https://doi.org/10.1007/s40840-021-01122-x
Sharp Bounds of the Hermitian Toeplitz Determinants for Some Classes of Close-to-Convex Functions
Adam Lecko1 ·Barbara ´Smiarowska1
Received: 26 January 2021 / Revised: 17 March 2021 / Accepted: 5 April 2021 / Published online: 16 April 2021
© The Author(s) 2021
Abstract
Sharp upper and lower bounds of the Hermitian Toeplitz determinants of the second and third orders are found for various subclasses of close-to-convex functions.
Keywords Hermitian Toeplitz determinant·Univalent function·Close-to-convex functions·Carathéodory class
Mathematics Subject Classification 30C45·30C50
1 Introduction and Definitions
Let D := {z ∈ C : |z| < 1}andD := {z ∈ C : |z| ≤ 1},and forr > 0,let Tr := {z ∈ C : |z| = r}andT := T1. Denote byHbe the class of all analytic functions f inDand byAthe subclass ofHwith f normalized such that f(0)=0 and f(0)=1,so that f(z)is of the form
f(z)= ∞ n=1
anzn, a1:=1, z∈D. (1)
LetSbe the subclass ofAconsisting of univalent functions.
Communicated by V. Ravichandran.
B
Adam Leckoalecko@matman.uwm.edu.pl Barbara ´Smiarowska
b.smiarowska@matman.uwm.edu.pl
1 Department of Complex Analysis, Faculty of Mathematics and Computer Science, University of Warmia and Mazury in Olsztyn, ul. Słoneczna 54, 10-710 Olsztyn, Poland
Forq,n ∈N,consider the matrixTq,n(f)with f ∈Agiven by (1) defined by
Tq,n(f):=
⎡
⎢⎢
⎢⎣
an an+1 . . .an+q−1
an+1 an . . .an+q−2
... ... ... ...
an+q−1an+q−2. . . an
⎤
⎥⎥
⎥⎦,
whereak :=ak.In the case whenanis a real number,Tq,n(f)is called an Hermitian Toeplitz matrix.
In recent years, a great many papers have been devoted to the estimation of deter- minants whose entries are coefficients of functions inA or its subclasses. Hankel matrices, i.e., square matrices which have constant entries along the reverse diagonal and the generalized Zalcman functional Jm,n(f):=am+n−1−aman, m,n∈N,are of particular interest (see, e.g., [5,6,8,13,15,16,18–20,25]). Also of interest are the determinants of symmetric Toeplitz matrices, the study of which was initiated in [1].
In [9,11,14], research was investigated into the study of Hermitian Toeplitz deter- minants whose entries are the coefficients of functions in subclasses ofA.
In this paper, we continue this research by computing the sharp upper and lower bounds of the second- and third-order Hermitian Toeplitz determinants over some subclasses of close-to-convex functions, but first noting that the following general result was proved in [11].
Theorem 1 ([11]) LetF be a subclass ofAsuch that{f ∈ F : a2 = 0} = ∅and A2(F):=max{|a2| : f ∈F}exists. Then
1−A22(F)≤detT2,1(f)≤1.
Both inequalities are sharp.
We next define the classes of close-to-convex functions considered in this paper. First denote byS∗the subclass ofSconsisting of the starlike functions, i.e., f ∈S∗if and only if f ∈Aand
Rez f(z)
f(z) >0, z∈D.
A function f ∈Ais called close-to-convex if there existg∈S∗andδ ∈Rsuch that Reeiδz f(z)
g(z) >0, z∈D. (2)
The classCof all close-to-convex functions (which is necessarily a subclassS), was introduced by Kaplan [12] (see also [10, Vol. II, p. 3]), where the following geometrical interpretation was given:f ∈Ais close-to-convex if and only if there are no sections of the curve f(Tr),for everyr ∈(0,1),in which tangent vector turns backward through an angle not less than π (cf. [10, Vol. II, p. 4]). Lewandowski [22,23] proved that
the class of close-to-convex functions is identical with the class of linearly accessible functions introduced by Biernacki [3].
