Computational Methods and Function TheoryVolume 12 (2012), No. 1, 167–172

A Reverse cos πρ Theorem and a Question of Fryntov

Peter C. Fenton and John Rossi

(Communicated by Alexandre Eremenko)

Abstract. The authors continue their work on reverse Denjoy theorems, prov-ing a reverse cos πρ theorem. The theorem is connected to a question of Fryn-tov on entire functions with gaps.

Keywords. Subharmonic function, entire function, cosπρ Theorem, gap se-ries.

2000 MSC. 30D15.

1. Introduction

In a recent series of papers [5, 6, 7] the authors showed what they termed ReverseDenjoy Theorems. Specifically let C1 and C2 be two simple curves joining 0 to∞, non-intersecting in the finite plane except at 0 and enclosing a domain D.Let Dr = D ∩C(0, r), where C(0, r) is the circle centred at 0 with radius r, andsuppose that, for all large r,

(1) meas Er ≤ 2α,

where Er = {θ : reiθ ∈ Dr} and 0 < α < π. Let u be a subharmonic function inthe plane, and denote B(r, u) = sup|z|=r u(z) and AD(r, u) = infz∈Dr

u(z). Thenwe have the following result [5].

Theorem 1. If

(2) u(z) = B(|z|, u)

for all large z ∈ C1 ∪ C2 and

(3) AD(r, u) = O(1) as r → ∞then

(4) lim infr→∞

B(r, u)

rπ/(2α)> 0.

Received October 28, 2011, in revised form December 5, 2011.Published online January 21, 2012.

ISSN 1617-9447/$ 2.50 c© 2012 Heldermann Verlag

168 P. C. Fenton and J. Rossi CMFT

The term Reverse Denjoy Theorem comes from the fact that in Theorem 1, uis assumed large on the boundary of D and small inside. A logical question toask is whether there is a reverse analog to the cos πρ Theorem or more correctlyto Baernstein’s generalization of that theorem [2]. Our main result answers thisquestion in the affirmative.

Theorem 2. Suppose that u is subharmonic in the plane and that (1) and (2)hold. Then given any λ satisfying 0 < αλ ≤ π/2, either

AD(r, u) > cos(αλ)B(r, u)

for certain arbitrarily large values of r or

(5) limr→∞

B(r, u)

rλ

exists and is either positive or +∞.

Using the same technique as in [5], it is enough to prove Theorem 2 in the caseu(z) = B(|z|, u) for all z outside D. Noting that then

AD(r, u) = A(r, u) := inf0≤θ≤2π

u(reiθ),

Theorem 2 will follow directly from the following result.

Theorem 3. Suppose that u is subharmonic in the plane and that G is an openset in which

(6) m{θ : reiθ ∈ G

} ≥ 2(π − α)

for all large r and for some number α satisfying 0 < α < π. Given any λsatisfying 0 < αλ ≤ π/2, either

(7) A(r, u) > cos(αλ)B(r, u)

for certain arbitrarily large values of r or (5) holds.

We remark that Theorem 3 both generalizes Theorem 1 (where λ = π/(2α)) andimproves it with the addition of the regularity conclusion (5). It should be pos-sible, as in the aforementioned theorem of Baernstein, to replace the restrictionmade on λ by the weaker one, 0 < αλ < π. We are unable to do this.

We note that Theorem 3 is closely related to a question of Fryntov regardinggap series. Let f(z) =

∑∞j=0 ajz

j be an entire function or order λ > 0 for whichthe maximal density of the non-zero coefficients, Δ, satisfies Δλ < 1. Here themaximal density is defined by

Δ = sup0<η<1

lim supn→∞

N(n) − N(ηn)

(1 − η)n,

12 (2012), No. 1 A Reverse cos πρ Theorem and a Question of Fryntov 169

where N(n) is the number of non-zero aj with j ≤ n. A conjecture of Kovari’s [10]states that if f has order at most λ > 0 and λΔ < 1, then

lim supr→∞

A(r, log |f |)B(r, log |f |) ≥ cos (πλΔ).

