Some new inequalities of Simpson’s type for functions whose third derivatives in absolute value at some powers are strongly \((s,m)\)- convex in the second sense are provided. An application to the Simpson’s quadrature rule is also provided.
Theorem 1. Let \(f”:I\subseteq\mathbb{R}\to\mathbb{R}\) be an absolutely continuous function on \(I^\circ\) such that \(f”’\in L_1([a,b])\), where \(a, b\in I\) with \(a< b\). If \(|f'''|\) is quasiconvex on \([a,b]\), then the following inequality holds: \begin{align} \nonumber &\bigg|\int_a^b(t)dt-\frac{b-a}{6}\bigg[f(a)+4f\Big(\frac{a+b}{2}\Big)+f(b)\bigg]\bigg|\\ \nonumber &\leq\frac{(b-a)^4}{1152}\bigg[\max\bigg\{|f'''(a)|,\bigg|f'''\bigg(\frac{a+b}{2}\bigg)\bigg|\bigg\}+\max\bigg\{\bigg|f'''\bigg(\frac{a+b}{2}\bigg)\bigg|, |f'''(b)|\bigg\}\bigg]. \end{align}
Theorem 2. Let \(f”:I\subseteq\mathbb{R}\to\mathbb{R}\) be an absolutely continuous function on \(I^\circ\) such that \(f”’\in L_1([a,b])\), where \(a, b\in I\) with \(a< b\). If \(|f”’|^q\) is quasiconvex on \([a,b]\) for some fixed \(q>1\), then the following inequality holds: \begin{align} \nonumber &\bigg|\int_a^b(t)dt-\frac{b-a}{6}\bigg[f(a)+4f\Big(\frac{a+b}{2}\Big)+f(b)\bigg]\bigg|\leq\frac{2^{-1/p}(b-a)^4}{48}\bigg(\frac{\Gamma(p+1)\Gamma(2p+1)}{\Gamma(3p+2)}\bigg)^{1/p}\\ \nonumber &\times\bigg[\bigg(\max\bigg\{|f”'(a)|^q,\bigg|f”’\bigg(\frac{a+b}{2}\bigg)\bigg|^q\bigg\}\bigg)^{1/q} +\bigg(\max\bigg\{\bigg|f”’\bigg(\frac{a+b}{2}\bigg)\bigg|^q, |f”'(b)|^q\bigg\}\bigg)^{1/q}\bigg], \end{align} where \(\frac{1}{p}+\frac{1}{q}=1\) and \(\Gamma(\cdot)\) is the gamma function (see Definition 8).
Definition 3.[3] A function \(f: [0, b]\to\mathbb{R}, b>0\), is said to be \(m\)-convex, for \(m\in[0, 1]\), if \begin{align} \nonumber f(tx+m(1-t)y)\leq tf(x)+m(1-t)f(y), \end{align} for all \(x, y\in[0, b]\) and \(t\in[0,1]\).
In [4], Özdemir et al. presented following Simpsion’s type inequality for functions whose third derivatives in absolute value at some powers are \(m\)-convex.Theorem 4. Let \(f:I \subset[0, b^*] \to\mathbb{R}\) be a three times differentiable function on \(I^\circ\) such that \(f”’\in L_1([a, b])\), where \(a, b\in I^\circ\) with \(a> b, b^*>0\). If \(|f”’|^q\) is \(m\)-convex, for \(m\in(0,1]\) and \(q > 1\), then the following inequality holds: \begin{align} \nonumber &\bigg|\int_a^{mb}(t)dt-\frac{mb-a}{6}\bigg[f(a)+4f\Big(\frac{a+b}{2}\Big)+f(mb)\bigg]\bigg| \leq\frac{2^{-1/p}(mb-a)^4}{96}\bigg(\frac{\Gamma(p+1)\Gamma(2p+1)}{\Gamma(3p+2)}\bigg)^{1/p}\\ \nonumber &\times\bigg[\bigg(\frac{|f”'(a)|^q+3m|f”'(b)|^q}{4}\bigg)^{1/q}+\bigg(\frac{3|f”'(a)|^q+m|f”'(b)|^q}{4}\bigg)^{1/q}\bigg] \end{align} where \(\frac{1}{p}+\frac{1}{q}=1\) and \(\Gamma(\cdot)\) is the gamma function (see Definition 8).
