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\fancyhead[LO]{\textbf{Finite Field Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality}}
\fancyhead[RE]{\textbf{K. Mahesh Krishna}}


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{\bf\large{{Finite Field Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality}}}\\
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\textbf{K. Mahesh Krishna}\\
School of Mathematics and Natural Sciences\\
Chanakya University Global Campus\\
NH-648, Haraluru Village\\
Devanahalli Taluk, 	Bengaluru  North District\\
Karnataka State, 562 110, India\\
Email: kmaheshak@gmail.com\\

Date: \today
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\textbf{Abstract}: Hudzik and Landes  \textit{[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 \textit{[Bull. Aust. Math. Soc., 2006]} and by Kato, Saito and Tamura \textit{[Math. Inequal. Appl., 2007]}. We derive a finite field version of Hudzik-Landes-Dragomir-Kato-Saito-Tamura inequality.\\
\textbf{Keywords}:  Normed linear space, Triangle inequality, Finite field, Sub-modulus, Sub-norm.\\
\textbf{Mathematics Subject Classification (2020)}:  12E20, 46B20.\\

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\section{Introduction}

Let $\mathcal{X}$ be a normed linear space (NLS) over scalar field. Definition of the norm has the triangle inequality 
\begin{align}\label{TI}
\|x+y\|\leq \|x\|+\|y\|, \quad \forall x, y \in \mathcal{X}.
\end{align}
In 1992, Hudzik and Landes derived a breakthrough generalization of Inequality (\ref{TI}) which is valid for any two nonzero elements in a NLS \cite{HUDZIKLANDES}.
\begin{theorem} \cite{HUDZIKLANDES} \label{HKT} (\textbf{Hudzik-Landes Inequlaity})
Let $\mathcal{X}$ be a NLS. Then for all $x, y \in \mathcal{X}\setminus\{0\}$, 
\begin{align}\label{HL}
\|x+y\|\leq \|x\|+\|y\|-\left(2-\left\|\frac{x}{\|x\|}+\frac{y}{\|y\|}\right\|\right)\min\{\|x\|, \|y\|\}.
\end{align}
\end{theorem}
In 2006, Maligranda independently derived Inequality (\ref{HL}) \cite{MALIGRANDA}.
It is natural to look for a generalization of Inequality (\ref{HL}) to more than two non-zero vectors. This is done independently by Dragomir in 2006 \cite{DRAGOMIR} and by Kato, Saito and Tamura in 2007 \cite{KATOSAITOTAMURA}.
\begin{theorem} \cite{KATOSAITOTAMURA, DRAGOMIR} \label{HLKSTT}
(\textbf{Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality})
Let $\mathcal{X}$ be a NLS and $n \in \mathbb{N}$. Then for all $x_1, \dots, x_n \in \mathcal{X}\setminus\{0\}$, 	we have 
\begin{align*}
\left\|\sum_{j=1}^{n}x_j\right\|\leq \sum_{j=1}^{n}\|x_j\|-\left(n-\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|\right)\min_{1\leq k \leq n} \|x_k\|.
\end{align*}	
\end{theorem}
It is natural and important to ask what are finite field  versions of Theorems \ref{HKT} and \ref{HLKSTT}? We answer the question by deriving finite field version of Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality (Theorem \ref{NAHLDKSTT}).





