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

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Abstract

Hudzik and Landes [Math. Ann., 1992] derived a major generalization of the triangle inequality for two nonzero elements in normed linear spaces, which was extended to finitely many nonzero elements independently by Dragomir [Bull. Aust. Math. Soc., 2006] and by Kato, Saito and Tamura [Math. Inequal. Appl., 2007]. We derive a finite field version of Hudzik-Landes-Dragomir-Kato-Saito-Tamura inequality.

Introduction

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

\lVert x+y\rVert\leq\lVert x\rVert+\lVert y\rVert,\quad\forall x,y\in\mathcal{X}. \tag*{(1)}

In 1992, Hudzik and Landes derived a breakthrough generalization of Inequality (1) which is valid for any two nonzero elements in a NLS [2].

Theorem 1.1. [2] (Hudzik-Landes Inequality) Let $\mathcal{X}$ be a NLS. Then for all $x,y\in\mathcal{X}\setminus\{0\}$ ,

\lVert x+y\rVert\leq\lVert x\rVert+\lVert y\rVert-\left(2-\left\lVert\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right\rVert\right)\min\{\lVert x\rVert,\lVert y\rVert\}. \tag*{(2)}

In 2006, Maligranda independently derived Inequality (2) [4]. 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 [1] and by Kato, Saito and Tamura in 2007 [3].

Theorem 1.2. [1, 3] (Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality) Let $\mathcal{X}$ be a NLS and $n\in\mathbb{N}$ . Then for all $x_1,\ldots,x_n\in\mathcal{X}\setminus\{0\}$ , we have

\left\lVert\sum_{j=1}^{n}x_j\right\rVert\leq\sum_{j=1}^{n}\lVert x_j\rVert-\left(n-\left\lVert\sum_{j=1}^{n}\frac{x_j}{\lVert x_j\rVert}\right\rVert\right)\min_{1\leq k\leq n}\lVert x_k\rVert.

It is natural and important to ask what are finite field versions of Theorems Theorems 1.1 and 1.2? We answer the question by deriving finite field version of Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality (Theorem 2.9).

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

Definition 2.1. Let $F$ be a (finite) field. A map $|\cdot|: F \to[0,\infty)$ is said to be sub-modulus or sub-valued if the following conditions hold.

(i) If $\lambda\in F$ is such that $|\lambda| = 0$ , then $\lambda= 0$ .

(ii) $|\lambda\mu| \leq|\lambda||\mu|$ for all $\lambda,\mu\in F$ .

(iii) $|\lambda+\mu| \leq|\lambda|+|\mu|$ for all $\lambda,\mu\in F$ .

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

Example 2.2. Let $p$ be a prime, let $\mathbb{Z}_p := \{0,1,\ldots,p-1\}$ be the standard field of integers modulo $p$ . Given $n \in\mathbb{Z}_p$ , we define

|n| := \text{unique natural number }a\text{ such that }0 \leq a \leq p-1\text{ and }n \equiv a \pmod{p}.

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

Definition 2.3. Let $\mathcal{X}$ be a vector space over a sub-modulus field $F$ . A map $\|\cdot\|:\mathcal{X}\to[0,\infty)$ is said to be sub-norm if the following conditions hold.

(i) If $x \in\mathcal{X}$ is such that $\|x\| = 0$ , then $x = 0$ .

(ii) $\|\lambda x\| \leq|\lambda|\|x\|$ for all $\lambda\in F$ , for all $x \in\mathcal{X}$ .

(iii) $\|x+y\| \leq\|x\|+\|y\|$ for all $x,y \in\mathcal{X}$ .

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

Example 2.4. Let $p \in[1,\infty)$ . Let $F$ be a sub-modulus field. For $d \in\mathbb{N}$ , let $F^d$ be the standard vector space. We define

\left\|(a_j)_{j=1}^{d}\right\|_p := \left(\sum_{j=1}^{d}|a_j|^p\right)^{\frac{1}{p}}, \qquad\forall(a_j)_{j=1}^{d}\in F^d.

Then $\|\cdot\|_p$ is a sub-norm on $F^d$ .

