EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 16, No. 2, 2023, 1290-1301
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Hankel Determinant and Toeplitz Determinant
on the Class of Bazilevič Functions Related

to the Bernoulli Lemniscate

Ni Made Asih1,2,∗, Sa’adatul Fitri1, Ratno Bagus Edy Wibowo1, Marjono1

1 Department of Mathematics, Faculty of Mathematics and Natural Sciences,
University of Brawijaya, Jl. Veteran Malang 65145, Indonesia,
2 Department of Mathematics, Faculty of Mathematics and Natural Sciences,
University of Udayana, Indonesia

Abstract. In this papers, we investigate the Hankel determinant and Toeplitz determinant for
the class Bazilevič Function B1(α, δ) related to the Bernoulli Lemniscate function on the unit
disk D = {z : |z| < 1} and obtain the upper bounds of the determinant H2(1), H2(2), T2(1), and
investigate H2(1) using coefficients invers function. We used lemma from Charateodory-Toeplitz
and Libera about sharp inequalities for functions with positive real part.

2020 Mathematics Subject Classifications: 30C45, 30C50, 30C55, 30C80
Key Words and Phrases: Coefficients, Bazilevič functions, Bernoulli Lemniscate, Subordina-
tion, Hankel determinant, Toeplitz determinant.

1. Introduction

Let S denotes the class of analytic univalent function f defined on the unit disk
D = {z : |z| < 1}, and normalized by f(0) = 0 and f

′
(0) = 1, given by

f(z) = z +

∞∑
n=2

anz
n. (1)

Let P denotes the class of analytic p and satisfies the condition Re(p(z)) > 0 for
z ∈ D = {z : |z| < 1}, p ∈ P gives,

p(z) = 1 +
∞∑
n=1

pnz
n, n = 1, 2, 3, .. (2)

where pn is the positive real part [1].

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4772

Email addresses: madeasih2@student.ub.ac.id (N.M. Asih), saadatulfitri@ub.ac.id (S. Fitri),
rbagus@ub.ac.id (R.B.E. Wibowo), marjono@ub.ac.id (Marjono)

https://www.ejpam.com 1290 © 2023 EJPAM All rights reserved.



N.M. Asih et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1290-1301 1291

Definition 1. Let f ∈ S and satisfying the condition f(0) = 1 and f
′
(0) = 0. The

function f ∈ B1(α, δ) for α ≥ 0 and δ > 0 if and only if,

[
f

′
(z) f(z)α−1

zα−1

]
≺

√
1 + z =: ξ(z), for z ∈ D and ξ(0) = 1, (3)

where the branch of the square root is chosen to be ξ(0) = 1, the set ξ(D) lies in the region
bounded the right loop of the Bernoulli Lemniscate function is (x2+y2)2−a2(x2−y2)) = 0,
see [2], [14]. We say that an analytic function f is subordinate to an analytic function g,
and write f(z) ≺ g(z), if and only if there exists a function ω, analytic in D, such that

ω(0) = 0, |ω(z)| < 1 for |z| < 1 and f(z) = g(ω(z)), where ω(z) =
δp(z)− 1

δp(z) + 1
. The form of

the Lemniscate Bernoulli will be depended on the value of positive real δ. The following
picture shows the Bazilevič function B1(α, δ) related to Bernoulli Lemniscate function.

Figure 1: Bazilevič B1(α, δ) subordination Bernoulli Lemniscate

From (3) we obtain initial coefficients which are used to determine the Hankel deter-
minant and Toeplitz determinant for the sharp boundaries. The q-th Hankel determinant
is denoted by Hq(n), where q ≥ 1 and n ≥ 1 of functions f was stated by Noonan and
Thomas [12] as,

Hq(n) =

∣∣∣∣∣∣∣∣
an an + 1 .... an+q+1

an + 1 an + 2 ... an + q
... ... ... ...

an+q−1 an + q ... an+2q−2

∣∣∣∣∣∣∣∣ (4)

Since f ∈ S, a1 = 1, in particular we have H2(1) as follow,

H2(1) =

∣∣∣∣a1 a2
a2 a3

∣∣∣∣ = (a1a3 − a22).

