Abstract. In this paper we shall give some inequalities for convex-dominated functions which improve the well-known results of Jensen, Fuchs, Jensen - Steffensen, Peckarié, Barlow -Marshall-Proschan and Vasić-Mijalković.

We shall introduce the following class of functions.
Definition 1. Let $g:I\to \mathbb{R}$$g:I\to \mathbb{R}$g:I rarrRg: I \rightarrow \mathbb{R}$g:I\to \mathbb{R}$ be a given convex function on interva ${}^{1}I$${}^{1}I$^(1)I{ }^{1} I${}^{1}I$ from $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$. The real function $f:I\to R$$f:I\to R$f:I rarr Rf: I \rightarrow R$f:I\to R$ is called g-convex-dominated on $I$$I$II$I$ if the following condition is satisfied:

for all $x,y$$x,y$x,yx, y$x,y$ in $I$$I$II$I$ and $\lambda \in [0,1]$$\lambda \in [0,1]$lambda in[0,1]\lambda \in[0,1]$\lambda \in [0,1]$.
The next simple characterization of convex-dominated functions is valid.

Lemma 1. Let $g$$g$gg$g$ be a convex function on $I$$I$II$I$ and $f:I\to \mathbb{R}$$f:I\to \mathbb{R}$f:I rarrRf: I \rightarrow \mathbb{R}$f:I\to \mathbb{R}$. Then the following statements are equivalent :
(i) $f$$f$ff$f$ is $g$$g$gg$g$-convex-dominated on I ;
(ii) $g-f$$g-f$g-fg-f$g-f$ and $g+f$$g+f$g+fg+f$g+f$ are convex on $I$$I$II$I$;
(iii) there exists two convex mappings $h$$h$hh$h$, $l$$l$ll$l$ on $I$$I$II$I$ such that $f=1/2(h-$$f=1/2(h-$f=1//2(h-f=1 / 2(h-$f=1/2(h-$ and $g=1/2(h+l)$$g=1/2(h+l)$g=1//2(h+l)g=1 / 2(h+l)$g=1/2(h+l)$.

Proof. "(i) ⇔ (ii)". Condition (1) is equivalent to $\lambda (g(x)-f(x))+(1-\lambda )(g(y)-g(y))⩾g(\lambda x+(1-\lambda )y)-f(\lambda x+(1-\lambda )y)$$\lambda (g(x)-f(x))+(1-\lambda )(g(y)-g(y))⩾g(\lambda x+(1-\lambda )y)-f(\lambda x+(1-\lambda )y)$lambda(g(x)-f(x))+(1-lambda)(g(y)-g(y)) >= g(lambda x+(1-lambda)y)-f(lambda x+(1-lambda)y)\lambda(g(x)-f(x))+(1-\lambda)(g(y)-g(y)) \geqslant g(\lambda x+(1-\lambda) y)-f(\lambda x+(1-\lambda) y)$\lambda (g(x)-f(x))+(1-\lambda )(g(y)-g(y))⩾g(\lambda x+(1-\lambda )y)-f(\lambda x+(1-\lambda )y)$ and
$\lambda (g(x)+f(x))+(1-\lambda )(g(y)+f(y))⩾g(\lambda x+(1-\lambda )y)+f(\lambda x+(\mathbb{1}-\lambda )y)$$\lambda (g(x)+f(x))+(1-\lambda )(g(y)+f(y))⩾g(\lambda x+(1-\lambda )y)+f(\lambda x+(\mathbb{1}-\lambda )y)$lambda(g(x)+f(x))+(1-lambda)(g(y)+f(y)) >= g(lambda x+(1-lambda)y)+f(lambda x+(1-lambda)y)\lambda(g(x)+f(x))+(1-\lambda)(g(y)+f(y)) \geqslant g(\lambda x+(1-\lambda) y)+f(\lambda x+(\mathbb{1}-\lambda) y)$\lambda (g(x)+f(x))+(1-\lambda )(g(y)+f(y))⩾g(\lambda x+(1-\lambda )y)+f(\lambda x+(\mathbb{1}-\lambda )y)$ for all $x,y$$x,y$x,yx, y$x,y$ in $I$$I$II$I$ and $\lambda \in [0,1]$$\lambda \in [0,1]$lambda in[0,1]\lambda \in[0,1]$\lambda \in [0,1]$, i.e., $g-f$$g-f$g-fg-f$g-f$ and $g+f$$g+f$g+fg+f$g+f$ are convex on $I$$I$II$I$ iff (1) holds.
"(ii) ⇔ (iii)". It's obvious.