Giveng∈S∗andδ∈R,letCδ(g)be the subclass ofCof all f satisfying (2). The four classesC0(gi),i =1, . . . ,4,where
g1(z):= z
1−z2, g2(z):= z
(1−z)2, g3(z):= z 1+z+z2 and
g4(z):= z
1−z, z∈D,
are particularly interesting and have been studied by various authors (e.g., [2,7,17]). In [14], the sharp bounds of the second- and third-order Hermitian Toeplitz determinants were found for the classesC0(g1)andC0(g2).In this paper, we will do the same for the other two classes, i.e., forC0(g3)=:F1andC0(g4)=:F2which f in view of (2) satisfy the conditions
Re 1+z+z2
f(z)
>0, z∈D, (3)
and
Re
(1−z)f(z)
>0, z∈D, (4)
respectively. We note here that in [21] the classesC(δ, ξ1, ξ2),whereδ∈(−π/2, π/2), ξ1, ξ2∈D,of univalent functions were introduced by generalizing Robertson’s con- dition for convexity in the direction of the imaginary axis [26]. In particular, the class C(0, (−1−√
3i)/2, (−1+√
3i)/2)is identical to the classF1and the classC(0,0,1) is identical to the classF2.A geometrical property of functions in classesC(δ, ξ1, ξ2) relating to the hyperbolic or parabolic family of arcs related toξ1andξ2was presented in [21].
LetPbe the class of allp∈Hof the form p(z)=1+∞
n=1
cnzn, z∈D, (5)
having a positive real part inD.
2 Lemmas
In the proof of our main result, we will use the following lemma, see ([4], [24, p.
166]).
Lemma 1 If p∈Pis of the form(5), then
|cn| ≤2, n ∈N. (6)
Moreover,
c1=2ζ1 (7)
and
c2=2ζ12+2(1− |ζ1|2)ζ2 (8) for someζi ∈D, i∈ {1,2}.
Forζ1∈T, there is a unique function p∈Pwith c1as in(7), namely p(z)= 1+ζ1z
1−ζ1z, z∈D.
Forζ1∈Dandζ2∈T, there is a unique function p∈P with c1and c2as in(7)and (8), namely
p(z)= 1+(ζ1ζ2+ζ1)z+ζ2z2
1+(ζ1ζ2−ζ1)z−ζ2z2, z∈D. (9) 3 The ClassF1
Let f ∈F1be the form (1). Then by (3) there exists p∈P of the form (5) such that
1+z+z2
f(z)= p(z), z∈D. (10)
Substituting (1) and (5) into (10) and equating the coefficients, we obtain a2= 1
2(−1+c1) , a3=1
3(−c1+c2) . (11)
Hence, using (6) it follows that A2(F1)=3/2 with f1∈F1satisfying
1+z+z2
f1(z)=1−z
1+z, z∈D, i.e.,
f1(z)= z
0
1−ξ
1+2ξ+2ξ2+ξ3dξ =z−3
2z2+ · · · , z∈D.
Also f2∈F1such that
1+z+z2
f2(z)= 1
1−z, z∈D, i.e.,
f2(z)= z
0
dξ
1−ξ3 =z+1
4z4+ · · ·, z∈D, (12) serves as the extreme function, sincea2=0. Thus, Theorem1gives
Theorem 2 If f ∈F1, then
−5
4 ≤detT2,1(f)≤1. Both inequalities are sharp.
We now find the upper and lower bounds of detT3,1(f)in the classF1, first noting that
detT3,1(f)=
1 a2 a3
a2 1 a2
a3a2 1 =2 Re
a22a3
−2|a2|2− |a3|2+1. (13)
Theorem 3 If f ∈F1,then
−1≤detT3,1(f)≤1. (14) Both inequalities are sharp.
Proof We first find the upper bound.
By (11) and (6), we see that|a2| ≤3/2 and|a3| ≤4/3.Since Re(a22a3)≤ |a2|2|a3|, it follows from (13) that
detT3,1(f)≤F(|a2|,|a3|) , (15) where
F(x,y):=2x2y−2x2−y2+1, (x,y)∈[0,3/2]×[0,4/3].
Observe now that the point(1,1)is the unique solution in(0,3/2)×(0,4/3)of the system of equations
∂F
∂x =4x(y−1)=0,
∂F
∂y =2(x2−y)=0.
However,
∂2F
∂x2(1,1)∂2F
∂y2(1,1)− ∂F
∂x∂y(1,1) 2
= −16<0, so(1,1)is a saddle point ofF.
We now considerFon the boundary of[0,3/2] × [0,4/3].
(1) On the sidex=0,
F(0,y)=1−y2≤1, 0≤y≤ 4 3.
