# Integral Means Inequalities, Convolution, and Univalent Functions

Authors: Daniel Girela; Cristóbal González
Journal: Journal of Function Spaces
 (2019)

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## Abstract

We use the Baernstein star-function to investigate several questions about the integral means of the convolution of two analytic functions in the unit disc. The theory of univalent functions plays a basic role in our work.


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## Body

## 1. Introduction

LetD={z∈C:z<1} and T={z∈C:z=1} denote the open unit disc and the unit circle in the complex plane C. We let also Hol(D) be the space of all analytic functions in D endowed with the topology of uniform convergence in compact subsets.If0≤r<1 and f∈Hol(D), we set (1)Mpr,f=∫-ππfreitpdt2π1/p,if0<p<∞,M∞r,f=supz=rfz.For0<p≤∞, the Hardy space Hp consists of those f∈Hol(D) such that(2)fHp=defsup0≤r<1Mpr,f<∞. We refer to [1] for the theory of Hp-spaces.Iff,g∈Hol(D),(3)fz=∑n=0∞anzn,gz=∑n=0∞bnznz∈D, the (Hadamard) convolution (f⋆g) of f and g is defined by(4)f⋆gz=∑n=0∞anbnzn,z∈D. We have the following integral representation: (5)f⋆gz=12πi∫ξ=rfzξgξdξξ,z<r<1,(see [2, p. 11]). The convolution operation ⋆ makes Hol(D) into a commutative complex algebra with an identity(6)Iz=11-z=∑n=0∞zn,z∈D.We refer to [2] for the theory of the convolution of analytic functions and its connections with geometric function theory.Following [3], we shall say that a function F∈Hol(D) isbound preserving if for every f∈H∞ we have that f⋆F∈H∞ and(7)f⋆FH∞≤fH∞.Sheil-Small [3, Theorem 1.3] (see also [2, p. 123]) proved that a function F∈Hol(D) is bound preserving if and only if there exists a complex Borel measure μ on T with μ≤1 such that (8)Fz=∫Tdμξ1-zξ,z∈D.The measure μ is a probability measure if and only if F isconvexity preserving; that is, for any f∈Hol(D) the range of f⋆F is contained in the closed convex hull of the range of f [2, pp. 123, 124].It turns out that ifF is bound preserving and 1≤p≤∞, then for every f∈Hp we have that f⋆F∈Hp and(9)f⋆FHp≤fHp.Actually, the following stronger result holds.Theorem 1.
Suppose thatf,F∈Hol(D) with F being bound preserving. Then(10)Mpr,f⋆F≤Mpr,f,0<r<1,whenever 1≤p≤∞.Proof.
SinceF is bound preserving, there exists a complex Borel measure μ on T with μ≤1 such that(11)Fz=∫Tdμξ1-zξ=∑n=0∞∫Tξndμξzn,z∈D.If f(z)=∑n=0∞‍anzn (z∈D), we have (12)f⋆Fz=∑n=0∞an∫Tξndμξzn=∫T∑n=0∞anξnzndμξ=∫Tfξzdμξ,z∈D.This immediately yields (10) for p=∞. Now, if 1≤p<∞, using Minkowski’s integral inequality we obtain (13)Mpr,f⋆F=12π∫-ππ∫Tfrξeiθdμξpdθ1/p≤12π∫-ππ∫Tfrξeiθdμξpdθ1/p≤∫T12π∫-ππfrξeiθpdθ1/pdμξ=∫TMpr,fdμξ≤Mpr,f.


