.7) for c<0. Lemma 3.6. Let n c 0. Then there is a constant C = Cc) such that { dsζ) ε c if c<0, C z ζ n+c log + ε ) δz if c = 0, Q ε z) for z B and 0 < ε <. Proof. Let z B and 0 < ε <. We may assume δ z < ε; otherwise the integral is simply 0. Let I c denote the integral under consideration. Choose a positive integer N such that 2 N ε δ z < 2 N+ ε. Then we have by 3.8) I c N j=0 E j dsζ) N N 2 z ζ j ε) c = ε c 2 cj, n+c j=0 j=0 where E j = Q 2 j ε z) \ Q 2 j ε z). Note that N log ε ) δz for c = 0. Also, note that the last summation above is a part of a convergent geometric series for c<0. So, the lemma holds. The following lemma is a localized version of Proposition 3.5 for c<0. Lemma 3.7. Let s 0 and n c <0. Then there is a constant C = Cc,s) such that log δ ξ s z ξ n++c dvξ) Cε c log ε s R ε z) for z B and 0 < ε < 2.

320 B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 Proof. Let z B, 0< ε < 2 and assume δ z < ε to avoid triviality. Let I c denote the integral under consideration. Note that ξ ε for ξ R ε z). Thus, by integration in polar coordinates, we have I c ε Q ε rz) dsζ) rz ζ n++c log ) s dr. r For c<, we have by Lemma 3.6 and Lemma 3.4 I c ε c ε 0 log r ) s dr ε c log ε s, as desired. Next, for c =, we have by Lemma 3.6 I c = ε ε ε ε )] [log + ε log r z log ε ) log ) s dr r r log ) log ) s dr r εr 0 0 0 ε log ε s, ) s dr r log ){ log ε s + log ) s } dr r r as desired. Finally, for <c<0, we have by 3.7) and Lemma 3.4 I c ε ε 0 r z ) +c r +c ε c log ε s, log log r ) s dr ) s dr r as desired. The proof is complete. We now estimate the integrals I c,s.forz = w, note that the integrals I c,s z, z) reduce to those considered in Proposition 3.5. It is more subtle and complicated to estimate the integrals I c,s in two variables. We have the following estimates for those integrals. In our application s will be an integer.

B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 32 Proposition 3.8. Given s 0 and c n, there exists a constant C = Cc,s) such that for z, w B. { δ c w I c,s z, w) C Φ 0,s+z, w) if c>0, Φ c,s+ z, w) if c 0, Proof. Fix s 0 and c n. Let z, w B. Since I n+),s z, w) + log δ w s )J 0,s z), we have the lemma for c = n by Proposition 3.5. So, assume c> n. Put L c,s z, w) = B log δ ξ s z ξ n+ w ξ n++c dvξ) for simplicity. Then, since + logδ ξ δ w ) s + log δ w s + log δ ξ s,wehave I c,s z, w) + log δ w s )L c,0 z, w) + L c,s z, w) 3.9) and thus we need to estimate L c,s s. We claim that { δ c w L c,s z, w) Φ 0,s+z, w) Φ c,s+ z, w) if c>0, if c 0. 3.0) Having this estimate, we easily conclude the lemma by 3.9). It remains to prove 3.0). Decompose B into two pieces E and E 2 given by and consider corresponding integrals E ={ξ B : 4 w ξ z w }, E 2 ={ξ B : 4 w ξ < z w } log δ ξ I j := Ej s dvξ) z ξ n+ w ξ n++c for j =, 2. For the integral I we have I z w n++c B log δ ξ s z ξ dvξ) + log δ z s+ ; 3.) n+ z w n++c

322 B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 the second equivalence holds by Proposition 3.5. This shows that the integral I satisfies the desired estimates. We now turn to the estimate of the integral I 2. Note that we have z w /2 z ξ /2 + w ξ /2 by Proposition 5..2 of [5]. This gives the estimate z ξ z w, ξ E 2 and therefore we have I 2 log δ ξ z w E2 s n+ dvξ). 3.2) w ξ n++c Consider the case c 0. It follows from the above that I 2 J c,s w) z w n+, which implies that I 2 also satisfies the desired estimate for c 0 by Proposition 3.5. Next, consider the case c<0. In this case we have by 3.2) and Lemma 3.7 I 2 z w n++c log ) 4 s + log δ zδ w ) s z w z w n++c, where we used 2.5) to get the second inequality. This completes the proof. 4. Zero products Note that if a function u h is continuous on B W and vanishes on W for some boundary open set W, then u is harmonic across W. This is a consequence of the well-known reflection principle. See, for example, Theorem 4..5 of [3]. The study of harmonic functions has nothing to do with complex structure of C n. So, we introduce real-variable notation. As usual, we identify points z,...,z n ) C n with x,x 2,...,x 2n,x 2n ) R 2n where Rz j = x 2j and Iz j = x 2j for each j. With this convention we let D j = x j, j =,...,2n

B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 323 and let Ru denote the radial derivative of u. More explicitly, we let Ru = 2n j= x j D j u for C -functions u. We will use two elementary facts about the radial derivatives. One is the fact that R commutes with linear transformations on R 2n. More explicitly, Ru ρ) = Ru) ρ 4.) for each linear transformation ρ on R 2n. The other one is the fact that R preserves harmonicity, which is easily seen by a straightforward calculation. Finally, we put for simplicity. e =, 0,...,0) R 2n C n Proposition 4.. Suppose that u is a function harmonic on B and continuous on B W for some boundary open set W. If both u and Ru vanish on W, then u = 0 on B. Proof. Suppose that both u and Ru vanish on W. Since u = 0onW, u extends to an harmonic function across W by the reflection principle, as mentioned above. We claim that R 2 u = 0 on W. 4.2) In order to see this let ζ W, choose a rotation ρ = ρ ζ on R 2n with ρζ) = e, and put ũ = u ρ. Then ũ vanishes on ρw which is a boundary neighborhood of e. Let z = x,z ) where z = x 2,...,x 2n ). Define vz ) = ũ z 2,z ). Then, v vanishes near 0, because ũ vanishes on ρw. Hence, a routine calculation gives 0 = D 2 j v0 ) = D 2 j ũe) D ũe) for j 2. Also, a straightforward calculation yields R 2 ũe) = Rũe) + D2 ũe). Now, since Rũe) = Ruζ) = 0by4.), we have 0 = 2n j=2 D 2 j v0 ) = 2n )D ũe) 2n j=2 D 2 j ũe)

324 B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 = 2n )D ũe) + D ũe) 2 ) Δũe) = 0 = 2n 2)Rũe) + R 2 ũe) D ũe) = Rũe) ) = R 2 ũe). Thus, applying 4.) twice, we have 0 = R 2 ũe) = R 2 ũ ρ)ζ) = R 2 ũ ρ)ζ) = R 2 uζ). Now, since ζ W is arbitrary, we have 4.2) as desired. What we have shown so far is the following: u = Ru = 0 on W R 2 u = 0 on W. Note that Ru is harmonic, because u is. Thus, applying the above with Ru in place of u, we see that R 3 u = 0onW. Repeating the same argument, we see that R j u = 0 on W for all j 0. Now, let N denote the outward normal differentiation along B. Note that R, when restricted to B, is the same as N. Thus, we see from 4.) that N commutes with rotations. Also, given any C -function ψ near e, it is not hard to see that D j ψe) can be written as a linear combination of Rψe),...,R j ψe) for each j. So, given j and ζ W, we see that N j uζ) can be written as a linear combination of Ruζ),...,R j uζ). Consequently, we conclude that N j uζ) = 0 for each j. It follows from real-analyticity that utζ) = 0 for all t sufficiently close to. This shows that the vanishing property of u on W extends to some open set in B and therefore u = 0 on the whole B. The proof is complete. Remark. It seems worth mentioning that our uniqueness result Proposition 4. has some connection with a local Hopf lemma obtained by Baouendi and Rothschild in [6]. Consider a function u harmonic on B and continuous on B W for some boundary open set W. Then the Local Hopf lemma of Baouendi and Rothschild asserts that if u 0 onw and u vanishes of infinite order in the normal direction at some ζ W, then u vanishes near ζ in W and along the radius ending at ζ. Hence, if, in addition, u vanishes of infinite order in arbitrary direction) at ζ, then u must be identically 0. In case u already vanishes in W, Proposition 4. shows that, while the normal derivative of u is assumed to vanish globally) in W, the vanishing order can be weakened from infinity to just. We introduce more notation. First, we let σz) = 2 Rz ) n z j 2 4.3) j=2

B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 325 for z B. Note that σz) = O z ). Next, for 0 t < and z B, welet αt) = t, βz) = β t z) = tz for simplicity. For integers k, m with k m, we use the same notation P m = P m α, β) for either the zero polynomial or any nonzero polynomial in α, β with no term of degree lower than m. Finally, given a positive integer d, let F d denote the class of all functions f on [0, ) B such that ft,z) C + log t)δ z d tz n+, 0 t <, z B for some constant C depending on f. Note that we have T σ ψ t F 2 for functions ψ t such that ψ t z) = O λ t n+2 ). This follows from the fact σz) =O z ) and thus T σ λ t n+2 F 2 by Proposition 3.8. Now, before turning to the proof of our results, we prove two more lemmas. Lemma 4.2. Suppose that u h CB W) vanishes on W for some boundary neighborhood W of e. Then for z B W near e. uz) = D ue) 2 σz) + O z 3/2) Proof. As in the proof of Proposition 4., u is harmonic across W. So, we similarly let z = x,z ) where z = x 2,...,x 2n ) and define vz ) = u z 2,z ). Then, v vanishes near 0, because u vanishes on W. Thus, by a straightforward calculation, we have and It follows that 0 = D j v0 ) = D j ue), 0 = D 2 j v0 ) = D 2 j ue) D ue) j = 2,...,2n, 0 = D i D j v0 ) = D i D j ue), i, j = 2,...,2n, i = j. uz) = D ue) 2 σz) + D2 ue) 2 x ) 2 + D2 2 ue) x2 2 2 + 2n j=2 D D j ue) x )x j + O e z 3 ) 2

326 B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 for z = x,...,x 2n ) B near e. Note that x 2 z and x j =O x ) /2) for each j 3. It follows that e z =O x ) /2) and we conclude the lemma. The proof is complete. Lemma 4.3. Let k, d be positive integers with k>n+d. Then there exist a polynomial P d+ and a function f d F d such that for 2 <t< where b k,d = Proof. Let have 2 T d σ λk t = b k,d λ k d t + λ k t Pd+ + O λ t n+2 ) + f d t, ) 4.4) k ) k d). <t<. Consider k>nfor a moment. By Lemma 3.2 and 3.2) we T w λ k t = tλk t + t nb k, )λ k t = tλ k t + nb k, )λ k t + λ k t P + λ k t P. Put η = 2 z z and γ = z. Note that α+γ = αγ+β. Thus, a little manipulation yields T η λ k t = α + γ)λk t nb k, )λ k t + λ k t P = nb k, λ k t + αγλ k t + λ k t P2. 4.5) Note that γ = β α)/t. Also, note that α tz. So, it is clear that αγλ k t = O λ t n+2 ) for k n + 4. Meanwhile, for k>n+ 4, we have Therefore, we see from 4.5) that Meanwhile, we have T σ η λ k t = αγλ k t = αβ α)λk t k n 5 j=0 α j + αk n 4 t = λ k t P2 + α k n 3 β α)λ k t t = λ k t P2 + O λ t n+2 ). T η λ k t = nb k, λ k t + λ k t P2 + O λ t n+2 ), k > n. n j=2 T wj 2λk t = n)b k, λ k t + λ k t P2, k > n+,

B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 327 by Lemma 3.3 and 3.2). Combining these two estimates, we have T σ λ k t = b k, λ k t + λ k t P2 + O λ t n+2 ), k > n +, 4.6) and thus the lemma holds for d =. We now proceed by induction on d. So, assume that d and the lemma holds for d with k>n+ d. So, we have 4.4). Now, assume k>n+ d + and consider the case d +. Apply T σ to both sides of 4.4) once more. For the first term in the right side of 4.4), we have by 4.6) b k,d T σ λ k d t = b k,d+ λ k d t + λ k d t P 2 + O λ t n+2 ) = b k,d+ λ k d+) t + λ k t Pd+)+ + O λ t n+2 ), which is of desired form. Also, T σ takes the last two terms in the right side of 4.4) into F d+ by Proposition 3.8. For the second term in the right side of 4.4), note that λ k t Pd+ is a linear combination of terms of the type α i λ k j t, i + j d +. We only need to consider terms with k j>n+ 2; otherwise terms are of O λ t n+2 ). For such a term we have T σ [α i λ k j t ]=b k j, α i λ k j t + α i λ k j t P 2 + O λ t n+2 ) = λ k t P+i+j + O λ t n+2 ) = λ k t Pd+)+ + O λ t n+2 ) by 4.6). This shows that 4.4) also holds for d + and the induction is complete. The proof is complete. We are now ready for the proofs of Theorems.,.2 and.3. First, we prove Theorem.. Proof of Theorem.. Assume T u T u2 = 0. Then, since u and u 2 are both continuous on B W by assumption, we have 0 = T u T u2 ) = u u 2 on W B by Proposition 2.. There are two cases to consider: i) Both u and u 2 vanish everywhere on W and ii) either u or u 2 does not vanish on some boundary open subset of W. In case of i) we have u,u 2 Lip ζ) for some ζ W by Lemma 4.2. Thus, the case i) is contained in Theorem.3 to be proved below.

328 B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 So, we may assume ii). Note that we may further assume that u does not vanish on some boundary open set, still denoted by W, because otherwise we can use the adjoint operator T u T u2 ) = T u2 T u.wenowhaveu 2 = 0onW. Assume e W without loss of generality. By Proposition 4. we may further assume that Ru 2 vanishes nowhere on W. This will lead us to a contradiction. Let c = u e) = 0 and c 2 = D u 2 e) 2 = 0. Let e = u c and e 2 = u 2 c 2 σ where σ is the function introduced in 4.3). Then we have 0 = T u T u2 = T c +e T c2 σ+e 2 = c c 2 T σ + c 2 T e T σ + T u T e2 and thus c c 2 T σ = c 2 T e T σ + T u T e2. 4.7) Now we apply each sides of the above to the same test functions λ k t k>n+ 2 and derive a contradiction. First, we have by 4.6) with an integer T σ λ k t z) = k λk t z) { + O tz )}, t, so that T σ λ k t te) = k λk t te) { + O t )} t) k 4.8) as t. Next, note that e 2 z) =O z 3/2 ) by Lemma 4.2. Also, note that k 2 3 >n+. Thus an application of Proposition 3.8 gives T e2 λ k t z) and thus an application of Proposition 3.5 gives B dvξ) tξ k 3/2 z ξ n+ Φ 0,z, te) t) k 3/2 n T u T e2 λ k t te) J n+, te) log t) t) k 3/2 n t) k 3/2 = o) t) k 4.9) as t. Finally we estimate T e T σ λ k t. Write T σλ k t = f + g where f = k λk t and g = T σ λ k t f. First, consider T e f.givenε > 0 sufficiently small, let ωε) = sup{ u ξ) u e) :ξ B W, e ξ < ε}