(2) On the sidex=3/2,
F 3
2,y
= −7 2 +9
2y−y2≤F 3
2,4 3
= 13
18, 0≤ y≤ 4 3. (3) On the sidey=0,
F(x,0)=1−2x2≤1, 0≤x≤ 3 2. (4) On the sidey=4/3
F
x,4 3
= −7 9+2
3x2≤ 13
18, 0≤x≤ 3 2.
Therefore, the inequalityF(x,y)≤1 holds for all(x,y)∈ [0,3/2]×[0,4/3],which in view of (15) gives the upper bound.
For the lower bound, we substitute (7) and (8) into (11) to obtain a2=1
2(−1+2ζ1) , a3=1 3
−2ζ1+2ζ12+2
1− |ζ1|2 ζ2
withζi ∈D,i =1,2.Therefore, from (13) we have detT3,1(f)= 1
18(Ψ1+Ψ2) , (16)
where
Ψ1:=9−20|ζ1|2+16|ζ1|4−8(1− |ζ1|2)2|ζ2|2 (17) and
Ψ2:=30 Reζ1+6 Re ζ12
−32|ζ1|2Reζ1−8
1− |ζ1|2 Re
ζ1ζ2
+6
1− |ζ1|2
Reζ2+8
1− |ζ1|2 Re
ζ12ζ2
.
We now consider various cases.
A.Suppose thatζ1ζ2 = 0.Thenζ1 =reiθ andζ2= seiψ withr,s ∈ (0,1]and θ, ψ∈ [0,2π).Then
Ψ2=Ψ3+Ψ4, (18)
where
Ψ3:=30rcosθ+6r2cos 2θ−32r3cosθ
= −6r2+30rcosθ−32r3cosθ+12r2cos2θ (19)
and
Ψ4:= −8r s 1−r2
cos(θ−ψ)+6s 1−r2
cosψ+8r2s
1−r2
cos(2θ−ψ)
=2s(1−r2)
κ12+κ22sin(ψ+α),
(20) whereα∈Rsatisfies
cosα= κ1
κ12+κ22, sinα= κ2
κ12+κ22 (21)
with
κ1:= −4rsinθ+4r2sin 2θ, κ2:=3−4rcosθ+4r2cos 2θ. (22) Since sin(ψ+α)≥ −1 ands≤1, we have
Ψ4≥ −2
1−r2 κ12+κ22
= −2
1−r2 9−8r2+16r4−24rcosθ−32r3cosθ+48r2cos2θ.
Therefore, from (18)–(20), we obtain Ψ2=Ψ3+Ψ4
≥30rcosθ+6r2cos 2θ−32r3cosθ
−2
1−r2 9−8r2+16r4−24rcosθ−32r3cosθ+48r2cos2θ.
(23) Noting that|ζ2| ≤1 from (17) we have
Ψ1≥1−4r2+8r4. (24)
Thus, from (16), (23) and (24) it follows that
18 detT3,1(f)≥G(r,cosθ), r∈(0,1], θ∈ [0,2π), (25) where
G(t,x):=g1(t,x)−2(1−t2) g2(t,x) with
g1(t,x):=1−10t2+8t4+30t x−32t3x+12t2x2 and
g2(t,x):=9−8t2+16t4−24t x−32t3x+48t2x2 (26)
fort ∈ [0,1]andx∈ [−1,1].
LetΩ:= [0,1] × [−1,1].Now we will show that min{G(t,x):(t,x)∈Ω} = −18.
A1.We next deal with the critical points ofG in the interior ofΩ, i.e., in(0,1)× (−1,1).Note thatg2(t,x)≥0.Moreover,g2(t,x)=0 holds only fort =√
3/2 and x=√
3/3. A1.1.Whent =√
3/2 andx=√
3/3,from (17), (19)and (20) we have Ψ1=3−1
2|ζ2|2, Ψ3= 3
2, Ψ4=0, which gives
detT3,1(f)= − 1
36(−9+ |ζ2|2)≥ 2 9.
A1.2.We now consider caseg2(t,x) >0.DifferentiatingGwith respect toxyields 0= ∂G
∂x(t,x)= ∂g1
∂x (t,x)−(1−t2)(g2(t,x))−1/2∂g2
∂x(t,x). (27) A1.2.a.Assume first that∂g1/∂x =0.Then from (27), it follows that∂g2/∂x =0, which is possible only fort =√
6/2>1 andx=√ 6/4.
