Applied Mathematics and Computation 249 (2014) 532–534

Contents lists available at ScienceDirect

Applied Mathematics and Computation

journal homepage: www.elsevier .com/ locate /amc

An integral inequality for cosine polynomials

http://dx.doi.org/10.1016/j.amc.2014.10.0860096-3003/� 2014 Elsevier Inc. All rights reserved.

⇑ Corresponding author.E-mail addresses: h.alzer@gmx.de (H. Alzer), allal.guessab@univ-pau.fr (A. Guessab).

Horst Alzer a, Allal Guessab b,⇑a Morsbacher Str. 10, 51545 Waldbröl, Germanyb Laboratoire de Mathématiques et de leurs Applications, UMR CNRS 4152, Université de Pau et des Pays de l’Adour, 64000 Pau, France

a r t i c l e i n f o

Keywords:Integral inequalityCosine polynomialConvex function

a b s t r a c t

Let

TnðxÞ ¼12

a0 þXn

k¼1

ak cosðkxÞ:

We prove that if ak ðk ¼ 0;1; . . . ;nÞ is an increasing sequence of real numbers, then for anyfunction f which is increasing and convex on the real line we have

1p

Z p

0fjTnðxÞj

Ln

� �dx P f

1nþ 1

Xn

k¼0

ak

!;

where

Ln ¼1p

Z p

0

12þXn

k¼1

cosðkxÞ�����

�����dx

denotes the Lebesgue constant. This extends and refines a result due to Fejes (1939).� 2014 Elsevier Inc. All rights reserved.

In what follows, we denote by Tn a cosine polynomial of degree n with real coefficients,

TnðxÞ ¼12

a0 þXn

k¼1

ak cosðkxÞ:

In 1938, Sidon [5] published the following inequality.

Proposition 1. Let ak ðk ¼ 0;1; . . . ;nÞ be non-negative real numbers. Then there exists a constant C such that

Z p

0jTnðxÞjdx > C � log n � min

06k6nak:

One year later, Fejes [1] proved a refinement of this result (see also [3, Section 2.2.2]).

Proposition 2. Let ak ðk ¼ 0;1; . . . ;nÞ be non-negative real numbers. Then,

1p

Z p

0jTnðxÞjdx P Ln � min

06k6nak; ð1Þ

H. Alzer, A. Guessab / Applied Mathematics and Computation 249 (2014) 532–534 533

where Ln denotes the Lebesgue constant

Ln ¼1p

Z p

0

12þXn

k¼1

cosðkxÞ�����

�����dx:

Setting a0 ¼ a1 ¼ � � � ¼ an ¼ 1 reveals that Ln is the best possible constant in (1).

Two questions arise:

(i) Is it possible to improve the lower bound in (1)?(ii) Does there exist a positive lower bound for

R p0 jTnðxÞjdx, if some of the coefficients are negative?

It is the aim of this note to show that in (1) we can replace the factor min06k6nak by the arithmetic meanPn

k¼0ak=ðnþ 1Þunder the assumption that the coefficients are increasing. This leads to an improvement of (1). Moreover, if a0 6 a1 6 � � � 6 an

andPn

k¼0ak > 0, then we obtain a positive lower bound forR p

0 jTnðxÞjdx. Actually, we prove a bit more. We present a lower

bound forR p

0 f ðjTnðxÞj=LnÞdx, where f is an increasing and convex function. In order to prove our result we need two classicalinequalities.

Tchebychef’s inequality. Let xk and yk ðk ¼ 0;1; . . . ;nÞ be real numbers with

x0 6 x1 6 � � � 6 xn and y0 6 y1 6 � � � 6 yn:

Then,

Xn

k¼0

xk �Xn

k¼0

yk 6 ðnþ 1ÞXn

k¼0

xkyk: ð2Þ

The sign of equality holds in (2) if and only if x0 ¼ � � � ¼ xn or y0 ¼ � � � ¼ yn.Jensen’s inequality. Let g be convex on ðc; dÞ and let h be integrable on ½a; b� with c < hðtÞ < d. Then,

g1

b� a

Z b

ahðtÞdt

!6

1b� a

Z b

ag hðtÞð Þdt:

Proofs can be found, for instance, in [4, Section 2.5] and [2, Section 6.14], respectively.Our general inequality can be stated as follows.

