 Research
 Open Access
 Published:
HermiteHadamard type and Fejér type inequalities for general weights (I)
Journal of Inequalities and Applications volume 2013, Article number: 170 (2013)
Abstract
In this paper, we establish some weighted versions of the HermiteHadamard type and Fejér type inequalities and from which generalize HermiteHadamard inequality, Fejér inequality and several results in (Dragomir in J. Math. Anal. Appl. 167:4956, 1992; Yang and Hong in Tamkang. J. Math. 28(1):3337, 1997; Yang and Tseng in J. Math. Anal. Appl. 239:180187, 1999; Yang and Tseng in Taiwan. J. Math. 7(3):433440, 2003).
MSC:26D15, 26A51.
1 Introduction
Throughout this paper, let a<b in ℝ, c<d in ℝ, f:[a,b]\to \mathbb{R} be convex, the weight function p:[a,b]\to [0,\mathrm{\infty}) be integrable and symmetric about the line s=\frac{a+b}{2}, the weight function {p}_{1}:[c,d]\to [0,\mathrm{\infty}) be integrable and symmetric about the line s=\frac{c+d}{2} and let the weight function g:[c,d]\to [a,b] be continuous and symmetric about the point (\frac{c+d}{2},g(\frac{c+d}{2})), that is, \frac{1}{2}[g(s)+g(c+ds)]=g(\frac{c+d}{2}) (s\in [c,d]). Define the following functions on [0,1]:
and
Remark 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)
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
is known as HermiteHadamard inequality.
See [2–8] and [9–16] for some results in which this famous integral inequality (1.1) is generalized, improved and extended.
Dragomir [2] established the following HermiteHadamard 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\in [0,1], we have
Theorem B Let the functions f, F be defined as in the first page. Then:

(1)
The function F is convex on [0,1], symmetric about \frac{1}{2}, F is decreasing on [0,\frac{1}{2}] and increasing on [\frac{1}{2},1], and we have
\underset{t\in [0,1]}{sup}F(t)=F(0)=F(1)=\frac{1}{ba}{\int}_{a}^{b}f(s)\phantom{\rule{0.2em}{0ex}}ds
and

(2)
We have
f\left(\frac{a+b}{2}\right)\le F\left(\frac{1}{2}\right);\phantom{\rule{2em}{0ex}}H(t)\le F(t),\phantom{\rule{1em}{0ex}}t\in [0,1].(1.3)
Yang and Hong [12] established the following HermiteHadamard 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\in [0,1], we have
In 1906, Fejér [8] established the following weighted generalization of HermiteHadamard inequality (1.1).
Theorem D Let the functions f, p be defined as in the first page. Then
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 AC 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\in [0,1], we have
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)
The function WF is convex on [0,1] and symmetric about \frac{1}{2}.

(2)
The function WF is decreasing on [0,\frac{1}{2}] and increasing on [\frac{1}{2},1],
\underset{t\in [0,1]}{sup}WF(t)=WF(0)=WF(1)={\int}_{a}^{b}f(s)p(s)\phantom{\rule{0.2em}{0ex}}ds(1.7)
and

(3)
We have:
f\left(\frac{a+b}{2}\right){({\int}_{a}^{b}p(s)\phantom{\rule{0.2em}{0ex}}ds)}^{2}\le WF\left(\frac{1}{2}\right)(1.9)
and
for all t\in [0,1].
In this paper, we establish some weighted versions of the HermiteHadamard 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 AF.
2 HermiteHadamard type inequalities for general weights
In this section, we establish some HermiteHadamard type inequalities for general weights, which generalize the HermiteHadamard inequality (1.1) and Theorems AC.
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 a\le A\le C\le D\le B\le b with A+B=C+D. Then
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\in [a,b] with A+B=C+D and CD\le AB. Then
Proof Without loss of generalization, we can assume that a\le A\le B\le b and a\le C\le D\le b. For CD\le AB, we have AB\le CD and DC\le BA. Hence, by the above inequalities and A+B=C+D, we get a\le A\le C\le D\le B\le b. 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)
We have
f\left(g\left(\frac{c+d}{2}\right)\right)\le \frac{1}{dc}{\int}_{c}^{d}f(g(s))\phantom{\rule{0.2em}{0ex}}ds.(2.1) 
(2)
As the function g is monotonic on [c,d], we obtain
\frac{1}{dc}{\int}_{c}^{d}f(g(s))\phantom{\rule{0.2em}{0ex}}ds\le \frac{f(g(c))+f(g(d))}{2}.(2.2)
Proof

