CONVERGENCE OF BIEBERBACH POLYNOMIALS IN DOMAINS WITH INTERIOR CUSPS 0 0 0 V. V. ANDRIEVSKII AND I. E. PRITSKER 2 n Abstract. We extend the results on the uniform convergence of a J Bieberbach polynomials to domains with certain interior zero an- 4 gles (outward pointing cusps), and show that they play a special 2 role in the problem. Namely, we construct a Keldysh-type example on the divergence of Bieberbach polynomials at an outward ] V pointing cusp and discuss the critical order of tangency at this in- teriorzeroangle,separatingtheconvergentbehaviorofBieberbach C polynomials from the divergent one for sufficiently thin cusps. . h t a m [ 1. Introduction 1 Let G be a bounded Jordan domain, z ∈ G. Define the Bergman v 0 2 space L (G) as the space of square integrable analytic functions with 2 2 norm 1 1/2 1 kfk := |f(z)|2dxdy . 0 2 0 (cid:18)ZZG (cid:19) 0 We also use the uniform norm on G in the sequel: / h at kfk∞ := sup|f(z)|. z∈G m The Bieberbach polynomial B (z),degB ≤ n, is the solution of the : n n v following extremal problem [9]: i X (1.1) kB′k = inf {kP′k : P (z ) = 0, P′(z ) = 1}, r n 2 n 2 n 0 n 0 a Pn∈Pn(C) where P (C) is the class of algebraic polynomials of degree at most n, n with complex coefficients. The conformal mapping ϕ : G → D := R0 {z : |z| < R }, normalized by ϕ(z ) = 0 and ϕ′(z ) = 1, solves the 0 0 0 same extremal problem in the class of all analytic functions f in G, 1991 Mathematics Subject Classification. 30E10,41A10, 30C40. Key words and phrases. Bieberbach polynomials, conformal mapping, uniform convergence. Research of both authors was supported in part by the National Science Foun- dation grant DMS-9707359. Research of the second author was also supported in part by the National Science Foundation grant DMS-9970659. 1 2 V. V. ANDRIEVSKII AND I. E. PRITSKER satisfying f(z ) = 0 and f′(z ) = 1. Here, R is the inner conformal 0 0 0 radius of the domain G with respect to z . Moreover, 0 (1.2) kϕ′ −B′k = inf {kϕ′ −P′k : P (z ) = 0, P′(z ) = 1} n 2 n 2 n 0 n 0 Pn∈Pn(C) (see, e.g., [14] or [11]). It is clear from (1.2) that (1.3) lim B′(z) = ϕ′(z) and lim B (z) = ϕ(z), z ∈ G, n n n→∞ n→∞ because polynomials are dense in L (G). A more delicate fact of the 2 uniform convergence of B (z) to ϕ(z) on G was first observed by n Keldyshin1939[14],forthedomainsGwithsufficientlysmoothbound- aries. He also constructed an example of a starlike domain, bounded by a piecewise analytic curve with one singular point, where Bieber- bach polynomials diverge. A considerable progress in the area has been achieved by Mergelyan [16], Suetin [25], Simonenko [22], Andrievskii [2]-[5] and Gaier [11]-[13]. In particular, Andrievskii [2] proved that the uniform convergence of Bieberbach polynomials holds on G, where G is any quasidisk, and Gaier [11]-[13] showed that the rate of this uniform convergence is quite close to the best possible rate in uniform polynomial approximation of the conformal mapping ϕ. It is well known that a quasiconformal curve does not allow zero angles (cusps). The first results on the uniform convergence of Bieber- bach polynomials in domains with cusps were obtained by Andrievskii [3]-[4]. Pritsker [20] developed his approach to improve these results for domains with