# Second Hankel Determinants for the Class of Typically Real Functions

Authors: Paweł Zaprawa
Journal: Abstract and Applied Analysis
 (2016)

---

## Abstract

We discuss the Hankel determinantsH2(n)=anan+2-an+12 for typically real functions, that is, analytic functions which satisfy the condition Im⁡zIm⁡f(z)≥0 in the unit disk Δ. Main results are concerned with H2(2) and H2(3). The sharp upper and lower bounds are given. In general case, for n≥4, the results are not sharp. Moreover, we present some remarks connected with typically real odd functions.


---

## Body

## 1. Introduction

LetΔ be the unit disk {z∈C:|z|<1} and let A be the family of all functions f analytic in Δ that have the Taylor series expansion f(z)=z+∑n=2∞anzn. In [1, 2] Pommerenke defined qth Hankel determinant for a function f as(1)Hqn=anan+1⋯an+q-1an+1an+2⋯an+q⋯⋯⋯⋯an+q-1an+q⋯an+2q-2,where n,q∈N.Recently, the Hankel determinant has been studied intensively by many mathematicians. The research was focused onH2(2) for various classes of univalent functions. The papers by Janteng et al. [3, 4], Lee et al. [5], Vamshee Krishna and Ramreddy [6], and Selvaraj and Kumar [7] are worth mentioning here. Janteng et al. derived the exact bounds of |H2(2)| for the classes: S∗ of star-like functions (|H2(2)|≤1), K of convex functions (|H2(2)|≤1/8), and R of functions whose derivative has a positive real part (|H2(2)|≤4/9). Lee et al. [5] investigated the Hankel determinant in the general class S∗(φ) of star-like functions with respect to a given function φ. This class was defined by Ma and Minda in [8]. In particular, Lee et al. obtained the results for the following classes: S∗(α) of star-like functions of order α (|H2(2)|≤(1-α)2), SL∗ of lemniscate star-like functions (|H2(2)|≤1/16; for the definition of SL∗, see [9]), and Sβ∗ of strongly star-like functions of order β (|H2(2)|≤β2). Vamshee Krishna and Ramreddy [6] generalized the result of Janteng et al. They gave the bound of |H2(2)| in the class K(α) of convex functions of order α. Selvaraj and Kumar [7] proved that the estimate of the second Hankel determinant for the class C of close-to-convex functions is the same as that for the class S∗. The question whether this bound is good for the class S of all univalent functions has no answer yet. One can find some other results in this direction in [10–14].Taking different set of parametersq and n, the research on the Hankel determinant is much more difficult. In [15] Hayami and Owa discussed H2(n) for functions f satisfying Ref(z)/z>α or Ref′(z)>α. On the other hand, Babalola [16] tried to estimate |H3(1)| for S∗, K, and R. Shanmugam et al. [17] discussed |H3(1)| for the class Mα of α-star-like functions defined by Mocanu in [18].In particular, ifq=2 and n=1 then H2(1) is known as a classical functional of Fekete-Szegö. A lot of papers have been devoted to the studies concerning this functional. Because H2(1) is not related to the subject of this paper, we omit recalling results obtained in this direction.The majority of results concerning the Hankel determinants were obtained for univalent functions. In this paper we discuss functions which, in general, are not univalent. We focus our investigation on typically real functions.