## 2. Star-Type Inequalities

The main purpose of this article is studying the possibility of extending Theorem1 to cover other integral means, at least for some special classes of functions. In order to do so, we shall use the method of the star-function introduced by A. Baernstein [4, 5].Ifu is a subharmonic function in D∖{0}, the function u∗ is defined by(14)u∗reiθ=supE=2θ∫Eureitdt,0<r<1,0≤θ≤π,where E denotes the Lebesgue measure of the set E. The basic properties of the star-function which make it useful to solve extremal problems are the following [5]. (i)
Ifu is a subharmonic function in D∖{0}, then the function u∗ is subharmonic in D+={z=reiθ:0<r<1,0<θ<π} and continuous in {z=reiθ:0<r<1,0≤θ≤π}.(ii)
Ifv is harmonic in D∖{0}, and it is a symmetric decreasing function on each of the circles {z=r} (0<r<1), then v∗ is harmonic in D+ and; in fact, v∗(reiθ)=∫-θθ‍v(reit)dt.The relevance of the star-function to obtain integral means estimates comes from the following result.Proposition A (see [5]).
Letu and v be two subharmonic functions in D. Then the following two conditions are equivalent: (i)
u∗≤v∗ in D+.(ii)
For every convex and increasing functionΦ:R→R, we have that (15)∫-ππΦureiθdθ≤∫-ππΦvreiθdθ,0<r<1.Proposition A yields the following result about analytic functions.Proposition B.
Letf and g be two nonidentically zero analytic functions in D. Then the following conditions are equivalent: (i)
log⁡f∗≤log⁡g∗ in D+.(ii)
For every convex and increasing functionΦ:R→R, we have that (16)∫-ππΦlog⁡freiθdθ≤∫-ππΦlog⁡greiθdθ,0<r<1.Since for anyp>0 the function Φ defined by Φ(x)=exp⁡(px) (x∈R) is convex and increasing, we deduce that if f and g are as in Proposition B and log⁡f∗≤log⁡g∗ in D+, then(17)Mpr,f≤Mpr,g,0<r<1,for all p>0.The main achievement in the use of the star-function by A. Baernstein in [5] was the proof that the Koebe function k(z)=z/(1-z)2 (z∈D) is extremal for the integral means of functions in the class S of univalent functions (see [1, 6] for the notation and results regarding univalent functions). Namely, Baernstein proved that if f∈S then(18)±log⁡f∗≤±log⁡k∗ and, hence,(19)∫-ππfreiθpdθ≤∫-ππkreiθpdθ,0<r<1, for all p∈R. In particular, we have that if f∈S and 0<p≤∞, then (20)Mpr,f≤Mpr,k,0<r<1.Subsequently the star-function has been used in a good number of papers to obtain bounds on the integral means of distinct classes of analytic functions (see, e.g., [7–12]).Coming back to convolution, the following questions arise in a natural way.Question 1.
Letf,g,F,G be analytic functions in D with F and G being symmetric decreasing on each of the circles {z=r} and suppose that(21)log⁡f∗≤log⁡F∗andlog⁡g∗≤log⁡G∗.Does it follow that (log⁡f⋆g)∗≤(log⁡F⋆G)∗?Question 2.
LetF and f be two analytic functions in D and suppose that F is bound preserving. Can we assert that log⁡f⋆F∗≤log⁡f∗?We shall show that the answer to these two questions is negative. Regarding Question1 we have the following result.Theorem 3.
There exist two functionsF1,F2∈Hol(D) with(22)log⁡Fj∗≤log⁡I∗,forj=1,2,and such that(23)theinequalitylog⁡F1⋆F2∗≤log⁡I⋆I∗doesnothold.
Here,I is the identity element of the convolution defined in (6); that is, I(z)=1/1-z(z∈D). Hence I⋆I=I.Proof.
Leth be an odd function in the class S with Taylor expansion(24)hz=z+a3z3+a5z5+…with a5>1. The existence of such h was proved by Fekete and Szegö (see [13, p. 104]). Set also(25)h1z=hzz=1+a3z2+a5z4+…,z∈D.It is well known that there exists a function H∈S such that h(z)=H(z2) (see [13, p. 64]). Set k2(z)=k(z2)=z/1-z2 and J(z)=k2(z)/z=1/1-z2 (z∈D). By Baernstein’s theorem we have log⁡H∗≤log⁡k∗, a fact which easily implies that log⁡h1∗≤log⁡J∗. Now, it is clear that J is subordinate to I and then, using [8, Lemma 2], we see that log⁡J∗≤log⁡I∗. Thus it follows that(26)log⁡h1∗≤log⁡I∗.For n=1,2,3,…, we define fn inductively as follows:(27)f1=h1andfn=fn-1⋆f1,forn≥2. In other words, fn=h1⋆⋯⋆h1︷(n). Clearly, (25) yields(28)fnz=1+a3nz2+a5nz4+…. Since a5>1, it follows that a5n→∞, as n→∞. This is equivalent to saying that(29)fn40⟶∞,asn→∞.Then it follows that the family {fn(4):n=1,2,3,…} is not a locally bounded family of holomorphic functions in D. Using [14, Theorem 16, p. 225] we see that the same is true for the family {fn:n=1,2,3,…}. Take p∈(0,1), then I∈Hp. Since a bounded subset of Hp is a locally bounded family [1, p. 36], it follows that(30)supn≥1fnHp=∞.Now, (30) implies that fnHp>IHp for some n. Using Proposition B, we see that this implies that (31)theinequalitylogfn∗≤log⁡I∗isnottrueforsomen.Let N be the smallest of all such n. Using (26) and the fact that f1=h1, it follows that N>1.
