Non-Archimedean Massera-Schaffer-Maligranda-Pecaric-Rajic Inequality

1. Introduction

Let $\mathcal{X}$ be a normed linear space (NLS). Recall that for $x,y\in \mathcal{X}\setminus \left\{0\right\}$, the Clarkson angle[1] between x and y is defined as

$$\begin{array}{c}\hfill \alpha [x,y]∥{\displaystyle \frac{x}{\parallel x\parallel }}-{\displaystyle \frac{y}{\parallel y\parallel }}∥.\end{array}$$

Important Massera-Schaffer inequality [2] says that

$$\begin{array}{c}\hfill \alpha [x,y]=∥{\displaystyle \frac{x}{\parallel x\parallel }}-{\displaystyle \frac{y}{\parallel y\parallel }}∥\le {\displaystyle \frac{2\parallel x-y\parallel }{max\{\parallel x\parallel ,\parallel y\parallel \}}},\forall x,y\in \mathcal{X}\setminus \left\{0\right\}.\end{array}$$

(1)

Maligranda improved Inequality (1) in 2006 [3].

Theorem 1.

[3,4] (Massera-Schaffer-Maligranda Inequality) Let $\mathcal{X}$ be a NLS. Then for all $x,y\in \mathcal{X}\setminus \left\{0\right\}$,

$$\begin{array}{c}\hfill \alpha [x,y]=∥{\displaystyle \frac{x}{\parallel x\parallel }}-{\displaystyle \frac{y}{\parallel y\parallel }}∥\le {\displaystyle \frac{\parallel x-y\parallel +|\parallel x\parallel -\parallel y\parallel |}{max\{\parallel x\parallel ,\parallel y\parallel \}}}\le {\displaystyle \frac{2\parallel x-y\parallel }{max\{\parallel x\parallel ,\parallel y\parallel \}}}.\end{array}$$

(2)

We note that, in 1964 Dunkl and Williams [5] independently showed that

$$\begin{array}{c}\hfill ∥{\displaystyle \frac{x}{\parallel x\parallel }}-{\displaystyle \frac{y}{\parallel y\parallel }}∥\le {\displaystyle \frac{4\parallel x-y\parallel }{\parallel x\parallel +\parallel y\parallel }},\forall x,y\in \mathcal{X}\setminus \left\{0\right\}.\end{array}$$

(3)

Note that

$$\begin{array}{c}\hfill {\displaystyle \frac{2\parallel x-y\parallel }{max\{\parallel x\parallel ,\parallel y\parallel \}}}\le {\displaystyle \frac{4\parallel x-y\parallel }{\parallel x\parallel +\parallel y\parallel }},\forall x,y\in \mathcal{X}\setminus \left\{0\right\}.\end{array}$$

Therefore Inequality (1) improves Inequality (3). Inequality (3) is famously known as Dunkl-Williams inequality in the literature. In 2007, Pecaric and Rajic extended Inequality (2) to finitely many nonzero elements [6].

Theorem 2.

[6] (Massera-Schaffer-Maligranda-Pecaric-Rajic Inequality) Let $\mathcal{X}$ be a NLS and $n\in \mathbb{N}$. Then for all $x_1,\dots ,x_n\in \mathcal{X}\setminus \left\{0\right\}$,

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

It is important to ask what are non-Archimedean versions of Theorems 1 and 2? We answer the question by deriving non-Archimedean Massera-Schaffer-Maligranda-Pecaric-Rajic Inequality (Theorem 4).

2. Non-Archimedean Massera-Schaffer-Maligranda-Pecaric-Rajic Inequality

Let $\mathbb{K}$ be a field. 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 [7]. Let $\mathcal{X}$ be a vector space over a non-Archimedean valued field $\mathbb{K}$ with valuation $|·|$. 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) [8]. Let $\mathcal{X}$ be a NALS. Let $x,y\in \mathcal{X}\setminus \left\{0\right\}$ with $\parallel x\parallel ,\parallel y\parallel \in \mathcal{X}$. We define the non-Archimedean Clarkson angle between x and y as

$$\begin{array}{c}\hfill \alpha [x,y]∥{\displaystyle \frac{x}{\parallel x\parallel }}-{\displaystyle \frac{y}{\parallel y\parallel }}∥.\end{array}$$

Non-Archimedean version of Theorem 1 now reads as follows.

Theorem 3.(Non-Archimedean Massera-Schaffer 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

$$\begin{array}{c}\hfill \alpha [x,y]=∥{\displaystyle \frac{x}{\parallel x\parallel }}-{\displaystyle \frac{y}{\parallel y\parallel }}∥\le {\displaystyle \frac{max\{\parallel x-y\parallel ,|\parallel x\parallel -\parallel y\parallel \left|\right\}}{max\{\parallel x\parallel ,\parallel y\parallel \}}}.\end{array}$$

Proof. 

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

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

and

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

Therefore

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

(4)

and

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

(5)

Inequalities (4) and (5) give

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

□

Note the additional assumption $\parallel x\parallel ,\parallel y\parallel \in \mathcal{X}$ in the previous theorem. This is necessary because, since the norm is a real number, we generally cannot guarantee that it belongs to the given non-Archimedean field. We now derive the non-Archimedean version of Theorem 2.

Theorem 4.(Non-Archimedean Massera-Schaffer-Maligranda-Pecaric-Rajic 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

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

Proof. 

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

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

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

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

□

3. Conclusions

(1)

In 1958, Massera and Schaffer derived a surprising upper bound for the Clarkson angle between two nonzero elements in a normed linear space [2].

(2)

In 2006, Maligranda improved Massera-Schaffer inequality [3].

(3)

In 2007, Pecaric and Rajic extended Maligranda inequality for finitely many nonzero elements [6].

(4)

In this article, we introduced the notion of non-Archimedean Clarkson angle and derived non-Archimedean version of Massera-Schaffer-Maligranda-Pecaric-Rajic inequality.