## 2. ClassT and the Hankel Determinants for a Selected Functions in T

A functionf∈A that satisfies the condition Im⁡zIm⁡f(z)≥0 for z∈Δ is called a typically real function. Let T denote the class of all typically real functions. Robertson [19] proved that f∈T if and only if there exists a probability measure μ on [-1,1] such that the following formula holds:(2)fz=∫-11z1-2zt+z2dμt.The coefficients of a function f(z)=z+∑n=2∞anzn∈T can be written as follows:(3)an=∫-11sin⁡narccos⁡tsin⁡arccos⁡tdμt=∫-11Un-1tdμt,n=1,2,….The functions Un(t), n=1,2,…, which appear in the above formula, are the well-known Chebyshev polynomials of the second kind.Since all coefficients off∈T are real we look for the lower and the upper bounds of H2(2) instead of the bound of |H2(2)|. At the beginning, let us look at a few examples.Example 1.
All the functionsft(z)=z/(1-2zt+z2), t∈[-1,1], are in S∗. Since ft(z)=z+2tz2+(4t2-1)z3+(8t3-4t)z4+⋯, we have H2(2)=-1 for each t∈[-1,1]. Moreover, H2(n)=Un-1(t)Un+1(t)-Un(t)2. This and the Turan identity for Chebyshev polynomials Un(t) result in H2(n)=-1 for each n=2,3,….Example 2.
For a functionf(z)=z(1+z2)/(1-z2)2 having the Taylor series expansion f(z)=z+3z3+5z5+⋯ there is H2(n)=-(n+1)2 for even n and H2(n)=(n+1)2-1 for odd n. In this case, the function f is not univalent; the bound of |H2(2)| is much greater than 1, the value of the second Hankel determinant for star-like functions or close-to-convex functions.Example 3.
Every Hankel determinantH2(n), n=1,2,…, for a function f(z)=log⁡(1/1-z)=z+(1/2)z2+(1/3)z3+⋯ is positive. Namely, H2(n)=1/n(n+1)2(n+2).For a given classA⊂A, we denote by Ωn(A), n≥1, the region of variability of three succeeding coefficients of functions in A, that is, the set {anf,an+1f,an+2f:f∈A}. As it is seen in (3), the coefficients of typically real functions are the Stieltjes integrals of the Chebyshev polynomials of the second kind with respect to a probability measure. Hence, Ωn(T) is the closed convex hull of the curve γ:[-1,1]∋t→(Un-1(t),Un(t),Un+1(t)) (see, e.g., [20]).Lemma 4.
The functionalT∋f→H2(n), n≥2, attains its extreme values on the boundary of Ωn(T).Proof.
The only critical point ofh(x,y,z)=xz-y2, where x=an, y=an+1, and z=an+2, is (0,0,0). But h(0,0,0)=0. Since h may be positive as well as negative for (x,y,z)∈Ωn(T), (see Examples 1 and 3), it means that the extreme values of h are attained on the boundary of Ωn(T).