Definition 5. [12] A function \(f: [0, \infty)\to\mathbb{R}\) is said to be \(s\)-convex in the second sense, for \(s\in[0, 1]\), if \begin{align} \nonumber f(tx+(1-t)y)\leq t^sf(x)+(1-t)^sf(y), \end{align} for all \(x, y\in[0,\infty)\) and \(t\in[0,1]\).
In [5], Özdemir et al. presented following Simpsion’s type inequality for functions whose third derivatives in absolute value at some powers are \(s\)-convex.Theorem 6. Let \(f:I \subset[0, \infty) \to\mathbb{R}\) be a three times differentiable function on \(I^\circ\) such that \(f”’\in L_1([a, b])\), where \(a, b\in I^\circ\) with \(a< b\). If \(|f”’|^q\) is \(s\)-convex in the second sense, for \(s\in (0, 1]\) and \(q > 1\), then the following inequality holds: \begin{align} \nonumber &\bigg|\int_a^b(t)dt-\frac{b-a}{6}\bigg[f(a)+4f\Big(\frac{a+b}{2}\Big)+f(b)\bigg]\bigg|\leq\frac{2^{-1/p}(b-a)^4}{48}\bigg(\frac{\Gamma(p+1)\Gamma(2p+1)}{\Gamma(3p+2)}\bigg)^{1/p}\\ \nonumber &\times\bigg[\bigg(\frac{1}{2^{s+1}(s+1)}|f”'(a)|^q+\frac{2^{s+1}-1}{2^{s+1}(s+1)}|f”'(b)|^q\bigg)\bigg|^q\bigg\}\bigg)^{1/q}\\ \nonumber &+\bigg(\frac{1}{2^{s+1}(s+1)}|f”'(b)|^q+\frac{2^{s+1}-1}{2^{s+1}(s+1)}|f”'(a)|^q\bigg)\bigg|^q\bigg\}\bigg)^{1/q}\bigg], \end{align} where \(\frac{1}{p}+\frac{1}{q}=1\) and \(\Gamma(\cdot)\) is the gamma function (see Definition 8).
Recently, Bracamonte et al. [13] introduced the concept of functions that are strongly \((s,m)\)-convex in the second sense as follows.Definition 7. [13] A function \(f:[0,\infty)\to\mathbb{R}\) is said to be strongly \((s,m)\)-convex function with modulus \(\mu\geq0\) in the second sense, for \((s,m)\in[0,1]\times[0,1]\), if \begin{align} \nonumber f(tx+m(1-t)y)\leq t^sf(x)+m(1-t)^sf(y)-\mu t(1-t)(x-y)^2 \end{align} for all \(x, y\in[0,\infty)\) and \(t\in[0,1]\).
Remark 1. If \(\mu=0\) in Definition 7, then we have the definition for \((s,m)\)-convex functions in the second sense. Choosing \(\mu=0\) and \(m=1\) gives the definition of \(s\)-convex functions. Also for \(\mu=0\) and \(s=1\), we have \(m\)-convex. If \(s\in[0,1]\) and \(m=1\), then we have the class of strongly \(s\)-convex functions. Also, if \(m\in[0,1]\) and \(s=1\), then we have the class of strongly \(m\)-convex functions.
