Bounded Point Evaluations for Certain P t (µ) Spaces. Thomas Len Miller. Department of Mathematics. Mississippi State University

Transcription

1 Bounded Point Evaluations for Certain P t (µ) Spaces Thomas Len Miller Department of Mathematics Mississippi State University Mississippi State, MS Wayne Smith Department of Mathematics University of Hawaii Honolulu, HI Liming Yang Department of Mathematics University of Hawaii Honolulu, HI Supported in part by NSF Grant DMS95397 and a seed-money grant from University of Hawaii Typeset by AMS-TEX

2 . Introduction. For a positive measure µ with compact support in the complex plane C and for t <, let P t (µ) denote the closure in L t (µ) of the polynomials in z. A point w in C is a bounded point evaluation for P t (µ) if there exists a constant M > 0 such that p(w) M p Lt (µ) for each polynomial p. We denote the set of bounded point evaluations for P t (µ) by bpe ( P t (µ) ). The existence of bounded point evaluations and mean polynomial approximation have received a great deal of attention, culminating in a result of Thomson, [27]: either P t (µ) = L t (µ) or P t (µ) has bounded point evaluations. In the latter case, P t (µ) admits a structure related to that of bpe ( P t (µ) ). If P t (µ) is irreducible, i.e., if P t (µ) does not split into the direct sum of nontrivial spaces P t (µ ) and P t (µ 2 ), then the set of bounded point evaluations for P t (µ) is a simply connected region whose closure contains the support of µ. Thomson s dichotomy does not enable one however to determine the bounded point evaluations for an arbitrary measure µ; indeed, such a characterization seems generally out of reach. Nevertheless, for some natural classes of measures, it is possible to study the structure of the bounded point evaluations, and much interesting analysis has resulted. The canonical result in this vein is of course Szegö s Theorem, which characterizes the set of bounded point evaluations for a measure µ with support on the unit circle D. In the case that µ is weighted area measure restricted to a bounded region, we mention work by Carleman [9], Keldy s [9], D zrba sjan and Saginjan [3, p.58], Havin [5], Havin and Maz ja [6, 7], Saginjan [24], Shapiro [25], and Brennan [6, 7]. Akeroyd [, 2, 3] has dealt with harmonic measures for some crescent like sets, and Hruscev [8], Kriete [20],

3 Trent [28], Vol berg [29, 30], and Kriete and MacCluer [2] are among those who have contributed in the case that µ consists of weighted Lebesgue measure on the unit circle and weighted area measure inside the unit disk. More recent work by Akeroyd [4, 5] is also relevant in this setting. The literature in this area is extensive, and our references above are by no means complete. In the present paper, we continue the study of bounded point evaluations for µ of the form dµ = h dm + W da D, where m and A D respectively denote Lebesgue measure on the unit circle and area measure on the unit disk D. Before stating our results, we need to introduce some notation. Let K be a compact subset of the unit circle D and let {J n } denote the components of D \ K. For each n, let I n denote the chord in the closed unit disk D with the same endpoints as J n. Denote by G n the region with boundary J n I n ; let Γ be the rectifiable Jordan curve K n {I n} and U the interior of Γ. Define the measure µ by dµ = h dm K + W da Gn where h is a nonnegative integrable function satisfying h dm > and W is K 2

4 a positive continuous function on each G n such that W L (A Gn ). It follows that G n = bpe ( P (W A Gn ) ). We should mention that if the function h is not -integrable over K, an argument similar to that in [5, Theorem 2.2] yields P t (µ) = P t (W da Gn ) L t (hm K ). In this case, zero is not a bounded point evaluation for P t (µ). For simplicity, in this section we just give our characterization in the case that the weight W is identically. It will be stated and proved for more general weights in the next sections. Let z 0 U and let ω be the harmonic measure for U at z 0. Denote by I n and J n the lengths of I n and J n, respectively. Theorem A. Define the measure µ by dµ = h dm K + da Gn, where h is a nonnegative integrable function satisfying h dm >. Then the following conditions are K equivalent: (a) z 0 is a bounded point evaluation for P t (µ); (b) bpe(p t (µ)) = D and P t (µ) is irreducible; (c) P t (µ) does not split; (d) ω(i n ) I n <. n= We note that the convergence of the series (d) is independent of z 0 U, since ω is comparable to harmonic measure at any other point z U. We remark that the convergence of this series is very sensitive. An example was shown to us by F. Nazarov for which the addition of a single point to K changes the series (d) from convergent to divergent. Condition (d) is related to the well known Carleson condition (introduced in [0]) on the set K, which is that n= J n J n <. 3

