%------------------------------------------------------------------------------ % 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} \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} \input{mathrsfs.sty} \begin{document} \setcounter{page}{85} \noindent\parbox{2.95cm}{\includegraphics*[keepaspectratio=true,scale=0.125]{AFA.jpg}} \noindent\parbox{4.85in}{\hspace{0.1mm}\\[1.5cm]\noindent \qquad Ann. Funct. Anal. 2 (2011), no. 2, 85--91\\ {\footnotesize \qquad \textsc{\textbf{$\mathscr{A}$}nnals of \textbf{$\mathscr{F}$}unctional \textbf{$\mathscr{A}$}nalysis}\\ \qquad ISSN: 2008-8752 (electronic)\\ \qquad URL: \textcolor[rgb]{0.00,0.00,0.99}{www.emis.de/journals/AFA/} }\\[.5in]} \title[On strongly $h$-convex]{On strongly $h$-convex functions} \author[H. Angulo, J. Gim\'{e}nez, A.M. Moros, K. Nikodem]{Hiliana Angulo$^1$, Jos\'{e} Gim\'{e}nez$^1$, Ana Milena Moros $^1$ \\ and Kazimierz Nikodem$^{*}$$^2$} \address{$^{1}$ Facultad de Ciencias, Dpto. de Matem\'aticas, Universidad de Los Andes, M\'erida, Venezuela.} \email{\textcolor[rgb]{0.00,0.00,0.84}{hiliana@ula.ve}} \email{\textcolor[rgb]{0.00,0.00,0.84}{jgimenez@ula.ve}} \email{\textcolor[rgb]{0.00,0.00,0.84}{milena13\_m@hotmail.com}} \address{$^{2}$ Department of Mathematics and Computer Science, University of Bielsko-Bia{\l}a, ul.\ Wil\-lo\-wa~2, 43-309 Bielsko-Bia{\l}a, Poland.} \email{\textcolor[rgb]{0.00,0.00,0.84}{knikodem@ath.bielsko.pl}} \dedicatory{{\rm Communicated by M. S. Moslehian}} \subjclass[2010]{Primary 26A51; Secondary 46C15, 39B62.} \keywords{Hermite--Hadamard inequality, $h$-convex function, strongly convex function, inner product space.} \date{Received: 30 September 2011; Accepted: 14 October 2011. \newline \indent $^{*}$ Corresponding author} \begin{abstract} We introduce the notion of strongly $h$-convex functions (defined on a normed space) and present some properties and representations of such functions. We obtain a characterization of inner product spaces involving the notion of strongly $h$-convex functions. Finally, a Hermite--Hadamard--type inequality for strongly $h$-convex functions is given. \end{abstract} \maketitle \section{Introduction} \noindent Let $I$ be an interval in $\mathbb{R}$ and $h: (0,1)\rightarrow (0,\infty)$ be a given function. Following Varo\u{s}anec \cite{SV07}, a function $f:I \rightarrow \mathbb{R}$ is said to be \emph{$h$-convex} if \begin{equation}\label{ecu1} f(tx+(1-t)y)\leq h(t)f(x)+h(1-t)f(y) \end{equation} for all $x, y \in I$ and $t \in (0,1)$. This notion unifies and generalizes the known classes of convex functions, $s$ - convex functions, Godunova-Levin functions and $P$-functions, which are obtained by putting in (\ref{ecu1}) $h(t)=t$, $h(t)=t^{s}$, $h(t)=\frac{1}{t}$, and $h(t)=1$, respectively. Many properties of them can be found, for instance, in \cite{B78, BV09, DPP95, PR99, SSO10, SSY08, SV07}. Recall also that a function $f: I \rightarrow \mathbb{R}$ is called \emph{strongly convex with modulus $c>0$}, if \begin{equation*} f(tx+(1-t)y) \leq t f(x)+ (1-t) f(y) - ct(1-t)(x-y)^{2} \end{equation*} for