Now, let $P(I)$$P(I)$P(I)P(I)$P(I)$ be the linear space of all real valued functions defined on $I$$I$II$I$ and $J:F(I)\to \mathbb{R}$$J:F(I)\to \mathbb{R}$J:F(I)rarrRJ: F(I) \rightarrow \mathbb{R}$J:F(I)\to \mathbb{R}$ be a functional satisfying the properties:
(J1) $J(\alpha f+\beta g)=\alpha J(f)+\beta J(g)$$J(\alpha f+\beta g)=\alpha J(f)+\beta J(g)$J(alpha f+beta g)=alpha J(f)+beta J(g)J(\alpha f+\beta g)=\alpha J(f)+\beta J(g)$J(\alpha f+\beta g)=\alpha J(f)+\beta J(g)$ for all $\alpha ,\beta \in \mathbb{R}$$\alpha ,\beta \in \mathbb{R}$alpha,beta inR\alpha, \beta \in \mathbb{R}$\alpha ,\beta \in \mathbb{R}$ and $f,g\in F(I)$$f,g\in F(I)$f,g in F(I)f, g \in F(I)$f,g\in F(I)$;
(J 2) $J(f)⩾0$$J(f)⩾0$J(f) >= 0J(f) \geqslant 0$J(f)⩾0$ for all convex function $f$$f$ff$f$ on $I$$I$II$I$.
The following lemma plays a very important role in the sequel.
LEMAR 2. Let J be a functional satisfying conditions (J1), (J2). Then for every convex function $g$$g$gg$g$ and for every $g$$g$gg$g$-convex-dominated function $f$$f$ff$f$ on $I$$I$II$I$, the following inequality holds:

Proof. Let $g$$g$gg$g$ be a convex function and $f$$f$ff$f$ be $g$$g$gg$g$-convex-dominated on $I$$I$II$I$. By Lemma, 1 it follows that $g-f$$g-f$g-fg-f$g-f$ and $g+f$$g+f$g+fg+f$g+f$ are convex on $I$$I$II$I$. Then

which gives

Since $J(g)⩾0$$J(g)⩾0$J(g) >= 0J(g) \geqslant 0$J(g)⩾0$, inequality (2) is proven.
Corollary 2.1. Let $f\in C^2[a,b],h\in O[a,b],h⩾0$$f\in C^2[a,b],h\in O[a,b],h⩾0$f inC^(2)[a,b],h in O[a,b],h >= 0f \in C^{2}[a, b], h \in O[a, b], h \geqslant 0$f\in C^2[a,b],h\in O[a,b],h⩾0$ and

Then for all functional $J$$J$JJ$J$ having the properties (J1), (J2), the following inequalities hold :

and
(4)

COROLLABY 2.2. Let $f,J$$f,J$f,Jf, J$f,J$ be as above and $M:=\underset{t\in [n,b]}{sup}|f^{\prime \prime }(t)|$$M:=\underset{t\in [n,b]}{sup} \left|f^{\prime \prime }(t)\right|$M:=s u p_(t in[n,b])|f^('')(t)|M:=\sup _{t \in[n, b]}\left|f^{\prime \prime}(t)\right|$M:=\underset{t\in [n,b]}{sup}|f^{\prime \prime }(t)|$
Then the following inequality is valid:

where $e(x)=x$$e(x)=x$e(x)=xe(x)=x$e(x)=x$ on the interval $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$.
The above corollaries follow by Lemma 2 observing that:
${\int }_a({\int }_a^{t}h(s)\mathrm{d}s)\mathrm{d}t$${\int }_a \left({\int }_a^t h(s)\mathrm{d}s\right)\mathrm{d}t$int_(a)(int_(a)^(t)h(s)ds)dt\int_{a}\left(\int_{a}^{t} h(s) \mathrm{d} s\right) \mathrm{d} t${\int }_a({\int }_a^{t}h(s)\mathrm{d}s)\mathrm{d}t$ is convex, $f$$f$ff$f$ is $\int ({\int }_a^{t}h(s)\mathrm{d}s)\mathrm{d}t$$\int \left({\int }_a^t h(s)\mathrm{d}s\right)\mathrm{d}t$int(int_(a)^(t)h(s)ds)dt\int\left(\int_{a}^{t} h(s) \mathrm{d} s\right) \mathrm{d} t$\int ({\int }_a^{t}h(s)\mathrm{d}s)\mathrm{d}t$ - convex-dominated; $\int \left({\int }_a^t|f^{\prime \prime }(t)|\mathrm{d}s\right)\mathrm{d}t$$\int \left({\int }_a^t \left|f^{\prime \prime }(t)\right|\mathrm{d}s\right)\mathrm{d}t$int(int_(a)^(t)|f^('')(t)|ds)dt\int\left(\int_{a}^{t}\left|f^{\prime \prime}(t)\right| \mathrm{d} s\right) \mathrm{d} t$\int \left({\int }_a^t|f^{\prime \prime }(t)|\mathrm{d}s\right)\mathrm{d}t$ is convex, $f$$f$ff$f$ is $\left({\int }_a^t|f^{\prime \prime }(s)|\mathrm{d}s\right)\mathrm{d}t$$\left({\int }_a^t \left|f^{\prime \prime }(s)\right|\mathrm{d}s\right)\mathrm{d}t$(int_(a)^(t)|f^('')(s)|ds)dt\left(\int_{a}^{t}\left|f^{\prime \prime}(s)\right| \mathrm{d} s\right) \mathrm{d} t$\left({\int }_a^t|f^{\prime \prime }(s)|\mathrm{d}s\right)\mathrm{d}t$ - convex -
dominated and $1/2M{e}^2$$1/2M{e}^2$1//2Me^(2)1 / 2 M e^{2}$1/2M{e}^2$ is convex and $f$$f$ff$f$ is $1/2M{e}^2$$1/2M{e}^2$1//2Me^(2)1 / 2 M e^{2}$1/2M{e}^2$ - convex dominated on $[a,b]$$[a,b]$[a,b][a, b]$[a,b]$.

The following improvement of Jensen inequality holds.
Theorem 1. Let $g$$g$gg$g$ be a given convex function on $I$$I$II$I$ and $f:I\to R$$f:I\to R$f:I rarr Rf: I \rightarrow R$f:I\to R$ be $g$$g$gg$g$-convex-dominated. Then for every $x_i\in I,p_i⩾0(1⩽i⩽n)$$x_i\in I,p_i⩾0(1⩽i⩽n)$x_(i)in I,p_(i) >= 0(1 <= i <= n)x_{i} \in I, p_{i} \geqslant 0(1 \leqslant i \leqslant n)$x_i\in I,p_i⩾0(1⩽i⩽n)$ such that $P_n:=\sum _{i=1}^n{p}_i>0$$P_n:=\sum _{i=1}^n p_i>0$P_(n):=sum_(i=1)^(n)p_(i) > 0P_{n}:=\sum_{i=1}^{n} p_{i}>0$P_n:=\sum _{i=1}^n{p}_i>0$, we have the inequality :

Proof. Let us consider the functional:

Then $J$$J$JJ$J$ satisfy conditions (J1) and (J2) (by Jensen's inequality). Applying Lemma 2, we obtain inequality (6).

