Bull. Malays. Math. Sci. Soc. DOI 10.1007/s40840-015-0251-9

Approximating the Finite Hilbert Transform via Some Companions of Ostrowski’s Inequalities Wenjun Liu1 · Xingyue Gao1 · Yaqiong Wen1

Received: 29 December 2013 / Revised: 22 March 2014 © Malaysian Mathematical Sciences Society and Universiti Sains Malaysia 2015

Abstract The finite Hilbert transform is a helpful tool in fields like aerodynamics, the theory of elasticity, and other areas of the engineering sciences. In this paper, by using some companions of Ostrowski’s inequalities for absolutely continuous functions due to Dragomir, we give some new inequalities and approximations for the finite Hilbert transform, which may have the better error bounds than the known results. Keywords Ostrowski’s inequality · Absolutely continuous functions · L p space · Composite quadrature rule Mathematics Subject Classification

26D15 · 41A55 · 41A80

1 Introduction Cauchy principal value integrals (finite Hilbert transform) of the form  t−ε   b  b 1 f (τ ) f (τ ) f (τ ) 1 dτ := lim dτ + dτ . (T f ) (a, b; t) = P V π π ε→0+ a τ −t a τ −t t+ε τ − t (1.1) Communicated by Rosihan M. Ali, Dato’.

B

Wenjun Liu [email protected] Xingyue Gao [email protected] Yaqiong Wen [email protected]

1

College of Mathematics and Statistics, Nanjing University of Information Science and Technology, Nanjing 210044, China

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play an important role in many fields like aerodynamics, the theory of elasticity, and other areas of the engineering sciences. When the interval (a, b) coincides with the interval (−∞, ∞), this integral is called the Hilbert transform of f (τ ), which appears frequently in tomographic image reconstruction [33], signal processing, disordered alloys, magnetic properties of disordered alloys, and phase determination in brightfield electron microscopy [27]. For different approaches in approximating the finite Hilbert transform (1.1) including noninterpolatory, interpolatory, Gaussian, Chebychevian and spline methods, see for example the papers [6–8,10,12–15,18] and the references therein. In [10], Dragomir proved the following Ostrowski-type inequalities for absolutely continuous functions. Theorem 1.1 Let u : [a, b] → R be an absolutely continuous function on [a, b]. Then we have    b   u(x)(b − a) − u(t)dt   a ⎧

  1 a+b 2 ⎪ u [a,b],∞ , u  ∈ L ∞ [a, b], ⎪ 4 (b − a)2 + x − 2 ⎪ ⎪

⎪ ⎨ 1 1+ 1 1+ 1 (x − a) q + (b − x) q u  [a,b], p , u  ∈ L p [a, b], 1 ≤ (q+1) q (1.2) ⎪ ⎪ ⎪ p > 1, 1p + q1 = 1, ⎪ ⎪   ⎩1 (b − a) + x − a+b  u  [a,b],1 , u  ∈ L 1 [a, b], 2

2

where  · [a,b],r (r ∈ [1, ∞]) are the usual Lebesgue norms, i.e., for c < d,  h[c,d],∞ := ess sup |h(t)| and h[c,d],r := t∈[c,d]

d

|h(t)| dt r

 r1

, r ≥ 1.

c

Then, Dragomir pointed out a new method in approximating the finite Hilbert transform and proved the following estimates for the finite Hilbert transform via the above Ostrowski-type inequalities for absolutely continuous function. Theorem 1.2 Let f : [a, b] → R be a function such that its derivative f  : [a, b] → R is absolutely continuous on [a, b]. Then we have the inequality:       (T f )(a, b; t) − f (t) ln b − t − b − a [ f ; λt + (1 − λ)b, λt + (1 − λ)a]   π t −a π



⎧

  1 1 2 1 a+b 2 2 ⎪  f  [a,b],∞ , f  ∈ L ∞ [a, b], ⎪ 4 + λ− 2 4 (b − a) + t − 2 ⎪ ⎪



⎪ 1 1 1 1 1 ⎨ q 1 λ1+ q + (1 − λ)1+ q (t − a)1+ q + (b − t)1+ q  f  [a,b], p , ≤ (q+1) q π⎪ ⎪ ⎪ f  ∈ L p [a, b], p > 1, 1p + q1 = 1, ⎪ ⎪      ⎩  1  1 1 a+b    f [a,b],1 , f  ∈ L 1 [a, b] 2 + λ− 2 2 (b − a) + t − 2 (1.3)