Example 2.5. $p \in[1,\infty)$ . Let $F$ be a sub-modulus field. Define $c_{00}^{p}(\mathbb{N},F):=\{\{a_n\}_{n=1}^{\infty}:a_n\in F,\ \forall n\in\mathbb{N},\ a_n\ne0\text{ only for finitely many }n\}$ .

We define

\left\|\{a_n\}_{n=1}^{\infty}\right\|_p := \left(\sum_{n=1}^{\infty}|a_n|^p\right)^{\frac{1}{p}}, \qquad\forall\{a_n\}_{n=1}^{\infty}\in c_{00}^{p}(\mathbb{N},F).

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

Example 2.6. Let $F$ be a sub-modulus field. Define $c_{00}(\mathbb{N},F):=\{\{a_n\}_{n=1}^{\infty}:a_n\in F,\ \forall n\in\mathbb{N},\ a_n\ne0\text{ only for finitely many }n\}$ .

We define

\left\|\{a_n\}_{n=1}^{\infty}\right\|_{00}:=\max_{n\in\mathbb{N}}|a_n|,\qquad\forall\{a_n\}_{n=1}^{\infty}\in c_{00}(\mathbb{N},F).

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

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

Theorem 2.7. (Finite Field Hudzik-Landes Inequality) Let $\mathcal{X}$ be a sub-normed linear space over $\mathbb{F}$ . Then for all $x,y \in\mathcal{X} \setminus\{0\}$ with $\lVert x\rVert,\lVert y\rVert\in\mathbb{F}$ it holds

\lVert x+y\rVert\le\min\left\{\lVert x\rVert\left(\left\lVert\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right\rVert+\left\lvert\frac{1}{\lVert x\rVert}-\frac{1}{\lVert y\rVert}\right\rvert\lVert y\rVert\right),\lVert y\rVert\left(\left\lVert\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right\rVert+\left\lvert\frac{1}{\lVert y\rVert}-\frac{1}{\lVert x\rVert}\right\rvert\lVert x\rVert\right)\right\}.

Proof. Let $x,y \in\mathcal{X} \setminus\{0\}$ with $\lVert x\rVert,\lVert y\rVert\in\mathbb{F}$ . Then

\begin{aligned} \lVert x+y\rVert&= \left\lVert\lVert x\rVert\left(\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right)+\left(1-\frac{\lVert x\rVert}{\lVert y\rVert}\right)y\right\rVert\\ &\le\lVert x\rVert\left\lVert\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right\rVert+\left\lVert\left(1-\frac{\lVert x\rVert}{\lVert y\rVert}\right)y\right\rVert\\ &\le\lVert x\rVert\left\lVert\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right\rVert+\left\lvert1-\frac{\lVert x\rVert}{\lVert y\rVert}\right\rvert\lVert y\rVert\\ &\le\lVert x\rVert\left\lVert\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right\rVert+\lVert x\rVert\left\lvert\frac{1}{\lVert x\rVert}-\frac{1}{\lVert y\rVert}\right\rvert\lVert y\rVert. \end{aligned}

and

\begin{aligned} \lVert x+y\rVert&= \left\lVert\left(1-\frac{\lVert y\rVert}{\lVert x\rVert}\right)x+\lVert y\rVert\left(\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right)\right\rVert\\ &\le\left\lVert\left(1-\frac{\lVert y\rVert}{\lVert x\rVert}\right)x\right\rVert+\left\lVert\lVert y\rVert\left(\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right)\right\rVert\\ &\le\left\lvert1-\frac{\lVert y\rVert}{\lVert x\rVert}\right\rvert\lVert x\rVert+\lVert y\rVert\left\lVert\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right\rVert\\ &\le\lVert y\rVert\left\lvert\frac{1}{\lVert y\rVert}-\frac{1}{\lVert x\rVert}\right\rvert\lVert x\rVert+\lVert y\rVert\left\lVert\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right\rVert. \end{aligned}

