Hypernorm on hypervector spaces over a hyperfield

Abstract

Hypernorm is a generalization of the notion of a norm on a vector space over a field. In this paper, we consider a hypervector space $(\mathbb {V}, +)$ over a hyperfield, where $+$ is a hyperoperation, and prove that the hypernorm is continuous. We show that the natural linear transformation from $\mathbb {V}$ to $\dfrac{\mathbb {V}}{Z}$ is continuous and open for all closed subhyperspaces Z of $\mathbb {V}$. We prove $BL(\mathbb {V},\mathbb {W}),$ the set of all bounded linear transformations from $\mathbb {V}$ to $\mathbb {W}$ is a hyper-Banach space whenever $\mathbb {W}$ is complete. Furthermore, we obtain that in a hyper-Banach space $\mathbb {V}$ if $\lbrace \mu _n \rbrace$ is a sequence of continuous linear transformations with $\lbrace /\mu _n(u)/ \rbrace$ is bounded for every $u \in \mathbb {V},$ then $\lbrace \Vert \mu _n\Vert \rbrace$ is bounded. In the sequel, we prove several properties of hypernorm and linear transformations on hypernormed spaces.

1 Introduction

The concept of hyperoperation is a generalization of the binary operation in a classical algebraic system. Algebraic hyperstructures are the extensions of classical algebraic structures, where the composition of two elements is again a set. Let $\mathcal {P}^{*}(\mathbb {S})$ be the collection of non-empty subsets of $\mathbb {S}\ne \emptyset .$ A hyperoperation on $\mathbb {S}$ is a mapping $\circ :\mathbb {S} \times \mathbb {S }\rightarrow \mathcal {P}^{*}(\mathbb {S}),$ and $(H,\circ )$ is called a hypergropoid. A hypergroupoid is a semihypergroup if $(\mathbb {S},\circ )$ is a hypergroupoid which satisfies

$$\begin{aligned} (s_1\circ s_2)\circ s_3=s_1\circ (s_2\circ s_3) \text {, that is, } \bigcup _{v\in s_1\circ s_2} v\circ s_3=\bigcup _{u\in s_2\circ s_3}s_1\circ u \end{aligned}$$

for all $s_1,s_2,s_3$ in $\mathbb {S}.$

Al-Tahan and Davvaz [4] characterized $\lambda$-constacylic codes over hyperfields in terms of their generating polynomial. Mirvakili et al. [9] constructed regular hyperrings from hyperrings using $\alpha ^*$ relations. Mirvakili and Davvaz [8] studied the properties of rings obtained from $\mathcal {U}$-relation and showed that the derieved hyperring is a normal hyperideal. Mirvakili et al. [10] introduce a new relation so that the quotient ring obtained with respect to this relation is finitely generated. Pallavi et al. [11] and Harikrishnan et al. [12] explored several properties related to 2-absorbing hyperideals in hyperlattices. Hypervector space is formed by considering at least one operation as hyperoperation in a vector space over a field. A simple generalization of vector spaces over fields is the one given by Tallini [17] in which scalar multiplication is a hyperoperation. In this structure, Ameri and Dehghan [5] have discussed notions like linearly independent set, and basis, and explored results on certain classes of hypervector spaces. Raja and Vaezpour [14] have proved results on continuity and convexity. Later, Taghavi and Hosseinzadeh [18] considered the weak notions of this structure and studied operators using the idea of essential points. They proved several fundamental results like the Hanh Banach theorem [19] and discussed the spectrum of Banach Algebra [20]. AL-Tahan and Davvaz [2] generalized the hypervector space by considering canonical hypergroup over a hyperfield and established rank-nullity theorem. In [3], the authors defined the norm of an operator by providing the link between continuous and bounded operators. In [13], the notions like direct sums, complements, etc., of subhyperspace, linear transformation between two hypervector spaces were introduced by Pallavi et al., and isomorphism theorems were proved. Section 2 provides the preliminary notions and results connected to hypervector spaces over a hyperfield. In Sect. 3, we obtain results on open sets in a hypernormed space. We extend the linear functional $\mu$ on a subhyperspace $\mathbb {W}$ of a real hypervector space $\mathbb {V}$ to a linear functional $\mu '$ to a subhyperspaces $W\subsetneq W'$ of $\mathbb {V}.$ In Sect. 4, we prove that the natural linear transformation from $\mathbb {V}$ to $\dfrac{\mathbb {V}}{Z}$ is continuous and open for all closed subhyperspace Z of $\mathbb {V}$. Next, we prove that $BL(\mathbb {V},\mathbb {W}),$ the set of all bounded linear transformations from $\mathbb {V}$ to $\mathbb {W}$ is a hyper-Banach space whenever $\mathbb {W}$ is complete. Finally, we establish that in a hyper-Banach space $\mathbb {V}$ over a hyperfield K,  and if $\lbrace \mu _n \rbrace$ is a sequence of continuous linear functionals with $\lbrace /\mu _n(u)/ \rbrace$ is bounded for every $u \in \mathbb {V},$ then $\lbrace \Vert \mu _n\Vert \rbrace$ is bounded. We refer to [2, 13] for the basics of hypervector spaces and [7] for the basics of metric spaces.

2 Preliminaries

In this section, we discuss the necessary concepts used in this paper.

Definition 2.1

[6] A semihypergroup $(H,\circ )$ is a hypergroup if $\,\text {for all}\, x\in H$, we have

$$\begin{aligned} x\circ H=H\circ x=H. \end{aligned}$$

Definition 2.2

[6] A hypergroup $(H,+)$ is called a quasicanonical hypergroup if

• 1. there is $0\in H\,$such that $\,u+0=0+u=u,\,\text {for all}\, u\in H;$

• 2. for every $u\in H,$ there exists one and only one $u'\in H$ such that $\, 0\in (u+u')\cap (u'+u),$ $u'$ is denoted by $-u;$

• 3. $w\in u+v\,\text {implies} \,v\in -u+w$  and  $u\in w-v,$ for all $u,v,w \in H.$

A canonical hypergroup is a hypergroup $(H,+)$ which is quasicanonical and commutative.

Definition 2.3

[6] A weak hyperring is an algebraic hyperstructure $(R,+,\cdot )$ which satisfies the following axioms:

• 1. $(R,+)$ is a canonical hypergroup;

• 2. $(R,\cdot )$ is a semigroup with zero as a bilaterally absorbing element, that is, $u\cdot 0=0\cdot u=0;$

• 3. $u(v+w)\subseteq uv+uw$ and $(u+v)w \subseteq uw+vw,$

for all $u,v,w\in R.$

A weak hyperring is called a Krasner hyperring if equality holds in Condition 3 of Definition 2.3. A Krasner hyperring $(R,+,\cdot )$ is called commutative (with unit element) if $(R,\cdot )$ is a commutative semigroup (with unit element).

Definition 2.4

[6] A Commutative Krasner hyperring with unity is called a hyperfield if $(R\setminus \{0\},\cdot )$ is a group.

Example 2.5

[3] Let $\mathbb {T}=\mathbb {R}\cup \{-\infty \}$ with the following hyperoperations

$$s \oplus t = \left\{ {\begin{array}{ll}{\max \{ s,t\} ,}&{{\rm if} s \ne t;}\\{[ - \infty ,s],}&{otherwise}\end{array}} \right.,$$

and

$$\begin{aligned} s\odot t=s+t \end{aligned}$$

for all $s,t\in \mathbb {T}.$ Then $(\mathbb {T},\oplus ,\odot )$ is a hyperfield.

Example 2.6

[21] Let $V=\mathbb {R}_{\ge 0}$ set of all non-negative real numbers. We define the following hyperoperations on V.

$$\begin{aligned}&r_1 \bigtriangleup r_2= \{r\in V:|r_1-r_2| \le r\le r_1+r_2\},\\ &r_1 \bigtriangledown r_2= {\left\{ \begin{array}{ll} \max \{r_1,r_2\}, & {\rm if}\, r_1 \ne r_2;\\ \,[0,r_1], & {\rm otherwise} \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} r_1\odot r_2=r_1r_2 \end{aligned}$$

for all $r_1,r_2\in V.$ Then $(V,\triangle ,\odot )$ and $(V,\triangledown ,\odot )$ are hyperfields, called triangle and ultra triangle hyperfields respectively.