certain interior zero angles (outward pointing cusps). The interior zero angles seem to play a special role in this problem, as Keldysh’s counterexample, although being implicit, but gives impres- sion that his piecewise analytic boundary curve has an outward cusp as its only singular point (see [14]). We confirm this by constructing an example on the divergence of Bieberbach polynomials at an outward pointing cusp (see Theorem 2.2 and its proof). An interesting problem arising here is to find the critical order of tangency at this interior zero angle, separating the convergent behavior of Bieberbach polynomials, exhibited below in Theorem 2.1, from the divergent one for sufficiently thin cusps. This would give a rather complete answer to the old question on the geometry of domains with uniform convergence of Bieberbach polynomials. Introducingtheareaorthonormalpolynomials{K (z)}∞ ,suchthat n n=0 1, m = n, (1.4) K (z)K (z)dxdy = m n 0, m 6= n, ZZG (cid:26) BIEBERBACH POLYNOMIALS IN DOMAINS WITH INTERIOR CUSPS 3 one can find the following representation for Bieberbach polynomials [10, p. 34]: n−1 z K (z ) K (t)dt k 0 k (1.5) B (z) = Xk=0 Zz0 , n ∈ N. n n−1 |K (z )|2 k 0 k=0 X This gives a constructive method for generating Bieberbach polynomials via the Gram-Schmidt orthonormalization process and for numeri- cal approximation of the conformal mapping ϕ (see [10]). In addition, (1.5) indicates the connection with the Bergman kernel function [8] ∞ ϕ′(z) (1.6) K(z,z ) = K (z )K (z) = , z,z ∈ G. 0 k 0 k πR2 0 k=0 0 X It is clear from (1.6) and ϕ′(z ) = 1 that 0 ∞ 1 (1.7) K(z ,z ) = |K (z )|2 = . 0 0 k 0 πR2 0 k=0 X Thus, many problems on the convergence of Bieberbach polynomials are equivalent to those on the convergence of the integrated bilinear series of (1.6). 2. Convergence and Divergence Results Let τ be a conformal mapping of the unit disk D onto a quasidisk (cf. [15]). We say that a Jordan arc γ is quasianalytic if γ = τ([−1,1]) for such a mapping τ (quasianalytic arcs were introduced in [5]). It is known that a quasianalytic arc is rectifiable and quasismooth, i.e., it satisfies the following chord-arc condition, by Lavrentiev: |γ(z ,z )| ≤ M |z −z |, ∀z ,z ∈ γ, 1 2 1 2 1 2 where |γ(z ,z )| is the length of a subarc γ(z ,z ) ⊂ γ, with the end- 1 2 1 2 points z ,z , and M ≥ 1 is a constant, depending only on γ. A Jordan 1 2 curve is said to be piecewise quasianalytic, if it consists of a finite number of quasianalytic arcs. Suppose that G is bounded by a piecewise quasianalytic curve L = ∂G, with the quasianalytic arcs joining at the points {z }m ⊂ L. Two j j=1 quasianalytic arcs L ⊂ L and L ⊂ L, meeting at z , form an xp- j j+1 j type interior zero angle, if there exists a neighborhood of z such that j in a local coordinate system, with the origin at z , we have j (x,y) ∈ L ⇒ c xP ≤ y ≤ c xp j 1 2 4 V. V. ANDRIEVSKII AND I. E. PRITSKER and (x,y) ∈ L ⇒ −c xp ≤ y ≤ −c xP, j+1 2 1 where P ≥ p > 1 and c ,c > 0. With these notations, our result on 1 2 the convergence of Bieberbach polynomials is stated below. Theorem 2.1. If ∂G is piecewise quasianalytic, with xp-type interior zero angles at the joint points, then there exist q = q(G), r = r(G), 0 < q,r < 1, and C = C(G) > 0 such that (2.1) kϕ−B k ≤ C qnr, n ∈ N. n ∞ It is worth noting that one cannot have r = 1 in Theorem 2.1, as this would imply that ϕ is analytic on G, by a well known result (see, e.g., [10, p. 27]), which is obviously not the case (cf. Theorem 3.1 below). A companion divergence result is based on Keldysh’s construction in [14], but its geometry is made more explicit here. Theorem 2.2. There exists a domain with piecewise smooth boundary and one outward pointing cusp, such that Bieberbach polynomials diverge at this cusp. Furthermore, the boundaryof this domainis analytic outside of any neighborhood of the cusp point. One might speculate that the interior zero angle in Theorem 2.2 has an exponential order of tangency at the cusp point. It would be very interesting to find the critical order of tangency at this interior zero angle, separating the convergent behavior of Bieberbach polynomials in Theorem 2.1, from the divergent one in Theorem 2.2. Remark 2.3. It is also interesting to note that the boundary of the domain in Theorem 2.2 (or in Keldysh’s counterexample) cannot be piecewise analytic, as is claimed in [14]. In other words, the boundary arc, with endpoints meeting at the only irregular boundary point, cannot be an analytic arc, which is the image of a segment under a mapping, analytic in a domain containing this segment inside. Indeed, if this arc is analytic then we can define the angle at the irregular boundary point, by using one-sided tangents. In the case this angle is non-zero, we see that the boundary of our domain is quasiconformal, so that the associated Bieberbach polynomials must converge uniformly by [2]. Thus divergence is only possible if we have a zero angle at the irregular point, which translates into an outward pointing cusp in Keldysh’s construction. However, an analytic arc can only form an xp-type zero angle, because it cannot have an arbitrarily high order of contact, asweshowinSection4. Hencewegettheuniformconvergence of Bieberbach polynomials again, by Theorem 2.1! BIEBERBACH POLYNOMIALS IN DOMAINS WITH INTERIOR CUSPS 5 3. Continuity and Differentiability of a Conformal Mapping at a Cusp The results on the behavior of ϕ at an interior zero angle, obtained in this paper, may be of independent interest. We summarize them below for convenience of the reader. Theorem 3.1. Suppose that G has an xp-type interior zero angle at z ∈ ∂G, which is formed by two quasianalytic arcs. Then there exist constants C,c > 0 such that c (3.1) |ϕ(t)−ϕ(z)| ≤ C exp − , t → z, t ∈ G. |t−z|p−1 (cid:18) (cid:19) Furthermore, ϕ(k)(t) (3.2) lim = 0, ∀ k,m ∈ N. t→z |t−z|m t∈G\{z} Corollary 3.2. Suppose that ∂G is piecewise quasianalytic, with xp- type interior zero angles at the joint points. Then ϕ ∈ C∞(G), i.e., (3.3) ϕ(k) ∈ C(G), ∀ k ∈ N. 4. Proofs LetC,c,c ,c ,... denotepositiveconstants, notnecessarily thesame 1 2 at different places. Writing a 4 b, we mean that a ≤ c b for a constant 1 c , which doesn’t depend on a and b. The relation a ∼ b indicates that 1 c b ≤ a ≤ c b, where c ,c are independent of a and b. 2 1 1 2 Our proofs heavily rely on the distortion properties of conformal and quasiconformalmappings, wherewestartwiththefollowinglemma(see Andrievskii [6, pp. 97-98]). Lemma 4.1. Let w = F(ζ) be a K-quasiconformal