SOME COEFFICIENT ESTIMATES FOR H^{p} FUNCTIONS

CATHERINE BENETEAU AND BORIS KORENBLUM

Abstract. We find the maximum modulus of the n-th Taylor coefficient c_{n} of a function in the unit ball of H^{p}, 1 ≤ p ≤ ∞, provided that c_{0} is fixed, and identify the corresponding extremal functions.

1. Introduction

We are interested in finding the maximum Taylor coefficient of a function in the unit ball of H^{p}, 1 ≤ p ≤ ∞, whose value c at the origin is fixed and in identifying the corresponding extremal functions.

The motivation for this problem comes in part from the Hausdorff-Young inequality and the Bohr phenomenon. The Hausdorff-Young inequality states that if f ∈ L^{p} = L^{p}(𝕋), where 𝕋 = {z ∈ ℂ : |z| = 1}, then

∥f∥_{L^{p}} ≥ ∥{f̂(n)}∥_{l^{q}}

for 1 ≤ p ≤ 2, and q such that \frac{1}{p} + \frac{1}{q} = 1. If p = 2, then q = 2 and we have equality; this is Parseval’s formula. If p > 2, the inequality is known to fail. In particular, if p = ∞, it is obviously not the case that

∥f∥_{∞} ≥ ∑_{n=−∞}^{∞} |f̂(n)|.

However, for f ∈ H^{∞} and r = \frac{1}{3},

∥f∥_{∞} ≥ ∑_{n=0}^{∞} |f̂(n)|r^{n}.

This is a classical theorem of H. Bohr ([3]) and can be shown from F. Wiener’s estimates [2]. One can ask whether such a phenomenon exists for H^{p}, 2 < p < ∞. Namely, does there exist 0 < r < 1 such that

Date: November 1, 2001.

1991 Mathematics Subject Classification. Primary: 30H05, 42A16.

The first author was supported in part by the University Research Council at Seton Hall University.

This paper is in final form and no version of it will be submitted for publication elsewhere.

∥f∥_{p} ≥ (∑_{n=0}^{∞} |f̂(n)|^{q}r^{n})^{\frac{1}{q}}?

Our estimates of the Taylor coefficients of H^{p} functions show that no such Bohr phenomenon holds when 2 < p < ∞.

Finally, our results give a new proof of a conjecture by Hummel, Scheinberg, and Zalcman (see [5], p. 189) related to the Krzyz problem for H^{p} functions in the case that n = 1. The conjecture states that for n ≥ 1

sup\{|f̂(n)| : f ∈ H^{p}, ||f||_{p} ≤ 1, f zero-free \} = (\frac{2}{e})^{\frac{1}{q}}

and gives the form of the corresponding extremal function.

2. The Main Problem and Known Cases

Problem. Given 1 ≤ p ≤ ∞, 0 ≤ c ≤ 1, and n ≥ 1, find

M_{p}(n, c) = max{|f̂(n)| : f ∈ H^{p}, ∥f∥_{p} ≤ 1, |f(0)| = c},

where f(z) = ∑_{j=0}^{∞} f̂(j)z^{j}, and identify the extemal functions.

Notice that if f is extremal for M_{p}(n, c), then so is e^{iα}f(e^{iβ}z). We will therefore consider extremal f so that f(0) = c and f̂(n) ≥ 0. This problem is known for certain cases. If p = 2, by Parseval’s identity, it is easy to see that M_{2}(n, c) = \sqrt{1 − c^{2}} and the corresponding unique extremal function is f_{ext}(z) = c + \sqrt{1 − c^{2}}z^{n}. When c = 0, for all 1 ≤ p ≤ ∞, M_{p}(n, 0) = 1 and f_{ext}(z) = z^{n}. When c = 1, M_{p}(n, 1) = 0, and f_{ext}(z) = 1. If p = ∞, M_{∞}(n, c) = 1−c^{2} (see [2]) and the corresponding unique extremal function is f_{ext}(z) = \frac{c+z^{n}}{1+cz^{n}} .