Hankel determinant H2(1) = |a3 − a22| is well known as Fekete Szegö function. Previous
research about Hankel determinant on Starlike function related to Bernoulli Lemniscate
function in [4] obtained one of them is Hankel determinant H2(2). The other researches



N.M. Asih et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1290-1301 1292

on Third Hankel determinant are studied in [5], [9]. Research by Thomas and Halim [15]
defined the symmetric Toeplitz determinant Tq(n) for q ≥ 1 and n ≥ 1 gives,

Tq(n) =

∣∣∣∣∣∣∣∣
an an + 1 .... an+q−1

an + 1 an ... an+q−2

... ... ... ...
an+q−1 an+q−2 ... an

∣∣∣∣∣∣∣∣ (5)

An example of second order of Toeplitz determinant is T2(1) with a1 = 1, is given by

T2(1) =

∣∣∣∣a1 a2
a2 a1

∣∣∣∣ = (a21 − a22).

The well known research about the contruction of Toeplitz matrices has previously studied
by (see [13] for more detail). In his work whose element are the coefficient f univalent
functions assosiated with q-derivative operator.

2. Preliminaries

We have some lemmas used to determine sharp inequalities boundaries of Hankel de-
terminant and Toeplitz determinant.

Lemma 1. [1], [3]. If p ∈ P analityc in D with p(z) = 1 +
∑∞

n=1 pnz
n for n ≥ 1 than

|pn| ≤ 2 (6)

For the p(z) = (1+z)/(1−z), this lemmas an know as inequlity Caratheodory Toeplitz.

Lemma 2. [7]. If p ∈ P analityc in D with p(z) = 1+
∑∞

n=1 pnz
n then for some complex

values x with |x| ≤ 1 and some complex values ρ with |ρ| ≤ 1,

2p2 = p21 + x(4− p21) (7)

4p3 = p31 + 2(4− p21)p1x− p1(4− p21)x
2 + 2(4− p21)(1− |x|2)ρ (8)

3. Results

Now, we state and prove the results from Hankel determinant and Toeplitz determinant
of our investigation.

Theorem 1. If f ∈ B1(α, δ) for 0 ≤ α ≤ 1 and 0 < δ ≤ 1 then

H2(1) ≤
3
√
2
√
δ + 2(1 + α)(2 + α)δ

√
1 + δ + 3

√
2δ3/2((3 + 2α))

2(2 + α)(1 + δ)7/2
,

and the inequality is sharp.



N.M. Asih et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1290-1301 1293

Proof. First consider from (3), we have initial coefficients a1, a2 and a3 by [10], with
a1 = 1, a2 and a3 gives,

a2 =
p1
√
α√

2(1 + δ)3/2
, (9)

a3 =
δ

8(2 + α)(1 + δ)7/8

(
4
√
2p2(1 + δ)2 − p21(

√
2 + 5

√
2)δ

+4
√
2δ2 − 2(−2 + α+ α2

√
δ
√
1 + δ)

)
. (10)

From (3) and (4), we can write Hankel determinant H2(1) gives,

H2(1) =

∣∣∣∣a1 a2
a2 a3

∣∣∣∣ = |a1a3 − a22|

=

∣∣∣∣∣ p2
√
δ√

2(2 + α)(1 + δ)7/2

+
p21
√
δ(
√
2 + 5

√
2δ + 4

√
2δ2 + 2(2 + 3α+ α2)

√
δ
√
1 + δ)

8(2 + α)(1 + δ)3/2

∣∣∣∣∣. (11)