Example 2.7

[21] Let $\mathbb {C}$ be the set of all complex numbers with the following hyperoperation:

$$\begin{aligned} x\diamond y={\left\{ \begin{array}{ll} \{x\}, &{}\, \text {if}\, |x|>|y|;\\ \{y\}, &{} \,\text {if}\, |y|> |x|;\\ \{c\in \mathbb {C}:c\in S_{xy}\}, &{} \,\text {if}\, x+y\ne 0\, \text {and}\,|x|=|y|;\\ \{c\in \mathbb {C}:|c|\le |x|\}, &{} \,\text {if} \,x+y=0, \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} x\odot y=xy, \end{aligned}$$

for all $x,y\in \mathbb {C}.$ Here, $S_{xy}$ is the shortest arc connecting x to y on the circle |x| as the absolute value. Then $(\mathbb {C},\diamond ,\odot )$ is a hyperfield called a Complex tropical hyperfield, and it is denoted by $\mathcal {T}_{\mathbb {C}} .$

Example 2.8

[21] Let $\mathcal {S}$ be the complex tropical hyperfield described in Example 2.7. Then the set $\mathfrak {S}=\lbrace z\in \mathbb {C}:\Vert z\Vert =1\rbrace \cup \{0\}$ inherits the hyperfield structure from $\mathcal {T}_{\mathbb {C}} .$ The hyperoperation addition is as follows:

$$\begin{aligned} z_1\diamond _{\mathfrak {S}} z_2=z_1\diamond z_2\cap \mathfrak {S}={\left\{ \begin{array}{ll} \{c\in \mathbb {C}:c\in S_{z_1z_2}\}, &{} \text {if}\, z_1+z_2\ne 0;\\ \mathfrak {S}, &{} \text {otherwise}, \end{array}\right. } \end{aligned}$$

for all $z_1,z_2\in \mathfrak {S}.$

Definition 2.9

[2] Let F be a hyperfield. A canonical hypergroup $(\mathbb {V},+)$ together with a map $\cdot :F\times \mathbb {V}\rightarrow \mathbb {V}$ is called a weak hypervector space over F if the following conditions hold:

• 1. $\alpha \cdot (u_1+u_2)\subseteq \alpha \cdot u_1 + \alpha \cdot u_2;$

• 2. $(\alpha +\beta )\cdot u_1\subseteq \alpha \cdot u_1+\beta \cdot u_1;$

• 3. $\alpha \cdot (\beta \cdot u_1)=(\alpha \beta )\cdot u_1;$

• 4. $\alpha \cdot (-u_1)=(-\alpha )\cdot u_1= -(\alpha \cdot u_1);$

• 5. $1\cdot u_1=u_1,$

for all $\alpha ,\beta \in F,u_1, u_2\in \mathbb {V}$ and $1\in F.$

A weak hypervector space is a hypervector space over F if (1) and (2) hold with equality.

Definition 2.10

[3] A hyperabsolute value of a hyperfield K is a function $/\cdot /:K\rightarrow \mathbb {R}_{\ge 0}$ satisfying the following conditions:

• 1. $/k/=0$ if and only if $k=0;$

• 2. $/kk'/ =/k//k'/;$

• 3. $\sup \{/k/:k_1\in k+k'\} \le /k/+ /k'/,$

for all $k,k'\in K.$

Example 2.11

[1] In Example 2.5, $(\mathbb {T},\oplus ,\odot )$ is a hyperfield. We define $/\cdot /$ on $\mathbb {T}$ as follows:

$$\begin{aligned} /v/={\left\{ \begin{array}{ll} 0, \,&{}\text {if} \,v=-\infty ;\\ e^v, \,&{} \text {otherwise.} \end{array}\right. } \end{aligned}$$

Then $/\cdot /$ is a hyperabsolute value of $\mathbb {T}.$ Furthermore, $/\cdot /$ is an onto map.

3 Hypernormed spaces

This section provides some examples of hypervector spaces and explores linear transformations on hypernormed spaces.

Throughout this paper, let K denote the hyperfield on which the hyperabsolute value defined is an onto map.

Example 3.1

In Example 2.6, the map $/\cdot /:\mathbb {V}\rightarrow \mathbb {R}$ defined by $/a/=a$ for all $a\in \mathbb {V}$ is a hyperabsolute value.

Example 3.2

For $\mathfrak {S}$ given in Example 2.8, the map $/\cdot /:\mathfrak {S}\rightarrow \mathbb {R}$ defined by $/z/=|z|,$ (the absolute value of a complex number) for all $z\in \mathfrak {S}$ is a hyperabsolute map on $\mathfrak {S}.$

Definition 3.3

[3] A hypernorm on a hypervector space $\mathbb {V}$ over a valued hyperfield K is a function $\Vert \cdot \Vert :\mathbb {V}\rightarrow \mathbb {R}$ satisfying the following conditions:

• 1. $\Vert v\Vert =0$ if and only if $v=\overline{0};$

• 2. $\Vert av\Vert = /a/\Vert v\Vert ;$

• 3. $sup \Vert z\Vert _{z\in v+w} \le \Vert v\Vert +\Vert w\Vert ,$

for all $v,w\in \mathbb {V},a\in K.$ Then $(V,\Vert .\Vert )$ is called a hypernormed space.

Example 3.4

[3] Let $\mathbb {V}$ be a finite dimensional hypervector space over a hyperfield and $\{u_1,u_2,\ldots u_p\}$ be a basis of $\mathbb {V}.$ Define $\Vert \cdot \Vert$ on $\mathbb {V}$ as follows:

$$\begin{aligned} \Vert v\Vert =\left\| \sum \limits _{i=1}^{p} \alpha _iu_i\right\| =\max \{/\alpha _i/:1\le i \le p\}. \end{aligned}$$

Then $\Vert \cdot \Vert$ is a norm on $\mathbb {V}.$

Example 3.5

[13] From Example 2.6, with the external operation $(a,v)\mapsto av,$ $(\mathbb {V},\triangle )$ is a weak hypervector space over $K=(\mathbb {V},\triangledown ,\odot ).$

Example 3.6

For the hypervector space discussed in Example 3.5, the map $\Vert \cdot \Vert :\mathbb {V}\rightarrow \mathbb {R}$ given by $\Vert v\Vert =v$ is a hypernorm on $\mathbb {V}.$

Example 3.7

Let K be a valued hyperfield and $\mathbb {V}=\mathcal {M}_{n\times m }$ denote the weak hypervector space of $n\times m$ matrices over the hyperfield K. The maps $\Vert \cdot \Vert _1,\Vert \cdot \Vert _2: \mathbb {V}\times \mathbb {V}\rightarrow \mathbb {R}$ defined by

• 1. $\Vert A\Vert _1=\max \limits _{1\le i\le n} \sum \limits _{1\le j\le m}/a_{ij}/,$

• 2. $\Vert A\Vert _2=\max \limits _{1\le j\le m} \sum \limits _{1\le i\le n}/a_{ij}/,$

for all $A\in \mathcal {M}_{n\times m}$ are norms on $\mathcal {M}_{n\times m }.$

In [13], the authors have shown that $(\mathbb {C},\diamond )$ is a weak hypervector space over the hyperfield $(\mathfrak {S},\diamond _{\mathfrak {S}},\odot ).$

Example 3.8

The map $\Vert \Vert :\mathbb {C}\rightarrow \mathbb {R}$ defined by $\Vert v\Vert =|v|,$ (the absolute value of a complex number) for all $v\in \mathbb {C},$ is a hypernorm on $(\mathbb {C},\diamond ) .$

Proposition 3.9

[15] Let $\mathbb {V}$ be a hypernormed space and $u,v,w\in \mathbb {V}.$ Then $\inf \{|t\Vert :t\in u-v\}\le \inf \{\Vert t\Vert :t\in u-w\}+\inf \{\Vert t\Vert :t\in w-v\} .$

Proposition 3.10

[15] Let $(\mathbb {V}, \Vert \cdot \Vert )$ be a hypernormed space and $u,v\in \mathbb {V}.$ Then $\inf \{\Vert z\Vert :z\in u-v\}\ge \,\bigl |\Vert u\Vert -\Vert v\Vert \bigl |.$

Remark 3.11

Let $(\mathbb {V},\Vert .\Vert )$ be a hypernormed space. We define $d:\mathbb {V}\times \mathbb {V}\rightarrow \mathbb {R}$ as follows:

$$\begin{aligned} d(v,w)=\inf \left\{ \Vert z\Vert :z\in v-w\right\} \end{aligned}$$

for all $v,w\in \mathbb {V}.$ Then the map d has the following properties

• 1. $d(v,v)=0;$

• 2. $d(v,w)=d(w,v);$

• 3. $d(v,w) \le d(v+z)+d(z+w);$

• 4. $d(v,w)\ge |\Vert v\Vert -\Vert w\Vert |.$ In particular, $d(v,w)=0$ implies $\Vert v\Vert =\Vert w\Vert .$

Now on wards we consider hypernormed space $\mathbb {V}$ in which for any $x,y\in \mathbb {V}$ with $\inf \{\Vert z\Vert :z\in x-y\}=0$ implies $x=y$ holds. In such a case, $d(x,y)=\inf \left\{ \Vert z\Vert :z\in x-y\right\}$ is a metric on $\mathbb {V}.$

Corollary 3.12

Let $(\mathbb {V},\Vert \cdot \Vert )$ be a hypernormed space. Then the map $\Vert \cdot \Vert$ is continuous.

