## 1. Introduction

A function f : I → ℝ is said to be convex function on I if the inequality

$f\left(\alpha x+\left(1-\alpha \right)y\right)\le \alpha f\left(x\right)+\left(1-\alpha \right)f\left(y\right),$

holds for all x, yI and α ∈ [0,1].

One of the most famous inequality for convex functions is so called Hermite-Hadamard's inequality as follows: Let f : I ⊆ ℝ → ℝ be a convex function defined on the interval I of real numbers and a, bI, with a < b. Then:

$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)dx\le \frac{f\left(a\right)+f\left(b\right)}{2}.$
(1.1)

In [1], the following theorem which was obtained by Dragomir and Agarwal contains the Hermite-Hadamard type integral inequality.

Theorem 1. Let f : I° ⊆ ℝ → ℝ be a differentiable mapping on I°, a, bI° with a < b. If |f'| is convex on [a, b], then the following inequality holds:

$\left|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du\right|\le \frac{\left(b-a\right)\left(|{f}^{\prime }\left(a\right)|+|{f}^{\prime }\left(b\right)|\right)}{8}.$
(1.2)

In [2] Kirmaci, Bakula, Özdemir and Pečarić proved the following theorem.

Theorem 2. Let f : I → ℝ, I ⊂ ℝ be a differentiable function on I° such that f'L [a, b], where a, bI, a < b. If |f'|qis concave on [a, b] for some q > 1, then:

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du\right|\\ \le \left(\frac{b-a}{4}\right){\left[\frac{q-1}{2q-1}\right]}^{\frac{q-1}{q}}\left(\left|{f}^{\prime }\left(\frac{a+3b}{4}\right)\right|+\left|{f}^{\prime }\left(\frac{3a+b}{4}\right)\right|\right).\end{array}$
(1.3)

In [3], Kirmaci obtained the following theorem and corollary related to this theorem.

Theorem 3. Let f : I° ⊂ ℝ → ℝ be a differentiable mapping on I°, a, bI° with a < b and let p > 1. If the mapping |f'|pis concave on [a, b], then we have

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left|f\left(ca+\left(1-c\right)b\right)\left(B-A\right)+f\left(a\right)\left(1-B\right)+f\left(b\right)A-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)dx\right|\\ \le \left(b-a\right)\left[K\left|{f}^{\prime }\left(\frac{aT+b\left(K-T\right)}{K}\right)\right|+M\left|{f}^{\prime }\left(\frac{aN+b\left(M-N\right)}{M}\right)\right|\right]\end{array}$

where

$\begin{array}{c}K=\frac{{A}^{2}+{\left(c-A\right)}^{2}}{2},T=\frac{{A}^{3}+{c}^{3}}{3}-\frac{A{c}^{2}}{2},M=\frac{{\left(B-c\right)}^{2}+{\left(1-B\right)}^{2}}{2},\\ N=\frac{{B}^{3}+{c}^{3}+1}{3}-\left(1+{c}^{2}\right)\frac{B}{2}.\end{array}$

Corollary 1. Under the assumptions of Theorem 3 with$A=B=c=\frac{1}{2}$, we have

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)dx\right|\\ \le \frac{\left(b-a\right)}{8}\left[\left|{f}^{\prime }\left(\frac{5a+b}{6}\right)\right|+\left|{f}^{\prime }\left(\frac{a+5b}{6}\right)\right|\right].\end{array}$
(1.4)

For recent results and generalizations concerning Hermite-Hadamard's inequality see [1]-[5] and the references given therein.

## 2. The New Hermite-Hadamard Type Inequalities

In order to prove our main theorems, we first prove the following lemma:

Lemma 1. Let f : I ⊆ ℝ → ℝ be a differentiable mapping on I°, where a, bI with a < b. If f'L [a, b], then the following equality holds:

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du\\ =\frac{{\left(x-a\right)}^{2}}{b-a}{\int }_{0}^{1}\left(t-1\right){f}^{\prime }\left(tx+\left(1-t\right)a\right)dt+\frac{{\left(b-x\right)}^{2}}{b-a}{\int }_{0}^{1}\left(1-t\right){f}^{\prime }\left(tx+\left(1-t\right)b\right)dt.\end{array}$