Next, applying Lemma (2) to (11), gives

H2(1) =

∣∣∣∣∣ (p21 + (4− p21)x)
√
δ√

2(2 + α)(1 + δ)7/2

+
p21
√
δ(
√
2 + 5

√
2δ + 4

√
2δ2 + 2(2 + 3α+ α2)

√
δ
√
1 + δ)

8(2 + α)(1 + δ)3/2

∣∣∣∣∣. (12)

By taing p1 = p and 0 ≤ p ≤ 2 and applying them to (12) it follows that,

H2(1) ≤ (p2 + (4− p2)|x|)
√
δ

2
√
2(2 + α)(1 + δ)3/2

+
p2
√
δ(
√
2 + 5

√
2δ + 4

√
2δ2 + 2(2 + 3α+ α2)

√
δ
√
1 + δ)

8(2 + α)(1 + δ)3/2

:= φ1(α, δ, p, |x|) (13)

From (13) then taking |x| ≤ 1 gives,

H2(1) ≤ p2 + (4− p2)
√
δ

2
√
2(2 + α)(1 + δ)3/2

+
p2(

√
δ(
√
2 + 5

√
2δ + 4

√
2δ2 + 2(2 + 3α+ α2)

√
δ
√
1 + δ)

8(2 + α)(1 + δ)7/2



N.M. Asih et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1290-1301 1294

=

√
δ(8

√
2(1 + δ)2

8(2 + α)(1 + δ)3/2

+
p2
√
δ(
√
2 + 5

√
2δ + 4

√
2δ2 + 2(2 + 3α+ α2)

√
δ
√
1 + δ)

8(2 + α)(1 + δ)7/2

:= φ1(α, δ, p) (14)

Next, we determine the derivative of φ(α, δ, p) with respect to p from (14)are we obtain,

φ
′
1(α, δ, p) =

2p
√
δ(
√
2 + 5

√
2δ + 4

√
2δ2 + 2(2 + 3α+ α2)

√
δ
√
1 + δ)

8(2 + α)(1 + δ)7/2
. (15)

Let the derivative of φ1(α, δ, p) with respect to p is φ
′
1(α, δ, p). Then, from (15), we

can show that φ
′
1 > 0 for 0 ≤ p ≤ 2. Hence, φ1 is an increasing monoton function. From

which we obtain

H2(1) ≤ φ1(α, δ, 2) =
3
√
2
√
δ + 2(1 + α)(2 + α)δ

√
1 + δ + 3

√
2δ3/2((3 + 2α))

2(2 + α)(1 + δ)7/2
.

The inequality is sharp when p1 = p2 = 2. The proof is completed.

Theorem 2. If f ∈ B1(α, δ) for α1 ≤ α ≤ 1 and 0 < δ ≤ 1 then

H2(2) ≤

(
δ(1 + δ)3/2)

6(2 + α)2(1 + δ)8

)[
(6
√
2(−1 + δ)(2 + α)2

√
δ + 30

√
2(−1 + α)

(2 + α)2δ3/2 + 24
√
2(−1 + α)(2 + α)2δ5/2 + 3(7 + 8α+ 2α2)√

1 + δ + (117 + 64α− 20α2 + 36α3 + 38α4 + 8α5)δ
√
1 + δ

+72(3 + 4α+ α2)δ2
√
1 + δ + 48(3 + 4α+ α2)δ3

√
1 + δ)

−12(1 + 5α+ 10α2 + 10α3 + 5α4)− (2 + α)2(1 + δ)5

]
,

with α1 = 0, 205 is real root of the equation x3 + 4x2 + 4x− 1 = 0,

and the inequality is sharp.