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for any t ∈ [a, b] and λ ∈ [0, 1), where [ f ; α, β] is the divided difference, i.e., [ f ; α, β] :=

f (α) − f (β) . α−β

Moreover, Dragomir obtained the following theorem in approximating the finite Hilbert transform of a differentiable function whose derivative is absolutely continuous. Theorem 1.3 Let f : [a, b] → R be a differentiable function such that its derivative f  is absolutely continuous on [a, b]. If λ = (λi )i=0,n−1 , λi ∈ [0, 1) (i = 0, n − 1) and  n−1  b−a  b−t a−t f ; t + (i + 1 − λi ) , t − (i + 1 − λi ) , Sn ( f ; λ, t) = πn n n i=0 (1.4) then we have   f (t) b−t (T f )(a; b, t) = ln + Sn ( f ; λ, t) + Rn ( f ; λ, t), (1.5) π t −a and the remainder Rn ( f ; λ, t) satisfies the estimate: |Rn ( f ; λ, t)| ⎧   n−1  1 2   ⎪ 1 1 1  1 2 ⎪ ⎪ λ  f [a,b],∞ , + − (b − a)2 + t − a+b i ⎪ n 4 n 2 4 2 ⎪ ⎪ i=0 ⎪ ⎪ ⎪ f  ∈ L ∞ [a, b], ⎪ ⎪ ⎪  n−1  q  q1 ⎪ 1+ q1 1 ⎨1 1+ q1 q 1  λi + (1 − λi ) ≤ n 1+ q1 ⎪ n (q+1) π⎪ i=0

⎪ 1 ⎪ 1+ q 1+ q1 ⎪ (t − a)  f  [a,b], p , + − t) (b ⎪ ⎪ ⎪ ⎪ ⎪ f  ∈ L p [a, b], p > 1, 1p + q1 = 1, ⎪ ⎪       ⎪ ⎩ 1 1 1 1 a+b     f [a,b],1 , f  ∈ L 1 [a, b]; n 2 + max λi − 2 2 (b − a) + t − 2 (1.6) ⎧ 1 2   ∞ ⎪ , f ∈ L [a, b], ⎪ 4n (b − a)  f [a,b],∞ 1⎨ 1+ q1 q   f [a,b], p , f  ∈ L p [a, b], p > 1, 1p + q1 = 1, (b − a) ≤ 1+ q1 π⎪ (q+1) ⎪ ⎩1  f  ∈ L 1 [a, b]. n (b − a) f [a,b],1 , (1.7)

Recently, some companions of Ostrowski’s inequality were established in [1– 3,11,25,28] (see also [4,5,16,17,19–24,26,29–32] for other related Ostrowski-type inequalities). In this paper, we shall provide some new inequalities and approximations for the finite Hilbert transform by the use of some companions of Ostrowski’s inequalities for absolutely continuous functions in various L p spaces due to Dragomir in [11] (see

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Lemma 2.1 below). For our purpose, we will use the idea of Dragomir in [10]. In Sect. 2, we will estimate the finite Hilbert transform on the interval [a, b]. Our result can give a smaller error bound than the similar result in [10] (see Remark 1 below). In Sect. 3, we will prove a quadrature formula for equidistant division of [a, b] in approximating the finite Hilbert transform.

2 Some Inequalities in Estimating the Finite Hilbert Transform For the sake of completeness, we state and prove the following lemma which provides some companions of Ostrowski’s inequalities for absolutely continuous functions in various L p spaces (see [9,11]). Lemma 2.1 Let u : [a, b] → R be an absolutely continuous function on [a, b]. Then, for all x ∈ a, a+b 2 , we have    b  u(x) + u(a + b − x)    u(t)dt (b − a) −   2 a