Proof

The proof follows from Proposition 3.10. $\square$

Definition 3.13

[3] For a hypernormed space $\mathbb {V},$ we define an open ball B(v, r) in $\mathbb {V}$ as

$$\begin{aligned} B(v,r)=\{w\in \mathbb {V}:d ( v,w)<r\} \end{aligned}$$

for all $v\in \mathbb {V},r>0.$

Definition 3.14

Let $\{v_n\}$ be a sequence in $\mathbb {V}.$ Then

• 1. $\{v_n\}$ is called Cauchy if for every $\epsilon >0$ there exists $N\in \mathbb {N}$ such that

  $$\begin{aligned} \inf \{\Vert z\Vert :z\in v_n-v_m\}<\epsilon \end{aligned}$$

  for every $m>n\ge N.$

• 2. $\{v_n\}$ is said to converge to $v\in \mathbb {V}$ if for every $\epsilon >0$ there exists $N\in \mathbb {N}$ such that

  $$\begin{aligned} \inf \{\Vert z\Vert :z\in v_n-v\}<\epsilon \end{aligned}$$

  for every $n\ge N.$

Definition 3.15

[13] Let $\mathbb {V}$ and $\mathbb {U}$ be two weak hypervector spaces over a hyperfield F. A map $\mu :\mathbb {V}\rightarrow \mathbb {U}$ is called a weak linear transformation if

• 1. $\mu (v+v')\subseteq \mu (v)+\mu (v');$

• 2. $\mu (\alpha v)=\alpha \mu (v),$

for all $v,v'\in \mathbb {V}$ and $\alpha \in F.$

A weak linear transformation $\mu$ is a linear transformation if Condition (1) holds with equality.

Proposition 3.16

[15] Every Cauchy sequence in a hypernormed space is bounded. In particular, every convergent sequence in a hypernormed space is bounded.

Theorem 3.17

Let $\mathbb {V}_1$ and $\mathbb {V}_2$ be hypernormed spaces and $\mu :\mathbb {V}_1\rightarrow \mathbb {V}_2$ be a linear transformation. Then $\mu$ is an open map if and only if there exists some $\gamma >0$ such that for every $y \in \mathbb {V}_2$ there is some $x\in \mathbb {V}_1$ with $\mu (x)=y$ and $\Vert x\Vert \le \gamma \Vert y\Vert .$

Proof

Let $\mu$ be an open map. Since $B_{\mathbb {V}_1}(0,1)$ is open in $\mathbb {V}_1,$ $\mu \big (B_{\mathbb {V}_1}\left( 0,1\right) \big )$ is open in $\mathbb {V}_2 .$ Now as $\mu \left( 0\right) \in \mu \big (B_{\mathbb {V}_1}\left( 0,1\right) \big ),$ there exists $\delta >0$ such that $\overline{B_{\mathbb {V}_2}\left( 0,\delta \right) }\subseteq \mu \big (B_{\mathbb {V}_1}\left( 0,1\right) \big ).$ For any $y\in \mathbb {V}_2$ with $0\ne y\in \mathbb {V}_2,$ $ky \in \overline{B_{\mathbb {V}_2}\left( 0,\delta \right) }$ where $/k/=\dfrac{\delta }{\Vert y\Vert }, k\in K.$ So there exists $x_1\in B_{\mathbb {V}_1}(0,1)$ such that $\mu (x_1) =ky.$ Take $x=\frac{1}{k}x_1.$ Then $\mu (x) =k \frac{1}{k} y=y$ and $\Vert x\Vert =\dfrac{\Vert x_1\Vert }{/k/}<\dfrac{\Vert y\Vert }{\delta }.$

Conversely, suppose that there exists $\gamma >0$ such that for every $y\in \mathbb {V}_2,$ there is some $x\in X$ with $\mu (x)=y$ and $\Vert x\Vert \le \gamma \Vert y\Vert .$ Let E be an open set in X,  and $x_0\in E.$ Then there exists $\delta >0$ such that $B_{\mathbb {V}_1}(x_0,\delta ) \subseteq E.$ Let $y\in \mathbb {V}_2$ with $d\big (y,\mu (x_0)\big ) <\dfrac{\delta }{\gamma }.$ That is, $\inf \{ \Vert z\Vert :z\in y-\mu (x_0)\}<\dfrac{\delta }{\gamma } .$ Then there exists $t\in y-\mu (x_0)$ such that $\Vert t\Vert <\dfrac{\delta }{\gamma }.$ The latter implies that, there exists $x\in \mathbb {V}_1$ such that $\mu (x)= t$ and $\Vert x\Vert \le \gamma \Vert t\Vert <\delta$ implies $\mu (x)\in y-\mu (x_0)$ which gives $y\in \mu (x) +\mu (x_0)=\mu (x+x_0).$ So there exists $x_1\in x+x_0$ such that $\mu (x_1)=y.$ Further, $x\in x_1-x_0 ,$ and so $d(x_1,x_0)\le \Vert x\Vert <\delta .$ Therefore, $x_1\in B_{\mathbb {V}_1}(x_0,\delta )\subseteq E.$ Having $y=\mu (x_1)\in \mu (E),$ implies that $B_{\mathbb {V}_2}\big (\mu (x_0,\frac{\delta }{\gamma })\big ) \subseteq \mu (E).$ $\square$

Theorem 3.18

Let $\mathbb {V}$ be a hypernormed space, and W be a subhyperspace of $\mathbb {V}.$ If W is open then for any $E\subseteq \mathbb {V},$ $W+E$ is open.

Proof

First, we show that for $x\in \mathbb {V},$ $r>0,$ $B(x,r)=x+rB(0,1).$ Now,

$$\begin{aligned} t\in B(x,r)&\iff \inf \{\Vert z\Vert :z\in t-x\}<r\\&\iff \text { there exists } u\in t-x \text { such that } \Vert u\Vert <r \text { and so } u\in rB(0,1)\\&\iff t\in u+x \subseteq x+rB(0,1). \end{aligned}$$

Now $x_1\in E.$ Since W is open, there exists $r>0$ such that $B(x,r)\subseteq W.$ Now

$$\begin{aligned} x_1+B(x,r)=x_1+x+rB(0,1) =\bigcup \limits _{t\in x_1+x}t+rB(0,1), \end{aligned}$$

is open. Also, $x_1+B(x,r)=x_1+x+rB(0,1)\subseteq E+W.$ Thus, $W+E$ is open. $\square$

Lemma 3.19

[2] Every linearly independent subset of $\mathbb {V}$ is contained in a basis of $\mathbb {V}$.

