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\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

$\begin{array}{cl}
    & f(tx+(1-t)y) \\
  = & g(tx+(1-t)y)+ c \| (tx+(1-t)y) \|^{2}\\
\leq & h(t)g(x)+h(1-t)g(y)+ c \| (tx+(1-t)y) \|^{2}\\
 =  & h(t)f(x)+h(1-t)f(y) - ch(t)\|x\|^{2}- ch(1-t)\|y\|^{2}+ c \| (tx+(1-t)y) \|^{2} \\
\leq & h(t)f(x)+h(1-t)f(y) - ct\|x\|^{2}- c(1-t)\|y\|^{2} \\
+ & c (t^{2}\|x\|^{2}+2t(1-t)<x\mid y>+ (1-t)^{2} \|y\|^{2})\\
= & h(t)f(x)+h(1-t)f(y) - ct(1-t)\|x-y\|^{2},
\end{array}$

\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 $a, b \in I$, $a<b$. These classical Hermite--Hadamard
  inequalities play an important role in convex analysis and there is  an extensive
    literature dealing with its applications, various generalizations
   and refinements (see for instance \cite{DP02, NPE06}, and the references therein).
   The following result is a counterpart of the Hermite--Hadamard inequalities for strongly $h$-convex functions.
\begin{theorem}
let  $h:(0,1) \rightarrow (0,\infty)$ be a given function. If a
function $f: I \rightarrow \mathbb{R}$ is Lebesgue integrable and
strongly $h$-convex with modulus \ $c>0$, 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}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


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\vspace{2ex}

\end{thebibliography}

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