On the Jensen functional and superterzaticity

We consider a real valued function f defined on an interval $I\text{,}{x}_1\text{,}{x}_2\text{,}\dots \text{,}{x}_n\in I$ and $p_1\text{,}{p}_2\text{,}\dots \text{,}{p}_n\in \left(0\text{,}1\right)$ with ${\sum }_{i=1}^n{p}_i=1$ . The Jensen functional is defined by $$\mathcal{J}\left(f\text{,}\mathbf{p}\text{,}\mathbf{x}\right)={\displaystyle \sum _{i=1}^n}{p}_{i}f\left(x_i\right)-f\left({\displaystyle \sum _{i=1}^n}{p}_i{x}_i\right)$$ (see Dragomir, 2006).

A function f defined on an interval $I=[0\text{,}a)$ is superterzatic, if for each $\overline{x}\in I$ there exists a real number $C(\overline{x})$ such that $$\begin{array}{cc}\mathcal{J}(f\text{,}\mathbf{p}\text{,}\mathbf{x})\hfill & ⩾{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\left[(x_i-\overline{x})C(\overline{x})+\frac{f(|x_i-\overline{x}|)}{|x_i-\overline{x}|}\right]\hfill \\ \hfill & ={\displaystyle \sum _{i=1}^n}{p}_i{(x_i-\overline{x})}^{2}C(\overline{x})+{\displaystyle \sum _{i=1}^n}{p}_i{x}_i\frac{f(|x_i-\overline{x}|)}{|x_i-\overline{x}|}\hfill \end{array}$$ for all $x_i⩾0\text{,}i=1\text{,}\dots \text{,}n$ , and $p_i⩾0\text{,}i=1\text{,}\dots \text{,}n$ , with ${\sum }_{i=1}^n{p}_i=1$ , such that $\overline{x}={\sum }_{i=1}^n{p}_i{x}_i$ . We use the convention $f(0)/0=0$ .

(Abramovich et al., 2012) Let $f\left(x\right)=x^p\text{,}p⩾3$ . This function is superterzatic with $C(\overline{x})=p{\overline{x}}^{p-1}$ . The function $g\left(x\right)=x^3\log \frac{1}{x}$ is also subterzatic (i.e. the inequality in Definition 2 holds with reversed sign).

Assume that we have a real valued function f defined on an interval I, the real numbers $p_{\mathit{ij}}\text{,}i=1\text{,}\dots \text{,}k$ and $j=1\text{,}\dots \text{,}{n}_i$ such that $p_{\mathit{ij}}>0\text{,}{\sum }_{j=1}^{n_i}{p}_{\mathit{ij}}=1$ for all $i=1\text{,}\dots \text{,}k$ (we denote ${\mathbf{p}}_i=$ $\left(p_{i1}\text{,}{p}_{i2}\text{,}\dots \text{,}{p}_{{\mathit{in}}_i}\right)$ ), ${\mathbf{x}}_i=$ $\left(x_{i1}\text{,}{x}_{i2}\text{,}\dots \text{,}{x}_{{\mathit{in}}_i}\right)\in I^{n_i}$ for all $i=1\text{,}\dots \text{,}k$ and $\mathbf{q}=\left(q_1\text{,}\dots \text{,}{q}_k\right)\text{,}{q}_i>0$ such that ${\sum }_{i=1}^k{q}_i=1$ . We define the generalized Jensen functional by $${\mathcal{J}}_k\left(f\text{,}{\mathbf{p}}_1\text{,}\dots \text{,}{\mathbf{p}}_k\text{,}\mathbf{q}\text{,}{\mathbf{x}}_1\text{,}\dots \text{,}{\mathbf{x}}_k\right):={\displaystyle \sum _{j_1\text{,}\dots \text{,}{j}_k=1}^{n_1\text{,}\dots \text{,}{n}_k}}{p}_{1j_1}\dots p_{{\mathit{kj}}_k}f\left({\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}\right)-f\left({\displaystyle \sum _{i=1}^k}{q}_i{\displaystyle \sum _{j=1}^{n_i}}{p}_{\mathit{ij}}{x}_{\mathit{ij}}\right)\text{.}$$