Theorem 3.20

[13] Let $\mathbb {V}$ and $\mathbb {V'}$ be hypervector spaces and $\lbrace v_1, v_2,\ldots , v_n \rbrace$ be an ordered basis for $\mathbb {V}.$ If $\lbrace w_1 , w_2 ,\ldots , w_n \rbrace$ is an ordered set of elements in $\mathbb {V'}$ with at most one non-zero $w_i$ then there exists a unique weak linear transformation $\varphi$ from $\mathbb {V}$ to $\mathbb {W}$ such that $\varphi (v_i)=w_i$ for every $1\le i \le n.$

Proposition 3.21

Let $W\subsetneq \mathbb {V}$ be a subhyperspace of a finite-dimensional hypervector space, $u\in \mathbb {V} {\setminus } W ,$ and $\gamma \in K.$ Then there exists a weak linear functional $\mu$ on $\mathbb {V}$ such that $\mu$ vanishes on W and $\mu (u)=\gamma .$

Proof

Let B be a basis of W. Then $B\cup \{u\}$ is linearly independent and hence by Lemma 3.19, is contained in a basis $B'$ of $\mathbb {V}.$ Now by Theorem 3.20, there exists a weak linear functional $\mu$ such that $\mu (B)=\{0\}$ and $\mu (u)=\gamma .$ This gives $\mu (W)=\{0\}.$ $\square$

Proposition 3.22

Let $\mathbb {V}$ be a hypervector space over $\mathbb {R},$ and let W be a subhyperspace of $\mathbb {V}.$ If $\tau$ is a linear functional on W,  then $\tau$ can be extended to a linear functional $\tau '$ on $\mathbb {V}.$

Proof

Let $\mathcal {F}$ be the collection of $(S,\mu )$ such that $W\subseteq S$ is a subhyperspace of $\mathbb {V}$ and $\mu :S\rightarrow \mathbb {R}$ is a linear functional with $\mu (x)=\tau (x) \text { on }W .$ Since $(W,\tau )\in \mathcal {F},\,\,\mathcal {F}\ne \emptyset .$ Define a relation “$\le$” on $\mathcal {F}$ by $(S_1,\mu _1)\le (S_2,\mu _2)$ if and only if $S_1\subseteq S_2$ and $\mu _2(x)=\mu _1(x)$ for all $x\in S_1.$ Then $(\mathcal {F},\le )$ is a poset. Let $C=\lbrace (S_{\alpha },\mu _{\alpha }):\alpha \in \varDelta \rbrace$ be a chain in $\mathcal {F}.$ Since C is a chain, for any $\alpha ,\beta \in \varDelta$, either $(S_{\alpha },\mu _{\alpha })\le (S_{\beta },\mu _{\beta })$ or $(S_{\beta },\mu _{\beta })\le (S_{\alpha },\mu _{\alpha }).$ Take $S=\bigcup \lbrace S_{\alpha }:\alpha \in \varDelta \rbrace .$ Let $u_1,u_2\in S$ and $a,b\in \mathbb {R}.$ Then there exist $\alpha ,\beta$ such that $u_1\in S_{\alpha }$ and $u_2\in S_{\beta } .$ We may assume that $S_{\alpha }\subseteq S_{\beta }.$ Then $u_1\in S_{\alpha } \subseteq S_{\beta }.$ Since $S_{\beta }$ is a subhyperspace, $au_1+bu_2\subseteq S_{\beta }\subseteq S.$ Hence, S is a subhyperspace of $\mathbb {V}$. Since $W\subseteq S_{\beta }$ for each $\beta \in \varDelta ,$ $W\subseteq S .$ Now for each $u\in S,$ there exists $\beta \in \varDelta$ such that $u\in S_{\beta }.$ We define $\mu :S\rightarrow \mathbb {R}$ by $\mu (u)=\mu _{\beta }(u).$ Now if $u\in S_{\alpha }$ and $u\in S_{\beta } ,$ where $\alpha ,\beta \in \varDelta ,$ we may assume that $(S_{\alpha },\mu _{\alpha })\le (S_{\beta },\mu _{\beta }).$ Then $\mu _{\beta }(u)=\mu _{\alpha }(u) .$ So $\mu$ is well-defined. Let $u_1,u_2\in S .$ Then there exist $\alpha ,\beta$ such that $u_1\in S_{\alpha }$ and $u_2\in S_{\beta } .$ Then $u_1\in S_{\alpha } \subseteq S_{\beta }.$ Now $\mu (u_1+u_2)= \mu _{\beta }(au_1+bu_2)=a\mu _{\beta }(u_1)+b\mu _{\beta }(u_2)=a\mu (u_1)+b\mu (u_2).$ Therefore $\mu$ is a linear functional on S. Also, $\tau (w)=\mu _{\alpha }(w)=\mu (w) ,$ for all $w\in W.$ Thus $(S,\mu ) \in \mathcal {F}.$ Since for all $\alpha \in \varDelta ,$ $S_{\alpha }\subseteq S$ and $\mu _{\alpha }(u)=\mu (u)$ for all $u\in S_{\alpha },$ we have that $(S_{\alpha },\mu _{\alpha })\le (S,\mu ).$ Thus, C is bounded above. Since every non-empty chain in $\mathcal {F}$ is bounded above, by Zorn’s lemma, $\mathcal {F}$ has a maximal element, say $( S, \tau _1).$ Suppose that $S\subsetneq \mathbb {V} ,$ then there exists $u_0\in \mathbb {V}{\setminus } S.$ Take $S'=span(S\cup \lbrace u_0\rbrace ) .$ Then $u\in S'$ implies there exist $s\in S$ and $a\in \mathbb {R}$ such that $u\in s+au_0.$ Suppose $u\in s+au_0$ and $u\in s'+au'_0.$ Then

$$\begin{aligned} 0\in u-u \subseteq&s+au_0+ s'+au'_0 = s-s'+au_0-au'_0= s-s'+a(u_0-u'_0). \end{aligned}$$

So there exist $u''\in u-u'$ and $a''\in a-a'$ such that $0\in u''+a''u_0.$ Then $-u''=a''u_0.$ If $a''=0 ,$ then $a=a'$ and $u=u' .$ If $a''\ne 0 ,$ then $u_0=-\frac{1}{a''}u''$ and so $u_0\in S,$ a contradiction. So $a=a'$ and $u=u'.$ Hence, the representation is unique. Now we construct a linear functional $\tau '$ on $S'$ so that $\tau '(u)=\tau _1(u)$ on S. For any $u\in S',$ $u\in s+au_0$ for unique $s\in S,a\in \mathbb {R}.$ So we define $\tau '(u)=\tau _1(s)+a\gamma ,$ where $\gamma =\tau '(u_0)\in K.$ Clearly $\tau '$ is linear and $\tau '(u)=\tau _1(u)$ on $S'$. But then $(S',\tau ')\in \mathcal {F},$ and $(S,\tau _1)<(S',\tau ') ,$ a contradiction. Therefore, $S'=\mathbb {V}.$ $\square$

Proposition 3.23

Let $\mathbb {V}$ be a hypervector space over $\mathbb {R}, \,\,p:\mathbb {V}\rightarrow \mathbb {R}$ be a mapping such that $\sup \{p(t):t\in v+w\} \le p(v)+p(w) \,\,\text {and} \,\,p(av)=ap(v)$ for all $a>0$ and $v,w\in \mathbb {V}.$ Let $S\subsetneq \mathbb {V}$ be a subhyperspace of $\mathbb {V}.$ If $\mu$ is a linear functional on S such that for any $v,w\in W,$ $\sup \left\{ \mu (t):t\in v+w\right\} \le \sup \left\{ p(t):t\in v+w\right\} ,$ then for any $u_0\in \mathbb {V}\setminus S,$ $\mu$ can be extended to a linear functional $\mu '$ on $S'=S\oplus \mathbb {R} u_0$ such that for any $t\in s+av,$ $\mu '(t)\le \sup \left\{ p(t):t\in s+au_0\right\} .$

Proof

Take $S'=span(S\cup \lbrace u_0\rbrace ) .$ Then $u\in S'$ implies there exist $s\in S$ and $a\in \mathbb {R}$ such that $u\in s+au_0.$ As discussed in the proof of Proposition 3.22, we can see that the above representation is unique. Now we construct a linear functional $\mu '$ on $S'$ so that

• 1. $\mu '(u)=\mu _1(u)$ on S, 

• 2. for any $v,w\in S',$ $\mu '(t)\le \sup \big \{p(t):t\in v-w\big \}$ for all $t\in v-w.$

For any $u\in S',$ $u\in s+au_0$ for unique $s\in S,a\in \mathbb {R}.$ So we define $\mu '(u)=\mu _1(s)+a\gamma ,$ where $\gamma =\mu '(u_0).$ Clearly $\mu '$ is linear and $\mu '(u)=\mu _1(u)$ on $S'$. We choose $\gamma$ such that

$$\begin{aligned} \mu _1(u)= \mu _1(s)+a\gamma \le \sup \big \{p(t):t\in s+au_0\big \}. \end{aligned}$$

