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 [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)

We note that, in 2006, Maligranda independently derived Inequality (2) [2]. 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 [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 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.

• If $\lambda \in \mathbb{K}$ is such that $\left|\lambda \right|=0$, then $\lambda =0$.
• $\left|\lambda \mu \right|=\left|\lambda \right|\left|\mu \right|$ for all $\lambda ,\mu \in \mathbb{K}$.
• (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 [5]. 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.

• If $x\in \mathcal{X}$ is such that $\parallel x\parallel =0$, then $x=0$.
• $\parallel \lambda x\parallel =\left|\lambda \right|\parallel x\parallel$ for all $\lambda \in \mathbb{K}$, for all $x\in \mathcal{X}$.
• (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) [6]. We first derive non-Archimedean version of Inequality (2).

Theorem 3.

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

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

Proof. 

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

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

and

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

Therefore

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

(3)

and

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

(4)

Inequalities (3)and (4) give

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

□

Note the additional assumption $\parallel x\parallel ,\parallel y\parallel \in \mathcal{X}$ 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 and $n\in \mathbb{N}$. Then for all $x_1,\dots ,x_n\in \mathcal{X}\setminus \left\{0\right\}$ with $\parallel x_1\parallel ,\dots ,\parallel x_n\parallel \in \mathcal{X}$ it holds

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

Proof. 

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

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

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

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

□

Now we derive continuous version of Theorem 4.

Theorem 5.

Let $\mathcal{X}$ be a NALS 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 $\parallel \varphi \left(\alpha \right)\parallel \in \mathcal{X}$ for every $\alpha \in \Omega$. Then

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

Proof. 

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)∥,∥{\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 & \le 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)∥,\underset{\alpha \in \Omega }{sup}∥\left(1-{\displaystyle \frac{\parallel \varphi \left(\beta \right)\parallel }{\parallel \varphi \left(\alpha \right)\parallel }}\right)\varphi \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)∥,\underset{\alpha \in \Omega }{sup}|\parallel \varphi \left(\alpha \right)\parallel -\parallel \varphi \left(\beta \right)\parallel |\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{\parallel \varphi \left(\alpha \right)\parallel }{\parallel \varphi \left(\beta \right)\parallel }}-1\right|\right\}.\hfill \end{array}$$

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

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

□

3. Conclusions

• In 1992, Hudzik and Landes improved centuries old triangle inequality in normed linear spaces.
• In 2006, Dragomir extended Hudzik-Landes inequality for more than two vectors.
• In 2007, Kato, Saito and Tamura extended Hudzik-Landes inequality without knowing the work of Dragomir.
• In this article, we extended centuries old ultra-norm inequality.