Non-Archimedean Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality

1. Introduction

Let $\mathcal{X}$ be a normed linear space (NLS). From the definition of the norm, we have 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 [2].

Theorem 1.  

[2] (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)

We note that, in 2006, Maligranda independently derived Inequality (2) [4]. It is natural to ask 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 2.  

[1,3] (Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality) Let $\mathcal{X}$ be a NLS and $n\in \mathbb{N}$. Then for all $x_1,\dots ,x_n\in \mathcal{X}\setminus \left\{0\right\}$, 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 non-Archimedean versions of Theorems 1 and 2? We answer the question by deriving non-Archimedean version of Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality (Theorem 4).

2. Non-Archimedean Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality

Let $\mathbb{K}$ be a field. Recall that a map $|·|:\mathbb{K}\to [0,\infty )$ is said to be a non-Archimedean valuation if following conditions holds.

(i)

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

(ii)

$\left|\lambda \mu \right|=\left|\lambda \right|\left|\mu \right|$ for all $\lambda ,\mu \in \mathbb{K}$.

(iii)

(Ultra-triangle inequality) $|\lambda +\mu |\le max\left\{\right|\lambda |,|\mu \left|\right\}$ for all $\lambda ,\mu \in \mathbb{K}$.

In this case, $\mathbb{K}$ is called as non-Archimedean valued field [6]. Let $\mathcal{X}$ be a vector space over a non-Archimedean valued field $\mathbb{K}$ with valuation $|·|$. Recall that a map $\parallel ·\parallel :\mathcal{X}\to [0,\infty )$ is said to be a non-Archimedean norm if following conditions holds.

(i)

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

(ii)

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

(iii)

(Ultra-norm inequality) $\parallel x+y\parallel \le max\{\parallel x\parallel ,\parallel y\parallel \}$ for all $x,y\in \mathcal{X}$.

In this case, $\mathcal{X}$ is called as non-Archimedean linear space (NALS) [5]. We first derive non-Archimedean version of Inequality (2).

Theorem 3.(Non-Archimedean Hudzik-Landes Inequality) Let $\mathcal{X}$ be a NALS over $\mathbb{K}$. Then for all $x,y\in \mathcal{X}\setminus \left\{0\right\}$ with $\parallel x\parallel ,\parallel y\parallel \in \mathbb{K}$ it holds

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

Proof. 

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

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

and

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

Therefore

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

(3)

and

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

(4)

Inequalities (3)and (4) give

$$\parallel x+y\parallel \le min\left\{|\parallel x\parallel |max\left\{∥{\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}∥,\left|{\displaystyle \frac{1}{\parallel x\parallel }}-{\displaystyle \frac{1}{\parallel y\parallel }}\right|\parallel y\parallel \right\},\left|\parallel y\parallel \right|max\left\{∥{\displaystyle \frac{x}{\parallel x\parallel }}+{\displaystyle \frac{y}{\parallel y\parallel }}∥,\left|{\displaystyle \frac{1}{\parallel x\parallel }}-{\displaystyle \frac{1}{\parallel y\parallel }}\right|\parallel x\parallel \right\}\right\}.$$

□

Note the additional assumption $\parallel x\parallel ,\parallel y\parallel \in \mathbb{K}$ 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 non-Archimedean field. Now we derive non-Archimedean version of Theorem 2.

Theorem 4.(Non-Archimedean Hudzik-Landes-Dragomir-Kato-Saito-Tamura Inequality) Let $\mathcal{X}$ be a NALS over $\mathbb{K}$ and $n\in \mathbb{N}$. Then for all $x_1,\dots ,x_n\in \mathcal{X}\setminus \left\{0\right\}$ with $\parallel x_1\parallel ,\dots ,\parallel x_n\parallel \in \mathbb{K}$ it holds

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

Proof. 

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

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

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

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

□

Now we derive continuous version of Theorem 4.

Theorem 5.  

Let $\mathcal{X}$ be a NALS over $\mathbb{K}$ and $(\Omega ,\mu )$ be a non-Archimedean measure space. Let $\varphi :\Omega \to \mathcal{X}\setminus \left\{0\right\}$ be a measurable function such that $\parallel \varphi \left(\alpha \right)\parallel \in \mathbb{K}$ for every $\alpha \in \Omega$. Then

$$\begin{array}{cc}\hfill ∥{\int }_{\Omega }\varphi \left(\alpha \right)d\mu \left(\alpha \right)∥& \le \underset{\beta \in \Omega }{inf}\left\{|\parallel \varphi \left(\beta \right)\parallel |max\left\{∥{\int }_{\Omega }{\displaystyle \frac{\varphi \left(\alpha \right)}{\parallel \varphi \left(\alpha \right)\parallel }}d\mu \left(\alpha \right)∥,\underset{\alpha \in \Omega }{sup}\left|{\displaystyle \frac{1}{\parallel \varphi \left(\alpha \right)\parallel }}-{\displaystyle \frac{1}{\parallel \varphi \left(\beta \right)\parallel }}\right|\parallel \varphi \left(\alpha \right)\parallel \right\}\right\}.\hfill \end{array}$$

Proof. 