(3.1)

We note that for any $s_1,s_2\in S,$

$$\begin{aligned} \mu _1(s_1)+\mu _1(s_2)&=\mu _1(s_1+s_2)\le \sup \big \{p(t):t\in s_1+s_2\big \}\,\Big (\text {by hypothesis}\Big )\\&\le \sup \big \{p(t):t\in s_1+u_0-u_0+s_2\big \} \\&=\sup \big \{p(t):t\in v+w,v\in s_1+u_0,w\in s_2-u_0\big \}\\&\le \sup \big \{p(v)+p(w):v\in s_1+u_0,w\in s_2-u_0\big \}\\&\big ( \text {since}\, \sup \big \{p(t):t\in v+w\big \}\le p(v)+p(w)\big )\\&=\sup \big \{ p(v):v\in s_1+u_0\big \} +\sup \big \{p(w):w\in s_2-u_0\big \}. \end{aligned}$$

$\text {That is,}\, \mu _1(s_1)+\mu _1(s_2)\le \sup \big \{p(t):t\in s_1+u_0\big \} +\sup \big \{p(t):t\in s_2-u_0\big \}.$ Then $\mu _1(s_2)-\sup \big \{p(t):t\in s_2-u_0\big \}\le \sup \big \{p(t):t\in s_1+u_0\big \} -\mu _1(s_1).$ By fixing $s_1$ and varying $s_2$ in S,  we get $\sup \big \{p(t):t\in s_1+u_0\big \} -\mu _1(s_1)$ is an upperbound for the set $\Big \{\mu _1(s_2)-\sup \big \{p(t):t\in s_2-u_0\big \}:s_2\in S\Big \}$ for each $s_1\in S ,$ and so

$$\begin{aligned} \sup \Big \{\mu _1(s_2)-\sup \big \{p(t):t\in s_2-u_0\big \}:s_2\in S\Big \}\le \sup \big \{p(t):t\in s_1+u_0\big \} -\mu _1(s_1) \end{aligned}$$

for each $s_1\in S.$ Take $\alpha =\sup \Big \{\mu _1(s_2)-\sup \big \{p(t):t\in s_2-u_0\big \}:s_2\in S\Big \}.$ Then $\alpha \le \sup \big \{p(t):t\in s_1+u_0\big \} -\mu _1(s_1)$ for all $s_1\in S.$ and so $\alpha \le \inf \Big \{\sup \big \{p(t):t\in s_1+u_0\big \} -\mu _1(s_1):s_1\in S\Big \}=\beta$ (say). Choose $\gamma$ so that $\alpha \le \gamma \le \beta .$ Then for each $s_1,s_2\in S,$

$$\begin{aligned} \mu _1(s_2) -\sup \big \{p(t):t\in s_2-u_0\big \}\le \gamma \le \sup \big \{p(t):t\in s_1+u_0\big \} -\mu _1(s_1). \end{aligned}$$

(3.2)

With this $\gamma ,$ we show that the inequality in (3.1) is satisfied. First, we show that $\mu '(u)\le p(u)$ for all $u\in S'.$ Let $u\in S'.$ Then $u\in s+au_0$ for unique $s\in S$ and $a\in \mathbb {R}.$

• Case 1 $a=0 .$ Then $\mu '(u)=\mu _1(s)+0\gamma =\mu _1(s)\le p(s)=p(u).$

• Case 2 $a>0 .$ Then $\mu '(u)=\mu _1(s)+a\gamma .$ Now from the RHS of the inequality (3.2), we get

  $$\begin{aligned} \gamma&\le \sup \left\{ p(t):t\in \frac{s}{a}+u_0\right\} -\mu _1\left( \frac{s}{a}\right) \\&= \sup \left\{ p\left( \frac{t}{a}\right) :t\in s+au_0 \right\} -\mu _1\left( \frac{s}{a}\right) \\&= \frac{1}{a}\sup \left\{ p(t):t\in s+au_0 \right\} -\mu _1\left( \frac{s}{a}\right) . \end{aligned}$$

  $\text {That is},\, a\gamma \le \sup \left\{ p(t):t\in s+au_0 \right\} -a\mu _1\left( \frac{s}{a}\right) ,$ and so,

  $$\begin{aligned} \,\mu '(u)=\mu _1(s)+a\gamma \le \sup \left\{ p(t):t\in s+au_0 \right\} \le p(u). \end{aligned}$$

• Case 3 $a<0 .$ Then $\mu '(u)=\mu _1(s)+a\gamma .$ Now from the LHS of the inequality (3.2), we get

  $$\begin{aligned} \gamma&\ge \mu _1\left( \frac{s}{-a}\right) -\sup \left\{ p(t):t\in \frac{s}{(-a)}-u_0\right\} \\&= \mu _1\left( \frac{s}{-a}\right) -\sup \left\{ p\left( \frac{t}{-a}\right) :t\in s+au_0 \right\} \\&= \mu _1\left( \frac{s}{(-a)}\right) -\frac{1}{(-a)}\sup \left\{ p(t):t\in s+au_0 \right\} . \end{aligned}$$

  $\text {That is},\, -a\gamma \ge \mu _1\left( s\right) -\sup \left\{ p(t):t\in s+au_0 \right\} \text {, which implies, }$

  $$\begin{aligned} a\gamma +\mu _1\left( s\right) \le \sup \left\{ p(t):t\in s+au_0 \right\} \end{aligned}$$

  $\text {and so},\,\mu '(u)=\mu _1(u)+a\gamma \le \sup \left\{ p(t):t\in s+au_0 \right\} .$

Therefore, in all the cases we get $\mu '(u)\le \sup \left\{ p(t):t\in s+au_0 \right\} .$ $\square$

Definition 3.24

Let $\mathbb {V}$ be a hypernormed space and $\emptyset \ne Z\subseteq \mathbb {V},$ we define $d(v,Z)=\inf \{d(v,z):z\in Z\}=\inf \big \lbrace \inf \{\Vert t\Vert :t\in v-z\}:z\in Z\big \rbrace$.

Theorem 3.25

Let W be a closed proper subhyperspace of $\mathbb {V}$ and $\epsilon >0 .$ Then there exists $w_{\epsilon } \in \mathbb {V}$ with $\Vert w_{\epsilon }\Vert \le 1$ such that $\inf \{d(w,w_{\epsilon }):w\in W\}>1-\epsilon .$

Proof

Let $x\in \mathbb {V}\setminus W.$ Put $k=\inf \{d(x,w):w\in W\}.$ If $k=\inf \{d(x,w):w\in W\} =0,$ then for every $n\in \mathbb {N},$ there exists $w_n$ such that $d(x,w_n)<\frac{1}{n}.$ Then $\lim \limits _{n\rightarrow \infty }w_n=x,$ and since W is closed, $x\in Y,$ a contradiction. So $k>0.$ Let $\epsilon >0.$ Then $k(1+\epsilon )>k.$ Now $k=\inf \{d(x,w):w\in W\}<k(1+\epsilon ).$ So there exists $w_0\in W$ such that $d(x,w_0)<k(1+\epsilon ) .$ That is,

$$\begin{aligned} \inf \{\Vert z\Vert :z\in x-w_0\}< k(1+\epsilon ). \end{aligned}$$

Then there exists $z\in x-w_0$ such that $k\le \Vert z\Vert <k(1+\epsilon ) .$ Let $\gamma$ be such that $/\gamma ^n/\le \Vert z\Vert \le \Vert z\Vert /\gamma ^{n+1}/ .$ Put $\beta = \gamma ^n$ and $w_{\epsilon }=\frac{1}{\beta }z.$ Then $\Vert w_{\epsilon }\Vert \le 1 .$ Let $w\in W.$ Now

$$\begin{aligned} \inf \{\Vert s\Vert :s\in w-w_{\epsilon }\}&=\inf \{\Vert s\Vert :s\in w-\frac{1}{\beta }z\}=\inf \left\{ \Vert s\Vert :s\in \dfrac{\beta w-z}{\beta }\right\} . \end{aligned}$$

Since $z\in x-w_0 ,$ $\beta w-z \subseteq \beta w-x-w_0 =w_0+\beta w-x,$ and so,

$$\begin{aligned} \inf \{\Vert s\Vert :s\in w-w_{\epsilon }\}&=\dfrac{1}{/\beta /}\inf \{\Vert s\Vert :s\in \beta w-z\}\\&\ge \dfrac{1}{/\beta /}\inf \{\Vert s\Vert :s\in w_0+\beta w-x\}. \end{aligned}$$

As $w_0+\beta w\subseteq W,$

$$\begin{aligned} \inf \{\Vert t\Vert :t\in w_0+\beta w-x\}=\inf \{\Vert t\Vert :t\in s-x,s\in w_0+\beta w\}\ge k. \end{aligned}$$

Therefore,

$$\begin{aligned} d(w,w_{\epsilon })= \inf \{\Vert s\Vert :s\in w-w_{\epsilon }\}&\ge \dfrac{1}{/\beta /}k\ge \dfrac{k}{1+\epsilon }>1-\epsilon . \end{aligned}$$

$\square$

4 $BL\left( \mathbb {V},\mathbb {W}\right)$ in hypernormed spaces

In this section, we show that the set of all bounded linear transformations from $\mathbb {V}$ to $\mathbb {W},$ $BL\left( \mathbb {V},\mathbb {W}\right)$ is a hypernormed space, and study some of its properties.

