ISSN 0001-4346, Mathematical Notes, 2012, Vol. 91, No. 5, pp. 714–724. © Pleiades Publishing, Ltd., 2012. Original Russian Text © A. A. Pekarskii, 2012, published in Matematicheskie Zametki, 2012, Vol. 91, No. 5, pp. 761–772.
Approximation to the Function z α by Rational Fractions in a Domain with Zero External Angle A. A. Pekarskii* Byelorussian State University, Minsk Received March 8, 2010; in final form, March 18, 2011
Abstract—Rational approximations to the function z α , α ∈ R \ Z, were studied by Newman, Gonchar, Bulanov, Vyacheslavov, Andersson, Stahl, and others. The present paper deals with the order of best rational approximations to this function in a domain with zero external angle and vertex at the point z = 0. In particular, the obtained results show that the conditions imposed on the boundary of the domain in the Jackson-type inequality proved by the author in 2001 for the best rational approximations in Smirnov spaces cannot be weakened significantly. DOI: 10.1134/S0001434612050136 Keywords: best uniform rational approximation, polynomial approximation, Smirnov space, analytic function, rational function, rectifiable Jordan boundary, Lavrentiev curve.
1. INTRODUCTION. MAIN RESULTS The following notation will be used. Let K be a compact set in the complex plane C, let C(K) be the space of functions continuous on K, and let CA (K) be the subspace of C(K) consisting of functions that are analytic at the interior points of K (if there are such points). For f ∈ C(K), we introduce the norm f C(K) = max{|f (z)| : z ∈ K}. We assume that Rn is the set of rational functions of degree at most n (n = 0, 1, 2, . . .) and introduce Rn (f )C(K) = inf{f − rn C(K) : r ∈ Rn }, which is the best uniform rational approximation of f ∈ C(K). If f ∈ CA (K), then we also use the notation f CA (K) and Rn (f )CA (K) . For α ∈ R \ Z, we consider the function fα (z) = z α which is analytic in the domain C \ (−∞, 0] and satisfies the condition that fα (x) > 0 for x ∈ (0, +∞). Uniform rational approximations to the function fα (x), α ∈ (0, +∞) \ N, on the interval [0, 1] were studied (see, e.g., [1]) by Newman, Gonchar, Vyacheslavov, Stahl, and others. Stahl [2] obtained the following strongest result: Rn (fα )C[0,1] ∼ 4α+1 |sin πα|e−2π
√
αn
,
n → ∞.
Andersson [3] and the author [4] studied best uniform approximations to this function in a closed disk D, where D = {z : |z − 1| < 1}. To be precise, the lower bound in the following relation was obtained by Andersson, and the upper one, by the author, Rn (fα )CA (D) e−π
√
2αn
n → ∞.
,
For 0 < β < π, we introduce the domain Ωβ = {z : |z| < 1, |arg z| < π − β} with external angle equal to 2β at the point z = 0. In [4], the following relation was also obtained: √
Rn (fα )CA (Ωβ ) e−2 *
παβn
E-mail:
[email protected]
714
,
n → ∞.
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In the present paper, we consider the best approximations to the function fα in a domain with zero external angle at the point z = 0. Namely, for γ > 1, in the plane of the complex variable z = x + iy, we introduce the domain Gγ = {x + iy : x2 + y 2 < 1, |y| > |x|γ for x ≤ 0}. Theorem 1. Let α ∈ (0, +∞) \ N, and let γ > 1. Then there are positive constants c1 > 0 and c2 > 0 depending only on α and γ and satisfying the condition that, for n = 2, 3, 4, . . ., α/(γ−1) 1 log n α/(γ−1) c1 ≤ Rn (fα )CA (Gγ ) ≤ c2 . n n For 0 < p < ∞, we also consider the best approximations to fα in the Lebesgue spaces Lp [0, 1], Hardy spaces Hp (D), and Smirnov spaces Ep (Q). The definitions of these spaces are given below in Secs. 2 and 3. The best approximations in these spaces are defined in the same way as in the spaces C(K) and CA (K), i.e., the norm · C(K) is replaced by the quasinorms · Lp [0,1] , · Hp (D) , and · Ep (Q) . Andersson [3] obtained the following relations for 1 < p < ∞ and α ∈ (−1/p, +∞) \ Z: 1 1/(2p) n , n → ∞, exp −2π α+ Rn (fα )Lp [0,1] n p 1 1/(2p) n , n → ∞. exp −π 2 α + Rn (fα )Hp (D) n p For the Ep -approximations in the domain Gγ , we have the following theorem. Theorem 2. If 0 < p < ∞, α ∈ (−1/p, ∞) \ Z, and γ > 1, then (α+1/p)/(γ−1) 1 , Rn (fα )Ep (Gγ ) n
n → ∞.