5 Notice that ω(i n ) C I n and thus the Carleson condition implies that bpe(p t (µ)) = D for the measure µ in Theorem A. We will present an example showing that the Carleson condition is in fact strictly stronger than this. Theorem B. There is a compact subset K of the unit circle with m(k) > 0 such that K satisfies Theorem A (d) but n= J n J n =. Sufficient conditions for P t (µ) to be irreducible will be stated and proved in 2, and the case that P t (µ) splits will be considered in 3. Theorem A will be an immediate consequence, as the hypotheses for these theorems will be satisfied when the weight W is identically. Theorem B will be proved in P t (µ) is irreducible. Let I n, J n, G n, K and U be defined as in the introduction. Define the measure µ by dµ = h dm K + W da Gn where h is a nonnegative integrable function satisfying h dm > and W is a K positive continuous function on each G n such that W L (A Gn ). Let α t,n (λ) be the norm of the kernel function of P t (W A Gn ) corresponding to λ G n ; i.e., ( ) α t,n (λ) = sup p(λ) p t t W da, G n where the supremum is taken over all nonzero polynomials. For an open set G, we use δ G (z) to denote the distance of z from the boundary of G. Theorem 2.. Suppose that K satisfies n= ω(i n ) I n < (2.) 4

6 and that t < is such that for some s, C > 0 α t,n (λ) Cδ Gn (λ) s for every n and λ G n. Then P t (µ) is irreducible and bpe ( P t (µ) ) = D. Recall that ω is harmonic measure for U at a point z 0 U, and that convergence of the series (2.) is independent of the choice of z 0. A typical example of a weight W to which the theorem applies is given by W (z) = ( z 2 ) α, where α is greater than ; see Corollary 2.3 below. A region G is said to satisfy a θ wedge condition if there exists r > 0 and θ (0, ) such that, for every w G, a closed circular sector of radius r and opening θπ lies in Ḡ, with vertex at w. In particular, it is clear that U satisfies a θ wedge condition for some θ > 0. We now fix such a θ. Let J n U \ G n be the arc of the circle connecting the endpoints of I n and at the angle θ/2 to I n at each endpoint. We now fix z 0 U such that J n separates z 0 from I n, for all n. Choose a Riemann mapping ϕ from the unit disk D onto U such that ϕ(z 0 ) = 0, and let ψ be the inverse of ϕ. We will denote by C or c absolute constants that may change from one step to the next. Similarly, C(z 0 ) will denote a quantity that depends at most on z 0, etc. Lemma 2.2. For each n, there exists a smooth curve γ n in G n that joins the endpoints of J n such that γ n δ Gn (z) dω V C θ 2 ω(i n ) I n. Here V is the region bounded by γ, where γ = n γ n K and ω V for V at z 0. is the harmonic measure Proof. Let α n = ψ(i n ) and β n = ψ(j n). Let a n, b n be the endpoints of J n, and set A n = ψ(a n ) and B n = ψ(b n ). 5