⎧

 1 3a+b 2 2 ⎪ u  [a,b],∞ , u  ∈ L ∞ [a, b], ⎪ 8 (b − a) + 2 x − 4 ⎪  ⎪ ⎪ 1  1+ q1 ⎨ 1+ q1 1 + 2 q a+b u  [a,b], p , u  ∈ L p [a, b], 2 −x ≤ (q+1) q1 2 (x − a) ⎪ ⎪ ⎪ p > 1, 1p + q1 = 1, ⎪ ⎪    ⎩1 3a+b   u [a,b],1 , u  ∈ L 1 [a, b], 4 (b − a) + x − 4 (2.1) where  · [a,b],r (r ∈ [1, ∞]) are the usual Lebesgue norms. Proof Using the integration by parts formula, we have  b u(x) + u(a + b − x) (b − a) − u(t)dt 2 a   a+b−x   b  x a+b   t− u (t)dt + (t − a)u (t)dt + (t − b)u  (t)dt = 2 a x a+b−x (2.2) for any x ∈ [a, b]. Taking the modulus, we have    b  u(x) + u(a + b − x)   (b − a) − u(t)dt   2 a   x  a+b−x      a + b        ≤ (t − a) u (t) dt + t − 2  u (t) dt a x  b   + (b − t) u  (t) dt := M(x). a+b−x

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Now, it is obvious that M(x) ≤ u  [a,x],∞



x

a

(t − a)dt + u  [x,a+b−x],∞

   b   t − a + b  dt + u  [a+b−x,b],∞ (b − t)dt  2  x a+b−x  2 a+b (x − a)2 + u  [x,a+b−x],∞ − x dt = u  [a,x],∞ 2 2 

a+b−x

(x − a)2 + u  [a+b−x,b],∞ 2    2 a+b (x − a)2 (x − a)2  + −x + ≤ u [a,b],∞ 2 2 2     1 3a + b 2 (b − a)2 + 2 x − = u  [a,b],∞ , 8 4 proving the first part of (2.1). Using Hölder’s inequality, we may write: 



M(x) ≤ u [a,x], p

x

1

a





+ u [x,a+b−x], p

x

+ u [a+b−x,b], p

≤ u [a,b], p

1+ q1

(q + 1)

+ u [a+b−x,b], p

b

q  q1    t − a + b  dt  2 

(b − t) dt q

 q1

a+b−x

(x − a)





a+b−x





= u  [a,x], p

q

(t − a) dt q

1 q

+ u  [x,a+b−x], p ⎣2

(x − a)

1 (q + 1)

⎡ 1 q

1+ q1 ⎤ 2 ⎦ 1 (q + 1) q

 a+b

−x

1+ q1 1

(q + 1) q  1 q

2(x − a)

1+ q1

+2

1 q



a+b −x 2

1+ 1  q

,

proving the second part of (2.1). Finally, we observe that

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  a+b − x u  [x,a+b−x],1 + (x − a)u  [a+b−x,b],1 M(x) ≤ (x − a)u  [a,x],1 + 2   a+b − x u  [a,b],1 ≤ max x − a, 2     3a + b  1  u  [a,b],1 , = (b − a) + x − 4 4 



and the lemma is proved.

The best inequalities we can get from (2.1) are embodied in the following corollary. Corollary 2.1 With the assumption in Lemma 2.1, we have     b   u 3a+b + u  a+3b   4 4 (b − a) − u(t)dt     2 a ⎧1 2  u  ∈ L ∞ [a, b], ⎪ 8 (b − a) u [a,b],∞ , ⎪ ⎪ 1 ⎨ 1+ 1 (b − a) q u  [a,b], p , u  ∈ L p [a, b], p > 1, ≤ 4(q+1) q1 ⎪ ⎪ ⎪ ⎩1  u  ∈ L 1 [a, b]. 4 (b − a)u [a,b],1 ,

1 p

+

1 q

= 1,

(2.3)

Using the above inequalities (2.1), we may point out the following result in estimating the finite Hilbert transform. Theorem 2.1 Let f : [a, b] → R be a function such that its derivative f  : [a, b] → R is absolutely continuous on [a, b]. Then we have the inequality:     (T f )(a, b; t) − f (t) ln b − t  π t −a

 b − a [ f ; λt + (1 − λ)b, λt + (1 − λ)a] + [ f ; λb + (1 − λ)t, λa + (1 − λ)t]  −  π 2



⎧

  1 3 2 1 a+b 2 2 ⎪  f  [a,b],∞ , f  ∈ L ∞ [a, b], ⎪ 8 + 2 λ − 4 4 (b − a) + t − 2 ⎪ ⎪