\section{Finite Field Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality}
We begin by introducing the notion of sub-modulus field. 
\begin{definition}
	Let $\mathbb{F}$ be a  (finite) field. A map $|\cdot|: \mathbb{F}\to [0, \infty)$ is said to be sub-modulus or sub-valued if the following conditions hold.
	\begin{enumerate}[\upshape (i)]
		\item If $\lambda  \in \mathbb{F}$ is such that $|\lambda|=0$, then $\lambda=0$.
		\item $|\lambda \mu|\leq |\lambda||\mu|$ for all $\lambda, \mu \in \mathbb{F}$.
		\item $|\lambda +\mu|\leq |\lambda|+|\mu|$ for all $\lambda, \mu \in \mathbb{F}$.
	\end{enumerate} 
	In this case, we say that $\mathbb{F}$ is a  sub-modulus or sub-valued field. 
\end{definition}
\begin{example}
	Let $p$ be a prime, let $\mathbb{Z}_p\coloneqq \{0, 1, \dots, p-1\}$ be the standard field of integers modulo $p$. Given $n \in \mathbb{Z}_p$, we define 
	\begin{align*}
		|n|\coloneqq \text{unique natural number } a \text{ such that } 0\leq a \leq p-1 \text{ and }n \equiv a (\text{mod }p).
	\end{align*}
\end{example}
We next introduce the notion of sub-normed linear space.
\begin{definition}
	Let $\mathcal{X}$ be a vector space over a sub-modulus field $\mathbb{F}$. A map $\|\cdot\|: \mathcal{X}\to [0, \infty)$ is said to be sub-norm if the following conditions hold.	
	\begin{enumerate}[\upshape (i)]
		\item If $x \in \mathcal{X}$ is such that $\|x\| =0$, then $x=0$.
		\item $\|\lambda x\|\leq |\lambda|\|x\|$ for all $ \lambda \in \mathbb{F}$, for all $x \in \mathcal{X}$.
		\item $\|x+y\|\leq \|x\|+\|y\|$ for all $x,y \in \mathcal{X}$.
	\end{enumerate} 
	In this case, we say $\mathcal{X}$ is a sub-normed linear space. 
\end{definition}
\begin{example}
	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{align*}
		\|(a_j)_{j=1}^d\|_p\coloneqq \left(\sum_{j=1}^{d}|a_j|^p\right)^\frac{1}{p}, \quad \forall (a_j)_{j=1}^d \in \mathbb{F}^d.
	\end{align*}
	Then $\|\cdot\|_p$ is a sub-norm on $\mathbb{F}^d$.
\end{example}
\begin{example}
	$p\in [1, \infty)$.	Let $\mathbb{F}$ be a sub-modulus field. Define 
	\begin{align*}
		c_{00}^p(\mathbb{N}, \mathbb{F})\coloneqq\left\{\{a_n\}_{n=1}^\infty: a_n \in \mathbb{F}, \forall n \in \mathbb{N}, a_n \neq 0 \text{ only for finitely many } n\right\}. 
	\end{align*}
	We define 
		\begin{align*}
		\|\{a_n\}_{n=1}^\infty\|_p\coloneqq \left(\sum_{n=1}^{\infty}|a_n|^p\right)^\frac{1}{p}, \quad \forall \{a_n\}_{n=1}^\infty \in 	c_{00}^p(\mathbb{N}, \mathbb{F}).
	\end{align*}
	Then $\|\cdot\|_p$ is a sub-norm on $c_{00}^p(\mathbb{N}, \mathbb{F})$.
\end{example}
\begin{example}
	Let $\mathbb{F}$ be a sub-modulus field. Define 
	\begin{align*}
		c_{00}(\mathbb{N}, \mathbb{F})	\coloneqq\left\{\{a_n\}_{n=1}^\infty: a_n \in \mathbb{F}, \forall n \in \mathbb{N}, a_n \neq 0 \text{ only for finitely many } n\right\}. 
	\end{align*}
	We define 
	\begin{align*}
		\|\{a_n\}_{n=1}^\infty\|_{00}\coloneqq \max_{n\in \mathbb{N}}|a_n|, \quad \forall \{a_n\}_{n=1}^\infty\in c_{00}(\mathbb{N}, \mathbb{F}).
	\end{align*}
	Then $\|\cdot\|_{00}$ is a sub-norm on  $c_{00}(\mathbb{N}, \mathbb{F}).$	
\end{example}
 We now derive finite field  version of Inequality (\ref{HL}).
\begin{theorem} (\textbf{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 $\|x\|, \|y\| \in \mathbb{F}$ it holds
\begin{align*}
