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For 0 < p ≤∞,let H denote the classical Hardy space of the unit disc. We consider the extremal problem of maximizing the modulus of the kth Taylor coefﬁcient of a function f ∈ H which satisﬁes f ≤ 1 and f (0) = t for some 0 ≤ t ≤ 1. In particular, we provide a complete solution to this problem for k = 1 and 0 < p < 1. We also study F. Wiener’s trick, which plays a crucial role in various coefﬁcient-related extremal problems for Hardy spaces. Keywords Hardy spaces · Extremal problems · Coefﬁcient estimates Mathematics Subject Classiﬁcation Primary 30H10; Secondary 42A05 1 Introduction Let H denote the classical Hardy space of analytic functions in the unit disc D = {z ∈ C :|z| < 1}. Suppose that k is a positive integer. For 0 < p ≤∞ and 0 ≤ t ≤ 1, consider the extremal problem Communicated by Dmitri Khavinson. Sigrid Grepstad is supported by Grant 275113 of the Research Council of Norway. Sarah May Instanes is supported by the Olav Thon Foundation through the StudForsk program. B Sigrid Grepstad sigrid.grepstad@ntnu.no Ole Fredrik Brevig obrevig@math.uio.no Sarah May Instanes sarahmin@stud.ntnu.no Department of Mathematics, University of Oslo, 0851 Oslo, Norway Department of Mathematical Sciences, Norwegian University of Science and Technology (NTNU), No. 7491, Trondheim, Norway Published online: 12 September 2022 O. F. Brevig et al. (k) f (0) ( p, t ) = sup Re : f ≤ 1 and f (0) = t . (1) k H k! By a standard normal families argument, there are extremals f ∈ H attaining the supremum in (1) for every k ≥ 1 and every 0 ≤ t ≤ 1. A general framework for a class of extremal problems for H which includes (1) has been developed by Havinson [8], Kabaila [9], Macintyre and Rogosinski [11] and Rogosinski and Shapiro [14]. A particular consequence of this theory is that the structure of the extremals is well- known (see Lemma 4 below). For our extremal problem, it can be deduced directly from Parseval’s identity that √ √ 2 2 (2, t ) = 1 − t and that the unique extremal is f (z) = t + 1 − t z . Similarly, the Schwarz–Pick inequality (see e.g. [15, VII.17.3]) shows that (∞, t ) = 1 − t and that the unique extremal is f (z) = (t + z)/(1 + tz). This served as the starting point for Beneteau and Korenblum [1], who studied the extremal problem (1)inthe range 1 ≤ p ≤∞. We will enunciate their results in Sects. 4 and 5,but fornow we present a brief account of their approach. Theﬁrststepin[1] is to compute ( p, t ) and identify an extremal function. This is achieved by interpolating between the two cases p = 2 and p =∞ mentioned above, facilitated by the inner-outer factorization of H functions. It follows from the argument that the extremal function thusly obtained is unique. The second step in [1] is to show that ( p, t ) = ( p, t ) for every k ≥ 2using k 1 a trick attributed to F. Wiener [2], which we shall now recall. Set ω = exp(2π i /k) and suppose that f (z) = a z . F. Wiener’s trick is based on the transform n≥0 k−1 ∞ kn W f (z) = f (ω z) = a z . (2) k kn j =0 n=0 p p The triangle inequality yields that W f ≤ f for f ∈ H if 1 ≤ p ≤∞. k H H Hence, if f is an extremal function for ( p, t ), then f (z) = f (z ) is an extremal 1 1 k 1 function for ( p, t ) and consequently ( p, t ) = ( p, t ). Note that this argument k k 1 does not guarantee that the extremal f is unique for ( p, t ). k k We are interested in the extremal problem (1)for 0 < p < 1 and whether the extremal identiﬁed using F. Wiener’s trick above for 1 ≤ p ≤∞ is unique. We shall obtain the following general result, which may be of independent interest. Theorem 1 Fix k ≥ 2 and suppose that 0 < p ≤∞. Let W denote the F. Wiener transform (2). The inequality 1/ p−1 W f p ≤ max k , 1 f p k H H is sharp. Moreover, equality is attained if and only if (a) f ≡ 0 when 0 < p < 1, (b) W f = f when 1 < p < ∞. 123 p F. Wiener’s Trick and an Extremal Problem for H The upper bound in the estimate is easily deduced from the triangle inequality. Hence, the novelty of Theorem 1 is that the inequality is sharp for 0 < p < 1, and the statements (a) and (b). In Sect. 3, we also present examples of functions in H and H which attain equality in Theorem 1, but for which W f = f . However, we will conversely establish that if both f and W f are inner functions, then f = W f . k k To illustrate the role played by the F. Wiener transform in various coefﬁcient related extremal problems, we ﬁrst recall that the estimate W f ≤ f was originally k ∞ ∞ used by F. Wiener to resolve a problem posed by Bohr [2] and compute the so-called Bohr radius for H . We also know from [12, Sect. 1.7] that the Krzyz˙ conjecture on the maximal magnitude of the kth coefﬁcient in the power series expansion of a non-vanishing function with f = 1 is equivalent to the assertion that if f is an extremal for the corresponding extremal problem, then f = W f . As far as we are aware, the Krzyz˙ conjecture remains open for k ≥ 6. Theorem 1 shows that the extremal for ( p, t ) is unique when 1 < p < ∞.We shall see in Sect. 5 that the extremal problem ( p, t ) with k ≥ 2 and 1 ≤ p ≤∞ has a unique extremal except for when p = 1 and 0 ≤ t < 1/2. In the range 0 < p < 