The problem of approximating the function fα in the domain Gγ is considered in connection with the following Jackson-type theorem proved by the author in [5]. Assume that Q is a simply connected bounded domain with rectifiable Jordan boundary ∂Q, the function f is analytic in Q, 0 < p ≤ ∞, s ∈ N, 1/σ = s + 1/p, and f (s) ∈ Eσ (Q). Then (under the conditions imposed on ∂Q below) we have both f ∈ Ep (Q) and c n = s, s + 1, s + 2, . . . , (1.1) Rn (f )Ep (Q) ≤ s f (s) Eσ (Q) , n where c > 0 is independent of f . The inequality (1.1) was obtained in [5] under the following conditions on ∂Q: ∂Q is a Lavrentiev curve for 0 < p < ∞; ∂Q is an Alper or Radon curve for p = ∞. We note that these domains have neither internal nor external zero angles. In [5], it was also shown that if the domain Q has a zero internal angle, then, for the above values of p, s, and σ (except for the case p = ∞ and s = 1) the implication f (s) ∈ Eσ (Q) ⇒ f ∈ Ep (Q)
(1.2)
does not hold, and hence inequality (1.1) is not satisfied either. In [6], it was shown that the implication (1.2) holds for a wide class of domains Q with zero external angles. In particular, (1.2) holds for the domain Q = Gγ . By Theorems 1 and 2, for the domain Q = Gγ and the function 1 (γ − 1)s − 1 \ Z, with α ∈ − , f = fα p p (s)
inequality (1.1) does not hold despite the inclusion fα ∈ Eσ (Gγ ). MATHEMATICAL NOTES Vol. 91
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Apparently, for 0 < p < ∞, the class of domains Q for which (1.1) is satisfied coincides with the set of domains whose boundary ∂Q is a Lavrentiev curve. In the case p = ∞, a similar question apparently has a more complicated answer, and we do not venture to make any assumptions. The problem of generalizing the polynomial Jackson theorem to complex domains, which was posed by S. M. Nikolskii and was studied in detail by V. K. Dzyadyk, V. I. Belyi, V. V. Andrievskii, E. M. Dyn’kin, N. A. Shirokov, and others. The same problem is considered in Belyi’s survey [7]. We also note Andrievskii’s preprint [8], where some results about the best polynomial approximations to analytic functions in domains with zero external angles were obtained. It follows from the above that the lower bounds in Theorems 1 and 2 are the most interesting. Therefore, our main attention is paid to their proofs. The proofs of the upper bounds are only outlined. A preliminary version of this paper was presented in [9]. 2. AN EXTREMUM PROBLEM FOR RATIONAL FUNCTIONS The proof of Theorems 1 and 2 is based on Theorems 3 and 4, which are of interest in themselves. Some additional notation is required to formulate Theorems 3 and 4. Let I be a closed set in C that belongs to some locally rectifiable curve or a finite family of such curves. We assume that |I|, which is the linear Lebesgue measure of the set I, is finite and positive. By Lp (I), 0 < p ≤ ∞, we denote the Lebesgue space of complex measurable functions f on I equipped with the finite quasinorm (the norm for 1 ≤ p ≤ ∞) ˆ 1/p 1 p |f (z)| |dz| , 0 < p < ∞, f Lp (I) = |I| I f L∞ (I) = ess sup|f (z)|. z∈I
If f ∈ C(I), then, obviously, f L∞ (I) = f C(I) . By L0 (I) we denote the space of functions f that are measurable on I and satisfy the condition ˆ −∞ ≤ log |f (z)| |dz| < ∞. I
