An integral inequality for cosine polynomials
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Transcript of An integral inequality for cosine polynomials
Applied Mathematics and Computation 249 (2014) 532–534
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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: [email protected] (H. Alzer), [email protected] (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 p0jTnðxÞjdx > 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,
Xnk¼0
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
12pn0 < 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.