(1)
Using simple techniques of integration, we have the identity
\frac{1}{dc}{\int}_{c}^{d}f(g(s))\phantom{\rule{0.2em}{0ex}}ds=\frac{1}{dc}{\int}_{c}^{\frac{c+d}{2}}[f(g(s))+g(c+ds)]\phantom{\rule{0.2em}{0ex}}ds.(2.3)
Next, using g(s)+g(c+ds)=2g(\frac{c+d}{2}) and
in Lemma 2, we obtain
where s\in [c,d]. Integrating the above inequality over s on [c,\frac{c+d}{2}], dividing both sides by dc and using the above identity, we obtain the inequality (2.1).

(2)
For the monotonicity of g, we have g(s)g(c+ds)\le g(c)g(d) for all s\in [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+ds))\le f(g(c))+f(g(d)),(2.5)
where s\in [c,d]. Integrating the above inequality over s on [c,\frac{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 HermiteHadamard inequality (1.1).
Theorem 2 Let the functions f, g, {H}_{g} be defined as in the first page. Then:

(1)
The function {H}_{g} is convex on [0,1].

(2)
The function {H}_{g} is increasing on [0,1] and for all t\in [0,1], we have
f\left(g\left(\frac{c+d}{2}\right)\right)={H}_{g}(0)\le {H}_{g}(t)\le {H}_{g}(1)=\frac{1}{dc}{\int}_{c}^{d}f(g(s))\phantom{\rule{0.2em}{0ex}}ds.(2.6)
Proof (1) It is easily observed from the convexity of f that the function {H}_{g} is convex on [0,1].

(2)
Using simple techniques of integration, we have the following identity:
\begin{array}{rl}{H}_{g}(t)=& \frac{1}{dc}{\int}_{c}^{\frac{c+d}{2}}[f(tg(s)+(1t)g\left(\frac{c+d}{2}\right))\\ +f(tg(c+ds)+(1t)g\left(\frac{c+d}{2}\right))]\phantom{\rule{0.2em}{0ex}}ds\end{array}
for all t\in [0,1]. Let {t}_{1}<{t}_{2} in [0,1]. Since g(s)+g(c+ds)=2g(\frac{c+d}{2}) (s\in [c,d]), we obtain
and
for all s\in [c,\frac{c+d}{2}]. Therefore, by Lemma 2, the following inequality holds for all s\in [c,\frac{c+d}{2}]:
where A={t}_{2}g(s)+(1{t}_{2})g(\frac{c+d}{2}), B={t}_{2}g(c+ds)+(1{t}_{2})g(\frac{c+d}{2}), C={t}_{1}g(s)+(1{t}_{1})g(\frac{c+d}{2}) and {t}_{1}g(c+ds)+(1{t}_{1})g(\frac{c+d}{2}). Integrating the above inequality over s on [c,\frac{c+d}{2}], dividing both sides by dc and using the above identity, we have
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)
In Theorem 2, the inequality (2.6) refines the inequality (2.1).

(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\in [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)
The function {P}_{g} is convex on [0,1].

(2)
The function {P}_{g} is increasing on [0,1] and, for all t\in [0,1], we have
\frac{1}{dc}{\int}_{c}^{d}f(g(s))\phantom{\rule{0.2em}{0ex}}ds={P}_{g}(0)\le {P}_{g}(t)\le {P}_{g}(1)=\frac{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].