7 Claim : δ D (z) c min{ z A n 2/θ, z B n 2/θ }, z β n. Since U satisfies a θ wedge condition, from Theorem in [22] we get that ϕ(z ) ϕ(z 2 ) C z z 2 θ. (2.2) On the other hand, since U is convex, ψ (z) is bounded [23, p 225], so z z 2 C ϕ(z ) ϕ(z 2 ). (2.3) Now let z β n, and assume without loss of generality that z B n z A n. Then, using (2.2) and (2.3), there exists z D such that δ D (z) = z z c ϕ(z) ϕ(z ) /θ cδ U (ϕ(z)) /θ c ϕ(z) b n 2/θ c z B n 2/θ. Claim 2: There exists a C 2 curve Γ n ψ(g 0 n), where G 0 n is bounded by J n and I n, that joins A n and B n such that Γ = Γ n ψ(k) is a C 2 curve and δ ψ(g 0 n )(z) c min{ z A n 2/θ, z B n 2/θ }, z Γ n. To see the claim, let F be a Riemann map from D to the upper half plane R 2 + with F (0) = i and F mapping the middle point of α to. For n 2, let α n = F (α n ) and β n = F (β n ). Let A n = F (A n ) and B n = F (B n ). Then it follows from Claim that δ R 2 + (z) c min{ z A n 2/θ, z B n 2/θ }, z β n. Let Γ n = {z = x + iy : y = c (x A n )2/θ (B n x)2/θ (A n B n )2/θ, B n x C n}. It is easy to check that Γ = i=2 Γ n F (ψ(k)) α is a C 2 curve and Γ n F (ψ(g 0 n)). Let Γ n = F (Γ n). Let F 2 be another Riemann map from D to R 2 + with F 2 (0) = i and F 2 mapping the middle point of D \ α to. Using the same method as above, we can construct Γ as above 6

8 such that Γ F 2 ( D \ α ) is smooth. Let Γ = F 2 (Γ ) and let Γ = n= Γ n ψ(k). Clearly, Γ satisfies the conditions of the claim. Let M n = ϕ(γ n ). Let γ n be the reflection of M n with respect to I n and γ = γ n K. From the construction, γ D K. Let V be the region bounded by γ. By the Schwarz reflection principle, we see that ψ extends to a Riemann map from V to ψ(v ) where the boundary of ψ(v ) is the reflection of Γ with respect to the unit circle. Hence the boundary of ψ(v ) is a C 2 curve. Claim 3: For z D we have c θ δ D (z) δ U (ϕ(z)) C δ D (z). To see this, let z D such that δ D (z) = z z. Using (2.2) and (2.3), we have δ D (z) c ϕ(z) ϕ(z ) /θ cδ U (w) /θ, and similarly, δ U (w) cδ D (z). Taking arithms gives the claim. Claim 4: c θ In fact, using (2.2) and (2.3), we get ω(i n ) I n C ω(i n ). c I n ω(i n ) = α n C θ I n. Let ω 0 be the harmonic measure for Ω = ψ(v ) at zero. Let z be the reflection of the point z with respect to I n and Γ 0 n be the reflection of Γ n with respect to the unit circle. Now γ n δ Gn (z) dω V C γ n δ U (z ) dω V C γ n where the last step is from Claim 3. Using Claim 2, we see that γ n δ D (ψ(z )) dω V C θ γ n δ D (ψ(z )) dω V, ( ) ψ(z ) A n + ψ(z dω V. (2.4) ) B n 7

9 Now, working with just the first term of this last integral, we estimate γ n ψ(z ) A n dω V C γ n ψ(z) A n dω V = C Γ n w A n dω 0, where the change of variable w = ψ(z) was used. Since the boundary of Ω is C 2, we see that w A n is comparable to the arc length s(w) from A n to w on Γ n, and dω 0 is comparable to ds. Therefore, Γ n w A n dω 0 C Γ n s(w) ds(w) C A n B n A n B n C α n α n. But α n = ω(i n ) and so from Claim 4 and the last two displays we get that γ n ψ(z ) A n dω V Cω(I n ) ω(i n ) C θ ω(i n ) I n. Clearly the same estimate applies to the term in (2.4) involving B n, and hence γ n δ Gn (z) dω V C θ 2 ω(i n ) I n, as required. Proof of Theorem 2.. Define { δ ts h 0 = Gn (z), for z γ n min(, h), for z K. Since V D, we have, for each K, that ω V ( ) = ω V (z 0, ) ω D (z 0, ) c m( )/( z 0 ). Then using Lemma 2.2 we get h 0 dω V =ts δ Gn dω V + γ n Cts ω(i n ) I n + Cts ω(i n ) I n + >, K {h } c z 0 c z 0 hdω V K {h } K hdm hdm c z 0 hdm 8