Proof. We note that

$\begin{array}{ll}\hfill J& =\frac{{\left(x-a\right)}^{2}}{b-a}{\int }_{0}^{1}\left(t-1\right){f}^{\prime }\left(tx+\left(1-t\right)a\right)dt\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\frac{{\left(b-x\right)}^{2}}{b-a}{\int }_{0}^{1}\left(1-t\right){f}^{\prime }\left(tx+\left(1-t\right)b\right)dt.\phantom{\rule{2em}{0ex}}\end{array}$

Integrating by parts, we get

$\begin{array}{ll}\hfill J& =\frac{{\left(x-a\right)}^{2}}{b-a}\left[{\left(t-1\right)\frac{f\left(tx+\left(1-t\right)a\right)}{x-a}|}_{0}^{1}-{\int }_{0}^{1}\frac{f\left(tx+\left(1-t\right)a\right)}{x-a}dt\right]\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\frac{{\left(b-x\right)}^{2}}{b-a}\left[{\left(1-t\right)\frac{f\left(tx+\left(1-t\right)b\right)}{x-b}|}_{0}^{1}+{\int }_{0}^{1}\frac{f\left(tx+\left(1-t\right)b\right)}{x-b}dt\right]\phantom{\rule{2em}{0ex}}\\ =\frac{{\left(x-a\right)}^{2}}{b-a}\left[\frac{f\left(a\right)}{x-a}-\frac{1}{{\left(x-a\right)}^{2}}{\int }_{a}^{x}f\left(u\right)du\right]\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\frac{{\left(b-x\right)}^{2}}{b-a}\left[-\frac{f\left(b\right)}{x-b}+\frac{1}{{\left(x-b\right)}^{2}}{\int }_{b}^{x}f\left(u\right)du\right]\phantom{\rule{2em}{0ex}}\\ =\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du.\phantom{\rule{2em}{0ex}}\end{array}$

Using the Lemma 1 the following results can be obtained.

Theorem 4. Let f : I ⊆ ℝ → ℝ be a differentiable mapping on I° such that f'L [a, b], where a, bI with a < b. If |f'| is convex on [a, b], then the following inequality holds:

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left|\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du\right|\\ \le \frac{{\left(x-a\right)}^{2}}{b-a}\left[\frac{|{f}^{\prime }\left(x\right)|+2|{f}^{\prime }\left(a\right)|}{6}\right]+\frac{{\left(b-x\right)}^{2}}{b-a}\left[\frac{|{f}^{\prime }\left(x\right)|+2|{f}^{\prime }\left(b\right)|}{6}\right]\end{array}$

for each x ∈ [a, b].

Proof. Using Lemma 1 and taking the modulus, we have

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left|\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du\right|\\ \le \frac{{\left(x-a\right)}^{2}}{b-a}{\int }_{0}^{1}\left(1-t\right)|{f}^{\prime }\left(tx+\left(1-t\right)a\right)|dt\\ \phantom{\rule{1em}{0ex}}+\frac{{\left(b-x\right)}^{2}}{b-a}{\int }_{0}^{1}\left(1-t\right)|{f}^{\prime }\left(tx+\left(1-t\right)b\right)|dt.\end{array}$

Since |f'| is convex, then we get

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left|\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du\right|\\ \le \frac{{\left(x-a\right)}^{2}}{b-a}{\int }_{0}^{1}\left(1-t\right)\left[t|{f}^{\prime }\left(x\right)|+\left(1-t\right)|{f}^{\prime }\left(a\right)|\right]dt\\ \phantom{\rule{1em}{0ex}}+\frac{{\left(b-x\right)}^{2}}{b-a}{\int }_{0}^{1}\left(1-t\right)\left[t|{f}^{\prime }\left(x\right)|+\left(1-t\right)|{f}^{\prime }\left(b\right)|\right]dt\\ =\frac{{\left(x-a\right)}^{2}}{b-a}\left[\frac{|{f}^{\prime }\left(x\right)|+2|{f}^{\prime }\left(a\right)|}{6}\right]+\frac{{\left(b-x\right)}^{2}}{b-a}\left[\frac{|{f}^{\prime }\left(x\right)|+2|{f}^{\prime }\left(b\right)|}{6}\right]\end{array}$

which completes the proof. □

Corollary 2. In Theorem 4, if we choose$x=\frac{a+b}{2}$we obtain

$\left|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du\right|\le \frac{b-a}{12}\left(|{f}^{\prime }\left(a\right)|+\left|{f}^{\prime }\left(\frac{a+b}{2}\right)\right|+|{f}^{\prime }\left(b\right)|\right).$

Remark 1. In Corollary 2, using the convexity of |f'| we have

$\left|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du\right|\le \frac{b-a}{8}\left(|{f}^{\prime }\left(a\right)|+|{f}^{\prime }\left(b\right)|\right)$

which is the inequality in (1.2).