A1.2.b.Assume next that∂g1/∂x =0.Then we can write (27) as
g2(t,x)1/2=
1−t2∂g2
∂x (t,x)
∂g1
∂x(t,x) = 4
1−t2 −3−4t2+12t x
15−16t2+12t x , (28)
or equivalently, by substituting (26), as
Φ =Φ(t,x):=3840t8−12800t7x+24576t6x2−23040t5x3+6912t4x4
−9600t6+16128t5x−21888t4x2+13824t3x3+10112t4
−2784t3x+1152t2x2−6216t2−1008t x+1881=0.
(29) Now note that by (28) the inequality
0< (g2(t,x))1/2 4
1−t2 = −3−4t2+12t x
15−16t2+12t x (30)
holds for
0<t <
3
2 and 4t2+3
12t <x <16t2−15 12t .
DifferentiatingGwith respect tot, we have
∂G
∂t (t,x)=∂g1
∂t (t,x)+4t(g2(t,x))1/2− 1−t2
(g2(t,x))−1/2∂g2
∂t (t,x). (31) By (28) and (31), we get
∂G
∂t (t,x)= − 72t
12xt−4t2−3 12xt−16t2+15H(t,x),
where for(t,x)∈(0,1)×(−1,1), H(t,x):=
−3+4t2−4t x 9−18t2+8t4+6t x+8t3x−24t2x2 .
Therefore, each critical point ofGsatisfies
−3+4t2−4t x=0 (32) or
9−18t2+8t4+6t x+8t3x−24t2x2=0. (33) I.Assume that (32) holds, thenx = x(t)=(−3+4t2)/(4t).Thus, by (29) we see that
Φ
t,−3+4t2 4t
= −8(2t+1)(2t−1)
4t2−5 2t2−3 2
=0
occurs only whent=1/2.Thus,x=x(1/2)= −1.However, it can be seen that the right side of (30) equals to−2 fort =1/2 andx = −1,which means that then the inequality (30) is not true. Therefore,Gdoes not have critical point in the interior of Ωin the case of (32).
II.Suppose now that (33) is satisfied. Since Eq. (33) is a quadratic inxwith:=
225−408t2+208t4>0, t∈(0,1),it has two roots, namely xi =xi(t)= 3+4t2+(−1)i+1√
225−408t2+208t4
24t , i =1,2. (34)
a.Letx=x1.ThenΦ(t,x1(t))=0 is equivalent to the equation
−561+2204t2−2736t4+1088t6 225−408t2+208t4
=11169−48552t2+80288t4−59520t6+16640t8. (35) Squaring both sides of (35) leads to
2304γ1(t)γ22(t)=0, (36)
where fort∈(0,1),
γ1(t):=289−408t2+208t4 and
γ2(t):=9−27t2+26t4−8t6= −(1−t)(1+t)
3−4t2 3−2t2 .
We see that there is a unique roott =√
3/2 of the Eq. (36), which also satisfies (35).
Then by (34),x1(√
3/2)=√
3/3.However, this case was discussed in A1.1.
b.Letx=x2.ThenΦ(t,x2(t))=0 is equivalent to the equation
−
−561+2204t2−2736t4+1088t6 225−408t2+208t4
=11169−48552t2+80288t4−59520t6+16640t8. (37) Squaring both sides of (37) yields again the Eq. (36) having a unique roott =√
3/2, which does not satisfy (37)
A2.It therefore remains to considerGon the boundary ofΩ. (1) On the sidet =0,
G(0,x)≡ −5, x∈ [−1,1].
(2) On the sidet =1,
G(1,x)= −1−2x+12x2≥G
1, 1 12
= −13
12, x ∈ [−1,1].
(3) On the sidex = −1,
G(t,−1)= −5−38t+40t3+16t4=:1(t), t∈ [0,1].
Since1(t)=0 occurs only whent=1/2 and1(1/2)=168>0, we have 1(t)≥1(1/2)= −18, t ∈ [0,1].
(4) On the sidex =1,
G(t,1)= −5+38t−40t3+16t4=:2(t), t ∈ [0,1].
Since 2(t) = 0 occurs only when t = (19−√
57)/16 ∈ [0,1] and2((19−
√57)/16)=(57−27√
57)/2<0, we have
2(t)≥2(0)= −5, t ∈ [0,1].
B.Suppose thatζ1=0.Then
Ψ1=9−8|ζ2|2, Ψ2=6 Reζ2, and therefore,
detT3,1(f)= 1
18(9+6 Reζ2−8|ζ2|2)≥ −5
18. (38)
C.Suppose thatζ2=0 andζ1=reiθ =0,wherer∈(0,1]andθ∈ [0,2π).Then Ψ1=9−20|ζ1|2+16|ζ1|4=9−20r2+16r4
and
Ψ2=30 Reζ1+6 Re ζ12
−32|ζ1|2Reζ1=30rcosθ+6r2cos 2θ−32r3cosθ.
