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 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, $\frac{1}{2}[g(s)+g(c+d-s)]=g(\frac{c+d}{2})$ ($s\in [c,d]$). Define the following functions on $[0,1]$:

$$\begin{array}{c}H(t)=\frac{1}{b-a}{\int }_a^{b}f(ts+(1-t)\frac{a+b}{2})ds;\hfill \\ H_g(t)=\frac{1}{d-c}{\int }_c^{d}f(tg(s)+(1-t)g\left(\frac{c+d}{2}\right))ds;\hfill \\ WH(t)={\int }_a^{b}f(ts+(1-t)\frac{a+b}{2})p(s)ds;\hfill \\ W{H}_g(t)={\int }_c^{d}f(tg(s)+(1-t)g\left(\frac{c+d}{2}\right))p_1(s)ds;\hfill \\ F(t)=\frac{1}{{(b-a)}^2}{\int }_a^b{\int }_a^{b}f(ts+(1-t)u)dsdu;\hfill \\ F_g(t)=\frac{1}{{(d-c)}^2}{\int }_c^d{\int }_c^{d}f(tg(s)+(1-t)g(u))dsdu;\hfill \\ WF(t)={\int }_a^b{\int }_a^{b}f(ts+(1-t)u)p(s)p(u)dsdu;\hfill \\ W{F}_g(t)={\int }_c^d{\int }_c^{d}f(tg(s)+(1-t)g(u))p_1(s)p_1(u)dsdu;\hfill \\ P(t)=\frac{1}{2(b-a)}{\int }_a^b[f(\left(\frac{1+t}{2}\right)a+\left(\frac{1-t}{2}\right)s)\hfill \\ \phantom{P(t)=}+f(\left(\frac{1+t}{2}\right)b+\left(\frac{1-t}{2}\right)s)]ds;\hfill \\ P_g(t)=\frac{1}{2(d-c)}{\int }_c^d[f((1-t)g\left(\frac{s+c}{2}\right)+tg(c))\hfill \\ \phantom{P_g(t)=}+f((1-t)g\left(\frac{s+d}{2}\right)+tg(d))]ds;\hfill \\ WP(t)=\frac{1}{2}{\int }_a^b[f(\left(\frac{1+t}{2}\right)a+\left(\frac{1-t}{2}\right)s)p\left(\frac{s+a}{2}\right)\hfill \\ \phantom{WP(t)=}+f(\left(\frac{1+t}{2}\right)b+\left(\frac{1-t}{2}\right)s)p\left(\frac{s+b}{2}\right)]ds\hfill \end{array}$$

and

$$\begin{array}{c}W{P}_g(t)=\frac{1}{2}{\int }_c^d[f((1-t)g\left(\frac{s+c}{2}\right)+tg(c))p_1\left(\frac{s+c}{2}\right)\hfill \\ \phantom{W{P}_g(t)=}+f((1-t)g\left(\frac{s+d}{2}\right)+tg(d))p_1\left(\frac{s+d}{2}\right)]ds.\hfill \end{array}$$

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

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f(s)ds\le \frac{f(a)+f(b)}{2}$$

(1.1)

is known as Hermite-Hadamard 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 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 $t\in [0,1]$, we have

$$f\left(\frac{a+b}{2}\right)=H(0)\le H(t)\le H(1)=\frac{1}{b-a}{\int }_a^{b}f(s)ds.$$

(1.2)

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

  $$\underset{t\in [0,1]}{sup}F(t)=F(0)=F(1)=\frac{1}{b-a}{\int }_a^{b}f(s)ds$$

and

$$\underset{t\in [0,1]}{inf}F(t)=F\left(\frac{1}{2}\right)=\frac{1}{{(b-a)}^2}{\int }_a^b{\int }_a^{b}f\left(\frac{s+u}{2}\right)dsdu.$$

• (2) We have

  $$f\left(\frac{a+b}{2}\right)\le F\left(\frac{1}{2}\right);H(t)\le F(t),t\in [0,1].$$

