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The asymptotic representations on the norm of the Fourier operators
J. Math. Anal. Appl. 307 (2005) 579–584 www.elsevier.com/locate/jmaa
The asymptotic representations on the norm of the Fourier operators ✩ Tingfan Xie, Linsen Xie ∗ Department of Mathematics, Lishui Normal College, Lishui, Zhejiang Province 323000, People’s Republic of China Received 6 June 2003 Available online 9 April 2005 Submitted by L. Grafakos
Abstract
In this paper by a new method we first get 0π | sint t| dt = π2 ln n + C + O(n−2 ). Based on it, we then obtain a new asymptotic representation on the norm Ln of the Fourier operators as follows: Ln = 42 ln n + C ∗ + O(n−1 ). π 2004 Elsevier Inc. All rights reserved. Keywords: Asymptotic representation; Fourier operator; Lebesgue constant; Euler constant
1. Introduction In Fourier analysis, the integral π In =
| sin nt| dt t
(n = 1, 2, . . .)
0
appears frequently. Its simple estimate is ✩
Supported by Zhejiang Provincial Natural Science Foundation of China.
* Corresponding author.
E-mail address: [email protected] (L. Xie). 0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2004.01.051
580
T. Xie, L. Xie / J. Math. Anal. Appl. 307 (2005) 579–584
In = O(1) ln(n + 2), in which, and in the sequel, O(1) is uniform for n. Recently G.L. He [1] obtained 2 2C 1 1 2 ln n + +r < − , − < In − π π π 2n nπ π where r = 0 sint t dt, and C = (0.5772 . . .) is the Euler constant. Obviously, this inequality cannot obtain the asymptotic representation of In . A natural problem is that whether there exists a constant C such that 2 In = ln n + C + o(1). π Studying the problem is one of our aims in this paper. Writing π ∞ sin t | sin t| 2C − 1 + dt − 2 dt, C = π t t3 0
π
we shall prove 2 (1) In = ln n + C + O(n−2 ). π Here it is unexpected that the remainder is not O(n−1 ). The integral In is closely related with the norm of the partial sum operator for Fourier series, the Fourier operator called for short, or the Lebesgue constant π 2n+1 sin 2 t 2 dt. Ln = π 2 sin 2t 0
The well-known estimate for Ln is (see [2]) 4 Ln = 2 ln n + O(1). π For easy application, T.J. Rivlin [3] pointed out that 4 Ln < 2 ln n + 3. π G.L. He [1] improved it to 4 1 < Ln − 2 ln n < 2. π The second aim of this paper is to establish the asymptotic representation of Ln . Our result is 4 Ln = 2 ln n + C ∗ + O(n−1 ), (2) π where 2 2 4 ∗ C = C + ln . π π π We divide this paper into four sections. We prove (1) in Section 2, and (2) in Section 3. In Section 4 we will give some related numerical calculation.
T. Xie, L. Xie / J. Math. Anal. Appl. 307 (2005) 579–584
581
2. The asymptotic representation of In One of our main results is Theorem 1. Let C be the Euler constant, and π ∞ sin t | sin t| 2C − 1 + dt − 2 C = dt. π t t3 π
0
Then 2 ln n + C + O(n−2 ). π
In =
Proof. Obviously, n−1 π sin t dt. In = t + kπ k=0 0
For any k 1, we have π π 1 1 1 sin t sin t dt = + −2 dt. t + kπ k k+1 π (t + kπ)3 0
0
Therefore for n 2, n−1 1 1 sin t | sin t| 2 1 − + + dt − 2 In = dt. π k π nπ t t3 π
k=1
k=1
k
= ln n + C −
(3)
π
0
It is well known that n−1 1
nπ
∞ 1 1 − + · · · , 2k 2 3k 3
(4)
k=n
where
∞ 1 1 C= − + ··· , 2k 2 3k 3 k=1
which is the Euler constant. Applying the following inequality, which can be established easily, ∞ 1 1 1 1 1 1 1 − + + ··· < − + 2, − − 2 <− 2n 3n 2n 6n 2k 2 3k 3 4k 4 k=n
and by (3) and (4), we get 2C − 1 2 + In = ln n + π π
π 0
sin t dt − 2 t
nπ π
| sin t| dt + Rn , t3
(5)
582
T. Xie, L. Xie / J. Math. Anal. Appl. 307 (2005) 579–584
where 1 1 < Rn < 2 . 2 3n π 6n π Writing that −
2C − 1 C = + π
π
sin t dt − 2 t
0
∞
| sin t| dt, t3
π
we have 2 ln n + C + O(n−2 ). π This completes the proof of Theorem 1. In =
2
3. The asymptotic representation of Ln As is known, the partial sum operator for Fourier series is 1 Sn (f, x) = π
π f (x + t) −π
sin 2n+1 2 t 2 sin 2t
dt.
