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Hermite-Hadamard type and Fejér type inequalities for general weights (I)

Abstract

In this paper, we establish some weighted versions of the Hermite-Hadamard type and Fejér type inequalities and from which generalize Hermite-Hadamard inequality, Fejér inequality and several results in (Dragomir in J. Math. Anal. Appl. 167:49-56, 1992; Yang and Hong in Tamkang. J. Math. 28(1):33-37, 1997; Yang and Tseng in J. Math. Anal. Appl. 239:180-187, 1999; Yang and Tseng in Taiwan. J. Math. 7(3):433-440, 2003).

MSC:26D15, 26A51.

1 Introduction

Throughout this paper, let a<b in , c<d in , f:[a,b]R be convex, the weight function p:[a,b][0,) be integrable and symmetric about the line s= a + b 2 , the weight function p 1 :[c,d][0,) be integrable and symmetric about the line s= c + d 2 and let the weight function g:[c,d][a,b] be continuous and symmetric about the point ( c + d 2 ,g( c + d 2 )), that is, 1 2 [g(s)+g(c+ds)]=g( c + d 2 ) (s[c,d]). Define the following functions on [0,1]:

H ( t ) = 1 b a a b f ( t s + ( 1 t ) a + b 2 ) d s ; H g ( t ) = 1 d c c d f ( t g ( s ) + ( 1 t ) g ( c + d 2 ) ) d s ; W H ( t ) = a b f ( t s + ( 1 t ) a + b 2 ) p ( s ) d s ; W H g ( t ) = c d f ( t g ( s ) + ( 1 t ) g ( c + d 2 ) ) p 1 ( s ) d s ; F ( t ) = 1 ( b a ) 2 a b a b f ( t s + ( 1 t ) u ) d s d u ; F g ( t ) = 1 ( d c ) 2 c d c d f ( t g ( s ) + ( 1 t ) g ( u ) ) d s d u ; W F ( t ) = a b a b f ( t s + ( 1 t ) u ) p ( s ) p ( u ) d s d u ; W F g ( t ) = c d c d f ( t g ( s ) + ( 1 t ) g ( u ) ) p 1 ( s ) p 1 ( u ) d s d u ; P ( t ) = 1 2 ( b a ) a b [ f ( ( 1 + t 2 ) a + ( 1 t 2 ) s ) P ( t ) = + f ( ( 1 + t 2 ) b + ( 1 t 2 ) s ) ] d s ; P g ( t ) = 1 2 ( d c ) c d [ f ( ( 1 t ) g ( s + c 2 ) + t g ( c ) ) P g ( t ) = + f ( ( 1 t ) g ( s + d 2 ) + t g ( d ) ) ] d s ; W P ( t ) = 1 2 a b [ f ( ( 1 + t 2 ) a + ( 1 t 2 ) s ) p ( s + a 2 ) W P ( t ) = + f ( ( 1 + t 2 ) b + ( 1 t 2 ) s ) p ( s + b 2 ) ] d s

and

W P g ( t ) = 1 2 c d [ f ( ( 1 t ) g ( s + c 2 ) + t g ( c ) ) p 1 ( s + c 2 ) W P g ( t ) = + f ( ( 1 t ) g ( s + d 2 ) + t g ( d ) ) p 1 ( s + d 2 ) ] d s .

Remark 1

  1. (1)

    Let c=a, d=b and the function g(s)=s on [a,b]. Then the functions H g (t)=H(t), F g (t)=F(t) and P g (t)=P(t) on [0,1].

  2. (2)

    Let c=a, d=b and let the functions g(s)=s and p 1 (s)=p(s) on [a,b]. Then the functions W H g (t)=WH(t), W F g (t)=WF(t) and W P g (t)=WP(t) on [0,1].

In 1893, Hadamard [1] established the following inequality.

If the function f is defined as above, then

f ( a + b 2 ) 1 b a a b f(s)ds f ( a ) + f ( b ) 2
(1.1)

is known as Hermite-Hadamard inequality.

See [28] and [916] for some results in which this famous integral inequality (1.1) is generalized, improved and extended.

Dragomir [2] established the following Hermite-Hadamard type inequalities related to the functions H, F, which refine the first inequality of (1.1).