## 3. Bounds ofH2(2) in T

In [21] Ma proved so-called generalized Zalcman conjecture for the class T:(4)anam-an+m-1≤n+1,m=2,n=2,4,6,…,m+1,n=2,m=2,4,6,…,n-1m-1,otherwise.We apply this result to prove the following.Theorem 5.
Iff∈T then |H2(2)|≤9.Proof.
The result of Ma and the triangle inequality result in(5)a2a4-a32≤a2a4-a5+a5-a32≤5+4=9.This result is sharp; the equality holds forf(z)=z(1+z2)/(1-z2)2. Furthermore, we can see the following.Corollary 6.
ForT one has(6)min⁡H22:f∈T=-9.For our next theorem let us cite two results. First one is the obvious conclusion from the Carathéodory theorem and the Krein-Milman theorem. We assume thatX is a compact Hausdorff space and(7)Jμ=∫XJtdμt.Theorem A (see [22, Thm. 1.40]).
IfJ:X→Rn is continuous then the convex hull of J(X) is a compact set and it coincides with the set {Jμ:μ∈PX,supp(μ)≤n}.In the above, the symbolsPX and supp(μ) stand for the set of probability measures on X and the cardinality of the support of μ, respectively.It means thatμ is atomic measure having at most n steps. More precise information about the relation between the measure and the convex hull is presented in the following theorem. In what follows, 〈a→,b→〉 means the scalar product of a→ and b→.Theorem B (see [22, Thm. 1.49]).
LetJ:[α,β]→Rn be continuous. Suppose that there exists a positive integer k, such that for each nonzero p in Rn the number of solutions of any equation 〈J(t)→,p→〉=const, α≤t≤β, is not greater than k. Then, for every μ∈P[α,β] such that Jμ belongs to the boundary of the convex hull of J([α,β]) the following statements are true: (1)
Ifk=2m then
(a)
supp(μ)≤m, or(b)
supp(μ)=m+1 and {α,β}⊂supp(μ).(2)
Ifk=2m+1 then
(a)
supp(μ)≤m, or(b)
supp(μ)=m+1 and one of the points α and β belongs to supp(μ).This theorem, in slightly modified version, was published in [23] as Lemma 2.PuttingJ(t)=[U1(t),U2(t),U3(t)], t∈[-1,1], and p→=[p1,p2,p3], we can see that any equation of the form(8)p1U1t+p2U2t+p3U3t=const,t∈-1,1is equivalent to W3(t)=const, where W3(t) is a polynomial of degree 3. Hence, (8) has at most 3 solutions. According to Theorem B, the boundary of the convex hull of J([-1,1]) is determined by atomic measures μ for which support consists of at most 2 points. Moreover, one of them has to be −1 or 1. We have proved the following.Lemma 7.
The boundary ofΩ2(T) consists of points (a2,a3,a4) that correspond to the following functions:(9)fz=αz1-2zt+z2+1-αz1-z2,α∈0,1,t∈-1,1or(10)fz=αz1-2zt+z2+1-αz1+z2,α∈0,1,t∈-1,1.Now, we are ready to prove the following.Theorem 8.
ForT one has(11)maxH22:f∈T=1.Proof.
By Lemma7, it is enough to take functions given by (9) or (10). Consider the following:
(I) Function (9) has the series expansion(12)fz=z+21-α+αtz2+31-α+4t2-1αz3+41-α+2t3-tαz4+⋯.Hence, H2(2)=g1(α,t), where(13)g1α,t=81-α+αt1-α+2t3-tα-4t2α-4α+32α∈0,1,t∈-1,1.
From(14)∂g1∂α=81-t222-αt2+1-2α,∂g1∂t=16αα-14t2-3t+3,it follows that the critical points of g1 are as follows: (0,-1), (0,1), (1,-1), (1,1), (1,-1/2), (1,1/2), and (1/2,0). Among these points, only (1/2,0) lies inside the set [0,1]×[-1,1].
Ifα=0 or α=1 then functions (9) coincide with ft from Example 1. If t=1 then f(z)=z/(1-z)2. In each case H2(2)=-1. For t=-1, function (9) takes the form f(z)=α(z/(1+z)2)+(1-α)(z/(1-z)2). Then H2(2)=8(1-2α)2-9≤-1.
Ifα=1/2 and t=0 we have H2(2)=1. It means that the greatest value of H2(2) for functions given by (9) is equal to 1. The extremal function is(15)fz=12z1+z2+z1-z2=z+z2+z3+2z4+3z5+⋯.
(II) For functions (10), H2(2) is equal to g2(α,t), where(16)g2α,t=8-1+α+αt-1+α+2t3-tα-4t2α-4α+32α∈0,1,t∈-1,1.Moreover, g2(α,t)=g1(α,-t). Taking into account the symmetry of the range of variability of t, we obtain the same result as above also for functions defined by (10). The extremal function is(17)fz=12z1+z2+z1+z2=z-z2+z3-2z4+3z5+⋯.