Let $f\text{,}{\mathbf{p}}_i\text{,}{\mathbf{x}}_i\text{,}\mathbf{q}$ be as in Definition 3. If f is superterzatic then we have $${\mathcal{J}}_k\left(f\text{,}{\mathbf{p}}_1\text{,}\dots \text{,}{\mathbf{p}}_k\text{,}\mathbf{q}\text{,}{\mathbf{x}}_1\text{,}\dots \text{,}{\mathbf{x}}_k\right)⩾{\sum }_{j_1\text{,}\dots \text{,}{j}_k=1}^{n_1\text{,}\dots \text{,}{n}_k}{p}_{1j_1}\dots p_{{\mathit{kj}}_k}{\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}\left[\left({\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}-\overline{x}\right)C(\overline{x})+\frac{f\left(\left|{\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}-\overline{x}\right|\right)}{\left|{\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}-\overline{x}\right|}\right]\text{,}$$ where $\overline{x}={\sum }_{i=1}^k{q}_i{\sum }_{j=1}^{n_i}{p}_{\mathit{ij}}{x}_{\mathit{ij}}$ .

Since $${\displaystyle \sum _{j_1\text{,}\dots \text{,}{j}_k=1}^{n_1\text{,}\dots \text{,}{n}_k}}{p}_{1j_1}\dots p_{{\mathit{kj}}_k}=1$$ and $$\overline{x}={\displaystyle \sum _{j_1\text{,}\dots \text{,}{j}_k=1}^{n_1\text{,}\dots \text{,}{n}_k}}{p}_{1j_1}\dots p_{{\mathit{kj}}_k}{\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}\text{,}$$ we use Definition 2 and the conclusion follows. □