Theorem 5. Let f : I ⊆ ℝ → ℝ be a differentiable mapping on I° such that f'L [a, b], where a, bI with a < b. If$|{f}^{\prime }{}^{{|}^{\frac{p}{p-1}}}$is convex on [a, b] and for some fixed q > 1, then the following inequality holds:

$\begin{array}{l}\phantom{\rule{0.5em}{0ex}}|\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du|\\ \le {\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{2}\right)}^{\frac{1}{q}}\\ \phantom{\rule{0.5em}{0ex}}×\left[\frac{{\left(x-a\right)}^{2}{\left[\mid {f}^{\prime }\left(a{\right)\mid }^{q}+\mid {f}^{\prime }\left(x{\right)\mid }^{q}\right]}^{\frac{1}{q}}+{\left(b-x\right)}^{2}{\left[\mid {f}^{\prime }\left(x{\right)\mid }^{q}+\mid {f}^{\prime }\left(b{\right)\mid }^{q}\right]}^{\frac{1}{q}}}{b-a}\right]\end{array}$

for each x ∈ [a, b] and$q=\frac{p}{p-1}$.

Proof. From Lemma 1 and using the well-known Hölder integral inequality, we have

$\begin{array}{l}\phantom{\rule{0.5em}{0ex}}|\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du|\\ \le \frac{{\left(x-a\right)}^{2}}{b-a}{\int }_{0}^{1}\left(1-t\right)\mid {f}^{\prime }\left(tx+\left(1-t\right)a\right)\mid dt\\ \phantom{\rule{0.5em}{0ex}}+\frac{{\left(b-x\right)}^{2}}{b-a}{\int }_{0}^{1}\left(1-t\right)\mid {f}^{\prime }\left(tx+\left(1-t\right)b\right)\mid dt\\ \le \frac{{\left(x-a\right)}^{2}}{b-a}{\left({\int }_{0}^{1}{\left(1-t\right)}^{p}dt\right)}^{\frac{1}{p}}{\left({\int }_{0}^{1}\mid {f}^{\prime }\left(tx+\left(1-t\right)a{\right)\mid }^{q}dt\right)}^{\frac{1}{q}}\\ \phantom{\rule{0.5em}{0ex}}+\frac{{\left(b-x\right)}^{2}}{b-a}{\left({\int }_{0}^{1}{\left(1-t\right)}^{p}dt\right)}^{\frac{1}{p}}{\left({\int }_{0}^{1}\mid {f}^{\prime }\left(tx+\left(1-t\right)b{\right)\mid }^{q}dt\right)}^{\frac{1}{q}}.\end{array}$

Since $|{f}^{\prime }{}^{{|}^{\frac{p}{p-1}}}$ is convex, by the Hermite-Hadamard's inequality, we have

${\int }_{0}^{1}\mid {f}^{\prime }\left(tx+\left(1-t\right)a{\right)\mid }^{q}dt\le \frac{\mid {f}^{\prime }\left(a{\right)\mid }^{q}+\mid {f}^{\prime }\left(x{\right)\mid }^{q}}{2}$

and

${\int }_{0}^{1}\mid {f}^{\prime }\left(tx+\left(1-t\right)b{\right)\mid }^{q}dt\le \frac{\mid {f}^{\prime }\left(b{\right)\mid }^{q}+\mid {f}^{\prime }\left(x{\right)\mid }^{q}}{2},$

so

$\begin{array}{l}\phantom{\rule{0.5em}{0ex}}|\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du|\\ \le {\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{2}\right)}^{\frac{1}{q}}\\ \phantom{\rule{0.5em}{0ex}}×\left[\frac{{\left(x-a\right)}^{2}{\left[\mid {f}^{\prime }\left(a{\right)\mid }^{q}+\mid {f}^{\prime }\left(x{\right)\mid }^{q}\right]}^{\frac{1}{q}}+{\left(b-x\right)}^{2}{\left[\mid {f}^{\prime }\left(x{\right)\mid }^{q}+\mid {f}^{\prime }\left(b{\right)\mid }^{q}\right]}^{\frac{1}{q}}}{b-a}\right]\end{array}$