## 4. Bounds ofH2(3) in T

The proof of the following theorem is obvious.Theorem 9.
Ifn is odd then(18)max⁡H2n:f∈T=nn+2.Hence, one has the following.Corollary 10.
ForT one has(19)max⁡H23:f∈T=15.In similar way, as it was done for Lemma7, one can prove the following.Lemma 11.
The boundary ofΩ3(T) consists of points (a3,a4,a5) that correspond to the following functions:(20)fz=αz1-2zt1+z2+1-αz1-2zt2+z2,α∈0,1,t1,t2∈-1,1or(21)fz=αz1+z2+βz1-z2+1-α-βz1-2zt+z2,α,β∈0,1,α+β≤1,t∈-1,1.Theorem 12.
ForT one has(22)min⁡H23:f∈T=-43+896=-3.51….Proof.
By Lemma11, it suffices to discuss functions given by (20) or (21). Consider the following:
(I) For functions (20), we have (23)a3=4t12-1α+4t22-11-α,a4=8t13-4t1α+8t23-4t21-α,a5=16t14-12t12+1α+16t24-12t22+11-α,and, hence, applying the Turan identity, H2(3)=g3(α,t1,t2), where(24)g3α,t1,t2=-α2-1-α2+2α1-α8t1-t221-t121-t22-1,α∈0,1,t1,t2∈-1,1.
The expression in brackets is greater than or equal to −1 for allt1,t2∈[-1,1]. Hence,(25)H23≥-α2-1-α2-2α1-α=-1.
(II) If functionf is given by (21) then (26)a3=3α+β+4t2-11-α-β,a4=4β-α+8t3-4t1-α-β,a5=5α+β+16t4-12t2+11-α-β.Using the Turan identity, it follows that H2(3)=g4(α,β,t), where(27)g4α,β,t=-1+2α+β-2α+β2+64αβ+21-α-βqt,qt=α+β24t4-8t2-1+α-β32t3-16t,under the assumptions α,β∈0,1, α+β≤1, and t∈[-1,1].
Letα and β be fixed. Since(28)∂q∂t=86t2-1tα+β+α-β,the critical points of q are as follows: -1/6, 1/6, and (β-α)/(α+β). It is easily seen that all these points are in [-1,1]. Therefore,(29)min⁡qt:t∈-1,1=min⁡q-1,q-16,q16,q1,qβ-αα+β.
Fort=-1 or t=1, the functions given by (21) have the form(30)fz=α~z1+z2+β~z1-z2,α~+β~=1.One can show directly from formula (30) that(31)H23=-1+64α~1-α~≥-1.For t=(β-α)/(α+β), there is(32)H23=-1+64αβ4α+βαβ+β-α2α+β3;hence,(33)H23≥-1.If t=-1/6 or t=1/6 then H2(3) is equal to(34)H23=-1+2α+β-2α+β2+64αβ+1-α-β-103α+β-6436α-βor(35)H23=-1+2α+β-2α+β2+64αβ+1-α-β-103α+β+6436α-β,respectively. Without loss of generality, we can assume that α≥β. Then, while looking for the minimum value of H2(3), we can restrict the research to the first stated above case (since expression (35) is not less than expression (34)).
Transforming (35), we obtain(36)H23=-1+64αβ-43α+β1-α-β+6436β-α1-α-β.Taking the smallest possible β (i.e., β=0) the second and the forth component of this expression will not increase. The value of the third component does not depend only on β; in fact, it depends on α+β. For this reason, we can take β=0. Combining these facts, it yields that(37)H23≥-1-43α1-α-6436α1-α.The smallest value of the right hand side of this inequality is achieved for α=1/2. In this case,(38)H23≥-1-13-1636=-43-896=-3.51….
Combining two parts of the proof we obtain the conclusion of the theorem. Furthermore, the above shows that the extremal functions are(39)fz=12z1+z2+z1-2zt0+z2,fz=12z1-z2+z1+2zt0+z2,where t0=1/6.