Let $f\text{,}{\mathbf{p}}_i\text{,}{\mathbf{x}}_i$ and $\mathbf{q}$ be as in Definition 3 and the positive real numbers $r_{\mathit{ij}}\text{,}i=1\text{,}\dots \text{,}k$ and $j=1\text{,}\dots \text{,}{n}_i$ such that ${\sum }_{j=1}^{n_i}{r}_{\mathit{ij}}=1$ for all $i=1\text{,}\dots \text{,}k$ . We denote $$\begin{array}{cc}{\mathbf{r}}_i=& \left(r_{i1}\text{,}{r}_{i2}\text{,}\dots \text{,}{r}_{{\mathit{in}}_i}\right)\mathrm{for}\mathrm{all}i=1\text{,}\dots \text{,}k\text{,}\\ m=& \underset{1⩽j_1⩽n_1\dots 1⩽j_k⩽n_k}{\min }\left\{\frac{p_{1j_1}\dots p_{{\mathit{kj}}_k}}{r_{1j_1}\dots r_{{\mathit{kj}}_k}}\right\}\text{,}\\ M=& \underset{1⩽j_1⩽n_1\dots 1⩽j_k⩽n_k}{\max }\left\{\frac{p_{1j_1}\dots p_{{\mathit{kj}}_k}}{r_{1j_1}\dots r_{{\mathit{kj}}_k}}\right\}\text{.}\end{array}$$ If f is superterzatic, then: $${\mathcal{J}}_k\left(f\text{,}{\mathbf{p}}_1\text{,}\dots \text{,}{\mathbf{p}}_k\text{,}\mathbf{q}\text{,}{\mathbf{x}}_1\text{,}\dots \text{,}{\mathbf{x}}_k\right)-m{\mathcal{J}}_k\left(f\text{,}{\mathbf{r}}_1\text{,}\dots \text{,}{\mathbf{r}}_k\text{,}\mathbf{q}\text{,}{\mathbf{x}}_1\text{,}\dots \text{,}{\mathbf{x}}_k\right)⩾{\displaystyle \sum _{j_1\text{,}\dots \text{,}{j}_k=1}^{n_1\text{,}\dots \text{,}{n}_k}}\left(p_{1j_1}\dots p_{{\mathit{kj}}_k}-{\mathit{mr}}_{1j_1}\dots r_{{\mathit{kj}}_k}\right){\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}\times \left[\left({\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}-\overline{x}\right)C\left(\overline{x}\right)+\frac{f\left(\left|{\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}-\overline{x}\right|\right)}{\left|{\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}-\overline{x}\right|}\right]+m{\displaystyle \sum _{i=1}^k}{q}_i{\displaystyle \sum _{j=1}^{n_i}}{r}_{\mathit{ij}}{x}_{\mathit{ij}}\times \left[{\displaystyle \sum _{i=1}^k}{q}_i{\displaystyle \sum _{j=1}^{n_i}}\left(r_{\mathit{ij}}-p_{\mathit{ij}}\right)x_{\mathit{ij}}C\left(\overline{x}\right)+\frac{f\left(\left|\check{x}-\overline{x}\right|\right)}{\left|\check{x}-\overline{x}\right|}\right]$$ and $$M{\mathcal{J}}_k\left(f\text{,}{\mathbf{r}}_1\text{,}\dots \text{,}{\mathbf{r}}_k\text{,}\mathbf{q}\text{,}{\mathbf{x}}_1\text{,}\dots \text{,}{\mathbf{x}}_k\right)-{\mathcal{J}}_k\left(f\text{,}{\mathbf{p}}_1\text{,}\dots \text{,}{\mathbf{p}}_k\text{,}\mathbf{q}\text{,}{\mathbf{x}}_1\text{,}\dots \text{,}{\mathbf{x}}_k\right)⩾{\displaystyle \sum _{j_1\text{,}\dots \text{,}{j}_k=1}^{n_1\text{,}\dots \text{,}{n}_k}}\left({\mathit{Mr}}_{1j_1}\dots r_{{\mathit{kj}}_k}-p_{1j_1}\dots p_{{\mathit{kj}}_k}\right){\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}\times \left[\left({\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}-\check{x}\right)C\left(\check{x}\right)+\frac{f\left(\left|{\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}-\check{x}\right|\right)}{\left|{\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}-\check{x}\right|}\right]+{\displaystyle \sum _{i=1}^k}{q}_i{\displaystyle \sum _{j=1}^{n_i}}{r}_{\mathit{ij}}{x}_{\mathit{ij}}\times \left[{\displaystyle \sum _{i=1}^k}{q}_i{\displaystyle \sum _{j=1}^{n_i}}\left(r_{\mathit{ij}}-p_{\mathit{ij}}\right)x_{\mathit{ij}}C\left(\check{x}\right)+\frac{f\left(\left|\check{x}-\overline{x}\right|\right)}{\left|\check{x}-\overline{x}\right|}\right]\text{.}$$ where $\overline{x}={\sum }_{i=1}^k{q}_i{\sum }_{j=1}^{n_i}{p}_{\mathit{ij}}{x}_{\mathit{ij}}$ and $\check{x}={\sum }_{i=1}^k{q}_i{\sum }_{j=1}^{n_i}{r}_{\mathit{ij}}{x}_{\mathit{ij}}$ .

The first inequality. Obviously $${\mathcal{J}}_k\left(f\text{,}{\mathbf{p}}_1\text{,}\dots \text{,}{\mathbf{p}}_k\text{,}\mathbf{q}\text{,}{\mathbf{x}}_1\text{,}\dots \text{,}{\mathbf{x}}_k\right)-m{\mathcal{J}}_k\left(f\text{,}{\mathbf{r}}_1\text{,}\dots \text{,}{\mathbf{r}}_k\text{,}\mathbf{q}\text{,}{\mathbf{x}}_1\text{,}\dots \text{,}{\mathbf{x}}_k\right)={\sum }_{j_1\text{,}\dots \text{,}{j}_k=1}^{n_1\text{,}\dots \text{,}{n}_k}\left(p_{1j_1}\dots p_{{\mathit{kj}}_k}-{\mathit{mr}}_{1j_1}\dots r_{{\mathit{kj}}_k}\right)f\left({\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}\right)+\mathit{mf}\left({\displaystyle \sum _{i=1}^k}{q}_i{\displaystyle \sum _{j=1}^{n_i}}{r}_{\mathit{ij}}{x}_{\mathit{ij}}\right)-f\left({\displaystyle \sum _{i=1}^k}{q}_i{\displaystyle \sum _{j=1}^{n_i}}{p}_{\mathit{ij}}{x}_{\mathit{ij}}\right)\text{.}$$

