On the existence of m-norms in vector spaces over valued fields

Abstract

Gähler (Untersuchungen über verallgemeinerte m-metrische Räume. I”. German. Math Nachr 40, 165–189, 1969) investigated m-metric spaces and in particular m-normed spaces over the field of reals. Here the existence of m-norms will be investigated and this in the more general setting of vector spaces over arbitrary non-trivial valued fields.

1 Introduction and motivation

Looking for (more and more exotic) generalizations of the notion of a normed space when investigating stability questions for functional equations the so-called m-normed spaces appeared. Examples of papers in this direction are for instance [4, 5, 7, 8] as well as [2] and [6].

Now a quite natural question arises, namely how special are vector spaces admitting an m-norm. The main source for the following investigations is [9] where the notion of an m-normed space is a special case of that of an m-metric space.

It is interesting to note that most of the recent papers on m-normed spaces refer to a paper from 1989 [11] for the definition. It is even more interesting that the author of this paper mentions Siegfried Gähler’s paper from 1969 and even that the paper was written under the supervision of Gähler.

In [9, 1.2, p. 169] the following definition is given when $K={\mathbb {R}}$.

Definition 1

Let X be a vector space over a non-trivially valued field K (with absolute value $\vert \quad \vert$) and $m\in {\mathbb {N}}$. Then $\left\| ~\right\| :X^m\rightarrow {\mathbb {R}}$ is called m-norm, if

• 1. $\left\| x_1,x_2,\ldots ,x_m\right\| =0$ iff $x_1,x_2,\ldots ,x_m$ are linearly dependent.

• 2. $\left\| \alpha x_1,x_2,\ldots ,x_m\right\| =\left| \alpha \right| \left\| x_1,x_2,\ldots ,x_m\right\|$ for all $x_1,x_2,\ldots ,x_m\in X$ and all $\alpha \in K$.

• 3. $\left\| x_{\pi (1)},x_{\pi (2)},\ldots ,x_{\pi (m)}\right\| =\left\| x_1,x_2,\ldots ,x_m\right\|$ for all $x_1,x_2,\ldots ,x_m\in X$ and every permutation $\pi$ of ${1,2,\ldots ,m}$

• 4. $\left\| x'_1+x''_1,x_2,\ldots ,x_m\right\| \le \left\| x'_1,x_2,\ldots ,x_m\right\| +\left\| x''_1,x_2,\ldots ,x_m\right\|$ for all $x'_1,x''_1,x_2,\ldots ,x_m\in X$.

Remark 1

$\left\| ~\right\| \ge 0$ since $0=\left\| 0,x_2,\ldots ,x_m\right\| =\left\| x_1-x_1,x_2,\ldots ,x_m\right\| \le \left\| x_1,x_2,\ldots ,x_m\right\| +\left\| -x_1,x_2,\ldots ,x_m\right\| =2\left\| x_1,x_2,\ldots ,x_m\right\|$

Remark 2

• 1. For $\dim (X)<m$ we have $\left\| ~\right\| =0$ since every m-tuple of vectors in X must be linearly dependent.

• 2. ([9, Satz 3]) In the case $\dim (X)=m$ (und $K={\mathbb {R}}$), $||\quad ||$ is an m-norm iff there is some $\alpha >0$ such that

  $$\begin{aligned} \begin{aligned}&\left\| x_1,x_2,\ldots ,x_m\right\| =\\&\quad \alpha \cdot \left| {\det }(x_1,x_2,\ldots ,x_m)\right| \text { for all } x_1,x_2,\ldots ,x_m\in X \end{aligned} \end{aligned}$$

  (1)

• 3. 1-norms are norms (in the usual sense).

Remark 3

Using [9, Satz 6] for arbitrary real normed spaces X with $\dim (X)\ge m$ leads to the following result.

Let ${\mathcal {B}}:=\{f\in X'\mid \left\| f\right\| \le 1\}\subseteq X'$ be the unit ball in the dual $X'$ of X (where $\left\| f\right\| :=\sup _{\left\| x\right\| \le 1}\left| f(x)\right|$).

Then

$$\begin{aligned} \left\| x_1,x_2,\ldots ,x_m\right\| := \sup _{(f_1,f_2,\ldots ,f_m)\in {\mathcal {B}}^m}\left| \det ({\mathcal {C}}_{f_1,f_2,\ldots ,f_m}^{x_1,x_2,\ldots ,x_m})\right| \end{aligned}$$

(2)

describes an m-norm on X.