Motivated by the above results, the goal of this paper is to provide some new Simpson’s type inequality for functions whose third derivatives in absolute value at certain powers are strongly \((s,m)\)-convex functions. We complete this section be recalling the definitions of the gamma, beta and incomplete Beta functions.Definition 8. The gamma function is given by \begin{align} \nonumber \Gamma(x)=\int_0^\infty t^{x-1}e^{-x}dx, ~~~~~~Re(x)>0, \end{align} the (complete) beta function is given by \begin{align} \nonumber B(x, y)=\int_0^1 t^{x-1}(1-t)^{y-1}dx, ~~~~~~Re(x)>0~~\mbox{and}~~Re(y)>0, \end{align} and the incomplete beta function is given by \begin{align} \nonumber B_a(x, y)=\int_0^a t^{x-1}(1-t)^{y-1}dx, ~~~~~~Re(x)>0, Re(y)>0~~\mbox{and}~~0< a< 1. \end{align}
Remark 2. The gamma and beta function satisfies the following properties:
Lemma 9.[2] Let \(f: I\to\mathbb{R}\) be a three times differentiable function on \(I^\circ\) such that \(f”’\in L_1([a,b])\), for \(a, b\in I\) and \(a< b\). Then the following equality holds; \begin{align} \nonumber \int_a^bf(x)dx&-\frac{b-a}{6}\Big[f(a)+4f\Big(\frac{a+b}{2}\Big)+f(b)\Big]=(b-a)^4\int_0^1p(t)f'''(ta+(1-t)b)dt, \end{align} where \[p(t)= \begin{cases} \frac{1}{6}t^2(t-\frac{1}{2}) & 0\leq t\leq \frac{1}{2}, \\ \frac{1}{6} (t-1)^2(t-\frac{1}{2}) & \frac{1}{2}\leq t\leq 1. \end{cases} \]
Theorem 10. Let \(f: [0,\infty)\to\mathbb{R}\) be a three times differentiable function on \((0,\infty)\) such that \(f”’\in L_1([a,b])\), for \(0\leq a< b\). If \(|f'''|\) is strongly \((s,m)\)-convex with modulus \(\mu\geq0\), for \((s,m)\in(0,1]\times(0,1]\), then the following inequality holds: \begin{align} \nonumber &\bigg|\int_a^bf(x)dx-\frac{b-a}{6}\Big[f(a)+4f\Big(\frac{a+b}{2}\Big)+f(b)\Big]\bigg|\leq\frac{(b-a)^4}{6}\bigg[\frac{2^{-s-4}}{(s+3)(s+4)}\bigg(\Big|f'''(a)\Big|+\Big|f'''(b)\Big|\bigg)\\ \nonumber &+\frac{m2^{-s-4}\Big(s^2+11s+2^{s+4}(s-2)+34\Big)}{(s+1)(s+2)(s+3)(s+4)}\bigg(\Big|f'''\Big(\frac{a}{m}\Big)\Big|+\Big|f'''\Big(\frac{b}{m}\Big)\Big|\bigg)\frac{\mu}{960}\bigg(\Big(\frac{b}{m}-a\Big)^2+\Big(b-\frac{a}{m}\Big)^2\bigg)\bigg]. \end{align}
