F. Wiener's trick and an extremal problem for $H^p$

For $0<p \leq \infty$, let $H^p$ denote the classical Hardy space of the unit disc. We consider the extremal problem of maximizing the modulus of the $k$th Taylor coefficient of a function $f \in H^p$ which satisfies $\|f\|_{H^p}\leq1$ and $f(0)=t$ for some $0 \leq t \leq 1$. 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.


Introduction
Let H p 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 By a standard normal families argument, there are extremals f ∈ H p 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 p which includes (1) has been developed by Havinson [8], Kabaila [9], Macintyre-Rogosinski [11] and Rogosinski-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 Φ k (2, t) = √ 1 − t 2 and that the unique extremal is f (z) = t + √ 1 − t 2 z k . Similarly, the Schwarz-Pick inequality (see e.g. [15,VII.17.3]) shows that Φ 1 (∞, t) = 1 − t 2 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) in the range 1 ≤ p ≤ ∞. We will enunciate their results in Section 4 and Section 5, but for now we present a brief account of their approach.
The first step in [1] is to compute Φ 1 (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 p functions. It follows from the argument that the extremal function thusly obtained is unique.
The second step in [1] is to show that Φ k (p, t) = Φ 1 (p, t) for every k ≥ 2 using a trick attributed to F. Wiener [2], which we shall now recall. Set ω k = exp(2πi/k) and suppose that f (z) = n≥0 a n z n . F. Wiener's trick is based on the transform a kn z kn .
The triangle inequality yields that Hence, if f 1 is an extremal function for Φ 1 (p, t), then f k (z) = f 1 (z k ) is an extremal function for Φ k (p, t) and consequently Φ k (p, t) = Φ 1 (p, t). Note that this argument does not guarantee that the extremal f k is unique for Φ k (p, t).
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 k denote the F. Wiener transform (2). The inequality

equality is attained if and only if
(a) f ≡ 0 when 0 < p < 1, 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 Section 3, we also present examples of functions in H 1 and H ∞ which attain equality in Theorem 1, but for which W k f = f . However, we will conversely establish that if both f and W k f are inner functions, then f = W k f .
To illustrate the role played by the F. Wiener transform in various coefficient related extremal problems, we first recall that the estimate W k f ∞ ≤ f ∞ was originally used by F. Wiener to resolve a problem posed by H. Bohr [2] and compute the so-called Bohr radius for H ∞ . We also know from [12,Sec. 1.7] that the Krzyż 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 k f . As far as we are aware, the Krzyż conjecture remains open for k ≥ 6.
Theorem 1 shows that the extremal for Φ k (p, t) is unique when 1 < p < ∞. We shall see in Section 5 that the extremal problem Φ k (p, t) with k ≥ 2 and 1 ≤ p ≤ ∞ has a unique extremal except for when p = 1 and 0 ≤ t < 1/2.
In Section 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 p < 1/2 such that the extremal for Φ 1 (p, t p ) is not unique, thereby confirming the above-mentioned conjectures.
Another difference between 0 < p < 1 and 1 ≤ p ≤ ∞ appears when we consider k ≥ 2. Recall that in the latter case, we have Φ k (p, t) = Φ 1 (p, t) for every k ≥ 2 and every 0 ≤ t ≤ 1. In the former case, we only get from Theorem 1 that Theorem 1 also shows that the upper bound in (3) is attained if and only if t = 1, since trivially Φ 1 (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 − 1 in (3) cannot be improved as k → ∞. In the 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 Section 2, we collect some preliminary results pertaining to H p and the structure of extremals for Φ k (p, t). Section 3 is devoted to F. Wiener's trick and the proof of Theorem 1. A complete solution to the extremal problem Φ 1 (p, t) for 0 < p ≤ ∞ and 0 ≤ t ≤ 1 is presented in Section 4. In Section 5, we consider Φ k (p, t) for k ≥ 2 and 1 ≤ p ≤ ∞ and study when the extremal is unique. Section 6 contains some remarks on Φ k (p, t) for k ≥ 2 and 0 < p < 1. Appendix A contains the proof of a crucial lemma needed to resolve the extremal problem Φ 1 (p, t) for 0 < p < 1.
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.