Theorem A Let the functions f, H be defined as in the first page. Then the function H is convex, increasing on [0,1], and for all t[0,1], we have

f ( a + b 2 ) =H(0)H(t)H(1)= 1 b a a b f(s)ds.
(1.2)

Theorem B Let the functions f, F be defined as in the first page. Then:

  1. (1)

    The function F is convex on [0,1], symmetric about 1 2 , F is decreasing on [0, 1 2 ] and increasing on [ 1 2 ,1], and we have

    sup t [ 0 , 1 ] F(t)=F(0)=F(1)= 1 b a a b f(s)ds

and

inf t [ 0 , 1 ] F(t)=F ( 1 2 ) = 1 ( b a ) 2 a b a b f ( s + u 2 ) dsdu.
  1. (2)

    We have

    f ( a + b 2 ) F ( 1 2 ) ;H(t)F(t),t[0,1].
    (1.3)

Yang and Hong [12] established the following Hermite-Hadamard type inequality related to the function P, which refines the second inequality of (1.1).

Theorem C Let the functions f, P be defined as in the first and second pages. Then the function P is convex, increasing on [0,1], and for all t[0,1], we have

1 b a a b f(s)ds=P(0)P(t)P(1)= f ( a ) + f ( b ) 2 .
(1.4)

In 1906, Fejér [8] established the following weighted generalization of Hermite-Hadamard inequality (1.1).

Theorem D Let the functions f, p be defined as in the first page. Then

f ( a + b 2 ) a b p(s)ds a b f(s)p(s)ds f ( a ) + f ( b ) 2 a b p(s)ds
(1.5)

is known as the Fejér inequality.

Yang and Tseng [13, 16] established the following Fejér type inequalities related to the functions WH, WP, WF and which generalize Theorems A-C and refine Fejér inequality (1.5).

Theorem E [13]

Let the functions f, p, WH, WP be defined as in the first and second pages. Then the functions Hg, Pg are convex and increasing on [0,1], and for all t[0,1], we have

f ( a + b 2 ) a b g ( s ) d s = W H ( 0 ) W H ( t ) W H ( 1 ) = a b f ( s ) p ( s ) d s = W P ( 0 ) W P ( t ) W P ( 1 ) = f ( a ) + f ( b ) 2 a b p ( s ) d s .
(1.6)

Theorem F [16]

Let the functions f, p, WH, WF be defined as in the first and second pages. Then we have the following results:

  1. (1)

    The function WF is convex on [0,1] and symmetric about 1 2 .

  2. (2)

    The function WF is decreasing on [0, 1 2 ] and increasing on [ 1 2 ,1],

    sup t [ 0 , 1 ] WF(t)=WF(0)=WF(1)= a b f(s)p(s)ds
    (1.7)

and

inf t [ 0 , 1 ] WF(t)=WF ( 1 2 ) = a b a b f ( s + u 2 ) p(s)p(u)dsdu.
(1.8)
  1. (3)

    We have:

    f ( a + b 2 ) ( a b p ( s ) d s ) 2 WF ( 1 2 )
    (1.9)

and

WH(t) a b p(s)dsWF(t)
(1.10)

for all t[0,1].

In this paper, we establish some weighted versions of the Hermite-Hadamard type and Fejér type inequalities related to the functions H g , F g , P g , W H g , W F g , W P g , which generalize the inequality (1.1) and Theorems A-F.

2 Hermite-Hadamard type inequalities for general weights

In this section, we establish some Hermite-Hadamard type inequalities for general weights, which generalize the Hermite-Hadamard inequality (1.1) and Theorems A-C.

In order to prove the results in this paper, we need the following lemmas.

Lemma 1 (see [9])

Let the function f be defined as in the first page and let aACDBb with A+B=C+D. Then

f(C)+f(D)f(A)+f(B).

The assumptions in Lemma 1 can be weakened as in the following lemma.

Lemma 2 Let the function f be defined as in the first page and let A,B,C,D[a,b] with A+B=C+D and |CD||AB|. Then

f(C)+f(D)f(A)+f(B).

Proof Without loss of generalization, we can assume that aABb and aCDb. For |CD||AB|, we have ABCD and DCBA. Hence, by the above inequalities and A+B=C+D, we get aACDBb. Thus, the proof is completed by Lemma 1. □

Now, we are ready to state and prove our new results.