For f ∈ L0 (I), we define the geometric average ˆ 1 log |f (z)| |dz| . f L0 (I) = exp |I| I Obviously, the functional · L0 (I) on L0 (I) is not a quasinorm. If I is the interval [a, b], then, instead of C([a, b]) and Lp ([a, b]), we respectively write C[a, b] and Lp [a, b]. In Theorem 3 and in what follows, we use the function 1 x > 0. λ(x) = log(1 + x2 ) + 2x arctan , x Theorem 3. Let δ > 0, and let I be a closed set in R with 0 < |I| < ∞; let I ± be the projections of the set I on the straight lines Im z = ±δ. Then, for any rational function rn ≡ 0 of degree at most n, the following inequality holds: rn L0 (I + ) 4δ . ≤ exp nλ rn L0 (I − ) |I| To prove Theorem 3, we need Lemma 1 and Theorem 4. Lemma 1. Let δ > 0, and let I be a closed set in R with 0 < |I| < ∞. Then the following inequality holds for any ξ ∈ C: ˆ x + iδ − ξ |I| 4δ dx ≤ λ . (2.1) log x − iδ − ξ 2 |I| I
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Proof. If Im ξ > 0, then |x + iδ − ξ| < |x − iδ − ξ|
for all
x ∈ R,
and hence inequality (2.1) holds in this case. Therefore, we can assume that ξ = α − iβ, where α ∈ R and β ≥ 0. Thus, it suffices to prove the inequality ˆ 1 4δ (x − α)2 + (δ + β)2 |I| λ (2.2) log dx ≤ 2 2 2 I (x − α) + (δ − β) 2 |I| for all α ∈ R and β ≥ 0. The integrand in (2.2) is an even function with respect to the straight line x = α and decreases on the semiaxis [α, +∞). Therefore, the integral in the left-hand side of (2.2) takes its maximum for I = [α − γ, α + γ], Thus, it remains to prove the inequality ˆ J(β) :=
γ=
|I| . 2
t2 + (δ + β)2 2δ log 2 dt ≤ γλ 2 t + (δ − β) γ
γ 0
(2.3)
for all γ > 0 and β ∈ [0, +∞). The function J(β) is continuous on [0, +∞), J(0) = J(+∞) = 0, and J(β) > 0 for 0 < β < ∞. Hence J(β) attains its maximum for some β ∈ (0, +∞). We shall show that (2.4)
max J(β) = J(δ).
0<β<∞
The integral in (2.3) can be differentiated with respect to the parameter β ∈ [0, δ) ∪ (δ, +∞). Using this fact, it is easy to see that γ γ + arctan , β ∈ [0, δ) ∪ (δ, +∞). J (β) = 2 arctan δ+β δ−β Hence J (β) > 0
for 0 ≤ β < δ
J (β) < 0
and
for δ < β < +∞.
Thus, the proof of (2.4) is complete. Further, it is easy to show that ˆ γ γ |I| 4δ t2 + 4δ2 4δ2 4δ arctan = λ . log dt = γ log 1 + 2 + J(δ) = 2 t γ γ 2δ 2 |I| 0 This completes the proof of Lemma 1. Theorem 4. Let δ > 0, and let I be a closed set in R with 0 < |I| < ∞. Then, for any algebraic polynomial pn ≡ 0 of degree at most n, the following inequality holds: ˆ pn (x + iδ) n|I| 4δ dx ≤ λ . log pn (x − iδ) 2 |I| I Proof. Since the inequality is obvious for n = 0, we assume that n ∈ N. Without loss of generality, we assume that the degree of pn (z) is equal to n. Then the polynomial pn (z) can be written as pn (z) = a
n
(z − ξk ),
k=1
where a = 0 and ξ1 , ξ2 , . . . , ξn are zeros of pn (z) written with their multiplicity taken into account. It only remains to use Lemma 1. MATHEMATICAL NOTES Vol. 91
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Proof of Theorem 3. For n = 0, this inequality is obvious, namely, both of its sides are equal to 1. Let n ≥ 1, and let rn (z) = pn (z)/qn (z), where pn (z) and qn (z) are algebraic polynomials of degree at most n. We have ˆ ˆ ˆ rn (x + iδ) pn (x + iδ) q n (x − iδ) dx. dx = log dx + log log rn (x − iδ) pn (x − iδ) q (x + iδ) I
I
I
n
Obviously, Theorem 4 can be applied to both terms in the right-hand side of the last relation. Therefore, we have ˆ rn (x + iδ) 4δ dx ≤ n|I|λ , log rn (x − iδ) |I| I which is equivalent to the statement of Theorem 3. Remark 1. As follows from the proofs, Theorems 3 and 4 are sharp. For example, the inequality in Theorem 3 becomes an equality for I = [−γ, γ] and z + iδ n . rn (z) = z − iδ Corollary 1. Let δ > 0, and let I be a closed set on R with 0 < |I| < ∞; let I ± be the projections of I on the straight lines Im z = ±δ. Then, for any rational function rn ≡ 0 of degree at most n, the following inequality holds: 4πδn minz∈I + |rn (z)| ≤ exp . maxz∈I − |rn (z)| |I| Proof. It suffices to note that λ(x) ≤ πx