The proof is finished.
Remark $1^{\circ }$$1^{\circ }$1^(@)1^{\circ}$1^{\circ }$. Let $f,h$$f,h$f,hf, h$f,h$ be as in Corollary 2.1. Then we can put in (6) $g=H$$g=H$g=Hg=H$g=H$ or $g=F$$g=F$g=Fg=F$g=F$ where

$2^{\circ }$$2^{\circ }$2^(@)2^{\circ}$2^{\circ }$. If $f\in {\mathcal{C}}^2[a,b]$$f\in {\mathcal{C}}^2[a,b]$f inC^(2)[a,b]f \in \mathscr{C}^{2}[a, b]$f\in {\mathcal{C}}^2[a,b]$ and $M:=\underset{t\in [a,b]}{sup}|f^{\prime \prime }(t)|$$M:=\underset{t\in [a,b]}{sup} \left|f^{\prime \prime }(t)\right|$M:=s u p_(t in[a,b])|f^('')(t)|M:=\sup _{t \in[a, b]}\left|f^{\prime \prime}(t)\right|$M:=\underset{t\in [a,b]}{sup}|f^{\prime \prime }(t)|$, then the following inequality is valid:

where $x_i\in [a,b]$$x_i\in [a,b]$x_(i)in[a,b]x_{i} \in[a, b]$x_i\in [a,b]$ and $p_i(1⩽i⩽n)$$p_i(1⩽i⩽n)$p_(i)(1 <= i <= n)p_{i}(1 \leqslant i \leqslant n)$p_i(1⩽i⩽n)$ are as above (see also Theorem 1 from [1]).

Now, we shall give an improvement of Fuchs generalization of the Majorization theorem (see [3]). This result can be written in the following form :

Theorem 2. Let $a_1⩾\dots ⩾a_s,b_1⩾\dots ⩾b_s$$a_1⩾\dots ⩾a_s,b_1⩾\dots ⩾b_s$a_(1) >= dots >= a_(s),b_(1) >= dots >= b_(s)a_{1} \geqslant \ldots \geqslant a_{s}, b_{1} \geqslant \ldots \geqslant b_{s}$a_1⩾\dots ⩾a_s,b_1⩾\dots ⩾b_s$ and $q_1,\dots ,q_s$$q_1,\dots ,q_s$q_(1),dots,q_(s)q_{1}, \ldots, q_{s}$q_1,\dots ,q_s$ be real numbers such that:

If $g$$g$gg$g$ is convex on $I$$I$II$I$ and $f$$f$ff$f$ is $g$$g$gg$g$-convex dominated on $I$$I$II$I$, then the following inequality holds:

Proof. Let consider the functional:

Then $J$$J$JJ$J$ satisfies conditions (J1) and (J2) (by Fuchs' inequality see alsoTheorem B from [4]). Applying Lemma 2, we deduce inequality (8).

Remarks $3^{\circ }$$3^{\circ }$3^(@)3^{\circ}$3^{\circ }$. Let $f,g,h,H,F$$f,g,h,H,F$f,g,h,H,Ff, g, h, H, F$f,g,h,H,F$ be as in Remark $1^{\circ }$$1^{\circ }$1^(@)1^{\circ}$1^{\circ }$, then in (8) we can put $g=H$$g=H$g=Hg=H$g=H$ оr $g=F$$g=F$g=Fg=F$g=F$.
4. Let $f\in C^2[a,b]$$f\in C^2[a,b]$f inC^(2)[a,b]f \in C^{2}[a, b]$f\in C^2[a,b]$ and $M:=\underset{t\in [a,b]}{sup}|f^{\prime \prime }(t)|$$M:=\underset{t\in [a,b]}{sup} \left|f^{\prime \prime }(t)\right|$M:=s u p_(t in[a,b])|f^('')(t)|M:=\sup _{t \in[a, b]}\left|f^{\prime \prime}(t)\right|$M:=\underset{t\in [a,b]}{sup}|f^{\prime \prime }(t)|$, then the following inequality holds :