Since $$\overline{x}={\sum }_{i=1}^k{q}_i{\displaystyle \sum _{j=1}^{n_i}}{p}_{\mathit{ij}}{x}_{\mathit{ij}}={\displaystyle \sum _{j_1\text{,}\dots \text{,}{j}_k=1}^{n_1\text{,}\dots \text{,}{n}_k}}\left(p_{1j_1}\dots p_{{\mathit{kj}}_k}-{\mathit{mr}}_{1j_1}\dots r_{{\mathit{kj}}_k}\right){\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}+m{\displaystyle \sum _{i=1}^k}{q}_i{\displaystyle \sum _{j=1}^{n_i}}{r}_{\mathit{ij}}{x}_{\mathit{ij}}$$ and $${\displaystyle \sum _{j_1\text{,}\dots \text{,}{j}_k=1}^{n_1\text{,}\dots \text{,}{n}_k}}\left(p_{1j_1}\dots p_{{\mathit{kj}}_k}-{\mathit{mr}}_{1j_1}\dots r_{{\mathit{kj}}_k}\right)+m=1\text{,}$$ the conclusion follows by Lemma 1.

The second inequality. One has $$M{\mathcal{J}}_k\left(f\text{,}{\mathbf{r}}_1\text{,}\dots \text{,}{\mathbf{r}}_k\text{,}\mathbf{q}\text{,}{\mathbf{x}}_1\text{,}\dots \text{,}{\mathbf{x}}_k\right)-{\mathcal{J}}_k\left(f\text{,}{\mathbf{p}}_1\text{,}\dots \text{,}{\mathbf{p}}_k\text{,}\mathbf{q}\text{,}{\mathbf{x}}_1\text{,}\dots \text{,}{\mathbf{x}}_k\right)={\sum }_{j_1\text{,}\dots \text{,}{j}_k=1}^{n_1\text{,}\dots \text{,}{n}_k}\left({\mathit{Mr}}_{1j_1}\dots r_{{\mathit{kj}}_k}-p_{1j_1}\dots p_{{\mathit{kj}}_k}\right)f\left({\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}\right)+f\left({\displaystyle \sum _{i=1}^k}{q}_i{\displaystyle \sum _{j=1}^{n_i}}{p}_{\mathit{ij}}{x}_{\mathit{ij}}\right)-\mathit{Mf}\left({\displaystyle \sum _{i=1}^k}{q}_i{\displaystyle \sum _{j=1}^{n_i}}{r}_{\mathit{ij}}{x}_{\mathit{ij}}\right)\text{.}$$

Since $$\check{x}={\sum }_{i=1}^k{q}_i{\displaystyle \sum _{j=1}^{n_i}}{r}_{\mathit{ij}}{x}_{\mathit{ij}}={\displaystyle \sum _{j_1\text{,}\dots \text{,}{j}_k=1}^{n_1\text{,}\dots \text{,}{n}_k}}\frac{{\mathit{Mr}}_{1j_1}\dots r_{{\mathit{kj}}_k}-p_{1j_1}\dots p_{{\mathit{kj}}_k}}{M}{\displaystyle \sum _{i=1}^k}{q}_i{x}_{{\mathit{ij}}_i}+\frac{1}{M}{\displaystyle \sum _{i=1}^k}{q}_i{\displaystyle \sum _{j=1}^{n_i}}{r}_{\mathit{ij}}{x}_{\mathit{ij}}$$ and $${\displaystyle \sum _{j_1\text{,}\dots \text{,}{j}_k=1}^{n_1\text{,}\dots \text{,}{n}_k}}\frac{{\mathit{Mr}}_{1j_1}\dots r_{{\mathit{kj}}_k}-p_{1j_1}\dots p_{{\mathit{kj}}_k}}{M}+\frac{1}{M}=1\text{,}$$ we may apply again Lemma 1 and the conclusion follows. □