(2)
Using simple techniques of integration, we have the following identity:
\begin{array}{rl}{P}_{g}(t)=& \frac{1}{dc}{\int}_{c}^{\frac{c+d}{2}}[f(tg(c)+(1t)g(s))\\ +f(tg(d)+(1t)g(c+ds))]\phantom{\rule{0.2em}{0ex}}ds\end{array}
for all t\in [0,1]. Let {t}_{1}<{t}_{2} in [0,1]. Since g(s)+g(c+ds)=2g(\frac{c+d}{2}) (s\in [c,d]) and the monotonicity of g on [c,d], we obtain
and
for all s\in [c,\frac{c+d}{2}]. Therefore, by Lemma 2, the following inequality holds for all s\in [c,\frac{c+d}{2}]:
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,\frac{c+d}{2}], dividing both sides by dc and using the above identity, we have
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)
In Theorem 3, the inequality (2.8) refines the inequality (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\in [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)
The function {F}_{g} is convex on [0,1] and symmetric about \frac{1}{2}.

(2)
The function {F}_{g} is decreasing on [0,\frac{1}{2}] and increasing on [\frac{1}{2},1],
\underset{t\in [0,1]}{sup}{F}_{g}(t)={F}_{g}(0)={F}_{g}(1)=\frac{1}{dc}{\int}_{c}^{d}f(g(s))\phantom{\rule{0.2em}{0ex}}ds(2.10)
and

(3)
We have:
{H}_{g}(t)\le {F}_{g}(t)\phantom{\rule{1em}{0ex}}(t\in [0,1])(2.12)
and
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
from which we get that the function {F}_{g} is symmetric about \frac{1}{2}.

(2)
Let {t}_{1}<{t}_{2} in [0,\frac{1}{2}]. Then {t}_{2}+(1{t}_{2})={t}_{1}+(1{t}_{1}), {t}_{2}(1{t}_{2})\le {t}_{1}(1{t}_{1}) and by Lemma 2, we obtain
\frac{1}{2}[{F}_{g}({t}_{2})+{F}_{g}(1{t}_{2})]\le \frac{1}{2}[{F}_{g}({t}_{1})+{F}_{g}(1{t}_{1})].(2.14)
Using the symmetry of {F}_{g}, we have
From (2.14)(2.16), we obtain that the function {F}_{g} is decreasing on [0,\frac{1}{2}]. Since the function {F}_{g} is symmetric about \frac{1}{2} and the function {F}_{g} is decreasing on [0,\frac{1}{2}], we obtain that the function {F}_{g} is increasing on [\frac{1}{2},1]. Using the symmetry and monotonicity of {F}_{g}, we derive the inequalities (2.10) and (2.11).

(3)
Using the substitution rules for integration, we have the identity
\begin{array}{rl}{F}_{g}(t)=& \frac{1}{{(dc)}^{2}}{\int}_{c}^{d}{\int}_{c}^{\frac{c+d}{2}}[f(tg(s)+(1t)g(u))\\ +f(tg(s)+(1t)g(c+du))]\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\end{array}
for all t\in [0,1]. Let t\in [0,1]. Since g(u)+g(c+du)=2g(\frac{c+d}{2}) (u\in [c,d]), we obtain
and
for all s\in [c,d] and u\in [c,\frac{c+d}{2}]. Therefore, by Lemma 2, the following inequality holds for all s\in [c,d] and u\in [c,\frac{c+d}{2}]:
where A=tg(s)+(1t)g(u), B=tg(s)+(1t)g(c+du) and C=D=tg(s)+(1t)g(\frac{c+d}{2}). Dividing the above inequality by {(dc)}^{2}, integrating it over s on [c,d], over u on [c,\frac{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\in [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 DF.
Theorem 5 Let the functions f, g, {p}_{1} be defined as in the first page. Then:

(1)
We have
f\left(g\left(\frac{c+d}{2}\right)\right){\int}_{c}^{d}{p}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds\le {\int}_{c}^{d}f(g(s)){p}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds.(3.1) 
(2)
As the function g is monotonic on [c,d], we obtain
{\int}_{c}^{d}f(g(s)){p}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds\le \frac{f(g(c))+f(g(d))}{2}{\int}_{c}^{d}{p}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds.(3.2)
Proof (1) Using simple techniques of integration and the hypothesis of {p}_{1}, we have the identities
and
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,\frac{c+d}{2}] and using the above identities, we obtain the inequality (3.1).