  (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 $t\in [0,1]$, we have

$$\frac{1}{b-a}{\int }_a^{b}f(s)ds=P(0)\le P(t)\le P(1)=\frac{f(a)+f(b)}{2}.$$

(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

$$f\left(\frac{a+b}{2}\right){\int }_a^{b}p(s)ds\le {\int }_a^{b}f(s)p(s)ds\le \frac{f(a)+f(b)}{2}{\int }_a^{b}p(s)ds$$

(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 $t\in [0,1]$, we have

$$\begin{array}{rl}f\left(\frac{a+b}{2}\right){\int }_a^{b}g(s)ds& =WH(0)\le WH(t)\le WH(1)\\ ={\int }_a^{b}f(s)p(s)ds\\ =WP(0)\le WP(t)\le WP(1)\\ =\frac{f(a)+f(b)}{2}{\int }_a^{b}p(s)ds.\end{array}$$

(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) 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 $[\frac{1}{2},1]$,

  $$\underset{t\in [0,1]}{sup}WF(t)=WF(0)=WF(1)={\int }_a^{b}f(s)p(s)ds$$

  (1.7)

and

$$\underset{t\in [0,1]}{inf}WF(t)=WF\left(\frac{1}{2}\right)={\int }_a^b{\int }_a^{b}f\left(\frac{s+u}{2}\right)p(s)p(u)dsdu.$$

(1.8)

• (3) We have:

  $$f\left(\frac{a+b}{2}\right){({\int }_a^{b}p(s)ds)}^2\le WF\left(\frac{1}{2}\right)$$

  (1.9)

and

$$WH(t){\int }_a^{b}p(s)ds\le WF(t)$$

(1.10)

for all $t\in [0,1]$.

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 $a\le A\le C\le D\le B\le b$ with $A+B=C+D$. Then

$$f(C)+f(D)\le f(A)+f(B).$$

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 $A,B,C,D\in [a,b]$ with $A+B=C+D$ and $|C-D|\le |A-B|$. Then

$$f(C)+f(D)\le f(A)+f(B).$$

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 $|C-D|\le |A-B|$, we have $A-B\le C-D$ and $D-C\le B-A$. 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

  $$f\left(g\left(\frac{c+d}{2}\right)\right)\le \frac{1}{d-c}{\int }_c^{d}f(g(s))ds.$$

  (2.1)

• (2) As the function g is monotonic on $[c,d]$, we obtain

  $$\frac{1}{d-c}{\int }_c^{d}f(g(s))ds\le \frac{f(g(c))+f(g(d))}{2}.$$

  (2.2)

Proof

• (1) Using simple techniques of integration, we have the identity

  $$\frac{1}{d-c}{\int }_c^{d}f(g(s))ds=\frac{1}{d-c}{\int }_c^{\frac{c+d}{2}}[f(g(s))+g(c+d-s)]ds.$$

  (2.3)

Next, using $g(s)+g(c+d-s)=2g(\frac{c+d}{2})$ and

$$|g\left(\frac{c+d}{2}\right)-g\left(\frac{c+d}{2}\right)|\le |g(s)-g(c+d-s)|$$

in Lemma 2, we obtain

$$2f\left(g\left(\frac{c+d}{2}\right)\right)\le f(g(s))+f(g(c+d-s)),$$

(2.4)

where $s\in [c,d]$. Integrating the above inequality over s on $[c,\frac{c+d}{2}]$, dividing both sides by $d-c$ and using the above identity, we obtain the inequality (2.1).

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

  $$f(g(s))+f(g(c+d-s))\le f(g(c))+f(g(d)),$$

  (2.5)

where $s\in [c,d]$. Integrating the above inequality over s on $[c,\frac{c+d}{2}]$, dividing both sides by $d-c$ 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) 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

  $$f\left(g\left(\frac{c+d}{2}\right)\right)=H_g(0)\le H_g(t)\le H_g(1)=\frac{1}{d-c}{\int }_c^{d}f(g(s))ds.$$

  (2.6)