In addition, it is easily seen that for all p ≥ 1, M_{p}(n, c) = M_{p}(1, c). For suppose f is a solution to the given extremal problem. Consider

f̃(z) = \frac{1}{n} ∑_{k=0}^{n−1} f(e^{\frac{2πk}{n} i}z^{\frac{1}{n}})

= ∑_{j=0}^{∞} f̂(jn)z^{j}.

Then f̃ ∈ H^{p}, ∥f̃∥_{p} ≤ ∥f∥_{p}, f̃(0) = f(0), and f̃′(0) = f̂(n). Therefore the extremal values of f̂(n) are the same as the ones calculated in the case when n = 1.

All other cases are unknown. Therefore we consider the extremal problem stated above for values of p such that 1 ≤ p < ∞ (p ≠ 2) and c such that 0 < c < 1. In the following sections, we consider only such values of p and c. In proving the main result, we prove some intermediate theorems which are of independent interest.

3. Statement of the Main Results

Theorem 3.1. If 2^{− \frac{1}{p}} ≤ c ≤ 1, then

M_{p}(n, c) = \frac{2}{p}c^{1− \frac{p}{2}} \sqrt{1 − c^{p}}

and the corresponding extremal function is

f(z) = (c^{\frac{p}{2}} + \sqrt{1 − c^{p}}z^{n})^{\frac{2}{p}} .

Theorem 3.2. If 0 < c < 2^{− \frac{1}{p}} , then the zero-free function f such that ∥f∥_{p} ≤ 1 and |f(0)| = c that maximizes |f′(0)| is

f(z) = 2^{− \frac{1}{p}} (1 + z)^{\frac{2}{p}} (2^{\frac{1}{p}} c)^{\frac{1−z}{1+z}}

and

f′(0) = c(\frac{2}{p} + log \frac{1}{2^{\frac{2}{p}} c^{2}} ).

Theorem 3.3. If 0 < c < 2^{− \frac{1}{p}} , then

M_{p}(n, c) = (\frac{2}{p} − 1)cv + \frac{c}{v}

and the corresponding extremal function is

f(z) = \frac{c}{v} (1 + vz^{n})^{\frac{2}{p} −1}(v + z^{n})

where v is the unique root ( 0 < v ≤ 1) of v^{p} − c^{p} = c^{p}v^{2}. In particular, for p = 1 and 0 < c < \frac{1}{2}, M_{1}(n, c) = 1 and f(z) = c + z^{n} + cz^{2n}.

4. Intermediate Results

In the following theorems, we will take 0 < c < 1.

Proposition 4.1. The inner function G that maximizes ReG′(0) if G(0) = c is

G(z) = \frac{c + z}{1 + cz}

and

G′(0) = 1 − c^{2}.

Proof. This follows directly from the H^{∞} case, since the solution to the maximal problem in that situation is in fact the above inner function.

□ Proposition 4.2. The singular function S(z) that maximizes ReS′(0) if S(0) = c is

S(z) = c^{\frac{1−z}{1+z}}

and

S′(0) = 2c log \frac{1}{c}.

Proof. Since S is a singular function, there exists a singular positive measure μ such that

S(z) = exp(− ∫_{−π}^{π} \frac{e^{it} + z}{e^{it} − z} dμ(t)).

Therefore

S′(z) = S(z)(− ∫_{−π}^{π} \frac{2e^{it}}{(e^{it} − z)^{2}} dμ(t)).

In particular,

S′(0) = c(− ∫_{−π}^{π} 2e^{−it}dμ(t)).

Therefore

ReS′(0) = c(− ∫_{−π}^{π} 2 cos tdμ(t))
≤ 2c ∫_{−π}^{π} dμ(t)
= 2c log \frac{1}{c}

Notice that this maximum is in fact attained when μ has full mass at t = π and thus

S(z) = c^{\frac{1−z}{1+z}} .
□

Proposition 4.3. If 0 < c < 2^{− \frac{1}{p}} , then there is no outer function f that maximizes Ref′(0) under the restrictions ∥f∥_{p} ≤ 1 and f(0) = c.

Proof. For simplicity, we will consider p = 1, the proof for p > 1 being similar. Let f ∈ H^{1} be an outer function such that ∥f∥_{1} ≤ 1 and f(0) = c. Then

f(z) = exp(\frac{1}{2π} ∫_{−π}^{π} \frac{e^{it} + z}{e^{it} − z} φ(t)dt)

where φ(t) = log |f(e^{it})|. Taking derivatives and real parts, we can conclude that

Ref′(0) = 2c\frac{1}{2π} ∫_{−π}^{π} φ(t) cos tdt.

