Finite Field Massera-Schaffer-Maligranda-Pecaric-Rajic Inequality

1. Introduction

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

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

In 1958, Massera and Schaffer derived following upper bound for the Clarkson angle [2].

Theorem 1.

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

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

(1)

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

Theorem 2.

[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)

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)

Since

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

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 3.

[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 finite field versions of Theorems 2 and 3? We answer the question by deriving finite field Massera-Schaffer-Maligranda-Pecaric-Rajic Inequality (Theorem 5).

2. Finite Field Massera-Schaffer-Maligranda-Pecaric-Rajic Inequality

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

Definition 1.

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

(i)

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

(ii)

$\left|\lambda \mu \right|\le \left|\lambda \right|\left|\mu \right|$ for all $\lambda ,\mu \in \mathbb{F}$.

(iii)

$|\lambda +\mu |\le \left|\lambda \right|+\left|\mu \right|$ for all $\lambda ,\mu \in \mathbb{F}$.

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

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

Definition 2.

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

(i)

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

(ii)

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

(iii)

$\parallel x+y\parallel \le \parallel x\parallel +\parallel y\parallel$ for all $x,y\in \mathcal{X}$.

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

Let $\mathcal{X}$ be a sub-normed linear space over $\mathbb{F}$. Let $x,y\in \mathcal{X}\setminus \left\{0\right\}$ with $\parallel x\parallel ,\parallel y\parallel \in \mathbb{F}$. We define the non-Archimedean Clarkson angle between x and y as

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

Finite field version of Theorem 2 now reads as follows.

Theorem 4.(Finite Field Massera-Schaffer Inequality) Let $\mathcal{X}$ be a sub-normed linear space over $\mathbb{F}$. Then for all $x,y\in \mathcal{X}\setminus \left\{0\right\}$ with $\parallel x\parallel ,\parallel y\parallel \in \mathbb{F}$ it holds

$$\begin{array}{c}\hfill \alpha [x,y]=∥{\displaystyle \frac{x}{\parallel x\parallel }}-{\displaystyle \frac{y}{\parallel y\parallel }}∥\le min\left\{\left|{\displaystyle \frac{1}{\parallel x\parallel }}\right|\left(\parallel x-y\parallel +\left|1-{\displaystyle \frac{\parallel x\parallel }{\parallel y\parallel }}\right|\parallel y\parallel \right),\left|{\displaystyle \frac{1}{\parallel y\parallel }}\right|\left(\parallel x-y\parallel +\left|{\displaystyle \frac{\parallel y\parallel }{\parallel x\parallel }}-1\right|\parallel x\parallel \right)\right\}.\end{array}$$

Proof. 

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

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

and

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

Therefore

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

(4)

and

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

(5)

Inequalities (4) and (5) give

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

□

Note the additional assumption $\parallel x\parallel ,\parallel y\parallel \in \mathbb{F}$ in the previous theorem. This is required because the norm, being a real number, cannot generally be guaranteed to belong to the specified sub-modulus field. We now derive the finite field version of Theorem 3.

Theorem 5.(Finite Field Massera-Schaffer-Maligranda-Pecaric-Rajic Inequality) Let $\mathcal{X}$ be a sub-normed linear space over $\mathbb{F}$ and $n\in \mathbb{N}$. Then for all $x_1,\dots ,x_n\in \mathcal{X}\setminus \left\{0\right\}$ with $\parallel x_1\parallel ,\dots ,\parallel x_n\parallel \in \mathbb{F}$ it holds

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

Proof. 

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

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

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

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

□

3. Conclusions

1.

In 1958, Massera and Schaffer derived an 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 finite field Clarkson angle and derived finite field version of Massera-Schaffer-Maligranda-Pecaric-Rajic inequality.