Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 4, Issue 4, Article 78, 2003
SOME NEW INEQUALITIES FOR TRIGONOMETRIC POLYNOMIALS WITH
SPECIAL COEFFICIENTS
ŽIVORAD TOMOVSKI
FACULTY OF M ATHEMATICS AND NATURAL S CIENCES
D EPARTMENT OF M ATHEMATICS
S KOPJE 1000 MACEDONIA.
tomovski@iunona.pmf.ukim.edu.mk
Received 21 July, 2003; accepted 14 November, 2003
Communicated by H. Bor
A BSTRACT. Some new inequalities for certain trigonometric polynomials with complex semiconvex and complex convex coefficients are given.
Key words and phrases: Petrović inequality, Complex trigonometric polynomial, Complex semi-convex coefficients, Complex
convex coefficients.
2000 Mathematics Subject Classification. 26D05, 42A05.
1. I NTRODUCTION AND P RELIMINARIES
Petrović [4] proved the following complementary triangle inequality for sequences of complex numbers {z1 , z2 , . . . , zn } .
Theorem A. Let α be a real number and 0 < θ < π2 . If {z1 , z2 , . . . , zn } are complex numbers
such that α − θ ≤ arg zν ≤ α + θ, ν = 1, 2, . . . , n, then
n
n
X X
zν ≥ (cos θ)
|zν |.
ν=1
For
Let
Then,
0 < θ < π2
∆λn = λn
ν=1
denote by K (θ) the cone K (θ) = {z : |arg z| ≤ θ} .
− λn+1 , for n = 1, 2, 3, . . . , where {λn } is a sequence of complex numbers.
∆2 λn = ∆ (∆λn ) = ∆λn − ∆λn+1 = λn − 2λn+1 + λn+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. Let x 6= 2kπ for k = 0, ±1, ±2, . . .
ISSN (electronic): 1443-5756
c 2003 Victoria University. All rights reserved.
101-03
2
Ž IVORAD T OMOVSKI
(1) Let {bk } be apositive
nondecreasing sequence and {uk } a sequence of complex numbers
uk
such that ∆ bk ∈ K (θ) . Then
m
X
1
1 bm
1 uk f (kx) ≤ x 1+
|um | +
|un | , (∀n, m ∈N, m > n) .
cos θ
cos θ bn
sin 2
k=n
(2) Let {bk } be a positive nondecreasing sequence and {uk } a sequence of complex numbers
such that ∆ (uk bk ) ∈ K (θ) . Then
m
X
1
1 bm
1 |un | +
|um | , (∀n, m ∈N, m > n) .
uk f (kx) ≤ x 1+
cos θ
cos θ bn
sin 2
k=n
Here f (x) = sin x or f (x) = cos x.
the results of Theorem B were given by the author in [5] for sums of type
PSimilarly,
m
k
(−1)
u
k f (kx) , where again f (x) = sin x or f (x) = cos x.
k=n
Mitrinović and Pečarić (see [2, 3]) proved the following inequalities for cosine and sine polynomials with nonnegative coefficients.
Theorem C. Let x 6= 2kπ for k = 0, ±1, ±2, ..
(1) Let {b
a positive nondecreasing sequence and {ak } a nonnegative sequence such
k } be
−1
that ak bk is a decreasing sequence. Then
m
X
an b m
ak f (kx) ≤ x ,
bn
sin 2
k=n
(∀n, m ∈N, m > n) .
(2) Let {bk } be a positive nondecreasing sequence and {ak } a nonnegative sequence such
that {ak bk } is an increasing sequence. Then
m
X
am b m
ak f (kx) ≤ x ,
b
sin
n
2
k=n
(∀n, m ∈N, m > n) .
Heref (x) = sin x or f (x) = cos x.