1 with k = 1, the extremal problem (1) has been studied −1/ p by Connelly [4, Sect. 4], who resolved the problem in the cases 0 ≤ t < 2 and −1/ p 1/ p−1/2 2 p(2 − p) < t ≤ 1. Connelly also states conjectures on the behavior −1/ p −1/ p 1/ p−1/2 of ( p, t ) in the range 2 ≤ t ≤ 2 p(2 − p) . The conjectures are based on numerical analysis (see [4, Sect. 5]). In Sect. 4, we will extend Connelly’s result to the full range 0 ≤ t ≤ 1. Our result demonstrates that for each 0 < p < 1 there is a unique 0 < t < 1/2 such that the extremal for ( p, t ) is not unique, thereby conﬁrming the above-mentioned 1 p conjectures. Brevig and Saksman [3] have recently studied the extremal problem (k) f (0) ( p) = sup Re : f p ≤ 1 k H k! for 0 < p < 1. It is observed in [3, Sect. 5.3] that ( p) = max ( p, t ).In k 0≤t ≤1 k particular, the maxima of ( p, t ) for 0 ≤ t ≤ 1is 1/ p p 2 ( p) = 1 − √ 2 p(2 − p) 1/ p and this is attained for t = (1 − p/2) . From the main result in [1], it is easy to see that t → ( p, t ) is a decreasing function from ( p, 0) = 1to ( p, 1) = 0 1 1 1 when 1 ≤ p ≤∞. Similarly, our main result shows that ( p, t ) is increasing from ( p, 0) = 1 to the maxima mentioned above, then decreasing to ( p, 1) = 0. 1 1 Figure 1 contains the plot of t → ( p, t ) for several values 0 < p ≤∞, which illustrates this difference between 0 < p < 1 and 1 ≤ p ≤∞. 123 O. F. Brevig et al. Fig. 1 Plot of the curves t → ( p, t ) for p = 1/2, p = 1, p = 2and p =∞ Another difference between 0 < p < 1 and 1 ≤ p ≤∞ appears when we consider k ≥ 2. Recall that in the latter case, we have ( p, t ) = ( p, t ) for every k ≥ 2 and k 1 every 0 ≤ t ≤ 1. In the former case, we only get from Theorem 1 that 1/ p−1 ( p, t ) ≤ ( p, t ) ≤ k ( p, t ). (3) 1 k 1 Theorem 1 also shows that the upper bound in (3) is attained if and only if t = 1, since trivially ( p, 1) = 0 for every 0 < p ≤∞. However, by adapting an example due to Hardy and Littlewood [7], it is easy to see that if 0 < p < 1 and 0 ≤ t < 1 are ﬁxed, then the exponent 1/ p − 1in(3) cannot be improved as k →∞.Inthe ﬁnal section of the paper, we present some evidence that the lower bound in (3) can be attained for sufﬁciently large t,if k ≥ 2 and 0 < p < 1 are ﬁxed. Organization The present paper is organized into ﬁve additional sections and one appendix. In Sect. 2, we collect some preliminary results pertaining to H and the structure of extremals for ( p, t ). Section 3 is devoted to F. Wiener’s trick and the proof of Theorem 1. A complete solution to the extremal problem ( p, t ) for 0 < p ≤∞ and 0 ≤ t ≤ 1 is presented in Sect. 4. In Sect. 5, we consider ( p, t ) for k ≥ 2 and 1 ≤ p ≤∞ and study when the extremal is unique. Section 6 contains some remarks on ( p, t ) for k ≥ 2 and 0 < p < 1. “Appendix A” contains the proof of a crucial lemma needed to resolve the extremal problem ( p, t ) for 0 < p < 1. 123 p F. Wiener’s Trick and an Extremal Problem for H 2 Preliminaries Recall that for 0 < p < ∞, the Hardy space H consists of the analytic functions f in D for which the limit of integral means 2π dθ i θ p f = lim | f (re )| 2π r →1 is ﬁnite. H is the space of bounded analytic functions in D, endowed with the norm f = sup | f (z)|. It is well-known (see e.g. [6]) that H is a Banach space |z|<1 when 1 ≤ p ≤∞ and a quasi-Banach space when 0 < p < 1. In the Banach space range 1 ≤ p ≤∞, the triangle equality is p p p f + g ≤ f +g . (4) H H H The Hardy space H is strictly convex when 1 < p < ∞, which means that it is impossible to attain equality in (4) unless g ≡ 0or f = λg for a non-negative constant λ. H is not strictly convex for p = 1 and p =∞, so in this case there are other ways to attain equality in (4). In the range 0 < p < 1, the triangle inequality takes the form p p p f + g ≤ f +g , (5) p p p H H H so here H is not even locally convex [5]. Our ﬁrst goal is to establish that the triangle inequality (5) is not attained unless f ≡ 0or g ≡ 0. This result is probably known to experts, but we have not found it in the literature. If f ∈ H for some 0 < p ≤∞, then the boundary limit function ∗ i θ i θ f (e ) = lim f (re ) (6) r →1 ∗ p p exists for almost every θ. Moreover, f ∈ L = L ([0, 2π ]) and 1/ p 2π dθ ∗ ∗ i θ p p f = f = f (e ) H L 2π ∗ i θ if 0 < p < ∞ and f ∞ = ess sup | f (e )|. For simplicity, we henceforth omit the asterisk and write f = f with the limit (6) in mind. Lemma 2 Fix 0 < p < 1 and suppose that f , g ∈ H .If p p p f + g = f +g p p p H H H then either f ≡ 0 or g ≡ 0. 