Now let us consider its norm, i.e., Lebesgue constant π 2n+1 sin 2 t 2 Ln = dt, π 2 sin 2t 0
which can also be written as π 2n+1 π sin 2 t 1 2 2 1 2n + 1 sin dt + t dt. − Ln = π t π t 2 2 sin 2t 0
(6)
0
It is easy to show that the function 0, t = 0, g(t) = 1 1 π sin t − t , 0 < t 2 ,
is monotone increasing in the interval 0, π2 , and π π2 0 < g (t) < 0t . 8 2
(7)
Obviously, 2 π
π 0
π/2 2 1 1 2n + 1 t dt = g(t)sin(2n + 1)t dt. t − t sin 2 π 2 sin 2 0
(8)
T. Xie, L. Xie / J. Math. Anal. Appl. 307 (2005) 579–584
Let tk = 2 π
kπ 2(2n+1) ;
583
we have
t π/2 2n k+1 2 g(t)sin(2n + 1)t dt = g(t)sin(2n + 1)t dt. π
(9)
k=0 tk
0
π It is easy to see that there exists ξk ∈ 0, 2(2n+1) such that tk+1 g(t)sin(2n + 1)t dt = g(tk + ξk ) tk
1 . 2n + 1
Therefore, using (7) we obtain tk+1 tk+1 2 g(t)sin(2n + 1)t dt = g(t) dt + αn,k , π tk
tk
where |αn,k | <
1 π π2 . 8 (2n + 1)2 2
By (6), (8) and (9), we have 2 Ln = π
π 2n+1 sin t 2
t
4 dt + 2 π
0
π/2
1 1 − dt + αn , sin t t
(10)
0
where |αn | <
π2 1 · . 8 2n + 1
(11)
Since 2 π
(n+1/2)π
nπ
2 | sin t| 2 4 dt = − − 2 2 3 t π nπ (n + 1/2) π
(n+1/2)π
| sin t| dt, t
nπ
using 2 π
π 2n+1 sin t 2
t 0
2 dt = π
(n+1/2)π
| sin t| dt, t
0
and the discussion in Section 2, we have π 2n+1 π sin 2 t 4 sin t 4C − 2 2 2 dt = 2 ln n + dt + βn , + π t π t π π2 0
0
(12)
584
T. Xie, L. Xie / J. Math. Anal. Appl. 307 (2005) 579–584
where 2 8 4 βn = − − 2 2 3 π nπ (2n + 1) π
(n+1/2)
| sin t| 2 dt + Rn , 3 π t
(13)
π
which implies 4 βn = − π
∞
| sin t| 1 . dt + O n t3
(14)
π
Therefore, (10) and (12) implies that 4 4C − 2 2 + Ln = 2 ln n + 2 π π π
π
4 sin t dt + 2 t π
0
π/2
1 1 − dt + αn + βn . sin t t
(15)
0
Applying the well-known equality π/2
1 1 4 − dt = ln , sin t t π
0
we obtain the following theorem from (15), (11) and (14) immediately. Theorem 2. Let 4C − 2 2 + C = 2 π π ∗
π
4 4 sin t 4 dt + 2 ln − t π π π
0
∞
| sin t| dt. t3
π
Then Ln =
4 ln n + C ∗ + O(n−1 ). π2
This is the second main result in this paper.
Acknowledgment The authors express their sincere thanks to the referees for their valuable comments and suggestions.
References [1] G.L. He, A remark on the norm of the Fourier operators, in press. [2] T.F. Xie, S.P. Zhou, Real Function Approximation Theory, Hangzhou University Press, Hangzhou, 1998. [3] T.J. Rivlin, The Chebyshev Polynomials, Wiley, New York, 1974.