Proof. Based on equation (3), we have initial coefficients a2, and a3 in equation (9)
and (10) respectively while a4 is,

a4 =

(
1

48(2 + α)(1 + α)9/2

)[√
α(24

√
2p3(2 + α)(1 + δ)3 − 12p1p2(1 + δ)

(2α2
√
δ
√
1 + δ + 2(

√
2 + 5

√
2δ + 4

√
2δ2 − 3

√
δ
√
1 + δ) + α(

√
2 + 5

√
2δ

+4
√
2δ2 − 4

√
δ
√
1 + δ)) + p31(14

√
2α3δ + 4

√
2α4δ + 6(

√
2 + 7

√
2δ



N.M. Asih et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1290-1301 1295

+12
√
2δ2 + 8

√
2δ3 − 3

√
δ
√
1 + δ − 12δ3/2

√
1 + δ) + δ2(−4

√
2δ

+6
√
δ
√
1 + δ + 24δ3/2

√
1 + δ) + α(3

√
2− 11

√
2δ + 36

√
2δ2

+24
√
2δ3 + 12δ

√
1 + δ + 48δ3/2

√
1 + δ)))

]
. (16)

We can write Hankel determinant H2(2) as,

H2(2) =

∣∣∣∣a2 a3
a3 a4

∣∣∣∣ = |a2a4 − a23|

=

∣∣∣∣∣
(

p1p3δ

2(1 + δ)3

)
−
(
p22δ(1 + 5α+ 10α2 + 16α3 + 5α4)

2(2 + α)2(1 + δ)8

)

+

(
p41δ

96(2 + α)2(1 + δ)13/2

)[
(6
√
2(−1 + δ)(2 + α)2

√
δ + 30

√
2

(−1 + α)(2 + α)2δ3/2 + 24
√
2(−1 + α)(2 + α)2δ5/2 + 3(7 + 8α+ 2α2)√

1 + δ + (117 + 64α− 20α2 + 36α3 + 38α4 + 8α5)δ
√
1 + δ

+72(3 + 4α+ α2)δ2
√
1 + δ + 48(3 + 4α+ α2)δ3

√
1 + δ)

]

−
(

p2δ

4(2 + α)2(1 + δ)17/2

)[
(2p2δ

5
√
1 + δ + p21

√
2(−1 + α)(2 + α)2

√
δ(1 + δ)4 + (3 + 4α+ α2)(1 + δ)9/2 + 4(3 + 4α+ α2)(1 + δ)9/2)

]∣∣∣∣∣. (17)

Applying (17), Lemma 2 and taking p1 = p so that 0 ≤ p ≤ 2 gives,

H2(2) =

∣∣∣∣∣ δ(4− p2)x2

8(2 + α)2(1 + δ)3
+

2pδ(4− p2)(1− x2)ρ

8(1 + δ)3 + (3 + 4α+ α2

+

(
p2

8(2 + α)2(1 + δ)17/2

)[
(4− p2)x

(√
2(−1 + α)(2 + α)2

√
δ)

√
1 + δ + 2(3 + 4α+ α2)δ

√
1 + δ − 2(3 + 4α+ α2)δ2

√
1 + δ

)]

+

(
p4δ

96(2 + α)2(1 + δ)13/2

)[
(6
√
2(−1 + δ)(2 + α)2

√
δ + 30

√
2(−1 + α)

(2 + α)2δ3/2 + 24
√
2(−1 + α)(2 + α)2δ5/2 + 3(7 + 8α+ 2α2)√

1 + δ + (117 + 64α− 20α2 + 36α3 + 38α4 + 8α5)δ
√
1 + δ

+72(3 + 4α+ α2)δ2
√
1 + δ + 48(3 + 4α+ α2)δ3

√
1 + δ)



N.M. Asih et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1290-1301 1296

−12(1 + 5α+ 10α2 + 10α3 + 5α4)− (2 + α)2(1 + δ)5

]∣∣∣∣∣
:= φ1(α, δ, p, x, ρ). (18)