⎪ 1 1  1 1 1 1 ⎨ q 1 2(1 − λ)1+ q + 2 q 1 − λ 1+ q (t − a)1+ q + (b − t)1+ q 2 ≤ (q+1) q π⎪ ⎪    ∈ L p [a, b], p > 1, 1 + 1 = 1, ⎪ , f  f ⎪ [a,b], p p q ⎪     ⎩  1  3 1 t − a+b   f  [a,b],1 , + (b − a) + f  ∈ L 1 [a, b] λ − 4 4 2 2

(2.4) for any t ∈ [a, b] and λ ∈ [ 21 , 1), where [ f ; α, β] is the divided difference. Proof Since f  is bounded on [a, b], it follows that f is Lipschitzian on [a, b] and thus the finite Hilbert transform exists everywhere in (a, b). As in [10], for the function f 0 : (a, b) → R, f 0 (t) = 1, t ∈ (a, b), we have   1 b−t (T f )(a, b; t) = ln , t ∈ (a, b), π t −a

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then obviously

(T f )(a, b; t) −

   b 1 f (τ ) − f (t) f (t) b−t = PV ln dτ. π t −a π τ −t a

(2.5)

Now, if we choose in (2.1), u = f  , x = λc + (1 − λ)d, λ ∈ [ 21 , 1], c, d ∈ [a, b], then we get        f (d) − f (c) − f (λc + (1 − λ)d) + f (λd + (1 − λ)c) (d − c)   2

⎧  2 ⎪ (d − c)2 18 + 2 λ − 43  f  [c,d],∞ , f  ∈ L ∞ [a, b], ⎪ ⎪   ⎪ 1 ⎪ ⎪ ⎨ |d−c|1+ q 2(1 − λ)1+ q1 + 2 q1  1 − λ 1+ q1  f    p [c,d], p , f ∈ L [a, b], 1 2 ≤ q (q+1) ⎪ ⎪ ⎪ p > 1, 1p + q1 = 1, ⎪ ⎪ ⎪    ⎩ |d − c| 41 + λ − 43   f  [c,d],1 , f  ∈ L 1 [a, b], which is equivalent to    f (d) − f (c) f  (λc + (1 − λ)d) + f  (λd + (1 − λ)c)   −   d −c 2

⎧  2 ⎪ |d − c| 18 + 2 λ − 43  f  [c,d],∞ , ⎪ ⎪   ⎪ 1 ⎪ ⎪ ⎨ |d−c| q 2(1 − λ)1+ q1 + 2 q1  1 − λ 1+ q1  f   [c,d], p , 2 ≤ (q+1) q1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪   ⎩  1  3 4 + λ − 4  f [c,d],1 ,

f  ∈ L ∞ [a, b], f  ∈ L p [a, b], p > 1, f 

∈

1 1 p + q = L 1 [a, b].

1,

(2.6) Using (2.6), we may write   b 1 f (τ ) − f (t)  PV dτ π τ −t a   b  1   [ f (λt + (1 − λ)τ ) + f (λτ + (1 − λ)t)]dτ  − PV 2π a ⎧

b 2  1 1 ⎪ P V a |t − τ |  f  [t,τ ],∞ dτ, + 2 λ − 43 f  ∈ L ∞ [a, b], ⎪ π 8 ⎪   ⎪ 1 ⎪ 1  1 b 1+ ⎨1 1 1+ 1 2(1 − λ) q + 2 q 21 − λ q P V a |t − τ | q  f  [t,τ ], p dτ, ≤ π (q+1) q1 ⎪ ⎪ ⎪ f  ∈ L p [a, b], p > 1, 1p + q1 = 1, ⎪ ⎪   b  ⎩ 1 1  + λ − 3  P V  f [t,τ ],1 dτ, f  ∈ L 1 [a, b]; π

4

4

a

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⎧

b   1 1 3 2 ⎪  f [a,b],∞ P V a |t − τ | dτ, f  ∈ L ∞ [a, b], ⎪ π 8 + 2 λ − 4 ⎪  ⎪ ⎪ 1  1 b 1+ 1 ⎪ 1+ 1 ⎨1 1 2(1 − λ) q + 2 q 21 − λ q  f  [a,b], p P V a |t − τ | q dτ, 1 π ≤ (q+1) q ⎪ ⎪ f  ∈ L p [a, b], p > 1, 1p + q1 = 1, ⎪ ⎪