(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,\frac{c+d}{2}] and using the above identities, we obtain the inequality (3.2). This completes the proof. □
Remark 6

(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)
Let the function {p}_{1}(s)\equiv \frac{1}{dc} 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)
The function W{H}_{g} is convex on [0,1].

(2)
The function W{H}_{g} is increasing on [0,1] and, for all t\in [0,1], we have
\begin{array}{rcl}f\left(g\left(\frac{c+d}{2}\right)\right){\int}_{c}^{d}{p}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds& =& W{H}_{g}(0)\\ \le & W{H}_{g}(t)\\ \le & W{H}_{g}(1)={\int}_{c}^{d}f(g(s)){p}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds.\end{array}(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].

(2)
Using simple techniques of integration and the hypothesis of {p}_{1}, we have the following identity:
\begin{array}{rl}W{H}_{g}(t)=& {\int}_{c}^{\frac{c+d}{2}}[f(tg(s)+(1t)g\left(\frac{c+d}{2}\right))\\ +f(tg(c+ds)+(1t)g\left(\frac{c+d}{2}\right))]{p}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds\end{array}
for all t\in [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,\frac{c+d}{2}] and using the above identity, we obtain
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)
In Theorem 6, the inequality (3.5) refines the inequality (3.1).

(2)
Let the function {p}_{1}(s)\equiv \frac{1}{dc} 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)
The function W{P}_{g} is convex on [0,1].

(2)
The function W{P}_{g} is increasing on [0,1] and, for all t\in [0,1], we have
\begin{array}{rcl}{\int}_{c}^{d}f(g(s)){p}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds& =& W{P}_{g}(0)\\ \le & W{P}_{g}(t)\\ \le & W{P}_{g}(1)=\frac{f(g(c))+f(g(d))}{2}{\int}_{c}^{d}{p}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds\end{array}(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].

(2)
Using simple techniques of integration and the hypothesis of {p}_{1}, we have the following identity:
\begin{array}{rl}W{P}_{g}(t)=& {\int}_{c}^{\frac{c+d}{2}}[f(tg(c)+(1t)g(s))\\ +f(tg(d)+(1t)g(c+ds))]{p}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds\end{array}
for all t\in [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,\frac{c+d}{2}] and using the above identity, we obtain
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)
In Theorem 7, the inequality (3.6) refines the inequality (3.2).

(2)
Let the function {p}_{1}(s)\equiv \frac{1}{dc} 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)
The function W{F}_{g} is convex on [0,1] and symmetric about \frac{1}{2}.

(2)
The function W{F}_{g} is decreasing on [0,\frac{1}{2}] and increasing on [\frac{1}{2},1],
\underset{t\in [0,1]}{sup}W{F}_{g}(t)=W{F}_{g}(0)=W{F}_{g}(1)={\int}_{c}^{d}f(g(s)){p}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds
and

(3)
We have
W{H}_{g}(t){\int}_{c}^{d}{p}_{1}(s)\phantom{\rule{0.2em}{0ex}}ds\le W{F}_{g}(t)\phantom{\rule{1em}{0ex}}(t\in [0,1])(3.7)
and
Proof (1)(2) Proceeding as in the proof of Theorem 4, the parts (1) and (2) hold.

