INTRODUCTION ANDPRELIMINARIES Petrovi´c [4] proved the following complementary triangle inequality for sequences of com- plex numbers{z1, z2

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Volume 4, Issue 4, Article 78, 2003

SOME NEW INEQUALITIES FOR TRIGONOMETRIC POLYNOMIALS WITH SPECIAL COEFFICIENTS

ŽIVORAD TOMOVSKI

FACULTY OFMATHEMATICS ANDNATURALSCIENCES

DEPARTMENT OFMATHEMATICS

SKOPJE1000 MACEDONIA.

tomovski@iunona.pmf.ukim.edu.mk

Received 21 July, 2003; accepted 14 November, 2003 Communicated by H. Bor

ABSTRACT. Some new inequalities for certain trigonometric polynomials with complex semi- convex and complex convex coefficients are given.

Key words and phrases: Petrovi´c inequality, Complex trigonometric polynomial, Complex semi-convex coefficients, Complex convex coefficients.

2000 Mathematics Subject Classification. 26D05, 42A05.

1. INTRODUCTION ANDPRELIMINARIES

Petrovi´c [4] proved the following complementary triangle inequality for sequences of com- plex numbers{z₁, z₂, . . . , z_n}.

Theorem A. Letα be a real number and0 < θ < ^π₂.If{z₁, z₂, . . . , z_n}are complex numbers such thatα−θ≤argz_ν ≤α+θ, ν= 1,2, . . . , n,then

n

X

ν=1

z_ν

≥(cosθ)

n

X

ν=1

|z_ν|.

For0< θ < ^π₂ denote byK(θ)the coneK(θ) ={z :|argz| ≤θ}.

Let∆λ_n = λ_n−λ_n+1,forn = 1,2,3, . . . , where{λ_n}is a sequence of complex numbers.

Then,

∆²λ_n= ∆ (∆λ_n) = ∆λ_n−∆λ_n+1 =λ_n−2λ_n+1+λ_n+2, n = 1,2,3, . . .

The author Tomovski (see [5]) proved the following inequality for cosine and sine polynomials with complex-valued coefficients.

Theorem B. Letx6= 2kπfork = 0,±1,±2, . . .

ISSN (electronic): 1443-5756

c 2003 Victoria University. All rights reserved.

101-03

(1) Let{b_k}be a positive nondecreasing sequence and{u_k}a sequence of complex numbers such that∆

u_k bk

∈K(θ).Then

m

X

k=n

u_kf(kx)

≤ 1

sin^x₂

1 + 1 cosθ

|u_m|+ 1 cosθ

bm

b_n |u_n|

, (∀n, m∈N, m > n).

(2) Let{b_k}be a positive nondecreasing sequence and{u_k}a sequence of complex numbers such that∆ (u_kb_k)∈K(θ).Then

m

X

k=n

u_kf(kx)

≤ 1

sin^x₂

1 + 1 cosθ

|u_n|+ 1 cosθ

bm

b_n |u_m|

, (∀n, m∈N, m > n).

Heref(x) = sinxorf(x) = cosx.

Similarly, the results of Theorem B were given by the author in [5] for sums of type Pm

k=n(−1)^ku_kf(kx),where againf(x) = sinxorf(x) = cosx.

Mitrinovi´c and Peˇcari´c (see [2, 3]) proved the following inequalities for cosine and sine poly- nomials with nonnegative coefficients.

Theorem C. Letx6= 2kπfork = 0,±1,±2, ..

(1) Let{bk}be a positive nondecreasing sequence and{ak}a nonnegative sequence such that

a_kb⁻¹_k is a decreasing sequence. Then

m

X

k=n

a_kf(kx)

≤ a_n sin^x₂

b_m

b_n

, (∀n, m∈N, m > n).