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

• (2) Using simple techniques of integration, we have the following identity:

  $$\begin{array}{rl}{H}_g(t)=& \frac{1}{d-c}{\int }_c^{\frac{c+d}{2}}[f(tg(s)+(1-t)g\left(\frac{c+d}{2}\right))\\ +f(tg(c+d-s)+(1-t)g\left(\frac{c+d}{2}\right))]ds\end{array}$$

for all $t\in [0,1]$. Let $t_1<t_2$ in $[0,1]$. Since $g(s)+g(c+d-s)=2g(\frac{c+d}{2})$ ($s\in [c,d]$), we obtain

$$\begin{array}{r}[t_{1}g(s)+(1-t_1)g\left(\frac{c+d}{2}\right)]+[t_{1}g(c+d-s)+(1-t_1)g\left(\frac{c+d}{2}\right)]\\ =[t_{2}g(s)+(1-t_2)g\left(\frac{c+d}{2}\right)]+[t_{2}g(c+d-s)+(1-t_2)g\left(\frac{c+d}{2}\right)]\end{array}$$

and

$$\begin{array}{r}|[t_{1}g(s)+(1-t_1)g\left(\frac{c+d}{2}\right)]-[t_{1}g(c+d-s)+(1-t_1)g\left(\frac{c+d}{2}\right)]|\\ =t_1|g(s)-g(c+d-s)|\\ \le t_2|g(s)-g(c+d-s)|\\ =|[t_{2}g(s)+(1-t_2)g\left(\frac{c+d}{2}\right)]-[t_{2}g(c+d-s)+(1-t_2)g\left(\frac{c+d}{2}\right)]|\end{array}$$

for all $s\in [c,\frac{c+d}{2}]$. Therefore, by Lemma 2, the following inequality holds for all $s\in [c,\frac{c+d}{2}]$:

$$\begin{array}{r}f(t_{1}g(s)+(1-t_1)g\left(\frac{c+d}{2}\right))+f(t_{1}g(c+d-s)+(1-t_1)g\left(\frac{c+d}{2}\right))\\ \le f(t_{2}g(s)+(1-t_2)g\left(\frac{c+d}{2}\right))+f(t_{2}g(c+d-s)+(1-t_2)g\left(\frac{c+d}{2}\right)),\end{array}$$

(2.7)

where $A=t_{2}g(s)+(1-t_2)g(\frac{c+d}{2})$, $B=t_{2}g(c+d-s)+(1-t_2)g(\frac{c+d}{2})$, $C=t_{1}g(s)+(1-t_1)g(\frac{c+d}{2})$ and $t_{1}g(c+d-s)+(1-t_1)g(\frac{c+d}{2})$. Integrating the above inequality over s on $[c,\frac{c+d}{2}]$, dividing both sides by $d-c$ and using the above identity, we have

$$H_g(t_1)\le H_g(t_2).$$

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

• (2) In Theorem 2, let $c=a$, $d=b$and 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

  $$\frac{1}{d-c}{\int }_c^{d}f(g(s))ds=P_g(0)\le P_g(t)\le P_g(1)=\frac{f(g(c))+f(g(d))}{2}$$

  (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:

  $$\begin{array}{rl}{P}_g(t)=& \frac{1}{d-c}{\int }_c^{\frac{c+d}{2}}[f(tg(c)+(1-t)g(s))\\ +f(tg(d)+(1-t)g(c+d-s))]ds\end{array}$$

for all $t\in [0,1]$. Let $t_1<t_2$ in $[0,1]$. Since $g(s)+g(c+d-s)=2g(\frac{c+d}{2})$ ($s\in [c,d]$) and the monotonicity of g on $[c,d]$, we obtain

$$\begin{array}{c}|g(s)-g(c+d-s)|\le |g(c)-g(d)|,\hfill \\ [t_{1}g(c)+(1-t_1)g(s)]+[t_{1}g(d)+(1-t_1)g(c+d-s)]\hfill \\ =[t_{2}g(c)+(1-t_2)g(s)]+[t_{2}g(d)+(1-t_2)g(c+d-s)]\hfill \end{array}$$

and

$$\begin{array}{r}|[t_{1}g(c)+(1-t_1)g(s)]-[t_{1}g(d)+(1-t_1)g(c+d-s)]|\\ =|t_1[g(c)-g(d)]+(1-t_1)[g(s)-g(c+d-s)]|\\ =t_1|g(c)-g(d)|+(1-t_1)|g(s)-g(c+d-s)|\\ \le t_1|g(c)-g(d)|+(1-t_1)|g(s)-g(c+d-s)|\\ =|[t_{2}g(c)+(1-t_2)g(s)]-[t_{2}g(d)+(1-t_2)g(c+d-s)]|\end{array}$$