Theorem 1 Let the functions f, g be defined as in the first page. Then:

  1. (1)

    We have

    f ( g ( c + d 2 ) ) 1 d c c d f ( g ( s ) ) ds.
    (2.1)
  2. (2)

    As the function g is monotonic on [c,d], we obtain

    1 d c c d f ( g ( s ) ) ds f ( g ( c ) ) + f ( g ( d ) ) 2 .
    (2.2)

Proof

  1. (1)

    Using simple techniques of integration, we have the identity

    1 d c c d f ( g ( s ) ) ds= 1 d c c c + d 2 [ f ( g ( s ) ) + g ( c + d s ) ] ds.
    (2.3)

Next, using g(s)+g(c+ds)=2g( c + d 2 ) and

|g ( c + d 2 ) g ( c + d 2 ) ||g(s)g(c+ds)|

in Lemma 2, we obtain

2f ( g ( c + d 2 ) ) f ( g ( s ) ) +f ( g ( c + d s ) ) ,
(2.4)

where s[c,d]. Integrating the above inequality over s on [c, c + d 2 ], dividing both sides by dc and using the above identity, we obtain the inequality (2.1).

  1. (2)

    For the monotonicity of g, we have |g(s)g(c+ds)||g(c)g(d)| for all s[c,d]. Using the above inequality and g(s)+g(c+ds)=g(c)+g(d) in Lemma 2, we obtain

    f ( g ( s ) ) +f ( g ( c + d s ) ) f ( g ( c ) ) +f ( g ( d ) ) ,
    (2.5)

where s[c,d]. Integrating the above inequality over s on [c, c + d 2 ], dividing both sides by dc and using the inequality (2.3), we obtain the inequality (2.2). This completes the proof. □

Remark 2 In Theorem 1, let c=a, d=b and the function g(s)=s on [a,b]. Then Theorem 1 reduces to the Hermite-Hadamard inequality (1.1).

Theorem 2 Let the functions f, g, H g be defined as in the first page. Then:

  1. (1)

    The function H g is convex on [0,1].

  2. (2)

    The function H g is increasing on [0,1] and for all t[0,1], we have

    f ( g ( c + d 2 ) ) = H g (0) H g (t) H g (1)= 1 d c c d f ( g ( s ) ) ds.
    (2.6)

Proof (1) It is easily observed from the convexity of f that the function H g is convex on [0,1].

  1. (2)

    Using simple techniques of integration, we have the following identity:

    H g ( t ) = 1 d c c c + d 2 [ f ( t g ( s ) + ( 1 t ) g ( c + d 2 ) ) + f ( t g ( c + d s ) + ( 1 t ) g ( c + d 2 ) ) ] d s

for all t[0,1]. Let t 1 < t 2 in [0,1]. Since g(s)+g(c+ds)=2g( c + d 2 ) (s[c,d]), we obtain

[ t 1 g ( s ) + ( 1 t 1 ) g ( c + d 2 ) ] + [ t 1 g ( c + d s ) + ( 1 t 1 ) g ( c + d 2 ) ] = [ t 2 g ( s ) + ( 1 t 2 ) g ( c + d 2 ) ] + [ t 2 g ( c + d s ) + ( 1 t 2 ) g ( c + d 2 ) ]

and

| [ t 1 g ( s ) + ( 1 t 1 ) g ( c + d 2 ) ] [ t 1 g ( c + d s ) + ( 1 t 1 ) g ( c + d 2 ) ] | = t 1 | g ( s ) g ( c + d s ) | t 2 | g ( s ) g ( c + d s ) | = | [ t 2 g ( s ) + ( 1 t 2 ) g ( c + d 2 ) ] [ t 2 g ( c + d s ) + ( 1 t 2 ) g ( c + d 2 ) ] |

for all s[c, c + d 2 ]. Therefore, by Lemma 2, the following inequality holds for all s[c, c + d 2 ]:

f ( t 1 g ( s ) + ( 1 t 1 ) g ( c + d 2 ) ) + f ( t 1 g ( c + d s ) + ( 1 t 1 ) g ( c + d 2 ) ) f ( t 2 g ( s ) + ( 1 t 2 ) g ( c + d 2 ) ) + f ( t 2 g ( c + d s ) + ( 1 t 2 ) g ( c + d 2 ) ) ,
(2.7)

where A= t 2 g(s)+(1 t 2 )g( c + d 2 ), B= t 2 g(c+ds)+(1 t 2 )g( c + d 2 ), C= t 1 g(s)+(1 t 1 )g( c + d 2 ) and t 1 g(c+ds)+(1 t 1 )g( c + d 2 ). Integrating the above inequality over s on [c, c + d 2 ], dividing both sides by dc and using the above identity, we have

