Applied Mathematics and Computation 147 (2004) 137–146 www.elsevier.com/locate/amc
Inequalities for differentiable mappings and applications to special means of real numbers and to midpoint formula U gur S. Kirmaci Department of Mathematics, K.K. Education Faculty, Atat€urk University, 25240, Erzurum, Turkey
Abstract Some inequalities are presented here for differentiable convex mappings, using Hermite–HadamardÕs integral inequality holding for convex functions. Also, some applications to special means of real numbers are given. Finally, some error estimates for the midpoint formula are obtained. Ó 2002 Elsevier Inc. All rights reserved. Keywords: Hermite–Hadamard inequality; Convex functions; Special means; Midpoint formula
1. Introduction Let f : I R ! R be a convex function on the interval I of real numbers and a; b 2 I with a < b. The inequality f
aþb 2
1 6 ba
Z
b
f ðxÞ dx 6 a
f ðaÞ þ f ðbÞ 2
ð1:1Þ
is known as Hermite–HadamardÕs inequality for convex functions [1]. Some results connected with the right part of (1.1) was given in [2]. We shall deduce some inequalities connected with the left part of (1.1).
E-mail address:
[email protected] (U.S. Kirmaci). 0096-3003/$ - see front matter Ó 2002 Elsevier Inc. All rights reserved. doi:10.1016/S0096-3003(02)00657-4
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U.S. Kirmaci / Appl. Math. Comput. 147 (2004) 137–146
For several recent results concerning Hermite–HadamardÕs inequality, we refer the reader to [3–6].
2. The results We first prove the following lemma: Lemma 2.1. Let f : I R ! R be a differentiable mapping on I , a; b 2 I (I is the interior of I) with a < b. If f 0 2 L½a; b, then we have Z b 1 aþb f ðxÞ dx f ¼ ðb aÞ ba a 2 "Z Z 1=2
tf ðta þ ð1 tÞbÞ dt þ
#
1
0
0
0
ðt 1Þf ðta þ ð1 tÞbÞ dt
1=2
Proof. By integration by parts, we deduce Z
1=2
tf 0 ðta þ ð1 tÞbÞ dt þ
0
Z
1
ðt 1Þf 0 ðta þ ð1 tÞbÞ dt 1=2
1=2 Z 1=2 f ðta þ ð1 tÞbÞ f ðta þ ð1 tÞbÞ t dt ab ab 0 0 1 Z 1 f ðta þ ð1 tÞbÞ f ðta þ ð1 tÞbÞ þ ðt 1Þ dt ab ab 1=2 1=2 Z b 1 1 aþb ¼ f f ðxÞ dx 2 ba 2 ðb aÞ a ¼
where, we have used the change of the variable x ¼ ta þ ð1 tÞb, t 2 ½0; 1. Hence, we have the conclusion. Now, we prove three theorems. Theorem 2.2. Let f : I R ! R be a differentiable mapping on I , a; b 2 I with a < b. If jf 0 j is convex on ½a; b, then we have Z 1 ba
b
f ðxÞ dx f a
a þ b b a 0 0 6 8 ðjf ðaÞj þ jf ðbÞjÞ 2
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139
Proof. Using Lemma 2.1 and the convexity of jf 0 j, it follows that Z b 1 a þ b f ðxÞ dx f ba 2 a "Z # Z 1 1=2 0 0 tf ðta þ ð1 tÞbÞ dt þ ðt 1Þf ðta þ ð1 tÞbÞ dt ¼ ðb aÞ 0 1=2 "Z # Z 1=2
1
jtjjf 0 ðta þ ð1 tÞbÞj dt þ
6 ðb aÞ
0
6 ðb aÞ
"Z
jt 1jjf 0 ðta þ ð1 tÞbÞj dt
1=2 1=2
ðt2 jf 0 ðaÞj þ ð1 tÞtjf 0 ðbÞjÞ dt 0
þ
Z
#
1 2
0
0
ðð1 tÞtjf ðaÞj þ ð1 tÞ jf ðbÞjÞ dt 6 1=2
ba 0 ðjf ðaÞj þ jf 0 ðbÞjÞ 8
where we have used the facts that Z
1=2
t2 dt ¼
0
Z
1 ; 24
1
ð1 tÞ2 dt ¼
1=2
Z
1=2
ð1 tÞt dt ¼
0
Z
1
1=2
ð1 tÞt dt ¼
1 12
and
1 24
This concludes the proof.