Proof. Using Lemma 9 and the strong \((s, m)\)-convexity of \(|f”’|\), we have \begin{align*} \nonumber &\bigg|\int_a^bf(x)dx-\frac{b-a}{6}\Big[f(a)+4f\Big(\frac{a+b}{2}\Big)+f(b)\Big]\bigg|\leq(b-a)^4\int_0^1|p(t)||f”'(ta+(1-t)b)|dt\\ \nonumber &=\frac{(b-a)^4}{6}\bigg[\int_0^{\frac{1}{2}}\Big|t^2\Big(t-\frac{1}{2}\Big)\Big|\Big|f”'(ta+(1-t)b)\Big|dt+\int_{\frac{1}{2}}^1\Big|(t-1)^2\Big(t-\frac{1}{2}\Big)\Big|\Big|f”'(ta+(1-t)b)\Big|dt\bigg]\\ \nonumber &=\frac{(b-a)^4}{6}\bigg[\int_0^{\frac{1}{2}}t^2\Big(\frac{1}{2}-t\Big)\Big|f”'(ta+(1-t)b)\Big|dt+\int_0^{\frac{1}{2}}t^2\Big(\frac{1}{2}-t\Big)\Big|f”'(tb+(1-t)a)\Big|dt\bigg]\\ \nonumber &\leq\frac{(b-a)^4}{6}\bigg[\int_0^{\frac{1}{2}}t^2\Big(\frac{1}{2}-t\Big)\Big(t^s\Big|f”'(a)\Big|+m(1-t)^s\Big|f”’\Big(\frac{b}{m}\Big)\Big|-\mu t(1-t)\Big(\frac{b}{m}-a\Big)^2 \Big)dt\\ \nonumber &+\int_0^{\frac{1}{2}}t^2\Big(\frac{1}{2}-t\Big)\Big(t^s\Big|f”'(b)\Big|+m(1-t)^s\Big|f”’\Big(\frac{a}{m}\Big)\Big|-\mu t(1-t)\Big(b-\frac{a}{m}\Big)^2\Big)dt\bigg]\\ \nonumber &=\frac{(b-a)^4}{6}\bigg[\Big|f”'(a)\Big|\int_0^{\frac{1}{2}}t^{s+2}\Big(\frac{1}{2}-t\Big)dt\\\nonumber&+m\Big|f”’\Big(\frac{b}{m}\Big)\Big|\int_0^{\frac{1}{2}}t^2\Big(\frac{1}{2}-t\Big)(1-t)^sdt -\mu\Big(\frac{b}{m}-a\Big)^2\int_0^{\frac{1}{2}}t^{3}\Big(\frac{1}{2}-t\Big)(1-t)dt\\ \nonumber &+\Big|f”'(b)\Big|\int_0^{\frac{1}{2}}t^{s+2}\Big(\frac{1}{2}-t\Big)dt+m\Big|f”’\Big(\frac{a}{m}\Big)\Big|\int_0^{\frac{1}{2}}t^2\Big(\frac{1}{2}-t\Big)(1-t)^sdt -\mu\Big(b-\frac{a}{m}\Big)^2\int_0^{\frac{1}{2}}t^{3}\Big(\frac{1}{2}-t\Big)(1-t)dt\bigg]\\ \nonumber &=\frac{(b-a)^4}{6}\bigg[\frac{2^{-s-4}}{(s+3)(s+4)}\bigg(\Big|f”'(a)\Big|+\Big|f”'(b)\Big|\bigg) +\frac{m2^{-s-4}\Big(s^2+11s+2^{s+4}(s-2)+34\Big)}{(s+1)(s+2)(s+3)(s+4)}\\&\times\bigg(\Big|f”’\Big(\frac{a}{m}\Big)\Big|+\Big|f”’\Big(\frac{b}{m}\Big)\Big|\bigg) -\frac{\mu}{960}\bigg(\Big(\frac{b}{m}-a\Big)^2+\Big(b-\frac{a}{m}\Big)^2\bigg)\bigg], \end{align*} where
Theorem 11. Let \(f: [0,\infty)\to\mathbb{R}\) be a three times differentiable function on \((0,\infty)\) such that \(f”’\in L_1([a,b])\), for \(0\leq a< b\). If \(|f”’|^q\) is strongly \((s,m)\)-convex with modulus \(\mu\geq0\), for \((s,m)\in(0,1]\times(0,1]\) and \(q>1\), then the following inequality holds; \begin{align} \nonumber &\bigg|\int_a^bf(x)dx-\frac{b-a}{6}\Big[f(a)+4f\Big(\frac{a+b}{2}\Big)+f(b)\Big]\bigg| \leq\frac{(b-a)^4}{48(2^{1/p})}\Big(\frac{\Gamma(2p+1)\Gamma(p+1)}{\Gamma(3p+2)}\Big)^{\frac{1}{p}}\bigg[\Big(\frac{2^{-s-1}}{s+1}\Big|f”'(a)\Big|^q\\\nonumber &+\frac{m(1-2^{-s-1})}{s+1}\Big|f”’\Big(\frac{b}{m}\Big)\Big|^q -\frac{\mu}{12}\Big(\frac{b}{m}-a\Big)^2\Big)^{\frac{1}{q}}+\Big(\frac{2^{-s-1}}{s+1}\Big|f”'(b)\Big|^q+\frac{m(1-2^{-s-1})}{s+1}\Big|f”’\Big(\frac{a}{m}\Big)\Big|^q-\frac{\mu}{12}\Big(b-\frac{a}{m}\Big)^2\Big)^{\frac{1}{q}}\bigg], \end{align} where \(\frac{1}{p}+\frac{1}{q}=1\).