⎪  b   ⎪ ⎩ 1 1 + λ − 3  P V  t  f   dτ +  f  dτ , f  ∈ L 1 [a, b]; [τ,t],1 [t,τ ],1 a t π 4 4

⎧

   2 1 1 3 2 ⎪  f [a,b],∞ 41 (b − a)2 + t − a+b , f  ∈ L ∞ [a, b], ⎪π 8 +2 λ− 4 2 ⎪   ⎪ ⎪ 1  1+ 1 ⎪ 1+ 1 ⎨1 1 2(1 − λ) q + 2 q 21 − λ q  f  [a,b], p 1 π ≤ (q+1) q

⎪ ⎪ 1+ 1 1+ 1 q ⎪ (t − a) q + (b − t) q , f  ∈ L p [a, b], p > 1, 1p + q1 = 1, × q+1 ⎪ ⎪ ⎪      ⎩ 1  1  3 1 a+b   f [a,b],1 ,  f  ∈ L 1 [a, b]. π 4 + λ− 4 2 (b − a) + t − 2 (2.7)

Since (note that λ = 1) (see [10])  b 1 PV [ f  (λt + (1 − λ)τ ) + f  (λτ + (1 − λ)t)]dτ 2π a  t−ε  b  1 [ f  (λt + (1 − λ)τ ) + f  (λτ + (1 − λ)t)]dτ lim + = 2π ε→0+ a t+ε  1 − f (λt + (1 − λ)a + f (λt + (1 − λ)b = 2π 1−λ  − f (λa + (1 − λ)t + f (λb + (1 − λ)t + λ b − a [ f ; λt + (1 − λ)b, λt + (1 − λ)a] + [ f ; λb + (1 − λ)t, λa + (1 − λ)t] . = π 2 Using (2.5) and (2.7), we deduce the desired result (2.4).



The best inequality one may obtain from (2.4) is embodied in the following corollary. Corollary 2.2 With the assumption of Theorem 2.1, for λ = 43 , one has the inequality    3b+t 3a+t      3t+a  + f; 4 , 4  , f (t) b−t b − a f ; 3t+b  4 4 ln − (T f )(a, b; t) −    π t −a π 2

⎧

 1 1 a+b 2 2 ⎪  f  [a,b],∞ , f  ∈ L ∞ [a, b], ⎪ 8 4 (b − a) + t − 2 ⎪

⎪ 1 1 ⎪ 1+ 1+ ⎪ q (t − a) q + (b − t) q  f  [a,b], p , f  ∈ L p [a, b], 1 ⎨ qp 1+ q1 4 (q+1) ≤ π⎪ ⎪ p > 1, 1p + q1 = 1, ⎪ ⎪ ⎪ ⎪    ⎩ 1 1 a+b   | f [a,b],1 , f  ∈ L 1 [a, b] 4 2 (b − a) + t − 2 (2.8) for any t ∈ [a, b].

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Remark 1 As stated in [10], the best estimator one may obtain from (1.3) is, for λ = 21 ,        (T f )(a, b; t) − f (t) ln b − t − b − a f ; t + b , t + a    π t −a π 2 2

⎧

2  1 1 ⎪  f  [a,b],∞ , (b − a)2 + t − a+b f  ∈ L ∞ [a, b], ⎪ 4 4 2 ⎪ ⎪

⎪ 1+ 1 1+ 1 1⎨ p q (t − a) q + (b − t) q  f  [a,b], p , f  ∈ L p [a, b], 1+ q1 ≤ q 2 (q+1) π⎪ ⎪ ⎪ p > 1, 1p + q1 = 1, ⎪ ⎪     ⎩1 1 (b − a) + t − a+b  | f  [a,b],1 , f  ∈ L 1 [a, b]. 2

2

2

(2.9) We note that inequality (2.8) we obtained here gives a new estimate of the finite Hilbert transform and a smaller error bound than that of (2.9).