Theorem 2.3. Let f : I R ! R be a differentiable mapping on I , a; b 2 I p=ðp1Þ with a < b, and let p > 1. If the mapping jf 0 j is convex on ½a; b, then we have Z 1 ba 6
b
f ðxÞ dx f a
a þ b 2
1=p ðp1Þ=p ba 4 p=ðp1Þ p=ðp1Þ jf 0 ðaÞj þ 3jf 0 ðbÞj 16 pþ1 ðp1Þ=p
p=ðp1Þ p=ðp1Þ þ 3jf 0 ðaÞj þ jf 0 ðbÞj
ð2:1Þ
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Proof. Using Lemma 2.1 and H€ olderÕs integral inequality, we deduce Z b 1 a þ b f ðxÞ dx f ba 6 ðb aÞ 2 "Za Z 1=2
tjf 0 ðta þ ð1 tÞbÞj dt þ
0
2
6 ðb aÞ4
þ
jt 1jjf 0 ðta þ ð1 tÞbÞj dt
1=2
Z
1=p Z
1=2 p
1=q
1=2 q
0
jf ðta þ ð1 tÞbÞj dt
t dt 0
Z
#
1
0
!1=p
1
Z
p
jt 1j dt
1=2
1
!1=q 3 5 jf ðta þ ð1 tÞbÞj q
0
1=2
q
where 1=p þ 1=q ¼ 1. Using the convexity of jf 0 j , we obtain Z
1=2 q
0
jf ðta þ ð1 tÞbÞj dt 6
Z
0
1=2 q
q
½tjf 0 ðaÞj þ ð1 tÞjf 0 ðbÞj dt
0 q
q
jf 0 ðaÞj þ 3jf 0 ðbÞj ¼ 8
ð2:2Þ
and Z
1 q
jf 0 ðta þ ð1 tÞbÞj dt 6
Z
1=2
1 q
q
½tjf 0 ðaÞj þ ð1 tÞjf 0 ðbÞj dt 1=2
3jf 0 ðaÞjq þ jf 0 ðbÞjq ¼ 8
ð2:3Þ
Further, we have Z
1=2 p
t dt ¼
0
Z
1 p
jt 1j dt ¼
1=2
Z
1
1=2
p
ð1 tÞ dt ¼
1 ðp þ 1Þ2pþ1
A combination of (2.2)–(2.4) gives the required inequality (2.1).
ð2:4Þ
Theorem 2.4. Let f : I R ! R be a differentiable mapping on I , a; b 2 I p=ðp1Þ with a < b and let p > 1. If the mapping jf 0 j is convex on ½a; b, then we have Z 1 ba
b
f ðxÞ dx f a
1=p a þ b ba 4 ðjf 0 ðaÞj þ jf 0 ðbÞjÞ 6 2 4 pþ1
U.S. Kirmaci / Appl. Math. Comput. 147 (2004) 137–146
141
Proof. We consider the inequality (2.1) i.e., Z 1 ba
1=p a þ b ba 4 p=ðp1Þ jf 0 ðaÞj 6 2 16 pþ1 a ðp1Þ=p ðp1Þ=p
p=ðp1Þ p=ðp1Þ p=ðp1Þ þ 3jf 0 ðaÞj þ 3jf 0 ðbÞj þ jf 0 ðbÞj b
f ðxÞ dx f
Let a1 ¼ jf 0 ðaÞjq , b1 ¼ 3jf 0 ðbÞjq , a2 ¼ 3jf 0 ðaÞjq , b2 ¼ jf 0 ðbÞjq . Here 0 6 ðp 1Þ= p < 1, for p > 1. Using the fact that, n n n X X X s ðak þ bk Þ 6 ask þ bsk k¼1
k¼1
k¼1
for (0 6 s < 1), a1 ; a2 ; . . . ; an P 0, b1 ; b2 ; . . . ; bn P 0, we obtain Z 1 ba
1=p a þ b ba 4 f ðxÞ dx f 4ðjf 0 ðaÞj 6 2 16 pþ1 a 1=p ba 4 þ jf 0 ðbÞjÞ 6 ðjf 0 ðaÞj þ jf 0 ðbÞjÞ 4 pþ1 b
This concludes the proof.
