Every n-dimensional normed space is the space ${\mathbb{R}}^n$ endowed with a normal norm

Abstract

Recently, Alonso showed that every two-dimensional normed space is isometrically isomorphic to a generalized Day-James space introduced by Nilsrakoo and Saejung. In this paper, we consider the result of Alonso for n-dimensional normed spaces.

MSC:46B20.

A norm $\parallel \cdot \parallel$ on ${\mathbb{R}}^2$ is said to be absolute if $\parallel (x,y)\parallel =\parallel (|x|,|y|)\parallel$ for all $(x,y)\in {\mathbb{R}}^2$, and normalized if $\parallel (1,0)\parallel =\parallel (0,1)\parallel =1$. The set of all absolute normalized norms on ${\mathbb{R}}^2$ is denoted by $A{N}_2$. Bonsall and Duncan [1] showed the following characterization of absolute normalized norms on ${\mathbb{R}}^2$. Namely, the set $A{N}_2$ of all absolute normalized norms on ${\mathbb{R}}^2$ is in a one-to-one correspondence with the set ${\mathrm{\Psi }}_2$ of all convex functions ψ on $[0,1]$ satisfying $max\{1-t,t\}\le \psi (t)\le 1$ for all $t\in [0,1]$ (cf. [2]). The correspondence is given by the equation $\psi (t)=\parallel (1-t,t)\parallel$ for all $t\in [0,1]$. Note that the norm ${\parallel \cdot \parallel }_{\psi }$ associated with the function $\psi \in {\mathrm{\Psi }}_2$ is given by

$${\parallel (x,y)\parallel }_{\psi }=\{\begin{array}{cc}(|x|+|y|)\psi (\frac{|y|}{|x|+|y|}),\hfill & \text{if }(x,y)\ne (0,0),\hfill \\ 0,\hfill & \text{if }(x,y)=(0,0).\hfill \end{array}$$

The Day-James space ${\ell }_p\text{-}{\ell }_q$ is defined for $1\le p,q\le \mathrm{\infty }$ as the space ${\mathbb{R}}^2$ endowed with the norm

$${\parallel (x,y)\parallel }_{p,q}=\{\begin{array}{cc}{\parallel (x,y)\parallel }_p,\hfill & \text{if }xy\ge 0,\hfill \\ {\parallel (x,y)\parallel }_q,\hfill & \text{if }xy\le 0.\hfill \end{array}$$

James [3] considered the space ${\ell }_p\text{-}{\ell }_q$ with $p^{-1}+q^{-1}=1$ as an example of a two-dimensional normed space where Birkhoff orthogonality is symmetric. Recall that if x, y are elements of a real normed space X, then x is said to be Birkhoff-orthogonal to y, denoted by $x{\perp }_{B}y$, if $\parallel x+\lambda y\parallel \ge \parallel x\parallel$ for all $\lambda \in \mathbb{R}$. Birkhoff orthogonality is homogeneous, that is, $x{\perp }_{B}y$ implies $\alpha x{\perp }_B\beta y$ for any real numbers α and β. However, Birkhoff orthogonality is not symmetric in general, that is, $x{\perp }_{B}y$ does not imply $y{\perp }_{B}x$. More details about Birkhoff orthogonality can be found in Birkhoff [4], Day [5, 6] and James [3, 7, 8].

In 2006, Nilsrakoo and Saejung [9] introduced and studied generalized Day-James spaces ${\ell }_{\phi }\text{-}{\ell }_{\psi }$, where ${\ell }_{\phi }\text{-}{\ell }_{\psi }$ is defined for $\phi ,\psi \in {\mathrm{\Psi }}_2$ as the space ${\mathbb{R}}^2$ endowed with the norm

$${\parallel (x,y)\parallel }_{\phi ,\psi }=\{\begin{array}{cc}{\parallel (x,y)\parallel }_{\phi },\hfill & \text{if }xy\ge 0,\hfill \\ {\parallel (x,y)\parallel }_{\psi },\hfill & \text{if }xy\le 0.\hfill \end{array}$$

Recently, Alonso [10] showed that every two-dimensional normed space is isometrically isomorphic to a generalized Day-James space. In this paper, we consider the result of Alonso for n-dimensional spaces.