3 A Quadrature Formula for Equidistant Divisions The following lemma is of interest in itself. Lemma 3.1 Let u : [a, b] → R be an absolutely continuous function on [a, b]. Then for all n ≥ 1, λi ∈ [ 21 , 1) (i = 0, ..., n − 1) and t, τ ∈ [a, b] with t = τ , we have the inequality      n−1  1  τ + u t + (i + λi ) τ −t 1  u t + (i + 1 − λi ) τ −t   n n u(s)ds −    τ − t t n 2 i=0   ⎧ n−1 2  ⎪ |t−τ | 1 ⎪ u  [t,τ ],∞ , λi − 43 + n2 u  ∈ L ∞ [a, b], ⎪ n 8 ⎪ ⎪ i=0 ⎪ ⎪ ⎨  1 1  n−1  1  1 q q  q 1+ q1 |t−τ | 1 1 1+ q ≤ q 2(1 − λi ) + 2 λi − 2 1 n ⎪ ⎪ (q+1) q n i=0 ⎪ ⎪ ⎪ ⎪ u   , u  ∈ L p [a, b], ⎪   ⎩ 1  1 [t,τ ], p  3 u  ∈ L 1 [a, b], n 4 + max λi − 4 u [t,τ ],1 , (3.1) where  · [t,τ ],r (r ∈ [1, ∞]) are the usual Lebesgue norms, p > 1,

1 p

+

1 q

= 1.

Proof Consider the equidistant division of [t, τ ] (if t < τ ) or [τ, t] (if τ < t) given by τ −t , i = 0, n. E n : xi = t + i · (3.2) n τ −t 1 Then the points ξi = λi [t + i · τ −t n ] + (1 − λi )[t + (i + 1) · n ] (λi ∈ [ 2 , 1], i = 0, n − 1) are between xi and xi+1 . We observe that we may write for simplicity ξi = t + (i + 1 − λi ) τ −t n (i = 0, n − 1). We also have

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τ −t , xi + xi+1 − ξi = t + (i + λi )  n  3xi + xi+1 τ −t 3 ξi − = − λi , 4 n 4 τ −t ξi − xi = (1 − λi ) n and   xi + xi+1 1 τ −t − ξi = λi − 2 2 n for any i = 0, n − 1. If we apply the inequality (2.1) on the intervals [xi , xi+1 ] and the intermediate points ξi (i = 0, n − 1), then we may write that       τ − t u(t + (i + 1 − λ ) τ −t ) + u(t + (i + λ ) τ −t )  xi+1 i n i n   u(s)ds  −   n  2 xi

⎧

 2 2 2 (t−τ ) 1 (t−τ ) 3 ⎪ u  [xi ,xi+1 ],∞ , u  ∈ L ∞ [a, b], ⎪ 8 n2 + 2 n2 4 − λi ⎪   ⎪ ⎪ 1+ 1 1+ q1  1 1 ⎪ 1+ q1 |t−τ | q ⎨ 1 1 1+ q q |t−τ | u  [xi ,xi+1 ], p , λ 2 − λ + 2 − (1 ) i i 1 2 1+ q1 1+ q1 ≤ (q+1) q n n ⎪ ⎪ u  ∈ L p [a, b], p > 1, 1p + q1 = 1, ⎪ ⎪

⎪ ⎪ ⎩ 1 |τ −t| + |τ −t|  3 − λ  u   u  ∈ L 1 [a, b]. i [x ,x ],1 , 4

n

n

4

i

i+1

(3.3) Summing (3.3), we get      n−1  τ + u t + (i + λi ) τ −t τ − t  u t + (i + 1 − λi ) τ −t   n n u(s)ds −    t  n 2 i=0 ⎧ n−1

  ⎪ (t−τ )2  1 3 2 ⎪ u [xi ,xi+1 ],∞ , + 2 λ − u  ∈ L ∞ [a, b], ⎪ i 2 8 4 ⎪ n ⎪ i=0 ⎪   ⎪ 1 ⎪ n−1 ⎪ ⎨ |t−τ |1+ q  2 (1 − λ )1+ q1 + 2 q1 λ − 1 1+ q1 u   i i [xi ,xi+1 ], p , 2 ≤ (q+1) q1 n 1+ q1 i=0 (3.4) ⎪  ∈ L p [a, b], p > 1, 1 + 1 = 1, ⎪ ⎪ u ⎪ p q ⎪ ⎪ n−1 ⎪   1   ⎪ |τ −t| 3 ⎪   ⎩ n u  ∈ L 1 [a, b]. 4 + λi − 4 u [xi ,xi+1 ],1 , i=0