Thus,
18 detT3,1(f)=G(r,cosθ), r∈(0,1], θ∈ [0,2π), where
G(t,x):=9−26t2+16t4+30t x−32t3x+12t2x2 fort ∈(0,1]andx∈ [−1,1].Set
xw := −15−16t2
12t , t ∈(0,1].
Note that−1<xw occurs fort > (√
69−3)/8=0.663328· · · ,andxw<1 holds fort ∈(0,1).Hence, fort∈
(√
69−3)/8,1
we have G(t,x)≥G(t,xw)= −39
4 +14t2−16
3 t4=:φ1(t).
Sinceφ1is increasing in (√
69−3)/8,1 ,so
φ1(t)≥ − 1 32
123+3√ 69
= −4.622495· · ·
for t ∈ (√
69−3)/8,1
. Further, xw < −1 occurs fort ∈ 0, (√
69−3)/8 . Hence,
G(t,x)≥G(t,−1)=9−30t−14t2+32t3+16t4=:φ2(t).
Sinceφ2(t)=0 have a unique roott0=(−2+√
19)/4=0.589724· · · andφ2(t0)= 152>0, then
φ2(t)≥φ2
1 4
−2+√ 19
= −81
16= −5.062500· · ·, t ∈
0,1 8
√ 69−3
.
Summarizing, form Parts A–C it follows the lower bound in (14).
We discuss now sharpness of (14). The function f2 defined by (12), for which a2 = a3 = 0, is extremal for the upper bound in (14). It is observed from (16), (23), (24) and (25) that equality for the lower bound in (14) holds when the following conditions are satisfied:
r =1
2, cosθ= −1, s=1, sin(ψ+α)= −1, (39) whereαis determined by the condition (21) withκ1andκ2given in (22). Thus,θ=π, α=π/2 andψ =π.Consequently,ζ1= −1/2 andζ2= −1,which in view of (9) holds for the function
p(z)= 1−z2
1+z+z2, z∈D,
in the classP.Therefore, the extremal function f in the classF1for the lower bound in (14) satisfies (10) with p given as above, having the coefficientsa2 = −1 and
a3=0.
4 The ClassF2
Let f ∈F2be the form (1). Then by (4) there exists p∈P of the form (5) such that
(1−z)f(z)= p(z), z∈D. (40)
Putting the series (1) and (5) into (40) by equating the coefficients, we get a2= 1
2(1+c1), a3= 1
3(1+c1+c2). (41) Hence and by (6), it follows that A2(F2)=3/2 with the extremal function f1∈F2
such that
(1−z)f1(z)= 1+z
1−z, z∈D. (42)
Observe also thata2=0 for the function f2∈F2such that (1−z)f2(z)= 1
1+z, z∈D.
Therefore, by Theorem1we have
Theorem 4 If f ∈F2, then
−5
4 ≤detT2,1(f)≤1. Both inequalities are sharp.
Now we estimate detT3,1(f)for functions in the classF2. Theorem 5 If f ∈F2,then
detT3,1(f)≤ 11
9 . (43)
The inequality is sharp.
Proof By (41) and (6), we see that|a2| ≤ 3/2 and|a3| ≤ 5/3.As in the proof of Theorem3, the inequality (15) holds with the function
F(x,y):=2x2y−2x2−y2+1, (x,y)∈ [0,3/2] × [0,5/3].
Repeating argumentation in the proof of Theorem3, we see that the functionF does not have any relative maxima in(0,3/2)×(0,5/3).
We considerFon the boundary of[0,3/2] × [0,5/3].
(1) On the sidex=0,
F(0,y)=1−y2≤1, 0≤y≤ 5 3. (2) On the sidex=3/2,
F 3
2,y
= −7 2 +9
2y−y2≤F 3
2,5 3
= 11
9 , 0≤ y≤ 5 3. (3) On the sidey=0,
F(x,0)=1−2x2≤1, 0≤x≤ 3 2. (4) On the sidey=5/3,
F
x,5 3
= −16 9 +4
3x2≤ F 3
2,5 3
= 11
9 , 0≤x ≤ 3 2.
Therefore, the inequality F(x,y)≤ 11/9 holds for all(x,y) ∈ [0,3/2] × [0,5/3]
which in view of (15) shows (43).
