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A PERTURBED TRAPEZOID INEQUALITY IN TERMS OF THE THIRD AND APPLICATIONS

N.S. BARNETT AND S.S. DRAGOMIR

Abstract. Some error estimates in terms of the p−norms of the third deriva- tive for the remainder in a perturbed trapezoid formula are given. Applications to composite quadrature formulae, for the expectation of a random variable and for Hermite-Hadamard in Information Theory are pointed out.

1. Introduction In [4], by the use of Gr¨uss’ inequality, the authors have obtained the following perturbed trapezoid inequality. Theorem 1. Let f :[a, b] → R be a twice differentiable on (a, b) and asume that γ := inf f 00 (x) > −∞ and Γ := sup f 00 (x) < ∞. x∈(a,b) x∈(a,b) Then we have the inequality

1 Z b b − a (b − a)2 0 0 (1.1) f (x) dx − [f (a) + f (b)] + [f (b) − f (a)] b − a a 2 12 (b − a)3 ≤ (Γ − γ) . 12 Using a finer argument based on a pre-Gr¨uss inequality, Cerone and Dragomir [2, p. 121] improved the above result as follows. Theorem 2. Let f have the properties of Theorem 1. Then

1 Z b b − a (b − a)2 0 0 (1.2) f (x) dx − [f (a) + f (b)] + [f (b) − f (a)] b − a a 2 12 1 ≤ √ (b − a)3 (Γ − γ) . 24 5 The main aim of the present work is to obtain some bounds for the left part of (1.2) in terms of the p−norms of f 000 assuming that the function f is twice differentiable on (a, b) and that the is absolutely continuous on (a, b). A number of applications are also pointed out.

Date: January 11, 2001. 1991 Subject Classification. Primary 26D15; Secondary 26D10, 41A55. Key words and phrases. Trapezoid inequality. 1 2 N.S. BARNETT AND S.S. DRAGOMIR

2. A Perturbed Trapezoid Formula The following representation lemma holds.

Lemma 1. Let f :[a, b] → R be such that the second derivative is absolutely continuous on [a, b]. Then we have the equality:

Z b f (a) + f (b) (b − a)2 f (t) dt − (b − a) + [f 0 (b) − f 0 (a)](2.1) a 2 12 1 Z b Z b Z t  t + s  = u − f 000 (u) du (t − s) dtds. 4 (b − a) a a s 2

Proof. Integrating by parts, we have

Z b Z b Z t  t + s  I : = u − f 000 (u) du (t − s) dtds a a s 2 Z b Z b f 00 (t) + f 00 (s) Z t  = (t − s) − f 00 (u) du (t − s) dtds a a 2 s " # Z b Z b f 00 (t)(t − s)2 + f 00 (s)(t − s)2 = − (f 0 (t) − f 0 (s)) (t − s) dtds a a 2 " # 1 Z b Z b Z b Z b = f 00 (t)(t − s)2 dtds + f 00 (s)(t − s)2 dtds 2 a a a a Z b Z b − (f 0 (t) − f 0 (s)) (t − s) dtds. a a

By symmetry,

Z b Z b Z b Z b J := f 00 (t)(t − s)2 dtds = f 00 (s)(t − s)2 dtds, a a a a and using Korkine’s identity or direct computation, we have

Z b Z b K : = (f 0 (t) − f 0 (s)) (t − s) dtds a a " Z b Z b Z b # = 2 (b − a) f 0 (t) tdt − f 0 (t) dt · tdt . a a a