which completes the proof. □

Corollary 3. In Theorem 5, if we choose$x=\frac{a+b}{2}$we obtain

$\begin{array}{l}\phantom{\rule{0.5em}{0ex}}|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du|\\ \le \frac{b-a}{4}{\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{2}\right)}^{\frac{1}{q}}\\ \phantom{\rule{0.5em}{0ex}}×\left[{\left(\mid {f}^{\prime }\left(a{\right)\mid }^{q}+{|{f}^{\prime }\left(\frac{a+b}{2}\right)|}^{q}\right)}^{\frac{1}{q}}+{\left(\mid {f}^{\prime }\left(b{\right)\mid }^{q}+{|{f}^{\prime }\left(\frac{a+b}{2}\right)|}^{q}\right)}^{\frac{1}{q}}\right]\\ \le \frac{b-a}{2}{\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{2}\right)}^{\frac{1}{q}}\left(\mid {f}^{\prime }\left(a\right)\mid +\mid {f}^{\prime }\left(b\right)\mid \right).\end{array}$

The second inequality is obtained using the following fact:${\sum }_{k=1}^{n}{\left({a}_{k}+{b}_{k}\right)}^{s}\le {\sum }_{k=1}^{n}{\left({a}_{k}\right)}^{s}+{\sum }_{k=1}^{n}{\left({b}_{k}\right)}^{s}$for (0 ≤ s < 1), a1, a2, a3,⋯, a n ≥ 0; b1, b2, b3,⋯, b n ≥ 0 with$0\le \frac{p-1}{p}<1$, for p > 1.

Theorem 6. Let f : I ⊆ ℝ → ℝ be a differentiable mapping on I° such that f'L [a, b], where a, bI with a < b. If |f'| q is concave on [a, b], for some fixed q > 1, then the following inequality holds:

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left|\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du\right|\\ \le {\left[\frac{q-1}{2q-1}\right]}^{\frac{q-1}{q}}\left[\frac{{\left(x-a\right)}^{2}\left|{f}^{\prime }\left(\frac{a+x}{2}\right)|+{\left(b-x\right)}^{2}|{f}^{\prime }\left(\frac{b+x}{2}\right)\right|}{b-a}\right]\end{array}$

for each x ∈ [a, b].

Proof. As in Theorem 5, using Lemma 1 and the well-known Hölder integral inequality for q > 1 and $p=\frac{q}{q-1}$, we have

$\begin{array}{l}\phantom{\rule{0.5em}{0ex}}|\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du|\\ \le \frac{{\left(x-a\right)}^{2}}{b-a}{\int }_{0}^{1}\left(1-t\right)\mid {f}^{\prime }\left(tx+\left(1-t\right)a\right)\mid dt\\ \phantom{\rule{0.5em}{0ex}}+\frac{{\left(b-x\right)}^{2}}{b-a}{\int }_{0}^{1}\left(1-t\right)\mid {f}^{\prime }\left(tx+\left(1-t\right)b\right)\mid dt\\ \le \frac{{\left(x-a\right)}^{2}}{b-a}{\left({\int }_{0}^{1}{\left(1-t\right)}^{\frac{q}{q-1}}dt\right)}^{\frac{q-1}{q}}{\left({\int }_{0}^{1}\mid {f}^{\prime }\left(tx+\left(1-t\right)a{\right)\mid }^{q}dt\right)}^{\frac{1}{q}}\\ \phantom{\rule{0.5em}{0ex}}+\frac{{\left(b-x\right)}^{2}}{b-a}{\left({\int }_{0}^{1}{\left(1-t\right)}^{\frac{q}{q-1}}dt\right)}^{\frac{q-1}{q}}{\left({\int }_{0}^{1}\mid {f}^{\prime }\left(tx+\left(1-t\right)b{\right)\mid }^{q}dt\right)}^{\frac{1}{q}}.\end{array}$

Since |f'| q is concave on [a, b], we can use the Jensen's integral inequality to obtain:

$\begin{array}{ll}\hfill {\int }_{0}^{1}|{f}^{\prime }\left(tx+\left(1-t\right)a\right){|}^{q}dt& ={\int }_{0}^{1}{t}^{0}|{f}^{\prime }\left(tx+\left(1-t\right)a\right){|}^{q}dt\phantom{\rule{2em}{0ex}}\\ \le \left({\int }_{0}^{1}{t}^{0}dt\right){\left|{f}^{\prime }\left(\frac{1}{{\int }_{0}^{1}{t}^{0}dt}{\int }_{0}^{1}\left(tx+\left(1-t\right)a\right)dt\right)\right|}^{q}\phantom{\rule{2em}{0ex}}\\ ={\left|{f}^{\prime }\left(\frac{a+x}{2}\right)\right|}^{q}\phantom{\rule{2em}{0ex}}\end{array}$