H g ( t 1 ) H g ( t 2 ).

Thus, the function H g is increasing on [0,1] and from which the inequality (2.6) holds. This completes the proof. □

Remark 3

  1. (1)

    In Theorem 2, the inequality (2.6) refines the inequality (2.1).

  2. (2)

    In Theorem 2, let c=a, d=band the function g(s)=s on [a,b]. Then the functions H g (t)=H(t) (t[0,1]) and Theorem 1 reduces to Theorem A.

Theorem 3 Let the functions f, g, P g be defined as in the first and second pages. Then:

  1. (1)

    The function P g is convex on [0,1].

  2. (2)

    The function P g is increasing on [0,1] and, for all t[0,1], we have

    1 d c c d f ( g ( s ) ) ds= P g (0) P g (t) P g (1)= f ( g ( c ) ) + f ( g ( d ) ) 2
    (2.8)

as the function g is monotonic on [c,d].

Proof (1) It is easily observed from the convexity of f that the function P g is convex on [0,1].

  1. (2)

    Using simple techniques of integration, we have the following identity:

    P g ( t ) = 1 d c c c + d 2 [ f ( t g ( c ) + ( 1 t ) g ( s ) ) + f ( t g ( d ) + ( 1 t ) g ( c + d s ) ) ] d s

for all t[0,1]. Let t 1 < t 2 in [0,1]. Since g(s)+g(c+ds)=2g( c + d 2 ) (s[c,d]) and the monotonicity of g on [c,d], we obtain

| g ( s ) g ( c + d s ) | | g ( c ) g ( d ) | , [ t 1 g ( c ) + ( 1 t 1 ) g ( s ) ] + [ t 1 g ( d ) + ( 1 t 1 ) g ( c + d s ) ] = [ t 2 g ( c ) + ( 1 t 2 ) g ( s ) ] + [ t 2 g ( d ) + ( 1 t 2 ) g ( c + d s ) ]

and

| [ t 1 g ( c ) + ( 1 t 1 ) g ( s ) ] [ t 1 g ( d ) + ( 1 t 1 ) g ( c + d s ) ] | = | t 1 [ g ( c ) g ( d ) ] + ( 1 t 1 ) [ g ( s ) g ( c + d s ) ] | = t 1 | g ( c ) g ( d ) | + ( 1 t 1 ) | g ( s ) g ( c + d s ) | t 1 | g ( c ) g ( d ) | + ( 1 t 1 ) | g ( s ) g ( c + d s ) | = | [ t 2 g ( c ) + ( 1 t 2 ) g ( s ) ] [ t 2 g ( d ) + ( 1 t 2 ) g ( c + d s ) ] |

for all s[c, c + d 2 ]. Therefore, by Lemma 2, the following inequality holds for all s[c, c + d 2 ]:

f ( t 1 g ( c ) + ( 1 t 1 ) g ( s ) ) + f ( t 1 g ( d ) + ( 1 t 1 ) g ( c + d s ) ) f ( t 2 g ( c ) + ( 1 t 2 ) g ( s ) ) + f ( t 2 g ( d ) + ( 1 t 2 ) g ( c + d s ) )
(2.9)

where A= t 2 g(c)+(1 t 2 )g(s), B= t 2 g(d)+(1 t 2 )g(c+ds), C= t 1 g(c)+(1 t 1 )g(s) and t 1 g(d)+(1 t 1 )g(c+ds). Integrating the above inequality over s on [c, c + d 2 ], dividing both sides by dc and using the above identity, we have

P g ( t 1 ) P g ( t 2 ).