Then, I = J − K. Since

Z b Z b ! J = f 00 (t) (t − s)2 ds dt a a " # 1 Z b Z b = f 00 (t)(b − t)3 dt + (t − a)3 f 00 (t) dt 3 a a PERTURBED TRAPEZOID INEQUALITY 3 " 1 b Z b 0 3 2 0 = f (t)(b − t) + 3 (b − t) f (t) dt 3 a a # b Z b 0 3 2 0 + f (t)(t − a) − 3 (t − a) f (t) dt a a " " # 1 b Z b 0 3 2 = −f (a)(b − a) + 3 f (t)(b − t) − 2 (b − t) f (t) dt 3 a a " ## b Z b 0 3 2 +f (b)(b − a) − 3 f (t)(t − a) − 2 (t − a) f (t) dt a a " " # 1 Z b = [f 0 (b) − f 0 (a)] (b − a)3 + 3 −f (a)(b − a)2 + 2 (b − t) f (t) dt 3 a " Z b ## − 3 f (b)(b − a)2 − 2 (t − a) f (t) dt a 1 f (a) + f (b) Z b = [f 0 (b) − f 0 (a)] (b − a)3 − (b − a)2 + 2 (b − a) f (t) dt 3 2 a and

" " b Z b # 2 2 # b − a K = 2 (b − a) f (t) t − f (t) dt − [f (b) − f (a)] a a 2 " " # # Z b a + b = 2 (b − a) f (b) b − f (a) a − f (t) dt − (b − a)[f (b) − f (a)] a 2 " # Z b a + b a + b = 2 (b − a) f (b) b − f (a) a − f (t) dt − f (b) + f (a) a 2 2 Z b = (b − a)2 [f (a) + f (b)] − 2 (b − a) f (t) dt, a then

1 Z b I = [f 0 (b) − f 0 (a)] (b − a)3 − 2 [f (a) + f (b)] (b − a)2 + 4 (b − a) f (t) dt 3 a

Dividing by 4 (b − a) we deduce the desired equality (2.1).

The following perturbed version of the trapezoid inequality holds. 4 N.S. BARNETT AND S.S. DRAGOMIR

Theorem 3. Let f :[a, b] → R be such that the second derivative is absolutely continuous on [a, b]. Then we have

Z b f (a) + f (b) (b − a)2 0 0 (2.2) f (t) dt − (b − a) + [f (b) − f (a)] a 2 12

 1 R b R b |t − s|3 kf 000k dtds if f 000 ∈ L [a, b];  16(b−a) a a [t,s],∞ ∞    b b 1  1 R R 2+ q 000 000 ≤ 1 |t − s| kf k[t,s],p dtds if f ∈ Lp [a, b] , q a a  8(q+1) (b−a)  p > 1, 1 + 1 = 1;  p q  1 R b R b 2 000 8(b−a) a a (t − s) kf k[t,s],1 dtds

4  (b−a) 000 000  160 kf k[a,b],∞ if f ∈ L∞ [a, b];    1  2 3+  q (b−a) q 000 000 ≤ 1 kf k[a,b],p if f ∈ Lp [a, b] ,  4(3q+1)(4q+1)(q+1) q  1 1  p > 1, p + q = 1;  3  (b−a) 000 48 kf k[a,b],1 ,

1 l where khk := R d |h (u)|l du if l ≥ 1 and khk = ess sup |h (u)| . [c,d],l c [c,d],∞ u∈[c,d] (u∈[d,c])

Proof. Denote

1 Z b Z b Z t  t + s  R (f; a, b) := u − f 000 (u) du (t − s) dtds. 4 (b − a) a a s 2 As Z t   Z t t + s 000 000 t + s u − f (u) du ≤ kf k[t,s],∞ u − du s 2 s 2 (t − s)2 = kf 000k , 4 [t,s],∞

Z t   t + s 000 u − f (u) du s 2 1 Z t q q 000 t + s ≤ kf k[t,s],p u − du s 2 1 1+ q |t − s| 000 1 1 = 1 kf k[t,s],q , p > 1, + = 1 2 (q + 1) q p q and Z t   t + s 000 t + s 000 u − f (u) du ≤ sup u − kf k[t,s],1 s 2 u∈[t,s] 2 (u∈[t,s]) PERTURBED TRAPEZOID INEQUALITY 5 then we can state that Z t   t + s 000 (2.3) u − f (u) du s 2  2 (t−s) kf 000k if f 000 ∈ L [a, b];  4 [t,s],∞ ∞    1+ 1  |t−s| q 000 000 ≤ 1 kf k[t,s],q if f ∈ Lp [a, b] , 2(q+1) q  1 1  p > 1, + = 1;  p q  |t−s| 000 2 kf k[t,s],1 . Taking the modulus of R (f; a, b) we get, by (2.3), |R (f; a, b)| Z b Z b Z t   1 t + s 000 ≤ |t − s| u − f (u) du dtds 4 (b − a) a a s 2