Proof. Using Lemma 9, the Hölder’s inequality and the strong \((s, m)\)-convexity of \(|f”’|^q\), we have \begin{align} \nonumber &\bigg|\int_a^bf(x)dx-\frac{b-a}{6}\Big[f(a)+4f\Big(\frac{a+b}{2}\Big)+f(b)\Big]\bigg|\leq(b-a)^4\int_0^1|p(t)||f”'(ta+(1-t)b)|dt\end{align} \begin{align}\nonumber &=\frac{(b-a)^4}{6}\bigg[\int_0^{\frac{1}{2}}\Big|t^2\Big(t-\frac{1}{2}\Big)\Big|\Big|f”'(ta+(1-t)b)\Big|dt +\int_{\frac{1}{2}}^1\Big|(t-1)^2\Big(t-\frac{1}{2}\Big)\Big|\Big|f”'(ta+(1-t)b)\Big|dt\bigg]\\ \nonumber &=\frac{(b-a)^4}{6}\bigg[\int_0^{\frac{1}{2}}t^2\Big(\frac{1}{2}-t\Big)\Big|f”'(ta+(1-t)b)\Big|dt +\int_0^{\frac{1}{2}}t^2\Big(\frac{1}{2}-t\Big)\Big|f”'(tb+(1-t)a)\Big|dt\bigg]\\ \nonumber &\leq\frac{(b-a)^4}{6}\bigg[\Big(\int_0^{\frac{1}{2}}\Big[t^2\Big(\frac{1}{2}-t\Big)\Big]^pdt\Big)^{\frac{1}{p}}\Big(\int_0^{\frac{1}{2}}\Big|f”'(ta+(1-t)b)\Big|^qdt\Big)^{\frac{1}{q}} \\ \nonumber &+\Big(\int_0^{\frac{1}{2}}\Big[t^2\Big(\frac{1}{2}-t\Big)\Big]^pdt\Big)^{\frac{1}{p}}\Big(\int_0^{\frac{1}{2}}\Big|f”'(tb+(1-t)a)\Big|^qdt\Big)^{\frac{1}{q}}\bigg]\\ \nonumber &\leq\frac{(b-a)^4}{6}\Big(\int_0^{\frac{1}{2}}\Big[t^2\Big(\frac{1}{2}-t\Big)\Big]^pdt\Big)^{\frac{1}{p}}\bigg[\Big(\int_0^{\frac{1}{2}}\Big(t^s\Big|f”'(a)\Big|^q+m(1-t)^s\Big|f”’\Big(\frac{b}{m}\Big)\Big|^q -\mu t(1-t)\Big(\frac{b}{m}-a\Big)^2\Big)dt\Big)^{\frac{1}{q}}\\ \nonumber &+\Big(\int_{\frac{1}{2}}^1\Big(t^s\Big|f”'(b)\Big|^q+m(1-t)^s\Big|f”’\Big(\frac{a}{m}\Big)\Big|^q-\mu t(1-t)\Big(b-\frac{a}{m}\Big)^2\Big)dt\Big)^{\frac{1}{q}}\bigg]\\ \nonumber &=\frac{(b-a)^4}{6}\Big(\int_0^{\frac{1}{2}}\Big[t^2\Big(\frac{1}{2}-t\Big)\Big]^pdt\Big)^{\frac{1}{p}}\bigg[\Big(\Big|f”'(a)\Big|^q\int_0^{\frac{1}{2}}t^sdt+m\Big|f”’\Big(\frac{b}{m}\Big)\Big|^q\int_0^{\frac{1}{2}}(1-t)^sdt -\mu\Big(\frac{b}{m}-a\Big)^2\\ \nonumber & \times\int_0^{\frac{1}{2}}t(1-t)dt\Big)^{\frac{1}{q}} +\Big(\Big|f”'(b)\Big|^q\int_0^{\frac{1}{2}}t^sdt+m\Big|f”’\Big(\frac{a}{m}\Big)\Big|^q\int_0^{\frac{1}{2}}(1-t)^sdt-\mu\Big(b-\frac{a}{m}\Big)^2\int_0^{\frac{1}{2}}t(1-t)dt\Big)^{\frac{1}{q}}\bigg]\\ \nonumber &=\frac{(b-a)^4}{48(2^{1/p})}\Big(\frac{\Gamma(2p+1)\Gamma(p+1)}{\Gamma(3p+2)}\Big)^{\frac{1}{p}}\bigg[\Big(\frac{2^{-s-1}}{s+1}\Big|f”'(a)\Big|^q+\frac{m(1-2^{-s-1})}{s+1}\Big|f”’\Big(\frac{b}{m}\Big)\Big|^q\\ \nonumber &-\frac{\mu}{12}\Big(\frac{b}{m}-a\Big)^2\Big)^{\frac{1}{q}}+\Big(\frac{2^{-s-1}}{s+1}\Big|f”'(b)\Big|^q+\frac{m(1-2^{-s-1})}{s+1}\Big|f”’\Big(\frac{a}{m}\Big)\Big|^q-\frac{\mu}{12}\Big(b-\frac{a}{m}\Big)^2\Big)^{\frac{1}{q}}\bigg], \end{align} where \begin{align} \nonumber \int_0^{\frac{1}{2}}\Big[t^2\Big(\frac{1}{2}-t\Big)\Big]^pdt&=\frac{1}{2^{3p+1}}B(2p+1,p+1)=\frac{\Gamma(2p+1)\Gamma(p+1)}{2^{3p+1}\Gamma(3p+2)}, \end{align} \begin{align} \nonumber \int_0^{\frac{1}{2}}t^sdt&=\frac{2^{-s-1}}{s+1} \end{align} and \begin{align} \nonumber \int_0^{\frac{1}{2}}(1-t)^sdt&=\frac{1-2^{-s-1}}{s+1}. \end{align} This completes the proof.