We now have a variational problem, where we need to find

max \frac{1}{2π} ∫_{−π}^{π} φ(t) cos tdt

under the constraints \frac{1}{2π} ∫_{−π}^{π} e^{φ(t)}dt = 1 and \frac{1}{2π} ∫_{−π}^{π} φ(t)dt = log c < 0.

If this maximum equals μ and is attained when φ(t) = φ_{0}(t), then φ_{0} also solves the following dual variational problem: find

min \frac{1}{2π} ∫_{−π}^{π} e^{φ(t)}dt

under the constraints \frac{1}{2π} ∫_{−π}^{π} φ(t) cos tdt = μ and \frac{1}{2π} ∫_{−π}^{π} φ(t)dt = log c, because the above minimum is then equal to 1. To see this, suppose that for some φ(t) = φ_{1}(t) satisying the constraints of the dual problem \frac{1}{2π} ∫_{−π}^{π} e^{φ_{1}(t)}dt < 1. Then there is some s > 0 such that the function φ_{2}(t) = φ_{1}(t) + s cos t satisfies \frac{1}{2π} ∫_{−π}^{π} φ_{2}(t) cos tdt = μ + \frac{s}{2}, \frac{1}{2π} ∫_{−π}^{π} φ_{2}(t)dt = log c, and \frac{1}{2π} ∫_{−π}^{π} e^{φ_{2}(t)}dt = 1, contrary to the assumed maximality of μ in the original variational problem.

The constraints in the dual problem say that in the real Fourier series of φ the two coefficients are fixed, namely the constant term and the coefficient of cos t. By a standard variational argument we obtain that, whenever a bounded function h(t) satisfies

∫_{−π}^{π} h(t)dt = ∫_{−π}^{π} h(t) cos tdt = 0,

the following must hold:

(\frac{d}{ds} ∫_{−π}^{π} e^{φ_{0}(t)+sh(t)}dt)_{s=0} = ∫_{−π}^{π} e^{φ_{0}(t)}h(t)dt = 0.

In particular, this is true for h(t) = sin nt (n > 0) and h(t) = cos nt (n ≠ 0, 1), which implies that e^{φ_{0}(t)} = A + B cos t, φ_{0}(t) = log(A + B cos t), where A and B are constants, and φ_{0}(t) is the extremal function for the original variational problem.

Since \frac{1}{2π} ∫_{−π}^{π} e^{φ_{0}(t)}dt = 1, A = 1. Therefore 0 ≤ B < 1. Consider now the function

k(γ) = \frac{1}{2π} ∫_{−π}^{π} log(1 + γ cos t)dt

for 0 ≤ γ ≤ 1. Notice that k(B) = log c. Also k′(γ) < 0, k(0) = 0 and k(1) = − log 2. This forces c ≥ \frac{1}{2}, which is a contradiction. Therefore no such constant B exists, and there is no solution to the extremal problem. □

5. Proofs of the Main Results

We begin by considering the main problem for functions whose value at the origin is not close to 0 as stated in Theorem 3.1.

Proof. Let 2^{− \frac{1}{p}} ≤ c < 1, n = 1, and consider f ∈ H^{p} such that f(0) = c and ∥f∥_{p} ≤ 1. Write f(z) = B(z)F (z), where B is a Blaschke product and where F has no zeros. Let B(0) = v > 0 and F (0) = u > 0. Note that uv = c. Since B ∈ H^{∞} and ∥B∥_{∞} = 1, F. Wiener’s estimates ([2]) show that

|B′(0)| ≤ 1 − v^{2}.

Note that F ∈ H^{p}, ∥F ∥_{p} = ∥f∥_{p} = 1, and F has no zeros. Define G(z) = F (z)^{\frac{p}{2}} . Then ∥G∥_{2}^{2} = ∥F ∥_{p}^{p} = 1 and G(0) = u^{\frac{p}{2}} . Therefore, by the case p = 2,

|G′(0)| ≤ \sqrt{1 − u^{p}}.