 1 R b R b |t − s|3 kf 000k dtds if f 000 ∈ L [a, b];  4 a a [t,s],∞ ∞    b b 1 1  1 R R 2+ q 000 000 ≤ 1 |t − s| kf k[t,s],p dtds if f ∈ Lp [a, b] , q a a 4 (b − a)  2(q+1)  p > 1, 1 + 1 = 1;  p q  1 R b R b 2 000 2 a a |t − s| kf k[t,s],1 dtds which proves the first inequality in (2.2). Now, consider the double integral Z b Z b Z b "Z t Z b # m m m Im : = |t − s| dtds = (t − s) ds + (s − t) ds dt a a a a t " # Z b (t − a)m+1 + (b − t)m+1 2 (b − a)m+2 = dt = a m + 1 (m + 1) (m + 2) for all m > 0. Using the above calculation for Im, we have:- Z b Z b Z b Z b 3 000 000 3 |t − s| kf k[t,s],∞ dtds ≤ kf k[a,b],∞ |t − s| dtds a a a a (b − a)5 = · kf 000k , 10 [a,b],∞

b b b b Z Z 1 Z Z 1 2+ q 000 000 2+ q |t − s| kf k[t,s],p dtds ≤ kf k[a,b],p |t − s| dtds a a a a 4+ 1 2q2 (b − a) q = · kf 000k (3q + 1) (4q + 1) [a,b],p and Z b Z b Z b Z b 2 000 000 2 (t − s) kf k[t,s],1 dtds ≤ kf k[a,b],1 (t − s) dtds a a a a (b − a)4 = · kf 000k , 6 [a,b],1 6 N.S. BARNETT AND S.S. DRAGOMIR which give the last part of (2.2).

3. Applications to Composite Quadrature Formulae

Consider the division In : a = x0 ≤ x1 ≤ x2 ≤ · · · ≤ xn−1 ≤ xn = b and define   hi := xi+1 − xi i = 0, n − 1 and ν (In) := max hi|i = 0, n − 1 . If

n−1 n−1 1 X 1 X (3.1) P (f; I ) := [f (x ) + f (x )] h − h2 [f 0 (x ) − f 0 (x )] n n 2 i i+1 i 12 i i+1 i i=0 i=0 is the perturbed trapezoid formula associated with the absolutely continuous func- tion f :[a, b] → R, then we may state the following theorem. Theorem 4. Let f :[a, b] → R be such that the second derivative is absolutely continuous on [a, b]. Then, for a given division In, we have:

Z b (3.2) f (t) dt = Pn (f; In) + Rn (f; In) , a where Pn (f; In) is given in (3.1) and the remainder Rn (f; In) satisfies the estimate:

 n−1  1 000 P 4 000  160 kf k[a,b],∞ hi if f ∈ L∞ [a, b];  i=0    1  q  2 n−1  q 000 P 3q+1 (3.3) |Rn (f; In)| ≤ 1 kf k[a,b],p hi q  4(3q+1)(4q+1)(q+1) i=0  if f 000 ∈ L [a, b] , p > 1, 1 + 1 = 1;  p p q    1 3 000 48 [ν (h)] kf k[a,b],1 .