all $x, y \in I$ and $t \in (0,1)$. Strongly convex functions have been introduced by Polyak \cite{BP66}, and they play an important role in optimization theory and mathematical economics. Various properties and applications of them can be found in the literature (see, for instance, \cite{HUL01, MN10, NP11, POL96, JV82, JV83} and the references therein). In this paper we introduce the notion of strongly $h$-convex functions defined in normed spaces and present some examples and properties of them. In particular we obtain a representation of strongly $h$-convex functions in inner product spaces and, using the methods of \cite{NP11}, we give a characterization of inner product spaces, among normed spaces, that involves the notion of strongly $h$-convex function. Finally, a version of Hermite--Hadamard-type inequalities for strongly $h$-convex functions is presented. This result generalizes the Hermite--Hadamard-type inequalities obtained in \cite{MN10} for strongly convex functions, and for $c=0$, coincides with the classical Hermite--Hadamard inequalities, as well as the corresponding Hermite--Hadamard-type inequalities for h-convex functions, $s$-convex functions, Godunova-Levin functions and $P$-functions presented in \cite{SSY08, DF97, DPP95}, respectively. \section{Some basic properties and representations} In what follows $(X, \|\cdot\|)$ denotes a real normed space, $D$ stands for a convex subset of $X$, $h: (0,1)\rightarrow (0,\infty)$ is a given function and $c$ is a positive constant. We say that a function $f:D \rightarrow \mathbb{R}$ is \emph{strongly h-convex with modulus $c$} if \begin{equation}\label{ecu2} f(tx+(1-t)y)\leq h(t)f(x)+h(1-t)f(y)- ct(1-t)\| x-y \| ^{2} \end{equation} for all $x, y \in D$ and $t \in (0,1)$. In particular, if $f$ satisfies (\ref{ecu2}) with $h(t)=t$, $h(t)=t^{s}$ $(s\in (0,1))$, $h(t)=\frac{1}{t}$, and $h(t)=1$, then $f$ is said to be strongly convex, strongly s-convex, strongly Godunova-Levin functions and strongly $P$-function, respectively. The notion of $h$-convex function corresponds to the case $c=0$. We start with two lemmas which give some relationships between strongly $h$-convex functions and $h$-convex functions in the case where $X$ is a real inner product space (that is, the norm $\| \cdot \|$ is induced by an inner product: $ \| x\|^2 := < x \mid x>$). \begin{lemma} Let $(X, \parallel \cdot \parallel)$ be a real inner product space, D be a convex subset of $X$ and $c >0$. Assume that $h:(0,1) \rightarrow (0,\infty)$ satisfies the condition \begin{equation}\label{ecu3} h(t) \geq t, \qquad t \in (0, 1).