Analogously,

${\int }_{0}^{1}|{f}^{\prime }\left(tx+\left(1-t\right)b\right){|}^{q}dt\le {\left|{f}^{\prime }\left(\frac{b+x}{2}\right)\right|}^{q}.$

Combining all the obtained inequalities, we get

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left|\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du\right|\\ \le {\left[\frac{q-1}{2q-1}\right]}^{\frac{q-1}{q}}\left[\frac{{\left(x-a\right)}^{2}|{f}^{\prime }\left(\frac{a+x}{2}\right)|+{\left(b-x\right)}^{2}|{f}^{\prime }\left(\frac{b+x}{2}\right)|}{b-a}\right]\end{array}$

which completes the proof. □

Remark 2. In Theorem 6, if we choose$x=\frac{a+b}{2}$we have

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du\right|\\ \le {\left[\frac{q-1}{2q-1}\right]}^{\frac{q-1}{q}}\left(\frac{b-a}{4}\right)\left(\left|{f}^{\prime }\left(\frac{3a+b}{4}\right)\right|+\left|{f}^{\prime }\left(\frac{a+3b}{4}\right)\right|\right)\end{array}$

which is the inequality in (1.3).

Theorem 7. Let f : I ⊆ ℝ → ℝ be a differentiable mapping on I° such that f'L [a, b], where a, bI with a < b. If |f'| q is convex on [a, b] and for some fixed q ≥ 1, then the following inequality holds:

$\begin{array}{l}\phantom{\rule{0.5em}{0ex}}|\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du|\\ \le \frac{1}{2}{\left(\frac{1}{3}\right)}^{\frac{1}{q}}\left[\frac{{\left(x-a\right)}^{2}{\left[\mid {f}^{\prime }\left(x{\right)\mid }^{q}+2\mid {f}^{\prime }\left(a{\right)\mid }^{q}\right]}^{\frac{1}{q}}+{\left(b-x\right)}^{2}{\left[\mid {f}^{\prime }\left(x{\right)\mid }^{q}+2\mid {f}^{\prime }\left(b{\right)\mid }^{q}\right]}^{\frac{1}{q}}}{b-a}\right]\end{array}$

for each x ∈ [a, b].

Proof. Suppose that q ≥ 1. From Lemma 1 and using the well-known power-mean inequality, we have

$\begin{array}{l}\phantom{\rule{0.5em}{0ex}}|\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du|\\ \le \frac{{\left(x-a\right)}^{2}}{b-a}{\int }_{0}^{1}\left(1-t\right)\mid {f}^{\prime }\left(tx+\left(1-t\right)a\right)\mid dt\\ \phantom{\rule{0.5em}{0ex}}+\frac{{\left(b-x\right)}^{2}}{b-a}{\int }_{0}^{1}\left(1-t\right)\mid {f}^{\prime }\left(tx+\left(1-t\right)b\right)\mid dt\\ \le \frac{{\left(x-a\right)}^{2}}{b-a}{\left({\int }_{0}^{1}\left(1-t\right)dt\right)}^{1-\frac{1}{q}}{\left({\int }_{0}^{1}\left(1-t\right)\mid {f}^{\prime }\left(tx+\left(1-t\right)a{\right)\mid }^{q}dt\right)}^{\frac{1}{q}}\\ \phantom{\rule{0.5em}{0ex}}+\frac{{\left(b-x\right)}^{2}}{b-a}{\left({\int }_{0}^{1}\left(1-t\right)dt\right)}^{1-\frac{1}{q}}{\left({\int }_{0}^{1}\left(1-t\right)\mid {f}^{\prime }\left(tx+\left(1-t\right)b{\right)\mid }^{q}dt\right)}^{\frac{1}{q}}.\end{array}$