Let us consider $\mathbf{x}=\left(x_1\text{,}{x}_2\text{,}\dots \text{,}{x}_n\right)\in I^n\text{,}\mathbf{p}=\left(p_1\text{,}{p}_2\text{,}\dots \text{,}{p}_n\right)\in {\left(0\text{,}1\right)}^n$ with ${\sum }_{i=1}^n{p}_i=1\text{,}\mathbf{q}=\left(q_1\text{,}{q}_2\text{,}\dots \text{,}{q}_k\right)$ such that $q_i>0\text{,}{\sum }_{i=1}^k{q}_i=1$ ( $1⩽k⩽n$ ) and $\mathbf{r}=\left(r_1\text{,}{r}_2\text{,}\dots \text{,}{r}_n\right)\in {\left(0\text{,}1\right)}^n$ with ${\sum }_{i=1}^n{r}_i=1$ . We put $$\begin{array}{cc}m\hfill & ={\displaystyle \underset{1⩽i_1\text{,}\dots \text{,}{i}_k⩽n}{\min }}\left\{\frac{p_{i_1}\dots p_{i_k}}{r_{i_1}\dots r_{i_k}}\right\}\text{,}\hfill \\ M\hfill & ={\displaystyle \underset{1⩽i_1\text{,}\dots i_k⩽n}{\max }}\left\{\frac{p_{i_1}\dots p_{i_k}}{r_{i_1}\dots r_{i_k}}\right\}\text{.}\hfill \end{array}$$ We define $${\mathcal{J}}_k\left(f\text{,}\mathbf{p}\text{,}\mathbf{q}\text{,}\mathbf{x}\right)≔{\displaystyle \sum _{i_1\text{,}\dots \text{,}{i}_k=1}^n}{p}_{i_1}\dots p_{i_k}f\left({\displaystyle \sum _{j=1}^k}{q}_j{x}_{i_j}\right)-f\left({\displaystyle \sum _{i=1}^n}{p}_i{x}_i\right)\text{.}$$ If f is superterzatic, then $${\mathcal{J}}_k\left(f\text{,}\mathbf{p}\text{,}\mathbf{q}\text{,}\mathbf{x}\right)-m{\mathcal{J}}_k\left(f\text{,}\mathbf{r}\text{,}\mathbf{q}\text{,}\mathbf{x}\right)⩾{\sum }_{i_1\text{,}\dots \text{,}{i}_k=1}^n\left(p_{i_1}\dots p_{i_k}-{\mathit{mr}}_{i_1}\dots r_{i_k}\right){\displaystyle \sum _{j=1}^k}{q}_j{x}_{i_j}\times \left[\left({\displaystyle \sum _{j=1}^k}{q}_j{x}_{i_j}-\overline{x}\right)C\left(\overline{x}\right)+\frac{f\left(\left|{\displaystyle \sum _{j=1}^k}{q}_j{x}_{i_j}-\overline{x}\right|\right)}{\left|{\displaystyle \sum _{j=1}^k}{q}_j{x}_{i_j}-\overline{x}\right|}\right]+m{\displaystyle \sum _{i=1}^n}{r}_i{x}_i\left[{\displaystyle \sum _{j=1}^n}\left(r_j-p_j\right)x_{j}C\left(\overline{x}\right)+\frac{f\left(\left|\check{x}-\overline{x}\right|\right)}{\left|\check{x}-\overline{x}\right|}\right]$$ and $$M{\mathcal{J}}_k\left(f\text{,}\mathbf{r}\text{,}\mathbf{q}\text{,}\mathbf{x}\right)-{\mathcal{J}}_k\left(f\text{,}\mathbf{p}\text{,}\mathbf{q}\text{,}\mathbf{x}\right)⩾{\sum }_{i_1\text{,}\dots \text{,}{i}_k=1}^n\left({\mathit{Mr}}_{i_1}\dots r_{i_k}-p_{i_1}\dots p_{i_k}\right){\displaystyle \sum _{j=1}^k}{q}_j{x}_{i_j}\times \left[\left({\displaystyle \sum _{j=1}^k}{q}_j{x}_{i_j}-\check{x}\right)C\left(\check{x}\right)+\frac{f\left(\left|{\displaystyle \sum _{j=1}^k}{q}_j{x}_{i_j}-\check{x}\right|\right)}{\left|{\displaystyle \sum _{j=1}^k}{q}_j{x}_{i_j}-\check{x}\right|}\right]+{\displaystyle \sum _{i=1}^n}{r}_i{x}_i\left[{\displaystyle \sum _{j=1}^n}\left(r_j-p_j\right)x_{j}C\left(\check{x}\right)+\frac{f\left(\left|\check{x}-\overline{x}\right|\right)}{\left|\check{x}-\overline{x}\right|}\right]\text{,}$$ where $\overline{x}={\sum }_{j=1}^n{p}_j{x}_j$ and $\check{x}={\sum }_{j=1}^n{r}_j{x}_j$ .