Writing F (z) = G(z)^{\frac{2}{p}} , we get

|F′(0)| ≤ \frac{2}{p}(u^{\frac{p}{2}} )^{\frac{2}{p} −1}\sqrt{1 − u^{p}}
= \frac{2}{p}u^{1− \frac{p}{2}} \sqrt{1 − u^{p}}

Putting the two estimates above together gives

|f′(0)| = |B′(0)F (0) + B(0)F′(0)|
≤ (1 − v^{2})u + v \frac{2}{p}u^{1− \frac{p}{2}} \sqrt{1 − u^{p}}
= c(\frac{1}{v} − v) + \frac{2}{p}c^{1− \frac{p}{2}} \sqrt{v^{p} − c^{p}}

Therefore, define

φ(v) = c(\frac{1}{v} − v) + \frac{2}{p}c^{1− \frac{p}{2}} \sqrt{v^{p} − c^{p}}

and find max φ(v) for c ≤ v ≤ 1. Note that

φ′(v) = c(−\frac{1}{v^{2}} − 1) + \frac{2}{p}c^{1− \frac{p}{2}} \frac{1}{2} \frac{1}{\sqrt{v^{p} − c^{p}}} pv^{p−1}
= −c(1 + \frac{1}{v^{2}} ) + c^{1− \frac{p}{2}} \frac{v^{p−1}}{\sqrt{v^{p} − c^{p}}}

Therefore,

φ′(v) ≥ 0 ⇔ c^{− \frac{p}{2}} \frac{v^{p−1}}{\sqrt{v^{p} − c^{p}}} ≥ 1 + \frac{1}{v^{2}}
⇔ (1 + v^{2})^{2}(c^{p})^{2} − v^{p}(1 + v^{2})^{2}c^{p} + v^{2(p+1)} ≥ 0
⇔ (c^{p} − \frac{v^{p}}{1 + v^{2}} )(c^{p} − \frac{v^{2+p}}{1 + v^{2}} ) ≥ 0

When c ≥ 2^{− \frac{1}{p}} , φ′(v) ≥ 0 and the maximum of φ(v) is obtained at v = 1 :

φ(1) = \frac{2}{p}c^{1− \frac{p}{2}} \sqrt{1 − c^{p}}.

In that case, the function

f(z) = (c^{\frac{p}{2}} + \sqrt{1 − c^{p}}z)^{\frac{2}{p}}

is an element of H^{p} with norm 1 such that f(0) = c and f′(0) = φ(1).

When n > 1, use the function f̃ described in Section 2. Since f(z) = f̃(z^{n}), we obtain the extremal function

f(z) = (c^{\frac{p}{2}} + \sqrt{1 − c^{p}}z^{n})^{\frac{2}{p}}

with the same maximal n-th Taylor coefficient as in the case n = 1. □

Notice that f is a zero-free function, and therefore Theorem 3.1 also solves the extremal problem for zero-free H^{p} functions whose value at the origin is not too close to 0. Let us now consider zero-free functions in H^{p} whose value at the origin are small, as stated in Theorem 3.2.

Proof. Let 0 < c < 2^{− \frac{1}{p}} and let f ∈ H^{p} be a non-zero function such that f(0) = c and ∥f∥_{p} ≤ 1 for which |f′(0)| is maximal. Write f(z) = S(z)F (z) where S is a singular function and F is an outer function. Writing S(0) = u and F (0) = v, notice that by Proposition 4.3, v ≥ 2^{− \frac{1}{p}} . Using the estimates given by Proposition 4.2 and Theorem 3.1, we get that

|f′(0)| ≤ v2u log \frac{1}{u} + u\frac{2}{p}v^{1− \frac{p}{2}} \sqrt{1 − v^{p}}
= 2c log \frac{v}{c} + \frac{2c}{p}\sqrt{\frac{1}{v^{p}} − 1}
= φ(v)

One can easily show that φ(v) is decreasing on [2^{− \frac{1}{p}} , 1] and therefore attains its maximum at v = 2^{− \frac{1}{p}} . Therefore u = c2^{\frac{1}{p}} , and the function

f(z) = (2^{\frac{1}{p}} c)^{\frac{1−z}{1+z}} 2^{− \frac{1}{p}} (1 + z)^{\frac{2}{p}}

is a zero-free function such that f(0) = c, ∥f∥_{p} = 1 and

f′(0) = φ(2^{− \frac{1}{p}}) = c(\frac{2}{p} + log \frac{1}{2^{\frac{2}{p}} c^{2}} ).