From (18), then for some |ρ| ≤ 1 gives

H2(2) ≤ δ(4− p2)|x|2

8(2 + α)2(1 + δ)3
+

2pδ(4− p2)(1− |x|2)
8(1 + δ)3

+

(
p2

8(2 + α)2(1 + δ)17/2

)[
(4− p2)|x|

(√
2(−1 + α)(2 + α)2

√
δ

+(3 + 4α+ α2)
√
1 + δ + 2(3 + 4α+ α2)δ

√
1 + δ

−2(3 + 4α+ α2)δ2
√
1 + δ

)]

+

(
p4δ

96(2 + α)2(1 + δ)8

)[
(6
√
2(−1 + δ)(2 + α)2

√
δ + 30

√
2(−1 + α)

(2 + α)2δ3/2 + 24
√
2(−1 + α)(2 + α)2δ5/2 + 3(7 + 8α+ 2α2)√

1 + δ + (117 + 64α− 20α2 + 36α3 + 38α4 + 8α5)δ
√
1 + δ

+72(3 + 4α+ α2)δ2
√
1 + δ + 48(3 + 4α+ α2)δ3

√
1 + δ)

−12(1 + 5α+ 10α2 + 10α3 + 5α4)− (2 + α)2(1 + δ)5

]
:= φ1(α, δ, p, |x|). (19)

Now we check the derivative of φ1(α, δ, p, |x|) with respect to |x| from (18),

φ
′
1(α, δ, p, |x|) =

δ(4− p2)2|x|
4(2 + α)2(1 + δ)3

− pδ(4− p2)|x|
2(1 + δ)3

+

(
p2

8(2 + α)2(1 + δ)17/2

)[
(4− p2)

(√
2(−1 + α)(2 + α)2

√
δ

+(3 + 4α+ α2)
√
1 + δ + 2(3 + 4α+ α2)δ

√
1 + δ

−2(3 + 4α+ α2)δ2
√
1 + δ

)]
(20)

Since φ
′
1(α, δ, p, |x|) ≥ 0 when α1 ≤ α ≤ 1 and 0 < δ ≤ 1, then φ1 is increasing

monoton function. So that the maximum value of φ1(α, δ, p, |x|) is provided when |x| = 1
or

H2(2) ≤ δ(4− p2)

8(2 + α)2(1 + δ)3
+

(
p2

8(2 + α)2(1 + δ)17/2

)[
(4− p2)

(√
2(−1 + α)



N.M. Asih et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1290-1301 1297

(2 + α)2
√
δ + (3 + 4α+ α2)

√
1 + δ + 2(3 + 4α+ α2)δ

√
1 + δ

−2(3 + 4α+ α2)δ2
√
1 + δ

)]

+

(
p4δ

96(2 + α)2(1 + δ)8

)[
(6
√
2(−1 + δ)(2 + α)2

√
δ + 30

√
2(−1 + α)

(2 + α)2δ3/2 + 24
√
2(−1 + α)(2 + α)2δ5/2 + 3(7 + 8α+ 2α2)√

1 + δ + (117 + 64α− 20α2 + 36α3 + 38α4 + 8α5)δ
√
1 + δ

+72(3 + 4α+ α2)δ2
√
1 + δ + 48(3 + 4α+ α2)δ3

√
1 + δ)

−12(1 + 5α+ 10α2 + 10α3 + 5α4)− (2 + α)2(1 + δ)5

]
:= φ1(α, δ, p). (21)

Next, the derivative of φ1(α, δ, p) with respect to p from (21) is,

φ
′
1(α, δ, p) = − pδ(4− p2)

2(2 + α)2(1 + δ)3
−
(

p3

4(2 + α)2(1 + δ)17/2

)(√
2(−1 + α)

(2 + α)2
√
δ + (3 + 4α+ α2)

√
1 + δ + 2(3 + 4α+ α2)δ

√
1 + δ

−2(3 + 4α+ α2)δ2
√
1 + δ

)
+

(
p(4− p2)

4(2 + α)2(1 + δ)17/2

)(√
2(−1 + α)

(2 + α)2
√
δ + (3 + 4α+ α2)