Therefore

\lVert x+y\rVert\le\lVert x\rVert\left(\left\lVert\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right\rVert+\left\lvert\frac{1}{\lVert x\rVert}-\frac{1}{\lVert y\rVert}\right\rvert\lVert y\rVert\right) \tag*{(3)}

and

\lVert x+y\rVert\le\lVert y\rVert\left(\left\lVert\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right\rVert+\left\lvert\frac{1}{\lVert y\rVert}-\frac{1}{\lVert x\rVert}\right\rvert\lVert x\rVert\right). \tag*{(4)}

Inequalities (3) and (4) give

\lVert x+y\rVert\le\min\left\{\lVert x\rVert\left(\left\lVert\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right\rVert+\left\lvert\frac{1}{\lVert x\rVert}-\frac{1}{\lVert y\rVert}\right\rvert\lVert y\rVert\right),\lVert y\rVert\left(\left\lVert\frac{x}{\lVert x\rVert}+\frac{y}{\lVert y\rVert}\right\rVert+\left\lvert\frac{1}{\lVert y\rVert}-\frac{1}{\lVert x\rVert}\right\rvert\lVert x\rVert\right)\right\}.

Note the additional assumption $\lVert x\rVert,\lVert y\rVert\in\mathbb{F}$ 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 2.8. Consider $\mathbb{Z}_3$ . Let $x=1$ , $y=2$ . Then $\lvert x+y\rvert=\lvert1+2\rvert=\lvert3\rvert=0$ and $\lvert x\rvert+\lvert y\rvert=1+2=3$ . We now find

\begin{aligned} \lvert x\rvert\left(\left\lvert\frac{x}{\lvert x\rvert}+\frac{y}{\lvert y\rvert}\right\rvert+\left\lvert\frac{1}{\lvert x\rvert}-\frac{1}{\lvert y\rvert}\right\rvert\lvert y\rvert\right) &=\lvert1\rvert\left(\left\lvert\frac{1}{\lvert1\rvert}+\frac{2}{\lvert2\rvert}\right\rvert+\left\lvert\frac{1}{\lvert1\rvert}-\frac{1}{\lvert2\rvert}\right\rvert\lvert2\rvert\right)\\ &=\lvert1\rvert\left(\left\lvert\frac{1}{1}+\frac{2}{2}\right\rvert+\left\lvert\frac{1}{1}-\frac{1}{2}\right\rvert2\right)\\ &=\lvert1+1\rvert+\lvert1-2\rvert2\\ &=\lvert2\rvert+\lvert1\rvert+\lvert2\rvert2\\ &=2+\lvert2\rvert2\\ &=2+2\cdot2\\ &=2+1=0 \end{aligned}

and

\begin{aligned} \lvert y\rvert\left(\left\lvert\frac{x}{\lvert x\rvert}+\frac{y}{\lvert y\rvert}\right\rvert+\left\lvert\frac{1}{\lvert x\rvert}-\frac{1}{\lvert x\rvert}\right\rvert\lvert x\rvert\right) &=\lvert2\rvert\left(\left\lvert\frac{1}{\lvert1\rvert}+\frac{2}{\lvert2\rvert}\right\rvert+\left\lvert\frac{1}{\lvert2\rvert}-\frac{1}{\lvert1\rvert}\right\rvert\lvert1\rvert\right)\\ &=\lvert2\rvert\left(\left\lvert\frac{1}{1}+\frac{2}{2}\right\rvert+\left\lvert\frac{1}{2}-\frac{1}{1}\right\rvert\right)\\ &=2(\lvert1+1\rvert+\lvert2-1\rvert)\\ &=2(\lvert2\rvert+\lvert1\rvert)\\ &=2(2+1)=0. \end{aligned}

Therefore

0=\lvert x+y\rvert=\min\{0,0\}<\lvert x\rvert+\lvert y\rvert=3.

Now we derive finite field version of Theorem 1.2.