Let $\varphi :\Omega \to \mathcal{X}\setminus \left\{0\right\}$ be a measurable function such that $\parallel \varphi \left(\alpha \right)\parallel \in \mathbb{K}$ for every $\alpha \in \Omega$. Let $\beta \in \Omega$ be fixed. Then

$$\begin{array}{cc}\hfill ∥{\int }_{\Omega }\varphi \left(\alpha \right)d\mu \left(\alpha \right)∥& =∥{\int }_{\Omega }{\displaystyle \frac{\parallel \varphi \left(\beta \right)\parallel \varphi \left(\alpha \right)}{\parallel \varphi \left(\alpha \right)\parallel }}d\mu \left(\alpha \right)+{\int }_{\Omega }\left(1-{\displaystyle \frac{\parallel \varphi \left(\beta \right)\parallel }{\parallel \varphi \left(\alpha \right)\parallel }}\right)\varphi \left(\alpha \right)d\mu \left(\alpha \right)∥\hfill \\ \hfill & \le max\left\{∥{\int }_{\Omega }{\displaystyle \frac{\parallel \varphi \left(\beta \right)\parallel \varphi \left(\alpha \right)}{\parallel \varphi \left(\alpha \right)\parallel }}d\mu \left(\alpha \right)∥,∥{\int }_{\Omega }\left(1-{\displaystyle \frac{\parallel \varphi \left(\beta \right)\parallel }{\parallel \varphi \left(\alpha \right)\parallel }}\right)\varphi \left(\alpha \right)d\mu \left(\alpha \right)∥\right\}\hfill \\ \hfill & =max\left\{|\parallel \varphi \left(\beta \right)\parallel |∥{\int }_{\Omega }{\displaystyle \frac{\varphi \left(\alpha \right)}{\parallel \varphi \left(\alpha \right)\parallel }}d\mu \left(\alpha \right)∥,\left|\parallel \varphi \left(\beta \right)\parallel \right|∥{\int }_{\Omega }\left({\displaystyle \frac{1}{\parallel \varphi \left(\beta \right)\parallel }}-{\displaystyle \frac{1}{\parallel \varphi \left(\alpha \right)\parallel }}\right)\varphi \left(\alpha \right)d\mu \left(\alpha \right)∥\right\}\hfill \\ \hfill & =|\parallel \varphi \left(\beta \right)\parallel |max\left\{∥{\int }_{\Omega }{\displaystyle \frac{\varphi \left(\alpha \right)}{\parallel \varphi \left(\alpha \right)\parallel }}d\mu \left(\alpha \right)∥,∥{\int }_{\Omega }\left({\displaystyle \frac{1}{\parallel \varphi \left(\beta \right)\parallel }}-{\displaystyle \frac{1}{\parallel \varphi \left(\alpha \right)\parallel }}\right)\varphi \left(\alpha \right)d\mu \left(\alpha \right)∥\right\}\hfill \\ \hfill & \le |\parallel \varphi \left(\beta \right)\parallel |max\left\{∥{\int }_{\Omega }{\displaystyle \frac{\varphi \left(\alpha \right)}{\parallel \varphi \left(\alpha \right)\parallel }}d\mu \left(\alpha \right)∥,\underset{\alpha \in \Omega }{sup}∥\left({\displaystyle \frac{1}{\parallel \varphi \left(\beta \right)\parallel }}-{\displaystyle \frac{1}{\parallel \varphi \left(\alpha \right)\parallel }}\right)\varphi \left(\alpha \right)∥\right\}\hfill \\ \hfill & =|\parallel \varphi \left(\beta \right)\parallel |max\left\{∥{\int }_{\Omega }{\displaystyle \frac{\varphi \left(\alpha \right)}{\parallel \varphi \left(\alpha \right)\parallel }}d\mu \left(\alpha \right)∥,\underset{\alpha \in \Omega }{sup}\left|{\displaystyle \frac{1}{\parallel \varphi \left(\beta \right)\parallel }}-{\displaystyle \frac{1}{\parallel \varphi \left(\alpha \right)\parallel }}\right|\parallel \varphi \left(\alpha \right)\parallel \right\}.\hfill \end{array}$$

By varying $\beta$ and taking infimum in the previous inequality gives

$$\begin{array}{c}\hfill ∥{\int }_{\Omega }\varphi \left(\alpha \right)d\mu \left(\alpha \right)∥\le \underset{\beta \in \Omega }{inf}\left\{|\parallel \varphi \left(\beta \right)\parallel |max\left\{∥{\int }_{\Omega }{\displaystyle \frac{\varphi \left(\alpha \right)}{\parallel \varphi \left(\alpha \right)\parallel }}d\mu \left(\alpha \right)∥,\underset{\alpha \in \Omega }{sup}\left|{\displaystyle \frac{1}{\parallel \varphi \left(\alpha \right)\parallel }}-{\displaystyle \frac{1}{\parallel \varphi \left(\beta \right)\parallel }}\right|\parallel \varphi \left(\alpha \right)\parallel \right\}\right\}.\end{array}$$

□

3. Conclusions

(1)

In 1992, Hudzik and Landes improved centuries old triangle inequality in normed linear spaces [2].

(2)

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

(3)

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

(4)

In this article, we extended centuries old ultra-norm inequality.