mapping of the plane onto itself, such that F(∞) = ∞, ζ ∈ C, w = F(ζ ) (j = j j j 1,2,3), and |w −w | ≤ c |w −w |. Then |ζ −ζ | ≤ c |ζ −ζ | and 1 2 1 1 3 1 2 2 1 3 1/K K w −w ζ −ζ w −w 1 3 1 3 1 3 (4.1) c ≤ ≤ c , 3 4 w −w ζ −ζ w −w (cid:12) 1 2(cid:12) (cid:12) 1 2(cid:12) (cid:12) 1 2(cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) where c = c (c(cid:12) ,K), j =(cid:12) 2,3,4(cid:12). (cid:12) (cid:12) (cid:12) j j (cid:12)1 (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) We next introduce the arc l ⊂ D, with the endpoints at ±1, as q,a the union of two arcs {w = x + iy : y = (x + 1)q/a, −1 < x ≤ 0} and {w = x + iy : y = (1 − x)q/a, 0 ≤ x < 1}, q,a > 1. If τ is 6 V. V. ANDRIEVSKII AND I. E. PRITSKER the conformal map of D, defining a quasianalytic arc of the boundary L ⊂ ∂G with the cusp points z = τ(−1) and z = τ(1), then we set 1 1 2 L := τ(l ). q,a q,a We assume here that τ is extended to a K-quasiconformal homeomor- phism of the complex plane onto itself, with infinity as a fixed point. It is clear that L is an arc connecting z and z in the exterior of q,a 1 2 G. We give estimates for the distance d(ζ,L ) from ζ ∈ L to L , 1 q,a 1 q,a > 1, and also for the distance d(ζ,L ) from ζ ∈ L to L . q,a 1 q,a Lemma 4.2. If ζ ∈ L then q,a (4.2) min|ζ −z |(q−1)K2+1 4 d(ζ,L ) 4 min|ζ −z |(q−1)/K2+1. j 1 j j=1,2 j=1,2 Similarly, if ζ ∈ L then 1 (4.3) min|ζ −z |(q−1)K2+1 4 d(ζ,L ) 4 min|ζ −z |(q−1)/K2+1. j q,a j j=1,2 j=1,2 Proof. Set t = τ−1(ζ) and x = ℜt. Assume that x ≥ 0. Since |t−x| < |t−1|, we obtain that K K t−1 ζ −z |ζ −z | |t−1|1−q ∼ 4 2 ≤ 2 , t−x ζ −τ(x) d(ζ,L) (cid:12) (cid:12) (cid:12) (cid:12) (cid:18) (cid:19) (cid:12) (cid:12) (cid:12) (cid:12) by Lemma 4.1. There(cid:12)fore (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) d(ζ,L) 4 |t−1|(q−1)/K|ζ −z |. 2 Applying (4.1) again, with ζ = z , ζ = ζ and ζ = z , we have 1 2 2 3 1 (4.4) |ζ −z |K 4 |t−1| 4 |ζ −z |1/K, 2 2 which gives the right hand side of (4.2), by the previous inequality. Let ζ′ ∈ L be such that d(ζ,L) = |ζ − ζ′|, and set t′ = τ−1(ζ′). Assume that ℜt′ ≥ 0. Since |ζ −ζ′| ≤ |ζ −z |, Lemma 4.1 yields 2 K K |ζ −z | ζ −z t−1 t−1 2 = 2 4 ≤ ∼ |t−1|(1−q)K. d(ζ,L) ζ −ζ′ t−t′ t−x (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) Combining the ab(cid:12)ove est(cid:12)ima(cid:12)te wit(cid:12)h (4.4(cid:12)), we (cid:12)also prove the left hand (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) side of (4.2). The estimates in (4.3) are obtained by an analogous argument. Remark 4.3. If we consider l := {z¯ : z ∈ l } ⊂ D and set L := q,a q,a q,a τ(l ), then (4.2) holds for any ζ ∈ L as well. q,a q,a We now construct an analytic extension of the conformal mapping ˜ ϕ into a domain G containing G. BIEBERBACH POLYNOMIALS IN DOMAINS WITH INTERIOR CUSPS 7 Lemma 4.4. Suppose that ∂G is piecewise quasianalytic, with xp-type interior zero angles at the joint points. Then the mapping ϕ can be continued conformally into a domain G˜, with the rectifiable boundary ∂G˜ that consists of quasismooth arcs L , connecting the cusp points q,a {z }m , such that G ⊂ G˜ and ∂G∩∂G˜ = {z }m . j j=1 j j=1 Furthermore, there exist constants C,c > 0 such that c (4.5) |ϕ(z)−ϕ(z )| ≤ C exp − , z ∈ G˜, j |z −z |p−1 (cid:18) j (cid:19) where j = 1,... ,m. Proof. Let L , j = 1,2,... ,m, be the quasianalytic arc of L = ∂G, j connectingz andz ,whichisdefinedbythecorrespondingconformal j j+1 map τ . Denote the domain, bounded by the arcs l and l , by S . j q,a q,a q,a It follows from (4.2) and Remark 4.3 that τ (S )∩τ (S ) = ∅ for j 6= k, j q,a k q,a provided we choose a > 0 to be sufficiently small and q to satisfy (q −1)/K2 +1 > P. On defining m G˜ := G∪ τ (S ) , j q,a " # j=1 [ weextendtheconformalmappingϕintoG˜ usingthestandardreflection principle: R2 (4.6) ϕ(z) := 0 , z ∈ τ (S )\G, j q,a ϕ τ τ−1(z) j j h (cid:16) (cid:17)i where j = 1,... ,m. Wenextproceedtoproving(4.5),whereweusethemethodofmoduli of curve families (the method of extremal length). There is no loss of generality in assuming that z = 0 ∈ ∂G and that G∩D (0) ⊂ W for j R some R > 0, where the wedge W is defined by W := {w = x+iy : |y| < c xp,x > 0}, with p > 1. Fix a point a ∈ G and set d = min(|a|/2,R). 1 Consider a point z ∈ G, such that |z| < d, and a family of curves Γ, separating points 0 and z from the point a in G. We need to estimate the module of Γ, denoted by m(Γ), from below. This is accomplished with the help of an auxiliary family of curves Γ′, which consists of the circular arcs γ(r) := {|w| = r} ∩ W, with the radius r varying from c |z| to d, c |z| < r < d. Since the boundary arcs L and L , meeting 2 2 j j+1 at z = 0, are quasismooth, we can choose c such that each curve from j 2 8 V. V. ANDRIEVSKII AND I. E. PRITSKER Γ′ contains a curve from Γ. It follows by the comparison principle (see Theorem 4-1 in [1, p. 54]) that (4.7) m(Γ) ≥ m(Γ′). Let θ(r) be the angular measure of the arc γ(r) ∈ Γ′. It is known (cf. Theorem 2.6 in [18, p. 77]) that d dr (4.8) m(Γ′) = . rθ(r) Zc2|z| Using a simple estimate θ(r) ≤ 2c rp−1, p > 1, we conclude by (4.7) 1 and (4.8) that d dr 1 1 1 m(Γ) ≥ = − . 2c rp 2(p−1)c (c |z|)p−1 dp−1 Zc2|z| 1 1 (cid:18) 2 (cid:19) Hence c (4.9) |ϕ(z)−ϕ(0)| 4 e−πm(Γ) 4 exp − , z ∈ G, |z|p−1 (cid:18) (cid:19) ˜ by Theorem 1 of [7, p. 290] (see also [6, p. 34]). In the case z ∈ G\G, we obtain from (4.6) and (4.9) that c |ϕ(z)−ϕ(0)| 4 exp − ,  p−1 τ τ−1(z) j j    (cid:12) (cid:16) (cid:17)(cid:12)  withadifferentc. Applying(4.1)with(cid:12)ζ = 0, ζ =(cid:12)τ τ−1(z) , ζ = z (cid:12) 1 2 (cid:12) j j 3 and w = −1, w = τ−1(z), w = τ−1(z), we conclud(cid:16)e that (cid:17) 1 2 j 3 j τ τ−1(z) ∼ |z|, j j (cid:12) (cid:16) (cid:17)(cid:12) which implies (4.5) by the previous inequality. (cid:12) (cid:12) (cid:12) (cid:12) Lemma 4.5. Let G be a Jordan domain, which is symmetric in the real axis, and let z = 0 ∈ G. Assume that ξ ∈ ∂G is real and G ⊂ {z : 0 ℜ z < ξ}. If the conformal mapping ϕ is not analytic on G, then (4.10) limsup|B (x)| = ∞, ∀ x > ξ. n n→∞ Proof. Suppose to the contrary of (4.10) that there exists x > ξ such 0 that the sequence {B (x )}∞ is bounded. Then we obtain from (1.5) n 0 n=0 and (1.7) that the following sequence is also bounded: n−1 x0 K (0) K (t)dt, n ∈ N. k k k=0 Z0 X BIEBERBACH POLYNOMIALS IN DOMAINS WITH INTERIOR CUSPS 9 This implies, in turn, that x0 (4.11) K (0) K (t)dt < C, n ∈ N, n n (cid:12) Z0 (cid:12) (cid:12) (cid:12) for some constant(cid:12) C > 0. Note th(cid:12)at the orthonormal polynomials K (z) have