Preliminaries
Recall that for 0 < p < ∞, the Hardy space H p consists of the analytic functions f in D for which the limit of integral means is finite. H ∞ is the space of bounded analytic functions in D, endowed with the norm f H ∞ = sup |z|<1 |f (z)|. It is well-known (see e.g. [6]) that H p is a Banach space when 1 ≤ p ≤ ∞ and a quasi-Banach space when 0 < p < 1.
In the Banach space range 1 ≤ p ≤ ∞, the triangle equality is The Hardy space H p is strictly convex when 1 < p < ∞, which means that it is impossible to attain equality in (4) unless g ≡ 0 or f = λg for a non-negative constant λ. H p 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 so here H p is not even locally convex [5]. Our first goal is to establish that the triangle inequality (5) is not attained unless f ≡ 0 or g ≡ 0. This result is probably known to experts, but we have not found it in the literature. If f ∈ H p for some 0 < p ≤ ∞, then the boundary limit function exists for almost every θ. Moreover, f * ∈ L p = L p ([0, 2π]) and 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 p . If Proof. We begin by looking at equality in the triangle inequality for L p in the range 0 < p < 1.
Here we have We used the elementary estimate |z + w| p ≤ |z| p + |w| p for complex numbers z, w and 0 < p < 1. It is easily verified that this estimate is attained if and only if if and only if f (e iθ )g(e iθ ) = 0 for almost every θ. It is well-known (see [6,Thm. 2.2]) that the only function h ∈ H p whose boundary limit function (6) vanishes on a set of positive measure is h ≡ 0. Hence we conclude that either f ≡ 0 or 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.
, again for sufficiently small ε > 0, so g H p < 1. In both cases we find that which contradicts the extremality of f for Φ k (p, t).
Let (n j ) k j=1 denote a sequence of distinct non-negative integers and let (w j ) k j=1 denote a sequence of complex numbers. A special case of the Carathéodory-Fejér problem is to determine the infimum of f H p over all f ∈ H p which satisfy If f is an extremal for the Carathéodory-Fejér problem (7), then there are complex numbers |λ j | ≤ 1 for j = 1, . . . , k and a constant C such that for some 0 ≤ l ≤ k, and the strict inequality |λ j | < 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-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 Φ k (p, t) have to be of the same form.
Proof. Suppose that f is extremal for Φ k (p, t) and consider the Carathéodory-Fejér problem with conditions We claim that f is an extremal for the Carathéodory-Fejér problem (9). If it is not, then there must be some f ∈ H p with f H p < 1 which satisfies (9). However, this contradicts Lemma 3. Hence the extremal is of the stated form by (8).

We begin by giving two examples showing that
Example 5. Let k ≥ 2 and consider f (z) = (1 + z) 2k in H 1 . By the binomial theorem, we find that By another application of the binomial theorem and a well-known identity for the central binomial coefficient, we find that Consequently, f H ∞ = 4 and here the supremum is attained for z = ω j 2k for j = 0, 1, . . . , 2k − 1.
Proof of Theorem 1. It follows from the triangle inequality (4) that In the range 0 < p < 1, we get from the triangle inequality (5) the estimate (10) and (11), we have established that This is trivially attained for f (z) = z k when 1 ≤ p ≤ ∞. We need to show that the upper bound k 1/p−1 cannot be improved when 0 < p < 1 to finish proof of the first part of the theorem. Let ε > 0 and consider from which we conclude that Furthermore, By (12) we find that Hence, the constant k 1/p−1 in (11) cannot be replaced by any smaller quantity.
We next want to show that (a) and (b) holds. For a function f ∈ H p , define f j (z) = f (ω j k z) for j = 0, 1, . . . , k − 1 and recall that f H p = f j H p . We begin with (a). Suppose that W k f H p = k 1/p−1 f H p , which we can reformulate as By Lemma 2, the triangle inequality can be attained if and only if at least k − 1 of the k functions f j are identically equal to zero. Evidently this is possible if and only if f ≡ 0. For We need to prove that W k f = f . If f ≡ 0 there is nothing to do. As in the proof of (a), we note that W k f H p = f H p can be reformulated as Viewing H p as a subspace of L p , the strict convexity of the latter implies that there are non-negative constants λ j for j = 1, 2, . . . , k − 1 such that We shall only look at f = λ 1 f 1 which for f (z) = n≥0 a n z n is equivalent to ∞ n=0 a n z n = λ 1 ∞ n=0 a n ω n k z n .
Using W k on this identity we get This is only possible if λ 1 = 1 or W k f ≡ 0. The latter implies that f ≡ 0 since W k f H p = f H p by assumption. Therefore we can restrict our attention to the case λ 1 = 1. For all integers n that are not a multiple of k, we now find that a n = λ 1 ω n k a n =⇒ a n = 0, since λ 1 = 1 and ω n k = 1. Hence W k f = f as desired.
Recall that a function f ∈ H p is called inner if |f (e iθ )| = 1 for almost every θ. We shall require the following simple result later on.