for x > 0,
min |rn (z)| ≤ rn L0 (I + ) ,
z∈I +
max |rn (z)| ≥ rn L0 (I − ) ,
z∈I −
and to use Theorem 3. Remark 2. Corollary 1 can also be derived from the following result obtained by Gonchar [10]. Let E, F ⊂ C be nonintersecting closed sets of positive logarithmic capacity, and let C(E, F ) be the capacity of the condenser (E, F ). Then, for any rational function rn (z) ≡ 0 of degree at most n, the following inequality holds: n minz∈F |rn (z)| ≤ exp . maxz∈E |rn (z)| C(E, F ) 3. PROOF OF THE LOWER BOUNDS IN THEOREMS 1 AND 2 In the proof of the lower bounds in Theorems 1 and 2, the key role is played by Corollary 1 and by the fact that the point z = 0 is a point of branching of the function fα . This idea was first successfully used by Gonchar [11]. The following Lemma 2 is well known. For tn one can take the Taylor polynomial of the function f . Lemma 2. Assume that z0 ∈ C, 0 < ρ1 < ρ2 < ∞, a function f is analytic in the disk |z − z0 | ≤ ρ2 , and M = max{|f (z)| : |z − z0 | ≤ ρ2 }. Then, for any n ∈ N, there is an algebraic polynomial tn of degree at most n such that n M ρ1 ρ1 for |z − z0 | ≤ ρ1 . |f (z) − tn (z)| ≤ ρ2 − ρ1 ρ2 We assume that, just as in Sec. 1, γ > 1. For each n ∈ N, we introduce sn = n−1/(γ−1) and the γ ± interval In = [−s n , −(1/2)sn ]. By In we denote the projections of In on the straight lines Im z = ±sn . We set Kn = In+ In− and recall that the function fα (z) = z α , α ∈ R \ Z, was introduced in Sec. 1. The lower bounds in Theorems 1 and 2 for the best approximations will be obtained from the following lemma. MATHEMATICAL NOTES Vol. 91 No. 5 2012
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Lemma 3. If 0 < p ≤ ∞, α ∈ R \ Z, and γ > 1, then Rn (fα )Lp (Kn ) n−α/(γ−1) ,
n → ∞,
(3.1)
where the positive constants hidden by the symbol are independent of n. The upper bound in (3.1) is obvious: Rn (fα )Lp (Kn ) ≤ max |fα (z)| ≤ 2|α| · n−α/(γ−1) . z∈Kn
This bound is presented in (3.1) only for completeness. Proof of the lower bound in (3.1) for p = ∞. Let rn ∈ Rn be an element of the best uniform approximation to fα on Kn . For brevity, we set Rn = Rn (fα )C(Kn ) . Thus, we have |fα (z) − rn (z)| ≤ Rn
for z ∈ Kn .
(3.2)
By gα (z) we denote the function which is defined in the half-plane Re z < 0 and is the analytic continuation of fα (z) from the third quarter. To apply Lemma 2, we introduce the notation 3 3 1 ρ1 = sn , ρ2 = sn . z0 = − sn , 4 8 2 It is easy to see that the inequalities |α| 9 −|α| α sn ≤ |gα (z)| ≤ sαn (3.3) 4 8 hold for |z − z0 | ≤ ρ2 . In what follows, since Kn belongs to the disk |z − z0 | ≤ ρ1 for n ≥ 4, we assume that, without loss of generality, n ≥ 4. According to Lemma 2 and in view of the right-hand side in (3.3), there is an algebraic polynomial tn (z) of degree at most n such that |α| n 9 3 sαn , z ∈ Kn . (3.4) |gα (z) − tn (z)| ≤ 3 8 4 It follows from (3.2) and (3.4) that the inequality
|α| n 9 3 sαn |[fα (z) − gα (z)] − r2n (z)| ≤ Rn + 3 8 4
(3.5)
holds for z ∈ Kn and r2n (z) = rn (z) − tn (z). For z ∈ In− , we have fα (z) = gα (z), and hence, by (3.5), |α| n 9 3 sαn , z ∈ In− . (3.6) |r2n (z)| ≤ Rn + 3 8 4 But if z ∈ In+ , then with the left-hand inequality in (3.3) taken into account, we obtain fα (z) − 1 ≥ 2 · 4−|α| sαn |sin πα|. |fα (z) − gα (z)| = |gα (z)| · gα (z) In turn, it follows from (3.5) and (3.7) that |α| n 9 3 −|α| |sin πα| − 3 sαn − Rn , |r2n (z)| ≥ 2 · 4 8 4
z ∈ In+ .