 Proof. If we apply Theorem 3 to the intervals [xi, xi+1] i = 0, n − 1 we get

Z xi+1 2 f (xi) + f (xi+1) hi 0 0 f (t) dt − · hi + [f (xi+1) − f (xi)] xi 2 12  4 hi 000 000  kf k if f ∈ L∞ [a, b];  160 [xi,xi+1],∞    3+ 1  q2h q ≤ i kf 000k if f 000 ∈ L [a, b] , 1 [xi,xi+1],p p  4 (3q + 1) (4q + 1) (q + 1) q   p > 1, 1 + 1 = 1;  p q  h3  i kf 000k . 48 [xi,xi+1],1 PERTURBED TRAPEZOID INEQUALITY 7

Using the generalised triangle inequality, we obtain

(3.4) |Rn (f; In)|

n−1 Z xi+1 2  X f (xi) + f (xi+1) h = f (t) dt − · h + i [f 0 (x ) − f 0 (x )] 2 i 12 i+1 i i=0 xi

n−1 Z xi+1 2 X f (xi) + f (xi+1) hi 0 0 ≤ f (t) dt − · hi + [f (xi+1) − f (xi)] 2 12 i=0 xi  n−1 1 P 4 000 000  h kf k if f ∈ L∞ [a, b];  160 i [xi,xi+1],∞  i=0    2 n−1 3+ 1 q P q 000 000 ≤ 1 h kf k if f ∈ L [a, b] , i [xi,xi+1],p p 4(3q+1)(4q+1)(q+1) q i=0  1 1  p > 1, + = 1;  p q  n−1  1 P h3 kf 000k .  48 i [xi,xi+1],1 i=0 As n−1 n−1 X X h4 kf 000k ≤ kf 000k h4, i [xi,xi+1],∞ [a,b],∞ i i=0 i=0 then by (3.4) we deduce the first inequality in (3.3). Using H¨older’s discrete inequality, we may write

n−1 1 1 Z xi+1  p X 3+ q 000 p hi |f (t)| dt i=0 xi 1 1 n−1 ! q n−1 1 !p! p 1 Z xi+1  p X (3+ q )q X 000 p ≤ hi × |f (t)| dt i=0 i=0 xi 1 n−1 ! q n−1 1 Z xi+1  p X 3q+1 X 000 p = hi × |f (t)| dt i=0 i=0 xi 1 n−1 ! q X 3q+1 000 = hi kf k[a,b],p , i=0 which proves the second inequality in (3.3). Finally, we observe that

n−1 n−1 X 3 X 3 h3 kf 000k ≤ [ν (h)] kf 000k = [ν (h)] kf 000k , i [xi,xi+1],1 [xi,xi+1],1 [a,b],1 i=0 i=0 and the theorem is proved.

In practical applications, it is useful to consider an equidistant partitioning b − a E : x := a + i · , i = 0, . . . , n. n i n 8 N.S. BARNETT AND S.S. DRAGOMIR

In this case the perturbed trapezoid formula becomes:

n−1 b − a X   b − a  b − a (3.5) P (f) : = f a + i · + f a + (i + 1) · n 2n n n i=0 (b − a)2 − [f 0 (b) − f 0 (a)] . 12n2 Consequently, the following corollary holds. Corollary 1. If f is as in Theorem 4, then we have

Z b (3.6) f (t) dt = Pn (f) + Rn (f) , a where Pn (f) is given in (3.5) and the remainder Rn (f) satisfies the estimate

(3.7) |Rn (f)|

 4  (b − a) 000 000  kf k[a,b],∞ if f ∈ L∞ [a, b];  160n3    3+ 1  q2 (b − a) q  000 1 kf k[a,b],p ≤ 4 (3q + 1) (4q + 1) (q + 1) q n3  000 1 1  if f ∈ Lp [a, b] , p > 1, + = 1;  p q    (b − a)3  kf 000k .  48n3 [a,b],1 Remark 1. It is important to note that the perturbed trapezoid formula contains, in 2 (b−a) 0 0 addition to the classical trapezoid formula, the term − 12n2 [f (b) − f (a)], which can be calculated simply when the of the end-points a and b are known. As can be seen in formula (3.7), the order of the new formula is 3, while the order of the classical trapezoid formula is only 2. Remark 2. Atkinson [1] terms the quadrature rule (2.2) a connected trapezoidal rule and obtains it using an asymptotic error estimate approach which does not provide an expression for the error bound. He does, however, state that the corrected trapezoidal rule is O h4 compared with O h2 for the trapezoidal rule.