Proof.
Since |W k f (e iθ )| = |f (e iθ )| = 1 for almost every θ, we get from (2) that The equality on the right hand side of (13) is possible if and only if f (e iθ ) = f 1 (e iθ ) = · · · = f k−1 (e iθ ) for almost every θ. As in the proof of Theorem 1 (b), we find that f = W k f .

The extremal problem
In the present section, we resolve the extremal problem (1) in the case k = 1 completely. 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 ≤ ∞.

Lemma 10.
If α 1 < α < α 2 and α 2 < α < 1 produce the same t = f 1 (0) in (16), then the quantity f ′ 1 (0) from (20) is maximized by α. Proof. Since α and α give the same t = f 1 (0) in (20), we only need to prove that Fix α 1 < α < α 2 . The unique number α 2 < ξ < 1 such that is increasing for x > α 2 it is sufficient to prove that ξ > α to obtain (24). Since x → x (1 + x 2 ) 1/p is decreasing for x > α 2 , we see that ξ > α if and only if Here we used that α and α give the same t = f 1 (0) in (16) on the left hand side and the identity ξ = α 2 2 /α on the right hand side. We now substitute α = α 2 √ x for 0 < x < 1 to obtain the equivalent inequality Actually, we only need to consider (α 1 /α 2 ) 2 < x < 1, but the same proof works for 0 < x < 1. We raise both sides of (25) to the power p, multiply by x 1−p and rearrange to get the equivalent inequality F (x) > 0 where 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 which completes the proof.
By Lemma 10, we now only need to compare f ′ (16) and (17), we find that Next, we consider the equation f ′ 1 (0) = f ′ 2 (0) with β as in (26). Inspecting (20) and (21) and dividing by t, we get the equation We square both sides, multiply by p 2 and rearrange to find that (27) is equivalent to the equation F p (α) = 0, where (14) is the unique extremal for Φ 1 (p, t).
The proof of Lemma 11 is a rather laborious calculus exercise, which we postpone to Appendix A below. Let α p be as in Lemma 11 and define Note that 2 −1/p < t p < 2 −1/p √ p(2 − p) 1/p−1/2 by the fact that α 1 < α p < α 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 p be as in (29) and and an extremal is and an extremal is The extremals are unique for 0 ≤ t = t p ≤ 1. The only extremals for Φ 1 (p, t p ) are the functions given in (i) and (ii).
Theorem 12 extends [4, Thm. 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 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 p 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.
and then decreasing to Φ 1 (p, 1) = 0. Figure 2. Plot of the curve p → t p . Points (p, t) above and below the curve correspond to the cases (i) and (ii) of Theorem 12, respectively. The estimates 2 −1/p < t p < 2 −1/p √ p(2 − p) 1/p−1/2 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]. 5. The extremal problem Φ k (p, t) for k ≥ 2 and 1 ≤ p ≤ ∞ We begin by recalling how F. Wiener's trick was used in [1] to obtain the solution to the extremal problem Φ k (p, t) for k ≥ 2 from Theorem 8.
If f 1 is the extremal function for Φ 1 (p, t), then f k (z) = f 1 (z k ) is an extremal function for Φ k (p, t).
Proof. Suppose that f is an extremal for Φ k (p, t).
we conclude that W k f is also an extremal for Φ k (p, t). Thus we may restrict our attention to extremals f k of the form f k (z) = f (z k ) for f ∈ H p . The stated claims now follow at once from Theorem 8, since 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 Φ k (p, t) unique? Note that while Theorem 14 provides an extremal f k (z) = f 1 (z k ) where f 1 denotes the 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 p and Lemma 3 directly. Since H p 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 consequence of the general theory developed in [8,11,14].
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 Consequently, W k f is also of the form (30). In particular, since both f and W k f are inner, we get from Lemma 7 that f = W k f . From the definition of W k , we know that f (z) = W k f (z) = g(z k ) for some analytic function g. This shows that the only possibility in (30) is 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 Φ k (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 The constant C > 0 satisfies where j 1 , j 2 , . . . , j k take only the values 0 and 1. Evidently g H 2 = h H 2 = 1. Set A l = |α 1 · · · α l | and B l = |α l+1 · · · α k |. By keeping only the terms j = 0 and j = k we obtain the trivial estimate We will adapt an argument due to F. Riesz [13] to get some additional information on the relationship between g and h. Write c j z j and note that |b 0 | = t/|c 0 | = t/C. By the Cauchy product formula we find that Suppose that g ∈ H 2 satisfies | g(0)| = t/C and g H 2 ≤ 1. Define f = gh. The Cauchy-Schwarz inequality shows that f H 1 ≤ 1, so the extremality of f implies that | a k | ≤ |a k |. Inspecting (32) and using the Cauchy-Schwarz inequality, we find that the optimal g must therefore satisfy where we used that h H 2 = 1. Using that c 0 = C, we compare the coefficients for z k in (33) with the definition of g, to find that Next we insert t = C 2 A l from the definition of f = gh and |c k | 2 = C 2 A 2 l B 2 l from the definition of h to obtain The additional information we require is encoded in the equation on the right hand side of (34). Suppose that l ≥ 1. Evidently A l < 1, since |α j | < 1 for j = 1, . . . , l by Lemma 4. It follows that the second factor on the right hand side of (34) can never be 0, since the trivial estimate (31) implies that (35) .
From the right hand side of (34) we thus find that B l = 1, which shows that C 2 < 1/(2A 2 l ) by (35). Since t = C 2 A l , we conclude that 0 ≤ t < 1/2. By the contrapositive, we have established that if 1/2 ≤ t ≤ 1, then the extremal for Φ k (1, t) has l = 0. In this case A 0 = 1 by definition, which shows that C = √ t. The right hand side of (34) becomes Returning to the definition of h we find that |c 0 | 2 = t and |c k | 2 = tB 2 0 . Consequently, Since 1/2 ≤ t ≤ 1, we find that both B 0 = 1 and B 2 0 = 1/t − 1 will imply that c j = 0 for j = 1, . . . , k − 1. Thus h(z) = √ t + √ 1 − t z k . When l = 0 we have g = h, which shows that the unique extremal is which is of the form f k (z) = f 1 (z k ) as claimed.
Proof of Theorem 15 (c). In the case 0 ≤ t < 1/2, we know from Theorem 8 and Theorem 14 that Φ k (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 to satisfy the conditions f H 1 = 1, f (0) > 0 and f (k) (0) > 0. If the resulting function satisfies f (0) = t, then it is an extremal for Φ k (p, t) and every extremal is obtained in this way. (This can be established similarly to the case (b) above.) 6. The extremal problem Φ k (p, t) for k ≥ 2 and 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 Evidently, Ψ 0 (p) = 1 for every 0 < p ≤ ∞ and the unique extremal is f (z) = 1. Recall (from [3] or [9]) that the extremals for Ψ k 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, Sec. 5] states that the extremal for Ψ k (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).
Assume that the above-mentioned conjecture from [3] holds. This assumption yields that the extremal for Φ k (p, 0) has precisely one zero in D and the extremal for the t which maximizes Φ k (p, t) does not vanish in D. Note that the extremal for Φ k (p, 1), which is f (z) = 1, does not vanish in D.
Question 1. Suppose that 0 < p < 1. Is it true that the extremal for Φ k (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 Φ k (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).