\end{equation} If $g: D \rightarrow \mathbb{R}$ is $h$-convex, then $f: D \rightarrow \mathbb{R}$ defined by $f(x)= g(x)+c \| x \|^{2}, \qquad x \in D$ is strongly $h$-convex with modulus $c$. \end{lemma} \begin{proof} Assume that $g$ is $h$-convex. Then $\begin{array}{cl} & f(tx+(1-t)y) \\ = & g(tx+(1-t)y)+ c \| (tx+(1-t)y) \|^{2}\\ \leq & h(t)g(x)+h(1-t)g(y)+ c \| (tx+(1-t)y) \|^{2}\\ = & h(t)f(x)+h(1-t)f(y) - ch(t)\|x\|^{2}- ch(1-t)\|y\|^{2}+ c \| (tx+(1-t)y) \|^{2} \\ \leq & h(t)f(x)+h(1-t)f(y) - ct\|x\|^{2}- c(1-t)\|y\|^{2} \\ + & c (t^{2}\|x\|^{2}+2t(1-t)+ (1-t)^{2} \|y\|^{2})\\ = & h(t)f(x)+h(1-t)f(y) - ct(1-t)\|x-y\|^{2}, \end{array}$ \vspace{2ex} which shows that $f$ is strongly $h$-convex with modulus $c$. \end{proof} In a similar way we can prove the next lemma \begin{lemma} let $(X, \| \cdot \|)$ be a real inner product space, D be a convex subset of $X$ and $c >0$. Assume that $h:(0,1) \rightarrow (0,\infty)$ satisfies the condition \begin{equation*} h(t) \leq t, \qquad t \in (0, 1). \end{equation*} If $f: D \rightarrow \mathbb{R}$ is strongly $h$-convex with modulus $c$, then there exists an $h$-convex function $g: D \rightarrow \mathbb{R}$ such that $f(x)= g(x)+c \| x \|^{2}$, where $x \in D$. \end{lemma} \begin{remark}For strongly convex functions (i.e. if $f$ satisfies (\ref{ecu2}) with $h(t)=t$, $t\in(0,1)$) defined on a convex subset $D$ of an inner product space $X$ the following characterization holds (see \cite{NP11, JV83}, cf. also \cite[Prop 1.12]{HUL01} for the case $X=\mathbb{R}^{n}$): A function $f: D \rightarrow \mathbb{R}$ is strongly convex with modulus $c$ if and only if $g= f - c \| \cdot \|^{2}$ is convex. This result follows also from Lemma 1 and Lemma 2 above. However, an analogous characterization is not true for arbitrary $h$. \end{remark} \begin{example} Let $h(t):=1$, $t\in (0,1)$. Then $f: [-1,1] \rightarrow \mathbb{R}$ defined by $f(x):= 1$, $x \in [-1,1]$, is strongly $h$-convex with modulus $c=1.$ Indeed, for every $x, y \in [-1,1]$ and $t\in (0,1)$ we have $$f(tx+(1-t)y) =1 \leq 2- t(1-t)(x-y)^{2}= f(x)+f(y)-t(1-t)(x-y)^{2}.$$ However, $g(x):= f(x)-x^{2}$ is not $h$-convex. For instance, $$g\bigg(\frac{1}{2}(-1)+\frac{1}{2}1\bigg)=1 > 0 =g(-1)+ g(1).$$ Now, let $h(t):=t^{2}$, $t\in (0,1)$. Then $g: [-1,1] \rightarrow \mathbb{R}$ given by $g(x):=1$, $x \in [-1,1]$, is $h$-convex, but $f(x):=g(x)+x^{2}$, $x \in [-1,1]$, is not strongly $h$-convex with modulus 1. For instance, $$f\bigg(\frac{1}{2}(-1)+\frac{1}{2}1\bigg)=1 > 0 = \frac{1}{4}f(-1)+ \frac{1}{4}f(1)- \frac{1}{4}(1+1)^{2}.$$ \end{example} \begin{remark} Condition (\ref{ecu3}) is satisfied, for instance, for the following functions defined in $(0,1)$: $h_{1}(t)=t$, $h_{2}(t)=t^{s}$ $(s \in (0,1))$, $h_{3}(t)=\frac{1}{t}$, $h_{4}(t)=1$. Thus, if a function $g: I \rightarrow \mathbb{R}$ is convex, $s$-convex, a Godunova-Levin function or a $P$-function, then by Lemma 1, $f: I \rightarrow \mathbb{R}$ given by $f(x)=g(x)+cx^{2}$ is strongly $h$-convex with $h=h_{i}$, respectively. \end{remark} \begin{remark} We can easily check that if a function $g: D \rightarrow [0,\infty)$, defined on a convex subset $D$ of a normed space $X$, is convex then it is $h$-convex with any $h:(0,1) \rightarrow(0,\infty)$ satisfying (\ref{ecu3}). Therefore, if $X$ is an inner product space then, by Lemma 1, $f: D \rightarrow [0,\infty)$ given by $f(x)= g(x)+ c \|x\|^{2}$ is strongly $h$-convex. \end{remark} \section{A characterization of inner product spaces via strong $h$-convexity} The assumption that $X$ is an inner product space in Lemma 1 is essential. Moreover, it appears that the fact that for every $h$-convex function $g:X \rightarrow \mathbb{R}$ the function $f=g+\|\cdot\|^{2}$ is strongly $h$-convex characterizes inner product spaces among normed spaces. Similar characterizations of inner product spaces by strongly convex and strongly midconvex functions are presented in \cite{NP11}. \begin{theorem} Let $(X,\|\cdot\|)$ be a real normed space. Assume that $h:(0,1)\rightarrow \mathbb{R}$ satisfies (\ref{ecu3}) and $h(\frac{1}{2})=\frac{1}{2}$. The following conditions are equivalent: \vspace{1ex} \begin{enumerate} \item[1.] $(X,\|\cdot\|)$ is an inner product space; \vspace{1ex} \item[2.] For every $c>0$ and for every $h$-convex function $g:D\rightarrow \mathbb{R} $ defined on a convex subset $D$ of $X$, the function $f=g+c\|\cdot\|^{2}$ is strongly $h$-convex with modulus $c$; \vspace{1ex} \item[3.] $\|\cdot\|^{2}: X \rightarrow \mathbb{R}$ is strongly $h$-convex with modulus $1$. \end{enumerate} \end{theorem} \begin{proof} The implication $ 1 \Rightarrow 2$ follows be Lemma 1. \\ To see that $ 2 \Rightarrow 3$ take $g=0$. Clearly, $g$ is $h$-convex, whence $f=c\| \cdot \|^{2}$ is strongly $h$-convex with modulus $c$. Consequently, $\| \cdot\|^{2}$ is strongly $h$-convex with modulus 1. \\ To prove $ 3 \Rightarrow 1$ observe that by the strong $h$-convexity of $\| \cdot\|^{2}$ and the assumption $h(\frac{1}{2})=\frac{1}{2}$, we have \begin{equation*} \Big\| \frac{x+y}{2}\Big\|^{2} \leq \frac{1}{2}\|x\|^{2}+ \frac{1}{2}\|y\|^{2} -\frac{1}{4}\|x-y\|^{2} \end{equation*} \\ and hence \begin{equation}\label{ecu6} \| x+y\|^{2}+\| x-y\|^{2} \leq 2\|x\|^{2} + 2\|y\|^{2} \end{equation} \\ for all $x, y \in X.$ Now, putting $u=x+y$ and $v=x-y$ in (\ref{ecu6}) we get \begin{equation}\label{ecu7} 2\|u\|^{2} + 2\|v\|^{2} \leq \| u+v\|^{2}+\| u-v\|^{2} \end{equation} \\ for all $u,v \in X $ Conditions (\ref{ecu6}) and (\ref{ecu7}) mean that the norm $\| \cdot \|$ satisfies the parallelogram law, which implies, by the classical Jordan-Von Neumann theorem, that $(X,\| \cdot \| )$ is an inner product space. \end{proof} \section{Hermite--Hadamard-type Inequalities} It is known that if a function $f :I \rightarrow \mathbb{R}$ is convex then $$f\bigg(\frac{a+b}{2}\bigg) \leq \frac{1}{b-a} \int_{a}^{b}f(x)dx \leq \frac{f(a)+f(b)}{2} $$ for all $a, b \in I$, $a0$, then \begin{align}\label{H-H} \frac{1}{2h(\frac{1}{2})}\bigg[f\big(\frac{a+b}{2}\big)+\frac{c}{12}(b-a)^{2}\bigg] &\leq \frac{1}{b-a}\int_{a}^{b}f(x)dx \nonumber \\ & \leq(f(a)+f(b))\int_{0}^{1}h(t)dt - \frac{c}{6}(b- a)^{2} \end{align} for all $a, b \in I $, $a < b$ \end{theorem} \begin{proof} Fix $a, b \in I $, $a < b$, and take $u=ta + (1-t)b$, $v=(1-t)a + tb$. Then, the strong $h$-convexity of $f$ implies \begin{eqnarray*} f(\frac{a+b}{2})& = & f(\frac{u+v}{2}) \\ & \leq & h(\dfrac{1}{2})f(u) + h(\dfrac{1}{2})f(v)- \frac{c}{4}(u-v)^{2}\\ & = & h(\dfrac{1}{2})[f(ta+(1-t)b)+ f((1-t)a+tb)] \\&&- \dfrac{c}{4} ((2t-1)a +(1-2t)b)^{2}. \end{eqnarray*} Integrating the above inequality over the interval $(0,1),$ we obtain \vspace{2ex} $\begin{array}{cl} &\displaystyle{f\Big(\frac{a+b}{2}\Big)}\\ \leq & \displaystyle{h\Big(\frac{1}{2}\Big)\Bigg[\int_{0}^{1}f(ta+(1-t)b)dt+ \int_{0}^{1}f((1-t)a+tb)dt\Bigg]}\\ & - \displaystyle{\frac{c}{4}\int_{0}^{1}\Big((2t-1)a +(1-2t)b\Big)^{2}dt} \\ = & \displaystyle{h\Big(\frac{1}{2}\Big)\frac{2}{b-a}\int_{a}^{b}f(x)dx - \frac{c}{12}(b-a)^{2}} \end{array}$ \vspace{2ex} which gives the left-hand side inequality of \eqref{H-H}. For the proof of the right-hand side inequality of \eqref{H-H} we use inequality (2). Integrating over the interval $(0,1)$, we get \vspace{2ex} \begin{eqnarray*} \frac{1}{b-a} \int_{a}^{b} f(x)dx & = & \int_{0}^{1}f((1-t)a+tb)dt\\ &\leq & f(a)\int_{0}^{1} h(1-t)dt + f(b)\int_{0}^{1} h(t)dt \\&&- c (b-a)^{2}\int_{0}^{1} t(1-t)dt \\ & = & (f(a)+f(b))\int_{0}^{1} h(t)dt - \frac{c}{6}(b-a)^{2} \end{eqnarray*} \vspace{2ex} which gives the right-hand side inequality of \eqref{H-H}. \end{proof} \begin{remark} \begin{enumerate} \item In the case $c=0$ the Hermite--Hadamard-type inequalities \eqref{H-H} coincide with the Hermite--Hadamard-type inequalities for $h$-convex functions proved by Sarikaya, Saglam and Yildirim in \cite{SSY08}. \vspace{2ex} \item If $h(t)=t$, $t\in(0,1)$, then the inequalities \eqref{H-H} reduce to $$ f\Big(\frac{a+b}{2}\Big)+ \frac{c}{12}(b-a)^{2}\leq \frac{1}{b-a}\int_{a}^{b}f(x)dx \leq \frac{f(a)+f(b)}{2} - \frac{c}{6}(b-a)^{2}.$$ These Hermite--Hadamard-type inequalities for strongly convex functions have been proved by Merentes and Nikodem in \cite{MN10}. For $c=0$ we get the classical Hermite--Hadamard inequalities. \vspace{2ex} \item If $h(t)=t^{s}$, $t\in(0,1)$, then the inequalities \eqref{H-H} give $$ 2^{s-1}\Bigg[f\Big(\frac{a+b}{2}\Big)+ \frac{c}{12}(b-a)^{2}\Bigg] \leq \frac{1}{b-a}\int_{a}^{b}f(x)dx\leq \frac{f(a)+f(b)}{s+1} - \frac{c}{6}(b-a)^{2}.$$ For $c=0$ it reduces to the Hermite--Hadamard-type inequalities for $s$-convex functions proved by Dragomir and Fitzpatrik \cite{DF97}. \vspace{2ex} \item If $h(t)=\frac{1}{t}$, $t \in(0,1)$, then the inequalities \eqref{H-H} give $$\frac{1}{4}f\Big(\frac{a+b}{2}\Big)+\frac{c}{48}(b-a)^{2}\leq \frac{1}{b-a}\int_{a}^{b} f(x)dx \quad (\leq + \infty). $$ The case $c=0$ corresponds to the Hermite--Hadamard-type inequalities for Godunova--Levin functions obtained by Dragomir, Pe\u{c}ari\'{c} and Persson \cite{DPP95}. \vspace{2ex} \item If $h(t)=1$, $t \in(0,1)$, then the inequalities \eqref{H-H} reduce to $$ \frac{1}{2}f\Big(\frac{a+b}{2}\Big)+\frac{c}{24}(b-a)^{2} \leq \frac{1}{b-a}\int_{a}^{b} f(x)dx \leq f(a)+f(b)- \frac{c}{6}(b-a)^{2}. $$ In the case $c=0$ it gives the Hermite--Hadamard-type inequalities for $P$-convex functions proved by Dragomir, Pe\u{c}ari\'{c} and Persson in \cite{DPP95}. \end{enumerate} \end{remark} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \bibliographystyle{amsplain} \begin{thebibliography}{99} \bibitem{B78} W.W. 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