CARLESON MEASURES AND DOUGLAS QUESTION ON THE BERGMAN SPACE. Department of Mathematics, University of Toledo, Toledo, OH ANTHONY VASATURO

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1 CARLESON MEASURES AN OUGLAS QUESTION ON THE BERGMAN SPACE ŽELJKO ČUČKOVIĆ epartment of Mathematics, University of Toledo, Toledo, OH ANTHONY VASATURO epartment of Mathematics, University of Toledo, Toledo, OH ABSTRACT. Motivated by ouglas question about the invertibility of Toeplitz operators on the Hardy Space, we study a related question concerning the Berezin transform and averaging function of a Carleson measure for the weighted Bergman space of the disc. As a consequence, we obtain a necessary and sufficient condition for the invertibility of Toeplitz operators whose symbols are averaging functions of these Carleson measures. Keywords: Bergman space, Carleson measure, Toeplitz operator, Berezin transform 1. INTROUCTION Let be the open unit disc in the complex plane, and let da be the normalized Lebesgue area measure in the complex plane. Given α > 1, we write the weighted normalized Lebesgue area measure on as da α (z) := (α + 1)(1 z 2 ) α da(z). We denote the weighted Bergman space of the unit disc by A 2 α(), which is the set of all functions f holomorhic on which are square integrable on with respect to da α (z). H() will represent the set of all holomorphic functions on, and H () the set of all functions in H() L (). For a and z, the weighted Bergman kernel of the unit disc is given by K α a (z) := 1 (1 az) 2+α, addresses: zcuckovi@math.utoledo.edu, avasatu@rockets.utoledo.edu. 1

2 and the normalized weighted Bergman kernel of the disc is given by k α a(z) := (1 a 2 ) 2+α 2 (1 az) 2+α. It is well known that the Bergman space A 2 α() is a closed subspace of L 2 α(), so there is an orthogonal projection P α : L 2 α() A 2 α() known as the Bergman projection. The operator P α is an integral operator given by P α f (z) = Kw(z) α f (w)da α (w) = f (w) (1 zw) 2+α da α(w) for z. Given a function φ L (), we define the Toeplitz operator T α φ on A 2 α() with symbol φ by T α φ f := P α (φ f ) for f A 2 α(). From the integral representation of P α f (z), we may also write Tφ α as an integral operator as follows: Tφ α f (z) = Kw(z)φ(w) α f (w)da α (w) = f (w)φ(w) (1 zw) 2+α da α(w) for z. For a function φ L (), we define the weighted Berezin transform of φ, denoted φ α, as φ α (a) := Tφk α α a, k α a = φ(z) k α a(z) 2 da α (z) for a. It is also possible to define a Toeplitz operator with a finite, complex Borel measure symbol µ on. efine T α µ : H () H() 2

3 by the integral formula Tµ α f (z) := K α w(z) f (w)dµ(w). This operator is then called bounded on A 2 α() if there exists C > 0 such that Tµ α f 2 da α C f 2 da α for all f H. When T α µ is bounded on A 2 α(), we define the weighted Berezin transform of µ, denoted µ α, as µ α (a) := k α a(z) 2 dµ(z) for a. We say a finite, positive Borel measure µ on is a Carleson measure for A 2 α() if there exists a constant C > 0 such that f 2 dµ C f 2 da α for all f A 2 α(). The infimum over all such constants is called the Carleson constant of µ. The most well known and useful characterization of Carleson measures for the Bergman spaces is the following: Theorem 1. [3, p. 62] Let α > 1, and let r (0, 1). Suppose µ is a finite, positive Borel measure on. Then µ is a Carleson measure for A 2 α() if and only if there exists a constant K r > 0 such that sup a µ((a, r)) (1 a 2 ) 2+α = K r <. Here (a, r) represents the pseudohyperbolic disc centered at a with radius r. More information about the pseudohyperbolic metric is given below. Before we study the question of invertibility of Toeplitz operators with measure symbols, we would like to know when these operators are bounded. The following is well-known: Proposition 1. [10, p. 169] Let µ be a finite, positive Borel measure on. Then the following are equivalent. (1) T α µ is bounded on A 2 α(). 3

4 (2) µ α is bounded on. (3) µ is a Carleson measure for A 2 α(). Fubini s theorem directly gives us that if µ is Carleson for A 2 α(), then Tµ α f, g = f (z)g(z)dµ(z) for f, g H, as we might have expected. In fact, if µ is simply a positive measure such that T α µ is bounded on A 2 α(), then a simple approximation argument gives us that in fact Tµ α f, g = f (z)g(z)dµ(z) for all f, g A 2 α() [10, p. 169]. Lastly, for φ L (), given r (0, 1) and a, we define the averaging function φ α,r by φ α,r (a) := 1 φ(z)da α (z). A α ((a, r)) (a,r) Similarly, for a finite, positive Borel measure on, we define the averaging function µ α,r by µ α,r (a) := µ((a, r)) A α ((a, r)). A fundamental problem on spaces of holomorphic functions is determining necessary and sufficient conditions for the invertibility of Toeplitz operators on these spaces. On the Hardy space, invertible Toeplitz operators have been completely characterized [1], but this characterization is not derived from simple analytic or geometric properties of the symbol. In [2], using a homotopy argument, ouglas showed that if φ is a continuous function on the unit circle, then T φ is invertible if ˇφ δ on for some δ > 0, where ˇφ represents the harmonic extension of φ to the unit disc. He then posed the following question, to which we refer as ouglas question: Question 1. If φ L (T) and ˇφ δ on for some δ > 0, is T φ invertible on the Hardy space? Wolff, using martingale techniques, showed that in general, the answer to ouglas question is no [8]. From our definition of the Berezin transform on the Bergman space, we see that the Berezin transform of a function can 4

5 be defined on any reproducing kernel Hilbert space. This means that on the Hardy space, the Berezin transform of a function must coincide with the harmonic extension of the function. This leads to the following question for the Bergman space analogous to ouglas question on the Hardy space: Question 2. If φ L () and φ δ on for some δ > 0, is T φ invertible on the Bergman space? This question is answered positively for analytic symbols, conjugate analytic symbols, and real harmonic symbols in [6]. In these cases, φ = φ. On the Bergman space, Luecking found necessary and sufficient conditions for Toeplitz operators with bounded, nonnegative symbols to be invertible in [4]. Using Luecking s result, Zhao and Zheng recently obtained a positive result for this question and its converse in the case the symbol is nonnegative and bounded. The question remains open for harmonic symbols. In this paper, we look at Toeplitz operators with Carleson measure symbols on the Bergman space, and ask the following question: Question 3. If µ is a Carleson measure for the Bergman space and µ δ on for some δ > 0, what can be said about the Reverse Carleson properties of µ? In [9, p. 485], Zhao and Zheng found a bounded function on the unit disc whose Berezin transform was invertible in L (), but whose Toeplitz operator was not invertible on the Bergman space of the disc. Since, for φ L (), measures of the form φda are clearly Carleson measures for A 2 (), we cannot then say that T µ is invertible on the Bergman space just because µ is Carleson for A 2 () and µ δ on for some δ > 0. However, as a type of answer to Question 3, we show that the operator T µα,γ must be invertible for all γ (0, 1) after showing that the averaging function of µ must be invertible in L () for sufficiently large radii, and that µ must be what we have termed an almost Reverse Carleson measure for the weighted Bergman space. 2. PRELIMINARIES Let Aut() be the set of all automorphisms on. Given a, the special automorphism interchanging 0 and a is given by ϕ a (z) := a z 1 az for z. The pseudohyperbolic metric ρ on is defined as 5

6 ρ(z, a) := z a 1 za where z, a. It is easy to check that this is indeed a metric which is invariant under the action of automorphisms, meaning ρ(ϕ(z), ϕ(a)) = ρ(z, a) for all ϕ Aut() and for all z, a. Throughout this paper, for a and r (0, 1), (a, r) := {z : ρ(z, a) < r}, or the pseudohyperbolic disc centered at a with radius r. One very important property of the pseudohyperbolic metric is the strong triangle inequality. Namely, ρ satisfies ρ(w, a) ρ(w, z) 1 ρ(w, a)ρ(w, z) ρ(z, a) ρ(w, a) + ρ(w, z) 1 + ρ(w, a)ρ(w, z) for all a, z, w [3, p. 38]. Below are some basic facts we will need about pseudohyperbolic discs and the normalized Bergman kernel [3, p. 40, p. 42]. Lemma 3 follows from a change to polar coordinates. Lemma 1. For a, α > 1, and 0 < ɛ < 1, Lemma 2. Let ɛ > 0. Then for all z (a, ɛ). A α ((a, ɛ)) (1 a 2 ) 2+α. k α a(z) 2 Lemma 3. Let ɛ > 0 and α > 1. Then 1 (0,ɛ) 1 (1 a 2 ) 2+α da α (z) = (1 ɛ 2 ) α+1. The following definitions will be used in the proof of our main theorem. 6

7 efinition 1. If µ is a finite, positive, Borel measure on, we say µ is a Reverse Carleson measure for A 2 α() if there exists a constant C > 0 such that for all f A 2 α(). f 2 da α C f 2 dµ efinition 2. If µ is a finite, positive, Borel measure on, we say µ is an almost Reverse Carleson measure for A 2 α() if for every γ (0, 1), there exists a constant C γ > 0 such that for all f A 2 α(). f 2 da α C γ f 2 α,γdµ It is easy to see, using Lemma 1 and a well-known pointwise estimate (see [10, Proposition 4.13]), that every Reverse Carleson measure for A 2 α() is an almost Reverse Carleson measure for A 2 α(). We now list several well-known theorems that will be used in the proof of our main results. The following theorem is by Oleinik and Pavlov, and allows us to slightly strengthen the Carleson condition under the right circumstances. Theorem 2. [7] (Oleinik-Pavlov) Suppose µ is a Carleson measure for A 2 α(), and suppose E := {z : there exists a in the support of µ such that ρ(z, a) < 1 2 }. Then there exists C > 0 such that for all f A 2 α(). f 2 dµ C f 2 da α E The next theorem is a result from Luecking that we will use in the proof of Proposition 2 below. This theorem gives us sufficient conditions for the averaging function of a measure to be bounded below by a positive number on the unit disc. Theorem 3. [5, Theorem 4.3] (Luecking) Suppose µ is a measure on satisfying: 7

8 (1) There exists C 0 > 0 such that µ((a, 1 2 ) C 0(1 a ) 2+α for all a. (2) µ is Reverse Carleson for A 2 α(). Then there exist δ > 0 and ν < 1 such that µ((a, ν)) > δa α ((a, ν)) for all a. The next result also comes from Luecking and will be used in the proof of our main theorem. This result essentially gives us sufficient conditions for a measure being almost Reverse Carleson for the Bergman space. Theorem 4. [5, Theorem 4.5, Inequality 4.13] (Luecking) Let α > 1, and suppose µ is a positive measure on such that: (1) There exists C 0 > 0 such that µ((a, 1 2 ) C 0(1 a ) 2+α for all a. (2) There exist δ > 0 and ν < 1 such that µ((a, ν)) > δa α ((a, ν)) for all a. Then there exists γ 0 > 0 such that for all 0 < γ γ 0, there exists C γ > 0 such that for all f A 2 α(). f 2 da α C γ (a,γ) f (z) 2 da α (z) (1 a ) 2+α dµ(a) Remark 1: If this theorem is true for all γ γ 0, it must be true for all γ (0, 1) by Lemma 1. Indeed, if the above inequality holds for γ γ 0, then if γ 0 < γ < 1, we have C γ0 (a,γ 0 ) f (z) 2 da α (z) (1 a ) 2+α dµ(a) C γ0 C γ C γ0 C γ (a,γ ) f (z) 2 da α (z) A((a, γ dµ(a) )) (a,γ ) f (z) 2 da α (z) (1 a ) 2+α dµ(a). Since γ 0 is a fixed number dependent only on µ and α, we are finished. Finally, we state another theorem from Luecking, which gives us several necessary and sufficient conditions for a Toeplitz operator with a nonnegative, bounded symbol to be invertible on the Bergman space. Theorem 5. [4] (Luecking) Let α > 1, and let φ be a nonnegative, measurable function in L (). Then the following are equivalent. 8