(2) Let{b_k}be a positive nondecreasing sequence and{a_k}a nonnegative sequence such that{akbk}is an increasing sequence. Then

m

X

k=n

a_kf(kx)

≤ a_m sin^x₂

b_m

bn

, (∀n, m∈N, m > n).

Heref(x) = sinxorf(x) = cosx.

The special cases of these inequalities were proved by G.K. Lebed forb_k = k^s, s ≥ 0(see [1]). Similarly, the results of Theorem C, were given by Mitrinovi´c and Peˇcari´c in [2, 3] for sums of typePm

k=n(−1)^ka_kf(kx),where againf(x) = sinxorf(x) = cosx.

The sequence{u_k}is said to be complex semiconvex if there exists a coneK(θ), such that

∆²

uk

bk

∈K(θ)or∆²(u_kb_k)∈K(θ),where{b_k}is a positive nondecreasing sequence. For b_k = 1, the sequence{u_k}shall be called a complex convex sequence.

In this paper we shall give some estimates for cosine and sine polynomials with complex semi-convex and complex convex coefficients.

2. MAINRESULTS

Theorem 2.1. Let{z_k}be a sequence of complex numbers such thatA = max

n≤p≤q≤m

Pq j=p

Pj k=iz_k

. Further, let{b_k}be a positive nondecreasing sequence. If{u_k}is a sequence of complex num- bers such that∆²

uk

bk

∈K(θ),then

m

X

k=n

u_kz_k

≤A

|u_m|+b_m

1 + 1 cosθ

∆

um−1

bm−1

+ bm

cosθ

∆ un

b_n

,

(∀n, m∈N, m > n). Proof. Let us estimate the sumPm

k=nb_kz_k. Since

m

X

k=n

z_k

≤

m

X

j=n+1

j

X

k=n

z_k

≤A, we obtain

m

X

k=n

b_kz_k

=

b_n

m

X

k=n

z_k+

m

X

j=n+1 m

X

k=j

z_k

!

(b_j −bj−1)

≤b_n

m

X

k=n

z_k

+

m

X

j=n+1

m

X

k=j

z_k

(b_j −bj−1)

≤A(bn+bm−bn) = Abm. (*)

Then,

m

X

k=n

ukzk

=

m

X

k=n

u_k b_k (bkzk)

=

u_m b_m

m

X

k=n

b_kz_k+

m−1

X

j=n j

X

k=n

b_kz_k

!

∆ u_j

b_j

=

u_m b_m

m

X

k=n

bkzk+ ∆

u_m−1 bm−1

^m−1 X

j=n j

X

k=n

bkzk+

m−2

X

r=n

∆² u_r

b_r ^r

X

j=n j

X

k=n

bkzk

≤ |u_m| b_m

m

X

k=n

b_kz_k

+

∆

um−1

bm−1

m−1

X

j=n j

X

k=n

b_kz_k

+

m−2

X

r=n

∆² u_r

b_r

r

X

j=n j

X

k=n

b_kz_k

≤Ab_m|u_m|

b_m +Ab_m

∆

u_m−1 bm−1

+ Ab_m cosθ

m−2

X

r=n

∆² u_r

b_r

=A

|um|+bm

∆

u_m−1 bm−1

+ b_m cosθ

∆ u_n

b_n

−∆

u_m−1 bm−1

≤A

|u_m|+b_m

1 + 1 cosθ

∆

um−1

bm−1

+ b_m cosθ

∆ u_n

b_n

.

Theorem 2.2. Let{z_k}and{b_k}be defined as in Theorem 2.1. If{u_k}is a sequence of complex numbers such that∆²(u_kb_k)∈K(θ),then

m

X

k=n

ukzk

≤A

|un|+b⁻¹_n

1 + 1 cosθ

(|∆ (unbn)|+|∆ (um−1bm−1)|)

,

(∀n, m∈N, m > n). Proof. The sequence

b⁻¹_k ^m

k=nis nonincreasing, so from (*) we get

m

X

k=n

b⁻¹_k z_k

≤Ab⁻¹_n . Now, we have:

m

X

k=n

ukzk

=

m

X

k=n

(ukbk)b⁻¹_k zk

=

u_nb_n

m

X

k=n

b⁻¹_k z_k+

m

X

j=n+1 m

X

k=j

b⁻¹_k z_k

!