Theorem 12. Let \(f: [0,\infty)\to\mathbb{R}\) be a three times differentiable function on \((0,\infty)\) such that \(f”’\in L_1([a,b])\), for \(0\leq a< b\). If \(|f”’|^q\) is strongly \((s,m)\)-convex with modulus \(\mu\geq0\), for \((s,m)\in(0,1]\times(0,1]\) and \(q>1\), then the following inequality holds; \begin{align} \nonumber &\bigg|\int_a^bf(x)dx-\frac{b-a}{6}\Big[f(a)+4f\Big(\frac{a+b}{2}\Big)+f(b)\Big]\bigg|\leq\frac{(b-a)^4}{6}\Big(\frac{1}{192}\Big)^{\frac{1}{p}}\Bigg[\Bigg(\frac{2^{-s-4}}{(s+3)(s+4)}\Big|f”'(a)\Big|^q\\ \nonumber &+\frac{m2^{-s-4}\Big(s^2+11s+2^{s+4}(s-2)+34\Big)}{(s+1)(s+2)(s+3)(s+4)}\Big|f”’\Big(\frac{b}{m}\Big)\Big|^q-\frac{\mu}{960}\Big(\frac{b}{m}-a\Big)^2\Bigg)^{\frac{1}{q}}\\ \nonumber &+\Bigg(\frac{2^{-s-4}}{(s+3)(s+4)}\Big|f”'(b)\Big|^q+\frac{m2^{-s-4}\Big(s^2+11s+2^{s+4}(s-2)+34\Big)}{(s+1)(s+2)(s+3)(s+4)}\Big|f”’\Big(\frac{a}{m}\Big)\Big|^q -\frac{\mu}{960}\Big(b-\frac{a}{m}\Big)^2\Bigg)^{\frac{1}{q}}\Bigg], \end{align} where \(\frac{1}{p}+\frac{1}{q}=1\).
Proof. Using Lemma 9, the Hölder’s inequality and the strong \((s, m)\)-convexity of \(|f”’|^q\), we have\newpage
Theorem 13. Let \(f: [0,\infty)\to\mathbb{R}\) be a three times differentiable function on \((0,\infty)\) such that \(f”’\in L_1([a,b])\), for \(0\leq a< b\). If \(|f”’|^q\) is strongly \((s,m)\)-convex with modulus \(\mu\geq0\), for \((s,m)\in(0,1]\times(0,1]\) and \(q>1\), then the following inequality holds; \begin{eqnarray*} &&\bigg|\int_a^bf(x)dx-\frac{b-a}{6}\Big[f(a)+4f\Big(\frac{a+b}{2}\Big)+f(b)\Big]\bigg|\leq\frac{(b-a)^4}{24}\Big(\frac{1}{8(2p+1)(p+1)}\Big)^{\frac{1}{p}} \\&&\times\bigg[\Big(\frac{1}{2^{s+2}(s+1)(s+2)}\Big|f”'(a)\Big|^q+\frac{m(2s+2^{-s})}{4(s+1)(s+2)}\Big|f”’\Big(\frac{b}{m}\Big)\Big|^q-\frac{\mu}{64}\Big(\frac{b}{m}-a\Big)^2\Big)^{\frac{1}{q}}\\&&+\Bigg(\frac{1}{2^{s+2}(s+1)(s+2)} \Big|f”'(b)\Big|^q+\frac{m(2s+2^{-s})}{4(s+1)(s+2)}\Big|f”’\Big(\frac{a}{m}\Big)\Big|^q -\frac{\mu}{64}\Big(b-\frac{a}{m}\Big)^2\Bigg)^{\frac{1}{q}}\bigg], \end{eqnarray*} where \(\frac{1}{p}+\frac{1}{q}=1\).