where $a_i,b_i,q_i(1⩽i⩽\epsilon )$$a_i,b_i,q_i(1⩽i⩽\epsilon )$a_(i),b_(i),q_(i)(1 <= i <= epsi)a_{i}, b_{i}, q_{i}(1 \leqslant i \leqslant \varepsilon)$a_i,b_i,q_i(1⩽i⩽\epsilon )$ are as above.
Now, we shall give an improvement of Jensen-Steffensen inequality.
Theoresi 3. Lot $x$$x$xx$x$ and $p$$p$pp$p$ be two $n$$n$nn$n$-tuples of real numbers such that $x_i\in I(1⩽i⩽n,I$$x_i\in I(1⩽i⩽n,I$x_(i)in I(1 <= i <= n,Ix_{i} \in I (1 \leqslant i \leqslant n, I$x_i\in I(1⩽i⩽n,I$ is an interval from $\mathbb{R})$$\mathbb{R})$R)\mathbb{R})$\mathbb{R})$ and $P_n>0$$P_n>0$P_(n) > 0P_{n}>0$P_n>0$. Then the following sentences are equivalent :
(i) For every convex function $g:I\to \mathbb{R}$$g:I\to \mathbb{R}$g:I rarrRg: I \rightarrow \mathbb{R}$g:I\to \mathbb{R}$, for every $g$$g$gg$g$-convex-dominated function $f$$f$ff$f$ and for all monotonic n-tuple oc the inequality (6) holds;
(ii) $0⩽P_n⩽P_n$$0⩽P_n⩽P_n$0 <= P_(n) <= P_(n)0 \leqslant P_{n} \leqslant P_{n}$0⩽P_n⩽P_n$ for all $k=1,2,\dots ,n-1$$k=1,2,\dots ,n-1$k=1,2,dots,n-1k=1,2, \ldots, n-1$k=1,2,\dots ,n-1$.

Proof. "(i) ⇒ (ii)". Tt's obvious by Jensen-Steffensen theorem.
"(ii) ⇒ (i)". Let us consider the functional:

Then J werifies condifions (J1) and (J2) (by Jensen-Steffensen inequality; see for example [4], Theorem A). Apliying Lemma 2, we obtain (6).

Remarks $1^{\circ }$$1^{\circ }$1^(@)1^{\circ}$1^{\circ }$ and $2^{\circ }$$2^{\circ }$2^(@)2^{\circ}$2^{\circ }$ are also valid if $p_i(1⩽i⩽n)$$p_i(1⩽i⩽n)$p_(i)(1 <= i <= n)p_{i}(1 \leqslant i \leqslant n)$p_i(1⩽i⩽n)$ satisfies condition (ii) of the above theorem.

Wow, we shall give another result which improves Pecarie's theorem (see [4], Theorem 1):

THERREN 4. Let $x$$x$xx$x$ be nonincreasing n-tuple of real numbers, $x_i\in I(1⩽i⩽n)$$x_i\in I(1⩽i⩽n)$x_(i)in I(1 <= i <= n)x_{i} \in I (1 \leqslant i \leqslant n)$x_i\in I(1⩽i⩽n)$, p real $n$$n$nn$n$-tuple and exists $j\in (1,2,\dots ,n)$$j\in (1,2,\dots ,n)$j in(1,2,dots,n)j \in(1,2, \ldots, n)$j\in (1,2,\dots ,n)$ such that:

(if $x_1⩽\overline{x}$$x_1⩽\overline{x}$x_(1) <= bar(x)x_{1} \leqslant \bar{x}$x_1⩽\overline{x}$ the first condition in (10) is taken to be vacuous, if $x_n⩾\overline{x}$$x_n⩾\overline{x}$x_(n) >= bar(x)x_{n} \geqslant \bar{x}$x_n⩾\overline{x}$ the second condition in (10) is taken to be vacuous). If $\overline{x}\in I$$\overline{x}\in I$bar(x)in I\bar{x} \in I$\overline{x}\in I$, then for every.
convex function $g:I\to \mathbb{R}$$g:I\to \mathbb{R}$g:I rarrRg: I \rightarrow \mathbb{R}$g:I\to \mathbb{R}$ and for every $g$$g$gg$g$-convex dominated function $f$$f$ff$f$ : $:I\to \mathbb{R}$$:I\to \mathbb{R}$:I rarrR: I \rightarrow \mathbb{R}$:I\to \mathbb{R}$, we have :
(11) $g\left(\frac{1}{P_n}\sum _{i=1}^n{p}_i{x}_i\right)-\frac{1}{P_n}\sum _{i=1}^n{p}_{i}g\left(x_i\right)⩾|f\left(\frac{1}{P_n}\sum _{i=1}^n{p}_i{x}_i\right)-\frac{1}{P_n}{p}_{i}f\left(x_i\right)|$$g\left(\frac{1}{P_n}\sum _{i=1}^n p_i{x}_i\right)-\frac{1}{P_n}\sum _{i=1}^n p_{i}g\left(x_i\right)⩾\left|f\left(\frac{1}{P_n}\sum _{i=1}^n p_i{x}_i\right)-\frac{1}{P_n}{p}_{i}f\left(x_i\right)\right|$g((1)/(P_(n))sum_(i=1)^(n)p_(i)x_(i))-(1)/(P_(n))sum_(i=1)^(n)p_(i)g(x_(i)) >= |f((1)/(P_(n))sum_(i=1)^(n)p_(i)x_(i))-(1)/(P_(n))p_(i)f(x_(i))|g\left(\frac{1}{P_{n}} \sum_{i=1}^{n} p_{i} x_{i}\right)-\frac{1}{P_{n}} \sum_{i=1}^{n} p_{i} g\left(x_{i}\right) \geqslant\left|f\left(\frac{1}{P_{n}} \sum_{i=1}^{n} p_{i} x_{i}\right)-\frac{1}{P_{n}} p_{i} f\left(x_{i}\right)\right|$g\left(\frac{1}{P_n}\sum _{i=1}^n{p}_i{x}_i\right)-\frac{1}{P_n}\sum _{i=1}^n{p}_{i}g\left(x_i\right)⩾|f\left(\frac{1}{P_n}\sum _{i=1}^n{p}_i{x}_i\right)-\frac{1}{P_n}{p}_{i}f\left(x_i\right)|$.