For the function f1given by (42),a2 =3/2 anda3=5/3 which makes equality
in (43).
Theorem 6 If f ∈F2,then detT3,1(f)≥ 1
44
32−31√ 3
= −0.493035· · · (44)
The inequality is sharp.
Proof Substituting (7) and (8) into (41) yields a2=1
2(2ζ1+1) , a3= 1 3
1+2ζ1+2ζ12+2
1− |ζ1|2 ζ2
,
for someζi ∈D(i =1,2). Therefore, from (13) we get detT3,1(f)= 1
18(Ψ1+Ψ2) , (45)
where
Ψ1:=10−20|ζ1|2+16|ζ1|4−8
1− |ζ1|22
|ζ2|2 (46) and
Ψ2:= −26 Reζ1+10 Re ζ12
+32|ζ1|2Reζ1+8
1− |ζ1|2 Re
ζ1ζ2
−2
1− |ζ1|2
Reζ2+8(1− |ζ1|2)Re ζ12ζ2
.
A.Suppose thatζ1ζ2 = 0.Thus,ζ1 = reiθ andζ2 = seiψ withr,s ∈ (0,1] and θ, ψ∈ [0,2π).Then
Ψ2=2(Ψ3+Ψ4) , (47)
where
Ψ3:= −13rcosθ+5r2cos 2θ+16r3cosθ
= −5r2−13rcosθ+16r3cosθ+10r2cos2θ (48) and
Ψ4:=4r s
1−r2
cos(θ−ψ)−s
1−r2
cosψ+4r2s
1−r2
cos(2θ−ψ)
=s
1−r2 κ12+κ22sin(ψ+α),
(49) whereαis the quantity satisfying (21) with
κ1:=4rsinθ+4r2sin 2θ, κ2:= −1+4rcosθ+4r2cos 2θ. (50) Since sin(ψ+α)≥ −1 ands≤1, we have
Ψ4≥ −
1−r2 κ12+κ22
= −
1−r2 1+24r2+16r4−8rcosθ+32r3cosθ−16r2cos2θ.
Therefore, from (47), (48) and (49), we get 1
2Ψ2=Ψ3+Ψ4
≥ −13rcosθ+5r2cos 2θ+16r3cosθ
−
1−r2 1+24r2+16r4−8rcosθ+32r3cosθ−16r2cos2θ.
(51) Taking into account that|ζ2| ≤1 from (46) we have
Ψ1≥2
1−2r2+4r4
. (52)
Thus, from (45), (51) and (52) it follows that
9 detT3,1(f)≥G(r,cosθ), r∈(0,1], θ∈ [0,2π), (53) where
G(t,x):=g1(t,x)−
1−t2 g2(t,x) with
g1(t,x):=1−7t2+4t4−13t x+16t3x+10t2x2 and
g2(t,x):=1+24t2+16t4−8t x+32t3x−16t2x2 (54) fort ∈ [0,1]andx∈ [−1,1].
LetΩ:= [0,1] × [−1,1]and Θ:= 9
44
32−31√ 3
= −4.4373· · · .
Now we will show that
min{G(t,x):(t,x)∈Ω} =Θ.
A1. For this, we first we find the critical points of G in the interior ofΩ, i.e., in (0,1)×(−1,1).Sinceg2(t,x)= 0 holds only fort = (√
2−1)/2 andx = 1,it follows thatg2(t,x) >0 for(t,x)∈(0,1)×(−1,1).
DifferentiatingGwith respect toxyields 0= ∂G
∂x(t,x)= ∂g1
∂x (t,x)−1
2(1−t2)(g2(t,x))−1/2∂g2
∂x (t,x). (55)
A1.1Assume first that∂g1/∂x=0.Then by (55), it follows that∂g2/∂x=0,which is possible only fort =√
2/2 andx=√
2/4.From (53), we have
9 detT3,1(f)≥G √
2 2 ,
√2 4
= −17 8 −3
2
√2= −4.246320. . .
A1.2Assume now that∂g1/∂x=0.Then we can write the Eq. (55) as
g2(t,x)1/2=
1−t2∂g2
∂x(t,x) 2∂g1
∂x (t,x) =4
1−t2 −1+4t2−4t x
−13+16t2+20t x , (56)
or equivalently, by substituting (54), as
Φ =Φ(t,x):=3840t8+18944t7x+22528t6x2+2560t5x3−6400t4x4 +128t6−9472t5x−4992t4x2+5120t3x3−7552t4
−2336t3x+1600t2x2+3800t2−2000t x+153=0.