Definition 4.1

[3] Let $( \mathbb {V}_1,\Vert \cdot \Vert _1)$ and $( \mathbb {V}_2,\Vert \cdot \Vert _2)$ be two hypernormed spaces. Then $\mu \in L(\mathbb {V}_1,\mathbb {V}_2)$ is said to be bounded if there exists $M\in \mathbb {R}_{>0}$ such that $\Vert \mu (v)\Vert _2\le M\Vert v\Vert$ for all $v\in \mathbb {V}_1.$

Proposition 4.2

[3] Let $( \mathbb {V}_1,\Vert \cdot \Vert _1)$ and $( \mathbb {V}_2,\Vert \cdot \Vert _2)$ be two hypernormed spaces. Then $BL(\mathbb {V}_1,\mathbb {V}_2)$, the set of all bounded linear transformations is a subhyperspace of $L(\mathbb {V}_1,\mathbb {V}_2).$

Definition 4.3

[3] For $\mu \in Bl(\mathbb {V}_1,\mathbb {V}_2),$ we define $\Vert \mu \Vert$ as follows:

$$\begin{aligned} \Vert \mu \Vert =\sup \left\{ \dfrac{\Vert \mu (v)\Vert _2}{\Vert v\Vert _1}:v\in \mathbb {V}_1\setminus \{0\}\right\} . \end{aligned}$$

Remark 1

[3] For $\mu \in BL(\mathbb {V}_1,\mathbb {V}_2),$

• 1. $\Vert \mu \Vert =\sup \left\{ \Vert \mu (v)\Vert _2:v\in \mathbb {V}, \Vert v\Vert _1= 1\right\} ,$

• 2. $\Vert \mu \Vert =\inf \left\{ M\in \mathbb {R}_{>0}:\Vert \mu (v)\Vert _2\le M\Vert v\Vert _1 \text { for all } v\in \mathbb {V}_1\right\} .$

Proposition 4.4

[3] Let $(\mathbb {V},\Vert .\Vert _1)$ and $(\mathbb {W},\Vert .\Vert _2)$ be two hypernormed spaces over K and $\mu :\mathbb {V}\rightarrow \mathbb {W}$ be a linear transformation. Then

• 1. $\mu$ is bounded

• 2. $\mu$ is continuous

are equivalent.

Proposition 4.5

Let $(\mathbb {V},\Vert .\Vert _1)$ and $(\mathbb {W},\Vert .\Vert _2)$ be two hypernormed spaces over K and $\mu :\mathbb {V}\rightarrow \mathbb {W}$ be a linear transformation. Then

$$\begin{aligned} \Vert \mu \Vert = \sup \left\{ \Vert \mu (v)\Vert _2:v\in \mathbb {V}, \Vert v\Vert _1\le 1\right\} . \end{aligned}$$

Proof

Take $A= \sup \left\{ \dfrac{ \Vert \mu (v)\Vert _2}{\Vert v\Vert _1}:0\ne v\in \mathbb {V}\right\}$ and $B=\sup \left\{ \Vert \mu (v)\Vert _2:v\in \mathbb {V}, \Vert v\Vert _1\le 1\right\}$. Now $\dfrac{ \Vert \mu (v)\Vert _2}{\Vert v\Vert _1}\le A$ for all $0 \ne v\in \mathbb {V} .$ For $\Vert v\Vert _1\le 1,$ $\Vert \mu (v)\Vert _2\le \dfrac{ \Vert \mu (v)\Vert _2}{\Vert v\Vert _1}\le A .$ That is $\Vert \mu (v)\Vert _2\le A .$ So, $B \le A.$ Suppose that $B<A$. Then B is not a upper bound of the set $\left\{ \dfrac{ \Vert \mu (v)\Vert _2}{\Vert v\Vert _1}:0\ne v\in \mathbb {V}\right\} .$ Then there exists $v\in \mathbb {V}$ such that $\dfrac{ \Vert \mu (v)\Vert _2}{\Vert v\Vert _1}>B.$ Now we can choose $\beta \in K$ such that $/\beta /=\frac{1}{\Vert v\Vert } .$ Now $\Vert \beta v\Vert _1\le 1.$ Then $B\ge \Vert \mu (\beta v)\Vert _2 =/\beta / \Vert \mu (v)\Vert =\frac{\Vert \mu (v)\Vert }{\Vert v\Vert }>B,$ a contradiction. $\square$

Theorem 4.6

[16] Let $\mathbb {V}$ be a weak hypervector space and W be a closed subhyperspace of $\mathbb {V}.$ Then the map $\Vert \cdot \Vert :\dfrac{\mathbb {V}}{W}\rightarrow \mathbb {R}$ defined by

$$\begin{aligned} \Vert x+W\Vert =\inf \left\{ \sup \{\Vert t\Vert :t\in x+w\}:w\in W\right\} \end{aligned}$$

is a hypernorm on $\dfrac{\mathbb {V}}{W}.$

Proposition 4.7

If Z is a closed subhyperspace of $\mathbb {V},$ the quotient map Q from $\mathbb {V}$ to $\dfrac{ \mathbb {V}}{Z}$ is continuous and open.

Proof

For $v\in \mathbb {V},$

$$\begin{aligned} \Vert Q(v)\Vert =\Vert v+Z\Vert =&\inf \left\{ \sup \{\Vert t\Vert :t\in v+z:z\in Z\}\right\} \\ \le&\left\{ \Vert t\Vert :t\in v+0\right\} =\Vert v\Vert . \end{aligned}$$

That is, $\Vert Q(v)\Vert =\Vert v\Vert$ for all $v\in \mathbb {V}.$ Therefore, Q is continuous. As $\Vert v+Z\Vert < (1+\epsilon )\Vert v+Z\Vert ,$ there exists $z_0\in Z$ such that $\sup \left\{ \Vert t\Vert :t\in v+z_0\right\} <(1+\epsilon ) \Vert v+Z\Vert .$ Now $Q(v+z_0)=Q(v)+Q(z_0)=v+Z+z_0+Z=v+Z.$ Now take $\gamma =1+\epsilon .$ Then $\Vert v+z_0\Vert \le \gamma \Vert v+Z\Vert .$ Thus, by Theorem 3.17, Q is an open map. $\square$

For $\mu _1,\mu _2\in BL(\mathbb {V},\mathbb {W}),$ we denote,

$$\begin{aligned} \Vert (\mu _1+\mu _2)(x)\Vert =\sup \left\{ \Vert z\Vert _2:z\in \mu _1(x)+ \mu _2(x)\right\} \end{aligned}$$

and

$$\begin{aligned} \Vert \mu _1+\mu _2\Vert =\sup \left\{ \dfrac{\Vert (\mu _1+\mu _2)(x)\Vert }{\Vert x\Vert _1}:0\ne x\in \mathbb {V}\right\} . \end{aligned}$$

Proposition 4.8

[15] Let $\{v_n\}$ and $\{w_n\}$ be two convergent sequences such that $\lim \limits _{n\rightarrow \infty }v_n=v$ and $\lim \limits _{n\rightarrow \infty }w_n=w$. Then for each $t\in v+w ,$ there exists $t_n\in v_n+w_n$ such that $\lim \limits _{n\rightarrow \infty }t_n=t.$

Proposition 4.9

Let $(\mathbb {V}_1,\Vert .\Vert _1)$ and $(\mathbb {V}_2,\Vert .\Vert _2)$ be two weak hypernormed spaces over K and $BL(\mathbb {V}_1,\mathbb {V}_2)$ be the set of all weak bounded linear transformations from $\mathbb {V}_1$ to $\mathbb {V}_2.$ Then $BL(\mathbb {V}_1,\mathbb {V}_2)$ is a hypernormed space.