Proof. Using Lemma 9, the Hölder’s inequality and the strong \((s, m)\)-convexity of \(|f”’|^q\), we have \begin{align*} \nonumber &\bigg|\int_a^bf(x)dx-\frac{b-a}{6}\Big[f(a)+4f\Big(\frac{a+b}{2}\Big)+f(b)\Big]\bigg|\leq(b-a)^4\int_0^1|p(t)||f”'(ta+(1-t)b)|dt\\ \nonumber &=\frac{(b-a)^4}{6}\bigg[\int_0^{\frac{1}{2}}\Big|t^2\Big(t-\frac{1}{2}\Big)\Big|\Big|f”'(ta+(1-t)b)\Big|dt+\int_{\frac{1}{2}}^1\Big|(t-1)^2\Big(t-\frac{1}{2}\Big)\Big|\Big|f”'(ta+(1-t)b)\Big|dt\bigg]\end{align*} \begin{align}\label{D1}\nonumber &=\frac{(b-a)^4}{6}\bigg[\int_0^{\frac{1}{2}}t^2\Big(\frac{1}{2}-t\Big)\Big|f”'(ta+(1-t)b)\Big|dt+\int_0^{\frac{1}{2}}t^2\Big(\frac{1}{2}-t\Big)\Big|f”'(tb+(1-t)a)\Big|dt\bigg]\\ \nonumber &\leq\frac{(b -a)^4}{6}\Big(\int_0^{\frac{1}{2}}t^{2p}\Big(\frac{1}{2}-t\Big)dt\Big)^{\frac{1}{p}}\bigg[\Big(\int_0^{\frac{1}{2}}\Big(\frac{1}{2}-t\Big)\Big|f”'(ta+(1-t)b)\Big|^qdt\Big)^{\frac{1}{q}}\\ \nonumber&+\Big(\int_0^{\frac{1}{2}}\Big(\frac{1}{2}-t\Big)\Big|f”'(tb+(1-t)a)\Big|^qdt\Big)^{\frac{1}{q}}\bigg]\\ \nonumber &\leq\frac{(b-a)^4}{6}\Big(\int_0^{\frac{1}{2}}t^{2p}\Big(\frac{1}{2}-t\Big)dt\Big)^{\frac{1}{p}}\bigg[\Big(\int_0^{\frac{1}{2}}\Big(\frac{1}{2}-t\Big)\Big(t^s\Big|f”'(a)\Big|^q+m(1-t)^s\Big|f”’\Big(\frac{b}{m}\Big)\Big|^q-\mu t(1-t)\Big(\frac{b}{m}-a\Big)^2 \Big)dt\Big)^{\frac{1}{q}}\\ \nonumber &+\Big(\int_0^{\frac{1}{2}}\Big(\frac{1}{2}-t\Big)\Big(t^s\Big|f”'(b)\Big|^q+m(1-t)^s\Big|f”’\Big(\frac{a}{m}\Big)\Big|^q-\mu t(1-t)\Big(b-\frac{a}{m}\Big)^2 \Big)dt\Big)^{\frac{1}{q}}\bigg]\\ \nonumber &=\frac{(b-a)^4}{6}\Big(\int_0^{\frac{1}{2}}t^{2p}\Big(\frac{1}{2}-t\Big)dt\Big)^{\frac{1}{p}}\bigg[\Big(\Big|f”'(a)\Big|^q\int_0^{\frac{1}{2}}t^{s}\Big(\frac{1}{2}-t\Big)dt\\ \nonumber &+m\Big|f”’\Big(\frac{b}{m}\Big)\Big|^q\int_0^{\frac{1}{2}}\Big(\frac{1}{2}-t\Big)(1-t)^sdt-\mu\Big(\frac{b}{m}-a\Big)^2\int_0^{\frac{1}{2}}t(1-t)\Big(\frac{1}{2}-t\Big)dt\Big)^{\frac{1}{q}}\\ \nonumber &+\Big(\Big|f”'(b)\Big|^q\int_0^{\frac{1}{2}}t^{s}\Big(\frac{1}{2}-t\Big)dt+m\Big|f”’\Big(\frac{a}{m}\Big)\Big|^q\int_0^{\frac{1}{2}}\Big(\frac{1}{2}-t\Big)(1-t)^sdt -\mu\Big(b-\frac{a}{m}\Big)^2\int_0^{\frac{1}{2}}t(1-t)\Big(\frac{1}{2}-t\Big)dt\Big)^{\frac{1}{q}}\bigg]. \end{align} The desired