(3.7)
(3.8)
By (3.7) and (3.8), Corollary 1 can be applied to the function r2n (z) on Kn = In+ ∪ In− . Namely, we have |α| n 9 3 −|α| |sin πα| − 3 sαn − Rn 2·4 8 4 ≤ e8π . |α| n 9 3 Rn + 3 sαn 8 4 Obviously, this implies the lower bound in (3.1) for p = ∞. MATHEMATICAL NOTES Vol. 91
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Proof of the lower bound in (3.1) for 0 < p < ∞. We again use the notation introduced above. Let rn,p ∈ Rn be an element of the best Lp -approximation to the function fα on Kn . Then fα − rn,p Lp (Kn ) = Rn (fα )Lp (Kn ) .
(3.9)
For brevity, we set Rn,p = Rn (fα )Lp (Kn ) . It follows from (3.9) and the Chebyshev inequality that there is a closed set Kn ⊂ Kn and a constant c1 = c1 (p) > 0 such that |Kn | = (3/4)|Kn | and z ∈ Kn .
|fα (z) − rn,p (z)| ≤ c1 Rn,p ,
(3.10)
Then there is a closed set Jn ⊂ In with |Jn | = (1/2)|In | that satisfies the following condition. If Jn± denotes the projections of Jn on the straight lines Im z = ±sγn , then Jn± ⊂ Kn . By (3.10), we have |fα (z) − rn,p (z)| ≤ c1 Rn,p ,
z ∈ Jn+ ∪ Jn− .
Further, we follow the same reasoning as in the case p = ∞, but with In± replaced by Jn± and Rn replaced by c1 Rn,p . This proves the lower bound in (3.1) for 0 < p < ∞. Proof of the lower bound in Theorem 1. It follows from the maximum principle for the modulus of an analytic function that Rn (fα )CA (Gγ ) ≥ Rn (fα )C(Kn ) . It remains to use Lemma 3 with p = ∞ and α ∈ (0, +∞) \ N. In what follows, we use (see [12]) the Smirnov space Ep (Q), where 0 < p ≤ ∞, and Q is a simply connected bounded domain with rectifiable Jordan boundary ∂Q. By E∞ (Q) we denote the space of bounded analytic functions f in Q equipped with the norm f E∞ (Q) = sup |f (z)|. z∈Q
To define Ep (Q) for 0 < p < ∞, we consider {Qn }∞ n=1 , which is a sequence of simply connected domains Qn with rectifiable Jordan boundaries ∂Qn satisfying the following conditions: Qn ⊂ Qn+1 , Qn ⊂ Q for all n, and ∞
Qn = Q.