However,        n−1 n−1    1 1 3 2 3 2   λi − + 2 λi − n+2 u [xi ,xi+1 ],∞ ≤ u [t,τ ],∞ , 8 4 8 4 i=0

i=0

(3.5)

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Approximating the Finite Hilbert Transform... n−1 

 2 (1 − λi )

1+ q1

+2

1 q



i=0

1 λi − 2

1+ 1  q

u  [xi ,xi+1 ], p

1 n−1   1p 1+ 1 q  q   n−1  q 1 1 p 1+ q1  ≤ + 2 q λi − u [xi ,xi+1 ], p 2 (1 − λi ) 2

i=0

i=0

1 n−1   1+ 1 q  q  q 1 1 1 1+ = u  [t,τ ], p 2 (1 − λi ) q + 2 q λi − 2

(3.6)

i=0

and   n−1      1 3 3   u [xi ,xi+1 ],1 + λi −  u  [xi ,xi+1 ],1 ≤ + max λi −  4 4 4 4 i=0     1 3 + max λi −  u  [t,τ ],1 . = (3.7) 4 4

n−1   1 i=0

Now, using (3.4)–(3.7), we deduce the desired result (3.1).



We may now state the following theorem in approximating the finite Hilbert transform of a differentiable function whose derivative is absolutely continuous. Theorem 3.1 Let f : [a, b] → R be a differentiable function such that its derivative f  is absolutely continuous on [a, b]. If λ = (λi )i=0,n−1 , λi ∈ [ 21 , 1) (i = 0, n − 1),  n−1  b−a  b−t a−t f ; t + (i + 1 − λi ) , t + (i + 1 − λi ) (3.8) Sn ( f ; λ, t) = πn n n i=0

and  n−1  b−a  b−t a−t Tn ( f ; λ, t) = f ; t + (i + λi ) , t + (i + λi ) , πn n n

(3.9)

i=0

then we have (T f )(a, b; t) =

  f (t) b−t Sn ( f ; λ, t) + Tn ( f ; λ, t) ln + Rn ( f ; λ, t), + π t −a 2 (3.10)

and the reminder Rn ( f ; λ, t) satisfies the estimate:

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|Rn ( f ; λ, t)| ⎧   n−1 2 1   ⎪ 1 1 2 3 2 + t − a+b 2  f   ⎪ ⎪n 8+n λi − 4 (b − a) [a,b],∞ , ⎪ 4 2 ⎪ ⎪ i=0 ⎪ ⎪ ⎪ f  ∈ L ∞ [a, b], ⎪ ⎪ ⎪  n−1   1 ⎪ ⎪ 1  1+ q1 q q  ⎪ 1+ q1 q ⎪ 1 1 1 + 2 q λi − 2 2(1 − λi ) 1 ⎨n n 1+ 1 (q+1) q i=0 ≤

π⎪ 1+ 1 1+ 1 ⎪ × (t − a) q + (b − t) q  f  [a,b], p , ⎪ ⎪ ⎪ ⎪  p ⎪ ⎪ 1, 1 + 1 = 1, ⎪  f ∈ L [a, b], p> ⎪  1 p q  ⎪ 1 1 3 ⎪  ⎪ n 4 + max λi − 4  2 (b − a) + t − a+b ⎪ 2 ⎪ ⎪ ⎩   f [a,b],1 , f  ∈ L 1 [a, b]. (3.11) Proof Applying Lemma 3.1 for the function f  , we may write that    n−1    f (τ ) − f (t) 1  + f  t + (i + λi ) τ −t f t + (i + 1 − λi ) τ −t  n n  −     τ −t n 2 i=0 ⎧   n−1 2  ⎪ |t−τ | 1 ⎪ ⎪  f  [t,τ ],∞ , λi − 43 + n2 f  ∈ L ∞ [a, b], ⎪ n 8 ⎪ ⎪ i=0 ⎪ ⎪    1 ⎪ ⎪ 1  1 q q  ⎨ |t−τ | q1 1 n−1 1+ q1 1 1+ q q 2(1 − λi ) + 2 λi − 2 ≤ n(q+1) q1 n i=0 ⎪ ⎪ ⎪  f  [t,τ ], p , f  ∈ L p [a, b], ⎪ ⎪ ⎪ ⎪ ⎪ p > 1, 1p + q1 = 1, ⎪    ⎪ ⎩ 1 1 3   f  ∈ L 1 [a, b] n 4 + max λi − 4  f [t,τ ],1 , for any t, τ ∈ [a, b], t = τ . Taking the P V , we may write   b 1 f (τ ) − f (t)  PV dτ (3.12) π τ −t a    b  n−1 f t + (i + 1 − λi ) τ −t + f  t + (i + λi ) τ −t 1   n n − PV dτ   πn 2 a i=0   ⎧ n−1 b 2  ⎪ ⎪1 1+2 λi − 43 P V a |t − τ |  f  [t,τ ],∞ dτ, f  ∈ L ∞ [a, b], ⎪ n 8 n ⎪ ⎪ i=0 ⎪ ⎪  n−1   1 ⎪ ⎪ 1  1+ q1 q q ⎪  1+ q1 ⎨ 1 1 1 1 2(1 − λi ) + 2 q λi − 2 1 n ≤ n(q+1) q i=0 ⎪ π⎪ 1 b ⎪ ⎪ ×P V a |t − τ | q  f  [t,τ ], p dτ, f  ∈ L p [a, b], ⎪ ⎪ ⎪ ⎪ p > 1, 1p + q1 = 1, ⎪ ⎪    b  ⎩ 1 1 3 + max λi −  P V  f [t,τ ],1 dτ, f  ∈ L 1 [a, b]. n