The asymptotic representations on the norm of the Fourier operators ✩ Tingfan Xie, Linsen Xie ∗ Department of Mathematics, Lishui Normal College, Lishui, Zhejiang Province 323000, People’s Republic of China Received 6 June 2003 Available online 9 April 2005 Submitted by L. Grafakos
Abstract
In this paper by a new method we first get 0π | sint t| dt = π2 ln n + C + O(n−2 ). Based on it, we then obtain a new asymptotic representation on the norm Ln of the Fourier operators as follows: Ln = 42 ln n + C ∗ + O(n−1 ). π 2004 Elsevier Inc. All rights reserved. Keywords: Asymptotic representation; Fourier operator; Lebesgue constant; Euler constant
1. Introduction In Fourier analysis, the integral π In =
| sin nt| dt t
(n = 1, 2, . . .)
0
appears frequently. Its simple estimate is ✩
Supported by Zhejiang Provincial Natural Science Foundation of China.
* Corresponding author.
E-mail address: [email protected] (L. Xie). 0022-247X/$ – see front matter 2004 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2004.01.051
580
T. Xie, L. Xie / J. Math. Anal. Appl. 307 (2005) 579–584
In = O(1) ln(n + 2), in which, and in the sequel, O(1) is uniform for n. Recently G.L. He [1] obtained 2 2C 1 1 2 ln n + +r < − , − < In − π π π 2n nπ π where r = 0 sint t dt, and C = (0.5772 . . .) is the Euler constant. Obviously, this inequality cannot obtain the asymptotic representation of In . A natural problem is that whether there exists a constant C such that 2 In = ln n + C + o(1). π Studying the problem is one of our aims in this paper. Writing π ∞ sin t | sin t| 2C − 1 + dt − 2 dt, C = π t t3 0
π
we shall prove 2 (1) In = ln n + C + O(n−2 ). π Here it is unexpected that the remainder is not O(n−1 ). The integral In is closely related with the norm of the partial sum operator for Fourier series, the Fourier operator called for short, or the Lebesgue constant π 2n+1 sin 2 t 2 dt. Ln = π 2 sin 2t 0
The well-known estimate for Ln is (see [2]) 4 Ln = 2 ln n + O(1). π For easy application, T.J. Rivlin [3] pointed out that 4 Ln < 2 ln n + 3. π G.L. He [1] improved it to 4 1 < Ln − 2 ln n < 2. π The second aim of this paper is to establish the asymptotic representation of Ln . Our result is 4 Ln = 2 ln n + C ∗ + O(n−1 ), (2) π where 2 2 4 ∗ C = C + ln . π π π We divide this paper into four sections. We prove (1) in Section 2, and (2) in Section 3. In Section 4 we will give some related numerical calculation.
T. Xie, L. Xie / J. Math. Anal. Appl. 307 (2005) 579–584
581
2. The asymptotic representation of In One of our main results is Theorem 1. Let C be the Euler constant, and π ∞ sin t | sin t| 2C − 1 + dt − 2 C = dt. π t t3 π
0
Then 2 ln n + C + O(n−2 ). π
In =
Proof. Obviously, n−1 π sin t dt. In = t + kπ k=0 0
For any k 1, we have π π 1 1 1 sin t sin t dt = + −2 dt. t + kπ k k+1 π (t + kπ)3 0
0
Therefore for n 2, n−1 1 1 sin t | sin t| 2 1 − + + dt − 2 In = dt. π k π nπ t t3 π
k=1
k=1
k
= ln n + C −
(3)
π
0
It is well known that n−1 1
nπ
∞ 1 1 − + · · · , 2k 2 3k 3
(4)
k=n
where
∞ 1 1 C= − + ··· , 2k 2 3k 3 k=1
which is the Euler constant. Applying the following inequality, which can be established easily, ∞ 1 1 1 1 1 1 1 − + + ··· < − + 2, − − 2 <− 2n 3n 2n 6n 2k 2 3k 3 4k 4 k=n
and by (3) and (4), we get 2C − 1 2 + In = ln n + π π
π 0
sin t dt − 2 t
nπ π
| sin t| dt + Rn , t3
(5)
582
T. Xie, L. Xie / J. Math. Anal. Appl. 307 (2005) 579–584
where 1 1 < Rn < 2 . 2 3n π 6n π Writing that −
2C − 1 C = + π
π
sin t dt − 2 t
0
∞
| sin t| dt, t3
π
we have 2 ln n + C + O(n−2 ). π This completes the proof of Theorem 1. In =
2
3. The asymptotic representation of Ln As is known, the partial sum operator for Fourier series is 1 Sn (f, x) = π
π f (x + t) −π
sin 2n+1 2 t 2 sin 2t
dt.