n=1
Then f ∈ Ep (Q) if f is analytic in Q and sup f Lp (∂Qn ) < ∞
n∈N
at least for one sequence {Qn }∞ n=1 with the properties listed above. This definition of the space Ep (Q) was proposed by M. V. Keldysh and M. A. Lavrentiev. It is equivalent to the original definition given by Smirnov. If f ∈ Ep (Q), then, for almost all ζ ∈ ∂G, the function f has nontangentially (with respect to ∂G) limit values which we denote by f (ζ). For f ∈ Ep (Q), 0 < p ≤ ∞, we introduce the quasinorm (the norm for 1 ≤ p ≤ ∞) f Ep (Q) = f Lp (∂Q) . If Q is a disk, then Ep (Q) coincides with the Hardy space Hp (Q) [12], [13]. In Sec. 4, we also use the Smirnov space Ep (C \ K), where K is a continuum with rectifiable boundary ∂K. This space and the quasinorm in it are introduced similarly to Ep (Q). Moreover, in the case p = ∞, it is additionally assumed that f (∞) = 0. Lemma 4. Assume that 0 < p < ∞, γ > 1, and n ∈ N. If f ∈ Ep (Gγ ), then ˆ ˆ p |f (z)| |dz| ≥ c |f (z)|p |dz|, ∂Gγ
Kn
where c = c(γ) > 0. MATHEMATICAL NOTES Vol. 91 No. 5 2012
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This lemma follows from the Carleson embedding theorem [13]. Proof of the lower bound in Theorem 2. Let rn,p ∈ Rn be an element of the best Ep -approximation to fα in Gγ . By Lemma 4, we have |Kn | 1/p Rn (fα )Ep (Gγ ) ≥ c fα − rn,p Lp (Kn ) . |∂Gγ | It remains to use Lemma 3 with 0 < p < ∞ and α ∈ (−1/p, ∞) \ Z. 4. PROOF OF THE UPPER BOUNDS IN THEOREMS 1 AND 2 By Ω we denote the domain obtained by eliminating the half-interval (−2, 0] from the disk |z| < 2. The boundary ∂Ω of the domain Ω consists of the circle |ζ| = 2 and the upper [−2, 0]+ and lower [−2, 0]− intervals [−2, 0] of the real axis. For α ∈ (−1, +∞) \ Z, we use the Cauchy integral formula to obtain ˆ 1 fα (ζ) dζ , z ∈ Ω, fα (z) = 2πi ∂Ω ζ − z where ∂Ω is oriented in the positive direction. If ζ ∈ [−2, 0]+ , then arg ζ = π
and
ζ α = |ζ|α eαπi .
arg ζ = −π
and
ζ α = |ζ|α e−απi .
fα (z) = gα (z) +
sin πα ϕα (z), π
But if ζ ∈ [−2, 0]− , then
Therefore, we have
where
z ∈ Ω,
ˆ 1 fα (ζ)dζ , |z| < 2, gα (z) = 2πi |ζ|=2 ζ − z ˆ 0 |x|α dx , z ∈ [−2, 0]. ϕα (z) = −2 x − z
(4.1)
(4.2) (4.3)
To obtain the upper bounds in Theorems 1 and 2, we use (4.1). The function gα (z) is analytic in the disk |z| < 2, and, by Lemma 2, it can be uniformly approximated by polynomials in the disk |z| ≤ 1 with the rate of convergence of a geometric progression. Therefore, for 1 < p ≤ ∞, it suffices to show that ϕα (z) is a Markov function of special form and it can be Ep -approximated in Gγ with the corresponding rate of convergence. Then we must assume that α ∈ (−1/p, +∞). Thus, to obtain the upper bounds in Theorems 1 and 2 for 1 < p < ∞, it suffices to prove the following Lemmas 5 and 6. By c, c1 , c2 , . . . we shall denote some positive variables that are independent of these parameters. Lemma 5. Let α > 0, and let γ > 1. Then, for any n = 2, 3, 4, . . ., log n α/(γ−1) . Rn (ϕα )CA (Gγ ) ≤ c(α, γ) n Lemma 6. Assume that 1 < p < ∞, α ∈ (−1/p, ∞), and γ > 1. Then, for any n = 1, 2, 3, . . ., (α+1/p)/(γ−1) 1 . Rn (ϕα )Ep (Gγ ) ≤ c(p, α, γ) n MATHEMATICAL NOTES Vol. 91
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To prove Lemmas 5 and 6 and the subsequent Lemma 7, we introduce the sequence 1/(γ−1) 1 xk = −2 , k = 1, 2, 3, . . . . 1 + a(k − 1) Here a = a(γ) > 0 is chosen so that the disk + x x k k+1 ≤ xk+1 − xk z − 2 lies outside the domain Gγ for each k. It is easy to verify that such an a exists. Proof of Lemma 6. For each k = 1, 2, 3, . . ., we introduce ˆ xk+1 1/p−1 |x|α dx, τk = ρk xk
where ρk is the distance from the disk z − xk + xk+1 ≤ xk+1 − xk 2 2 to the domain Gγ . Simple calculations show that τk = O(k−(α+1/p)/(γ−1)−1/p ) as k → ∞. It remains to use Theorem 1 in [14] (also see Remark 1 to that theorem). The proof of Lemma 6 is complete. To prove Lemmas 5 and 7, we also need functions χ1 and χ2 defined on (−∞, 0]. Namely, χ1 (x) = 1
on (−∞, −2]
and
χ1 (x) = 0
on [−1, 0],
and χ1 (x) decreases on [−2, −1] and is infinitely differentiable on (−∞, 0]. We set χ2 (x) = 1 − χ1 (x)
for x ∈ (−∞, 0].