inequality follows from (5) by using the fact that $$ \int_0^{\frac{1}{2}}t^{2p}\Big(\frac{1}{2}-t\Big)dt=\frac{1}{2^{2p+2}(2p+1)(2p+2)}, $$ $$ \int_0^{\frac{1}{2}}t^{s}\Big(\frac{1}{2}-t\Big)dt=\frac{1}{2^{s+2}(s+1)(s+2)}, $$ $$ \int_0^{\frac{1}{2}}\Big(\frac{1}{2}-t\Big)(1-t)^{s}dt=\frac{2s+2^{-s}}{4(s+1)(s+2)}, $$ and $$ \int_0^{\frac{1}{2}}t(1-t)\Big(\frac{1}{2}-t\Big)dt=\frac{1}{64}. $$
Theorem 14. Let \(f: [0,\infty)\to\mathbb{R}\) be a three times differentiable function on \((0,\infty)\) such that \(f”’\in L_1([a,b])\), for \(0\leq a< b\). If \(|f”’|^q\) is strongly \((s,m)\)-convex with modulus \(\mu\geq0\), for \((s,m)\in(0,1]\times(0,1]\) and \(q>1\), then the following inequality holds;
Proof. Using Lemma 9, the Hölder’s inequality and the strong \((s, m)\)-convexity of \(|f”’|^q\), we have \begin{align*} \nonumber &\bigg|\int_a^bf(x)dx-\frac{b-a}{6}\Big[f(a)+4f\Big(\frac{a+b}{2}\Big)+f(b)\Big]\bigg|\leq(b-a)^4\int_0^1|p(t)||f”'(ta+(1-t)b)|dt\\ \nonumber &=\frac{(b-a)^4}{6}\bigg[\int_0^{\frac{1}{2}}\Big|t^2\Big(t-\frac{1}{2}\Big)\Big|\Big|f”'(ta+(1-t)b)\Big|dt+\int_{\frac{1}{2}}^1\Big|(t-1)^2\Big(t-\frac{1}{2}\Big)\Big|\Big|f”'(ta+(1-t)b)\Big|dt\bigg]\end{align*}
Proposition 15. Let \(f: [0,\infty)\to\mathbb{R}\) be a three times differentiable function on \((0,\infty)\) such that \(f”’\in L_1([a,b])\), for \(0\leq a< b\). If \(|f'''|\) is strongly \((s,m)\)-convex with modulus \(\mu\geq0\), for \((s,m)\in(0,1]\times(0,1]\), then for any division \(\mathcal{P}\) of \([a,b]\), the following inequality holds:\newpage \begin{align} \nonumber &\bigg|E_S(f,\mathcal{P})\bigg|\leq\frac{1}{6}\sum_{i=0}^{n-1}(x_{i+1}-x_i)^4\bigg[\frac{2^{-s-4}}{(s+3)(s+4)}\bigg(\Big|f'''(x_i)\Big|+\Big|f'''(x_{i+1})\Big|\bigg)\\ \nonumber &~+\frac{m2^{-s-4}\Big(s^2+11s+2^{s+4}(s-2)+34\Big)}{(s+1)(s+2)(s+3)(s+4)}\Bigg(\Big|f'''\Big(\frac{x_i}{m}\Big)\Big|+\Big|f'''\Big(\frac{x_{i+1}}{m}\Big)\Big|\Bigg)-\frac{\mu}{960}\bigg(\Big(\frac{x_{i+1}}{m}-x_i\Big)^2+\Big(x_{i+1}-\frac{x_i}{m}\Big)^2\bigg)\bigg]. \end{align}
Proof. First we observe that \begin{align} \nonumber E_S(f,\mathcal{P})=\sum_{i=0}^{n-1}\bigg[\int_{x_i}^{x_{i+1}}f(x)dx-\frac{f(x_i)+4f(x_i+h_i)+f(x_{i+1})}{3}h_i\bigg]. \end{align} It follows that