10 By Szegö s Theorem [4, p. 36], there is a constant C > 0 such that for each polynomial p, p(z 0 ) t C p t h 0 dω V. On the other hand, we are assuming that δg ts n (w) p(w) t C p(z) t W da(z) C G n D p(z) t dµ for w γ n. Therefore, p t h 0 dω V C p t δg ts n dω V + C γ n p t hdω V C p t dµ + C p t hdω V C D p t dµ. (2.5) So z 0 is a bounded point evaluation for P t (µ). Since, for every λ V, harmonic measure for V at λ is comparable to ω V, by Szegö s Theorem we have p(λ) t C(λ) p t h 0 dω V. Thus from (2.5) we see that V bpe(p t (µ)). Since G n = bpe(p t (W da Gn ) bpe(p t (µ)), we conclude that D bpe(p t (µ)). Now suppose that P t (µ) is not irreducible. By Thomson s theorem [27], there exists E K with m(e) > 0 such that L t (µ E ) is a summand of P t (µ). Since ω V (E) ω(e) > 0, from (2.5) we see that the characteristic function of E is a nonzero element of P t (h 0 dω V ). Hence, P t (h 0 dω V ) = P t (h 0 dω V E ) P t (h 0 dω V E c). This is a contradiction since, from Szegö s Theorem, bpe(p t (h 0 dω V E )) = bpe(p t (h 0 dω V E c)) =. 9

11 Corollary 2.3. If K satisfies the condition (2.) and W (z) = ( z 2 ) α for some α >, then P t (µ) is irreducible and bpe ( P t (µ) ) = D. Proof. Assume without loss of generality that α 0. Since p t is subharmonic for each polynomial p, we have for λ γ n, p(λ) t C δg 2 p(z) t da(z) n (λ) z λ 3 4 δ Gn (λ) Thus α t,n (λ) Cδ s G n, with s = 2+α t. 3. P t (µ) splits. C δ 2+α p(z) t ( z 2 ) α da(z). G n (λ) G n Theorem 3.. Suppose that W L p (A Gn ) for some p >. If K satisfies then n= ω(i n ) I n =, P t (µ) = L t (µ K ) ( P t (µ Gn )). If m(k) = 0, the proof of Theorem 3. will be similar to that of Theorem 5.7 in [7]. However if m(k) 0 the Cauchy transform that we will use is no longer in a Sobolev space, and therefore some other ideas will have to be used. Recall that ϕ is a Riemann map from D to U such that ϕ(0) = z 0. Define σ z0 (z) = z 0 z z 0 z. Lemma 3.2. There exists a constant 0 < c 0 < 2 such that for each r sufficiently close to, r <, there exits a smooth function τ r, 0 τ r on C satisfying 0 if z >, τ r (z) = if z ϕ(rd) \ { σ z0 (z) 2c 0 }, 0 if σ z0 (z) < c 0, 0

12 and such that τ r z (z) C(z 0) z, where C(z 0 ) is independent of r. Proof. Let r be close enough to so that σ z0 ϕ(rd) contains 2c 0 D. From Schwarz s lemma, we see that σ z0 ϕ(rd) rd. Let τ be a smooth function on R such that 0 τ, τ(x) = for x >, τ(x) = 0 for x < 0, and 0 τ (x) C. Define ( ) z τ if z > r r τr 0 (z) = if 2c 0 < z r τ(2 z 2c 0 ) if z 2c 0. It is easy to check the function τ r = τ 0 r σ z0 has the required properties. For q > and f L q (A), recall that the maximum function of f is defined by M f (λ) = sup f(z) da(z), r>0 A(O(λ, r)) O(λ,r) where O(λ, r) is the disk with center λ and radius r. It is well-known that f M f is bounded on L q (A); see for example [26, p.5]. For t, let t be the conjugate exponent of t, so that t + t =.