√
1 + δ + 2(3 + 4α+ α2)δ

√
1 + δ

−2(3 + 4α+ α2)δ2
√
1 + δ

)
+

(
p3δ(1 + δ)3/2)

24(2 + α)2(1 + δ)8

)[
(6
√
2(−1 + δ)

(2 + α)2
√
δ + 30

√
2(−1 + α)(2 + α)2δ3/2 + 24

√
2(−1 + α)

(2 + α)2δ5/2 + 3(7 + 8α+ 2α2)
√
1 + δ + (117 + 64α− 20α2

+36α3 + 38α4 + 8α5)δ
√
1 + δ + 72(3 + 4α+ α2)δ2

√
1 + δ

+48(3 + 4α+ α2)δ3
√
1 + δ)− 12(1 + 5α+ 10α2

+10α3 + 5α4)− (2 + α)2(1 + δ)5

]
(22)

From (22) we find the maximum value of φ1(α, δ, p) when 0 ≤ p ≤ 2. With elementary
calculus, we can show that φ1′ (α, δ, p) = 0 has three values of p but the only valid value
is p = 0 while the thers are not valid. Since φ1(α, δ, 0) ≤ φ1(α, δ, 2) for α1 ≤ α ≤ 1 and
0 < δ ≤ 1, then H2(2) ≤ φ1(α, δ, 2). The inequality is sharp when p1 = p2 = p3 = 2. The
proof is completed.

Theorem 3. If f ∈ B1(α, δ), for 0 ≤ α ≤ 0 and 0 < δ ≤ 1 then

T2(1) = |a21 − a22| ≤ 1, (23)



N.M. Asih et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1290-1301 1298

and the inequality is sharp.

Proof. Based on Definition 1, we have initial coefficients a1 = 1 and a2 (see (9)) and
we can write Toeplitz determinant T2(1) as

T2(1) =

∣∣∣∣a1 a2
a2 a1

∣∣∣∣ = |a21 − a22| = |1− p21
√
δ

(2 + δ)3
|

Since

(
1− p21

√
δ

(2 + δ)3

)
≥ 0, if δ ≥ 0 and 0 < p1 ≤ 2, we have,

T2(1) = 1− p21
√
δ

(2 + δ)3
:= φ(p1) (24)

The derivative of (24) is,

φ
′
(p1) =

−2p1
√
δ

(2 + δ)3
≤ 0 (25)

for all p1 ∈ [0, 2] and δ > 0. According to (25), φ(p1) is monoton decreasing function, so
the maximum value of φ(0) = 1. The inequality boundary is sharp for p1 = 0. The proof
is completed.

This research, we also obtain the upper bounds of the determinant Hankel H2(1) using
coefficients invers function.

Theorem 4. If f ∈ B1(α, δ) for 0 ≤ α ≤ 0 and 0 < δ ≤ 1 then

H2(1) = |A1A3 −A2
2| ≤

√
2
√
δ

(2 + α)(1 + δ)3/2
, (26)

and the inequality is sharp.

Proof. Let the coefficients on the inverse function are A1, A2 and A3 by [11] gives,

A2 = − p1
√
α√

2(1 + δ)3/2
, (27)

A3 =
1

8(2 + α)(1 + δ)7/8

(
−
√
δ(4

√
2p2(1 + δ)2

+p21(
√
2 + 5

√
2)δ + 4

√
2δ2 − 2(−2 + α+ α2

√
δ
√
1 + δ)

)
. (28)

From (4) we have H2(1),

H2(1) =

∣∣∣∣A1 A2

A2 A3

∣∣∣∣ = |A1A3 −A2
2|

=

∣∣∣∣∣− p2
√
δ√

2(2 + α)(1 + δ)3/2



N.M. Asih et al. / Eur. J. Pure Appl. Math, 16 (2) (2023), 1290-1301 1299

+
p21
√
δ(
√
2 + 5

√
2δ + 4

√
2δ2 + 2(−14− 5α+ α2)