Now let us consider its norm, i.e., Lebesgue constant π 2n+1 sin 2 t 2 Ln = dt, π 2 sin 2t 0
which can also be written as π 2n+1 π sin 2 t 1 2 2 1 2n + 1 sin dt + t dt. − Ln = π t π t 2 2 sin 2t 0
(6)
0
It is easy to show that the function 0, t = 0, g(t) = 1 1 π sin t − t , 0 < t 2 ,
is monotone increasing in the interval 0, π2 , and π π2 0 < g (t) < 0t . 8 2
(7)
Obviously, 2 π
π 0
π/2 2 1 1 2n + 1 t dt = g(t)sin(2n + 1)t dt. t − t sin 2 π 2 sin 2 0
(8)
T. Xie, L. Xie / J. Math. Anal. Appl. 307 (2005) 579–584
Let tk = 2 π
kπ 2(2n+1) ;
583
we have
t π/2 2n k+1 2 g(t)sin(2n + 1)t dt = g(t)sin(2n + 1)t dt. π
(9)
k=0 tk
0
π It is easy to see that there exists ξk ∈ 0, 2(2n+1) such that tk+1 g(t)sin(2n + 1)t dt = g(tk + ξk ) tk
1 . 2n + 1
Therefore, using (7) we obtain tk+1 tk+1 2 g(t)sin(2n + 1)t dt = g(t) dt + αn,k , π tk
tk
where |αn,k | <
1 π π2 . 8 (2n + 1)2 2
By (6), (8) and (9), we have 2 Ln = π
π 2n+1 sin t 2
t
4 dt + 2 π
0
π/2
1 1 − dt + αn , sin t t
(10)
0
where |αn | <
π2 1 · . 8 2n + 1
(11)
Since 2 π
(n+1/2)π
nπ
2 | sin t| 2 4 dt = − − 2 2 3 t π nπ (n + 1/2) π
(n+1/2)π
| sin t| dt, t
nπ
using 2 π
π 2n+1 sin t 2
t 0
2 dt = π
(n+1/2)π
| sin t| dt, t
0
and the discussion in Section 2, we have π 2n+1 π sin 2 t 4 sin t 4C − 2 2 2 dt = 2 ln n + dt + βn , + π t π t π π2 0
0
(12)
584
T. Xie, L. Xie / J. Math. Anal. Appl. 307 (2005) 579–584
where 2 8 4 βn = − − 2 2 3 π nπ (2n + 1) π
(n+1/2)
| sin t| 2 dt + Rn , 3 π t
(13)
π
which implies 4 βn = − π
∞
| sin t| 1 . dt + O n t3
(14)
π
Therefore, (10) and (12) implies that 4 4C − 2 2 + Ln = 2 ln n + 2 π π π
π
4 sin t dt + 2 t π
0
π/2
1 1 − dt + αn + βn . sin t t
(15)
0
Applying the well-known equality π/2
1 1 4 − dt = ln , sin t t π
0
we obtain the following theorem from (15), (11) and (14) immediately. Theorem 2. Let 4C − 2 2 + C = 2 π π ∗
π
4 4 sin t 4 dt + 2 ln − t π π π
0
∞
| sin t| dt. t3
π
Then Ln =
4 ln n + C ∗ + O(n−1 ). π2
This is the second main result in this paper.
Acknowledgment The authors express their sincere thanks to the referees for their valuable comments and suggestions.
References [1] G.L. He, A remark on the norm of the Fourier operators, in press. [2] T.F. Xie, S.P. Zhou, Real Function Approximation Theory, Hangzhou University Press, Hangzhou, 1998. [3] T.J. Rivlin, The Chebyshev Polynomials, Wiley, New York, 1974.