Since |f'| q is convex, therefore we have

$\begin{array}{l}\phantom{\rule{0.5em}{0ex}}{\int }_{0}^{1}\left(1-t\right)\mid {f}^{\prime }\left(tx+\left(1-t\right)a\right){\mid }^{q}dt\\ \le {\int }_{0}^{1}\left(1-t\right)\left[t\mid {f}^{\prime }\left(x{\right)\mid }^{q}+\left(1-t\right)\mid {f}^{\prime }\left(a{\right)\mid }^{q}\right]dt\\ =\frac{\mid {f}^{\prime }\left(x{\right)\mid }^{q}+2\mid {f}^{\prime }\left(a{\right)\mid }^{q}}{6}\end{array}$

Analogously,

${\int }_{0}^{1}\left(1-t\right)\mid {f}^{\prime }\left(tx+\left(1-t\right)b{\right)\mid }^{q}dt\le \frac{\mid {f}^{\prime }\left(x{\right)\mid }^{q}+2\mid {f}^{\prime }\left(b{\right)\mid }^{q}}{6}.$

Combining all the above inequalities gives the desired result. □

Corollary 4. In Theorem 7, choosing$x=\frac{a+b}{2}$and then using the convexity of |f'| q we have

$\begin{array}{l}\phantom{\rule{0.5em}{0ex}}|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du|\\ \le \left(\frac{b-a}{8}\right){\left(\frac{1}{3}\right)}^{\frac{1}{q}}\left[{\left(2\mid {f}^{\prime }\left(a{\right)\mid }^{q}+{|{f}^{\prime }\left(\frac{a+b}{2}\right)|}^{q}\right)}^{\frac{1}{q}}+{\left(2\mid {f}^{\prime }\left(b{\right)\mid }^{q}+{|{f}^{\prime }\left(\frac{a+b}{2}\right)|}^{q}\right)}^{\frac{1}{q}}\right]\\ \le \left(\frac{{3}^{1-\frac{1}{q}}}{8}\right)\left(b-a\right)\left(\mid {f}^{\prime }\left(a\right)\mid +\mid {f}^{\prime }\left(b\right)\mid \right).\end{array}$

Theorem 8. Let f : I ⊆ ℝ → ℝ be a differentiable mapping on I° such that f'L [a, b], where a, bI with a < b. If |f'| q is concave on [a, b], for some fixed q ≥ 1, then the following inequality holds:

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left|\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du\right|\\ \le \frac{1}{2}\left[\frac{{\left(x-a\right)}^{2}|{f}^{\prime }\left(\frac{x+2a}{3}\right)|+{\left(b-x\right)}^{2}|{f}^{\prime }\left(\frac{x+2b}{3}\right)|}{b-a}\right].\end{array}$

Proof. First, we note that by the concavity of |f'| q and the power-mean inequality,

we have

$|{f}^{\prime }\left(tx+\left(1-t\right)a\right){|}^{q}\ge t|{f}^{\prime }\left(x\right){|}^{q}+\left(1-t\right)|{f}^{\prime }\left(a\right){|}^{q}.$

Hence,

$|{f}^{\prime }\left(tx+\left(1-t\right)a\right)|\ge t|{f}^{\prime }\left(x\right)|+\left(1-t\right)|{f}^{\prime }\left(a\right)|,$

so |f'| is also concave.

Accordingly, using Lemma 1 and the Jensen integral inequality, we have

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left|\frac{\left(b-x\right)f\left(b\right)+\left(x-a\right)f\left(a\right)}{b-a}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du\right|\\ \le \frac{{\left(x-a\right)}^{2}}{b-a}{\int }_{0}^{1}\left(1-t\right)|{f}^{\prime }\left(tx+\left(1-t\right)a\right)|dt\\ \phantom{\rule{1em}{0ex}}+\frac{{\left(b-x\right)}^{2}}{b-a}{\int }_{0}^{1}\left(1-t\right)|{f}^{\prime }\left(tx+\left(1-t\right)b\right)|dt\\ \le \frac{{\left(x-a\right)}^{2}}{b-a}\left({\int }_{0}^{1}\left(1-t\right)dt\right)\left|{f}^{\prime }\left(\frac{{\int }_{0}^{1}\left(1-t\right)\left(tx+\left(1-t\right)a\right)dt}{{\int }_{0}^{1}\left(1-t\right)dt}\right)\right|\\ \phantom{\rule{1em}{0ex}}+\frac{{\left(b-x\right)}^{2}}{b-a}\left({\int }_{0}^{1}\left(1-t\right)dt\right)\left|{f}^{\prime }\left(\frac{{\int }_{0}^{1}\left(1-t\right)\left(tx+\left(1-t\right)b\right)dt}{{\int }_{0}^{1}\left(1-t\right)dt}\right)\right|\\ \le \frac{1}{2}\left[\frac{{\left(x-a\right)}^{2}\left|{f}^{\prime }\left(\frac{x+2a}{3}\right)\right|+{\left(b-x\right)}^{2}\left|{f}^{\prime }\left(\frac{x+2b}{3}\right)\right|}{b-a}\right].\end{array}$