Proof

For any $\mu \in BL(\mathbb {V}_1,\mathbb {V}_2),$ consider $\Vert \mu \Vert = \sup \left\{ \dfrac{ \Vert \mu (v)\Vert _2}{\Vert v\Vert _1}:0\ne v\in \mathbb {V}_1\right\}$. Clearly, for any $\mu \in BL(\mathbb {V}_1,\mathbb {V}_2),$ we have $\Vert \mu \Vert \ge 0.$ If $\mu =0$ then $\Vert \mu \Vert =0$. Suppose that $\Vert \mu \Vert =0 .$ Then $\sup \left\{ \dfrac{ \Vert \mu (v)\Vert _2}{\Vert v\Vert _1}:0\ne v\in \mathbb {V}_1\right\} =0 .$ This implies that $\dfrac{ \Vert \mu (v)\Vert _2}{\Vert v\Vert _1}\le 0$ for all $0\ne v\in \mathbb {V}_1$ implies that $\Vert \mu (v)\Vert _2\le 0$ for all $v\in \mathbb {V}_1.$ Then $\Vert \mu (v)\Vert _2= 0$ for all $v\in \mathbb {V}_1$ implies that $\mu (v) =0$ for all $v\in \mathbb {V}_1.$ Hence $\mu =0 .$ Now for any $v\in \mathbb {V}_1$ and $\alpha \in K,$

$$\begin{aligned} \Vert (\alpha \mu )(v)\Vert _2 =\Vert \alpha \mu (v)\Vert _2=/\alpha / \Vert \mu (v)\Vert _2 \le /\alpha / \Vert \mu \Vert \Vert v\Vert _1 \,\,\text {for all}\,\, v\in \mathbb {V}_1. \end{aligned}$$

That is,

$$\begin{aligned} \Vert \alpha \mu \Vert \le /\alpha / \Vert \mu \Vert . \end{aligned}$$

(4.1)

Suppose $\alpha =0$. Then $\alpha \mu =0,$ and so, $\Vert \alpha \mu \Vert =0.$ Assume that $\alpha \ne 0.$ Then $\dfrac{1}{\alpha }\in K.$ Replace $\alpha$ by $\dfrac{1}{\alpha }$ and $\mu$ by $\alpha \mu$ in inequality (4.1). Now $\Vert \dfrac{1}{\alpha }(\alpha \mu )\Vert \le /\dfrac{1}{\alpha }/ \Vert \alpha \mu \Vert ,$ which implies, $/\alpha /\Vert \mu \Vert \le \Vert \alpha \mu \Vert .$ Thus $/\alpha /\Vert \mu \Vert = \Vert \alpha \mu \Vert .$ Let $\mu _1,\mu _2\in BL(\mathbb {V}_1,\mathbb {V}_2) .$ Then for each $v\in \mathbb {V}_1,$ $(\mu _1+ \mu _2)(v)=\left\{ z:z\in \mu _1(v)+ \mu _2(v)\right\} .$ For each $z\in \mu _1(v)+ \mu _2(v),$ we have,

$$\begin{aligned} \sup \left\{ \Vert z\Vert _2:z\in \mu _1(v)+ \mu _2(v)\right\}&\le \Vert \mu _1(v)\Vert _2+ \Vert \mu _2(v)\Vert _2\\&\le \Vert \mu _1\Vert \Vert v\Vert _1+ \Vert \mu _2\Vert \Vert v\Vert _1\\&=(\Vert \mu _1\Vert + \Vert \mu _2\Vert )\Vert v\Vert _1. \end{aligned}$$

Then for all $0\ne v\in \mathbb {V}_1$, $\Vert (\mu _1+\mu _2)(v)\Vert \le (\Vert \mu _1\Vert + \Vert \mu _2\Vert )\Vert v\Vert _1,$ which implies,

$$\begin{aligned} \dfrac{\Vert (\mu _1+\mu _2)(v)\Vert }{\Vert v\Vert _1}\le \Vert \mu _1\Vert + \Vert \mu _2\Vert \end{aligned}$$

and so, $\sup \left\{ \dfrac{\Vert (\mu _1+\mu _2)(v)\Vert }{\Vert v\Vert _1}:0\ne v\in \mathbb {V}_1\right\} \le \Vert \mu _1\Vert + \Vert \mu _2\Vert .$ Thus, $\Vert \mu _1+\mu _2\Vert \le \Vert \mu _1\Vert + \Vert \mu _2\Vert .$ Hence $( BL(\mathbb {V}_1,\mathbb {V}_2),\Vert \cdot \Vert )$ is a hypernormed space. $\square$

We denote $\Vert (\mu _1-\mu _2)(v)\Vert =\inf \left\{ \Vert z\Vert _2:z\in \mu _1(v)- \mu _2(v)\right\} =d(\mu _1(v),\mu _2(v))$ and

$$\begin{aligned} \Vert \mu _1-\mu _2 \Vert = \inf \left\{ \dfrac{ \Vert (\mu _1-\mu _2)(v)\Vert }{\Vert v\Vert }:0\ne v\in V \right\} . \end{aligned}$$

Also,

$$\begin{aligned} \overline{d}(\mu _1,\mu _2)=\Vert \mu _1-\mu _2\Vert = \inf \left\{ \dfrac{d(\mu _1(v),\mu _2(v))}{\Vert v\Vert }:0\ne v\in V \right\} \end{aligned}$$

is a metric on $BL(\mathbb {V},\mathbb {W}).$

Definition 4.10

A hypernormed space $(\mathbb {V},\Vert \cdot \Vert )$ is said to be a hyper-Banach space if $\mathbb {V}$ is complete with respect to the hypernorm $\Vert \cdot \Vert .$

Proposition 4.11

If $(\mathbb {W},\Vert .\Vert _2)$ is a hyper-Banach space, then $BL(\mathbb {V},\mathbb {W})$ is also a hyper-Banach space.

Proof

Let $\{\mu _n \}$ be a Cauchy sequence in $( BL(\mathbb {V},\mathbb {W}),\Vert \cdot \Vert ).$ For any $x\in \mathbb {V}$, we show that $\mu _n(x)$ is a Cauchy sequence. Since $\{\mu _n \}$ is a Cauchy sequence, for $\dfrac{\epsilon }{\Vert x\Vert _1+1}>0$ there exists $N\in \mathbb {N}$ such that $\Vert \mu _n-\mu _m\Vert <\dfrac{\epsilon }{\Vert x\Vert _1+1}$ for all $m>n>N.$ Now for $m>n>N,$ $\Vert \mu _n(v)-\mu _m(v)\Vert \le \Vert \mu _n-\mu _m\Vert \Vert v\Vert _1<\dfrac{\epsilon }{\Vert v\Vert _1+1} \Vert v\Vert _1<\epsilon .$ Hence $\{\mu _n(v)\}$ is a Cauchy sequence in $\mathbb {W},$ and since $\mathbb {W}$ is complete, $\{\mu _n(v)\}$ converges, for each $v\in \mathbb {V}.$ Define $\mu :\mathbb {V}\rightarrow \mathbb {W}$ as $\mu (v) =\lim \limits _{n\rightarrow \infty }\mu _n(v) .$ Let $u,v\in \mathbb {V}, \alpha \in K.$

$$\begin{aligned} \mu (u+v)&= \lim \limits _{n\rightarrow \infty }\mu _n(u+v) \subseteq \lim \limits _{n\rightarrow \infty }\mu _n(u)+\lim \limits _{n\rightarrow \infty }\mu _n(v)(\text {by Proposition 4.8})\\&=\mu (u)+\mu (v). \end{aligned}$$