The special cases of these inequalities were proved by G.K. Lebed for bk = k s , s ≥ 0 (see
[1]). Similarly,
of Theorem C, were given by Mitrinović and Pečarić in [2, 3] for
P the results
k
sums of type m
(−1)
a
k f (kx) , where again f (x) = sin x or f (x) = cos x.
k=n
The
{uk } is said to be complex semiconvex if there exists a cone K (θ), such that
sequence
2 uk
∆ bk ∈ K (θ) or ∆2 (uk bk ) ∈ K (θ) , where {bk } is a positive nondecreasing sequence. For
bk = 1, the sequence {uk } 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.
J. Inequal. Pure and Appl. Math., 4(4) Art. 78, 2003
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S OME N EW I NEQUALITIES FOR T RIGONOMETRIC P OLYNOMIALS WITH S PECIAL C OEFFICIENTS
3
2. M AIN R ESULTS
P P
q
j
max j=p k=i zk .
Theorem 2.1. Let {zk } be a sequence of complex numbers such that A =
n≤p≤q≤m
Further, let {bk } be
positive nondecreasing sequence. If {uk } is a sequence of complex numa
2 uk
bers such that ∆ bk ∈ K (θ) , then
m
X
1
u
b
m−1
m
∆
+
∆ un ,
uk zk ≤ A |um | + bm 1 +
cos θ bm−1 cos θ bn k=n
(∀n, m ∈N, m > n) .
Proof. Let us estimate the sum
Since
Pm
k=n bk zk .
j
m
m X
X
X
z
≤
z
k
k ≤ A,
j=n+1 k=n
k=n
we obtain
m
m
m
X
X
X
bk zk = bn
zk +
j=n+1
k=n
k=n
m
m
X X
≤ bn zk +
k=n
(*)
m
X
!
zk
k=j
(bj − bj−1 )
m
X
zk (bj − bj−1 )
j=n+1 k=j
≤ A (bn + bm − bn ) = Abm .
Then,
m
m
X
X u
k
(bk zk )
uk zk = bk
k=n
k=n
! j
m
m−1
u X
X
X
uj m
=
bk zk +
bk zk ∆
bm
b
j
j=n
k=n
k=n
m−1 j
X
j
m
m−2
r X
u X
X
um−1 X X
u
m
r
=
bk zk + ∆
bk zk +
∆2
bk zk bm
bm−1 j=n k=n
br j=n k=n
r=n
k=n
r j
m−1 j
m
m−2
|um | X
um−1 X X
X 2 ur X X
≤
bk zk + ∆
bk zk +
∆
bk zk r=n
bm k=n
bm−1 j=n k=n
br j=n k=n
X
Abm m−2
|um |
u
ur m−1
2
+
≤ Abm
+ Abm ∆
∆
bm
bm−1
cos θ r=n
br u
b
u
u
m−1
m
n
m−1
+
∆
= A |um | + bm ∆
−
∆
bm−1 cos θ bn
bm−1 1
u
b
m−1
m
∆
+
∆ un .
≤ A |um | + bm 1 +
cos θ bm−1 cos θ bn J. Inequal. Pure and Appl. Math., 4(4) Art. 78, 2003
http://jipam.vu.edu.au/
4
Ž IVORAD T OMOVSKI
Theorem 2.2. Let {zk } and {bk } be defined as in Theorem 2.1. If {uk } is a sequence of complex
numbers such that ∆2 (uk bk ) ∈ K (θ) , then
m
X
1
−1
uk zk ≤ A |un | + bn 1 +
(|∆ (un bn )| + |∆ (um−1 bm−1 )|) ,
cos θ
k=n
(∀n, m ∈N, m > n) .
m
Proof. The sequence b−1
is nonincreasing, so from (*) we get
k
k=n
m
X
−1
b−1
z
k ≤ Abn .
k
k=n
Now, we have:
m
m
X
X
uk zk = z
(uk bk ) b−1
k
k
k=n
k=n
!