	\|x+y\|&\leq \min \left\{\big|\|x\|\big|\left(\left\|\frac{x}{\|x\|}+\frac{y}{\|y\|}\right\|, \left|\frac{1}{\|x\|}-\frac{1}{\|y\|}\right|\|y\|\right), \big|\|y\|\big|\left(\left\|\frac{x}{\|x\|}+\frac{y}{\|y\|}\right\|+ \left|\frac{1}{\|y\|}-\frac{1}{\|x\|}\right|\|x\|\right)\right\}.
\end{align*}
\end{theorem}
\begin{proof}
	Let $x, y \in \mathcal{X}\setminus\{0\}$ with $\|x\|, \|y\| \in \mathbb{F}$. Then 
		\begin{align*}
		\|x+y\|&=\left\|\|x\|\left(\frac{x}{\|x\|}+\frac{y}{\|y\|}\right)+\left(1-\frac{\|x\|}{\|y\|}\right)y\right\|\\
		&\leq\left\|\|x\|\left(\frac{x}{\|x\|}+\frac{y}{\|y\|}\right)\right\|+ \left\|\left(1-\frac{\|x\|}{\|y\|}\right)y\right\|\\
		&\leq \big|\|x\|\big|\left\|\frac{x}{\|x\|}+\frac{y}{\|y\|}\right\|+ \left|1-\frac{\|x\|}{\|y\|}\right|\|y\|\\
		&\leq \big|\|x\|\big|\left\|\frac{x}{\|x\|}+\frac{y}{\|y\|}\right\|+ \big|\|x\|\big|\left|\frac{1}{\|x\|}-\frac{1}{\|y\|}\right|\|y\|
	\end{align*}
and 
	\begin{align*}
	\|x+y\|&=\left\|\left(1-\frac{\|y\|}{\|x\|}\right)x+\|y\|\left(\frac{x}{\|x\|}+\frac{y}{\|y\|}\right)\right\|\\
	&\leq \left\|\left(1-\frac{\|y\|}{\|x\|}\right)x\right\|+ \left\|\|y\|\left(\frac{x}{\|x\|}+\frac{y}{\|y\|}\right)\right\|\\
		&\leq \left|1-\frac{\|y\|}{\|x\|}\right|\|x\|+\big|\|y\|\big|\left\|\frac{x}{\|x\|}+\frac{y}{\|y\|}\right\|\\
	&\leq\big|\|y\|\big|\left|\frac{1}{\|y\|}-\frac{1}{\|x\|}\right|\|x\|+ \big|\|y\|\big|\left\|\frac{x}{\|x\|}+\frac{y}{\|y\|}\right\|.
\end{align*}
Therefore 
\begin{align}\label{P1}
\|x+y\|\leq	\big|\|x\|\big|\left(\left\|\frac{x}{\|x\|}+\frac{y}{\|y\|}\right\|+ \left|\frac{1}{\|x\|}-\frac{1}{\|y\|}\right|\|y\|\right)
\end{align}
and 
\begin{align}\label{P2}
\|x+y\|\leq  \big|\|y\|\big|\left(\left\|\frac{x}{\|x\|}+\frac{y}{\|y\|}\right\|+ \left|\frac{1}{\|y\|}-\frac{1}{\|x\|}\right|\|x\|\right).
\end{align}
Inequalities (\ref{P1})and  (\ref{P2}) give 
\begin{align*}
	\|x+y\|\leq \min \left\{\big|\|x\|\big|\left(\left\|\frac{x}{\|x\|}+\frac{y}{\|y\|}\right\|+ \left|\frac{1}{\|x\|}-\frac{1}{\|y\|}\right|\|y\|\right), \big|\|y\|\big|\left(\left\|\frac{x}{\|x\|}+\frac{y}{\|y\|}\right\|+ \left|\frac{1}{\|y\|}-\frac{1}{\|x\|}\right|\|x\|\right)\right\}.
\end{align*}
\end{proof}
Note the additional assumption $\|x\|, \|y\| \in \mathbb{F}$ 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. 
\begin{example}
Consider $\mathbb{Z}_3$. Let $x=1, y=2$. Then $|x+y|=|1+2|=|3|=0$ and $|x|+|y|=1+2=3$. We now  find 
\begin{align*}
	\big||x|\big|\left(\left|\frac{x}{|x|}+\frac{y}{|y|}\right|+ \left|\frac{1}{|x|}-\frac{1}{|y|}\right||y|\right)&=\big||1|\big|\left(\left|\frac{1}{|1|}+\frac{2}{|2|}\right|+ \left|\frac{1}{|1|}-\frac{1}{|2|}\right||2|\right)\\
	&=\big|1\big|\left(\left|\frac{1}{1}+\frac{2}{2}\right|+ \left|\frac{1}{1}-\frac{1}{2}\right|2\right)\\
	&=\left|1+1\right|+ \left|1-2\right|2\\
		&=\left|2\right|+ \left|1+1\right|2\\
			&=2+ \left|2\right|2\\
			&=2+ 2\cdot 2\\
			&=2+1=0
\end{align*}
and 
\begin{align*}
	\big||y|\big|\left(\left|\frac{x}{|x|}+\frac{y}{|y|}\right|+ \left|\frac{1}{|x|}-\frac{1}{|x|}\right||x|\right)&=\big||2|\big|\left(\left|\frac{1}{|1|}+\frac{2}{|2|}\right|+ \left|\frac{1}{|2|}-\frac{1}{|1|}\right||1|\right)\\
	&=\big|2\big|\left(\left|\frac{1}{1}+\frac{2}{2}\right|+ \left|\frac{1}{2}-\frac{1}{1}\right|\right)\\
	&=2(\left|1+1\right|+ \left|2-1\right|)\\
	&=2(\left|2\right|+ \left|1\right|)\\
	&=2(2+ 1)=0.
\end{align*}
Therefore 
\begin{align*}
	0=|x+y|=\min\{0, 0\}<|x|+|y|=3.
\end{align*}
\end{example}