Thus, the function P g is increasing on [0,1] and from which the inequality (2.8) holds. This completes the proof. □

Remark 4

  1. (1)

    In Theorem 3, the inequality (2.8) refines the inequality (2.2).

  2. (2)

    In Theorem 3, let c=a, d=band the function g(s)=s on [a,b]. Then the functions P g (t)=P(t) (t[0,1]) and Theorem 3 reduces to Theorem C.

Theorem 4 Let the functions f, g, H g , F g be defined as in the first page. Then we have the following results:

  1. (1)

    The function F g is convex on [0,1] and symmetric about 1 2 .

  2. (2)

    The function F g is decreasing on [0, 1 2 ] and increasing on [ 1 2 ,1],

    sup t [ 0 , 1 ] F g (t)= F g (0)= F g (1)= 1 d c c d f ( g ( s ) ) ds
    (2.10)

and

inf t [ 0 , 1 ] F g ( t ) = F g ( 1 2 ) = 1 ( d c ) 2 c d c d f ( g ( s ) + g ( u ) 2 ) d s d u .
(2.11)
  1. (3)

    We have:

    H g (t) F g (t) ( t [ 0 , 1 ] )
    (2.12)

and

f ( g ( c + d 2 ) ) F g ( 1 2 ) .
(2.13)

Proof (1) It is easily observed from the convexity of f that the function F g is convex on [0,1].

By changing variables, we have

F g (t)= F g (1t),t[0,1]

from which we get that the function F g is symmetric about 1 2 .

  1. (2)

    Let t 1 < t 2 in [0, 1 2 ]. Then t 2 +(1 t 2 )= t 1 +(1 t 1 ), | t 2 (1 t 2 )|| t 1 (1 t 1 )| and by Lemma 2, we obtain

    1 2 [ F g ( t 2 ) + F g ( 1 t 2 ) ] 1 2 [ F g ( t 1 ) + F g ( 1 t 1 ) ] .
    (2.14)

Using the symmetry of F g , we have

F g ( t 1 )= 1 2 [ F g ( t 1 ) + F g ( 1 t 1 ) ] ,
(2.15)
F g ( t 2 )= 1 2 [ F g ( t 2 ) + F g ( 1 t 2 ) ]
(2.16)

From (2.14)-(2.16), we obtain that the function F g is decreasing on [0, 1 2 ]. Since the function F g is symmetric about 1 2 and the function F g is decreasing on [0, 1 2 ], we obtain that the function F g is increasing on [ 1 2 ,1]. Using the symmetry and monotonicity of F g , we derive the inequalities (2.10) and (2.11).

  1. (3)

    Using the substitution rules for integration, we have the identity

    F g ( t ) = 1 ( d c ) 2 c d c c + d 2 [ f ( t g ( s ) + ( 1 t ) g ( u ) ) + f ( t g ( s ) + ( 1 t ) g ( c + d u ) ) ] d u d s

for all t[0,1]. Let t[0,1]. Since g(u)+g(c+du)=2g( c + d 2 ) (u[c,d]), we obtain

2 [ t g ( s ) + ( 1 t ) g ( c + d 2 ) ] = [ t g ( s ) + ( 1 t ) g ( u ) ] + [ t g ( s ) + ( 1 t ) g ( c + d u ) ]

and

| [ t g ( s ) + ( 1 t ) g ( c + d 2 ) ] [ t g ( s ) + ( 1 t ) g ( c + d 2 ) ] | | [ t g ( s ) + ( 1 t ) g ( u ) ] [ t g ( s ) + ( 1 t ) g ( c + d u ) ] |

for all s[c,d] and u[c, c + d 2 ]. Therefore, by Lemma 2, the following inequality holds for all s[c,d] and u[c, c + d 2 ]:

2 f ( t g ( s ) + ( 1 t ) g ( c + d 2 ) ) f ( t g ( s ) + ( 1 t ) g ( u ) ) + f ( t g ( s ) + ( 1 t ) g ( c + d u ) ) ,
(2.17)

where A=tg(s)+(1t)g(u), B=tg(s)+(1t)g(c+du) and C=D=tg(s)+(1t)g( c + d 2 ). Dividing the above inequality by ( d c ) 2 , integrating it over s on [c,d], over u on [c, c + d 2 ] and using the above identity, we derive the inequality (2.12).