13 Lemma 3.3. Suppose that t and that g L t (µ) annihilates P t (µ). Define k on C by k Gn = gw χ Gn. Let H : U C be given by H(w) = ( w )N+ w z w k(z) z w da(z), where N is a positive integer. Then, there is a constant C(N) so that for all w U, H(w), for all j = 0,, 2,..., N. (w w )j+ k(z) da(z) (z w C(N)M k(w)( w ), )j+2 Proof. Since the proofs of the two inequalities are similar, we show only that H(w) C(N)M k (w)( w ). For w U, let δ w = w w and write H(w) = { z w <3δ w } = I + I 2. ( w )N+ w z w ( k(z) w )N+ z w da(z) + w { z w 3δ w } z w k(z) z w da(z) The lemma is established by the following estimates for I and I 2. I k= C(N) { 2 k 3δ w z w < 2 k δ w k= C(N)M k (w)( w ), 2 k 3δ w} ( w )N+ w z w { 2 k 3δ w z w < 2 k 3δ w} k(z) da(z) k(z) z w da(z) and I 2 k= k= {2 k 3δ w z w <2 k 3δ w } 2 (N+2)k δ w CM k (w)( w ). ( w )N+ w z w {2 k 3δ w z w <2 k 3δ w } k(z) z w da(z) k(z) da(z) 2

14 Suppose that g and H are as in the previous lemma. Let q satisfy and let t satisfy t + t = q. Then q t p t + p (< t ) (q )t q + p. Since W L p (A Gn ) and g L t (µ), it follows from Hölder s inequality that gw L q (A G n ) g L t (µ) ( W ) (q )t t q + da Hence, gw L q (A Gn ). From now on, we fix such a q with < q < 2 and q + q =. For 0 < r <, let ω r be the harmonic measure on U r = ϕ(rd) evaluated at z 0, i.e., ω r ϕ = 2πr dθ. By a change of variables, U r f dω r = 2πir f ( z ) ( ϕ (z) ) U r ϕ (z) dz, for all f continuous on U r. Lemma 3.4. With the notation above, for each ɛ, 0 < ɛ < q, there exists a constant C(N, z 0, µ) > 0 (depending on N, z 0 and µ) so that sup 0<r< H(z) ( z ) ɛ dω r(z) C(N, z 0, µ) g L t (µ). Proof. Let 0 < ɛ < q and let τ r be the smooth positive function constructed in Lemma 3.2. Using Green s formula [4, p 26], we have that H(z) ( z ) ɛ dω r(z) = 2πir = πr U r τ r (z)h(z) (ϕ (z)) ( z ) ɛ ϕ dz (z) ( ) τr (z)h(z) (ϕ (z)) ( z ) ɛ ϕ da(z). (z) U r 3

15 Since the function (ϕ (w)) ϕ (w) is bounded on the set {τ r > 0} U, we conclude that H(z) ( z ) ɛ dω r(z) C τ r(z) H(z) ( z ) ɛ da(z) for some constant C. by U r For a compactly supported finite Borel measure ν, the Cauchy transform of ν is defined ˆν(λ) = z λ dν(z). The function ˆν is locally integrable with respect to area measure. Let k r (w) = τ r (w)k(w) π ka(w) τ r w, It is well-known that k r A(w) = τ r (w) ka(w); see for example [, or 4, p.50 Lemma 0.]. Hence, the functions k r and k r A each have compact support and are each members of L q (A). Therefore, by the Calderon Zygmund theorem [8 or 26 p.35], grad( k r A) q C k r q, where grad f denotes the weak gradient of f. Thus k r A is in the Sobolev space W q = {u L q (A) : grad(u) L q (A)}. Notice that for w D, z w = z w N j=0 We write τ r (w)h(w) = τ r (w) ka(w) τ r (w) = k r A(w) + F (w)τ r (w). ( w )j w (w w )N+ z w + (z w )N+ z w. (3.) k(z) z w da(z) τ r (w) k(z) z w N j= ( w )j w z w da(z) From the construction of τ r, it follows that τ r F is in W q, and therefore so is τ rh. On the other hand, by Lemma 3.3, we get F (w) C(N)M k (w), for w 2. 4