for all $s\in [c,\frac{c+d}{2}]$. Therefore, by Lemma 2, the following inequality holds for all $s\in [c,\frac{c+d}{2}]$:

$$\begin{array}{r}f(t_{1}g(c)+(1-t_1)g(s))+f(t_{1}g(d)+(1-t_1)g(c+d-s))\\ \le f(t_{2}g(c)+(1-t_2)g(s))+f(t_{2}g(d)+(1-t_2)g(c+d-s))\end{array}$$

(2.9)

where $A=t_{2}g(c)+(1-t_2)g(s)$, $B=t_{2}g(d)+(1-t_2)g(c+d-s)$, $C=t_{1}g(c)+(1-t_1)g(s)$ and $t_{1}g(d)+(1-t_1)g(c+d-s)$. Integrating the above inequality over s on $[c,\frac{c+d}{2}]$, dividing both sides by $d-c$ and using the above identity, we have

$$P_g(t_1)\le P_g(t_2).$$

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

• (2) In Theorem 3, let $c=a$, $d=b$and 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 $[\frac{1}{2},1]$,

  $$\underset{t\in [0,1]}{sup}{F}_g(t)=F_g(0)=F_g(1)=\frac{1}{d-c}{\int }_c^{d}f(g(s))ds$$

  (2.10)

and

$$\begin{array}{rl}\underset{t\in [0,1]}{inf}{F}_g(t)& =F_g\left(\frac{1}{2}\right)\\ =\frac{1}{{(d-c)}^2}{\int }_c^d{\int }_c^{d}f\left(\frac{g(s)+g(u)}{2}\right)dsdu.\end{array}$$

(2.11)

• (3) We have:

  $$H_g(t)\le F_g(t)(t\in [0,1])$$

  (2.12)

and

$$f\left(g\left(\frac{c+d}{2}\right)\right)\le F_g\left(\frac{1}{2}\right).$$

(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

$$F_g(t)=F_g(1-t),t\in [0,1]$$

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

$$F_g(t_1)=\frac{1}{2}[F_g(t_1)+F_g(1-t_1)],$$

(2.15)

$$F_g(t_2)=\frac{1}{2}[F_g(t_2)+F_g(1-t_2)]$$

(2.16)

From (2.14)-(2.16), we obtain that the function $F_g$ is decreasing on $[0,\frac{1}{2}]$. 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

  $$\begin{array}{rl}{F}_g(t)=& \frac{1}{{(d-c)}^2}{\int }_c^d{\int }_c^{\frac{c+d}{2}}[f(tg(s)+(1-t)g(u))\\ +f(tg(s)+(1-t)g(c+d-u))]duds\end{array}$$

for all $t\in [0,1]$. Let $t\in [0,1]$. Since $g(u)+g(c+d-u)=2g(\frac{c+d}{2})$ ($u\in [c,d]$), we obtain

$$\begin{array}{r}2[tg(s)+(1-t)g\left(\frac{c+d}{2}\right)]\\ =[tg(s)+(1-t)g(u)]+[tg(s)+(1-t)g(c+d-u)]\end{array}$$

and

$$\begin{array}{r}|[tg(s)+(1-t)g\left(\frac{c+d}{2}\right)]-[tg(s)+(1-t)g\left(\frac{c+d}{2}\right)]|\\ \le |[tg(s)+(1-t)g(u)]-[tg(s)+(1-t)g(c+d-u)]|\end{array}$$

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 $u\in [c,\frac{c+d}{2}]$:

$$\begin{array}{r}2f(tg(s)+(1-t)g\left(\frac{c+d}{2}\right))\\ \le f(tg(s)+(1-t)g(u))+f(tg(s)+(1-t)g(c+d-u)),\end{array}$$

(2.17)

where $A=tg(s)+(1-t)g(u)$, $B=tg(s)+(1-t)g(c+d-u)$ and $C=D=tg(s)+(1-t)g(\frac{c+d}{2})$. Dividing the above inequality by ${(d-c)}^2$, 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 D-F.