Proof of Lemma 5. Let m ∈ N be such that xm > −1. Then ϕα (z) = ϕ1,α (z) + ϕ2,α (z), where
ˆ
xm
ϕ1,α (z) = x1 ˆ 0
ϕ2,α (z) = 2xm
z ∈ [−2, 0],
|x|α χ1 (x/|xm |) dx, x−z
z ∈ [x1 , xm ],
|x|α χ2 (x/|xm |) dx, x−z
z ∈ [2xm , 0].
To construct an approximating rational function for ϕ1,α (z), it is necessary to apply the scheme that we used in [5] and [14]. Namely, for z ∈ [x1 , xm ], we have m−1 m−1 ˆ xk+1 |x|α χ1 (x/|xm |) dx =: ψk,α (z). ϕ1,α (z) = x−z xk k=1
k=1
Each function ψk,α (z) is analytic in the domain xk+1 − xk + x x k k+1 > z − 2 2 and hence can be uniformly approximated in the domain z − xk + xk+1 ≥ xk+1 − xk 2 MATHEMATICAL NOTES Vol. 91 No. 5 2012
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by partial sums of the Laurent series at the rate of convergence of a geometric progression. Replacing each of the functions ψk,α in this domain by partial sum of order O(log m) of the Laurent series, we see that there is a rational function u(z) of degree O(m log m), and α/(γ−1) 1 ϕ1,α − uCA (Gγ ) ≤ . m It is also easy to show that α/(γ−1) 1 |ψ2,α (z)| ≤ c1 (α, γ) . sup m z∈C\[0,2xm ] This completes the proof of Lemma 5. To obtain the upper bound in Theorem 2 for 0 < p ≤ 1, it is necessary to slightly change formula (4.1). We introduce s = [1/p], which is the integer part of 1/p. If α ∈ (−1/p, ∞) \ Z, then α + s > −1 and, by (4.1)–(4.3), fα+s (z) = gα+s (z) + (−1)s
sin πα ϕα+s (z), π
z ∈ Ω.
Therefore, we have (s)
fα (z) =
(s)
πgα+s (z) + (−1)s sin πα · ϕα+s (z) , π(α + s)(α + s − 1) . . . (α + 1)
z ∈ Ω.
Thus, to obtain the upper bound in Theorem 2 for 0 < p ≤ 1, it suffices to prove the following lemma. Lemma 7. Let p ∈ (0, 1], s = [1/p], α ∈ (−1/p, +∞), and γ > 1. Then the inequality (α+1/p)/(γ−1) 1 (s) Rn (ϕα+s )Ep (Gγ ) ≤ c(p, α, γ) n holds for n = 2, 3, 4 . . . . Proof. Assume that there is m ∈ N such that xm > −1. Then (s)
ϕα+s (z) = s!(θ1,α (z) + θ2,α (z)), where
ˆ
xm
θ1,α (z) = x1 ˆ 0
θ2,α (z) = 2xm
z ∈ [−2, 0],
|x|α+s χ1 (x/|xm |) dx , (x − z)s+1
z ∈ [x1 , xm ],
|x|α+s χ2 (x/|xm |) dx , (x − z)s+1
z ∈ [2xm , 0].
According to the main result obtained in [15], we have θ2,α Ep (C\[2xm ,0])
(α+1/p)/(γ−1) 1 ≤ c1 (p, α, γ) . m
We apply the Carleson embedding theorem (see, e.g., [12], [13]) to the last inequality and obtain (α+1/p)/(γ−1) 1 . θ2,α Ep (Gγ ) ≤ c2 (p, α, γ) m To derive a similar bound for Rn (θ1,α )Ep (Gγ ) with n = O(m), it is necessary to proceed just as in the proof of Lemma 4.2 in [5]. The proof of Lemma 7 is complete. MATHEMATICAL NOTES Vol. 91
No. 5
2012
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PEKARSKII
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MATHEMATICAL NOTES Vol. 91 No. 5 2012