real c(cid:12)oefficients for any(cid:12)n ∈ N, because G is symmetric n about the real axis. Furthermore, we follow the usual convention in Gram-Schmidt orthonormalizationthat theleading coefficient of K (z) n is positive for any n ∈ N. It follows that each K (x) is real valued for n real x, and is positive for x → +∞. Since the zeros of K (z) are n contained in the convex hull of G (see [24, p. 31]), we conclude that K (x) has no zeros for x > ξ and that n (4.12) K (x) > 0, x > ξ, n ∈ N. n Using Theorem 1.1.4 of [24, p. 4], we obtain that (4.13) limsupkK k1/n ≤ 1 n ∞ n→∞ and (4.14) lim |K (x)|1/n = egΩ(x,∞) > 1, x > ξ, n n→∞ where g (x,∞) is the Green function of Ω := C\G with pole at ∞. Ω Combining (4.12)-(4.14) gives that x0 1/n liminf K (t)dt > 1 n n→∞ (cid:12)Z0 (cid:12) (cid:12) (cid:12) and that (cid:12) (cid:12) (cid:12) (cid:12) (4.15) limsup|K (0)|1/n < 1, n n→∞ by (4.11). Hence the conformal mapping ϕ must have an analytic continuation through ∂G, by (4.15) and Theorem 2.1 of [19], which contradicts our assumption. Proof of Theorem 2.1. We use a known method, based on the extremal property of Bieberbach polynomials (1.2). Namely, we first estimate the quantity kϕ′ − B′k , and then proceed to the uniform n 2 norm case, to prove (2.1). Recall that the conformal mapping ϕ can be continued, by Lemma 4.4, into a larger domain G˜, whose boundary consists of quasismooth arcs connecting the cusp points {z }m ⊂ ∂G. Let γ be a subarc of j j=1 j ∂G˜, with the endpoints z and z , and let ζ ∈ γ be a fixed point, j j+1 j j j = 1,... ,m. Note that ζ divides γ into γ1 and γ2, so that ∂G˜ = j j j j 10 V. V. ANDRIEVSKII AND I. E. PRITSKER m 2 γi. Since ∂G˜ is rectifiable, we have by Cauchy’s integral j=1 i=1 j formula that S S m 2 1 ϕ(ζ) 1 ϕ(ζ) ϕ(z) = dζ = dζ 2πi ζ −z 2πi ζ −z Z∂G˜ j=1 i=1 Zγji XX m 2 m 1 ϕ(ζ)−ϕ(z ) 1 ζ −z j+i−1 j−1 = dζ + ϕ(z )log , j 2πi ζ −z 2πi ζ −z j=1 i=1 Zγji j=1 j XX X for any z ∈ G, where we assume that ζ = ζ . It follows that 0 m m 2 1 ϕ(ζ)−ϕ(z ) (4.16) ϕ′(z) = j+i−1 dζ 2πi (ζ −z)2 j=1 i=1 Zγji XX m 1 1 1 + ϕ(z ) − , z ∈ G. j 2πi ζ −z ζ −z j=1 (cid:18) j j−1 (cid:19) X Observe that the second sum m 1 1 1 f(z) := ϕ(z ) − j 2πi ζ −z ζ −z j=1 (cid:18) j j−1 (cid:19) X represents a function, analytic on G. Hence there exists a sequence of polynomials {p }∞ and a number R > 1 such that n n=1 (4.17) kf −p k ≤ CR−n, n ∈ N, n ∞ (see, e.g., Theorem 4 in [10, p. 27]). Consequently, our problem of approximating ϕ′(z) by polynomials reduces to approximating functions of the following form ϕ(ζ)−ϕ(z ) j+i−1 g (z) := dζ, i,j (ζ −z)2 Zγji in view of (4.16). Furthermore, we can consider approximation in the uniform norm and then pass to L (G) norm, which suffices for our 2 purposes. We now set ϕ(ζ)−ϕ(z ) (4.18) g(z) := g (z) = 1 dζ, γ := γ1, 1,1 (ζ −z)2 1 Zγ and study the approximation of this function only, as the other functions g (z) are handled similarly. i,j Let Φ : Ω → D′ be a conformal map of Ω := C\G onto D′ := {w : |w| > 1}, satisfying the conditions Φ(∞) = ∞ and Φ′(∞) > 0. Define the level curves of Φ by L := {z ∈ Ω : |Φ(z)| = R}, R ≥ 1, R

Convergence of Bieberbach polynomials in domains with interior cusps

Similar Convergence of Bieberbach polynomials in domains with interior cusps