123 O. F. Brevig et al. Proof We begin by looking at equality in the triangle inequality for L in the range 0 < p < 1. Here we have 2π dθ i θ i θ f + g = f (e ) + g(e ) 2π 2π dθ p p i θ p i θ p ≤ | f (e )| +|g(e )| = f +g . p p L L 2π p p p We used the elementary estimate |z + w| ≤|z| +|w| for complex numbers z,w and 0 < p < 1. It is easily veriﬁed that this estimate is attained if and only if zw = 0. Consequently, p p p f + g = f +g p p p L L L i θ i θ if and only if f (e )g(e ) = 0 for almost every θ.Itiswell-known (see[6,Thm.2.2]) that the only function h ∈ H whose boundary limit function (6) vanishes on a set of positive measure is h ≡ 0. Hence we conclude that either f ≡ 0or g ≡ 0. Let us next establish a standard result on the structure of the extremals for the extremal problem (1). The ﬁrst step is the following basic result. Lemma 3 If f ∈ H is extremal for ( p, t ), then f = 1. k H Proof Suppose that f ∈ H is extremal for ( p, t ) but that f p < 1. For ε> 0, k H set g(z) = f (z) + εz . Note that g(0) = f (0) = t for any ε> 0. If 1 ≤ p ≤∞, then p p g ≤ f + ε< 1 H H for sufﬁciently small ε> 0. If 0 < p < 1, then p p g ≤ f + ε < 1, p p H H again for sufﬁciently small ε> 0, so g < 1. In both cases we ﬁnd that (k) (k) g (0) f (0) = + ε, k! k! which contradicts the extremality of f for ( p, t ). k k Let (n ) denote a sequence of distinct non-negative integers and let (w ) j j j =1 j =1 denote a sequence of complex numbers. A special case of the Carathéodory–Fejér problem is to determine the inﬁmum of f over all f ∈ H which satisfy (n ) (0) = w , (7) n ! 123 p F. Wiener’s Trick and an Extremal Problem for H for j = 1,..., k. Set k = max n .If f is an extremal for the Carathéodory– 1≤ j ≤k j Fejér problem (7), then there are complex numbers |λ |≤ 1for j = 1,..., k and a constant C such that l k λ − z j 2/ p f (z) = C 1 − λ z (8) 1 − λ z j =1 j =1 for some 0 ≤ l ≤ k, and the strict inequality |λ | < 1 holds for 0 < j ≤ l.In(8) and in similar formulas to follow, we adopt the convention that in the case l = 0 the ﬁrst product is empty and considered to be equal to 1. For 1 ≤ p ≤∞, this result is independently due to Macintyre and Rogosinski [11] and Havinson [8], while in the range 0 < p < 1 the result is due to Kabaila [9]. An exposition of these results can be found in [6, Ch. 8] and [10, pp. 82–85], respectively. Using Lemma 3, we can establish that the extremals of the extremal problem ( p, t ) have to be of the same form. Lemma 4 If f ∈ H is extremal for ( p, t ), then there are complex numbers |λ |≤ 1 k j for j = 1,..., k and a constant C such that l k λ − z j 2/ p f (z) = C 1 − λ z . 1 − λ z j =1 j =1 for some 0 ≤ l ≤ k, and the strict inequality |λ | < 1 holds for 0 < j ≤ l. Proof Suppose that f is extremal for ( p, t ) and consider the Carathéodory–Fejér problem with conditions (k) f (0) f (0) = t and = ( p, t ). (9) k! We claim that f is an extremal for the Carathéodory–Fejér problem (9). If it is not, then there must be some f ∈ H with f < 1 which satisﬁes (9). However, this contradicts Lemma 3. Hence the extremal is of the stated form by (8). 3 F. Wiener’s Trick Recall from (2) that if f (z) = a z and ω = exp(2π i /k), then n k n≥0 k−1 ∞ kn W f (z) = f (ω z) = a z . k kn j =0 n=0 We begin by giving two examples showing that W f p = f p may occur for k H H f such that W f = f when p = 1or p =∞. 123 O. F. Brevig et al. 2k 1 Example 5 Let k ≥ 2 and consider f (z) = (1 + z) in H . By the binomial theorem, we ﬁnd that 2k 2k f (z) = z , n=0 2k k 2k W f (z) = 1 + z + z . Note that f = W f since k ≥ 2. By another application of the binomial theorem and a well-known identity for the central binomial coefﬁcient, we ﬁnd that k 2k 1/2 2 f 1 = f = = . n k n=0 Moreover, 2π 2k dθ i θ ikθ = W f (e ) e ≤W f k k k 2π by the triangle inequality. Hence 2k 2k ≤W f ≤ f = , 1 1 H H k k so W f 1 = f 1. H H k 2 k 2 ∞ Example 6 Let k ≥ 2 and consider f (z) = (1+ z ) − z(1− z ) in H . It is clear that k 2 W f (z) = (1 + z ) = f (z) since k ≥ 2. Moreover W f = 4. The supremum k k H is attained for z = ω for j = 0, 1,..., k − 1. We next compute 2 2 kθ kθ i θ ikθ i θ ikθ ikθ 2 i θ 2 f (e ) = 1 + e − e 1 − e = 4e cos + e sin . 2 2 Consequently, f = 4 and here the supremum is attained for z = ω for 2k j = 0, 1,..., 2k − 1. Proof of Theorem 1 It follows from the triangle inequality (4) that p p W f ≤ f (10) k H H for every f ∈ H if 1 ≤ p ≤∞. In the range 0 < p < 1, we get from the triangle inequality (5) the estimate 1/ p−1 W f p ≤ k f p (11) k H H 123 p F. Wiener’s Trick and an Extremal Problem for H for every f ∈ H . Combining (10) and (11), we have established that 1/ p−1 p p W f ≤ max k , 1 f . k H H This is trivially attained for f (z) = z when 1 ≤ p ≤∞. We need to show that the 1/ p−1 upper bound k cannot be improved when 0 < p < 1 to ﬁnish proof of the ﬁrst part of the theorem. −1/ p Let ε> 0 and consider f (z) = (z − (1 + ε)) . Clearly f p →∞ as ε ε H ε → 0 . Moreover 2π 1 dθ f = i θ |e − (1 + ε)| 2π 1 dθ 6 dθ ≤ + i θ 2 |e − (1 + ε)| 2π θ 2π |θ |<π/k |θ |≥π/k 1 dθ 6k ≤ + , i θ 2 |e − (1 + ε)| 2π π |θ |<π/k from which we conclude that 1 dθ f = + O(1). (12) i θ |e − (1 + ε)| 2π |θ |<π/k Furthermore, k−1 k−1 i (θ +2πl/k) f e dθ W f = k ε 2π |θ −2π j /k|<π/k j =0 l=0 k−1 dθ 6k − p i (θ +2π j /k) ≥ k f e 2π π |θ −2π j /k|<π/k j =0 − p+3 1 dθ 6k − p+1 = k − . i θ 2 |e − (1 + ε)| 2π π |θ |<π/k By (12) we ﬁnd that W f k ε p H 1− p lim ≥ k . ε→0 f ε p 1/ p−1 Hence, the constant k in (11) cannot be replaced by any smaller quantity. We next want to show that (a) and (b) holds. For a function f ∈ H , deﬁne p p f (z) = f (ω z) for j = 0, 1,..., k − 1 and recall that f = f . j H j H 1/ p−1 p p We begin with (a). Suppose that W f = k f , which we can refor- k H H mulate as p p p p f + f + ··· + f = f + f + ··· + f . p p p p 0 1 k−1 0 1 k−1 H H H H 123 O. F. Brevig et al. By Lemma 2, the triangle inequality can be attained if and only if at least k − 1ofthe k functions f are identically equal to zero. Evidently this is possible if and only if f ≡ 0. p p For (b), we suppose that f ∈ H is such that W f = f . We need to k H H prove that W f = f .If f ≡ 0 there is nothing to do. As in the proof of (a), we note p p that W f = f can be reformulated as k H H f + f + ··· + f p = f p + f p + ··· + f p . 0 1 k−1 H 0 H 1 H k−1 H p p Viewing H as a subspace of L , the strict convexity of the latter implies that there are non-negative constants λ for j = 1, 2,..., k − 1 such that f = f = λ f = ... = λ f . 0 1 1 k−1 k−1 We shall only look at f = λ f which for f (z) = a z is equivalent to 1 1 n n≥0 ∞ ∞ n n n a z = λ a ω z . n 1 n n=0 n=0 Using W on this identity we get ∞ ∞ kn kn a z = λ a z . kn 1 kn n=0 n=0 This is only possible if λ = 1or W f ≡ 0. The latter implies that f ≡ 0 since 1 k p p W f = f by assumption. Therefore we can restrict our attention to the k H H case λ = 1. For all integers n that are not a multiple of k, we now ﬁnd that a = λ ω a ⇒ a = 0, n 1 n n since λ = 1 and ω = 1. Hence W f = f as desired. 1 k p i θ Recall that a function f ∈ H is called inner if | f (e )|= 1 for almost every θ. We shall require the following simple result later on. Lemma 7 If both f and W f are inner functions, then f = W f. k k i θ i θ Proof Since |W f (e )|=| f (e )|= 1 for almost every θ, we get from (2) that k−1 k−1 1 1 i θ i θ i θ 1 = W f (e ) = f (e ) = | f (e )|, (13) k j j k k j =0 j =0 where f (z) = f (ω z). The equality on the right hand side of (13) is possible if and only if i θ i θ i θ f (e ) = f (e ) = ... = f (e ) 1 k−1 123 p F. Wiener’s Trick and an Extremal Problem for H for almost every θ. As in the proof of Theorem 1 (b), we ﬁnd that f = W f . 4 The Extremal Problem 8 (p, t) for 0 <p ≤∞ In the present section, we resolve the extremal problem (1) in the case k = 1 com- pletely. We begin with the case 1 ≤ p ≤∞ which has been solved by Beneteau and Korenblum [1]. We give a different proof of their result based on Lemma 4, mainly to illustrate the differences between the cases 0 < p < 1 and 1 ≤ p ≤∞. Theorem 8 (Beneteau–Korenblum) Fix 1 ≤ p ≤∞ and consider (1) with k = 1. −1/ p (i) If 0 ≤ t < 2 ,let α denote the unique real number in the interval 0 ≤ α< 1 2 −1/ p such that t = α(1 + α ) . Then 1 2 ( p, t ) = 1 + − 1 α , 1/ p 2 p 1 + α and the unique extremal is 2/ p α + z (1 + αz) f (z) = . 1/ p 1 + αz 2 1 + α −1/ p (ii) If 2 ≤ t ≤ 1,let β denote the unique real number in the interval 0 ≤ β ≤ 1 2 −1/ p such that t = (1 + β ) . Then 1 2β ( p, t ) = , 1/ p 2 p 1 + β and the unique extremal is 2/ p (1 + βz) f (z) = . 1/ p 1 + β Proof Note that since k = 1, there are only two possibilities for the extremals in Lemma 4.Theyare 2/ p α + z (1 + αz) f (z) = , 0 ≤ α< 1, (14) 1/ p 1 + αz 2 1 + α 2/ p (1 + βz) f (z) = , 0 ≤ β ≤ 1. (15) 1/ p 1 + β 123 O. F. Brevig et al. Here we have made α, β ≥ 0 by rotations. Note that if p =∞, then f does not depend on β. Moreover, t = f (0) = , (16) 2 1/ p (1 + α ) t = f (0) = . (17) 2 1/ p (1 + β ) For 1 ≤ p ≤∞ it is easy to verify that the function → (18) 2 1/ p (1 + α ) −1/ p is strictly increasing on 0 ≤ α< 1 and maps [0, 1) to [0, 2 ). Similarly, for 1 ≤ p < ∞ we ﬁnd that the function → (19) 2 1/ p (1 + β ) −1/ p is strictly decreasing on 0 ≤ β ≤ 1 and maps [0, 1] to [2 , 1]. Consequently, −1/ p if 0 ≤ t < 2 , then the unique extremal is (14) with α given by (16), and if −1/ p 2 ≤ t ≤ 1, then the unique extremal is (15) with β given by (17). The proof is completed by computing 1 2 1 2 f (0) = 1 + α − 1 = t + α − 1 , (20) 1/ p p α p 1 + α 1 2β 2β f (0) = = t , (21) 2 1/ p (1 + β ) p p to obtain the stated expressions for ( p, t ) in (i) and (ii), respectively. Deﬁne α and β as functions of t implicitly through (16) and (17). Then α is increas- −1/ p −1/ p ingon0 ≤ t < 2 and β is decreasing on 2 ≤ t ≤ 1. Inspecting the left hand side of (20) and (21), we extract the following result. Corollary 9 If 1 ≤ p ≤∞, then the function t → ( p, t ) is decreasing and takes the values [0, 1]. In the range 0 < p < 1 a more careful analysis is required. This is due to the fact that the function (18) is increasing on the interval 0 ≤ α ≤ α and decreasing on the interval α ≤ α< 1, where α = . (22) 2 − p −1/ p −1/ p 1/ p−1/2 Inspecting (16), we conclude that for each 2 < t < 2 p(2 − p) there are two possible α-values which give the same t = f (0).Let α denote the 1 1 123 p F. Wiener’s Trick and an Extremal Problem for H unique real number in the interval (0, 1) such that 1 + α = 2α . (23) −1/ p Note that α gives the value t = 2 in (16). Lemma 10 If α <α <α and α < α< 1 produce the same t = f (0) in (16), then 1 2 2 1 the quantity f (0) from (20) is maximized by α. Proof Since α and α give the same t = f (0) in (20), we only need to prove that 1 α 1 α + > + . (24) 2 2 α α α α 2 2 Fix α <α <α . The unique number α <ξ < 1 such that 1 2 2 1 α 1 ξ + = + 2 2 α ξ α α 2 2 is ξ = α /α. Since the function 1 x → + is increasing for x >α it is sufﬁcient to prove that ξ> α to obtain (24). Since 2 1/ p (1 + x ) is decreasing for x >α , we see that ξ> α if and only if α ξ α > ⇐⇒ > . 2 1/ p 2 1/ p 2 1/ p 1/ p (1 + α ) (1 + ξ ) (1 + α ) 2 1 + Here we used that α and α give the same t = f (0) in (16) on the left hand side and the identity ξ = α /α on the right hand side. We now substitute α = α x for 0 < x < 1 to obtain the equivalent inequality x 1 > . (25) 2 1/ p 1/ p (1 + α x ) 1 + Actually, we only need to consider (α /α ) < x < 1, but the same proof works for 1 2 1− p 0 < x < 1. We raise both sides of (25) to the power p, multiply by x and rearrange 123 O. F. Brevig et al. to get the equivalent inequality F (x)> 0 where 1− p 2 2− p F (x ) = x − x + α 1 − x . Recalling that α = p/(2 − p), we compute − p 1− p − p−1 − p F (x )= 1 − (1 − p)x − px and F (x )= p(1 − p)x − p(1 − p)x . Since F (1) = F (1) = 0, we get from Taylor’s theorem that for every 0 < x < 1 there is some x <η < 1 such that F (η) p(1 − p) 2 − p −1 2 F (x ) = (x − 1) = η η − 1 (x − 1) > 0, 2 2 which completes the proof. By Lemma 10, we now only need to compare f (0) from (20)for α ≤ α ≤ α 1 2 with f (0) from (21)for β such that f (0) = t = f (0). Inspecting (16) and (17), we 1 2 ﬁnd that α 1 1 + α = ⇐⇒ β = − 1. (26) 1/ p 1/ p p 2 2 α 1 + α 1 + β Next, we consider the equation f (0) = f (0) with β as in (26). Inspecting (20) and 1 2 (21) and dividing by t, we get the equation 1 2 2β 2 1 + α + α − 1 = = − 1. (27) α p p p α We square both sides, multiply by p and rearrange to ﬁnd that (27) is equivalent to the equation F (α) = 0, where 2 −2 2 2 − p 2− p F (α) = p α + 2 p(2 − p) + (2 − p) α − 4 α + α − 1 . (28) Suppose that α ≤ α ≤ α .If 1 2 • F (α) > 0, then f from (14) is the unique extremal for ( p, t ). p 1 1 • F (α) = 0, then f from (14) and f from (15) are extremals for ( p, t ). p 1 2 1 • F (α) < 0, then f from (15) is the unique extremal for ( p, t ). p 2 1 Note that any solutions of F (α) = 0 with 0 <α <α are of no interest since this p 1 implies that β> 1by(26). Similarly, any solutions of F (α) = 0 with α <α < 1 p 2 can be ignored due to Lemma 10. The following result shows that there is only one solution, which is in the pertinent range. 123 p F. Wiener’s Trick and an Extremal Problem for H Lemma 11 Let F be as in (28). The equation F (α) = 0 has a unique solution, p p denoted α , on the interval (0, 1). Moreover, (a) if 0 <α <α , then F (α) > 0. p p (b) if α <α < 1, then F (α) < 0. p p (c) α <α <α where α and α are from (23) and (22), respectively. 1 p 2 1 2 The proof of Lemma 11 is a rather laborious calculus exercise, which we postpone to “Appendix A” below. Let α be as in Lemma 11 and deﬁne t = . (29) 1/ p 1 + α −1/ p −1/ p 1/ p−1/2 Note that 2 < t < 2 p(2 − p) by the fact that α <α <α . p 1 p 2 By the analysis above, Lemma 10 and Lemma 11, we obtain the following version of Theorem 8 in the range 0 < p < 1. Theorem 12 Fix 0 < p < 1 and consider (1) with k = 1. Let t be as in (29) and set α = p/(2 − p). (i) If 0 ≤ t ≤ t ,let α denote the unique real number in the interval 0 ≤ α< α p 2 2 −1/ p such that t = α(1 + α ) . Then 1 2 ( p, t ) = 1 + − 1 α , 1/ p 2 p 1 + α and an extremal is 2/ p α + z (1 + αz) f (z) = . 1/ p 1 + αz 1 + α (ii) If t ≤ t ≤ 1,let β denote the unique real number in the interval 0 ≤ β ≤ 1 such 2 −1/ p that t = (1 + β ) . Then 1 2β ( p, t ) = , 1/ p 2 p 1 + β and an extremal is 2/ p (1 + βz) f (z) = . 1/ p 1 + β The extremals are unique for 0 ≤ t = t ≤ 1. The only extremals for ( p, t ) are p 1 p the functions given in (i) and (ii). 123 O. F. Brevig et al. Fig. 2 Plot of the curve p → t . Points ( p, t ) above and below the curve correspond to the cases (i) and (ii) −1/ p −1/ p 1/ p−1/2 of Theorem 12, respectively. The estimates 2 < t < 2 p(2 − p) are represented by dotted curves. In the shaded area and in the range 1/2 ≤ t ≤ 1, Theorem 12 is originally due to Connelly [4] Theorem 12 extends [4, Theorem 4.1] to general 0 ≤ t ≤ 1. The analysis in [4]is similar to ours, and we are able to also identify the extremals in the range −1/ p −1/ p 1/ p−1/2 2 ≤ t ≤ 2 p(2 − p) due to Lemma 10 and Lemma 11. It is also demonstrated in [4, Thm. 4.1] that when p = 1/2 there must exist at least one value of 0 < t < 1 for which the extremal is not unique. Theorem 12 shows that there is precisely one such t and that this observation is not speciﬁc to p = 1/2, but in fact holds for any 0 < p < 1. Figure 2 shows the value t for which the extremal is not unique as a function of p. Inspecting Theorem 12, we get the following result similarly to how we extracted Corollary 9 from Theorem 8. Corollary 13 If 0 < p < 1, then the function t → ( p, t ) is increasing from ( p, 0) = 1 to 1/ p 1/ p p p 2 p, 1 − = 1 − √ 2 2 p(2 − p) and then decreasing to ( p, 1) = 0. 123 p F. Wiener’s Trick and an Extremal Problem for H 5 The Extremal Problem 8 (p, t) for k ≥ 2and 1 ≤ p ≤∞ We begin by recalling how F. Wiener’s trick was used in [1] to obtain the solution to the extremal problem ( p, t ) for k ≥ 2 from Theorem 8. Theorem 14 (Benetau–Korenblum) Let k ≥ 2 be an integer. For every 1 ≤ p ≤∞ and every 0 ≤ t ≤ 1, ( p, t ) = ( p, t ). k 1 If f is the extremal function for ( p, t ), then f (z) = f (z ) is an extremal function 1 1 k 1 for ( p, t ). p p Proof Suppose that f is an extremal for ( p, t ). Since W f ≤ f , k k H H (k) (k) f (0) (W f ) (0) f (0) = W f (0) and = , k! k! we conclude that W f is also an extremal for ( p, t ). Thus we may restrict our k k k p attention to extremals f of the form f (z) = f (z ) for f ∈ H . The stated claims k k p p now follow at once from Theorem 8, since f = f . k H H The purpose of the present section is to answer the following question. For which trios k ≥ 2, 1 ≤ p ≤∞ and 0 ≤ t ≤ 1 is the extremal for ( p, t ) unique? Note that while Theorem 14 provides an extremal f (z) = f (z ) where f denotes the k 1 1 extremal from (the statement of) Theorem 8, it might not be unique. In the case 1 < p < ∞ it follows at once from Theorem 1 (b) that this extremal is unique, although it is perhaps easier to use the strict convexity of H and Lemma 3 directly. Since H is not strictly convex for p = 1 and p =∞, these cases require further analysis. Note that the case (a) below is certainly known to experts as a conse- quence of the general theory developed in [8, 11, 14]. Theorem 15 Consider the extremal problem (1) for k ≥ 2 and 1 ≤ p ≤∞. (a) If 1 < p ≤∞, then the unique extremal is f (z) = f (z ). k 1 (b) If p = 1 and 1/2 ≤ t ≤ 1, then the unique extremal is f (z) = f (z ). k 1 (c) If p = 1 and 0 ≤ t < 1/2, then the extremals are the functions of the form f (z) = C λ − z 1 − λ z j j j =1 (k) with |λ |≤ 1 such that f 1 = 1,f (0) = t and f (0)> 0. Proof of Theorem 15(a) In view of the discussion above, we need only consider the case p =∞. By Lemma 4, we know that any extremal must be of the form λ − z i θ f (z) = e (30) 1 − λ z j =1 123 O. F. Brevig et al. for some 0 ≤ l ≤ k, constants λ ∈ D and θ ∈ R.If f is extremal for (∞, t ), then j k so is W f by Theorem 14. Consequently, W f is also of the form (30). In particular, k k since both f and W f are inner, we get from Lemma 7 that f = W f .Fromthe k k deﬁnition of W , we know that f (z) = W f (z) = g(z ) for some analytic function k k g. This shows that the only possibility in (30)is λ − z i θ f (z) = e 1 − λz for some λ ∈ D and θ ∈ R. The unique extremal has θ = π and λ =−t. Proof of Theorem 15(b) Suppose that f is extremal for (1, t ). By rotations, we extend our scope to functions f such that | f (0)|= t. In this case, we can use Lemma 4 and write f = gh for l k g(z) = C (z + α ) (1 + α z), j j j =1 j =l+1 h(z) = C (1 + α z). j =1 The constant C > 0 satisﬁes j j j 1 2 k = α α ··· α , 1 2 k j =0 j + j +···+ j = j 1 2 k where j , j ,..., j take only the values 0 and 1. Evidently g 2 =h 2 = 1. Set 1 2 k H H A =|α ··· α | and B =|α ··· α |. By keeping only the terms j = 0 and j = k l 1 l l l+1 k we obtain the trivial estimate 2 2 2 ≥ 1 +|α α ··· α | = 1 + A B . (31) 1 2 k l l We will adapt an argument due to F. Riesz [13] to get some additional information on the relationship between g and h. Write 2k k k j j j f (z) = a z , g(z) = b z and h(z) = c z j j j j =0 j =0 j =0 and note that |b |= t /|c |= t /C. By the Cauchy product formula we ﬁnd that 0 0 k k c b k 0 a = b c = t + b c . (32) k j k− j j k− j C |b | j =0 j =1 123 p F. Wiener’s Trick and an Extremal Problem for H Suppose that g ∈ H satisﬁes | g(0)|= t /C and g ≤ 1. Deﬁne f = gh.The Cauchy–Schwarz inequality shows that f 1 ≤ 1, so the extremality of f implies that | a |≤|a |. Inspecting (32) and using the Cauchy–Schwarz inequality, we ﬁnd k k that the optimal g must therefore satisfy t k 1 − t c C j g(z) = + c z , (33) k− j C |c | 1 −|c | k k j =1 whereweusedthat h 2 = 1. Using that c = C, we compare the coefﬁcients for z in (33) with the deﬁnition of g, to ﬁnd that t k t 1 − 1 − 2 2 C C 2 C = C α ⇒ = B . 2 2 1 −|c | 1 −|c | k k j =l+1 2 2 2 2 2 Next we insert t = C A from the deﬁnition of f = gh and |c | = C A B from l k l l the deﬁnition of h to obtain 2 2 2 2 2 2 1 − C A (1 − B )(1 − C A (1 + B )) l l l l = B ⇐⇒ = 0. (34) 2 2 2 2 2 2 1 − C A B 1 − C A B l l l l The additional information we require is encoded in the equation on the right hand side of (34). Suppose that l ≥ 1. Evidently A < 1, since |α | < 1for j = 1,..., l by Lemma 4. l j It follows that the second factor on the right hand side of (34) can never be 0, since the trivial estimate (31)impliesthat 1 1 C ≤ < . (35) 2 2 2 2 1 + A B A (1 + B ) l l l l From the right hand side of (34) we thus ﬁnd that B = 1, which shows that C < 2 2 1/(2 A ) by (35). Since t = C A , we conclude that 0 ≤ t < 1/2. By the contrapositive, we have established that if 1/2 ≤ t ≤ 1, then the extremal for (1, t ) has l = 0. In this case A = 1 by deﬁnition, which shows that C = t. k 0 The right hand side of (34) becomes 2 2 (1 − B )(1 − t (1 + B )) 0 0 = 0, 1 − tB so either B = 1or B = 1/t − 1. Returning to the deﬁnition of h we ﬁnd that 2 2 2 |c | = t and |c | = tB . Consequently, 0 k k−1 2 2 2 1 =h = t (1 + B ) + |c | . 2 j j =1 123 O. F. Brevig et al. Since 1/2 ≤ t ≤ 1, we ﬁnd that both B = 1 and B = 1/t − 1 will imply that c = 0 0 j for j = 1,..., k − 1. Thus h(z) = t + 1 − tz . When l = 0we have g = h, which shows that the unique extremal is √ √ 2 f (z) = t + 1 − tz , which is of the form f (z) = f (z ) as claimed. k 1 Proof of Theorem 15(c) In the case 0 ≤ t < 1/2, we know from Theorem 8 and Theorem 14 that (1, t ) = 1. See also Figure 1. The stated claim follows from Exercise 3 on page 143 of [6] by scaling and rotating the function f (z) = C λ − z 1 − λ z j j j =1 (k) to satisfy the conditions f 1 = 1, f (0)> 0 and f (0)> 0. If the resulting function satisﬁes f (0) = t, then it is an extremal for ( p, t ) and every extremal is obtained in this way. (This can be established similarly to the case (b) above.) 6 The Extremal Problem 8 (p, t) for k ≥ 2and 0 <p< 1 The purpose of this ﬁnal section is to record some observations pertaining to the extremal problem (1) in the unresolved case k ≥ 2 and 0 < p < 1. Suppose that k ≥ 0 and consider the related extremal problem (k) f (0) ( p) = sup Re : f ≤ 1 . k H k! Evidently, ( p) = 1 for every 0 < p ≤∞ and the unique extremal is f (z) = 1. Recall (from [3]or[9]) that the extremals for satisfy a structure result identical to Lemma 4. Note that the parameter l in Lemma 4 describes the number of zeroes of the extremal in D. Conjecture 1 from [3, Sect. 5] states that the extremal for ( p) does not vanish in D when 0 < p < 1. The conjecture has been veriﬁed in the cases k = 0, 1, 2 and for (k, p) = (3, 2/3). Let us now suppose that k ≥ 1. There are two obvious connections between the extremal problems and . Namely, k k ( p, 0) = ( p) and max ( p, t ) = ( p). k k−1 k k 0≤t ≤1 Assume that the above-mentioned conjecture from [3] holds. This assumption yields that the extremal for ( p, 0) has precisely one zero in D and the extremal for the t which maximizes ( p, t ) does not vanish in D. Note that the extremal for ( p, 1), k k which is f (z) = 1, does not vanish in D. 123 p F. Wiener’s Trick and an Extremal Problem for H Question 1 Suppose that 0 < p < 1. Is it true that the extremal for ( p, t ) has at most one zero in D? We have veriﬁed numerically that the question has an afﬁrmative answer for k = 2. Note that for 1 < p ≤∞, the extremal for ( p, t ) either has 0 or k zeroes in D by Theorem 15 (a). In the case p = 1, the extremal may have anywhere from 0 to k zeroes by Theorem 15 (b) and (c). As mentioned in the introduction, Theorem 1 yields the estimates 1/ p−1 ( p, t ) ≤ ( p, t ) ≤ k ( p, t ). 1 k 1 The upper bound is only attained if ( p, t ) = 0 which happens if and only if t = 1. Of course, since ( p, 1) = 0 the lower bound is also attained. Question 2 Fix k ≥ 2 and 0 < p < 1. Is there some t such that ( p, t ) = ( p, t ) 0 k 1 holds for every t ≤ t ≤ 1? By a combination of numerical and analytical computations, we have strong evi- dence that the question has an afﬁrmative answer for k = 2 and that in this case 1/ p t = 1 + . 2 − p Let us close by brieﬂy explaining our reasoning. We began by considering the case l = 0 in Lemma 4. Setting 2/ p−1 f = gh and arguing as in the proof of Theorem 15 (b) (see also [3]), we found if t ≥ t , then the only possible extremal for ( p, t ) with l = 0 is of the form f (z) = f (z ) 2 2 1 where f is the corresponding extremal for ( p, t ). Next, if l = 2 then (as in the case 1 1 −1/ p 1/ p−1/2 k = 1) we can only obtain t-values in the range 0 ≤ t ≤ 2 p(2 − p) . However, since −1/ p 1/ p−1/2 2 p(2 − p) < t for 0 < p < 1 we can ignore the case l = 2. The case l = 1 was excluded by numerical computations. Acknowledgements The authors extend their gratitude to Eero Saksman for a helpful discussion pertaining to Theorem 1. They also thank the referee for a careful reading of the paper. Funding Open access funding provided by NTNU Norwegian University of Science and Technology (incl St. Olavs Hospital - Trondheim University Hospital). Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included 123 O. F. Brevig et al. in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. Appendix A: Proof of Lemma 11 We will frequently appeal to the following corollary of Rolle’s theorem: Suppose that f is continuously differentiable on [a, b] and that f (x ) = 0 has precisely n solutions on (a, b). Then f (x ) = 0 can have at most n + 1 solutions on [a, b]. We are interested in solutions of the equation F (α) = 0 on the interval (0, 1), where we recall from (28) that 2 −2 2 2 − p 2− p F (α) = p α + 2 p(2 − p) + (2 − p) α − 4 α + α − 1 . The initial step in the proof of Lemma 11 is to identify the critical points of F on the interval 0 <α < 1. It turns out that there is only one. Lemma 16 Fix 0 < p < 1 and let F be as in (28). The equation F (α) = 0 has the unique solution α = α = 2 − p on 0 <α < 1. Proof We begin by computing 2 −3 2 − p−1 1− p F (α) =−2 p α + 2(2 − p) α + 4 pα − 4(2 − p)α . The solutions of the equation F (α) = 0on 0 <α < 1 do not change if we multiply 1+ p both sides by α /(4 − 2 p). Hence, we consider the equation G (α) = 0, where 1+ p 2 α p 2 p p−2 2+ p 2 G (α) = F (α) =− α + (2 − p)α + − 2α . 2(2 − p) 2 − p 2 − p Evidently, 2 p−4 2 p G (α) = α p α + (4 − p )α − 4 , 2 p−4 2 p and the sign of G (α) is the same as the sign of p α + (4 − p )α − 4. Since d 4 p − p 2 p−4 2 p p α + (4 − p )α − 4 = 0 ⇐⇒ α = , dα 4 − p 123 p F. Wiener’s Trick and an Extremal Problem for H and since G (1) = 0, we conclude that G changes sign at most once on 0 <α < 1. p p Since G (0) =−∞, this means that G (α) = 0 can have at most two solutions on p p (0, 1]. Hence F (α) = 0 can have at most two solutions on (0, 1]. It is easy to verify that these solutions are and α = 1, α = 2 − p and hence the proof is complete. We next want to demonstrate that F (α )> 0 and F (α )< 0 where α and α α 1 α 2 1 2 are from (23) and (22), respectively. Lemma 17 Fix 0 < p < 1.If α = p/(2 − p), then F (α )< 0. 2 p 2 Proof We begin reformulating the inequality F (α )< 0as H ( p)> 0, for p 2 2 − p 2 p/2 (2− p)/2 H ( p) =− α F (α ) = 2 − 1 + 2 p − p p (2 − p) . p 2 Since we have H (0) = H (1) = 0, it is sufﬁcient to prove that the function H has precisely one critical point on 0 < p < 1 and that it is strictly positive for some 0 < p < 1. We ﬁrst check that 3/4 16 − 7 · 3 H (1/2) = > 0. We then compute 1 + 2 p − p p p/2 (2− p)/2 H ( p) =− p (2 − p) 2(1 − p) + log . 2 2 − p The ﬁrst factor is non-zero, so we therefore need to check that the equation I ( p) = 0 has only one solution on 0 < p < 1, where 4(1 − p) p I ( p) = + log . 1 + 2 p − p 2 − p We compute 2 2 2 −4 3 − 2 p + p 2 2(1 − p) 3 p − 6 p + 1 I ( p) = + = . 2 2 2 p(2 − p) 2 1 + 2 p − p p(2 − p) 1 + 2 p − p Hence I ( p) = 0 has the unique solution p = 1 − 2/3 on the interval 0 < p < 1. Noting that I (0) =−∞ and I (1) = 0, we conclude by verifying that √ √ I ( p ) = 6 + log 5 − 2 6 > 0 123 O. F. Brevig et al. which demonstrates that I ( p) = 0 has a unique solution on 0 < p < 1. Lemma 18 Fix 0 < p < 1. Let α denote the unique solution of the equation 1 − p 2 2α + α = 0 on the interval (0, 1). Then F (α )> 0. p 1 − p 2− p Proof Using the equation deﬁning α , we see that α + α − 1 = 1. Hence, 1 1 2 2 F (α ) = + 2 p(2 − p) + (2 − p) α − 4 p 1 p 1 = + α (2 − p) + 2 − 1 ( p − α (2 − p)) . 1 1 α α 1 1 The ﬁrst two factors are strictly positive for every 0 <α < 1 and every 0 < p < 1. Consequently, F (α )> 0 if and only if α < p/(2 − p). The function p 1 1 p 2 J (α) = 1 − 2α + α 2− p satisﬁes J (0) = 1 and J (1) = 0. Moreover, J is strictly decreasing on (0, p ) p p p 2− p and strictly increasing on ( p , 1). Since α is the unique solution to J (α) = 0for 1 p 0 <α < 1, the desired inequality α < p/(2 − p) is equivalent to p 2 p p p 0 > J = 1 − 2 + . 2 − p 2 − p 2 − p In order to establish this inequality, we multiply by (2 − p) /2 on both sides to get the equivalent inequality K ( p)< 0, where 2 p 2− p K ( p) = 2 − 2 p + p − p (2 − p) . Our plan is to use Taylor’s theorem to write K (η) K ( p) = K (1) + K (1)( p − 1) + ( p − 1) where 0 < p <η < 1. The claim will follow if we can prove that K (1) = K (1) = 0 and K ( p)< 0for 0 < p < 1. Hence we compute p 2− p K ( p) =−2 + 2 p − p (2 − p) log , 2 − p p 2 p 2− p 2 K ( p) = 2 − p (2 − p) log + . 2 − p p(2 − p) Evidently, K (1) = K (1) = K (1) = 0. Hence we are done if we can prove that K is strictly increasing on 0 < p < 1. This will follow once we verify that both p 2 p 2− p 2 p (2 − p) and log + 2 − p p(2 − p) 123 p F. Wiener’s Trick and an Extremal Problem for H are strictly positive and strictly decreasing on 0 < p < 1. Strict positivity is obvious. The ﬁrst function is strictly decreasing since d p p 2− p p 2− p p (2 − p) = p (2 − p) log dp 2 − p and log( p/(2 − p)) < 0for 0 < p < 1. For the second function, we check that d p 2 4 p p p − 1 log + = log + < 0, 2 2 dp 2 − p p(2 − p) p 2 − p 2 − p (2 − p) where for the ﬁnal inequality we have again used that log( p/(2 − p)) < 0. We can ﬁnally wrap up the proof of Lemma 11. Proof of Lemma 11 By Lemma 16 we know that F (α) = 0 has precisely one solution for 0 <α < 1. Since F (0) =∞ and F (1) = 0, this implies that the equation p p F (α) = 0 can have at most one solution on the interval (0, 1). Lemma 17 shows that there is exactly one solution, since F (α )< 0. Let α denote this solution. Inspecting p 2 p the endpoints again, we ﬁnd that F (α) > 0for 0 <α <α and F (α) < 0for p p p α <α < 1. Using Lemma 17 again we conclude that α <α , while the inequality p p 2 α <α follows similarly from Lemma 18. 1 p References 1. 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Havinson, S.Y.: On some extremal problems of the theory of analytic functions. Moskov. Gos. Univ. Uˇcenye Zapiski Matematika 148(4), 133–143 (1951) 9. Kabaila, V.: On some interpolation problems in the class H for p < 1. Soviet Math. Dokl. 1, 690–692 (1960) 10. Khavinson, S.Y.: Two papers on extremal problems in complex analysis. Am. Math. Soc. Transl. 2(129), 1–114 (1986) 11. Macintyre, A.J., Rogosinski, W.W.: Extremum problems in the theory of analytic functions. Acta Math. 82, 275–325 (1950) 12. Martín, M.J., Sawyer, E.T., Uriarte-Tuero, I., Vukotic, ´ D.: The Krzyz˙ conjecture revisited. Adv. Math. 273, 716–745 (2015) 13. Riesz, F.: Über Potenzreihen mit vorgeschriebenen Anfangsgliedern. Acta Math. 42(1), 145–171 (1920) 14. Rogosinski, W.W., Shapiro, H.S.: On certain extremum problems for analytic functions. Acta Math. 90, 287–318 (1953) 123 O. F. Brevig et al. 15. Sarason, D.: Complex Function Theory, 2nd edn. American Mathematical Society, Providence (2007) Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional afﬁliations.
Computational Methods and Function Theory – Springer Journals
Published: Sep 12, 2022
Keywords: Hardy spaces; Extremal problems; Coefficient estimates; Primary 30H10; Secondary 42A05
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