First, we give a characterization of generalized Day-James spaces.

Proposition 1 Let $\parallel \cdot \parallel$ be a norm on ${\mathbb{R}}^2$. Then the space $({\mathbb{R}}^2,\parallel \cdot \parallel )$ is a generalized Day-James space if and only if ${\parallel \cdot \parallel }_{\mathrm{\infty }}\le \parallel \cdot \parallel \le {\parallel \cdot \parallel }_1$.

Proof If $({\mathbb{R}}^2,\parallel \cdot \parallel )$ is a generalized Day-James space, then one can easily have ${\parallel \cdot \parallel }_{\mathrm{\infty }}\le \parallel \cdot \parallel \le {\parallel \cdot \parallel }_1$. So, we assume that ${\parallel \cdot \parallel }_{\mathrm{\infty }}\le \parallel \cdot \parallel \le {\parallel \cdot \parallel }_1$. Let

$$\phi (t)=\parallel (1-t,t)\parallel \text{and}\psi (t)=\parallel (1-t,-t)\parallel$$

for all $t\in [0,1]$, respectively. Then, clearly, we have $\phi ,\psi \in {\mathrm{\Psi }}_2$ and $\parallel \cdot \parallel ={\parallel \cdot \parallel }_{\phi ,\psi }$. Hence, the space $({\mathbb{R}}^2,\parallel \cdot \parallel )$ is a generalized Day-James space. □

Motivated by this fact, we consider the following

Definition 2 A norm $\parallel \cdot \parallel$ on ${\mathbb{R}}^n$ is said to be normal if it satisfies ${\parallel \cdot \parallel }_{\mathrm{\infty }}\le \parallel \cdot \parallel \le {\parallel \cdot \parallel }_1$.

We recall some notions about multilinear forms. Let X be a real vector space. Then a real-valued function F on $X^n$ is said to be an n-linear form if it is linear separately in each variable, that is,

for each $i\in \{1,2,\dots ,n\}$. If $F:X^n\to \mathbb{R}$ is an n-linear form, then F is said to be alternating if

$$F(x_1,\dots ,x_i,x_{i+1},\dots ,x_n)=-F(x_1,\dots ,x_{i+1},x_i,\dots ,x_n)$$

for each $i\in \{1,2,\dots ,n\}$ or, equivalently, $F(x_1,x_2,\dots ,x_n)=0$ if $x_i=x_j$ for some i, j with $i\ne j$. Furthermore, F is said to be bounded if

$$\parallel F\parallel :=sup\{|F(x_1,x_2,\dots ,x_n)|:(x_1,x_2,\dots ,x_n)\in {(S_X)}^n\}<\mathrm{\infty },$$

where $S_X$ denotes the unit sphere of X. If F is bounded, then we have

$$|F(x_1,x_2,\dots ,x_n)|\le \parallel F\parallel \parallel x_1\parallel \parallel x_2\parallel \cdots \parallel x_n\parallel$$

for all $(x_1,x_2,\dots ,x_n)\in X^n$.

For our purpose, we give another simple proof of the following result of Day [5]. For each subset A of a normed space, let $[A]$ denote the closed linear span of A. If M, N are subspaces of a real normed space X, then M is said to be Birkhoff orthogonal to N, denoted by $M{\perp }_{B}N$, if $\parallel x+y\parallel \ge \parallel x\parallel$ for all $x\in M$ and all $y\in N$. In particular, $x{\perp }_{B}M$ denotes $[\{x\}]{\perp }_{B}M$.

Lemma 3 Let X be an n-dimensional real normed space. Then there exists a basis $\{e_1,e_2,\dots ,e_n\}$ for X such that $\parallel e_i\parallel =1$ and $e_i{\perp }_B[{\{e_k\}}_{k\ne i}]$ for all $i=1,2,\dots ,n$.