Here

$$\begin{aligned} {\mathcal {C}}_{f_1,f_2,\ldots ,f_m}^{x_1,x_2,\ldots ,x_m}:= (f_i(x_j))_{1\le i,j\le m} = \begin{pmatrix} f_1(x_1) &{} f_{1}(x_2)&{} \cdots &{} f_{1}(x_m) \\ f_2(x_1) &{} f_{2}(x_2)&{} \cdots &{} f_{2}(x_m) \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ f_m(x_1) &{} f_{m}(x_2)&{} \cdots &{} f_{m}(x_m) \end{pmatrix} \end{aligned}$$

(3)

is the so-called Casorati-Matrix.

Remark 4

It is shown in [13] that the dual space of a normed space over a valued field in general is not rich enough to separate a finite set of points. Thus a construction as above will not work in general.

Remark 5

From [9, Satz 7] it follows that for real inner product spaces X with inner product $\langle .,.\rangle$ and $\dim (X)\ge m$ there is another possibility to get an m-norm.

This m-norm is given by

$$\begin{aligned} \left\| x_1,x_2,\ldots ,x_m\right\| := \sqrt{{\det {(\langle x_i,x_j\rangle )_{1\le i,j\le m}}}}. \end{aligned}$$

(4)

Here

$$\begin{aligned} (\langle x_i,x_j\rangle )_{1\le i,j\le m} = \begin{pmatrix} \langle x_1,x_1\rangle &{} \langle x_1,x_2\rangle &{} \cdots &{} \langle x_1,x_m\rangle \\ \langle x_2,x_1\rangle &{} \langle x_2,x_2\rangle &{} \cdots &{} \langle x_2,x_m\rangle \\ \vdots &{} \vdots &{} \ddots &{} \vdots \\ \langle x_m,x_1\rangle &{} \langle x_m,x_2\rangle &{} \cdots &{} \langle x_m,x_m\rangle \end{pmatrix} \end{aligned}$$

(5)

denotes the Gram matrix for the vectors $x_1,x_2,\ldots ,x_m$.

2 Two answers for real spaces

Theorem 1

For every real vector space X and for all $m\in {\mathbb {N}}$ there is an m-norm defined on X.

Proof

First possibility: For $m=1$ we use the fact, that every real vector space X is normable. To see this we take a basis B X and define $||\quad ||$ by $\left\| \sum _{b\in B}\lambda _b b\right\| :=\max _{b\in B}\left| \lambda _b\right|$. It is easy to see that this has the properties of a norm. See for example [3, p. 279] where this was also used. For $m>1$ one may use Remark 3.

Second one: Given a basis B as of X as above, one immehadiately verifies that the definition $\langle \sum _{b\in B}\lambda _b b,\sum _{b\in B}\mu _{b} b\rangle :=\sum _{b\in B}\lambda _b\mu _b$ results in an inner product for X. Remark 5 gives an m-norm. $\square$

3 A general answer

Theorem 1 uses the existence of a basis for vector spaces, Remark 3 uses the Hahn-Banach Theorem. Thus the axiom of choice is involved and also that the field of scalars is the field of reals.

The answer given now applies to arbitrary vector spaces over arbitrary valued fields but also uses the axiom of choice insofar as applying the existence of a basis implies that it is enough to consider vector spaces of the form $X=K^{(B)}:=\{f\in K^B\mid \, \text {supp}(f)\text { is finite }\}$, where $\text {supp}(f):=f^{-1}(K\setminus \{0\})$.

Theorem 2

For any valued field K, for any vector space X over K and for any integer $m>0$ there is an m-norm defined on X.

For $X=K^{(B)}$ and $f=(f_1,f_2,\ldots ,f_m)\in X^m$ this m-norm is given by

$$\begin{aligned} \left\| f\right\| =\left\| f_1,f_2,\ldots ,f_m\right\| =\sum _{b\in B^m}\left| \det ({\mathcal {C}}^b_f)\right| . \end{aligned}$$

(6)

We start with a lemma. (Compare [1, p. 229], where the equivalent characterization of linearly dependent functions is given.)

Lemma 1

Let X be a set, K be a field and $m\in {\mathbb {N}}, m>0$. Then $f_1,f_2,\ldots ,f_m\in K^X$ are linearly independent iff there are $x_1,x_2,\ldots ,x_m\in X$ such that the Casorati matrix ${\mathcal {C}}_{f_1,f_2,\ldots ,f_m}^{x_1,x_2,\ldots ,x_m}$ is regular.

Proof of the lemma

If ${\mathcal {C}}_{f_1,f_2,\ldots ,f_m}^{x_1,x_2,\ldots ,x_m}$ is regular, the functions $f_1\big \vert _{X'},f_2\big \vert _{X'},\ldots ,f_m\big \vert _{X'}$ are linearly independent where $X':=\{x_1,x_2,\ldots ,x_m\}$. Thus the functions themselves are also linearly independent.