For $i=1\text{,}\dots \text{,}n$ , we consider $x_i\in I\text{,}{p}_i>0$ with ${\sum }_{i=1}^n{p}_i=1$ and $r_i>0$ with ${\sum }_{i=1}^n{r}_i=1$ . We denote $$m=\underset{i=1\text{,}\dots \text{,}n}{\min }\left\{\frac{p_i}{r_i}\right\}\text{,}M=\underset{i=1\text{,}\dots \text{,}n}{\max }\left\{\frac{p_i}{r_i}\right\}\text{.}$$

If f is a superterzatic function, then we have: $$\mathcal{J}\left(f\text{,}\mathbf{p}\text{,}\mathbf{x}\right)-m\mathcal{J}\left(f\text{,}\mathbf{r}\text{,}\mathbf{x}\right)⩾{\displaystyle \sum _{i=1}^n}\left(p_i-{\mathit{mr}}_i\right)x_i\left[\left(x_i-\overline{x}\right)C\left(\overline{x}\right)+\frac{f\left(\left|x_i-\overline{x}\right|\right)}{\left|x_i-\overline{x}\right|}\right]+m{\displaystyle \sum _{i=1}^n}{r}_i{x}_i\left[{\displaystyle \sum _{i=1}^n}\left(r_i-p_i\right)x_{i}C\left(\overline{x}\right)+\frac{f\left(\left|\check{x}-\overline{x}\right|\right)}{\left|\check{x}-\overline{x}\right|}\right]$$ and $$M\mathcal{J}\left(f\text{,}\mathbf{r}\text{,}\mathbf{x}\right)-\mathcal{J}\left(f\text{,}\mathbf{p}\text{,}\mathbf{x}\right)⩾{\displaystyle \sum _{i=1}^n}\left({\mathit{Mr}}_i-p_i\right)x_i\left[\left(x_i-\check{x}\right)C\left(\check{x}\right)+\frac{f\left(\left|x_i-\check{x}\right|\right)}{\left|x_i-\check{x}\right|}\right]+{\displaystyle \sum _{i=1}^n}{r}_i{x}_i\left[{\displaystyle \sum _{i=1}^n}\left(r_i-p_i\right)x_{i}C\left(\check{x}\right)+\frac{f\left(\left|\check{x}-\overline{x}\right|\right)}{\left|\check{x}-\overline{x}\right|}\right]\text{,}$$ where $\overline{x}={\sum }_{j=1}^n{p}_j{x}_j$ and $\check{x}={\sum }_{j=1}^n{r}_j{x}_j$ .