Then it is clear that (23) holds with F1=f1,F2=fN-1.We have the following result regarding Question2.Theorem 4.
There existf,F analytic and univalent in D such that F is convexity preserving and with the property that the inequality log⁡f⋆F∗≤log⁡f∗ does not hold.The following lemma will be used in the proof of Theorem4.Lemma 5.
Letf,F∈Hol((D) and suppose that F(0)=1, F is convexity preserving, and f and f⋆F are zero-free in D and satisfy the inequality log⁡f⋆F∗≤log⁡f∗. Then we also have that(32)log⁡1f⋆F∗≤log⁡1f∗.Proof.
Setu=log⁡f⋆F, v=log⁡f. Then u and v are harmonic in D, u(0)=v(0), and u∗≤v∗. Then it follows that, for 0<r<1 and 0≤θ≤π, (33)-u∗reiθ=supE=2θ∫E-ureitdt=supE=2θ-∫-ππureitdt+∫-π,π∖Eureitdt=-2πu0+u∗reiπ-θ=-2πv0+u∗reiπ-θ≤-2πv0+v∗reiπ-θ=-v∗reiθ.Hence, we have proved that (-u)∗≤(-v)∗ which is equivalent to (32).Proof of Theorem4.
Set(34)fz=11-z2=∑n=0∞n+1zn,Fz=1-12z,z∈D.Clearly, f and F are analytic, univalent, and zero-free in D. Also (35)f⋆Fz=1-z,z∈D.Hence f⋆F is also zero-free in D. Notice that 1/f⋆F∉H∞ and 1/f∈H∞. Then it follows that(36)theinequalitylog⁡1f⋆F∗≤log⁡1f∗doesnothold.Now, it is a simple exercise to check that(37)Fz=12π∫-ππ1-cos⁡θ1-eiθzdθand then it follows that F is convexity preserving. Then, using (36) and Lemma 5, it follows that the inequality log⁡f⋆F∗≤log⁡f∗ does not hold, as desired.We close the paper with a positive result, determining a class of univalent functionsZ such that (10) is true for all p>0, whenever f∈Z and F is convexity preserving.A domainD in C is said to be Steiner symmetric if its intersection with each vertical line is either empty, or is the whole line, or is a segment placed symmetrically with respect to the real axis. We let Z be the class of all functions f which are analytic and univalent in D with f(0)=0, f′(0)>0, and whose image is a Steiner symmetric domain. The elements of Z will be called Steiner symmetric functions. Using arguments similar to those used by Jenkins [15] for circularly symmetric functions, we see that a univalent function f with f(0)=0 and f′(0)>0 is Steiner symmetric if and only if it satisfies the following two conditions: (i) f is typically real and (ii) Ref is a symmetric decreasing function on each of the circles {z=r} (0<r<1). Then it follows that if f∈Z, then for every r∈(0,1), the domain f{z<r} is a Steiner symmetric domain and, hence, the function fr defined by fr(z)=f(rz) (z∈D) belongs to Z and it extends to an analytic function in the closed unit disc D¯. Now we can state our last result.Theorem 6.
Suppose thatf∈Z and let F be an analytic function in D which is convexity preserving. We have, for every p>0,(38)Mpr,f⋆F≤Mpr,f,0<r<1.Proof.
In view of Theorem1 we only need to prove (38) for 0<p<1. Let μ be the probability measure on T such that F(z)=∫T‍dμ(ξ)/1-zξ (z∈D). Then we have(39)f⋆Fz=∫Tfξzdμξ.Since F is convexity preserving, for 0<r<1, we have that fr⋆F(D¯) is contained in the closed convex hull of fr(D¯). This easily yields (40)minz∈DRefrz≤minz∈DRefr⋆Fz,maxz∈DRefr⋆Fz≤maxz∈DRefrz.By the remarks in the previous paragraph, we find that, for all r∈(0,1), fr belongs to Z and extends to an analytic function in the closed unit disc D¯. Finally, we claim that(41)Refr⋆F∗≤Refr∗,0<r<1.Once this is proved, using Proposition 6 of [10], we deduce that(42)Mpr,f⋆F=fr⋆FHp≤frHp=Mpr,f,0<p≤2,finishing our proof.
So we proceed to prove (41). Fix r∈(0,1) and set u=Re(fr⋆F), v=Refr. Using (39), we have, for 0<R<1 and 0<θ<π, (43)u∗Reiθ=supE=2θ∫EuReitdt=supE=2θ∫E∫TvReitξdμξdt=supE=2θ∫T∫EvReitξdtdμξ≤∫Tv∗Reiθdμξ=v∗Reiθ.



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*Source: 7817353-2019-03-03.xml*