Now we derive finite field  version of  Theorem \ref{HLKSTT}.
\begin{theorem} \label{NAHLDKSTT} (\textbf{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 $x_1, \dots, x_n \in \mathcal{X}\setminus\{0\}$ with $\|x_1\|, \dots, \|x_n\| \in \mathbb{F}$ it holds
\begin{align*}
\left\|\sum_{j=1}^{n}x_j\right\|&\leq\min_{1\leq k \leq n}\left\{\big|\|x_k\|\big|\left(\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+\sum_{1\leq j \leq n, j \neq k}\left|\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right|\|x_j\|\right)\right\}\\
&\leq \min_{1\leq k \leq n}\left\{\big|\|x_k\|\big|\left(\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+(n-1)\max_{1\leq j \leq n, j \neq k}\left|\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right|\|x_j\|\right)\right\}.
\end{align*}
\end{theorem}
\begin{proof}
Let $x_1, \dots, x_n \in \mathcal{X}\setminus\{0\}$ with $\|x_1\|, \dots, \|x_n\| \in \mathbb{F}$. Let $1\leq k \leq n$ be fixed. Then 
\begin{align*}
	\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\|\\
	&\leq\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\|\\
		&\leq\left\|\sum_{j=1}^{n}\frac{\|x_k\|x_j}{\|x_j\|}\right\|+ \left\|\|x_k\|\sum_{j=1}^{n}\left(\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right)x_j\right\|\\
	&\leq \big|\|x_k\|\big|\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+ \big|\|x_k\|\big| \left\|\sum_{j=1}^{n}\left(\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right)x_j\right\|\\
	&=\big|\|x_k\|\big|\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)\\
	&=\big|\|x_k\|\big|\left(\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+ \left\|\sum_{1\leq j \leq n, j \neq k}\left(\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right)x_j\right\|\right)\\
		&\leq \big|\|x_k\|\big|\left(\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+ \sum_{1\leq j \leq n, j \neq k}\left\|\left(\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right)x_j\right\|\right)\\
	&\leq \big|\|x_k\|\big|\left(\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+\sum_{1\leq j \leq n, j \neq k}\left|\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right|\|x_j\|\right).
\end{align*}
By varying $k$ and taking minimum in the right side of previous inequality gives 
\begin{align*}
	\left\|\sum_{j=1}^{n}x_j\right\|&\leq\min_{1\leq k \leq n}\left\{\big|\|x_k\|\big|\left(\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+\sum_{1\leq j \leq n, j \neq k}\left|\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right|\|x_j\|\right)\right\}\\
	&\leq \min_{1\leq k \leq n}\left\{\big|\|x_k\|\big|\left(\left\|\sum_{j=1}^{n}\frac{x_j}{\|x_j\|}\right\|+(n-1)\max_{1\leq j \leq n, j \neq k}\left|\frac{1}{\|x_k\|}-\frac{1}{\|x_j\|}\right|\|x_j\|\right)\right\}.
\end{align*}
\end{proof}


\section{Conclusions}
\begin{enumerate}
	\item In 1992, Hudzik and Landes improved triangle inequality for two nonzero elements in normed linear spaces  \cite{HUDZIKLANDES}.
	\item In 2006, Dragomir extended Hudzik-Landes inequality for more than two nonzero vectors \cite{DRAGOMIR}.
	\item In 2007, Kato, Saito and Tamura extended Hudzik-Landes inequality without knowing the work of Dragomir \cite{KATOSAITOTAMURA}.
	\item In this article, we derived finite field version of Hudzik-Landes-Dragomir-Kato-Saito-Tamura inequality.  
\end{enumerate}




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	\newblock Bounds for the normalised {Jensen} functional.
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	\bibitem{HUDZIKLANDES}
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	\bibitem{KATOSAITOTAMURA}
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	\newblock Sharp triangle inequality and its reverse in {Banach} spaces.
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\end{thebibliography}

 


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