m
m
m
X
X
X
−1
−1
bk zk (uj bj − uj−1 bj−1 )
= un bn
bk zk +
j=n+1
k=n
k=j
j
m−1
m
m
X
X
X
X
−1
2
= un bn
bk zk −
∆ (uj−1 bj−1 )
b−1
k zk
r=n k=r
j=n+1
k=n
m
m X
m
X
X
−1
−1 bk zk − ∆ (um−1 bm−1 )
bk zk + ∆ (un bn )
r=n k=r
k=n
j
m
m−1
m
X
X X
X
−1 2
≤ |un | bn b−1
∆
(u
b
)
b
z
+
z
j−1 j−1
k
k
k
k
r=n k=r
j=n+1
k=n
m
m X
m
X
X
−1
−1 + |∆ (un bn )| bk zk + |∆ (um−1 bm−1 )| bk zk r=n k=r
k=n
−1
≤ |un | bn Ab−1
n + Abn
m−1
X
2
∆ (uj−1 bj−1 ) + Ab−1
n |∆ (un bn )|
j=n+1
+ Ab−1
n |∆ (um−1 bm−1 )|
"
m−1
X
b−1
≤ A |un | + n ∆2 (uj−1 bj−1 )
cos θ j=n+1
b−1
n
b−1
n
+
|∆ (un bn )| +
|∆ (um−1 bm−1 )|
b−1
= A |un | + n |∆ (un bn ) − ∆ (um−1 bm−1 )|
cos θ
−1
+ b−1
|∆
(u
b
)|
+
b
|∆
(u
b
)|
n
n
m−1
m−1
n
n
1
−1
≤ A |un | + bn 1 +
(|∆ (un bn )| + |∆ (um−1 bm−1 )|) .
cos θ
J. Inequal. Pure and Appl. Math., 4(4) Art. 78, 2003
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S OME N EW I NEQUALITIES FOR T RIGONOMETRIC P OLYNOMIALS WITH S PECIAL C OEFFICIENTS
5
Lemma 2.3. For all p, q ∈ N, p < q, the following inequalities hold
q j
X X
q−p+2
ikx (2.1)
e ≤
, x 6= 2kπ, k = 0, ±1, ±2, . . . ,
2 sin2 x2
j=p k=l
q j
X X
q−p+2
(2.2)
, x 6= (2k + 1) π, k = 0, ±1, ±2, . . . .
(−1)k eikx ≤
2 x
2
cos
2
j=p k=l
Proof. It is sufficient to prove the first inequality, since the second inequality can be proved
analogously.
q j
q
X X
X
i(j−l+1)x
e
−
1
ikx
ilx
e =
e
ix
e −1 j=p k=l
j=p
q
1 X
1
eijx − (q − p + 1)
= ix
i(l−1)x
|e − 1| e
j=p
i(q−p+1)
e
− 1 q − p + 1
1
≤ + 2 sin x2 |eix − 1|
2 sin x2 q−p+1
q−p+2
2
≤
.
2 x +
2 x =
4 sin 2
2 sin 2
2 sin2 x2
By putting zk = 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 {bk } and {uk } be defined as in Theorem 2.1. Then
m
X
uk exp (ikx)
k=n
1
m−n+2
um−1 bm un ≤
|u
|
+
b
1
+
∆
+
∆
,
m
m
cos θ bm−1 cos θ bn 2 sin2 x2
(∀n, m ∈N, m > n) .
(ii) Let {bk } and {uk } be defined as in Theorem 2.2. Then
m
X
u
exp
(ikx)
k
k=n
m−n+2
1
−1
≤
|un | + bn 1 +
(|∆ (un bn )| + |∆ (um−1 bm−1 )|) ,
cos θ
2 sin2 x2
(∀n, m ∈N, m > n) .
In both cases x 6= 2kπ, k = 0, ±1, ±2, . . .