The second part is proved by induction on m.

For $m=1$ the functions $f_1,f_2,\ldots ,f_m$ are independent iff $f_1\not =0$. i.e., iff there is some $x_1\in X$ such that $f(x_1)\not =0$.

Now suppose that the claim is true for $m\ge 1$ and let $f_1,f_2,\ldots ,f_{m+1}$ be linearly independent. Then, by assumption, there are $x_1,x_2,\ldots ,x_m$, such that ${\mathcal {C}}_{f_1,f_2,\ldots ,f_m}^{x_1,x_2,\ldots ,x_m}$ is regular. Given $x\in X$ we now consider

$$\begin{aligned} {\mathcal {C}}_{f_1,f_2,\ldots ,f_m,f_{m+1}}^{x_1,x_2,\ldots ,x_m,x}= \begin{pmatrix} f_1(x_1) &{} f_{1}(x_2)&{} \cdots &{} f_{1}(x_m)&{} f_1(x) \\ f_2(x_1) &{} f_{2}(x_2)&{} \cdots &{} f_{2}(x_m)&{} f_2(x) \\ \vdots &{} \vdots &{} \ddots &{} \vdots &{}\vdots \\ f_m(x_1) &{} f_{m}(x_2)&{} \cdots &{} f_{m}(x_m) &{}f_m(x)\\ f_{m+1}(x_1)&{}f_{m+1}(x_2)&{}\cdots &{}f_{m+1}(x_m)&{}f_{m+1}(x) \end{pmatrix} \end{aligned}$$

By assumption $\alpha _{m+1}:=\det ({\mathcal {C}}_{f_1,f_2,\ldots ,f_m}^{x_1,x_2,\ldots ,x_m})\not =0$. Expanding $\det ({\mathcal {C}}_{f_1,f_2,\ldots ,f_m,f_{m+1}}^{x_1,x_2,\ldots ,x_m,x})$ with respect to the last column results in

$$\begin{aligned} \det ({\mathcal {C}}_{f_1,f_2,\ldots ,f_m,f_{m+1}}^{x_1,x_2,\ldots ,x_m,x})= \sum _{j=1}^{m+1}\alpha _j f_j(x) \end{aligned}$$

where the coefficients $\alpha _j$ do not depend on x. If this expression is 0 for all X, the functions $f_1,f_2,\ldots ,f_{m+1}$ will be linearly dependent, since $\alpha _{m+1}\not =0$. This is a contradiction. Accordingly the determinant in question must be different from 0 for some $x=x_{m+1}$. This implies the result. $\square$

Proof of Theorem 2

First we show that (6) is well-defined, i.e., we show that the sum is finite. To this aim let $B_0\subseteq B$ be finite, such that $\bigcup _{i=1}^m\text {supp}(f_i)\subseteq B_0$. Then $\sum _{b\in B^m}\left| \det ({\mathcal {C}}^b_f)\right| =\sum _{b\in B_0^m}\left| \det ({\mathcal {C}}^b_f)\right|$, since for $b\not \in B_0^m$ there is some $j_0$, such that $b_{j_0}\not \in B_0$. This implies $b_{j_0}\not \in \bigcup _{i=1}^m\text {supp}(f_i)$.Thus $f_i(b_{j_0})=0$ for all i implying that $\det ({\mathcal {C}}^b_f)=0$.

Properties 2. and 4. from Definition 1 are consequences of the linearity of the determinant function with respect to the first row, 3. follows from the fact that a permutation of the rows may at most change the sign of the determinant.

Now we come to property 1.: If $f_1,f_2,\ldots ,f_m$ are linearly dependent, all the expressions $\det ({\mathcal {C}}^b_f)=0$ vanish, thus $\left\| f_1,f_2,\ldots ,f_m\right\| =0$.

If, on the other hand, $f_1,f_2,\ldots ,f_m$ are linearly independent, Lemma 1 implies that at least one of the expressions $\det ({\mathcal {C}}^b_f)$ is different from 0. $\square$

Remark 6

When talking on this subject during a conference it was asked by Laczkovich Miklós whether the use of the axiom of choice (AC) is necessary to show that every (real) vector space is normable.

A very good answer in [12] shows that assuming a weaker axiom, the axiom of dependent choice (DC) (see [14]), only, the space $C({\mathbb {R}})$ of real valued continuous functions defined on ${\mathbb {R}}$ is not normable.

Thus anyway something stronger than DC must be assumed to get the property, that every real vector space is normable.