Proof Let $\{u_1,u_2,\dots ,u_n\}$ be a basis for X. Then each vector $x\in X$ is uniquely expressed in the form $x={\sum }_{k=1}^n{\alpha }_k(x)u_k$. Define the function F on $X^n$ by

$$F(x_1,x_2,\dots ,x_n)=\left|\begin{array}{cccc}{\alpha }_1(x_1)& {\alpha }_2(x_1)& \dots & {\alpha }_n(x_1)\\ {\alpha }_1(x_2)& {\alpha }_2(x_2)& \dots & {\alpha }_n(x_2)\\ ⋮& ⋮& \ddots & ⋮\\ {\alpha }_1(x_n)& {\alpha }_2(x_n)& \dots & {\alpha }_n(x_n)\end{array}\right|$$

for all $(x_1,x_2,\dots ,x_n)\in X^n$. Then it is easy to check that F is an alternating bounded n-linear form. Since F is jointly continuous on the compact subset ${(S_X)}^n$ of $X^n$, there exists $(e_1,e_2,\dots ,e_n)\in {(S_X)}^n$ such that

$$F(e_1,e_2,\dots ,e_n)=\parallel F\parallel >0.$$

For all $i=1,2,\dots ,n$ and all $({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n)\in {\mathbb{R}}^n$, we have

$$\begin{array}{rcl}\parallel F\parallel \parallel \sum _{k=1}^n{\alpha }_k{e}_k\parallel & \ge & \left|F(e_1,\dots ,e_{i-1},\sum _{k=1}^n{\alpha }_k{e}_k,e_{i+1},\dots ,e_n)\right|\\ =& |\sum _{k=1}^n{\alpha }_{k}F(e_1,\dots ,e_{i-1},e_k,e_{i+1},\dots ,e_n)|\\ =& |{\alpha }_{i}F(e_1,e_2,\dots ,e_n)|=\parallel F\parallel |{\alpha }_i|.\end{array}$$

Thus, we obtain

$$\parallel \sum _{k=1}^n{\alpha }_k{e}_k\parallel \ge |{\alpha }_i|=\parallel {\alpha }_i{e}_i\parallel ,$$

for all $i=1,2,\dots ,n$ and all $({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n)\in {\mathbb{R}}^n$. This means that $e_i{\perp }_B[{\{e_k\}}_{k\ne i}]$ for all $i=1,2,\dots ,n$. □

Now, we state the main theorem.

Theorem 4 Every n-dimensional normed space is isometrically isomorphic to the space ${\mathbb{R}}^n$ endowed with a normal norm.

Proof By Lemma 3, there exists an n-tuple $(e_1,e_2,\dots ,e_n)$ of elements of $S_X$ such that $e_i{\perp }_B[{\{e_k\}}_{k\ne i}]$ for all $i=1,2,\dots ,n$. Since $e_i{\perp }_B[{\{e_k\}}_{k\ne i}]$, we have

$$\parallel \sum _{k=1}^n{\alpha }_k{e}_k\parallel \ge \parallel {\alpha }_i{e}_i\parallel =|{\alpha }_i|,$$

for all $i=1,2,\dots ,n$ and all $({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n)\in {\mathbb{R}}^n$. Hence, we obtain

$$\parallel \sum _{k=1}^n{\alpha }_k{e}_k\parallel \ge max\{|{\alpha }_1|,|{\alpha }_2|,\dots ,|{\alpha }_n|\}$$

for all $({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n)\in {\mathbb{R}}^n$. From this fact, we note that $\{e_1,e_2,\dots ,e_n\}$ is linearly independent, that is, a basis for X.

Define the norm ${\parallel \cdot \parallel }_0$ on ${\mathbb{R}}^n$ by the formula

$${\parallel ({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n)\parallel }_0=\parallel \sum _{k=1}^n{\alpha }_k{e}_k\parallel$$

for all $({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n)\in {\mathbb{R}}^n$. Then, clearly, ${\parallel \cdot \parallel }_0$ is normal and X is isometrically isomorphic to the space $({\mathbb{R}}^n,{\parallel \cdot \parallel }_0)$. This completes the proof. □

Since the space ${\mathbb{R}}^2$ endowed with a normal norm is a generalized Day-James space by Proposition 1, we have the result of Alonso as a corollary.

Corollary 5 ([10])

Every two-dimensional real normed space is isometrically isomorphic to a generalized Day-James space.