Grüss-Type Inequalities for Vector-Valued Functions

Abstract

Grüss-type inequalities have been widely studied and applied in different contexts. In this work, we provide and prove vectorial versions of Grüss-type inequalities involving vector-valued functions defined on ${\mathbb{R}}^n$ for inner- and cross-products.

1. Introduction and Background

Grüss-type inequalities have been largely studied and applied to diverse frameworks. In [1], Grüss-type inequalities with multiple points for derivatives bounded by functions on time scales were analyzed. In [2], the Ostrowski–Grüss type inequality of the Chebyshev functional was analyzed and an application to the one-point integral formula was provided.

A number of authors have demonstrated some Grüss-type inequalities in one and several variables [3,4,5]. For more results on multivariate and multidimensional Grüss-type inequalities, we refer the reader to [6,7,8].

In 1935, Grüss [9] introduced his celebrated inequality, which reads

$$\begin{array}{c}\hfill \left|\frac{1}{(d-c)}{\int }_c^{d}H\left(t\right)K\left(t\right)\mathrm{d}t-\frac{1}{(d-c)}{\int }_c^{d}H\left(t\right)\mathrm{d}t\frac{1}{(d-c)}{\int }_c^{d}K\left(t\right)\mathrm{d}t\right|\le \frac{1}{4}\left(R_2-R_1\right)\left(S_2-S_1\right).\end{array}$$

(1)

The inequality presented in (1) is valid for integrable bounded functions $H,K:[c,d]\to \mathbb{R}$, such that $R_1\le H\left(t\right)\le R_2$ and $S_1\le K\left(t\right)\le S_2$, for all $t\in [c,d]$.

In [10], a Grüss-type inequality was proved for double integrals by means of

$$\begin{array}{c}\left|\frac{1}{\left({\beta }_2-{\alpha }_2\right)\left({\beta }_1-{\alpha }_1\right)}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_1}^{{\beta }_1}H\left(t,s\right)K\left(t,s\right)\mathrm{d}t\mathrm{d}s\right.\hfill \\ \left.-\frac{1}{\left({\beta }_2-{\alpha }_2\right)\left({\beta }_1-{\alpha }_1\right)}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_1}^{{\beta }_1}H\left(t,s\right)\mathrm{d}t\mathrm{d}s\frac{1}{\left({\beta }_2-{\alpha }_2\right)\left({\beta }_1-{\alpha }_1\right)}{\int }_{{\alpha }_2}^{{\beta }_2}{\int }_{{\alpha }_1}^{{\beta }_1}K\left(t,s\right)\mathrm{d}t\mathrm{d}s\right|\\ \le 4\left[U_1{V}_1\frac{{\left({\beta }_1-{\alpha }_1\right)}^{{\lambda }_1+{\delta }_1}}{\left({\lambda }_1+{\delta }_1+1\right)\left({\lambda }_1+{\delta }_1+2\right)}+U_1{V}_2\frac{{\left({\beta }_1-{\alpha }_1\right)}^{{\lambda }_1}{\left({\beta }_2-{\alpha }_1\right)}^{{\delta }_2}}{\left({\lambda }_1+1\right)\left({\delta }_2+1\right)\left({\lambda }_1+2\right)\left({\delta }_2+2\right)}\right.\\ \hfill \left.+U_2{V}_1\frac{{\left({\beta }_1-{\alpha }_1\right)}^{{\lambda }_2}{\left({\beta }_2-{\alpha }_2\right)}^{{\delta }_1}}{\left({\lambda }_2+1\right)\left({\delta }_1+1\right)\left({\lambda }_2+2\right)\left({\delta }_1+2\right)}+U_2{V}_2\frac{{\left({\beta }_2-{\alpha }_2\right)}^{{\lambda }_2+{\delta }_2}}{\left({\lambda }_2+{\delta }_2+1\right)\left({\lambda }_2+{\delta }_2+2\right)}\right],\end{array}$$

(2)

where $H,K:\left[{\alpha }_1,{\beta }_1\right]\times \left[{\alpha }_2,{\beta }_2\right]\to \mathbb{R}$ are two functions, with “×” representing here the standard Cartesian product, satisfying the conditions stated as

$$\begin{array}{c}\hfill \left|H\left(p,q\right)-H\left(s,t\right)\right|\le U_1{\left|p-s\right|}^{{\lambda }_1}+U_2{\left|q-t\right|}^{{\lambda }_2}\end{array}$$

and

$$\begin{array}{c}\hfill \left|K\left(p,q\right)-K\left(s,t\right)\right|\le V_1{\left|p-s\right|}^{{\delta }_1}+V_2{\left|q-t\right|}^{{\delta }_2},\end{array}$$

with $U_1,U_2>0$, ${\lambda }_1,{\lambda }_2\in (0,1]$, $V_1,V_2>0$, and ${\delta }_1,{\delta }_2\in (0,1]$.

Next, we present some notations and preliminaries. For $x_{\ell },y_{\ell }\in \mathbb{R}$ and each $1\le \ell \le n$, we consider the subset $\mathbb{E}:=\{(s_1,\cdots ,s_n):x_{\ell }\le s_{\ell }\le y_{\ell },1\le \ell \le n\}\subseteq {\mathbb{R}}^n$.

Now, for a real interval $[x_{\ell },y_{\ell }]$, with $\ell \in \{1,\cdots ,n\}$, we assume the partition

$$Q_{\ell }:=\left\{s_0^{\left(\ell \right)},s_1^{\left(\ell \right)},\cdots ,{s_{{\mu }_{\ell }}}^{\left(\ell \right)}\right\},$$

where $s_0^{(\ell )}<s_1^{(\ell )}<\cdots <s_n^{(\ell )}$, $s_0^{(\ell )}:=x_{\ell }$, $s_{{\mu }_{\ell }}^{(\ell )}:=y_{\ell }$ and $\Delta s_r^{(\ell )}:=s_r^{(\ell )}-s_{r-1}^{(\ell )}$, for all $r\in \{1,\cdots ,{\mu }_{\ell }\}$.

Let $Q:={\prod }_{\ell =1}^n{Q}_{\ell }$ be a partition of the n-dimensional interval $[\mathit{x},\mathit{y}]$$:={\prod }_{\ell =1}^n\left[x_{\ell },y_{\ell }\right]$, where $\mathit{x}:=(x_1,\cdots ,x_n)$ and $\mathit{y}:=(y_1,\cdots ,y_n)$. Then, we denote that

$$\begin{array}{c}\hfill {\int }_{\mathit{x}}^{\mathit{y}}f\left(\mathit{s}\right)\mathrm{d}\mathit{s}:={\int }_{x_n}^{y_n}\cdots {\int }_{x_1}^{y_1}f\left(s_1,\cdots ,s_n\right)\mathrm{d}{s}_1\cdots \mathrm{d}{s}_n,\end{array}$$

where $\mathit{s}:=(s_1,\cdots ,s_n)$. Now, for $j\in \{1,\cdots ,m\}$ and $\ell \in \{1,\cdots ,n\}$, let

$$\begin{array}{c}\hfill \Delta h_j\left(s_{\ell }^{\left(1\right)},s_{\ell }^{\left(2\right)},\cdots ,s_{\ell }^{\left(n\right)}\right):=h_j\left(s_{\ell }^{\left(1\right)},s_{\ell }^{\left(2\right)},\cdots ,s_{\ell }^{\left(n\right)}\right)-h_j\left(s_{\ell -1}^{\left(1\right)},s_{\ell -1}^{\left(2\right)},\cdots ,s_{\ell -1}^{\left(n\right)}\right).\end{array}$$

Note that the function $h_j(s_1,\cdots ,s_n)$ is said to be of bounded variation (in the Arzelà sense) [11,12], if there exists a positive number $A_j>0$ such that, for every partition on $\mathbb{E}$, we have

$$\begin{array}{c}\hfill \sum _{\ell =1}^n\left|\Delta h_j\left(s_{\ell }^{\left(1\right)},\cdots ,s_{\ell }^{\left(n\right)}\right)\right|\le A_j,\forall j\in \{1,\cdots ,m\}.\end{array}$$

Let $h_j$ be of bounded variation on $\mathbb{E}$, and $\sum \left(Q\right)$ denote the sum: ${\sum }_{\ell =1}^n|\Delta h_j\left(s_{\ell }^{\left(1\right)},\cdots ,s_{\ell }^{\left(n\right)}\right)|$, corresponding to the partition Q of $\mathbb{E}$. The number stated as

$$\begin{array}{c}\hfill \underset{\mathbb{E}}{\bigvee }\left(h_j\right):=sup\left\{\sum Q:Q\in \mathcal{Q}\left(\mathbb{E}\right)\right\}\end{array}$$

is called the total variation of $h_j$ on $\mathbb{E}$, where $\mathcal{Q}\left(\mathbb{E}\right)$ is the set for all partitions on $\mathbb{E}$.

A vector-valued function of several variables $\mathit{h}:\mathbb{E}\to {\mathbb{R}}^m$, with $\mathit{h}:=\left(h_1,\cdots ,h_m\right)$, is said to be of bounded variation if there exists a positive number $A>0$, such that, for every partition on $\mathbb{E}$, we obtain

$$\sum _{\ell =1}^n∥\Delta \mathit{h}\left(s_{\ell }^{\left(1\right)},\cdots ,s_{\ell }^{\left(n\right)}\right)∥\le A.$$

Here, $A:=\underset{1\le j\le m}{max}\left\{A_j\right\}$ and

$$\Delta \mathit{h}\left(s_{\ell }^{\left(1\right)},\cdots ,s_{\ell }^{\left(n\right)}\right):=\mathit{h}\left(s_{\ell }^{\left(1\right)},\cdots ,s_{\ell }^{\left(n\right)}\right)-\mathit{h}\left(s_{\ell -1}^{\left(1\right)},\cdots ,s_{\ell -1}^{\left(n\right)}\right).$$

Let $\mathit{h}$ be of bounded variation on $\mathbb{E}$. In addition, let $\sum \left(Q\right)$ denote the sum:

$$\sum _{i=1}^n\parallel \Delta \mathit{h}\left(s_{\ell }^{\left(1\right)},\cdots ,s_{\ell }^{\left(n\right)}\right)\parallel ,$$

corresponding to the partition Q of $\mathbb{E}$. Then, the number stated as

$$\begin{array}{c}\hfill \underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right):=sup\left\{\sum Q:Q\in \mathcal{Q}\left(\mathbb{E}\right)\right\}\end{array}$$

is called the total variation of $\mathit{h}$ on $\mathbb{E}$.

A function $h_j:\mathbb{E}\to \mathbb{R}$ is said to be of the Hölder type; that is, for each n-tuple $\mathit{s}:=\left(s_1,\cdots ,s_n\right)$ and $\mathit{t}:=\left(t_1,\cdots ,t_n\right)$ in $\mathbb{E}$, $h_j$ satisfies

$$\begin{array}{c}\hfill \left|h_j\left(\mathit{s}\right)-h_j\left(\mathit{t}\right)\right|\le \sum _{\ell =1}^n{A}_{\ell }{\left|s_{\ell }-t_{\ell }\right|}^r,\end{array}$$

(3)

for some $A_{\ell }>0$ and $r\in (0,1]$.

In general, a vector-valued function of several variables $\mathit{h}:\mathbb{E}\to {\mathbb{R}}^m$, with $\mathit{h}:=\left(h_1,\cdots ,h_m\right)$, is said to be of the Hölder type if, for each n-tuple $\mathit{s}:=\left(s_1,\cdots ,s_n\right)$ and $\mathit{t}:=\left(t_1,\cdots ,t_n\right)$ in $\mathbb{E}$, $\mathit{h}$ satisfies that

$$\begin{array}{c}\hfill \parallel \mathit{h}\left(\mathit{s}\right)-\mathit{h}\left(\mathit{t}\right)\parallel \le \sum _{\ell =1}^n{A}_{\ell }{\left|s_{\ell }-t_{\ell }\right|}^{r_{\ell }},\end{array}$$

(4)

if, and only if,

$$\begin{array}{c}\hfill \left|h_j\left(\mathit{s}\right)-h_j\left(\mathit{t}\right)\right|\le \sum _{\ell =1}^n{A}_{\ell }^{\left(j\right)}{\left|s_{\ell }-t_{\ell }\right|}^{r_{\ell }^{\left(j\right)}},\forall j\in \{1,\cdots ,m\},\end{array}$$

(5)

for some $A_{\ell }^{\left(j\right)}>0$ and $r_{\ell }^{\left(j\right)}\in (0,1]$. In our case, we have

$$A_{\ell }:=\underset{1\le j\le m}{max}\left\{A_{\ell }^{\left(j\right)}\right\},r_{\ell }:=\underset{1\le j\le m}{max}\left\{r_{\ell }^{\left(j\right)}\right\},\forall \ell \in \{1,\cdots ,n\}.$$

Consider vector-valued functions $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{R}}^m$. In this work, we study the Chebyshev functionals defined by

$$\begin{array}{c}\hfill {\mathcal{T}}_d\left(\mathit{h},\mathit{k}\right):=\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{s}\right)·\mathit{k}\left(\mathit{s}\right)\mathrm{d}\mathit{s}-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{s}\right)\mathrm{d}\mathit{s}·\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{s}\right)\mathrm{d}\mathit{s}\end{array}$$

(6)

and

$$\begin{array}{c}\hfill {\mathcal{T}}_c\left(\mathit{h},\mathit{k}\right):=\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{s}\right)\times \mathit{k}\left(\mathit{s}\right)\mathrm{d}\mathit{s}-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{s}\right)\mathrm{d}\mathit{s}\times \frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{s}\right)\mathrm{d}\mathit{s},\end{array}$$

(7)

where ${\mathcal{T}}_c\left(\mathit{h},\mathit{k}\right)$ is well defined only when $m=3$. This case only is considered in the paper. Note that, here, “×” and “·” denote the cross- and dot-products, respectively. We use the notation $\Delta :={\prod }_{\ell =1}^n\left(y_{\ell }-x_{\ell }\right)$.

The objective of this work is to prove vectorial versions of Grüss-type inequalities involving vector-valued functions defined on ${\mathbb{R}}^n$ for inner- and cross-products. In Section 2, we provide the main result of this investigation presenting vector versions, whereas Section 3 introduces the sharp inequality. In Section 4, some concluding remarks are stated.

2. The Vector Versions

In this section, we present our first main result by the following theorem.

Theorem 1.

Let $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{R}}^3$ be functions satisfying the conditions $\parallel \mathit{\gamma }\parallel \le \parallel \mathit{k}\left(\mathit{s}\right)\parallel \le \parallel \mathbf{\Gamma }\parallel$ and $\parallel \mathit{\varphi }\parallel \le \parallel \mathit{h}\left(\mathit{s}\right)\parallel \le \parallel \mathbf{\Phi }\parallel$, for all $\mathit{s}\in \mathbb{E}$ and for some real vectors $\mathit{\gamma },\mathbf{\Gamma },\mathit{\varphi },\mathbf{\Phi }\in {\mathbb{R}}^3$. Then, we have

$$\begin{array}{c}\hfill ∥{\mathcal{T}}_c\left(\mathit{h},\mathit{k}\right)∥\le \frac{1}{4}\left|sin\theta \right|\parallel \mathbf{\Phi }-\mathit{\varphi }\parallel \parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel \le \frac{1}{4}\parallel \mathbf{\Phi }-\mathit{\varphi }\parallel \parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel ,\end{array}$$

(8)

where $\theta \in \left[0,2\pi \right]$ is the angle between the vectors

$$\begin{array}{c}\hfill \mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t},\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}.\end{array}$$

Proof. 

We note that, for any two vectors $\mathit{p}:=(p_1,p_2,p_3)$ and $\mathit{q}:=(q_1,q_2,q_3)$ in ${\mathbb{R}}^3$, the cross-product is defined as

$$\begin{array}{c}\hfill \mathit{p}\times \mathit{q}:=\left|\begin{array}{ccc}{\mathit{e}}_1& {\mathit{e}}_2& {\mathit{e}}_3\\ p_1& p_2& p_3\\ q_1& q_2& q_3\end{array}\right|,\end{array}$$

where $|·|$ denotes the determinant of the $3\times 3$ given matrix. Moreover, ${\mathit{e}}_1,{\mathit{e}}_2,{\mathit{e}}_3$ are the unit coordinate vectors in ${\mathbb{R}}^3$, which are defined by ${\mathit{e}}_1:=(1,0,0)$, ${\mathit{e}}_2:=(0,1,0)$, and ${\mathit{e}}_3:=(0,0,1)$.

Let us recall the celebrated identity

$$\begin{array}{c}\hfill {∥\mathit{p}\times \mathit{q}∥}^2={∥\mathit{p}∥}^2{∥\mathit{q}∥}^2-{\left(\mathit{p}·\mathit{q}\right)}^2,\end{array}$$

(9)

for any vectors $\mathit{p},\mathit{q}\in {\mathbb{R}}^3$. The positive area formed by a parallelogram with sides $\mathit{p}$ and $\mathit{q}$ can be understood as the length of the cross-product. A celebrated definition of the dot-product is well-known and stated as

$$\begin{array}{c}\hfill \mathit{p}·\mathit{q}:=∥\mathit{p}∥∥\mathit{q}∥cos\theta .\end{array}$$

Thus, the identity presented in (9) can be written as

$$\begin{array}{c}\hfill {∥\mathit{p}\times \mathit{q}∥}^2={∥\mathit{p}∥}^2{∥\mathit{q}∥}^2\left(1-{cos}^2\theta \right),\end{array}$$

which is exactly

$$\begin{array}{c}\hfill ∥\mathit{p}\times \mathit{q}∥=∥\mathit{p}∥∥\mathit{q}∥\left|sin\theta \right|.\end{array}$$

(10)

Now, we observe that $\parallel \mathit{\gamma }\parallel \le \parallel \mathit{k}\left(\mathit{s}\right)\parallel \le \parallel \Gamma \parallel$ if, and only if, ${\gamma }_j\le k_j\left(\mathit{s}\right)\le {\Gamma }_j$, for each $j\in \{1,2,3\}$ and for all $\mathit{s}\in \mathbb{E}$, where $\parallel \mathit{\gamma }\parallel ={\left({\sum }_{j=1}^3{\gamma }_j^2\right)}^{1/2}$ and $\parallel \mathbf{\Gamma }\parallel ={\left({\sum }_{j=1}^3{\mathbf{\Gamma }}_j^2\right)}^{1/2}$.

Note that the cross-product version of the Korkine identity for vector-valued functions of several variables states that

$$\begin{array}{c}\hfill {\mathcal{T}}_c\left(\mathit{h},\mathit{k}\right)=\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left[\mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]\times \left[\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]\mathrm{d}\mathit{s}.\end{array}$$

(11)

Taking the Euclidean norm given in (11), employing the triangle inequality, and then considering the identity defined in (10), we get

$$\begin{array}{cc}\hfill ∥{\mathcal{T}}_c\left(\mathit{h},\mathit{k}\right)∥& =∥\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left[\mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]\times \left[\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]\mathrm{d}\mathit{s}∥\hfill \\ \hfill & \le \frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}∥\left[\mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]\times \left[\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]∥\mathrm{d}\mathit{s}\hfill \\ \hfill & =\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left\{∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\left|sin\theta \right|\right\}\mathrm{d}\mathit{s}.\hfill \end{array}$$

(12)

Now, we define

$$\begin{array}{cc}\hfill I& :=\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥}^2\mathrm{d}\mathit{s}\hfill \\ \hfill & =\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\sum _{j=1}^3{\left(k_j\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right)}^2\mathrm{d}\mathit{s}.\hfill \end{array}$$

Then, we get

$$\begin{array}{cc}\hfill \Delta ·I& ={\int }_{\mathit{x}}^{\mathit{y}}\sum _{j=1}^3\left(k_j^2\left(\mathit{s}\right)-2k_j\left(\mathit{s}\right)\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{t}\right)\mathrm{d}\mathit{t}+{\left(\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right)}^2\right)\mathrm{d}\mathit{s}\hfill \\ \hfill & ={\int }_{\mathit{x}}^{\mathit{y}}\sum _{j=1}^3\left(k_j^2\left(\mathit{s}\right)-2k_j\left(\mathit{s}\right)\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{t}\right)\mathrm{d}\mathit{t}+{\left(\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right)}^2\right)\mathrm{d}\mathit{s}\hfill \\ \hfill & =\sum _{j=1}^3\left\{{\int }_{\mathit{x}}^{\mathit{y}}{k}_j^2\left(\mathit{s}\right)\mathrm{d}\mathit{s}-2{\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{s}\right)\mathrm{d}\mathit{s}\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{t}\right)\mathrm{d}\mathit{t}+{\int }_{\mathit{x}}^{\mathit{y}}{\left(\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right)}^2\mathrm{d}\mathit{s}\right\}\hfill \\ \hfill & ={\displaystyle \sum _{j=1}^3}\left\{{\int }_{\mathit{x}}^{\mathit{y}}{k}_j^2\left(\mathit{s}\right)\mathrm{d}\mathit{s}-\frac{2}{\Delta }{\left({\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right)}^2+\frac{\prod _{\ell =1}^n\left(y_{\ell }-x_{\ell }\right)}{{\Delta }^2}{\left({\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right)}^2\right\}\hfill \\ \hfill & =\sum _{j=1}^3\left\{{\int }_{\mathit{x}}^{\mathit{y}}{k}_j^2\left(\mathit{s}\right)\mathrm{d}\mathit{s}-\frac{1}{\Delta }{\left({\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{s}\right)\mathrm{d}\mathit{s}\right)}^2\right\}\hfill \\ \hfill & ={\int }_{\mathit{x}}^{\mathit{y}}{\parallel \mathit{k}\left(\mathit{s}\right)\parallel }^2\mathrm{d}\mathit{s}-\frac{1}{\Delta }{∥{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{s}\right)\mathrm{d}\mathit{s}∥}^2.\hfill \end{array}$$

In addition, we have

$$\begin{array}{cc}\hfill I_j& :=\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{\left(k_j\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right)}^2\mathrm{d}\mathit{s},j\in \{1,2,3\},\hfill \\ \hfill & =\left({\Gamma }_j-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{s}\right)\mathrm{d}\mathit{s}\right)\left(\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{s}\right)\mathrm{d}\mathit{s}-{\gamma }_j\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left({\Gamma }_j-k_j\left(\mathit{s}\right)\right)\left(k_j\left(\mathit{s}\right)-{\gamma }_j\right)\mathrm{d}\mathit{s}.\hfill \end{array}$$

As ${\gamma }_j\le k_j\left(\mathit{s}\right)\le {\Gamma }_j$, for all $\mathit{s}\in \mathbb{E}$, hence

$$\begin{array}{c}\hfill \frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left({\Gamma }_j-k_j\left(\mathit{s}\right)\right)\left(k_j\left(\mathit{s}\right)-{\gamma }_j\right)\mathrm{d}\mathit{s}\ge 0,\end{array}$$

which implies that

$$\begin{array}{c}\hfill I_j\le \left({\Gamma }_j-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{s}\right)\mathrm{d}\mathit{s}\right)\left(\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{k}_j\left(\mathit{s}\right)\mathrm{d}\mathit{s}-{\gamma }_j\right).\end{array}$$

Using the most basic inequality stated as

$$\begin{array}{c}\hfill \left(Y-Z\right)\left(Z-X\right)\le \frac{1}{4}{\left(Y-X\right)}^2,\end{array}$$

which is valid for all $X,Y,Z\in \mathbb{R}$, we get

$$\begin{array}{c}\hfill I_j\le \frac{1}{4}{\left({\Gamma }_j-{\gamma }_j\right)}^2.\end{array}$$

(13)

However, since I$:={\sum }_{j=1}^3{I}_j$, from the Cauchy–Buniakowski–Schwarz integral inequality, it follows that

$$\begin{array}{c}\hfill I\ge {\left[\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}\right]}^2,\end{array}$$

which from (13) gives

$$\begin{array}{c}\hfill {\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}\le \frac{1}{2}\Delta \parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel .\end{array}$$

(14)

In a similar vein, we have

$$\begin{array}{c}\hfill {\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}\le \frac{1}{2}\Delta \parallel \mathbf{\Phi }-\mathit{\varphi }\parallel .\end{array}$$

(15)

Combining the inequalities established in (13), (14), and (15), we reach the required result. The last inequality holds, since $\left|sin\theta \right|\le 1$, and so the proof is complete. □

Theorem 2.

Let $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{R}}^m$ be functions satisfying the conditions $\parallel \mathit{\gamma }\parallel \le \parallel \mathit{k}\left(\mathit{s}\right)\parallel \le \parallel \mathbf{\Gamma }\parallel$ and $\parallel \mathit{\varphi }\parallel \le \parallel \mathit{h}\left(\mathit{s}\right)\parallel \le \parallel \mathbf{\Phi }\parallel$, for all $\mathit{s}\in \mathbb{E}$ and for some real vectors $\mathit{\gamma },\mathbf{\Gamma },\mathit{\varphi },\mathbf{\Phi }\in {\mathbb{R}}^m$. Then, we obtain

$$\begin{array}{c}\hfill \left|{\mathcal{T}}_d(\mathit{h},\mathit{k})\right|\le \frac{1}{4}\parallel \mathbf{\Phi }-\mathit{\varphi }\parallel \parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel \left|cos\theta \right|\le \frac{1}{4}\parallel \mathbf{\Phi }-\mathit{\varphi }\parallel \parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel ,\end{array}$$

(16)

where $\theta \in \left[0,2\pi \right]$ is the angle between the vectors

$$\begin{array}{c}\hfill \mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t},\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}.\end{array}$$

Proof. 

The dot-product version of the Korkine identity for vector-valued functions of several variables states that

$$\begin{array}{c}\hfill {\mathcal{T}}_d\left(\mathit{h},\mathit{k}\right):=\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left[\mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]·\left[\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]\mathrm{d}\mathit{s}.\end{array}$$

(17)

Taking the absolute value of the expression given in (17) and employing the triangle inequality, we get

$$\begin{array}{cc}\hfill \left|{\mathcal{T}}_d\left(\mathit{h},\mathit{k}\right)\right|& =\left|\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left[\mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]·\left[\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]\mathrm{d}\mathit{s}\right|\hfill \\ \hfill & \le \frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left|\left[\mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]·\left[\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]\right|\mathrm{d}\mathit{s}\hfill \\ \hfill & =\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left\{∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\left|cos\theta \right|\right\}\mathrm{d}\mathit{s}.\hfill \end{array}$$

(18)

Now, we proceed as in the proof of Theorem 1, but for all $j\in \{1,\dots ,m\}$. Then, we have

$$\begin{array}{c}\hfill {\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}\le \frac{1}{2}\Delta \parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel \end{array}$$

(19)

and

$$\begin{array}{c}\hfill {\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}\le \frac{1}{2}\Delta \parallel \mathbf{\Phi }-\mathit{\varphi }\parallel .\end{array}$$

(20)

Combining the inequalities established in (18), (19), and (20), we obtain the required result. The last inequality holds since $\left|cos\theta \right|\le 1$, and so the proof is complete. □

Theorem 3.

Let $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{R}}^3$ be such that $\mathit{h}$ satisfies

$$\begin{array}{c}\hfill \parallel \mathit{h}\left(\mathit{s}\right)-\mathit{h}\left(\mathit{t}\right)\parallel \le \sum _{\ell =1}^n{A}_{\ell }{\left|s_{\ell }-t_{\ell }\right|}^{r_{\ell }},\end{array}$$

where $A_{\ell }>0$, $r_{\ell }\in (0,1]$ and there exist real vectors $\mathit{\gamma },\mathbf{\Gamma }\in {\mathbb{R}}^3$, such that $\parallel \mathit{\gamma }\parallel \le \parallel \mathit{k}\left(\mathit{s}\right)\parallel \le \parallel \mathbf{\Gamma }\parallel$, for all $\mathit{s}\in \mathbb{E}$. Then, we have

$$\begin{array}{c}\hfill ∥{\mathcal{T}}_c\left(\mathit{h},\mathit{k}\right)∥\le \frac{1}{2^{n+1}}\left|sin\theta \right|\parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel ·\sum _{i=1}^n{A}_{\ell }{\left(y_{\ell }-x_{\ell }\right)}^{r_{\ell }}.\end{array}$$

(21)

Proof. 

A general Korkine identity for vector-valued functions of several variables states that

$$\begin{array}{c}\hfill {\mathcal{T}}_c\left(\mathit{h},\mathit{k}\right):=\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left[\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)\right]\times \left[\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]\mathrm{d}\mathit{s},\end{array}$$

(22)

where the summation is over all $\mathit{d}:=(d_1,\cdots ,d_n)\in {\mathbb{R}}^n$, such that $d_{\ell }\in \{x_{\ell },y_{\ell }\}$, for $1\le \ell \le n$.

Taking the Euclidean norm in the expression given in (22) and utilizing the triangle inequality, we get

$$\begin{array}{cc}\hfill ∥{\mathcal{T}}_c\left(\mathit{h},\mathit{k}\right)∥& =∥\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left[\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)\right]\times \left[\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]\mathrm{d}\mathit{s}∥\hfill \\ \hfill & \le \frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}∥\left[\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)\right]\times \left[\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]∥\mathrm{d}\mathit{s}\hfill \\ \hfill & =\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left\{\parallel \mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)\parallel \parallel \mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\parallel sin\theta \right\}\mathrm{d}\mathit{s}.\hfill \end{array}$$

(23)

From the proof of Theorem 1, we obtain

$$\begin{array}{c}\hfill {\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}\le \frac{1}{2}\Delta \parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel .\end{array}$$

(24)

Now, by assumption, we have

$$\begin{array}{c}\hfill ∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)∥\le \frac{1}{2^n}\sum _{\mathit{d}}\parallel \mathit{h}\left(\mathit{s}\right)-\mathit{h}\left(\mathit{d}\right)\parallel \le \frac{1}{2^n}\sum _{\ell =1}^n{A}_{\ell }{\left|s_{\ell }-d_{\ell }\right|}^{r_{\ell }}.\end{array}$$

Therefore, it follows that

$$\begin{array}{cc}\hfill \underset{\mathit{s}\in \mathbb{E}}{sup}∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)∥\le \frac{1}{2^n}\sum _{\mathit{d}}\underset{\mathit{s}\in \mathbb{E}}{sup}\parallel \mathit{h}\left(\mathit{s}\right)-\mathit{h}\left(\mathit{d}\right)\parallel & \le \frac{1}{2^n}\sum _{\ell =1}^n{A}_{\ell }\underset{\mathit{s}\in \mathbb{E}}{sup}{\left|s_{\ell }-d_{\ell }\right|}^{r_{\ell }}\hfill \\ \hfill & =\frac{1}{2^n}\sum _{\ell =1}^n{A}_{\ell }{\left(y_{\ell }-x_{\ell }\right)}^{r_{\ell }}.\hfill \end{array}$$

(25)

Combining the expressions formulated in (23), (24), and (25), we get the desired result established in (21). □

Theorem 4.

Let $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{R}}^m$ be such that $\mathit{h}$ satisfies

$$\begin{array}{c}\hfill \parallel \mathit{h}\left(\mathit{s}\right)-\mathit{h}\left(\mathit{t}\right)\parallel \le \sum _{\ell =1}^n{A}_{\ell }{\left|s_{\ell }-t_{\ell }\right|}^{r_{\ell }},\end{array}$$

where $A_{\ell }>0$, $r_{\ell }\in (0,1]$ and there exist real vectors $\mathit{\gamma },\mathbf{\Gamma }\in {\mathbb{R}}^m$, such that $\parallel \mathit{\gamma }\parallel \le \parallel \mathit{k}\left(\mathit{s}\right)\parallel \le \parallel \mathbf{\Gamma }\parallel$, for all $\mathit{s}\in \mathbb{E}$. Then, we get

$$\begin{array}{c}\hfill \left|{\mathcal{T}}_d\left(\mathit{h},\mathit{k}\right)\right|\le \frac{1}{2^{n+1}}\left|cos\theta \right|\parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel ·\sum _{\ell =1}^n{A}_{\ell }{\left(y_{\ell }-x_{\ell }\right)}^{r_{\ell }}.\end{array}$$

(26)

Proof. 

A general Korkine identity for vector-valued functions of several variables states that

$$\begin{array}{c}\hfill {\mathcal{T}}_d\left(\mathit{h},\mathit{k}\right):=\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left[\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)\right]·\left[\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]\mathrm{d}\mathit{s},\end{array}$$

(27)

where the summation is over all $\mathit{d}:=\left(d_1,\cdots ,d_n\right)\in {\mathbb{R}}^n$ such that $d_{\ell }\in \{x_{\ell },y_{\ell }\}$, for $1\le \ell \le n$.

Taking the absolute value in the expression given in (27) and utilizing the triangle inequality, we get

$$\begin{array}{cc}\hfill \left|{\mathcal{T}}_d\left(\mathit{h},\mathit{k}\right)\right|& =\left|\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left[\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)\right]·\left[\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]\mathrm{d}\mathit{s}\right|\hfill \\ \hfill & \le \frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left|\left[\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)\right]·\left[\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\right]\right|\mathrm{d}\mathit{s}\hfill \\ \hfill & =\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\left\{\parallel \mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)\parallel \parallel \mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\parallel \left|cos\theta \right|\right\}\mathrm{d}\mathit{s}.\hfill \end{array}$$

(28)

From the proof of Theorem 2, we have

$$\begin{array}{c}\hfill {\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}\le \frac{1}{2}\Delta \parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel .\end{array}$$

(29)

Now, by assumption, we have

$$\begin{array}{c}\hfill ∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)∥\le \frac{1}{2^n}\sum _{\mathit{d}}∥\mathit{h}\left(\mathit{s}\right)-\mathit{h}\left(\mathit{d}\right)∥\le \frac{1}{2^n}\sum _{\ell =1}^n{A}_{\ell }{\left|s_{\ell }-d_{\ell }\right|}^{r_{\ell }}.\end{array}$$

Therefore, it follows that

$$\begin{array}{cc}\hfill \underset{\mathit{s}\in \mathbb{E}}{sup}∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)∥\le \frac{1}{2^n}\sum _{\mathit{d}}\underset{\mathit{s}\in \mathbb{E}}{sup}∥\mathit{h}\left(\mathit{s}\right)-\mathit{h}\left(\mathit{d}\right)∥& \le \frac{1}{2^n}\sum _{\ell =1}^n{A}_{\ell }\underset{\mathit{s}\in \mathbb{E}}{sup}{\left|s_{\ell }-d_{\ell }\right|}^{r_{\ell }}\hfill \\ \hfill & =\frac{1}{2^n}\sum _{\ell =1}^n{A}_{\ell }{\left(y_{\ell }-x_{\ell }\right)}^{r_{\ell }}.\hfill \end{array}$$

(30)

Combining the expressions formulated in (28), (29), and (30), we obtain the desired result established in (26). □

3. Sharp Inequalities

In this section, we consider some sharp bounds for both Chebyshev functionals stated in (6) and (7). The following theorem holds.

Theorem 5.

Let $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{R}}^3$ be such that $\mathit{h}$ is of bounded variation on $\mathbb{E}$ and $\mathit{k}$ as in Theorem 1. Then, we have

$$\begin{array}{c}\hfill ∥{\mathcal{T}}_c\left(\mathit{h},\mathit{k}\right)∥\le \frac{1}{2^{n+1}}\left|sin\theta \right|·\parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel ·\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)\le \frac{1}{2^{n+1}}\parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel ·\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right),\end{array}$$

(31)

with the constant $1/2^{n+1}$ in the right-hand side of both inequalities being the best possible.

Proof. 

As in Theorem 1, note that

$$\begin{array}{c}\hfill ∥{\mathcal{T}}_c\left(\mathit{h},\mathit{k}\right)∥\le \left|sin\theta \right|\underset{\mathit{s}\in \mathbb{E}}{sup}∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)∥\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}.\end{array}$$

However, as there exist real vectors $\mathit{\gamma },\Gamma \in {\mathbb{R}}^3$, such that $\parallel \mathit{\gamma }\parallel \le \parallel \mathit{k}\left(\mathit{s}\right)\parallel \le \parallel \Gamma \parallel$ for all $\mathit{s}\in \mathbb{E}$, then

$$\begin{array}{c}\hfill {\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}\le \frac{1}{2}\Delta \parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel .\end{array}$$

Since $\mathit{h}$ is of bounded variation on $\mathbb{E}$, we have that

$$\begin{array}{c}\hfill \underset{\mathit{s}\in \mathbb{E}}{sup}∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)∥\le \frac{1}{2^n}\sum _{\mathit{d}}\underset{\mathit{s}\in \mathbb{E}}{sup}∥\mathit{h}\left(\mathit{s}\right)-\mathit{h}\left(\mathit{d}\right)∥\le \frac{1}{2^n}\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right).\end{array}$$

Merging the above inequalities, we get the required result given in (31). To prove that the sharpness of the expression stated in (31) holds for a constant $C>0$, that is,

$$\begin{array}{c}\hfill ∥{\mathcal{T}}_c\left(\mathit{h},\mathit{k}\right)∥\le C\parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel ·\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right),\end{array}$$

(32)

consider the vector-valued functions $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{R}}^3$ defined as $\mathit{h}:=(h,h,h)$ and $\mathit{k}:=(k,k,k)$ with

$$\begin{array}{c}\hfill h\left(\mathit{t}\right):=k\left(\mathit{t}\right)=\prod _{\ell =1}^n\mathrm{sign}\left(t_{\ell }-\frac{x_{\ell }+y_{\ell }}{2}\right).\end{array}$$

Note that $\mathit{h}:=(h,h,h)$ is of bounded variation on $\mathbb{E}$. Furthermore, we have

$$\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)=2^n,\parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel =2,{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}={\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}=0,{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}=\mathit{y}-\mathit{x}.$$

By using the expression given in (32), we get $1/2^{n+1}\le C$, which proves that $1/2^{n+1}$ is the best possible. Therefore, the proof is complete. □

Theorem 6.

Let $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{R}}^m$ be such that $\mathit{h}$ is of bounded variation on $\mathbb{E}$ and $\mathit{k}$ as in Theorem 1. Then, we have

$$\begin{array}{c}\hfill \left|{\mathcal{T}}_d\left(\mathit{h},\mathit{k}\right)\right|\le \frac{1}{2^{n+1}}\left|cos\theta \right|·\parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel ·\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)\le \frac{1}{2^{n+1}}\parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel ·\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right),\end{array}$$

(33)

with the constant $1/2^{n+1}$ in the right-hand side of both inequalities being the best possible.

Proof. 

As in Theorem 2, note that

$$\begin{array}{c}\hfill \left|{\mathcal{T}}_d\left(\mathit{h},\mathit{k}\right)\right|\le \left|cos\theta \right|·\underset{\mathit{s}\in \mathbb{E}}{sup}\parallel \mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)\parallel \frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\parallel \mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\parallel \mathrm{d}\mathit{s}.\end{array}$$

However, as there exist real vectors $\mathit{\gamma },\mathbf{\Gamma }\in {\mathbb{R}}^m$, such that $\parallel \mathit{\gamma }\parallel \le \parallel \mathit{k}\left(\mathit{s}\right)\parallel \le \parallel \mathbf{\Gamma }\parallel$ for all $\mathit{s}\in \mathbb{E}$, then

$$\begin{array}{c}\hfill {\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}\le \frac{1}{2}\Delta \parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel .\end{array}$$

Since $\mathit{h}$ is of bounded variation on $\mathbb{E}$, we have that

$$\begin{array}{c}\hfill \underset{\mathit{s}\in \mathbb{E}}{sup}∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)∥\le \frac{1}{2^n}\sum _{\mathit{d}}\underset{\mathit{s}\in \mathbb{E}}{sup}\parallel \mathit{h}\left(\mathit{s}\right)-\mathit{h}\left(\mathit{d}\right)\parallel \le \frac{1}{2^n}\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right).\end{array}$$

Merging the above inequalities, we get the required result given in (33). To prove that the sharpness of the expression stated in (33) holds for a constant $C>0$, that is,

$$\begin{array}{c}\hfill ∥{\mathcal{T}}_c\left(\mathit{h},\mathit{k}\right)∥\le C\parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel ·\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right),\end{array}$$

(34)

consider the vector-valued functions $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{R}}^m$ defined as $\mathit{h}:=(h,\cdots ,h)$ and $\mathit{k}:=(k,\cdots ,k)$ with

$$\begin{array}{c}\hfill h\left(\mathit{t}\right):=k\left(\mathit{t}\right)=\prod _{\ell =1}^n\mathrm{sign}\left(t_{\ell }-\frac{x_{\ell }+y_{\ell }}{2}\right).\end{array}$$

Note that $\mathit{h}:=(h,\cdots ,h)$ is of bounded variation on $\mathbb{E}$. Furthermore, we have

$$\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)=2^n,\parallel \mathbf{\Gamma }-\mathit{\gamma }\parallel =2,{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}={\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}=0,{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}=\mathit{y}-\mathit{x}.$$

By using the expression given in (34), we get $1/2^{n+1}\le C$, which proves that $1/2^{n+1}$ is the best possible. Therefore, the proof is complete. □

Theorem 7.

Let $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{C}}^3$ be such that $\mathit{h}$ is of bounded variation on $\mathbb{E}$ and $\mathit{k}$ is a Lebesgue integrable function on $\mathbb{E}$. Then, we get

$$\begin{array}{cc}\hfill ∥{\mathcal{T}}_c\left(\mathit{h},\mathit{k}\right)∥& \le \frac{1}{2^n}\left|sin\theta \right|\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}\hfill \\ \hfill & \le \frac{1}{2^n}\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s},\hfill \end{array}$$

(35)

with the constant $1/2^n$ in the right-hand side of both inequalities being the best possible.

Proof. 

As in Theorem 5, we note that

$$\begin{array}{c}\hfill ∥{\mathcal{T}}_c\left(\mathit{h},\mathit{k}\right)∥\le \left|sin\theta \right|\underset{\mathit{s}\in \mathbb{E}}{sup}\parallel \mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)\parallel \frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\parallel \mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\parallel \mathrm{d}\mathit{s}.\end{array}$$

Owing to the fact that $\mathit{h}$ is of bounded variation on $\mathbb{E}$, we have

$$\begin{array}{c}\hfill \underset{\mathit{s}\in \mathbb{E}}{sup}∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)∥\le \frac{1}{2^n}\sum _{\mathit{d}}\underset{\mathit{s}\in \mathbb{E}}{sup}\parallel \mathit{h}\left(\mathit{s}\right)-\mathit{h}\left(\mathit{d}\right)\parallel \le \frac{1}{2^n}\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right).\end{array}$$

Merging the obtained inequalities, we reach the required result given in (35). To prove that the sharpness of the expression stated in (35) holds for the constant $C_1>0$, that is,

$$\begin{array}{c}\hfill \left|\mathcal{T}\left(\mathit{h},\mathit{k}\right)\right|\le C_1\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s},\end{array}$$

(36)

consider the functions $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{R}}^3$, with $\mathit{h}:=(h,h,h)=\mathit{k}$ defined as

$$\begin{array}{c}\hfill h\left(\mathit{t}\right):=k\left(\mathit{t}\right)=\prod _{\ell =1}^n\mathrm{sign}\left(t_{\ell }-\frac{x_{\ell }+y_{\ell }}{2}\right).\end{array}$$

Observe that $\mathit{h}$ is of bounded variation on $\mathbb{E}$ and

$$\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)=2^n,{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)d\mathit{t}=0,{\int }_{\mathit{x}}^{\mathit{y}}\parallel \mathit{k}\left(\mathit{t}\right)\parallel \mathrm{d}\mathit{t}={\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}=\frac{1}{2^{2n}}\prod _{\ell =1}^n{\left(y_{\ell }-x_{\ell }\right)}^2.$$

By using the expression given in (36), we get $1/2^n\le C_1$, which proves that $1/2^n$ is the best possible for the formula presented in (35), and so the proof is complete. □

Theorem 8.

Let $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{C}}^m$ be such that $\mathit{h}$ is of bounded variation on $\mathbb{E}$ and $\mathit{k}$ is a Lebesgue integrable function on $\mathbb{E}$. Then, we get

$$\begin{array}{cc}\hfill \left|{\mathcal{T}}_d\left(\mathit{h},\mathit{k}\right)\right|& \le \frac{1}{2^n}\left|cos\theta \right|\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}\hfill \\ \hfill & \le \frac{1}{2^n}\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s},\hfill \end{array}$$

(37)

with the constant $1/2^n$ in the right-hand side of both inequalities being the best possible.

Proof. 

As in Theorem 6, we note that

$$\begin{array}{c}\hfill \left|\mathcal{T}\left(\mathit{h},\mathit{k}\right)\right|\le \left|cos\theta \right|·\underset{\mathit{s}\in \mathbb{E}}{sup}\parallel \mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)\parallel \frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\parallel \mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}\parallel \mathrm{d}\mathit{s}.\end{array}$$

Owing to the fact that $\mathit{h}$ is of bounded variation on $\mathbb{E}$, we have

$$\begin{array}{c}\hfill \underset{\mathit{s}\in \mathbb{E}}{sup}∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{2^n}\sum _{\mathit{d}}\mathit{h}\left(\mathit{d}\right)∥\le \frac{1}{2^n}\sum _{\mathit{d}}\underset{\mathit{s}\in \mathbb{E}}{sup}\parallel \mathit{h}\left(\mathit{s}\right)-\mathit{h}\left(\mathit{d}\right)\parallel \le \frac{1}{2^n}\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right).\end{array}$$

Merging the obtained inequalities, we reach the required result given in (37). To prove that the sharpness of the expression stated in (37) holds for the constant $C_1>0$, that is,

$$\begin{array}{c}\hfill \left|\mathcal{T}\left(\mathit{h},\mathit{k}\right)\right|\le C_1\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s},\end{array}$$

(38)

consider the functions $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{R}}^m$, with $\mathit{h}:=(h,\cdots ,h)=\mathit{k}$ defined as

$$\begin{array}{c}\hfill h\left(\mathit{t}\right):=k\left(\mathit{t}\right)=\prod _{\ell =1}^n\mathrm{sign}\left(t_{\ell }-\frac{x_{\ell }+y_{\ell }}{2}\right).\end{array}$$

Observe that $\mathit{h}$ is of bounded variation on $\mathbb{E}$ and

$$\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)=2^n,{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)d\mathit{t}=0,{\int }_{\mathit{x}}^{\mathit{y}}\parallel \mathit{k}\left(\mathit{t}\right)\parallel \mathrm{d}\mathit{t}={\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}=\frac{1}{2^{2n}}\prod _{\ell =1}^n{\left(y_{\ell }-x_{\ell }\right)}^2.$$

By using the expression given in (38), we get $1/2^n\le C_1$, which proves that $1/2^n$ is the best possible for the formula presented in (37), and so the proof is complete. □

The variance of the function $\mathit{h}:\mathbb{E}\to {\mathbb{C}}^m$, which is square integrable on $\mathbb{E}$ by ${\mathcal{K}}_d\left(\mathit{h}\right)$, is defined as

$$\begin{array}{cc}\hfill {\mathcal{K}}_d\left(\mathit{h}\right)& :={\left[{\mathcal{T}}_d\left(\mathit{h},\overline{\mathit{h}}\right)\right]}^{1/2}\hfill \\ \hfill & ={\left[\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}{∥\mathit{h}\left(\mathit{t}\right)∥}^2\mathrm{d}\mathit{t}-\frac{1}{\Delta }{∥{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥}^2\right]}^{1/2}\hfill \\ \hfill & ={\left[{\mathcal{T}}_c\left(\mathit{h},\overline{\mathit{h}}\right)\right]}^{1/2},(\mathrm{in} \mathrm{this} \mathrm{case} \mathrm{we} \mathrm{assume} m=3, \mathrm{that} \mathrm{is},\mathit{h}\in {\mathbb{C}}^3)\hfill \\ \hfill & ={\mathcal{K}}_c\left(\mathit{h}\right),\hfill \end{array}$$

(39)

where $\overline{\mathit{h}}$ denotes the complex conjugate function of $\mathit{h}$.

Corollary 1.

Let $\mathit{h}:\mathbb{E}\to {\mathbb{R}}^m$ and $\mathit{g}:\mathbb{E}\to {\mathbb{R}}^3$ be two functions which are of bounded variation on $\mathbb{E}$. Then, we have that

$$\begin{array}{c}\hfill {\mathcal{K}}_d\left(\mathit{h}\right)\le \frac{1}{2^n}\left|cos\theta \right|\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)\le \frac{1}{2^n}\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)\end{array}$$

(41)

and

$$\begin{array}{c}\hfill {\mathcal{K}}_c\left(\mathit{g}\right)\le \frac{1}{2^n}\left|sin\theta \right|\underset{\mathbb{E}}{\bigvee }\left(\mathit{g}\right)\le \frac{1}{2^n}\underset{\mathbb{E}}{\bigvee }\left(\mathit{g}\right),\end{array}$$

(42)

with the constant $1/2^n$ in the right-hand side of both inequalities being the best possible.

Proof. 

We prove (41). By applying Theorem 6 for $\mathit{k}=\overline{\mathit{h}}$, we get

$$\begin{array}{cc}\hfill {\mathcal{K}}_d^2\left(\mathit{h}\right)& \le \frac{1}{2^n}\left|cos\theta \right|\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}\hfill \\ \hfill & \le \frac{1}{2^n}\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}.\hfill \end{array}$$

(43)

By the Cauchy–Buniakowski–Schwarz inequality, we have

$$\begin{array}{c}\hfill \frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{h}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{h}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}\le {\mathcal{K}}_d\left(\mathit{h}\right).\end{array}$$

(44)

By combining the expressions given in (43) and (44), we obtain the required result. Now, if we choose $\mathit{h}:\mathbb{E}\to {\mathbb{R}}^m$, for $\mathit{h}:=(h,\cdots ,h)$, with

$$\begin{array}{c}\hfill h\left(\mathit{s}\right):=\prod _{\ell =1}^n\mathrm{sign}\left(s_{\ell }-\frac{x_{\ell }+y_{\ell }}{2}\right),\end{array}$$

then we establish the sharpness of the constant $1/2^n$ whose details are avoided. The proof of (42) goes similarly by applying Theorem 5, and the details are omitted. □

When both vector-valued functions have a bounded variation, we can now state the following theorem.

Theorem 9.

Let $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{R}}^m$ and $\mathit{f},\mathit{g}:\mathbb{E}\to {\mathbb{R}}^3$ be of bounded variation on $\mathbb{E}$. Then, we get that

$$\begin{array}{c}\hfill \left|{\mathcal{T}}_d\left(\mathit{h},\mathit{k}\right)\right|\le \frac{1}{2^{2n}}\left|cos\theta \right|\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)·\underset{\mathbb{E}}{\bigvee }\left(\mathit{k}\right)\le \frac{1}{2^{2n}}\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)·\underset{\mathbb{E}}{\bigvee }\left(\mathit{k}\right)\end{array}$$

(45)

and

$$\begin{array}{c}\hfill ∥{\mathcal{T}}_c\left(\mathit{f},\mathit{g}\right)∥\le \frac{1}{2^{2n}}\left|sin\theta \right|\underset{\mathbb{E}}{\bigvee }\left(\mathit{f}\right)·\underset{\mathbb{E}}{\bigvee }\left(\mathit{g}\right)\le \frac{1}{2^{2n}}\underset{\mathbb{E}}{\bigvee }\left(\mathit{f}\right)·\underset{\mathbb{E}}{\bigvee }\left(\mathit{g}\right),\end{array}$$

(46)

with the best possible constant in the right-hand side of both inequalities being $1/2^{2n}$.

Proof. 

From Theorem 6 and Corollary 1, it follows that

$$\begin{array}{cc}\hfill \left|{\mathcal{T}}_d\left(\mathit{h},\mathit{k}\right)\right|& \le \frac{1}{2^n}\left|cos\theta \right|\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)·\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}∥\mathit{k}\left(\mathit{s}\right)-\frac{1}{\Delta }{\int }_{\mathit{x}}^{\mathit{y}}\mathit{k}\left(\mathit{t}\right)\mathrm{d}\mathit{t}∥\mathrm{d}\mathit{s}\hfill \\ \hfill & \le \frac{1}{2^n}\left|cos\theta \right|\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)·{\mathcal{K}}_d\left(\mathit{k}\right)\le \frac{1}{2^{2n}}\left|cos\theta \right|\underset{\mathbb{E}}{\bigvee }\left(\mathit{h}\right)·\underset{\mathbb{E}}{\bigvee }\left(\mathit{k}\right).\hfill \end{array}$$

The case of equality is obtained in the expression given in (45) when $\mathit{h},\mathit{k}:\mathbb{E}\to {\mathbb{R}}^m$, for $\mathit{h}:=(h,\cdots ,h)=\mathit{k}$, with

$$\begin{array}{c}\hfill h\left(\mathit{s}\right):=k\left(\mathit{s}\right)=\prod _{\ell =1}^n\mathrm{sign}\left(s_{\ell }-\frac{x_{\ell }+y_{\ell }}{2}\right).\end{array}$$

The proof of (46) goes similarly to the previous case by applying Theorem 5 and Corollary 1, replacing ${\mathcal{K}}_d\left(\mathit{k}\right)$ by ${\mathcal{K}}_c\left(\mathit{g}\right)$ in the proof of (45), and the details are omitted. □

4. Concluding Remarks

In this work, we have proved vectorial versions of Grüss-type inequalities involving vector-valued functions defined on ${\mathbb{R}}^n$ for inner- and cross-products. These results can be helpful for different purposes and applications.