(u_jb_j−uj−1bj−1)

=

unbn m

X

k=n

b⁻¹_k zk−

m−1

X

j=n+1

∆²(uj−1bj−1)

j

X

r=n m

X

k=r

b⁻¹_k zk

+ ∆ (u_nb_n)

m

X

k=n

b⁻¹_k z_k−∆ (u_m−1b_m−1)

m

X

r=n m

X

k=r

b⁻¹_k z_k

≤ |u_n|b_n

m

X

k=n

b⁻¹_k z_k

+

m−1

X

j=n+1

∆²(uj−1bj−1)

j

X

r=n m

X

k=r

b⁻¹_k z_k

+|∆ (u_nb_n)|

m

X

k=n

b⁻¹_k z_k

+|∆ (um−1bm−1)|

m

X

r=n m

X

k=r

b⁻¹_k z_k

≤ |un|bnAb⁻¹_n +Ab⁻¹_n

m−1

X

j=n+1

∆²(uj−1bj−1)

+Ab⁻¹_n |∆ (unbn)|

+Ab⁻¹_n |∆ (um−1bm−1)|

≤A

"

|u_n|+ b⁻¹_n cosθ

m−1

X

j=n+1

∆²(uj−1bj−1)

+b⁻¹_n |∆ (unbn)|+b⁻¹_n |∆ (um−1bm−1)|

=A

|un|+ b⁻¹_n

cosθ |∆ (unbn)−∆ (um−1bm−1)|

+b⁻¹_n |∆ (unbn)|+b⁻¹_n |∆ (um−1bm−1)|

≤A

|u_n|+b⁻¹_n

1 + 1 cosθ

(|∆ (u_nb_n)|+|∆ (u_m−1b_m−1)|)

.

Lemma 2.3. For allp, q ∈N,p < q, the following inequalities hold

q

X

j=p j

X

k=l

e^ikx

≤ q−p+ 2

2 sin^{2 x}₂ , x6= 2kπ, k = 0,±1,±2, . . . , (2.1)

q

X

j=p j

X

k=l

(−1)^ke^ikx

≤ q−p+ 2

2 cos^{2 x}₂ , x6= (2k+ 1)π, k= 0,±1,±2, . . . . (2.2)

Proof. It is sufficient to prove the first inequality, since the second inequality can be proved analogously.

q

X

j=p j

X

k=l

e^ikx

=

q

X

j=p

e^ilxe^i(j−l+1)x−1 e^ix−1

= 1

|e^ix−1|

1 e^i(l−1)x

q

X

j=p

e^ijx−(q−p+ 1)

≤ 1

2 sin^x₂

e^i(q−p+1)−1

|e^ix−1| +q−p+ 1 2 sin^x₂

≤ 2

4 sin^2x₂ + q−p+ 1

2 sin^2x₂ = q−p+ 2 2 sin^{2 x}₂ .

By puttingz_k = exp (ikx)in Theorem 2.1 and Theorem 2.2 and using the inequality (2.1) of the above lemma, we have:

Theorem 2.4. (i) Let{b_k}and{u_k}be defined as in Theorem 2.1. Then

m

X

k=n

u_kexp (ikx)

≤ m−n+ 2 2 sin^2x₂

|u_m|+b_m

1 + 1 cosθ

∆

um−1

b_m−1

+ b_m cosθ

∆ u_n

b_n

, (∀n, m∈N, m > n). (ii) Let{b_k}and{u_k}be defined as in Theorem 2.2. Then

m

X

k=n

u_kexp (ikx)

≤ m−n+ 2 2 sin^{2 x}₂

|un|+b⁻¹_n

1 + 1 cosθ

(|∆ (unbn)|+|∆ (um−1bm−1)|)

,

(∀n, m∈N, m > n). In both casesx6= 2kπ, k= 0,±1,±2, . . .