(57)
Furthermore, note that by (56) the inequality 0< (g2(t,x))1/2
4
1−t2 = −1+4t2−4t x
−13+16t2+20t x (58)
is true for
0<t <
√2
2 and 4t2−1
4t <x<13−16t2 20t
or √
2
2 <t <1 and
x< 4t2−1
4t or x>−16t2−13 20t
. DifferentiatingGwith respect totyields
∂G
∂t (t,x)=∂g1
∂t (t,x)+2t(g2(t,x))1/2−1 2
1−t2
(g2(t,x))−1/2∂g2
∂t (t,x). (59) By (56) and (59), we get
∂G
∂t (t,x)= − 36t
−4xt+4t2−1 20xt+16t2−13H(t,x),
where
H(t,x):=
−3+4t2+4t x −11+6t2+8t4+2t x+24t3x+40t2x2 .
Therefore, each critical point ofGsatisfies
−3+4t2+4t x=0 (60) or
−11+6t2+8t4+2t x+24t3x+40t2x2=0. (61) I.Assume that (60) holds. Thenx=x(t)=(3−4t2)/(4t).Thus, by (57) we see that
Φ
t,3−4t2 4t
= −8(48t4−104t2+39)(2t2−1)2=0
occurs only whent = ˆti, i =1,2,where tˆ1:= 1
6
39−6√
13=0.6945· · · , tˆ2:=
√2
2 =0.7071· · · Thus,
x=xtˆ1
= ˆx1= −2+√ 13
39−6√
13 =0.3852· · · and
x=xtˆ2
= ˆx2=
√2
4 =0.3535· · · . However, it can be seen that
−1+4ˆt12−4tˆ1xˆ1
−13+16ˆt12+20ˆt1xˆ1
= −2<0,
which means that the inequality (58) is not satisfied fort = ˆt1andx= ˆx1. Note that the caset= ˆt2andx= ˆx2reduces to A1.1.
Therefore,Gdoes not have critical point in the interior ofΩ.
II.Suppose now that (61) is satisfied. Equation (61) as a quadratic equation ofxwith :=441−216t2−176t4>0, t∈(0,1)has two roots, namely
xi =xi(t)= −
1+12t2
+(−1)i+1√
441−216t2−176t4
40t , i =1,2. (62)
a.Letx=x1.ThenΦ(t,x1(t))=0 is equivalent to the equation −603−940t2−9936t4+7104t6 441−216t2−176t4
= −2313−96696t2+20384t4−44928t6+92928t8. (63)
Squaring the both sides of (63) leads to
518400γ1(t)γ22(t)=0, (64) where fort∈(0,1),
γ1(t):= −299+648t2+528t4 and
γ2(t):=1+t2−10t4+8t6=(1−t)(1+t)
1+4t2 1−2t2 .
Thus, there are two rootst1andt2in(0,1)of the Eq. (64), namely
t1:=
√2
2 =0.707· · ·, t2:= 1 66
−2673+2442√
3=0.59779· · · . (65) Fort =t1, by (62),x˜1:=x1(t1)=√
2/4 and this case was discussed in A1.1.
Fort =t2, by (62),
˜
x2:=x1(t2)= 21+13√ 3 2
−2673+2442√
3 =0.551477· · · . (66) It can be verified that(t˜ 2,x˜2) = 0 and the inequality (58) holds for t = t2 and x= ˜x2. Therefore,Ghas a critical point at(t2,x˜2).
b.Letx=x2.ThenΦ(t,x2(t))=0 is equivalent to the equation
−
−603−940t2−9936t4+7104t6 441−216t2−176t4
= −2313−96696t2+20384t4−44928t6+92928t8. (67) Squaring both sides of (67) yields again the Eq. (64) having rootst1andt2given by (65), which do not satisfy (67).
Therefore, by A, B1 and B2, the functionGhas a unique critical point at(t2,x˜2).
Denote
λ1:= ∂2G
∂t2 (t2,x˜2)= −30910188
351923 +90657386 1055769
√3=60.89649· · ·,
λ2:= ∂2G
∂t∂x(t2,x˜2)= −211055
1177 +396373 3531
√3,
λ3 = ∂2G
∂x2 (t2,x˜2)= −308371
1177 +551104 3531
√3.
Sinceλ1>0 and
λ1λ3−λ22= −7933032
1177 +4769748 1177
√3=279.02623· · ·>0,
the functionGhas a local minimum at(t2,x˜2).