3. Applications to special means As in [2], we shall consider the means for arbitrary real numbers a, b, a 6¼ b. We take 2 ; a; b 2 R n f0g þ b1 a aþb ; a; b 2 R Aða; bÞ ¼ 2 ba ; a; b 2 R n f0g Lða; bÞ ¼ ln jbj ln jaj nþ1
1=n b anþ1 Ln ða; bÞ ¼ ; ðn þ 1Þðb aÞ n 2 N ; n P 1; a; b 2 R; a < b H ða; bÞ ¼ 1
ðharmonic meanÞ ðarithmetic meanÞ ðlogarithmic meanÞ
ðgeneralized log-meanÞ
Now, using the results of Section 2 we give some applications to special means of real numbers.
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Proposition 3.1. Let a; b 2 R, a < b and n 2 N , n P 2. Then, we have jLnn ða; bÞ An ða; bÞj 6
nðb aÞ Aðjajn1 ; jbjn1 Þ 4
ð3:1Þ
Proof. The assertion follows from Theorem 2.2 applied for f ðxÞ ¼ xn , x 2 R. Proposition 3.2. Let a; b 2 R, a < b and n 2 N , n P 2. Then, we have, for all p>1 1=p nðb aÞ 4 jLnn ða; bÞ An ða; bÞj 6 16 pþ1 ðp1Þ=p ððn1ÞpÞ=ðp1Þ ððn1ÞpÞ=ðp1Þ
jaj þ 3jbj ðp1Þ=p þ 3jajððn1ÞpÞ=ðp1Þ þ jbjððn1ÞpÞ=ðp1Þ
ð3:2Þ
Proof. The assertion follows from Theorem 2.3 applied for f ðxÞ ¼ xn , x 2 R. Proposition 3.3. Let a; b 2 R, a < b and n 2 N , n P 2. Then we have, for all p>1 jLnn ða; bÞ An ða; bÞj 6 n
ba 2
4 pþ1
1=p Aðjaj
ðn1Þ
; jbj
ðn1Þ
Þ
ð3:3Þ
Proof. The assertion follows from Theorem 2.4 applied for f ðxÞ ¼ xn , x 2 R. Proposition 3.4. Let a; b 2 R, a < b and 0 62 ½a; b. Then, we have 1
jL ða; bÞ A1 ða; bÞj 6
ba Aðjaj2 ; jbj2 Þ 4
ð3:4Þ
Proof. The assertion follows from Theorem 2.2 applied for f ðxÞ ¼ 1=x, x 2 ½a; b.
U.S. Kirmaci / Appl. Math. Comput. 147 (2004) 137–146
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Proposition 3.5. Let a; b 2 R, a < b and 0 62 ½a; b. Then, we have, for p > 1 1=p ba 4 jL ða; bÞ A ða; bÞj 6 16 pþ1 ðp1Þ=p ð2pÞ=ðp1Þ ð2pÞ=ðp1Þ
jaj þ 3jbj 1
1
ð2pÞ=ðp1Þ
þ 3jaj
þ jbj
ð2pÞ=ðp1Þ
ðp1Þ=p
ð3:5Þ
Proof. The assertion follows from Theorem 2.3 applied for f ðxÞ ¼ 1=x, x 2 ½a; b. Proposition 3.6. Let a; b 2 R, a < b and 0 62 ½a; b. Then we have, for all p > 1 1
jL ða; bÞ A1 ða; bÞj 6
ba 2
4 pþ1
1=p
2
2
Aðjaj ; jbj Þ
ð3:6Þ
Proof. The assertion follows from Theorem 2.4 applied for f ðxÞ ¼ 1=x, x 2 ½a; b. Proposition 3.7. Let a; b 2 R n f0g, ða < bÞ, a1 > b1 . Then, we have, for n 2 N , n P 2 and all p > 1 (i) jLnn ðb1 ; a1 Þ H n ðb; aÞj 6 n
a1 b1 n1 n1 H 1 ðjaj ; jbj Þ 4
(ii) jLnn ðb1 ; a1 Þ H n ðb; aÞj 6
1=p a1 b1 4 pþ1 16 ðp1Þ=p
jbjðð1nÞpÞ=ðp1Þ þ 3jajðð1nÞpÞ=ðp1Þ ðp1Þ=p ðð1nÞpÞ=ðp1Þ ðð1nÞpÞ=ðp1Þ þ 3jbj þ jaj
(iii) 1
jL ðb1 ; a1 Þ H ðb; aÞj 6
a1 b1 2 2 H 1 ðjaj ; jbj Þ 4
