%------------------------------------------------------------------------------ % Here please write the date of submission of paper or its revisions: %------------------------------------------------------------------------------ % \documentclass[12pt, reqno]{amsart} \usepackage{amsmath, amsthm, amscd, amsfonts, amssymb, graphicx, color} \usepackage[bookmarksnumbered, colorlinks, plainpages]{hyperref} \hypersetup{colorlinks=true,linkcolor=red, anchorcolor=green, citecolor=cyan, urlcolor=red, filecolor=magenta, pdftoolbar=true} %\usepackage{draftwatermark} \usepackage{lineno} \textheight 22.5truecm \textwidth 14.5truecm \setlength{\oddsidemargin}{0.35in}\setlength{\evensidemargin}{0.35in} \setlength{\topmargin}{-.5cm} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{summary}[theorem]{Summary} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{problem}[theorem]{Problem} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \numberwithin{equation}{section} \usepackage{amscd} %\SetWatermarkText{Galley Proof}\SetWatermarkScale{4} \allowdisplaybreaks \begin{document} %\linenumbers \setcounter{page}{115} \begin{center}{\footnotesize Khayyam J. Math. 1 (2015), no. 1, 115--124}\\\end{center} \noindent\parbox{2.85cm}{\includegraphics*[keepaspectratio=true,scale=0.24]{KJM.jpg}} \vspace{0.5cm} \title[Inequalities for $\alpha-,m-,\left( \alpha ,m\right)-$logarithmically convex functions]{Some integral inequalities for $\alpha-,m-,\left( \alpha ,m\right)-$logarithmically convex functions} \author[M. TUN\c{C}, E. Y\"{U}KSEL]{MEVL\"{U}T TUN\c{C}$^1$$^{*}$, EBRU Y\"{U}KSEL$^2$} \address{$^{1}$ Department of Mathematics, Faculty of Science and Arts, Mustafa Kemal University, Hatay, 31000, Turkey.} \email{mevluttttunc@gmail.com} \address{$^{2}$ Department of Mathematics, Faculty of Science and Arts, A\u{g}r\i\ \.{I}brahim \c{C}e\c{c}en University, A\u{g}r\i , 04000, Turkey.} \email{yuksel.ebru90@hotmail.com} \dedicatory{\rm Communicated by S. Hejazian} \subjclass[2010]{Primary 26A15; Secondary 26A51, 26D10.} \keywords{ $\alpha-,m-,\left( \alpha ,m\right)-$logarithmically convex, Hadamard's inequality, H\"{o}lder's inequality, power mean inequality, Cauchy's inequality.} \date{Received: 12 November 2014; Revised: 15 December 2014; Accepted: 23 December 2014. \newline \indent $^{*}$ Corresponding author} \begin{abstract} In this paper, the authors establish some Hermite-Hadamard type inequalities by using elementary inequalities for functions whose first derivative absolute values are $\alpha $-, $m$-$,$ $\left( \alpha ,m\right) $% -logarithmically convex. \end{abstract} \maketitle \section{Introduction and preliminaries} In this section, we will present definitions and some results used in this paper. Let \ $f:I\subseteq %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion \rightarrow %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion $ be a convex mapping defined on the interval $I$\ of real numbers and $% a,b\in I$, with $a1$. If the new mapping $|f^{\prime }\left( x\right) |^{p/p-1}$ is convex on $[a,b]$, then% \begin{eqnarray} &&\left\vert \frac{f\left( a\right) +f\left( b\right) }{2}-\frac{1}{b-a}% \int_{a}^{b}f\left( x\right) dx\right\vert \\ &\leq &\frac{b-a}{2\left( p+1\right) ^{1/p}}\left[ \frac{\left\vert f^{\prime }\left( a\right) \right\vert ^{p/\left( p-1\right) }+\left\vert f^{\prime }\left( b\right) \right\vert ^{p/\left( p-1\right) }}{2}\right] ^{\left( p-1\right) /p}. \notag \end{eqnarray} \end{theorem} The aim of this paper is to establish some integral inequalities of Hermite-Hadamard type for $\alpha $-, $m$-$,$ $\left( \alpha ,m\right) $% -logarithmically convex functions. \section{Hadamard Type Inequalities} In order to prove our main theorems, we need the following lemma \cite{lmmm}. \begin{lemma} \label{l1}\cite{lmmm} Let $f:\ I\subset %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion \rightarrow %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion $ be a differentiable mapping on $I^{\circ }$\textit{, }$a,b\in $ $I^{\circ } $ with $a$ $<$ $b$. If $f^{\prime }\in $ $L\left[ a,b\right] ,$ then the following equality holds: \begin{eqnarray} &&\frac{f\left( a\right) +f\left( b\right) }{2}-\frac{1}{b-a}% \int_{a}^{b}f\left( x\right) dx \label{a} \\ &=&\frac{b-a}{2}\int_{0}^{1}\int_{0}^{1}\left[ f^{\prime }\left( ta+\left( 1-t\right) b\right) -f^{\prime }\left( sa+\left( 1-s\right) b\right) \right] \left( s-t\right) dtds. \notag \end{eqnarray} \end{lemma} A simple proof of this equality can be also done integrating by parts in the right hand side (see \cite{lmmm}). The next theorems gives a new result of the upper Hermite-Hadamard inequality for $\alpha $-, $m$-, $\left( \alpha ,m\right) $-logarithmically convex functions. \begin{theorem} \label{t1}Let$\ I\supset \left[ 0,\infty \right) $ be an open interval and let $f:\ I\rightarrow \left( 0,\infty \right) $ be a differentiable function on $I$ such that $f^{\prime }\in L\left( a,b\right) $ for $0\leq a1 $ with $\frac{1}{p}+\frac{1}{q}=1,$ then% \begin{eqnarray} &&\left\vert \frac{f\left( a\right) +f\left( b\right) }{2}-\frac{1}{b-a}% \int_{a}^{b}f\left( x\right) dx\right\vert \\ &\leq &\left\{ \begin{array}{cc} \left( b-a\right) \left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{m}\left( \frac{2}{\left( p+1\right) \left( p+2\right) }\right) ^{\frac{1}{p}},\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } & \eta =1 \\ \left( b-a\right) \left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{m}\left( \frac{2}{\left( p+1\right) \left( p+2\right) }\right) ^{\frac{1}{p}}\times \left( \frac{\eta \left( \alpha q,\alpha q\right) -1}{% \ln \eta \left( \alpha q,\alpha q\right) }\right) ^{\frac{1}{q}}, & \eta <1% \end{array}% \right. \notag \end{eqnarray}% where $\eta \left( \alpha ,\alpha \right) $ is same as Theorem \ref{t1}. \end{theorem} \begin{proof} Since $\left\vert f^{\prime }\right\vert ^{q}$ is an $\left( \alpha ,m\right) $-logarithmically convex on $\left[ 0,\frac{b}{m}\right] $, from Lemma \ref{l1} and the well known H\"{o}lder inequality, we have% \begin{eqnarray} && \label{y} \\ &&\left\vert \frac{f\left( a\right) +f\left( b\right) }{2}-\frac{1}{b-a}% \int_{a}^{b}f\left( x\right) dx\right\vert \notag \\ &\leq &\frac{b-a}{2}\int_{0}^{1}\int_{0}^{1}\left\vert \left( f^{\prime }\left( ta+\left( 1-t\right) b\right) \right) -\left( f^{\prime }\left( sa+\left( 1-s\right) b\right) \right) \right\vert \left\vert s-t\right\vert dtds \notag \\ &\leq &\frac{b-a}{2}\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert \left\vert f^{\prime }\left( a\right) \right\vert ^{t^{\alpha }}\left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{m\left( 1-t^{\alpha }\right) }dtds \notag \\ &&+\frac{b-a}{2}\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert \left\vert f^{\prime }\left( a\right) \right\vert ^{s^{\alpha }}\left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{m\left( 1-s^{\alpha }\right) }dtds \notag \\ &\leq &\frac{b-a}{2}\left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{m}\left( \int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert ^{p}dtds\right) ^{\frac{1}{p}} \notag \\ &&\times \left[ \left( \int_{0}^{1}\int_{0}^{1}\eta ^{qt^{\alpha }}dtds\right) ^{\frac{1}{q}}+\left( \int_{0}^{1}\int_{0}^{1}\eta ^{qs^{\alpha }}dtds\right) ^{\frac{1}{q}}\right] \notag \end{eqnarray}% If $\ \eta =1,$ by (\ref{1}), we obtain% \begin{eqnarray*} &&\left\vert \frac{f\left( a\right) +f\left( b\right) }{2}-\frac{1}{b-a}% \int_{a}^{b}f\left( x\right) dx\right\vert \\ &\leq &\left( b-a\right) \left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{m}\left( \int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert ^{p}dtds\right) ^{\frac{1}{p}} \\ &=&\left( b-a\right) \left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{m}\left( \frac{2}{\left( p+1\right) \left( p+2\right) }\right) ^{\frac{1}{p}} \end{eqnarray*}% If $\eta <1,$ by (\ref{1}), we obtain% \begin{eqnarray} &&\left\vert \frac{f\left( a\right) +f\left( b\right) }{2}-\frac{1}{b-a}% \int_{a}^{b}f\left( x\right) dx\right\vert \\ &\leq &\frac{b-a}{2}\left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{m}\left( \int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert ^{p}dtds\right) ^{\frac{1}{p}} \notag \\ &&\times \left[ \left( \int_{0}^{1}\int_{0}^{1}\eta ^{qt^{\alpha }}dtds\right) ^{\frac{1}{q}}+\left( \int_{0}^{1}\int_{0}^{1}\eta ^{qs^{\alpha }}dtds\right) ^{\frac{1}{q}}\right] \notag \\ &=&\left( b-a\right) \left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{m}\left( \frac{2}{\left( p+1\right) \left( p+2\right) }\right) ^{\frac{1}{p}}\times \left( \frac{\eta \left( \alpha q,\alpha q\right) -1}{% \ln \eta \left( \alpha q,\alpha q\right) }\right) ^{\frac{1}{q}} \notag \end{eqnarray} \textit{which completes the proof.} \end{proof} \begin{corollary} Let$\ I\supset \left[ 0,\infty \right) $ be an open interval and let $f:\ I\rightarrow \left( 0,\infty \right) $ be a differentiable function on $I$ such that $f^{\prime }\in L\left( a,b\right) $ for $0\leq a0$ with $\mu _{1}+\tau _{1}=1$ and $\mu _{2}+\tau _{2}=1$, then \begin{align} &\left\vert \frac{f\left( a\right) +f\left( b\right) }{2}-\frac{1}{b-a}% \int_{a}^{b}f\left( x\right) dx\right\vert \leq \frac{\left( b-a\right) }{2}% \left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{m} \\ &\times \left\{ \begin{array}{ll} \frac{2\mu _{1}^{3}}{\left( 2\mu _{1}+1\right) \left( \mu _{1}+1\right) }+% \frac{2\mu _{2}^{3}}{\left( 2\mu _{2}+1\right) \left( \mu _{2}+1\right) }% +\tau _{1}+\tau _{2}, & \eta =1 \\ \frac{2\mu _{1}^{3}}{\left( 2\mu _{1}+1\right) \left( \mu _{1}+1\right) }+% \frac{2\mu _{2}^{3}}{\left( 2\mu _{2}+1\right) \left( \mu _{2}+1\right) }% +\tau _{1}\frac{\eta \left( \frac{\alpha }{\tau _{1}},\frac{\alpha }{\tau_{1}}\right) -1}{\ln \eta \left( \frac{\alpha }{\tau _{1}},\frac{\alpha }{% \tau _{1}}\right) }+\tau _{2}\frac{\eta \left( \frac{\alpha }{\tau _{2}},% \frac{\alpha }{\tau _{2}}\right) -1}{\ln \eta \left( \frac{\alpha }{\tau _{2}},\frac{\alpha }{\tau _{2}}\right) }, & \eta <1\nonumber% \end{array}% \right. \end{align}% where $\eta \left( \alpha ,\alpha \right) $ is same as Theorem \ref{t1}. \end{theorem} % \begin{proof} Since $\left\vert f^{\prime }\right\vert ^{q}$ is an $\left( \alpha ,m\right) $-logarithmically convex on $\left[ 0,\frac{b}{m}\right] $, from Lemma \ref{l1}, we have% \begin{eqnarray} &&\left\vert \frac{f\left( a\right) +f\left( b\right) }{2}-\frac{1}{b-a}% \int_{a}^{b}f\left( x\right) dx\right\vert \label{f} \\ &\leq &\frac{\left( b-a\right) }{2}\int_{0}^{1}\int_{0}^{1}\left\vert \left( f^{\prime }\left( ta+\left( 1-t\right) b\right) \right) -\left( f^{\prime }\left( sa+\left( 1-s\right) b\right) \right) \right\vert \left\vert s-t\right\vert dtds \notag \\ &\leq &\frac{\left( b-a\right) }{2}\left[ \int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert \left\vert f^{\prime }\left( a\right) \right\vert ^{t^{\alpha }}\left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{m\left( 1-t^{\alpha }\right) }dtds\right] \notag \\ &&+\frac{\left( b-a\right) }{2}\left[ \int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert \left\vert f^{\prime }\left( a\right) \right\vert ^{s^{\alpha }}\left\vert f^{\prime }\left( \frac{b}{m}\right) \right\vert ^{m\left( 1-s^{\alpha }\right) }dtds\right] \notag \\ &=&\frac{\left( b-a\right) }{2}\left\vert f^{\prime }\left( \frac{b}{m}% \right) \right\vert ^{m}\left[ \int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert \eta ^{t^{\alpha }}dtds+\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert \eta ^{s^{\alpha }}dtds\right] \notag \end{eqnarray}% for all $t\in \left[ 0,1\right] .