Theorem. Let ak ðk ¼ 0;1; . . . ; nÞ be real numbers satisfying

a0 6 a1 6 � � � 6 an: ð3Þ

Then, for any function f which is increasing and convex on the real line we have

1p

Z p

0fjTnðxÞj

Ln

� �dx P f

1nþ 1

Xn

k¼0

ak

!: ð4Þ

If a0 < an and f is strictly increasing, then (4) holds with ‘‘>’’ instead of ‘‘P’’.

Proof. We follow the method of proof given in [1]. As shown in [1] we have

Ln ¼12

pn0 þXn

k¼1

pnk; ð5Þ

with

pnk ¼1p

Z p

0cosðkxÞsgn sin ðnþ 1=2Þxð Þdx ¼ 8ð2nþ 1Þ

p2

X1m¼0

1

ð2nþ 1Þ2ð2mþ 1Þ2 � 4k2 : ð6Þ

Applying jAj ¼ AsgnA P AsgnB we obtain

1p

Z p

0jTnðxÞjdx P

1p

Z p

0TnðxÞsgn sin ðnþ 1=2Þxð Þdx ¼ 1

2pn0a0 þ

Xn

k¼1

pnkak: ð7Þ

From the series representation in (6) we find

12

pn0 < pn1 < � � � < pnn: ð8Þ

Using (3) and (8) we conclude from Tchebychef’s inequality that

12

pn0a0 þXn

k¼1

pnkak P1

nþ 112

pn0 þXn

k¼1

pnk

!Xn

k¼0

ak; ð9Þ

534 H. Alzer, A. Guessab / Applied Mathematics and Computation 249 (2014) 532–534

with equality if and only if a0 ¼ an. Next, we combine (7), (9) and (5). This gives

1p

Z p

0

jTnðxÞjLn

dx P1

nþ 1

Xn

k¼0

ak: ð10Þ

Since f is increasing, we get

f1p

Z p

0

jTnðxÞjLn

dx� �

P f1

nþ 1

Xn

k¼0

ak

!: ð11Þ

An application of Jensen’s inequality leads to

1p

Z p

0fjTnðxÞj

Ln

� �dx P f

1p

Z p

0

jTnðxÞjLn

dx� �

:

Combining the last two inequalities yields (4). Moreover, if a0 < an and f is strictly increasing, then the inequalities (9)–(11)are strict, so that we obtain (4) with ‘‘>’’ instead of ‘‘P’’. h

Remark

1. If f is decreasing and concave, then the converse of (4) holds.2. The special case f ðxÞ ¼ x reveals that if a0 6 a1 6 � � � 6 an, then the following improvement of (1) is valid:

1p

Z p

0jTnðxÞjdx P Ln �

1nþ 1

Xn

k¼0

ak P Ln � min06k6n

ak:

References

[1] L. Fejes, Two inequalities concerning trigonometric polynomials, J. London Math. Soc. 14 (1939) 44–46.[2] G.H. Hardy, J.E. Littlewood, G. Pólya, Inequalities, Camb. Univ. Press, Cambridge, 1952.[3] G.V. Milovanovic, D.S. Mitrinovic, Th.M. Rassias, Topics in Polynomials: Extremal Problems, Inequalities, Zeros, World Sci, Singapore, 1994.[4] D.S. Mitrinovic, Analytic Inequalities, Springer, New York, 1970.[5] S. Sidon, Über Fourier-Koeffizienten, J. London Math. Soc. 13 (1938) 181–183.