√
δ
√
1 + δ)

8(2 + α)(1 + δ)7/2

∣∣∣∣∣. (29)

Applying (29), lemma 2, and taking p1 = p and 0 ≤ p ≤ 2 gives,

H2(1) =

∣∣∣∣∣− (p2 + (4− p2)x)
√
δ

2
√
2(2 + α)(1 + δ)3/2

+
p2
√
δ(
√
2 + 5

√
2δ + 4

√
2δ2 + 2(−14− 5α+ α2)

√
δ
√
1 + δ)

8(2 + α)(1 + δ)7/2

∣∣∣∣∣
=

∣∣∣∣∣ (4− p2)x
√
δ

2
√
2(2 + α)(1 + δ)3/2

+
p2
√
δ(3

√
2 + 9

√
2δ + 6

√
2δ2 + 2(−14− 5α+ α2)

√
δ
√
1 + δ)

8(2 + α)(1 + δ)7/2

∣∣∣∣∣ (30)

Case 1. When 0 ≤ α ≤ 1 and 0 < δ < δ1(α), with δ1(α) is real number root of the
equation

1 + (−383− 280α+ 6α2 + 20α3 − 2α4)x+ 24x2 + 16x3 = 0.

From (30), if |x| ≤ 1 then,

H2(1) ≤ (4− p2)|x|
√
δ

2
√
2(2 + α)(1 + δ)3/2

+
p2
√
δ(3

√
2 + 9

√
2δ + 6

√
2δ2 + 2(−14− 5α+ α2)

√
δ
√
1 + δ)

8(2 + α)(1 + δ)7/2

≤ 2
√
δ√

2(2 + α)(1 + δ)3/2

+
p2
√
δ(3

√
2 + 9

√
2δ + 6

√
2δ2 + 2(−14− 5α+ α2)

√
δ
√
1 + δ)

8(2 + α)(1 + δ)7/2

:= φ1(α, δ, p). (31)

Let the derivative of φ1(α, δ, p) with respect to p is φ′
1(α, δ, p). By solving φ′

1(α, δ, p) =
0, we obtain stationary point when p = 0. So we have two critical points p = 0 and p = 2.

Case 2. When 0 ≤ α ≤ 1 and δ1(α) ≤ δ ≤ 1.

From (30), if |x| ≤ 1 then,

H2(1) ≤ (4− p2)|x|
√
δ

2
√
2(2 + α)(1 + δ)3/2



REFERENCES 1300

−p2
√
δ(3

√
2 + 9

√
2δ + 6

√
2δ2 + 2(−14− 5α+ α2)

√
δ
√
1 + δ)

8(2 + α)(1 + δ)7/2

≤ 2
√
δ√

2(2 + α)(1 + δ)3/2

−p2
√
δ(3

√
2 + 9

√
2δ + 6

√
2δ2 + 2(−14− 5α+ α2)

√
δ
√
1 + δ)

8(2 + α)(1 + δ)7/2

:= φ1(α, δ, p) (32)

The same conclusion of case 1, let the derivative of φ1(α, δ, p) with respect to p is
φ′
1(α, δ, p). By solving φ′

1(α, δ, p) = 0, we obtain stationary point when p = 0. So we have
two critical points p = 0 and p = 2.

Since φ1(α, δ, 0) ≥ φ1(α, δ, 2), then H2(1) ≤ φ1(α, δ, 0) =

√
2
√
δ

(2 + α)(1 + δ)3/2
. The inequal-

ity is sharp when p1 = 0 and p2 = 2. The proof is completed.

Acknowledgements

Many thanks and appreciation for my supervisor Marjono and my co supervisor Sa‘adatul
Fitri and Ratno Bagus Edy Wibowo at the Department of Mathematics and Natural Sci-
ences Brawijaya University, for their support and guidance in completing this research and
this paper.

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