## 5. Bounds ofH2(n), n≥4, in T

It is easily seen thatH2(n)≤n(n+2) for any typically real function. By Theorem 9, this estimate is sharp providing that n is an odd integer. At the beginning of this section we will prove the following.Theorem 13.
ForT one has(40)min⁡anan+2:f∈T=-1.Proof.
The coefficients of the series expansion of functionf∈T can be written as follows:(41)an=∫0πsinnθsin⁡θdνθ,ν∈P0,π.Hence,(42)anan+2=∫0πsin⁡n+1θ-θsin⁡θdνθ·∫0πsin⁡n+1θ+θsin⁡θdνθ=∫0πsin⁡n+1θsin⁡θcos⁡θdνθ2-∫0πcos⁡n+1θdνθ2.Since(43)∫0πcos⁡n+1θdνθ≤∫0πdνθ=1,we obtain(44)anan+2≥-1.In order to prove that the estimate is sharp, let us take the measure ν for which support satisfies condition (n+1)θ=π. This measure corresponds to the function f(z)=z/(1-2zcos(π/(n+1))+z2).Observe thatanan+2=-1 holds not only for the measure stated above. Namely, the value −1 in (42) is taken also if (n+1)θ=kπ, where k is any positive integer less than or equal to n. From this we conclude that the support of the measure has n points θk=kπ/(n+1) with weights αk, k=1,2,…,n, such that ∑k=1nαk=1.The weights satisfy(45)∑k=1nαk-1k2=1.Indeed, if the support of ν consists of n points then f takes the form(46)fz=∑k=1nαkz1-2zcos⁡θk+z2.Using trigonometric identities we obtain(47)an=∑k=1nαk-1k+1,an+2=∑k=1nαk-1k,which results in (45).Connecting (45) and ∑k=1nαk=1 we conclude that f is of the form(48)fz=∑k=1,kisoddnαkz1-2zcos⁡kπ/n+1+z2or(49)fz=∑k=1,kisevennαkz1-2zcos⁡kπ/n+1+z2.It means that for even n the support of ν consists of n/2 points, and for even n the number of points of the support of ν is equal to (n+1)/2 or (n-1)/2.Taking into account|an+1|≤n+1 and Theorem 13, we obtain the following.Theorem 14.
ForT one has(50)H2n≥-n+12-1.Unfortunately, this bound is not sharp. However, the following can be conjectured.Conjecture 15.
For any positive integern, the following estimate H2(n)≥-(n+1)2 holds. Moreover, this bound is sharp for even n.This conjecture is supported by the facts that in the theorems concerningH2(2) and H2(3) the extremal functions are of the form(51)fz=12z1+z2+z1-2zt+z2,fz=12z1-z2+z1-2zt+z2,for appropriately taken t∈[-1,1]. The exact bounds of the Hankel determinants for these functions are collected in Table 1. They were obtained numerically.Table 1
The bounds of the Hankel determinants for functions defined by (51).
Bounds ofH2(n)
Equality in the lower bound for
Equality in the upper bound for
-
9
≤
H
2
(
2
)
≤
1
t
=
1
t
=
0
-
3.51
…
≤
H
2
(
3
)
≤
15
t
=
0,40
…
t
=
1
-
25
≤
H
2
(
4
)
≤
4.46
…
t
=
1
t
=
0.61
…
-
7.84
…
≤
H
2
(
5
)
≤
35
t
=
0,72
…
t
=
1
-
49
≤
H
2
(
6
)
≤
9.67
…
t
=
1
t
=
0.79
…


## 6. Remarks ConcerningH2(n) in T(2)

In classT we discuss subclass T(2) consisting of the functions which are odd. The definition of this class is (52)T2=f∈T:f-z=-fz,z∈Δ.For f∈T(2) the representation formula, similar to (2), is valid. Namely,(53)fz=∫-11z1+z21+z22-4z2t2dνt,ν∈P-1,1.Function f has the Taylor series expansion(54)fz=∑nisoddanzn,an=∫-11Un-1tdνt.The following inequalities are obvious:(55)-n+12≤H2n≤0forevenn,H2n≤n+12-1foroddn;equalities hold for f(z)=z(1+z2)/(1-z2)2.For a given classA⊂A, let us denote by Ψn(A), n≥1, the set {an,an+2:f∈A}. From (53) it follows that Ψn(T) is the closed convex hull of the curve(56)λ:-1,1∋t⟶Un-1t,Un+1t.From Theorem13 and from the equivalence(57)an,an+2∈ΨnT⟺an,an+2∈ΨnT2,we get(58)min⁡anan+2:f∈T2=-1.Hence, for odd n, we know that(59)min⁡H2n:f∈T2≥-1.The equality holds for functions (48) or (49) providing that αk=αn+1-k. Then, connecting the components of these formulae in pairs, we obtain(60)αkz1-2zcoskπ/n+1+z2+αn+1-kz1-2zcosn+1-kπ/n+1+z2=2αkz1+z21+z22-4z2cos2kπ/n+1.With help of the argument given in the proof of Theorem 13, we eventually obtain the odd functions for which anan+2=-1.



---

*Source: 3792367-2016-02-03.xml*