9 (1) T α φ is invertible on A2 α(). (2) There exist 0 < ɛ < 1, δ > 0, and r > 0 such that A(G (a, ɛ)) > δa((a, ɛ)) for all a, where G := {z : φ(z) > r}. (3) There exists a constant C > 0 such that φ(z) f (z) 2 da α (z) C for all f A 2 α(). f (z) 2 da α (z) In the next section, we state our main results and give some background and justifications for them. 3. MAIN RESULTS The following Proposition is amalgamation of easier results addressing the converse to Question 3 from a different perspective: namely, when we assume that T µ is invertible rather than T µα,γ. Part (1) is probably wellknown, but we state it here for completeness and for lack of reference as a characterization for the invertibility of T µ using the measure-theoretic properties defined above, given that µ is Carleson for the Bergman space. A similar statement to (1) was proven in [4] for measures of the form χ G da, where G is a measurable subset of. We show in parts (2) and (4) that T µ being invertible on the Bergman space means that both the Berezin transform and the averaging function of µ are bounded below provided the averages are taken over large enough discs. A similar statement to (2) was proven in [9] for measures of the form φda, where φ L (). Finally, in part (3), we show another likely well-known result: that the averaging function of µ being bounded below by a positive number on means that the Berezin transform of µ must share this property. This part is included because in Theorem 6 below, one step in the proof is to show the converse to this statement when the averaging function is taken over discs of large enough radii. Proposition 2. Let α > 1, and let µ be a Carleson for A 2 α(). Then: (1) T α µ is invertible on A 2 α() if and only if µ is Reverse Carleson for A 2 α(). (2) If T α µ is invertible on A 2 α(), then there exists δ > 0 such that µ α (a) δ for all a. (3) Let r (0, 1). If there exists δ r > 0 such that µ α,r (a) δ r for all a, then there exists δ > 0 such that µ α (a) δ for all a. 9

10 (4) If µ is Reverse Carleson for A 2 α(), then there exists δ r > 0 such that µ α,r (a) δ r for all a if r is sufficiently close to 1. Proof. Proof of (1): We begin by recalling from Proposition 1 that µ being Carleson for A 2 α() is equivalent to T α µ being a bounded operator on A 2 α(). To show the first statement, we begin by assuming that µ is Reverse Carleson for A 2 α(). This means there exists a constant C > 0 such that C f 2 2,α f 2 dµ = T α µ f, f T α µ f 2,α f 2,α for all f A 2 α(). Here the last inequality is just the Cauchy-Schwarz inequality. Thus Tµ α is bounded below on A 2 α(), so that it must be injective with closed range on A 2 α(). Since Tµ α is a positive operator, it must be selfadjoint. Since Tµ α is both self-adjoint and injective, it must have dense range in A 2 α(). Finally, since Tµ α is injective with closed, dense range, it must be invertible on A 2 α(). Conversely, if T α µ is invertible on A 2 α(), it must be bounded below on A 2 α(), so that if S is the unique, positive square root of T α µ, then S must also be invertible. Indeed, if TT 1 = I, then S(ST 1 ) = I, so S has a right inverse, and a similar proof shows S has a left inverse. Thus S is bounded below on A 2 α(), and so we have f 2 dµ = T α µ f, f = S f, S f = S f 2 2,α C f 2 2,α for some C > 0 for all f A 2 α(). This means precisely that µ is Reverse Carleson on A 2 α(), proving the first statement. Proof of (2): If we assume T α µ is invertible on A 2 α(), then it must be bounded below on A 2 α(). Since T α µ is also a positive operator, it must have a unique, positive, invertible square root S which is also bounded below on A 2 α(). Hence T α µ f, f = S 2 f, f = S f 2 2,α C f 2 2.α for all f A 2 α(). If we let a, and let f (z) = k a (z), then the above inequality becomes µ α (a) C since normalized Bergman kernels are unit 10

11 vectors in the Bergman space. Since a was arbitrary, this proves the second statement. Proof of (3): We assume, given r (0, 1), that µ((a, r)) A α ((a, r)) δ r > 0 for all a. By Lemma 1, this means that there exists a C r > 0 such that 1 C r (1 a 2 ) 2+α µ((a, r)) δ r for all a. By Lemma 2, we have that there exists a C r > 0 such that for all a. Hence (a,r) µ α (a) (a,r) C r C r k α a(z) 2 dµ(z) δ r k α a(z) 2 dµ(z) C r C r δ r > 0 for all a. This proves the third statement. Proof of (4) This follows from Theorem 3 and Remark 1 following Theorem 4. We now state our main theorem answering Question 3. The more interesting direction of the relationship between the Berezin transform of a Carleson measure with the averaging function of a Carleson measure is proven, and the concept of almost Reverse Carleson measures is used in the proof. Theorem 6. Let α > 1, 0 < r < 1, and let µ be a Carleson measure for A 2 α(). If there exists δ > 0 such that µ α (a) δ for all a, then there exists δ r > 0 such that µ α,r (a) δ r for all a and for r sufficiently close to 1. This implies that µ is an almost Reverse Carleson measure for A 2 α(). Furthermore, µ is an almost Reverse Carleson measure for A 2 α() if and only if T µα,γ is invertible on A 2 α() for all γ (0, 1). 11

12 Proof. Assuming the given conditions, for r (0, 1) and a, we have \(a,r) k α a(z) 2 dµ(z) + for some δ > 0. Our goal is to show that \(a,r) (a,r) k α a(z) 2 dµ(z) k α a(z) 2 dµ(z) δ can be made as small as we like independently of a if we push r close enough to 1. It is obvious that the restriction of a Carleson measure for A 2 α() to the complement of a pseudohyperbolic disc in is again a Carleson measure for A 2 α(). Setting f = k α a, and applying Theorem 2 to the restriction of our Carleson measure µ to \ (a, r), we arrive at the conclusion that \(a,r) for all a and r (0, 1), where k α a(z) 2 dµ(z) C a E a k α a(z) 2 da α (z) E a := {z : there exists w in supp(µ \(a,r) ) such that ρ(z, w) < 1 2 }. At this point, we will require the following Proposition. This Proposition was stated in [5] without proof, so we provide one here for lack of reference. Proposition 3. For 1 2 < r < 1, E a \ (a, r ), where r = r r. Proof of Proposition 3. Let a, and suppose z E a. Then there exists a w supp(µ \(a,r) ) such that and our assumption is that ρ(z, w) = z w 1 zw < 1 2, ρ(w, a) = w a 1 wa r. 12

13 We need to show that ρ(z, a) = z a 1 za r. By the stronger triangle inequality for the pseudohyperbolic metric, if we assume r > 1 2, we know that ρ(w, a) > ρ(w, z), hence ρ(w, a) ρ(w, z) 1 ρ(w, a)ρ(w, z) ρ(z, a). Letting x = ρ(w, a) and y = ρ(w, z), we transform the left hand side of the inequality and differentiate with respect to x as follows: This means that d ( x y ) = dx 1 xy 1 xy ( y(x y)) (1 xy) 2 = 1 y2 (1 xy) 2 > 0. ρ(w, a) ρ(w, z) 1 ρ(w, a)ρ(w, z) is an increasing function in the variable ρ(w, a), or that ρ(z, a) ρ(w, a) ρ(w, z) 1 ρ(w, a)ρ(w, z) r y 1 ry. ifferentiating the remaining expression with respect to y, we find that d ( r y ) = dy 1 ry which means that (1 ry) ( r)(r y) (1 ry) 2 = r ρ(w, z) 1 rρ(w, z) is a decreasing function in the variable ρ(w, z), or that ρ(z, a) 1 + r2 (1 ry) 2 < 0 ρ(w, a) ρ(w, z) 1 ρ(w, a)ρ(w, z) r y 1 ry r r 2 13

14 as desired. Continuing on with the proof of Theorem 6, we now know, if 1 2 < r < 1, that \(a,r) k α a(z) 2 dµ(z) C a = C a (1 \(a,r ) (0,r ) = C a (1 r 2 ) α+1 k α a(z) 2 da α (z) ) da α (z) for all a. Here, the first equality follows from a standard change of variables, and the last equality follows from Lemma 3. The proof of Theorem 2 shows us that the constant C from Theorem 2 depends only on the Carleson constant of µ. Since the Carleson constant of the restriction of a Carleson measure is clearly less than or equal to the Carleson constant of the original measure, we must have that C a C for all a. Hence \(a,r) k a (z) 2 dµ(z) C(1 r 2 ) α+1 for all a and for all 2 1 < r < 1. Note that as r 1, by definition, r 1. With this in mind, if we choose r 0 such that C(1 r 0 2)α+1 < 2 δ, then by our assumption that µ δ, we have that δ 2 (a,r) + k a (z) 2 dµ(z) k a (z) 2 dµ(z) + k a (z) 2 dµ(z) δ. \(a,r) (a,r) Therefore (a,r) for all a and for all r r 0. Hence 14 k a (z) 2 dµ(z) δ 2 > 0

15 2 2+α (1 a ) 2+α µ((a, r)) δ 2 = µ((a, r)) (1 a 2 ) 2+α δ 2 5+2α = µ α,r (a) δ C r 25+2α := δ r > 0 for all a and for all r r 0. Here, the first inequality comes from the fact that, given a, k α a(z) 2 22+α (1 a ) 2+α for all z, and the last line comes from Lemma 1. This proves the first part of Theorem 6. Since µ is Carleson, we have µ((a, 1 2 )) C 02 2+α (1 a ) 2+α for all a using Theorem 1. Thus, by Theorem 4 and Lemma 1, using the result we just proved, we must have that f 2 da α 2 2+α C γ (a,γ) f 2 da α (1 a 2 ) 2+α dµ(a) 22+α C γ f 2 α,γ(a)dµ(a) for all f A 2 α() and for all γ (0, 1). Thus µ is an almost Reverse Carleson measure for A 2 α(). We now show that T µ α,γ and then T µα,γ are invertible on A 2 α() for all γ (0, 1). To do so, we recall, by Theorem 1 and Lemma 1 together, that since µ is a Carleson measure of the Bergman space, µ α,γ must be a bounded, nonnegative function on for any γ > 0, so that by condition (3) in Theorem 5, to check that T µ α,γ is invertible on A 2 α() for all γ (0, 1) is equivalent to showing that there exists a constant C γ > 0 such that f 2 da α C γ µ α,γ f 2 da α for all f A 2 α() and for all γ (0, 1). We have just shown that if µ is a Carleson measure for the Bergman space such that µ is invertible in L (), then there exists a C γ > 0 such that 15

16 f (z) 2 da α (z) C γ (z,γ) f (w) 2 da α (w) dµ(z) A α ((z, γ)) for all f A 2 α() and for all γ (0, 1). Thus it suffices to check that there exists C γ > 0 such that (z,γ) f (w) 2 da α (w) dµ(z) C γ A α ((z, γ)) µ((z, γ)) A α ((z, γ)) f (z) 2 da α (z) for all f A 2 α() and for all γ (0, 1). To do this, recognize that (z,γ) f (w) 2 da α (w) dµ(z) A α ((z, γ)) = f (w) 2 1 (w,γ) A α ((z, γ)) dµ(z)da α(w) C γ f (w) 2 1 (w,γ) A α ((w, γ)) dµ(z)da α(w) = C γ f (w) 2 µ α,γ (w)da α (w) for all f A 2 α() and for all γ (0, 1), just as required. Here, the first equality came from Fubini s theorem and the fact that χ (w,γ) (z) = χ (z,γ) (w), and the first inequality came from Lemma 1 and the well known fact that (1 z 2 ) 2+α (1 w 2 ) 2+α for z (w, γ) [3, p. 41]. Thus T µ α,γ is invertible on A 2 α() for all γ (0, 1). Therefore, by condition (2) in Theorem 5, there exist 0 < ɛ < 1, δ > 0, and r > 0 such that A(G (a, ɛ)) > δa((a, ɛ)) for all a, where G := {z : µ α,γ (z) > r} = {z : µ α,γ (z) > r 2 }. Thus, by appealing again to condition (2) of Theorem 5, we have found 0 < ɛ < 1, δ > 0, and r = r 2 > 0 such that A(G (a, ɛ)) > δa((a, ɛ)) 16