Also, $\mu (\alpha u)= \lim \limits _{n\rightarrow \infty }\mu _n(\alpha u)= \lim \limits _{n\rightarrow \infty }\alpha \mu _n(u)=\alpha \lim \limits _{n\rightarrow \infty }\mu _n(u)=\alpha \mu (v).$ Hence $\mu$ is linear. Since $\left\{ \mu _n \right\}$ is a Cauchy sequence, by Proposition 3.16, $\left\{ \mu _n \right\}$ is bounded. So there exist $M>0$ such that $\Vert \mu _n\Vert \le M$ for all $n\in \mathbb {N}.$ Then,

$$\begin{aligned} \Vert \mu (u)\Vert =\Vert \lim \limits _{n\rightarrow \infty }\mu _n( u)\Vert _2\le \lim \limits _{n\rightarrow \infty }\Vert \mu _n\Vert \Vert u\Vert _1=\Vert u\Vert _1\lim \limits _{n\rightarrow \infty }\Vert \mu _n\Vert \le \Vert u\Vert _1 M. \end{aligned}$$

Hence $\mu \in BL(\mathbb {V},\mathbb {W}).$ Now to show, $\Vert \mu _n-\mu \Vert \rightarrow 0.$ Let $\epsilon >0.$ Since $\{\mu _n\}$ is Cauchy, for $\frac{\epsilon }{2} >0,$ there exists $N_1\in \mathbb {N}$ such that $\Vert \mu _n -\mu _m\Vert <\dfrac{\epsilon }{2}$ for all $n>m>N_1.$ Now, for $n>m>N_1,$

$$\begin{aligned} \Vert \mu _n -\mu _m\Vert =\sup \left\{ \dfrac{\inf \left\{ \Vert z\Vert _2:z\in \mu _n(u)- \mu _m(u)\right\} }{\Vert u\Vert _1}:0\ne u\in \mathbb {V}\right\} <\epsilon . \end{aligned}$$

For each $0\ne u\in \mathbb {V},$ $\dfrac{ \inf \left\{ \Vert z\Vert _2:z\in \mu _n(u)- \mu _m(u)\right\} }{\Vert u\Vert _1}<\epsilon \,\,\text {for all }\,\,0\ne u\in \mathbb {V}.$ Fixing m and letting $n\rightarrow \infty ,$ $\dfrac{ \inf \left\{ \Vert z\Vert _2:z\in \mu (u)- \mu _m(u)\right\} }{\Vert u\Vert _1}<\epsilon \,\,\text {for all }\,\,0\ne u\in \mathbb {V}.$ So, $\sup \left\{ \dfrac{\inf \left\{ \Vert z\Vert _2:z\in \mu (v)- \mu _m(v)\right\} }{\Vert v\Vert _1}:0\ne v\in \mathbb {V}\right\} <\epsilon .$ That is, $\Vert \mu _n-\mu _m \Vert <\epsilon .$ Hence, $\Vert \mu _n-\mu _m \Vert$ as $m\rightarrow \infty .$ Thus, $BL(\mathbb {V},\mathbb {W})$ is a hyper-Banach space. $\square$

Theorem 4.12

Let $(\mathbb {V},\Vert \cdot \Vert _1)$ and $(\mathbb {W},\Vert \cdot \Vert _2)$ be hyper-Banach spaces over a hyperfield K and $\{ \mu _n\}$ be a sequence of continuous linear transformations from $\mathbb {V}$ to $\mathbb {W}$ such that $\Vert \mu _n(u)\Vert _2$ is a bounded sequence for all $u\in \mathbb {V}.$ Then the sequence $\{\Vert \mu _{n}\Vert \}$ is bounded.

Proof

For each $m\in \mathbb {N},$ we define

$$\begin{aligned} X_m=\left\{ u\in \mathbb {V}:\Vert \mu _n(u)\Vert _2\le m \,\,\text {for all}\,\,n\in \mathbb {N} \right\} =\bigcap \limits _{n=1}^{\infty }\mu _{n}^{-1}\left( \overline{B(u_0,m)}\right) . \end{aligned}$$

Then $\mathbb {V}=\bigcup \limits _{m=1}^{\infty } X_{m}.$ Now by Baire’s category theorem for metric spaces, at least one $X_m$ contains a sphere. Let $u_0$ be an interior point of $X_m$. Then there exists $r>0$ such that $\overline{B(u_0,r)}\subseteq X_m.$ That means for any $u\in \overline{B(u_0,r)},$ $\inf \left\{ \Vert t\Vert _1:t\in u-u_0 \right\} \le r,$ and $\Vert \mu _n(u)\Vert _2\le m \,\,\text {for all}\,\, n\in \mathbb {N}.$

For $0\ne v\in \mathbb {V}$ with $\Vert v\Vert _1>r,$ we can find $\gamma \in K$ such that $/\gamma /= \dfrac{r}{2\Vert v\Vert _1} .$

Now $\gamma v\in \mathbb {V}$ and $\Vert \gamma v\Vert _1= \dfrac{r}{2}<r.$ Then $\gamma v\in \gamma v+u_0-u_0 .$ So there is $t\in \gamma v+u_0$ such that $\gamma v\in t-u_0.$ Now $\inf \{\Vert t'\Vert _1:t'\in t-u_0\}\le \Vert \gamma v\Vert _1< r$ and so $t\in \overline{B(u_0,r)}$ and $\Vert \mu _n(t)\Vert _2\le m \,\,\text {for all}\,\, n\in \mathbb {N}.$

Since $t\in \gamma v+u_0$ we get $\mu _n(t)\in \gamma \mu (v)+\mu (u_0) ,$ and so,

$$\begin{aligned} \Vert \gamma \mu _n(v)\Vert _2-\Vert \mu _n(u_0)\Vert _2&\le \inf \{\Vert s\Vert _1:s\in \mu _n(\gamma v)+\mu _n(u_0)\}\\&\le \Vert \mu _n(t)\Vert _2\le m \,\,\text {for all} \,\,n\in \mathbb {N} . \end{aligned}$$

Therefore, $\Vert \mu _n(v)\Vert _2\le \dfrac{ m+\Vert \mu _n(u_0)\Vert _2}{ /\gamma /}=\dfrac{(m+\Vert \mu _n(u_0)\Vert _2)2\Vert v\Vert _1}{r}$ for all $n\in \mathbb {N}.$ Since $\{\Vert \mu _n(u_0)\Vert _2\}$ is a bounded sequence, then there exists $m'>0$ such that $\Vert \mu _n(u_0) \Vert _2\le m'.$ This gives, $\Vert \mu _n(v)\Vert _2<\dfrac{2(m+m')\Vert v\Vert _1}{r}=M\Vert v\Vert _1(\text {say}).$ This implies,

$$\begin{aligned} \sup \left\{ \dfrac{ \Vert \mu _n(v)\Vert _2}{\Vert v\Vert _1} :0\ne v\in \mathbb {V}\right\} \le M \end{aligned}$$

which gives $\Vert \mu _n\Vert \le M$ for all $n\in \mathbb {N}.$

Thus, $\left\{ \Vert \mu _n\Vert \right\}$ is a bounded sequence. $\square$

The following result is an immediate consequence of Theroem 4.12.

Corollary 4.13

Let $(\mathbb {V},\Vert \cdot \Vert )$ be a hyper-Banach space over a hyperfield K such that the hyperabsolute value map defined on K is onto and $\{ \mu _n\}$ be a sequence of continuous linear functionals such that $\left\{ /\mu _n(u)/\right\}$ is a bounded sequence for all $u\in \mathbb {V}.$ Then the sequence $\{\Vert \mu _{n}\Vert \}$ is bounded.

5 Conclusion

In this paper, we have proved that the hypernorm in a hypervector space is continuous. The necessary and sufficient condition is obtained for a linear transformation between two hypernormed spaces to be open. In a hypernormed space $\mathbb {V},$ it was proved that if W is an open subhyperspace and E is any arbitrary subset, then $W+E$ is open. We have extended the linear functional $\mu$ on a subhyperspace $\mathbb {W}$ of a real hypervector space $\mathbb {V}$ to a linear functional $\mu ^{'}$ on $\mathbb {V}.$

Some open questions: As a future scope, one can attempt the following open problems in hypernormed spaces:

• 1. The hyperoperation $+$ on a hypervector space $\mathbb {V}$ is continuous or not?

• 2. An onto bounded linear transformation between two hyper-Banach spaces is open or not?

• 3. If a bounded linear transformation $\mu$ between two hypernormed spaces is bijective, then $\mu ^{-1}$ is a bounded linear transformation or not?