4. Applications for Expectation Let X be a random variable having the p.d.f., f :[a, b] → R and the cumulative distribution function F :[a, b] → [0, 1], i.e.,

Z x F (x) = f (t) dt, x ∈ [a, b] . a PERTURBED TRAPEZOID INEQUALITY 9

Theorem 5. With the above assumptions and if the p.d.f., f is differentiable on [a, b] and f 0 is absolutely continuous, then

a + b (b − a)2

(4.1) E (X) − − [f (b) − f (a)] 2 12 4  (b−a) 00 00  160 kf k[a,b],∞ if f ∈ L∞ [a, b];    1  2 3+  q (b−a) q 00 00 ≤ 1 kf k[a,b],p if f ∈ Lp [a, b] ,  4(3q+1)(4q+1)(q+1) q  1 1  p > 1, p + q = 1;  3  (b−a) 00 48 kf k[a,b],1 , where E (X) is the expectation of X.

Proof. Applying Theorem 3 for F , we may write that

Z b F (a) + F (b) (b − a)2

(4.2) F (t) dt − (b − a) + [f (b) − f (a)] a 2 12 4  (b−a) 00 00  160 kf k[a,b],∞ if f ∈ L∞ [a, b];    1  2 3+  q (b−a) q 00 00 ≤ 1 kf k[a,b],p if f ∈ Lp [a, b] ,  4(3q+1)(4q+1)(q+1) q  1 1  p > 1, p + q = 1;  3  (b−a) 00 48 kf k[a,b],1 , However, F (a) = 0, F (b) = 1 and

Z b F (t) dt = b − E (X) , a and then by (4.2) we obtain the desired inequality (4.1).

5. Applications for Hermite-Hadamard Divergence Assume that a set χ and the σ−finite measure µ is given. Consider the set of all n R o probability densities on µ to be Ω := p|p : χ → R, p (x) ≥ 0, χ p (x) dµ (x) = 1 . Csisz´ar f−divergence is defined as follows [3] Z  q (x)  (5.1) Df (p, q) := p (x) f dµ (x) , p, q ∈ Ω, χ p (x) where f is convex on (0, ∞). It is assumed that f (u) is zero and strictly convex at u = 1. By appropriately defining this convex function, various such as: Kullback-Leibler divergence, variation distance Dv, Hellinger discrimination DH , 2 χ −divergence Dχ2 , α−divergence Dα, Bhattacharyya distance DB, Harmonic dis- tance DHa, Jeffreys distance DJ , triangular discrimination D∆, etc. (see [5]). 10 N.S. BARNETT AND S.S. DRAGOMIR

In [6], Shioya and Da-te introduced the generalised Lin-Wong f−divergence 1 1  Df p, 2 p + 2 q and the Hermite-Hadamard (HH) divergence

q(x) Z R p(x) f (t) dt (5.2) Df (p, q) := p (x) 1 dµ (x) , p, q ∈ Ω, HH q(x) χ p(x) − 1 and, by the use of the Hermite-Hadamard inequality for convex functions, proved the following basic inequality

 1 1  1 (5.3) D p, p + q ≤ Df (p, q) ≤ D (p, q) , f 2 2 HH 2 f provided that f is convex and normalised, i.e., f (1) = 0. The following result holds.

Theorem 6. Let 0 ≤ r ≤ 1 ≤ R < ∞ and f :[r, R] → R be a twice differentiable function so that the second derivative f 00 :[r, R] → R is absolutely continuous on q(x) [r, R]. If p, q ∈ Ω and r ≤ p(x) ≤ R for a.e. x ∈ χ, then we have the inequality:

f 1 1 (5.4) D (p, q) − Df (p, q) + D(·−1)f 0(·) (p, q) HH 2 12

 Z 3 1 |q(x)−p(x)| 000  160 p2(x) kf k q(x) ,1 ,∞ dµ (x)  [ p(x) ]  χ    Z 2+ 1  β2 |q(x)−p(x)| β 000  kf k q(x) dµ (x) 1 1+ 1 [ ,1],α ≤ 4(3β+1)(4β+1)(β+1) β χ (p(x)) β p(x)  if α > 1, 1 + 1 = 1, f 000 ∈ L [r, R];  α β α    Z 2  1 (q(x)−p(x)) 000  kf k q(x) dµ (x)  48 p(x) [ ,1],1 χ p(x)  kf 000k  [r,R],∞ 000  D|χ|3 (p, q) if f ∈ L∞ [r, R];  160    β2 kf 000k  [r,R],α 000 ≤ D 2+ 1 (p, q) if f ∈ L [r, R] , 1 β α β |χ|  4 (3β + 1) (4β + 1) (β + 1)  α > 1, 1 + 1 = 1;  α β  kf 000k  [r,R],1  D 2 (p, q) ,  48 χ where

Z |q (x) − p (x)|m m D|χ| (p, q) := m−1 dµ (x) , m ∈ R, m > 0. χ [p (x)] PERTURBED TRAPEZOID INEQUALITY 11

Proof. We use the inequality (2.2) in the following version:

1 Z b f (a) + f (b) b − a 0 0 (5.5) f (t) dt − + [f (b) − f (a)] b − a a 2 12  |b − a|3  kf 000k if f 000 ∈ L [a, b];  [a,b],∞ ∞  160   1  2 2+ β  β |b − a| 000 000 ≤ 1 kf k[a,b],α if f ∈ Lα [a, b] ,  4 (3β + 1) (4β + 1) (β + 1) β  1 1  α > 1, + = 1;  α β  (b − a)2  kf 000k ,  48 [a,b],1 for either a < b or b < a. q(x) If we put in (5.5) a = 1 and b = p(x) , we get

q(x) R p(x)       1 f (t) dt 1 q (x) q (x) − p (x) 0 q (x) 0 (5.6) − f + f − f (1) q(x) − 1 2 p (x) 12p (x) p (x) p(x)  |q (x) − p (x)|3  000  3 kf k q(x) ,1 ,∞  160p (x) [ p(x) ]    1  2 2+ β  β |q (x) − p (x)| 000 ≤ 1 1 kf k q(x) 2+ [ p(x) ,1],α  4 (3β + 1) (4β + 1) (β + 1) β [p (x)] β    2  (q (x) − p (x)) 000  q(x)  2 kf k ,1 ,1 48p (x) [ p(x) ]

 |q (x) − p (x)|3  kf 000k  3 [r,R],∞  160p (x)    2+ 1  β2 |q (x) − p (x)| β ≤ kf 000k 1 2+ 1 [r,R],α  4 (3β + 1) (4β + 1) (β + 1) β [p (x)] β    2  (q (x) − p (x))  kf 000k .  48p2 (x) [r,R],1 If we multiply (5.6) by p (x) ≥ 0, integrate on χ and take into consideration that Z Z p (x) dµ (x) = q (x) dµ (x) = 1, χ χ then we get (5.4).

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[2] P. CERONE and S.S. DRAGOMIR, Trapezoidal type rules from an inequalities point of view, Analytic Computational Methods in Applied Mathematics, G. Anastassiou (Ed.), CRC press, N.Y., 2000, 65-134. [3] I. CSISZAR,´ Information-type measures of difference of probability distributions and indirect observations, Studia Math. Hungarica, 2 (1967), 299-318. [4] S.S. DRAGOMIR, P. CERONE and A. SOFO, Some remarks on the trapezoid rule in numerical integration, Indian J. Pure Appl. Math., 31(5), 475-494. [5] S.S. DRAGOMIR (Ed), Inequalities for Csisz´ar f−divergence in Information Theory, RGMIA Monographs, Victoria University, 2001 (http://rgmia.vu.edu.au/monographs/index.html). [6] H. SHIOYA and T. DA-TE, A generalisation of Ln divergence and the derivative of a new information divergence, Elec. and Comm. in Japan, 78 (7) (1995), 37-40. E-mail address: [email protected] URL: http://sci.vu.edu.au/staff/neilb.html

School of Communications and Informatics, Victoria University of Technology, PO Box 14428, Melbourne City MC 8001, Victoria, Australia. E-mail address: [email protected] URL: http://rgmia.vu.edu.au/SSDragomirWeb.html