17 for all a, where G is now thought of as {z : µ α,γ > r}. Therefore, by condition (2) of Theorem 5, T µα,γ is invertible on A 2 α() for all γ (0, 1). We now show the reverse implication is true. Given γ (0, 1), suppose µ is a Carleson measure for A 2 α() such that T µα,γ is invertible on A 2 α(). This means that T µα,γ is bounded below on A 2 α(), or that there exists an ɛ γ > 0 such that T µα,γ f ɛ γ f 2,α for all f A 2 α(). However, by applying well-known properties of the operator norm, we may expand this inequality to say that ɛ γ f 2,α T µα,γ f sup { µ α,γ f g da α }. g =1 By applying Hölder s inequality, we see that ( ) 1 sup { µ α,γ f g da α } µ α,γ f 2 2 ( ) 1 da α sup { µ α,γ g 2 2 da α } g =1 g =1 for all f A 2 α(). We now note that since µ is a Carleson measure for A 2 α(), µ α,γ K γ < by referring to Theorem 1 and Lemma 1. Thus we have that ( ) 1 µ α,γ f 2 2 ( ) 1 da α sup { µ α,γ g 2 2 ) 1 da α } (K γ µ α,γ f 2 2 da α g =1 for all f A 2 α(). Appealing to a Fubini s theorem argument similar to the one used earlier, 17

18 (K γ 1 µ α,γ (z) f (z) 2 2 da α (z)) = (K γ = (K γ (C γ K γ (z,γ) dµ(w) A α ((z, γ)) f (z) 2 da α (z) (w,γ) ) 1 2 f (z) 2 A α ((z, γ)) da α(z)dµ(w) (w,γ) f (z) 2 da α (z) dµ(w) A α ((w, γ)) ) 1 2 ) 1 2 for all f A 2 α(). Here the last inequality uses the fact that A α ((z, γ)) (1 z 2 ) 2+α (1 w 2 ) 2+α A α ((w, γ)) for z (w, γ). Putting this all together, this means that ɛ γ 2 f 2 2,α C γk γ (w,γ) f (z) 2 da α (z) dµ(w) A α ((w, γ)) for all f A 2 α(). Since γ was an arbitrary number in (0, 1), this means precisely that µ is an almost Reverse Carleson for A 2 α(). 4. CONCLUSION If µ is a Carleson measure for A 2 α() such that µ is also an almost Reverse Carleson measure for A 2 α(), what minimal extra conditions µ must satisfy in order for µ α,r to be invertible in L ()? REFERENCES [1] A. EVINATZ, Toeplitz operators on H 2 spaces, Trans. Amer. Math. Soc. 112 (1964), MR [2] R. G. OUGLAS, Banach algebra techniques in the theory of Toeplitz operators, American Mathematical Society, Providence, R.I., 1973, Expository Lectures from the CBMS Regional Conference held at the University of Georgia, Athens, Ga., June 12 16, 1972, Conference Board of the Mathematical Sciences Regional Conference Series in Mathematics, No. 15. MR [3] P. UREN and A. SCHUSTER, Bergman spaces, Mathematical Surveys and Monographs, vol. 100, American Mathematical Society, Providence, RI, MR , 6, 16 [4]. H. LUECKING, Inequalities on Bergman spaces, Illinois J. Math. 25 (1981), no. 1, MR , 8, 9 [5], Forward and reverse Carleson inequalities for functions in Bergman spaces and their derivatives, Amer. J. Math. 107 (1985), no. 1, MR , 8, 12 18

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