□

We now consider functions in H^{p} that can have zeros and whose value at the origin is small, as stated in Theorem 3.3.

Proof. Consider the case n = 1 and let f ∈ H^{p} be such that ∥f∥_{p} ≤ 1 and f(0) = c. Write f(z) = B(z)F (z) where B is a Blaschke product with B(0) = v > 0, and F is zero-free with F (0) = u.

Suppose first that u ≥ 2^{− \frac{1}{p}} . Then c ≤ v ≤ 2^{\frac{1}{p}} c, so by the proof of Theorem 3.1

|f′(0)| ≤ c(\frac{1}{v} − v) + \frac{2}{p}c^{1− \frac{p}{2}} \sqrt{v^{p} − c^{p}}
= φ(v)

and

φ′(v) ≥ 0 ⟺ (c^{p} − \frac{v^{p}}{1 + v^{2}} )(c^{p} − \frac{v^{2+p}}{1 + v^{2}} ) ≥ 0.

If c < 2^{− \frac{1}{p}} then φ′(v) = 0 has two solutions v_{1} and v_{2} in [c, 1] where v_{1} is the solution to v^{p} − c^{p} = c^{p}v^{2} and v_{2} is the solution to v^{2+p} − c^{p} = c^{p}v^{2}. It is not hard to show that

c ≤ v_{1} ≤ 2^{\frac{1}{p}} c ≤ v_{2} ≤ 1

and that φ attains its maximum in [c, 2^{\frac{1}{p}} c] at v_{1}. In that case, after simplification,

max φ(v) = (\frac{2}{p} − 1)cv_{1} + \frac{c}{v_{1}} .

Let us suppose now that 0 < u < 2^{− \frac{1}{p}} . In this case, 2^{\frac{1}{p}} c < v ≤ 1, and by Theorem 3.2, we know that

f′(0) ≤ u(1 − v^{2}) + vu((\frac{2}{p} + log \frac{1}{2^{\frac{2}{p}} u^{2}} )

= \frac{c}{v} − cv + \frac{2c}{p} − 2c log(\frac{2^{\frac{1}{p}} c}{v})
= ψ(v).

ψ is easily seen to be decreasing and therefore attains its maximum at 2^{\frac{1}{p}} c, and

ψ(2^{\frac{1}{p}} c) = 2^{− \frac{1}{p}} − c^{2}2^{\frac{1}{p}} + \frac{2c}{p}.

Comparing this value with φ(v_{1}), we conclude that φ(v_{1}) ≥ ψ(2^{\frac{1}{p}} c). Taking

f(z) = \frac{v + z}{1 + vz} ((\frac{c}{v})^{\frac{p}{2}} + \sqrt{1 − (\frac{c}{v})^{p}}z)^{\frac{2}{p}}
= (\frac{c}{v})(1 + vz)^{\frac{2}{p} −1}(v + z)

gives a function in H^{p} with norm 1 such that f(0) = c and

f′(0) = (\frac{2}{p} − 1)cv + \frac{c}{v} ,

where v is the unique solution (0 < v ≤ 1) of v^{p} − c^{p} = c^{p}v^{2}.

When n > 1, we use f̃ as described in Section 2 to show that M_{p}(n, c) = M_{p}(1, c) and the corresponding extremal function is

f(z) = (\frac{c}{v})(1 + vz^{n})^{\frac{2}{p} −1}(v + z^{n})

where v is such that v^{p} − c^{p} = c^{p}v^{2}. □

6. Corollaries

Corollary 6.1. If 1 ≤ p < ∞, ∥f∥_{p} = 1, |f(0)| < 2^{− \frac{1}{p}} and

|f′(0)| > |f(0)|(\frac{2}{p} + log \frac{1}{2^{\frac{2}{p}} |f(0)|^{2}} ),

then f has at least one zero in the open unit disk.