Remark 3. In Theorem 8, if we choose$x=\frac{a+b}{2}$we have

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left|\frac{f\left(a\right)+f\left(b\right)}{2}-\frac{1}{b-a}{\int }_{a}^{b}f\left(u\right)du\right|\\ \le \frac{b-a}{8}\left[\left|{f}^{\prime }\left(\frac{5a+b}{6}\right)\right|+\left|{f}^{\prime }\left(\frac{a+5b}{6}\right)\right|\right]\end{array}$

which is the inequality in (1.4).

## 3. Applications to Special Means

Recall the following means which could be considered extensions of arithmetic, logarithmic and generalized logarithmic from positive to real numbers.

1. (1)

The arithmetic mean:

$A=A\left(a,b\right)=\frac{a+b}{2};a,b\in ℝ$
2. (2)

The logarithmic mean:

$L\left(a,b\right)=\frac{b-a}{ln|b|-ln|a|};|a|\ne |b|,ab\ne 0,a,b\in ℝ$
3. (3)

The generalized logarithmic mean:

${L}_{n}\left(a,b\right)={\left[\frac{{b}^{n+1}-{a}^{n+1}}{\left(b-a\right)\left(n+1\right)}\right]}^{\frac{1}{n}};n\in ℤ\\left\{-1,0\right\},a,b\in ℝ,a\ne b$

Now using the results of Section 2, we give some applications to special means of real numbers.

Proposition 1. Let a, b ∈ ℝ, a < b, 0 ∉ [a, b] and n ∈ ℤ, |n| ≥ 2. Then, for all p > 1

1. (a)
$|A\left({a}^{n},{b}^{n}\right)-{L}_{n}^{n}\left(a,b\right)|\le |n|\left(b-a\right){\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{2}\right)}^{\frac{1}{q}}A\left(|a{|}^{n-1},|b{|}^{n-1}\right)$
(3.1)

and

1. (b)
$|A\left({a}^{n},{b}^{n}\right)-{L}_{n}^{n}\left(a,b\right)|\le |n|\left(b-a\right)\frac{{3}^{1-\frac{1}{q}}}{4}A\left(|a{|}^{n-1},|b{|}^{n-1}\right).$
(3.2)

Proof. The assertion follows from Corollary 3 and 4 for f (x) = xn , x ∈ ℝ, n ∈ ℤ, |n| ≥ 2. □

Proposition 2. Let a, b ∈ ℝ, a < b, 0 ∉ [a, b]. Then, for all q ≥ 1,

1. (a)
$|A\left({a}^{-1},{b}^{-1}\right)-{L}^{-1}\left(a,b\right)|\le \left(b-a\right){\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{2}\right)}^{\frac{1}{q}}A\left(|a{|}^{-2},|b{|}^{-2}\right)$
(3.3)

and

1. (b)
$|A\left({a}^{-1},{b}^{-1}\right)-{L}^{-1}\left(a,b\right)|\le \left(b-a\right)\left(\frac{{3}^{1-\frac{1}{q}}}{4}\right)A\left(|a{|}^{-2},|b{|}^{-2}\right).$
(3.4)

Proof. The assertion follows from Corollary 3 and 4 for $f\left(x\right)=\frac{1}{x}$. □

## 4. The Trapezoidal Formula

Let d be a division a = x0 < x1 < ... < xn - 1< x n = b of the interval [a, b] and consider the quadrature formula

${\int }_{a}^{b}f\left(x\right)dx=T\left(f,d\right)+E\left(f,d\right)$
(4.1)

where

$T\left(f,d\right)=\sum _{i=0}^{n-1}\frac{f\left({x}_{i}\right)+f\left({x}_{i+1}\right)}{2}\left({x}_{i+1}-{x}_{i}\right)$

for the trapezoidal version and E (f, d) denotes the associated approximation error.