If the inverse inequalities in (10) hold, then (6) holds.
The proof follows by a similar argument to that in the proof of the previous theorem using the result of J. E. Pecarić([4], Pheorem 1), We omit the details.

By Theorem 2 from [4] we also obtain :
Theorem 5. Let a and $p$$p$pp$p$ be two n-tuple of real mumbers such that $x_i\in I(1⩽i⩽n),\overline{x}\in I,P_n⩾0$$x_i\in I(1⩽i⩽n),\overline{x}\in I,P_n⩾0$x_(i)in I(1 <= i <= n), bar(x)in I,P_(n) >= 0x_{i} \in I(1 \leqslant i \leqslant n), \bar{x} \in I, P_{n} \geqslant 0$x_i\in I(1⩽i⩽n),\overline{x}\in I,P_n⩾0$. Then the following sentences are equivalent :
(i) inequality (11) holds for every conver function $g:I\to \mathbb{R}$$g:I\to \mathbb{R}$g:I rarrRg: I \rightarrow \mathbb{R}$g:I\to \mathbb{R}$, for every g-convea-dominated function $f:I\to Z$$f:I\to Z$f:I rarr Zf: I \rightarrow Z$f:I\to Z$ and for all monotonic n-tuple $x$$x$xx$x$;
(ii) there exists $m\in (1,2,\dots ,n)$$m\in (1,2,\dots ,n)$m in(1,2,dots,n)m \in(1,2, \ldots, n)$m\in (1,2,\dots ,n)$ such that $P_k⩽0(k<m)$$P_k⩽0(k<m)$P_(k) <= 0quad(k < m)P_{k} \leqslant 0 \quad(k<m)$P_k⩽0(k<m)$ and ${\overline{P}}_k⩽0(k>m)$${\overline{P}}_k⩽0(k>m)$bar(P)_(k) <= 0(k > m)\bar{P}_{k} \leqslant 0(k>m)${\overline{P}}_k⩽0(k>m)$, where ${\overline{P}}_k:P_n-P_{k-1}$${\overline{P}}_k:P_n-P_{k-1}$bar(P)_(k):P_(n)-P_(k-1)\bar{P}_{k}: P_{n}-P_{k-1}${\overline{P}}_k:P_n-P_{k-1}$.
Remark $5^{\circ }$$5^{\circ }$5^(@)5^{\circ}$5^{\circ }$. Let $f$$f$ff$f$ be as in Corollary 2.2. If $p,x$$p,x$p,xp, x$p,x$ satisfy conditions (10) or $x$$x$xx$x$ is a monotonic $n$$n$nn$n$-tuple and $p$$p$pp$p$ verifies (12), then the following inequality holds: