Finite Field Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality

1. Introduction

Let $\mathcal{X}$ be a normed linear space (NLS) over scalar field. Definition of the norm has the triangle inequality

$$\begin{array}{c}\hfill \parallel x+y\parallel \le \parallel x\parallel +\parallel y\parallel ,\forall x,y\in \mathcal{X}.\end{array}$$

(1)

In 1992, Hudzik and Landes derived a breakthrough generalization of Inequality (1) which is valid for any two nonzero elements in a NLS [1].

Theorem 1.

[1] (Hudzik-Landes Inequlaity) Let $\mathcal{X}$ be a NLS. Then for all $x,y\in \mathcal{X}\setminus \left\{0\right\}$,

$$\begin{array}{c}\hfill \parallel x+y\parallel \le \parallel x\parallel +\parallel y\parallel -\left(2-∥{\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}∥\right)min\{\parallel x\parallel ,\parallel y\parallel \}.\end{array}$$

(2)

In 2006, Maligranda independently derived Inequality (2) [2]. It is natural to look for a generalization of Inequality (2) to more than two non-zero vectors. This is done independently by Dragomir in 2006 [3] and by Kato, Saito and Tamura in 2007 [4].

Theorem 2.

[3,4] (Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality) Let $\mathcal{X}$ be a NLS and $n\in \mathbb{N}$. Then for all $x_1,\dots ,x_n\in \mathcal{X}\setminus \left\{0\right\}$, we have

$$\begin{array}{c}\hfill ∥\sum _{j=1}^n{x}_j∥\le \sum _{j=1}^n\parallel x_j\parallel -\left(n-∥\sum _{j=1}^n{\displaystyle \frac{x_j}{\parallel x_j\parallel }}∥\right)\underset{1\le k\le n}{min}\parallel x_k\parallel .\end{array}$$

It is natural and important to ask what are finite field versions of Theorems 1 and 2? We answer the question by deriving finite field version of Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality (Theorem 4).

2. Finite Field Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality

We begin by introducing the notion of sub-modulus field.

Definition 1.

Let $\mathbb{F}$ be a (finite) field. A map $|·|:\mathbb{F}\to [0,\infty )$ is said to be sub-modulus or sub-valued if the following conditions hold.

(i) 

If $\lambda \in \mathbb{F}$ is such that $\left|\lambda \right|=0$, then $\lambda =0$.

(ii) 

$\left|\lambda \mu \right|\le \left|\lambda \right|\left|\mu \right|$ for all $\lambda ,\mu \in \mathbb{F}$.

(iii) 

$|\lambda +\mu |\le \left|\lambda \right|+\left|\mu \right|$ for all $\lambda ,\mu \in \mathbb{F}$.

In this case, we say that $\mathbb{F}$ is a sub-modulus or sub-valued field.

Example 1.

Let p be a prime, let ${\mathbb{Z}}_p\{0,1,\dots ,p-1\}$ be the standard field of integers modulo p. Given $n\in {\mathbb{Z}}_p$, we define

$$\begin{array}{c}\hfill \left|n\right| := \mathit{unique}\mathit{natural}\mathit{number}\mathit{a}\mathit{such}\mathit{that}0\le \mathit{a}\le \mathit{p}-1\mathit{and}\mathit{n}\equiv \mathit{a}\left(\mathit{mod}p\right).\end{array}$$

We next introduce the notion of sub-normed linear space.

Definition 2.

Let $\mathcal{X}$ be a vector space over a sub-modulus field $\mathbb{F}$. A map $\parallel ·\parallel :\mathcal{X}\to [0,\infty )$ is said to be sub-norm if the following conditions hold.

(i) 

If $x\in \mathcal{X}$ is such that $\parallel x\parallel =0$, then $x=0$.

(ii) 

$\parallel \lambda x\parallel \le \left|\lambda \right|\parallel x\parallel$ for all $\lambda \in \mathbb{F}$, for all $x\in \mathcal{X}$.

(iii) 

$\parallel x+y\parallel \le \parallel x\parallel +\parallel y\parallel$ for all $x,y\in \mathcal{X}$.

In this case, we say $\mathcal{X}$ is a sub-normed linear space.

Example 2.

Let $p\in [1,\infty )$. Let $\mathbb{F}$ be a sub-modulus field. For $d\in \mathbb{N}$, let ${\mathbb{F}}^d$ be the standard vector space. We define

$$\begin{array}{c}\hfill \parallel {\left(a_j\right)}_{j=1}^d{\parallel }_p{\left(\sum _{j=1}^d{\left|a_j\right|}^p\right)}^{{\displaystyle \frac{1}{p}}},\forall {\left(a_j\right)}_{j=1}^d\in {\mathbb{F}}^d.\end{array}$$

Then ${\parallel ·\parallel }_p$ is a sub-norm on ${\mathbb{F}}^d$.

Example 3.

$p\in [1,\infty )$. Let $\mathbb{F}$ be a sub-modulus field. Define

$$\begin{array}{c}\hfill c_{00}^p(\mathbb{N},\mathbb{F})\left\{{\left\{a_n\right\}}_{n=1}^{\infty }:a_n\in \mathbb{F},\forall n\in \mathbb{N},a_n\ne 0\mathit{only}\mathit{for}\mathit{finitely}\mathit{many}n\right\}.\end{array}$$

We define

$$\begin{array}{c}\hfill \parallel {\left\{a_n\right\}}_{n=1}^{\infty }{\parallel }_p{\left(\sum _{n=1}^{\infty }{\left|a_n\right|}^p\right)}^{{\displaystyle \frac{1}{p}}},\forall {\left\{a_n\right\}}_{n=1}^{\infty }\in c_{00}^p(\mathbb{N},\mathbb{F}).\end{array}$$

Then ${\parallel ·\parallel }_p$ is a sub-norm on $c_{00}^p(\mathbb{N},\mathbb{F})$.

Example 4.

Let $\mathbb{F}$ be a sub-modulus field. Define

$$\begin{array}{c}\hfill c_{00}(\mathbb{N},\mathbb{F})\left\{{\left\{a_n\right\}}_{n=1}^{\infty }:a_n\in \mathbb{F},\forall n\in \mathbb{N},a_n\ne 0\mathit{only}\mathit{for}\mathit{finitely}\mathit{many}n\right\}.\end{array}$$

We define

$$\begin{array}{c}\hfill \parallel {\left\{a_n\right\}}_{n=1}^{\infty }{\parallel }_{00}\underset{n\in \mathbb{N}}{max}\left|a_n\right|,\forall {\left\{a_n\right\}}_{n=1}^{\infty }\in c_{00}(\mathbb{N},\mathbb{F}).\end{array}$$

Then ${\parallel ·\parallel }_{00}$ is a sub-norm on $c_{00}(\mathbb{N},\mathbb{F}).$

We now derive finite field version of Inequality (2).

Theorem 3.

(Finite Field Hudzik-Landes Inequality) Let $\mathcal{X}$ be a sub-normed linear space over $\mathbb{F}$. Then for all $x,y\in \mathcal{X}\setminus \left\{0\right\}$ with $\parallel x\parallel ,\parallel y\parallel \in \mathbb{F}$ it holds

$$\begin{array}{cc}\hfill \parallel x+y\parallel & \le min\left\{|\parallel x\parallel |\left(∥{\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}∥,\left|{\displaystyle \frac{1}{\parallel x\parallel }}-{\displaystyle \frac{1}{\parallel y\parallel }}\right|\parallel y\parallel \right),\left|\parallel y\parallel \right|\left(∥{\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}∥+\left|{\displaystyle \frac{1}{\parallel y\parallel }}-{\displaystyle \frac{1}{\parallel x\parallel }}\right|\parallel x\parallel \right)\right\}.\hfill \end{array}$$

Proof. 

Let $x,y\in \mathcal{X}\setminus \left\{0\right\}$ with $\parallel x\parallel ,\parallel y\parallel \in \mathbb{F}$. Then

$$\begin{array}{cc}\hfill \parallel x+y\parallel & =∥\parallel x\parallel \left({\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}\right)+\left(1-{\displaystyle \frac{\parallel x\parallel }{\parallel y\parallel }}\right)y∥\hfill \\ \hfill & \le ∥\parallel x\parallel \left({\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}\right)∥+∥\left(1-{\displaystyle \frac{\parallel x\parallel }{\parallel y\parallel }}\right)y∥\hfill \\ \hfill & \le |\parallel x\parallel |∥{\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}∥+\left|1-{\displaystyle \frac{\parallel x\parallel }{\parallel y\parallel }}\right|\parallel y\parallel \hfill \\ \hfill & \le |\parallel x\parallel |∥{\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}∥+|\parallel x\parallel |\left|{\displaystyle \frac{1}{\parallel x\parallel }}-{\displaystyle \frac{1}{\parallel y\parallel }}\right|\parallel y\parallel \hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \parallel x+y\parallel & =∥\left(1-{\displaystyle \frac{\parallel y\parallel }{\parallel x\parallel }}\right)x+\parallel y\parallel \left({\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}\right)∥\hfill \\ \hfill & \le ∥\left(1-{\displaystyle \frac{\parallel y\parallel }{\parallel x\parallel }}\right)x∥+∥\parallel y\parallel \left({\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}\right)∥\hfill \\ \hfill & \le \left|1-{\displaystyle \frac{\parallel y\parallel }{\parallel x\parallel }}\right|\parallel x\parallel +|\parallel y\parallel |∥{\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}∥\hfill \\ \hfill & \le |\parallel y\parallel |\left|{\displaystyle \frac{1}{\parallel y\parallel }}-{\displaystyle \frac{1}{\parallel x\parallel }}\right|\parallel x\parallel +|\parallel y\parallel |∥{\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}∥.\hfill \end{array}$$

Therefore

$$\begin{array}{c}\hfill \parallel x+y\parallel \le |\parallel x\parallel |\left(∥{\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}∥+\left|{\displaystyle \frac{1}{\parallel x\parallel }}-{\displaystyle \frac{1}{\parallel y\parallel }}\right|\parallel y\parallel \right)\end{array}$$

(3)

and

$$\begin{array}{c}\hfill \parallel x+y\parallel \le |\parallel y\parallel |\left(∥{\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}∥+\left|{\displaystyle \frac{1}{\parallel y\parallel }}-{\displaystyle \frac{1}{\parallel x\parallel }}\right|\parallel x\parallel \right).\end{array}$$

(4)

Inequalities (3) and (4) give

$$\begin{array}{c}\hfill \parallel x+y\parallel \le min\left\{|\parallel x\parallel |\left(∥{\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}∥+\left|{\displaystyle \frac{1}{\parallel x\parallel }}-{\displaystyle \frac{1}{\parallel y\parallel }}\right|\parallel y\parallel \right),\left|\parallel y\parallel \right|\left(∥{\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}∥+\left|{\displaystyle \frac{1}{\parallel y\parallel }}-{\displaystyle \frac{1}{\parallel x\parallel }}\right|\parallel x\parallel \right)\right\}.\end{array}$$

□

Note the additional assumption $\parallel x\parallel ,\parallel y\parallel \in \mathbb{F}$ in the previous theorem. The reason is that, since the norm is a real number, we generally do not have a guarantee that it belongs to the given sub-modulus field.

Example 5.

Consider ${\mathbb{Z}}_3$. Let $x=1,y=2$. Then $|x+y|=|1+2|=\left|3\right|=0$ and $\left|x\right|+\left|y\right|=1+2=3$. We now find

$$\begin{array}{cc}\hfill \left|\right|x\left|\right|\left(\left|{\displaystyle \frac{x}{\left|x\right|}}+{\displaystyle \frac{y}{\left|y\right|}}\right|+\left|{\displaystyle \frac{1}{\left|x\right|}}-{\displaystyle \frac{1}{\left|y\right|}}\right|\left|y\right|\right)& =\left|\right|1\left|\right|\left(\left|{\displaystyle \frac{1}{\left|1\right|}}+{\displaystyle \frac{2}{\left|2\right|}}\right|+\left|{\displaystyle \frac{1}{\left|1\right|}}-{\displaystyle \frac{1}{\left|2\right|}}\right|\left|2\right|\right)\hfill \\ \hfill & =\left|1\right|\left(\left|{\displaystyle \frac{1}{1}}+{\displaystyle \frac{2}{2}}\right|+\left|{\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{2}}\right|2\right)\hfill \\ \hfill & =\left|1+1\right|+\left|1-2\right|2\hfill \\ \hfill & =\left|2\right|+\left|1+1\right|2\hfill \\ \hfill & =2+\left|2\right|2\hfill \\ \hfill & =2+2·2\hfill \\ \hfill & =2+1=0\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \left|\right|y\left|\right|\left(\left|{\displaystyle \frac{x}{\left|x\right|}}+{\displaystyle \frac{y}{\left|y\right|}}\right|+\left|{\displaystyle \frac{1}{\left|x\right|}}-{\displaystyle \frac{1}{\left|x\right|}}\right|\left|x\right|\right)& =\left|\right|2\left|\right|\left(\left|{\displaystyle \frac{1}{\left|1\right|}}+{\displaystyle \frac{2}{\left|2\right|}}\right|+\left|{\displaystyle \frac{1}{\left|2\right|}}-{\displaystyle \frac{1}{\left|1\right|}}\right|\left|1\right|\right)\hfill \\ \hfill & =\left|2\right|\left(\left|{\displaystyle \frac{1}{1}}+{\displaystyle \frac{2}{2}}\right|+\left|{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{1}}\right|\right)\hfill \\ \hfill & =2(\left|1+1\right|+\left|2-1\right|)\hfill \\ \hfill & =2(\left|2\right|+\left|1\right|)\hfill \\ \hfill & =2(2+1)=0.\hfill \end{array}$$

Therefore

$$\begin{array}{c}\hfill 0=|x+y|=min\{0,0\}<\left|x\right|+\left|y\right|=3.\end{array}$$

Now we derive finite field version of Theorem 2.

Theorem 4.

(Finite Field Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality) Let $\mathcal{X}$ be a sub-normed linear space over $\mathbb{F}$ and $n\in \mathbb{N}$. Then for all $x_1,\dots ,x_n\in \mathcal{X}\setminus \left\{0\right\}$ with $\parallel x_1\parallel ,\dots ,\parallel x_n\parallel \in \mathbb{F}$ it holds

$$\begin{array}{cc}\hfill ∥\sum _{j=1}^n{x}_j∥& \le \underset{1\le k\le n}{min}\left\{|\parallel x_k\parallel |\left(∥\sum _{j=1}^n{\displaystyle \frac{x_j}{\parallel x_j\parallel }}∥+\sum _{1\le j\le n,j\ne k}\left|{\displaystyle \frac{1}{\parallel x_k\parallel }}-{\displaystyle \frac{1}{\parallel x_j\parallel }}\right|\parallel x_j\parallel \right)\right\}\hfill \\ \hfill & \le \underset{1\le k\le n}{min}\left\{|\parallel x_k\parallel |\left(∥\sum _{j=1}^n{\displaystyle \frac{x_j}{\parallel x_j\parallel }}∥+(n-1)\underset{1\le j\le n,j\ne k}{max}\left|{\displaystyle \frac{1}{\parallel x_k\parallel }}-{\displaystyle \frac{1}{\parallel x_j\parallel }}\right|\parallel x_j\parallel \right)\right\}.\hfill \end{array}$$

Proof. 

Let $x_1,\dots ,x_n\in \mathcal{X}\setminus \left\{0\right\}$ with $\parallel x_1\parallel ,\dots ,\parallel x_n\parallel \in \mathbb{F}$. Let $1\le k\le n$ be fixed. Then

$$\begin{array}{cc}\hfill ∥\sum _{j=1}^n{x}_j∥& =∥\sum _{j=1}^n{\displaystyle \frac{\parallel x_k\parallel x_j}{\parallel x_j\parallel }}+\sum _{j=1}^n\left(1-{\displaystyle \frac{\parallel x_k\parallel }{\parallel x_j\parallel }}\right)x_j∥\hfill \\ \hfill & \le ∥\sum _{j=1}^n{\displaystyle \frac{\parallel x_k\parallel x_j}{\parallel x_j\parallel }}∥+∥\sum _{j=1}^n\left(1-{\displaystyle \frac{\parallel x_k\parallel }{\parallel x_j\parallel }}\right)x_j∥\hfill \\ \hfill & \le ∥\sum _{j=1}^n{\displaystyle \frac{\parallel x_k\parallel x_j}{\parallel x_j\parallel }}∥+∥\parallel x_k\parallel \sum _{j=1}^n\left({\displaystyle \frac{1}{\parallel x_k\parallel }}-{\displaystyle \frac{1}{\parallel x_j\parallel }}\right)x_j∥\hfill \\ \hfill & \le |\parallel x_k\parallel |∥\sum _{j=1}^n{\displaystyle \frac{x_j}{\parallel x_j\parallel }}∥+\left|\parallel x_k\parallel \right|∥\sum _{j=1}^n\left({\displaystyle \frac{1}{\parallel x_k\parallel }}-{\displaystyle \frac{1}{\parallel x_j\parallel }}\right)x_j∥\hfill \\ \hfill & =|\parallel x_k\parallel |\left(∥\sum _{j=1}^n{\displaystyle \frac{x_j}{\parallel x_j\parallel }}∥+∥\sum _{j=1}^n\left({\displaystyle \frac{1}{\parallel x_k\parallel }}-{\displaystyle \frac{1}{\parallel x_j\parallel }}\right)x_j∥\right)\hfill \\ \hfill & =|\parallel x_k\parallel |\left(∥\sum _{j=1}^n{\displaystyle \frac{x_j}{\parallel x_j\parallel }}∥+∥\sum _{1\le j\le n,j\ne k}\left({\displaystyle \frac{1}{\parallel x_k\parallel }}-{\displaystyle \frac{1}{\parallel x_j\parallel }}\right)x_j∥\right)\hfill \\ \hfill & \le |\parallel x_k\parallel |\left(∥\sum _{j=1}^n{\displaystyle \frac{x_j}{\parallel x_j\parallel }}∥+\sum _{1\le j\le n,j\ne k}∥\left({\displaystyle \frac{1}{\parallel x_k\parallel }}-{\displaystyle \frac{1}{\parallel x_j\parallel }}\right)x_j∥\right)\hfill \\ \hfill & \le |\parallel x_k\parallel |\left(∥\sum _{j=1}^n{\displaystyle \frac{x_j}{\parallel x_j\parallel }}∥+\sum _{1\le j\le n,j\ne k}\left|{\displaystyle \frac{1}{\parallel x_k\parallel }}-{\displaystyle \frac{1}{\parallel x_j\parallel }}\right|\parallel x_j\parallel \right).\hfill \end{array}$$

By varying k and taking minimum in the right side of previous inequality gives

$$\begin{array}{cc}\hfill ∥\sum _{j=1}^n{x}_j∥& \le \underset{1\le k\le n}{min}\left\{|\parallel x_k\parallel |\left(∥\sum _{j=1}^n{\displaystyle \frac{x_j}{\parallel x_j\parallel }}∥+\sum _{1\le j\le n,j\ne k}\left|{\displaystyle \frac{1}{\parallel x_k\parallel }}-{\displaystyle \frac{1}{\parallel x_j\parallel }}\right|\parallel x_j\parallel \right)\right\}\hfill \\ \hfill & \le \underset{1\le k\le n}{min}\left\{|\parallel x_k\parallel |\left(∥\sum _{j=1}^n{\displaystyle \frac{x_j}{\parallel x_j\parallel }}∥+(n-1)\underset{1\le j\le n,j\ne k}{max}\left|{\displaystyle \frac{1}{\parallel x_k\parallel }}-{\displaystyle \frac{1}{\parallel x_j\parallel }}\right|\parallel x_j\parallel \right)\right\}.\hfill \end{array}$$

□

3. Conclusions

1.

In 1992, Hudzik and Landes improved triangle inequality for two nonzero elements in normed linear spaces [1].

2.

In 2006, Dragomir extended Hudzik-Landes inequality for more than two nonzero vectors [3].

3.

In 2007, Kato, Saito and Tamura extended Hudzik-Landes inequality without knowing the work of Dragomir [4].

4.

In this article, we derived finite field version of Hudzik-Landes-Dragomir-Kato-Saito-Tamura inequality.