16 It is easy to check that (τ r H) = πkτ r + ka(w) τ r w + F τ r + F (w) τ r = πkτ r + F τ r + H(w) τ r. Using a theorem from [26, p77], we have (τ r H ) q x ( τ rh ) q + y ( τ rh ) q x (τ rh) q + y (τ rh) q C (τ r H) q. Now using Lemmas 3.2 and 3.3, we see that H(z) ( z ) ɛ dω r(z) C U r (τ r H ) ( z ) ɛ da(z) + C U r C (τ r H ) q ( w ) ɛ q + C ( C ) (τ r H) q q da τ r H da(z) ( z ) +ɛ U r τ r H da(z) ( z ) +ɛ + C(N) τr (z)m k (z) ( z ) ɛ da(z) C(N) k q + C τ r H q + C(N) τ r F q + C(N) τ r M k q C(N, z 0 ) k q C(N, z 0, µ) g L t (µ). Here we used the fact that the Maximum Operator is bounded on L q (A). The lemma is now established. Lemma 3.5. Suppose that t < and g L t (µ). If g P t (µ), then there is a constant C so that sup 0<r< ĝµ(z) dω r (z) C(N, z 0 ) g L t (µ). 5

17 Proof. Using (3.), for g P t (µ) we get First notice that ĝµ(w) = z w z w = H(w) + K = H(w) + G(w). w j N j=0 z w z w ( w )j w z w g(z)dµ N j=0 ( w )j w z w g(z)dµ w z w j z w 2j P w, j N, where P w (z) is the Poisson kernel for w. We bound G as follows. N w w j G(w) P w (z) g(z) h(z) dm(z) + g(z) h(z) dm(z) z w j+ K K 2 N P w (z) g(z) h(z) dm(z) + K = 2 N u(w), P w g h dm K j= N w w j z w j z w w z w wz g(z) h(z) dm(z) K j= where u(w) is a positive harmonic function. By Lemma 3.4, we may conclude that ĝµ(w) dω r H(w) dω r + 2 N u(w)dω r C(N, z 0, µ) g L t (µ + 2N u(0) C(N, z 0, µ) g L t (µ). We now fix N such that N < θ. Let eiθ n be the midpoint of J n and let Σ n = {z D, J n 2 + θ n + I n N arg(z) J n 2 + θ n I n N } and let Γ nr = Σ n ϕ(rd). Lemma 3.6. Let G be as in the proof of the previous corollary. Then there exists an absolute constant C(N) > 0 such that for w Γ n we have G(w) C(N) J n g L t (µ). 6

18 Proof. From the definition of G, G(w) K w w N+ g(z) h(z) z w dz z w where g L t (µ). Since for w Γ n and z K we have w w C J n z w, and z w c J n N, the desired estimate is obtained. Proof of Theorem 3.. Let g L t (µ) be an annihilator of P t (µ) and ɛ < min( q, N ). Fix an M, we have ĝµ dω r M i= Γ ir M i= Γ ir ĝµ dω r + ĝµ dω r ĝµ ( z ) ɛ dω r + ɛ ( z ) dω r + M i= Γ ir ĝµ dω r. Hence, by Lemma 3.5 and the subharmonicity of ĝµ, we get ĝµ(z 0 ) M i= Γ ir ĝµ ( z ) ɛ dω r + ɛ ( z ) dω r + C(N, z 0, µ) g L t M i= Γ (µ). ir Thus, from Lemma 3.4 and letting r, we see that ĝµ(z 0 ) M i= Γ i G(z) ( z ) ɛ dω + ɛ ( z ) dω + C(N, z 0, µ) g L t M i= Γ (µ). i On the other hand, using Lemma 3.6 and the inequality Nɛ <, we have M i= Γ i G(z) M ( z ) ɛ dω J n Nɛ ω(γ i ) g L t (µ) C(N) g L t (µ). i= Hence, ĝµ(z 0 ) ɛ ( z ) dω + C(N, z 0, µ) g L t M i= Γ (µ) i M C(N, z 0, µ) g L t (µ) + C(N) ω(γ i ) I i. 7