Remark 7

(other m-norms) For all $p\in [1,\infty ]$

$$\begin{aligned} \left\| f\right\| _p=\left\| f_1,f_2,\ldots ,f_m\right\| _p=\root {}^{\displaystyle p} \of {\sum _{b\in B^m}\left| \det ({\mathcal {C}}^b_f)\right| ^p} \end{aligned}$$

(7)

defines an m-norm, where

$$\begin{aligned} \left\| f\right\| _\infty :=\lim _{p\rightarrow \infty }\left\| f\right\| _p=\max _{b\in B^m}\left| \det ({\mathcal {C}}^b_f)\right| . \end{aligned}$$

4 $\ell$-norms from m-norms

It was shown in [10, Theorem 2.1] that an m-normed space with dimension at least m also admits an $(m-1)$-norm. Going step by step then leads to Corollary 2.2 in that paper: Every m-normed space with dimension at least m admits $\ell$-norms for all $1\le \ell \le m$.

Now we present another proof for this fact. To this aim we need a lemma which for $\ell =m-1$ is used implicitly in the proof of Theorem 2.1 mentioned above.

Lemma 2

Let V be a vector space over the field K, let $1\le \ell <m$ and let $A=\{a_1,a_2,\ldots ,a_m\}\subseteq V$ have cardinality m: $\left| A\right| =m$. Furthermore let $X\subseteq V$ be linearly independent with $\left| X\right| =\ell$.

Assume that $X\cup C'$ is linearly dependent for all $C'\subseteq A$ with $\left| C'\right| =m-\ell$. Then A is linearly dependent.

Proof

Let $0\le s\le \ell +1$. Then we call s good, if there are some $B, C\subseteq A$ with $\left| B\right| =s, \left| C\right| =m-\ell -1$ such that $B\cap C=\emptyset$ and $B\subseteq {\mathcal {L}}(X\cup C)$ (where ${\mathcal {L}}(D)$ denotes the linear hull of D). Obviously 0 is good. So let $t=\max \{s\mid 0\le s\le \ell +1, s\text { is good}\}$.

In the case $t=\ell +1$ the corresponding B, C fulfill $\left| B\cup C\right| =\left| B\right| +\left| C\right| =\ell +1+(m-\ell -1)=m$. Thus $A=B\cup C\subseteq {\mathcal {L}}(X\cup C)=:U$ with $\dim (U)\le m-1$. Therefore A is linearly dependent as any m vectors in U do have this property.

Now suppose that $t<\ell +1$. Then for some $B, C\subseteq A$ with $\left| B\right| =t, \left| C\right| =m-\ell -1$ and $B\cap C=\emptyset$ we have $B\subseteq {\mathcal {L}}(X\cup C)$. Moreover $\left| B\cup C\right| =\left| B\right| +\left| C\right| =t+m-\ell -1<m$. Thus $A{\setminus }(B\cup C)\not =\emptyset$. Let $a\in A{\setminus }(B\cup C)$ and put $C':=C\cup \{a\}$. Then $\left| C'\right| =m-\ell$ and thus by assumption $X\cup C'$ is linearly dependent. Since X is linearly independent this implies that there is some $d\in C'$ with $d\in {\mathcal {L}}(X\cup (C'\setminus \{d\}))$.

If $d=a$ the set $B':=B\cup \{a\}$ would have $t+1$ elements. Moreover $B'\subseteq {\mathcal {L}}(X\cup (C'{\setminus }\{d\}))={\mathcal {L}}(X\cup C)$. Thus $t+1$ would be good in contradiction to the maximality of t.

Finally let $d\not =a$. Let $C'':=(C{\setminus }\{d\})\cup \{a\}$. Then $d\in {\mathcal {L}}(X\cup C'')$ and $C\setminus \{d\}\subseteq {\mathcal {L}}(X\cup C'')$ which implies that ${\mathcal {L}}(X\cup C)\subseteq {\mathcal {L}}(X\cup C'')$. Accordingly the set $B'=B\cup \{d\}$ is a subset of ${\mathcal {L}}(X\cup C'')$ with $t+1$ elements. Taking into account that $C''$ has $m-\ell -1$ elements and that $B'\cap C''=\emptyset$ this case also contradicts the choice of t. $\square$