Applying the known inequalities Rez ≤ |z| and Imz ≤ |z| for z ∈ C, we obtain the following result:

Theorem 2.5. Letx6= 2kπfork = 0,±1,±2, . . . .

(i) Let{b_k}and{u_k}be defined as in Theorem 2.1. Then

m

X

k=n

ukf(kx)

≤ m−n+ 2 2 sin^{2 x}₂

|um|+bm

1 + 1 cosθ

∆

um−1

bm−1

+ bm

cosθ

∆ un

b_n

,

(∀n, m∈N, m > n). (ii) Let{bk}and{uk}be defined as in Theorem 2.2. Then

m

X

k=n

u_kf(kx)

≤ m−n+ 2 2 sin^{2 x}₂

|u_n|+b⁻¹_n

1 + 1 cosθ

(|∆ (u_nb_n)|+|∆ (um−1bm−1)|)

,

(∀n, m∈N, m > n). Applying inequality (2.2) of Lemma 2.3, we obtain the following results:

Theorem 2.6. Letx 6= (2k+ 1)π fork = 0,±1,±2, . . . and letx 7→ f(x) be defined as in Theorem 2.5.

(i) If{b_k}and{u_k}are defined as in Theorem 2.1, then

m

X

k=n

(−1)^ku_kf(kx)

≤ m−n+ 2 2 cos^2x₂

|u_m|+b_m

1 + 1 cosθ

∆

um−1

bm−1

+ b_m cosθ

∆ u_n

b_n

, (∀n, m∈N, m > n). (ii) If{b_k}and{u_k}are defined as in Theorem 2.2, then

m

X

k=n

(−1)^kukf(kx)

≤ m−n+ 2 2 cos^{2 x}₂

|u_n|+b⁻¹_n

1 + 1 cosθ

(|∆ (u_nb_n)|+|∆ (um−1bm−1)|)

,

(∀n, m∈N, m > n). Forb_k= 1,we obtain the following theorem.

Theorem 2.7. Let{u_k}be a complex convex sequence.

(i) Ifx6= 2kπfork = 0,±1,±2, . . ., then we have:

m

X

k=n

u_kf(kx)

≤ m−n+ 2 2 sin^{2 x}₂

|u_m|+

1 + 1 cosθ

|∆u_m−1|+ 1

cosθ |∆u_n|

,

(∀n, m∈N, m > n). (ii) Ifx6= (2k+ 1)πfork = 0,±1,±2, . . ., then we have:

m

X

k=n

(−1)^kukf(kx)

≤ m−n+ 2 2 cos^{2 x}₂

|un|+

1 + 1 cosθ

(|∆un|+|∆um−1|)

,

(∀n, m∈N, m > n).

REFERENCES

[1] G.K. LEBED, On trigonometric series with coefficients which satisfy some conditions (Russian), Mat. Sb., 74 (116) (1967), 100–118

[2] D.S. MITRINOVI ´C AND J.E. PE ˇCARI ´C, On an inequality of G. K. Lebed’, Prilozi MANU, Od.

Mat. Tehn. Nauki, 12 (1991), 15–19

[3] J.E. PE ˇCARI ´C, Nejednakosti, 6 (1996), 175–179 (Zagreb).

[4] M. PETROVI ´C, Theoreme sur les integrales curvilignes, Publ. Math. Univ. Belgrade, 2 (1933), 45–

59

[5] Ž. TOMOVSKI, On some inequalities of Mitrinovi´c and Peˇcari´c, Prilozi MANU, Od. Mat. Tehn.

Nauki, 22 (2001), 21–28.