Theorem 5 Let the functions f, g, $p_1$ be defined as in the first page. Then:

• (1) We have

  $$f\left(g\left(\frac{c+d}{2}\right)\right){\int }_c^d{p}_1(s)ds\le {\int }_c^{d}f(g(s))p_1(s)ds.$$

  (3.1)

• (2) As the function g is monotonic on $[c,d]$, we obtain

  $${\int }_c^{d}f(g(s))p_1(s)ds\le \frac{f(g(c))+f(g(d))}{2}{\int }_c^d{p}_1(s)ds.$$

  (3.2)

Proof (1) Using simple techniques of integration and the hypothesis of $p_1$, we have the identities

$${\int }_c^{d}f(g(s))p_1(s)ds={\int }_c^{\frac{c+d}{2}}[f(g(s))+f(g(c+d-s))]p_1(s)ds$$

(3.3)

and

$${\int }_c^{\frac{c+d}{2}}{p}_1(s)ds=\frac{1}{2}{\int }_c^d{p}_1(s)ds.$$

(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 $[c,\frac{c+d}{2}]$ 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 $p_1(s)\equiv \frac{1}{d-c}$ 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) 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

  $$\begin{array}{rcl}f\left(g\left(\frac{c+d}{2}\right)\right){\int }_c^d{p}_1(s)ds& =& W{H}_g(0)\\ \le & W{H}_g(t)\\ \le & W{H}_g(1)={\int }_c^{d}f(g(s))p_1(s)ds.\end{array}$$

  (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:

  $$\begin{array}{rl}W{H}_g(t)=& {\int }_c^{\frac{c+d}{2}}[f(tg(s)+(1-t)g\left(\frac{c+d}{2}\right))\\ +f(tg(c+d-s)+(1-t)g\left(\frac{c+d}{2}\right))]p_1(s)ds\end{array}$$

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

$$W{H}_g(t_1)\le W{H}_g(t_2).$$

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

• (2) Let the function $p_1(s)\equiv \frac{1}{d-c}$ 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) 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

  $$\begin{array}{rcl}{\int }_c^{d}f(g(s))p_1(s)ds& =& W{P}_g(0)\\ \le & W{P}_g(t)\\ \le & W{P}_g(1)=\frac{f(g(c))+f(g(d))}{2}{\int }_c^d{p}_1(s)ds\end{array}$$

  (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:

  $$\begin{array}{rl}W{P}_g(t)=& {\int }_c^{\frac{c+d}{2}}[f(tg(c)+(1-t)g(s))\\ +f(tg(d)+(1-t)g(c+d-s))]p_1(s)ds\end{array}$$

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

$$W{P}_g(t_1)\le W{P}_g(t_2).$$

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

• (2) Let the function $p_1(s)\equiv \frac{1}{d-c}$ 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) 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 $[\frac{1}{2},1]$,

  $$\underset{t\in [0,1]}{sup}W{F}_g(t)=W{F}_g(0)=W{F}_g(1)={\int }_c^{d}f(g(s))p_1(s)ds$$

and

$$\underset{t\in [0,1]}{inf}W{F}_g(t)=W{F}_g\left(\frac{1}{2}\right)={\int }_c^d{\int }_c^{d}f\left(\frac{g(s)+g(u)}{2}\right)p_1(s)p_1(u)dsdu.$$

• (3) We have

  $$W{H}_g(t){\int }_c^d{p}_1(s)ds\le W{F}_g(t)(t\in [0,1])$$

  (3.7)

and

$$f\left(g\left(\frac{c+d}{2}\right)\right){({\int }_c^d{p}_1(s)ds)}^2\le W{F}_g\left(\frac{1}{2}\right).$$

(3.8)

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

  $$\begin{array}{rl}W{F}_g(t)=& {\int }_c^d{\int }_c^{\frac{c+d}{2}}[f(tg(s)+(1-t)g(u))\\ +f(tg(s)+(1-t)(c+d-u))]p_1(u)p_1(s)duds\end{array}$$

  (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 $p_1(s)\equiv \frac{1}{d-c}$ 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.