(3)
Using the substitution rules for integration and the hypothesis of {p}_{1}, we have the identity
\begin{array}{rl}W{F}_{g}(t)=& {\int}_{c}^{d}{\int}_{c}^{\frac{c+d}{2}}[f(tg(s)+(1t)g(u))\\ +f(tg(s)+(1t)(c+du))]{p}_{1}(u){p}_{1}(s)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}ds\end{array}(3.9)
for all t\in [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,\frac{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)
Theorem 8 refines the inequality (3.1).

(2)
Let the function {p}_{1}(s)\equiv \frac{1}{dc} on [c,d]. Then Theorem 8 reduces to Theorem 2.

(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.
References
Hadamard J: Étude sur les propriétés des fonctions entières en particulier d’une fonction considérée par Riemann. J. Math. Pures Appl. 1893, 58: 171–215.
Dragomir SS: Two mappings in connection to Hadamard’s inequalities. J. Math. Anal. Appl. 1992, 167: 49–56. 10.1016/0022247X(92)902334
Dragomir SS: A refinement of Hadamard’s inequality for isotonic linear functionals. Tamkang. J. Math. 1993, 24: 101–106.
Dragomir SS: On the Hadamard’s inequality for convex on the coordinates in a rectangle from the plane. Taiwan. J. Math. 2001, 5(4):775–788.
Dragomir SS: Further properties of some mapping associated with HermiteHadamard inequalities. Tamkang. J. Math. 2003, 34(1):45–57.
Dragomir SS, Cho YJ, Kim SS: Inequalities of Hadamard’s type for Lipschitzian mappings and their applications. J. Math. Anal. Appl. 2000, 245: 489–501. 10.1006/jmaa.2000.6769
Dragomir SS, Milošević DS, Sándor J: On some refinements of Hadamard’s inequalities and applications. Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. 1993, 4: 3–10.
Fejér L: Über die Fourierreihen, II. Math. Naturwiss Anz Ungar. Akad. Wiss. 1906, 24: 369–390. (In Hungarian)
Hwang DY, Tseng KL, Yang GS: Some Hadamard’s inequalities for coordinated convex functions in a rectangle from the plane. Taiwan. J. Math. 2007, 11(1):63–73.
Tseng KL, Hwang SR, Dragomir SS: On some new inequalities of HermiteHadamardFejér type involving convex functions. Demonstr. Math. 2007, XL(1):51–64.
Tseng KL, Yang GS, Hsu KC: On some inequalities of Hadamard’s type and applications. Taiwan. J. Math. 2009, 13(6B):1929–1948.
Yang GS, Hong MC: A note on Hadamard’s inequality. Tamkang. J. Math. 1997, 28(1):33–37.
Yang GS, Tseng KL: On certain integral inequalities related to HermiteHadamard inequalities. J. Math. Anal. Appl. 1999, 239: 180–187. 10.1006/jmaa.1999.6506
Yang GS, Tseng KL: Inequalities of Hadamard’s type for Lipschitzian mappings. J. Math. Anal. Appl. 2001, 260: 230–238. 10.1006/jmaa.2000.7460
Yang GS, Tseng KL: On certain multiple integral inequalities related to HermiteHadamard inequalities. Util. Math. 2002, 62: 131–142.
Yang GS, Tseng KL: Inequalities of HermiteHadamardFejér type for convex functions and Lipschitzian functions. Taiwan. J. Math. 2003, 7(3):433–440.
Acknowledgements
Dedicated to Professor Hari M Srivastava.
This research was partially supported by Grant NSC 1012115M156002.
Author information
Authors and Affiliations
Corresponding author
Additional information
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
The authors did not provide this information.
Rights and permissions
Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License ( https://creativecommons.org/licenses/by/2.0 ), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
About this article
Cite this article
Hwang, SR., Tseng, KL. & Hsu, KC. HermiteHadamard type and Fejér type inequalities for general weights (I). J Inequal Appl 2013, 170 (2013). https://doi.org/10.1186/1029242X2013170
Received:
Accepted:
Published:
DOI: https://doi.org/10.1186/1029242X2013170
Keywords
 HermiteHadamard inequality
 Fejér inequality
 convex function