Applying the known inequalities Re z ≤ |z| and Im z ≤ |z| for z ∈ C, we obtain the
following result:
Theorem 2.5. Let x 6= 2kπ for k = 0, ±1, ±2, . . . .
J. Inequal. Pure and Appl. Math., 4(4) Art. 78, 2003
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6
Ž IVORAD T OMOVSKI
(i) Let {bk } and {uk } be defined as in Theorem 2.1. Then
m
X
m−n+2
um−1 un 1
bm uk f (kx) ≤
|um | + bm 1 +
∆
+
∆
2
x
,
cos
θ
b
cos
θ
b
2
sin
m−1
n
2
k=n
(∀n, m ∈N, m > n) .
(ii) Let {bk } and {uk } be defined as in Theorem 2.2. Then
m
X
m−n+2
1
−1
uk f (kx) ≤
|un | + bn 1 +
(|∆ (un bn )| + |∆ (um−1 bm−1 )|) ,
2 x
cos
θ
2
sin
2
k=n
(∀n, m ∈N, m > n) .
Applying inequality (2.2) of Lemma 2.3, we obtain the following results:
Theorem 2.6. Let x 6= (2k + 1) π for k = 0, ±1, ±2, . . . and let x 7→ f (x) be defined as in
Theorem 2.5.
(i) If {bk } and {uk } are defined as in Theorem 2.1, then
m
X
k
(−1) uk f (kx)
k=n
1
um−1 bm m−n+2
un |u
|
+
b
1
+
∆
+
∆
,
≤
m
m
2 cos2 x2
cos θ bm−1 cos θ bn (∀n, m ∈N, m > n) .
(ii) If {bk } and {uk } are defined as in Theorem 2.2, then
m
X
(−1)k uk f (kx)
k=n
m−n+2
1
−1
≤
|un | + bn 1 +
(|∆ (un bn )| + |∆ (um−1 bm−1 )|) ,
2 cos2 x2
cos θ
(∀n, m ∈N, m > n) .
For bk = 1, we obtain the following theorem.
Theorem 2.7. Let {uk } be a complex convex sequence.
(i) If x 6= 2kπ for k = 0, ±1, ±2, . . . , then we have:
m
X
m−n+2
1
1
|um | + 1 +
uk f (kx) ≤
|∆um−1 | +
|∆un | ,
2 x
cos
θ
cos
θ
2
sin
2
k=n
(∀n, m ∈N, m > n) .
(ii) If x 6= (2k + 1) π for k = 0, ±1, ±2, . . . , then we have:
m
X
m−n+2
1
k
(−1) uk f (kx) ≤
|un | + 1 +
(|∆un | + |∆um−1 |) ,
2 x
2
cos
cos
θ
2
k=n
(∀n, m ∈N, m > n) .
J. Inequal. Pure and Appl. Math., 4(4) Art. 78, 2003
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S OME N EW I NEQUALITIES FOR T RIGONOMETRIC P OLYNOMIALS WITH S PECIAL C OEFFICIENTS
7
R EFERENCES
[1] G.K. LEBED, On trigonometric series with coefficients which satisfy some conditions (Russian),
Mat. Sb., 74 (116) (1967), 100–118
[2] D.S. MITRINOVIĆ AND J.E. PEČARIĆ, On an inequality of G. K. Lebed’, Prilozi MANU, Od.
Mat. Tehn. Nauki, 12 (1991), 15–19
[3] J.E. PEČARIĆ, Nejednakosti, 6 (1996), 175–179 (Zagreb).
[4] M. PETROVIĆ, Theoreme sur les integrales curvilignes, Publ. Math. Univ. Belgrade, 2 (1933), 45–
59
[5] Ž. TOMOVSKI, On some inequalities of Mitrinović and Pečarić, Prilozi MANU, Od. Mat. Tehn.
Nauki, 22 (2001), 21–28.
J. Inequal. Pure and Appl. Math., 4(4) Art. 78, 2003
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