From the inequalities (2.6), (2.12) and the monotonicity of H g , we derive the inequality (2.13).

This completes the proof. □

Remark 5 In Theorem 4, let c=a, d=b and the function g(s)=s on [a,b]. Then the functions F g (t)=F(t) (t[0,1]) and Theorem 4 reduces to Theorem B.

3 Fejér type inequalities for general weights

In this section, we establish some Fejér type inequalities for general weights which generalize Theorems D-F.

Theorem 5 Let the functions f, g, p 1 be defined as in the first page. Then:

  1. (1)

    We have

    f ( g ( c + d 2 ) ) c d p 1 (s)ds c d f ( g ( s ) ) p 1 (s)ds.
    (3.1)
  2. (2)

    As the function g is monotonic on [c,d], we obtain

    c d f ( g ( s ) ) p 1 (s)ds f ( g ( c ) ) + f ( g ( d ) ) 2 c d p 1 (s)ds.
    (3.2)

Proof (1) Using simple techniques of integration and the hypothesis of p 1 , we have the identities

c d f ( g ( s ) ) p 1 (s)ds= c c + d 2 [ f ( g ( s ) ) + f ( g ( c + d s ) ) ] p 1 (s)ds
(3.3)

and

c c + d 2 p 1 (s)ds= 1 2 c d p 1 (s)ds.
(3.4)

Proceeding as in the proof of Theorem 1, we also obtain the inequality (2.4). Multiplying the inequality (2.4) by p 1 (s), integrating it over s on [c, c + d 2 ] and using the above identities, we obtain the inequality (3.1).

  1. (2)

    Proceeding as in the proof of Theorem 1, we also obtain the inequality (2.5). Multiplying the inequality (2.5) by p 1 (s), integrating it over s on [c, c + d 2 ] and using the above identities, we obtain the inequality (3.2). This completes the proof. □

Remark 6

  1. (1)

    Let c=a, d=b and let the functions g(s)=s and p 1 (s)=p(s) on [a,b]. Then Theorem 5 reduces to Fejér inequality (1.5).

  2. (2)

    Let the function p 1 (s) 1 d c on [c,d]. Then Theorem 5 reduces to Theorem 1.

Theorem 6 Let the functions f, g, p 1 , W H g be defined as in the first page. Then:

  1. (1)

    The function W H g is convex on [0,1].

  2. (2)

    The function W H g is increasing on [0,1] and, for all t[0,1], we have

    f ( g ( c + d 2 ) ) c d p 1 ( s ) d s = W H g ( 0 ) W H g ( t ) W H g ( 1 ) = c d f ( g ( s ) ) p 1 ( s ) d s .
    (3.5)

Proof (1) It is easily observed from the convexity of f and the hypothesis of p 1 that the function W H g is convex on [0,1].

  1. (2)

    Using simple techniques of integration and the hypothesis of p 1 , we have the following identity:

    W H g ( t ) = c c + d 2 [ f ( t g ( s ) + ( 1 t ) g ( c + d 2 ) ) + f ( t g ( c + d s ) + ( 1 t ) g ( c + d 2 ) ) ] p 1 ( s ) d s

for all t[0,1].

Let t 1 < t 2 in [0,1]. Proceeding as in the proof of Theorem 2, we also obtain the inequality (2.7). Multiplying the inequality (2.7) by p 1 (s), integrating it over s on [c, c + d 2 ] and using the above identity, we obtain

W H g ( t 1 )W H g ( t 2 ).

Thus, the function W H g is increasing on [0,1] and from which the inequality (3.5) holds. This completes the proof. □

Remark 7

  1. (1)

    In Theorem 6, the inequality (3.5) refines the inequality (3.1).

  2. (2)

    Let the function p 1 (s) 1 d c on [c,d]. Then Theorem 6 reduces to Theorem 2.

Theorem 7 Let the functions f, g, p 1 , W P g be defined as in the first and second pages. Then:

  1. (1)

    The function W P g is convex on [0,1].

  2. (2)

    The function W P g is increasing on [0,1] and, for all t[0,1], we have

    c d f ( g ( s ) ) p 1 ( s ) d s = W P g ( 0 ) W P g ( t ) W P g ( 1 ) = f ( g ( c ) ) + f ( g ( d ) ) 2 c d p 1 ( s ) d s
    (3.6)

as the function g is monotonic on [c,d].