The conjecture is solved if min|z|=r |f(z)| is taken on a ray [8, 12] but in generalit is still open. We note that the key property of functions satisfying Δ < 1 isthat the set {θ : log |f(reiθ)| = (1 + O(1))B(r, log |f |)} intersects every intervalof length 2πΔ.

In [9] Fryntov considers a special case of Kovari’s conjecture, namely: if λΔ < 1/2then

(8) lim supr→∞

inf |z|=r,z∈γ log |f(z)|B(r, log |f |) ≥ cos (πλΔ),

where γ is a receding curve, that is a curve that goes to infinity and meets everycircle C(0, r) exactly once. Through very ingenious symmetrization methods, heproves a weaker version of (8) when λΔ < 1/3 and the right hand side is replacedby 2 cos(πλΔ) − 1.

Fryntov is able to translate Kovari’s conjecture into subharmonic terms as fol-lows. Let u be a subharmonic function in the plane for which the set

(9) {θ : u(reiθ) = B(r, u)}contains an interval of length 2(π−α), for some 0 < α < π/2. Assume also that0 ≤ B(r, u) ≤ rλ and B(1, u) = 1 for some λ < 1. Fryntov shows that if

(10) sup|z|=r, z∈γ

u(z)

B(r, u)≥ cos

(πλ

2

)

for any receding curve γ that joins 0 to ∞, then (8) is true. As Fryntov notes,his conjecture would follow from a positive answer to a problem posed by Ere-menko [3, Prob. 2.68] which is very similar to Theorem 3.

The conclusion of Theorem 3 is stronger than (10) (and Eremenko’s problem),in that no curves are involved, and in addition it includes the regularity conclu-sion (5). The hypothesis concerning the set on which u(z) = B(|z|, u) is partlyweaker than (9) — G∩C(0, r) need not consist of a single interval — but partlystronger since we need to assume that the set G is open. Because of this signif-icant additional assumption, whether (10) (or Eremenko’s problem), is true orfalse remains open.

2. Proof of Theorem 3

Suppose that u satisfies the hypotheses of Theorem 3 and that (7) fails for some λsatisfying 0 < αλ ≤ π/2, so that

(11) A(r, u) ≤ cos(αλ)B(r, u)

170 P. C. Fenton and J. Rossi CMFT

for all r ≥ R. For any fixed K > 0, (11) also holds for all r ≥ R if u is replacedby (u−K)+ = max{u−K, 0}, and in fact (11) holds for (u−K)+ for all r ≥ 0and (u − K)+(0) = 0 if K > 0 is large enough. It is enough, then, to provethat (5) follows if (11) holds for all r ≥ 0 and u(0) = 0.

We introduce the Baernstein function

u∗(reiθ) = supE

1

2π

∫E

u(reiω) dω,

where the supremum is taken over all sets of Lebesgue measure 2θ. Then u∗ issubharmonic in the upper half plane and continuous in its closure [1].

Lemma 1. Under the hypotheses of Theorem 1,

(12) Vα0(reiθ) = u∗(reiθ) − π − α0

πB(r, u)

is subharmonic for π − α0 < θ < π, for any α0 satisfying α < α0 < π.

Lemma 1 was obtained in [5] (see Lemma 2 and the argument that follows) underslightly different hypotheses — the set G was bounded by two curves on whichu(z) = B(|z|, u) — but in fact the only property of G that was used was that Gwas open. Let us note that the argument of [5, pp. 31–32] shows that Vα0(−r)is a convex function of log r.

Allowing α0 ↓ α in (12), we conclude that Vα is subharmonic for π − α < θ < πand that Vα(−r) is a convex function of log r.