B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 329 be the modulus of continuity of u at e. By continuity we have ωε) 0asε 0. Note that, since e ξ 2 ξ 2 + ξ 2 4 ξ,wehave In particular, we have e ξ 2 tξ ξ t) t). 4 tξ ε2 4 t) for e ξ ε. Accordingly, with ε > 0 independent of t) fixed, we have by Proposition 3.5 T e fte) = B e ξ) tξ + e ξ <ε k +n+ dvξ) e ξ ε e ξ) dvξ) tξ k+n ωε) t) k + [ε 2 /4 t)] k+n 4.0) for all t such that ε 2 /4 <t<. Next, consider T e g.by4.6) we have g λ t k 2. Thus we get by Proposition 3.5 recall k 2 >n>0) that T e gte) t) k 2 = o) t) k 4.) as t. Now, setting M = c c 2 T σ and R = c 2 T e T σ + T u T e2, we obtain from 4.7) 4.) that = Mλk t te) { t) k } Rλ k o) + ωε) + t te) [ε 2 /4 t)] k+n as t. So, first taking the limit t with ε > 0 fixed and then taking the limit ε 0, we have ωε), which is a contradiction. The proof is complete. Next, we prove Theorem.2.

330 B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 Proof of Theorem.2. Assume T u T un+3 = 0. Then, since each u j has a continuous extension to the whole boundary by assumption, we have T u T un+3 ) = u u n+3 = 0 on B by Proposition 2.. Since u u n+3 is continuous and vanishes everywhere on the boundary, there exists a boundary open set W B such that either u j ζ) = 0, ζ W, or u j = 0 on W holds for each j. Ifu 0, there is nothing to prove. So, we may assume that u vanishes nowhere on W. Suppose that there is a boundary open set W j W such that u j = Ru j = 0 on W j 4.2) that for some j. Since u j CB) is harmonic by assumption, Proposition 4. and 4.2) will lead us to conclude u j = 0onB, which completes the proof. Now, we assume that 4.2) dose not hold for all j and derive a contradiction. Since 4.2) does not hold, we may shrink if necessary) the set W to get a smaller boundary open set, still denoted by W, such that either i) u j ζ) = 0, ζ W, or ii) u j ζ) = 0, Ru j ζ) = 0, ζ W 4.3) holds for each j =, 2,...,n+ 3. We may further assume that e W; this causes no loss of generality by 4.). Note that Ru j e) = D u j e). We introduce more notation. In the rest of the proof we let 0 t < and z B represent arbitrary points. Recall that D u j e) = 0 by 4.3), in case u j e) = 0. Let d j = ifu j e) = 0 and d j = 0 otherwise. Note that d = 0. Now, define the major part of u j by { uj e) if d j = 0, m j := D u j e) 2 σ if d j =, and put e j = u j m j for each j. Note that e z =O z /2). Thus, by Lemma 4.2 and the Lipschitz continuity hypothesis, there is some ε 0 0, 2 ) such that ) e j z) = O z d j +ε 0 4.4)

B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 33 for each j. In connection with this estimate we notice that m j = O z d ) j and thus ) u j z) = O z d j 4.5) for each j. We introduce further notation. Put M = T m T mn+3 and R = T u T un+3 M for notational convenience. Then an inductive argument yields where R = n+3 j= R j, R j = T u T uj T ej T mj+ T mn+3 for each j. Note that R = M, because T u T un+3 = 0 by assumption. We will estimate the same expression Mλ k t = Rλ k t along the radius ending at e in two different ways and reach a contradiction. Before proceeding, we recall a consequence provided by Lemma 4.3. Note that t tz. Therefore, given positive integers k and m with k>n+ m, wehave by Lemma 4.3 T m σ λk t z) = b k,mλ k m t z) { + O tz +Et,z) )}, 4.6) where Et,z) = tz log t)δ z m. Now, put and d = d + +d n+3 p j = d + +d j, q j = d j + +d n+3 for each j =,...,n+ 3. Note that we have d, because u u n )e) = 0. Choose a positive integer k sufficiently large so that k>n+ d +. We first estimate Mλ k t. Let c j = u j e) if u j e) = 0 and c j = D u j e) 2 if u j e) = 0. Then we have M = ctσ d where c = c c n+3. It thus follows from 4.6) that Mλ k t te) = cb k,d { + o)}, c = 0 t) k d