A2.It remains to considerGin the boundary of. (1) On the sidet =0,
G(0,x)≡0> , x∈ [−1,1].
(2) On the sidet =1,
G(1,x)= −2+3x+10x2≥G
1,−3 20
= −89
40 > , x∈ [−1,1].
(3) On the sidex = −1, G(t,−1)=t
9+8t−12t2
≥G(0,−1)=0> , t∈ [0,1].
(4) On the sidex =1,
G(t,1)=1−13t+3t2+16t3+4t4−
1−t2 −1+4t+4t2=:(t).
Whent∈
0, (−1+√ 2)/2
=:I1, we have−1+4t+4t2≤0. Therefore,
(t)=t
−9+8t+12t2 ≥G
√ 2−1
2 ,1
= −√
2≈ −1.41421> , t∈ I1.
When t ∈
(−1+√ 2)/2,1
=: I2, we have −1+4t +4t2 ≥ 0 and (t) = (−2+t+2t2)(−1+8t+4t2). Since(t)=0 occurs only whent =1/2∈I2and (1/2)=80>0,we have
(t)≥ 1
2
= −4> , t∈ I2. B.Suppose thatζ1=0.Then
Ψ1=10−8|ζ2|2, Ψ2= −2 Re(ζ2) , and therefore,
detT3,1(f)= 1 18
10−8|ζ2|2−2 Re(ζ2)
≥0.
C.Suppose thatζ2=0 andζ1=0.Then
Ψ1=10−20|ζ1|2+16|ζ1|4, Ψ2= −26 Reζ1+10 Re(ζ12)+32|ζ1|2Reζ1. Thus, takingζ1=reiθ,wherer∈(0,1]andθ∈ [0,2π)we have
Ψ1=10−20r2+16r4 Ψ2= −26rcosθ+10r2cos 2θ+32r3cosθ.
Then
9 detT3,1(f)=G(r,cosθ), r ∈(0,1], θ ∈ [0,2π), where
G(t,x):=5−15t2+8t4−13t x+16t3x+10t2x2 fort ∈(0,1]andx∈ [−1,1].Set
xw =13−16t2
20t , t ∈(0,1].
Note that−1<xw occurs fort ∈(0,1)andxw <1 holds fort > (√
77−5)/8= 0.471870· · · .Hence, fort ∈
(√
77−5)/8,1
we have G(t,x)≥G(t,xw)= 31
40−23 5 t2+8
5t4=:φ1(t).
Sinceφ1is decreasing in (√
77−5)/8,1 ,
φ1(t)≥ −89
40 = −2.225, t ∈ 1
8 √
77−5 ,1
.
On the other hand,xw>1 occurs fort ∈ 0, (√
77−5)/8
.Hence, G(t,x)≥G(t,1)=5−13t−5t2+16t3+8t4=:φ2(t).
Sinceφ2is decreasing in(0, (√
77−5)/8), φ2(t)≥ 33
64 − 5 64
√77= −0.169919· · · , t∈
0,1 8
√ 77−5
.
Summarizing, form Parts A–C it follows that the inequality (44) holds.
It remains to show that the inequality (44) is sharp. It is observed from (45), (51), (52) and (53) that 9 detT3,1(f)=holds when the following conditions are satisfied:
r =t2, cosθ = ˜x2, s=1, sin(ψ+α)= −1, (68)
wheret2andx˜2are given by (65) and (66), and whereαis determined by the condition (21) withκ1andκ2given by (50).
Now setθ = Arccos(x˜2)so that it satisfies the second condition in (68). Then κ1 =3.309903· · · >0 andκ2 = −0.241293· · · <0. Thus, (21) is satisfied if we take
α= −Arccos
⎛
⎝ κ1
κ12+κ22
⎞
⎠= −0.0727716· · · .
Thus, if we put
ψ= 3π
2 −α=4.7851606· · · ,
thenψsatisfies the fourth condition in (68). Now consider a functionp˜which has the form (9) withζ1 =t2eiθ andζ2 =eiψ. Sinceζ1 ∈Dandζ2∈ T, from Lemma1it follows that p˜∈P, and so the extremal function f in the classF2for which equality
in (44) holds satisfies (40) with p:= ˜p.
Acknowledgements The authors thank the reviewers for theirs constructive comments and suggestions that helped to improve the clarity of this manuscript.
Funding None.
Declarations
Conflict of interest The authors declare that they have no conflict of interests.
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