Theorem 3

Let V be a vector space of dimension $m\ge 2$ over the field K with a non-trivial valuation and assume that $\left\| \,\right\| _m$ is an m-norm on V. Let furthermore A be a subset of m linearly independent vectors and let for any $x_1,x_2,\ldots ,x_\ell \in V$ and $B=\{b_1,b_2,\ldots ,b_{m-\ell }\}\subseteq A$ with $\left| B\right| =m-\ell$

$$\begin{aligned} \left\| x_1,x_2,\ldots ,x_\ell ,B\right\| :=\left\| x_1,x_2,\ldots ,x_\ell ,b_1,b_2,\ldots ,b_{m-\ell }\right\| _m. \end{aligned}$$

(8)

Then for any $1\le \ell <m$ we get an $\ell$-norm by defining

$$\begin{aligned} \left\| x_1,x_2,\ldots ,x_\ell \right\| _\ell :=\sum _{B\subseteq A,\left| B\right| =m-\ell } \left\| x_1,x_2,\ldots ,x_\ell ,B\right\| . \end{aligned}$$

(9)

Proof

Properties 2.–4. from Definition 1 immediately follow from the corresponding properties for $\left\| \,\right\| _m$. It is also clear that $\left\| x_1,x_2,\ldots ,x_\ell \right\| _\ell =0$ if $x_1,x_2,\ldots ,x_\ell$ are linearly dependent.

Now suppose that $\left\| x_1,x_2,\ldots ,x_\ell \right\| _\ell =0$. Then $\left\| x_1,x_2,\ldots ,x_\ell ,B\right\| =0$ for all $B\subseteq A$ with $\left| B\right| =m-\ell$. Then $x_1,x_2,\ldots ,x_\ell$ must be linearly dependent because otherwise Lemma 2 would result in the contradiction that A is linearly dependent. $\square$

5 m-metric spaces and an open question

In [9, p. 167] an m-metric is defined as follows.

Definition 2

Let X be a non-empty set and $m\in {\mathbb {N}}$. Then $d:X^{m+1}\rightarrow {\mathbb {R}}$ is called m-metric, if the following conditions are satisfied.

• 1. For all $a=(a_0,a_1,\ldots ,a_m)\in X^{m+1}$ we have $d(a)=0$, provided that $a_i=a_j$ for some $i\not = j$.

• 2. For all $\alpha , \beta \in X$ such that $\alpha \not =\beta$ there is some $a\in X^{m+1}$ such that $a_i=\alpha , a_j=\beta$ for some i, j and $d(a)\not =0$.

• 3. For all permutations $\pi$ of $\{0,1,\ldots ,m\}$ and all $a\in X^{m+1}$ we have $d(a)=d(a^\pi )$, where $a^\pi :=(a_{\pi (0)},a_{\pi (1)},\ldots ,a_{\pi (m)})$.

• 4. For all $a\in X^{m+1}$ and all $\alpha \in X$ we have

  $$\begin{aligned} d(a)\le \sum _{i=0}^m d(a^{(i,\alpha )}), \text { where }a_i^{(i,\alpha )}=\alpha \text { and } a_j^{(i,\alpha )}=a_j\text { if }j\not =i. \end{aligned}$$

Remark 8

For every m-normed space X with $\dim (X)\ge m$ the function d defined by $d(a_0,a_1,$...$,a_m):=\left\| a_1-a_0,a_2-a_0,\ldots ,a_{m}-a_0\right\|$ is an m-metric.

Remark 9

For every non-empty set X the set ${\mathcal {B}}(X,{\mathbb {R}}):=\{f\in {\mathbb {R}}^X\mid \, f\text { is bounded}\}$ becomes a real Banach space with $\left\| f\right\| :=\sup _{x\in X}\left| f(x)\right|$.

Given a metric (= 1-metric) d on X, there is an isometric embedding into ${\mathcal {B}}(X,{\mathbb {R}})$. This is for example given by $X\ni x\mapsto f_x$, where fixing some $x_0\in X$ we have $f_x(y):=d(y,x)-d(y,x_0)$.

Open problem It is easy to see, that

$$\begin{aligned} \left\| f_1,f_2,\ldots ,f_m\right\| :=\sup _{x\in X^m} \left| \det ({\mathcal {C}}^x_{f_1,f_2,\ldots ,f_m})\right| \end{aligned}$$

gives an m-norm for ${\mathcal {B}}(X,{\mathbb {R}})$.

Now the question:

Let d be an m-metric on X. Is there an injection $\iota :X\rightarrow {\mathcal {B}}(X,{\mathbb {R}})$ such that

$d(a_0,a_1,\ldots ,a_m)=\left\| \iota (a_1)-\iota (a_0),\iota (a_1)-\iota (a_0), \ldots ,\iota (a_m)-\iota (a_0)\right\|$ for all $a_0,a_1,\ldots ,a_m\in X$?

Or is something similar possible for other m-normed spaces associated with X?