Proof (1) It is easily observed from the convexity of f and the hypothesis of p 1 that the function W P g is convex on [0,1].

  1. (2)

    Using simple techniques of integration and the hypothesis of p 1 , we have the following identity:

    W P g ( t ) = c c + d 2 [ f ( t g ( c ) + ( 1 t ) g ( s ) ) + f ( t g ( d ) + ( 1 t ) g ( c + d s ) ) ] p 1 ( s ) d s

for all t[0,1].

Let t 1 < t 2 in [0,1]. Proceeding as in the proof of Theorem 3, we also obtain the inequality (2.9). Multiplying the inequality (2.9) by p 1 (s), integrating it over s on [c, c + d 2 ] and using the above identity, we obtain

W P g ( t 1 )W P g ( t 2 ).

Thus, the function W P g is increasing on [0,1] and from which the inequality (3.6) holds. This completes the proof. □

Remark 8

  1. (1)

    In Theorem 7, the inequality (3.6) refines the inequality (3.2).

  2. (2)

    Let the function p 1 (s) 1 d c on [c,d]. Then Theorem 7 reduces to Theorem 3.

Remark 9 Let c=a, d=b and let the functions g(s)=s and p 1 (s)=p(s) on [a,b]. Then Theorems 6 and 7 reduce to Theorem E.

Theorem 8 Let the functions f, g, p 1 , W H g , W F g be defined as in the first page. Then we have the following results:

  1. (1)

    The function W F g is convex on [0,1] and symmetric about 1 2 .

  2. (2)

    The function W F g is decreasing on [0, 1 2 ] and increasing on [ 1 2 ,1],

    sup t [ 0 , 1 ] W F g (t)=W F g (0)=W F g (1)= c d f ( g ( s ) ) p 1 (s)ds

and

inf t [ 0 , 1 ] W F g (t)=W F g ( 1 2 ) = c d c d f ( g ( s ) + g ( u ) 2 ) p 1 (s) p 1 (u)dsdu.
  1. (3)

    We have

    W H g (t) c d p 1 (s)dsW F g (t) ( t [ 0 , 1 ] )
    (3.7)

and

f ( g ( c + d 2 ) ) ( c d p 1 ( s ) d s ) 2 W F g ( 1 2 ) .
(3.8)

Proof (1)-(2) Proceeding as in the proof of Theorem 4, the parts (1) and (2) hold.

  1. (3)

    Using the substitution rules for integration and the hypothesis of p 1 , we have the identity

    W F g ( t ) = c d c c + d 2 [ f ( t g ( s ) + ( 1 t ) g ( u ) ) + f ( t g ( s ) + ( 1 t ) ( c + d u ) ) ] p 1 ( u ) p 1 ( s ) d u d s
    (3.9)

for all t[0,1]. Proceeding as in the proof of Theorem 4, we also obtain the inequality (2.17). Multiplying the inequality (2.17) by p 1 (u) p 1 (s), integrating it over s on [c,d], over u on [c, c + d 2 ] and using the identities (3.4) and (3.9), we obtain the inequality (3.7).

From the inequalities (3.5), (3.7) and the monotonicity of W H g , we derive the inequality (3.8).

This completes the proof. □

Remark 10

  1. (1)

    Theorem 8 refines the inequality (3.1).

  2. (2)

    Let the function p 1 (s) 1 d c on [c,d]. Then Theorem 8 reduces to Theorem 2.

  3. (3)

    Let c=a, d=b and the functions g(s)=s and p 1 (s)=p(s) on [a,b]. Then Theorem 8 reduces to Theorem F.

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Acknowledgements

Dedicated to Professor Hari M Srivastava.

This research was partially supported by Grant NSC 101-2115-M-156-002.

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Hwang, SR., Tseng, KL. & Hsu, KC. Hermite-Hadamard type and Fejér type inequalities for general weights (I). J Inequal Appl 2013, 170 (2013). https://doi.org/10.1186/1029-242X-2013-170

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Keywords

  • Hermite-Hadamard inequality
  • Fejér inequality
  • convex function