We follow the methods in [2, pp. 191–192] and [11, §3]. Given any R > 0,define H to be the harmonic function in {z : 0 < arg z < α, 0 < |z| < R} withboundary values H(r) = 0, H(reiα) = Vα(−r) and

H(Reiθ) =

⎧⎨⎩

B(R, u) if 0 < θ <α

2,

2B(R, u) ifα

2< θ < α.

We claim that there exist postitive constants C1, C2, depending only on α and λ,such that for 0 < r < s < R,

(13)

∫ s

r

A(t, u) − cos(αλ)B(t, u)

t1+λdt > C1

B(r, u)

rλ− C2

B(2s, u)

sλ.

To see this note that since u(0) = 0 and since Vα(−r) is a convex increasingfunction of log r, we may use [1, Prop. 7] to obtain [1, Eq. (29)], namely that for0 < r < s < R,

(14)

∫ s

r

Hθ(teiα) − cos(αλ)Hθ(t)

t1+λdt > K1

Hθ(r)

rλ− K2

B(2s, u)

sλ.

Here Hθ is the derivative with respect to θ and K1, K2 are positive constantsdepending only on α and λ.

12 (2012), No. 1 A Reverse cos πρ Theorem and a Question of Fryntov 171

Note that Vα is subharmonic and no greater than B(R, u) on the arc of the circleof radius R where it is defined. Furthermore

Vα(rei(π−α) = 0, H(reiα) = Vα(−r).

Thus we have that Vα(rei(θ+π−α)) ≤ H(reiθ) for all θ ∈ [0, α] and all r ∈ [0, R).Thus by definition of Vα, we obtain

(15) Hθ(r) ≥ (Vα)θ(r) =1

πB(r, u), Hθ(re

iα) ≤ 1

π(Vα)θ(−r) =

1

πA(r, u),

for 0 < r < s < R, for any R > 0. Since αλ < π/2, we can substitute (15)into (14) to obtain (13) for 0 < r < s < R and therefore for 0 < r < s < ∞,since R is arbitrary. From (11) and (13) we deduce that

(16) lim supr→∞

B(r, u)

rλ< ∞.

We now redefine H to be the harmonic function in {z : 0 < arg z < α} withboundary values H(r) = 0, H(reiα) = Vα(−r). From (16) and the fact thatαλ < π, H is well-defined and unique. Define h(z) = H(zγ) where γ = α/π.Thus h is harmonic in the upper half plane with

h(reiθ) =1

π

∫ ∞

0

Vα(−tγ)r sin θ

r2 + 2tr cos θ + t2dt

and

hθ(reiθ) =

1

π

∫ ∞

0

log

∣∣∣∣1 +reiθ

t

∣∣∣∣ dμ(t),

where μ(t) = γtγV ′α(−tγ). Since Vα(−r) is a convex function of log r, μ is a

positive Borel measure. Now hθ can be extended to be a subharmonic functionin the whole plane of order at most ρ = αλ/π < 1. Note that

hθ(r) = γHθ(rγ) ≥ 1

πγB(rγ).

By assuming (11), as we may for all r > 0 and by using (15), we obtain

hθ(−r) + hθ(r) = γ[Hθ(reiα) + Hθ(r

γ)] ≤ 1

πγ[1 + cos(αλ) + O(1)]B(rγ).

But cos(αλ) = cos(πρ). Thus

a(r) = hθ(−r) ≤ (cos(πρ) + O(1))hθ(r) = (cos(πρ) + O(1))b(r)

where a and b are the minimum and maximum of hθ on the circle of radius r.Thus hθ is extremal for the cos πρ Theorem and by a result of Drasin and Shea [4](cf. [11, pp. 135–136])

limr→∞

b(r)

rρ

exists and is either positive or +∞. This implies (5) and completes the proof ofTheorem 3.

172 P. C. Fenton and J. Rossi CMFT

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Peter C. Fenton E-mail: pfenton@maths.otago.ac.nzAddress: University of Otago, Dunedin, New Zealand.

John Rossi E-mail: rossij@math.vt.eduAddress: Virginia Tech, Blacksburg, Virginia, U.S.A.