$ Using the well known inequality $rt\leq \mu r^{\frac{1}{\mu }}+\tau t^{\frac{1}{\tau }},$ on the right side of (\ref% {f}), we get \begin{eqnarray} &&\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert \eta ^{t^{\alpha }}dtds+\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert \eta ^{s^{\alpha }}dtds \label{w} \\ &\leq &\mu _{1}\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert ^{\frac{1}{% \mu _{1}}}dtds+\tau _{1}\int_{0}^{1}\int_{0}^{1}\eta ^{\frac{t^{\alpha }}{% \tau _{1}}}dtds \notag \\ &&+\mu _{2}\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert ^{\frac{1}{\mu _{2}}}dtds+\tau _{2}\int_{0}^{1}\int_{0}^{1}\eta ^{\frac{s^{\alpha }}{\tau _{2}}}dtds \notag \end{eqnarray}% When $\eta =1,$ by (\ref{1}), we get \begin{eqnarray} &&\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert \eta ^{t^{\alpha }}dtds+\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert \eta ^{s^{\alpha }}dtds \label{e} \\ &\leq &\frac{2\mu _{1}^{3}}{\left( 2\mu _{1}+1\right) \left( \mu _{1}+1\right) }+\frac{2\mu _{2}^{3}}{\left( 2\mu _{2}+1\right) \left( \mu _{2}+1\right) }+\tau _{1}+\tau _{2} \notag \end{eqnarray}% When $\eta <1,$ by (\ref{1}), we get% \begin{eqnarray} && \label{r} \\ &&\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert \eta ^{t^{\alpha }}dtds+\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert \eta ^{s^{\alpha }}dtds \notag \\ &\leq &\mu _{1}\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert ^{\frac{1}{% \mu _{1}}}dtds+\tau _{1}\int_{0}^{1}\int_{0}^{1}\eta ^{\frac{t^{\alpha }}{% \tau _{1}}}dtds \notag \\ &&+\mu _{2}\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert ^{\frac{1}{\mu _{2}}}dtds+\tau _{2}\int_{0}^{1}\int_{0}^{1}\eta ^{\frac{s^{\alpha }}{\tau _{2}}}dtds \notag \\ &\leq &\mu _{1}\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert ^{\frac{1}{% \mu _{1}}}dtds+\mu _{2}\int_{0}^{1}\int_{0}^{1}\left\vert s-t\right\vert ^{% \frac{1}{\mu _{2}}}dtds \notag \\ &&+\tau _{1}\int_{0}^{1}\int_{0}^{1}\eta ^{\frac{\alpha t}{\tau _{1}}% }dtds+\tau _{2}\int_{0}^{1}\int_{0}^{1}\eta ^{\frac{\alpha s}{\tau _{2}}}dtds \notag \\ &=&\frac{2\mu _{1}^{3}}{\left( 2\mu _{1}+1\right) \left( \mu _{1}+1\right) }+% \frac{2\mu _{2}^{3}}{\left( 2\mu _{2}+1\right) \left( \mu _{2}+1\right) } \notag \\ &&+\tau _{1}\frac{\eta \left( \frac{\alpha }{\tau _{1}},\frac{\alpha }{\tau _{1}}\right) -1}{\ln \eta \left( \frac{\alpha }{\tau _{1}},\frac{\alpha }{% \tau _{1}}\right) }+\tau _{2}\frac{\eta \left( \frac{\alpha }{\tau _{2}},% \frac{\alpha }{\tau _{2}}\right) -1}{\ln \eta \left( \frac{\alpha }{\tau _{2}% },\frac{\alpha }{\tau _{2}}\right) } \notag \end{eqnarray} from (\ref{f})-(\ref{r}), which completes the proof. \end{proof} \begin{corollary} Under the assumptions of Theorem \ref{t4}, and $\mu =\mu _{1}=\mu _{2}>0,$ $% \tau =\tau _{1}=\tau _{2}>0$ with $\mu +\tau =1,$ then we have% \begin{eqnarray*} &&\left\vert \frac{f\left( a\right) +f\left( b\right) }{2}-\frac{1}{b-a}% \int_{a}^{b}f\left( x\right) dx\right\vert \\ &\leq &\frac{\left( b-a\right) }{2}\left\vert f^{\prime }\left( \frac{b}{m}% \right) \right\vert ^{m}\times \left\{ \begin{array}{cc} \frac{4\mu ^{3}}{\left( 2\mu +1\right) \left( \mu +1\right) }+2\tau , & \eta =1 \\ \frac{4\mu ^{3}}{\left( 2\mu +1\right) \left( \mu +1\right) }+2\tau \frac{% \eta \left( \frac{\alpha }{\tau },\frac{\alpha }{\tau }\right) -1}{\ln \eta \left( \frac{\alpha }{\tau },\frac{\alpha }{\tau }\right) }, & \eta <1% \end{array}% \right. \end{eqnarray*} \end{corollary} \bibliographystyle{amsplain} \begin{thebibliography}{99} \bibitem{bai} R.F. 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