Theorem 2.9. (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 $x_1,\ldots,x_n\in\mathcal{X}\setminus\{0\}$ with $\lVert x_1\rVert,\ldots,\lVert x_n\rVert\in\mathbb{F}$ it holds

\begin{aligned} \left\lVert\sum_{j=1}^{n}x_j\right\rVert &\leq\min_{1\leq k\leq n}\left\{\lVert x_k\rVert\left(\left\lVert\sum_{j=1}^{n}\frac{x_j}{\lVert x_j\rVert}\right\rVert+\sum_{\substack{1\leq j\leq n\\j\ne k}}\left\lvert\frac{1}{\lVert x_k\rVert}-\frac{1}{\lVert x_j\rVert}\right\rvert\lVert x_j\rVert\right)\right\}\\ &\leq\min_{1\leq k\leq n}\left\{\lVert x_k\rVert\left(\left\lVert\sum_{j=1}^{n}\frac{x_j}{\lVert x_j\rVert}\right\rVert+(n-1)\max_{\substack{1\leq j\leq n\\j\ne k}}\left\lvert\frac{1}{\lVert x_k\rVert}-\frac{1}{\lVert x_j\rVert}\right\rvert\lVert x_j\rVert\right)\right\}. \end{aligned}

Proof. Let $x_1,\ldots,x_n \in\mathcal{X} \setminus\{0\}$ with $\|x_1\|,\ldots,\|x_n\| \in\mathbb{F}$ . Let $1 \le k \le n$ be fixed. Then

\begin{aligned} \left\|\sum_{j=1}^{n} x_j\right\| &= \left\|\sum_{j=1}^{n} \frac{\|x_k\|}{\|x_j\|}x_j+\sum_{j=1}^{n}\left(1-\frac{\|x_k\|}{\|x_j\|}\right)x_j\right\| \\ &\le\left\|\sum_{j=1}^{n} \frac{\|x_k\|}{\|x_j\|}x_j\right\|+\left\|\sum_{j=1}^{n}\left(1-\frac{\|x_k\|}{\|x_j\|}\right)x_j\right\| \\ &\le\left\|\sum_{j=1}^{n} \frac{\|x_k\|}{\|x_j\|}x_j\right\|+\|x_k\|\left\|\sum_{j=1}^{n}\left(\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right)x_j\right\| \\ &\le\|x_k\|\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+\|x_k\|\left\|\sum_{j=1}^{n}\left(\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right)x_j\right\| \\ &= \|x_k\|\left(\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+\left\|\sum_{j=1}^{n}\left(\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right)x_j\right\|\right) \\ &= \|x_k\|\left(\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+\left\|\sum_{\substack{1\le j\le n,j\ne k}}\left(\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right)x_j\right\|\right) \\ &\le\|x_k\|\left(\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+\sum_{\substack{1\le j\le n,j\ne k}}\left\|\left(\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right)x_j\right\|\right) \\ &\le\|x_k\|\left(\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+\sum_{\substack{1\le j\le n,j\ne k}}\left|\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right|\|x_j\|\right). \end{aligned}

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

\begin{aligned} \left\|\sum_{j=1}^{n}x_j\right\| \le\min_{1\le k\le n}\left\{\|x_k\|\left(\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+\sum_{\substack{1\le j\le n,j\ne k}}\left|\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right|\|x_j\|\right)\right\} \\ &\le\min_{1\le k\le n}\left\{\|x_k\|\left(\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+(n-1)\max_{\substack{1\le j\le n,j\ne k}}\left|\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right|\|x_j\|\right)\right\}. \end{aligned}

Conclusions

In 1992, Hudzik and Landes improved triangle inequality for two nonzero elements in normed linear spaces [2].

In 2006, Dragomir extended Hudzik-Landes inequality for more than two nonzero vectors [1].

In 2007, Kato, Saito and Tamura extended Hudzik-Landes inequality without knowing the work of Dragomir [3].

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

References

[1]Sever S. Dragomir. Bounds for the normalised Jensen functional. Bull. Aust. Math. Soc., 74(3):471–478, 2006.DOI
[2]Henryk Hudzik and Thomas R. Landes. Characteristic of convexity of Köthe function spaces. Math. Ann., 294(1):117–124, 1992.DOI
[3]Mikio Kato, Kichi-Suke Saito, and Takayuki Tamura. Sharp triangle inequality and its reverse in Banach spaces. Math. Inequal. Appl., 10(2):451–460, 2007.DOI
[4]Lech Maligranda. Simple norm inequalities. Am. Math. Mon., 113(3):256–260, 2006.

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

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