Proof. This result follows directly from Theorem 3.2. □

Corollary 6.2. Let p ≥ 1. Then

sup\{|f′(0)| : f ∈ H^{p}, ||f||_{p} ≤ 1, f zero-free \} = (\frac{2}{e})^{\frac{1}{q}} .

Proof. If f ∈ H^{p} is non-zero and ∥f∥_{p} ≤ 1, then let f(0) = c for some 0 < c ≤ 1. By Theorems 3.1 and 3.2,

|f′(0)| ≤ \{ \begin{matrix}c(\frac{2}{p} + log _{\frac{1}{2^{\frac{2}{p}} c^{2}}} ) & 0 < c < 2^{− \frac{1}{p}} \\ \frac{2}{p}c^{1− \frac{p}{2}} \sqrt{1 − c^{p}} & 2^{− \frac{1}{p}} ≤ c ≤ 1 \end{matrix}

By varying c between 0 and 1, it is straightforward to show that the maximum derivative is obtained at c = \frac{1}{2}(\frac{2}{e})^{\frac{1}{q}} and the corresponding extremal function is

f(z) = (\frac{(1 + z)^{2}}{2} )^{\frac{1}{p}} (exp \frac{z − 1}{z + 1})^{\frac{1}{q}}

as conjectured by Hummel, Scheinberg, and Zalcman in [5]. □

This result was first obtained by Johnny Brown in [4] by a completely different method.

Corollary 6.3. Let 2 < p < ∞ and q be such that \frac{1}{p} + \frac{1}{q} = 1. Then there does not exist 0 < r < 1 such that

∥f∥_{p} = (∫_{−π}^{π} \frac{1}{2π} |f(e^{it})|^{p}dt)^{\frac{1}{p}} ≥ (∑_{n=0}^{∞} |f̂(n)|^{q}r^{n})^{\frac{1}{q}}

for every f ∈ H^{p}.

Proof. Consider the extremal function f in Theorem 3.1 for n = 1. We have ∥f∥_{p} = 1. For 2^{− \frac{1}{p}} ≤ c < 1, f(0) = c, and

f′(0) = \frac{2}{p}c^{1− \frac{p}{2}} \sqrt{1 − c^{p}} > A\sqrt{1 − c}

if c is close enough to 1, where A is a positive constant. Therefore, since q < 2,

∑_{k=0}^{∞} |f̂(k)|^{q}r^{k} ≥ c^{q} + A^{q}r(1 − c)^{\frac{q}{2}} > 1

for c sufficiently close to 1. □

Notice that this corollary implies that no Bohr phenomenon holds for H^{p} and L^{p} (2 < p < ∞.) This was also proved independently by L. Aizenberg [1] using a different method. It is known that the Bohr phenomenon does hold for real L^{∞}. For L^{∞}, the Bohr phenomenon does not hold as shown by an example of Liflyand’s [6]. Whether such a Bohr phenomenon holds for real L^{p} (2 < p < ∞) remains an open problem.

Acknowledgements. The authors thank Lawrence Zalcman for drawing their attention to [5] and Lev Aizenberg, Dima Khavinson, and Elijah Liflyand for helpful suggestions. Special thanks are due to the referee whose numerous suggestions improved the quality of the original version of the paper.

References

1. L. Aizenberg, private communication.

2. H. Boas and D. Khavinson, Vita: Friedrich Wilhelm Wiener, Mathematical Intelligencer (22) 2 (2000), 73-75.

3. H. Bohr, A theorem concerning power series, Proc. London Math. Soc. (2) 13 (1914), 1-5.

4. J. Brown, On a coefficient problem for non-vanishing H^{p} functions, Complex Variables 4 (1985), 253-265.

5. J. Hummel, S. Scheinberg and L. Zalcman, A coefficient problem for bounded nonvanishing functions, Journal d’Analyse Mathématique, 31 (1977), 169-190.

6. I. Liflyand, private communication.

Department of Mathematics and Computer Science, Seton Hall University, South Orange , New Jersey, 07079

E-mail address: beneteca@shu.edu

Department of Mathematics, The University at Albany, Albany, New York, 12222

E-mail address: borisko@math.albany.edu