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U.S. Kirmaci / Appl. Math. Comput. 147 (2004) 137–146
(iv) 1
jL ðb1 ; a1 Þ H ðb; aÞj 6
1=p a1 b1 4 pþ1 16 ðp1Þ=p
jbjð2pÞ=ðp1Þ þ 3jajð2pÞ=ðp1Þ
ð2pÞ=ðp1Þ
þ 3jbj
þ jaj
ð2pÞ=ðp1Þ
ðp1Þ=p
(v) jLnn ðb1 ; a1 Þ H n ðb; aÞj 6
a1 b1 2
4 pþ1
1=p
H 1 ðjaj
ðn1Þ
; jbj
ðn1Þ
Þ
(vi) 1
1
1
jL ðb ; a Þ H ðb; aÞj 6
a1 b1 2
4 pþ1
1=p
H 1 ðjaj2 ; jbj2 Þ
Proof. Making the substitutions a ! b1 , b ! a1 in the inequalities (3.1)– (3.6), we obtain desired inequalities, where A1 ða1 ; b1 Þ ¼ H ða; bÞ ¼ 2=ð1=a þ 1=bÞ, b1 < a1 (a < b). 4. The midpoint formula As in [6], let d be a division a ¼ x0 < x1 < < xn1 < xn ¼ b of the interval ½a; b and consider the quadrature formula Z b f ðxÞ dx ¼ T ðf ; dÞ þ Eðf ; dÞ ð4:1Þ a
where T ðf ; dÞ ¼
n1 X xi þ xiþ1 f ðxiþ1 xi Þ 2 i¼0
is the midpoint version and Eðf ; dÞ denotes the approximation error. Here, we derive some error estimates for midpoint formula. Proposition 4.1. Let f : I R ! R be a differentiable mapping on I , a; b 2 I with a < b. If jf 0 j is convex on ½a; b, then in (4.1), for every division d of ½a; b, we have jEðf ; dÞj 6
n1 1X 2 ðxiþ1 xi Þ ðjf 0 ðxi Þj þ jf 0 ðxiþ1 ÞjÞ 8 i¼0
U.S. Kirmaci / Appl. Math. Comput. 147 (2004) 137–146
145
Proof. On applying Theorem 2.2 on the subinterval ½xi ; xiþ1 (i ¼ 0; 1; . . . ; n 1) of the division d, we get Z
xiþ1
f ðxÞ dx f
xi
Hence, Z
a
b
x þ x ðxiþ1 xi Þ2 0 i iþ1 ðxiþ1 xi Þ 6 ðjf ðxi Þj þ jf 0 ðxiþ1 ÞjÞ 2 8
n1 Z xiþ1 x þx X i iþ1 f ðxÞ dx T ðf ; dÞ ¼ f ðxÞ dx f ðxiþ1 xi Þ i¼0 2 xi n1 Z xiþ1 x þx X i iþ1 f 6 f ðxÞ dx x Þ ðx iþ1 i 2 xi i¼0 6
n1 1X ðxiþ1 xi Þ2 ðjf 0 ðxi Þj þ jf 0 ðxiþ1 ÞjÞ 8 i¼0
Proposition 4.2. Let f : I R ! R be a differentiable mapping on I , a; b 2 I with a < b and let p > 1. If the mapping jf 0 jp=ðp1Þ is convex on ½a; b, then in (4.1), for every division d of ½a; b, we have jEðf ; dÞj 6
1=p X n1 1 4 2 ðxiþ1 xi Þ 16 p þ 1 i¼0 ðp1Þ=p p=ðp1Þ
jf 0 ðxi Þj þ 3jf 0 ðxiþ1 Þjp=ðp1Þ
0
þ 3jf ðxi Þj
p=ðp1Þ
0
þ jf ðxiþ1 Þj
p=ðp1Þ
ðp1Þ=p
Proof. The proof uses Theorem 2.3 and is similar to that of Proposition 4.1. Proposition 4.3. Let f : I R ! R be a differentiable mapping on I , a; b 2 I p=ðp1Þ with a < b and let p > 1. If the mapping jf 0 j is convex on ½a; b, then in (4.1), for every division d of ½a; b, we have 1 jEðf ; dÞj 6 4
4 pþ1
1=p X n1
ðxiþ1 xi Þ2 ðjf 0 ðxi Þj þ jf 0 ðxiþ1 ÞjÞ
i¼0
Proof. The proof uses Theorem 2.4 and is similar to that of Proposition 4.1.
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