Proposition 3. Let f : I ⊆ ℝ → ℝ be a differentiable mapping on I° such that f'L [a, b], where a, bI with a < b and$|{f}^{\prime }{}^{{|}^{\frac{p}{p-1}}}$is convex on [a, b], where p > 1. Then in (4.1), for every division d of [a, b], the trapezoidal error estimate satisfies

$|E\left(f,d\right)|\le {\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{2}\right)}^{\frac{1}{q}}\sum _{i=0}^{n-1}\frac{{\left({x}_{i+1}-{x}_{i}\right)}^{2}}{2}\left(|{f}^{\prime }\left({x}_{i}\right)|+|{f}^{\prime }\left({x}_{i+1}\right)|\right).$

Proof. On applying Corollary 3 on the subinterval [x i , xi+1] (i = 0, 1, 2,..., n - 1) of the division, we have

$\begin{array}{c}\phantom{\rule{1em}{0ex}}\left|\frac{f\left({x}_{i}\right)+f\left({x}_{i+1}\right)}{2}-\frac{1}{{x}_{i+1}-{x}_{i}}{\int }_{{x}_{i}}^{{x}_{i+1}}f\left(x\right)dx\right|\\ \le \frac{\left({x}_{i+1}-{x}_{i}\right)}{2}{\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{2}\right)}^{\frac{1}{q}}\left(|{f}^{\prime }\left({x}_{i}\right)|+|{f}^{\prime }\left({x}_{i+1}\right)|\right).\end{array}$

Hence in (4.1) we have

$\begin{array}{ll}\hfill \left|{\int }_{a}^{b}f\left(x\right)dx-T\left(f,d\right)\right|& =\left|\sum _{i=0}^{n-1}\left\{{\int }_{{x}_{i}}^{{x}_{i+1}}f\left(x\right)dx-\frac{f\left({x}_{i}\right)+f\left({x}_{i+1}\right)}{2}\left({x}_{i+1}-{x}_{i}\right)\right\}\right|\phantom{\rule{2em}{0ex}}\\ \le \sum _{i=0}^{n-1}\left|{\int }_{{x}_{i}}^{{x}_{i+1}}f\left(x\right)dx-\frac{f\left({x}_{i}\right)+f\left({x}_{i+1}\right)}{2}\left({x}_{i+1}-{x}_{i}\right)\right|\phantom{\rule{2em}{0ex}}\\ \le {\left(\frac{1}{p+1}\right)}^{\frac{1}{p}}{\left(\frac{1}{2}\right)}^{\frac{1}{q}}\sum _{i=0}^{n-1}\frac{{\left({x}_{i+1}-{x}_{i}\right)}^{2}}{2}\left(|{f}^{\prime }\left({x}_{i}\right)|+|{f}^{\prime }\left({x}_{i+1}\right)|\right)\phantom{\rule{2em}{0ex}}\end{array}$

which completes the proof. □

Proposition 4. Let f : I ⊆ ℝ → ℝ be a differentiable mapping on I° such that f'L [a, b], where a, bI with a < b. If |f'| q is concave on [a, b], for some fixed q > 1, Then in (4.1), for every division d of [a, b], the trapezoidal error estimate satisfies

$|E\left(f,d\right)|\le {\left(\frac{q-1}{2q-1}\right)}^{\frac{q-1}{q}}\sum _{i=0}^{n-1}\frac{{\left({x}_{i+1}-{x}_{i}\right)}^{2}}{4}\left(\left|{f}^{\prime }\left(\frac{3{x}_{i}+{x}_{i+1}}{4}\right)\right|+\left|{f}^{\prime }\left(\frac{{x}_{i}+3{x}_{i+1}}{4}\right)\right|\right).$

Proof. The proof is similar to that of Proposition 3 and using Remark 2. □

Proposition 5. Let f : I ⊆ ℝ → ℝ be a differentiable mapping on I° such that f'L [a, b], where a, bI with a < b. If |f'| q is concave on [a, b], for some fixed q ≥ 1, Then in (4.1), for every division d of [a, b], the trapezoidal error estimate satisfies

$|E\left(f,d\right)|\le \frac{1}{8}\sum _{i=0}^{n-1}{\left({x}_{i+1}-{x}_{i}\right)}^{2}\left(\left|{f}^{\prime }\left(\frac{5{x}_{i}+{x}_{i+1}}{6}\right)\right|+\left|{f}^{\prime }\left(\frac{{x}_{i}+5{x}_{i+1}}{6}\right)\right|\right).$

Proof. The proof is similar to that of Proposition 3 and using Remark 3. □