19 From our construction, we see that ω(i i \ Γ i ) C(z 0 ) I i \ Γ i C(z 0 ) I i N. On the other hand, U satisfies the θ wedge condition. Using Theorem in [22] and (2.2), since N > θ, we conclude that ω(i i ) c(z 0 ) I i θ 2ω(Ii \ Γ i ), for i sufficiently large. Hence, for such i, ω(i i ) 2ω(Γ i ), and therefore, M ĝµ(z 0 ) C(N, z 0, µ) g L t (µ) + C(N) ω(i i ) I i. Now letting M, we see that ĝµ(z 0 ) = 0. The same method shows that ĝµ(λ) = 0 for each λ U. Thus z λ P t (µ). Therefore, no point of U is a bounded point evaluation for P t (µ). Now the theorem follows from Thomson s theorem in [27]. Theorem A is a direct conclusion of Corollary 2.3 and Theorem 3., with the weight W identically. 4. Proof of Theorem B. In this section, we will construct an example for which the Carleson condition fails but n= ω(i n ) I n <. 8

20 Let {E n } with E n = 3 n be a sequence of arcs in the unit circle and {E n} be the corresponding chords whose positions will be chosen. Let {F n } be another sequence of arcs such that F n has an endpoint in common with E n. Let {F n} be the corresponding chords with F n = E n 2π E n. Now we can choose the positions of {E n } and {F n } such that (E n F n ) (E m F m ) = n m. Choose an integer N n such that F n N n F n 2. We divide F n into N n equal pieces denoted by {F nj } j Nn. Let {F nj } j N n be the corresponding chords. Let U be the region bounded by E n F nj ( D \ E n). Lemma 4.. Let U be as above and ω be the harmonic measure for U at zero. Then the following conditions hold: (i) N n F nj n= j= F nj =, that is, D \ ( E n F n ) does not satisfies the Carleson condition, and (ii) N n ω(f nj) F nj + n= j= n= ω(e n) E n <. Proof. Since it follows that (i) holds. N n F nj F nj c F n N n F n c, j= 9

21 Let G n be the region bounded by F n and F n. Let U n = U \Ḡn. Let ω n be the harmonic measure for U n at zero. We claim that ω n (F n) C F n E n. Let S n be the sector with center at the common endpoint of E n and F n and radius 2 such that F n E n S n. Let ω Sn be the harmonic measure for S n at zero. Since U n S n, ω n (F n) ω Sn (F n). Let d n be the common point of E n and F n. Then g(z) = (z d n ) π π ( En + Fn )/2 is a Riemann map from S n to a half disk with radius between 2 and 3. Hence, ω Sn (F n) C F n π π ( En + Fn )/2 C F n F n E n 2π C F n E n, and the claim is established. Clearly, ω( N n j= F nj) ω n (F n) C F n E n. 20

22 On the other hand, N n ω(f nj) F nj ω( N n j= F nj) N n F n C E n F n N n F n C E n. j= Therefore, condition (ii) holds. The following theorem is an immediate consequence of Theorem A and Lemma 4.. Theorem 4.2. Let U be as in Lemma 4. and let the measure µ satisfy the hypotheses of Theorem A. Then n= and bpe(p t (µ)) = D, for t <. I n I n = Acknowledgment We would like to thank James Brennan for pointing out an error in an earlier version of this paper. It was in correcting this error that we constructed the counterexample in Section 4 and we came to the final formulation of Theorem 3.. References [] J. Akeroyd, Polynomial approximation in the mean with respect to harmonic measure on crescents, Trans. Amer. Math. Soc., 303(987), [2] J. Akeroyd, Point evaluations and polynomial approximation in the mean with respect to harmonic measure, Proc. Amer. Math. Soc., 05(989), [3] J. Akeroyd, Polynomial approximation in the mean with respect to harmonic measure on crescents II, Mich. Math. J., 39 (992), [4] J. Akeroyd, Point evaluations for P t (µ) and the boundary of Support(µ), Mich. Math. J., to appear. [5] J. Akeroyd, An extension of Szegö s Theorem, Preprint. 2

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