332 B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 so that Mλ k t te). 4.7) t) k d Next, we estimate Rλ k t. So, fix j and consider R j. Note that So, we have T mj+ T mn+3 = c j+ c n+3 T q j+ σ. T mj+ T mn+3 λ k t λk q j+ t by 4.6). It follows from 4.4) and Proposition 3.8 that T ej T mj+ T mn+3 λ k t z) dvξ) B tξ k q j ε 0 z ξ n+ t) k n q j ε 0 Φ 0, z, te); 4.8) note that k n q j ε 0 d + q j ε 0 > 0 by the choice of k. Now, going one step further, we get by 4.5) and Proposition 3.8 that for j 2. Note that T uj T ej T mj+ T mn+3 λ k t z) Φ d j,2z, te) t) k n q j ε 0 p j j 2 n +, j = 2,...,n+ 3, 4.9) because d = 0; it is this step which requires the restriction on the number of factors in the product. Hence, by the same argument repeatedly using 4.9) and Proposition 3.5, we eventually obtain R j λ k t z) Φ p j,j z, te) t) k n q j ε 0 for each j 2. This also holds for j = by4.8) if we set p 0 = 0. So, evaluating at z = te, we therefore have R j λ k + log t) j + log t) n+3 t te) t) k d ε 0 t) k d ε. 0

B. Choe, H. Koo / Journal of Functional Analysis 233 2006) 307 334 333 This holds for arbitrary j. Accordingly, we have Rλ k t Now, we have by 4.7) and 4.20) + log t) n+3 te) t) k d ε. 4.20) 0 = Mλk t te) Rλ k t te) t) ε 0 + log t) n+3 ) as t, which is a contradiction. The proof is complete. Finally, we prove Theorem.3. The proof is exactly the same as that of Theorem.2 except at only one spot. We only indicate such a difference. Proof of Theorem.3. By a similar argument using Proposition 2., it is not hard to see that there is still a boundary open set W where u u 2 u n+2 vanishes and 4.3) holds. In addition, the local Lipschitz hypothesis allows us to assume e W and u j Lip ε e) for each j =, 2,...,n + 2. In the proof of Theorem.2 we were able to assume d = 0 under the global Lipschitz hypothesis, because the location of the boundary set W is of no significance. However, we cannot do the same under the present local Lipschitz hypothesis. That is, d = may well happen, which causes loss of a factor. So, the inequality 4.9) is no longer true and what we have now is p j j n + for 2 j n + 2. The rest of the proof is exactly the same. References [] P. Ahern, On the range of the Berezin transform, J. Funct. Anal. 25 2004) 206 26. [2] P. Ahern, Z. Cu cković, A theorem of Brown Halmos type for Bergman space Toeplitz operators, J. Funct. Anal. 87 200) 200 20. [3] S. Axler, P. Bourdon, W. Ramey, Harmonic Function Theory, Springer, New York, 992. [4] S. Axler, Z. Cu cković, Commuting Toeplitz operators with harmonic symbols, Integral Equations Operator Theory 4 99). [5] S. Axler, D. Zheng, Compact operators via the Berezin transform, Indiana Univ. Math. J. 47 998) 387 400. [6] M.S. Baouendi, L.P. Rothschild, A local Hopf lemma and unique continuation for harmonic functions, International Mathematics Research Notices, vol. 8, 993, pp. 245 25. [7] A. Brown, P. Halmos, Algebraic properties of Toeplitz operators, J. Reine Angew. Math. 23 964) 89 02. [8] B. Choe, Y.J. Lee, Pluriharmonic symbols of commuting Toeplitz operators, Illinois J. Math. 37 993) 424 436. [9] B. Choe, Y.J. Lee, K. Nam, D. Zheng, Products of Bergman space Toeplitz operators on the polydisk, preprint. [0] X. Ding, Products of Toeplitz operators on the polydisk, Integral Equations Operator Theory 45 2003) 389 403.

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