4

4

a

(3.13)

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However (see [10]),  b |t − τ |  f  [t,τ ],∞ dτ ≤ f  [a,b],∞ P V |t − τ | dτ a a   (t − a)2 + (b − t)2 = f  [a,b],∞ , 2  b  b 1 1   q |t | |t − τ | q dτ PV − τ  f [t,τ ], p dτ ≤ f [a,b], p P V 

b

PV

a

 = f  [a,b], p

a

1

(t + a) q

+1

1

+ (b − t) q



+1

1 q +1

1 1 q +1 +1 (t − a) q + (b − t) q  f  [a,b], p , = q +1  t   t  b  f  [t,τ ],1 dτ =P V  f  [τ,t],1 dτ +  f  [t,τ ],1 dτ PV a

a

a

≤ max{t − a, b − t} f  [a,b],1 ,     a + b  1   f  [a,b],1 , = (b − a) + t − 2 2 



and using the inequality (3.12) we obtain the desired estimate (3.11). Remark 2 For n = 1, we recapture the inequality (2.4).

For λi = 43 (i = 0, n − 1), the following particular case, which may be easily numerically implemented, holds. Corollary 3.1 Let f be as in Theorem 3.1. Define      n−1  1 b−t 1 a−t b−a  f;t + i + ,t + i + S M,n ( f ; t) = πn 4 n 4 n

(3.14)

     n−1  3 a−t b−a  3 b−t TM,n ( f ; t) = ,t + i + f;t + i + , πn 4 n 4 n

(3.15)

i=0

and

i=0

and the remainder R M,n ( f ; t) satisfies the estimate   b−t S M,n ( f ; t) + TM,n ( f ; t) f (t) ln + R M,n ( f ; t). + (T f )(a, b; t) = π t −a 2 (3.16)

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Then we have the representation:    R M,n ( f ; t) ⎧

 1 1 2 + t − a+b 2  f   ⎪ (b − a) [a,b],∞ , ⎪ 2 ⎪ 8n 4 1

1⎨ 1 q 1+ q 1+ q1 (2+2 )q 1 ≤ (t − a)  f  [a,b], p , + − t) (b n 1+ q1 π⎪ 1+ q1 ⎪ 4 (q+1) ⎪     ⎩ 1 1 a+b    f [a,b],1 , 4n 2 (b − a) + t − 2 for any t ∈ [a, b], where p > 1,

1 p

+

1 q

f  ∈ L ∞ [a, b], f  ∈ L p [a, b], f  ∈ L 1 [a, b] (3.17)

= 1.

Acknowledgments The authors would like to thank the anonymous referees and the editors for their suggestions which greatly improved the presentation of this paper. This work was supported by the National Natural Science Foundation of China (Grant No. 11301277), the Natural Science Foundation of Jiangsu Province (Grant No. BK20151523), the Six Talent Peaks Project in Jiangsu Province (Grant No. 2015XCL-020), and the Qing Lan Project of Jiangsu Province.

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