On bounded n-linear operators

J. Math. Comput. Sci. 8 (2018), No. 2, 196-215 https://doi.org/10.28919/jmcs/3446

ISSN: 1927-5307

ON BOUNDED n-LINEAR OPERATORS

S. ROMEN MEITEI, M.P. SINGH∗

Department of Mathematics, Manipur University , Canchipur 7951003, Manipur, India

Copyright c2018 Meitei and Singh. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract. In this paper, the various concepts of bounded and continuousn-linear operators in normed spaces as well as inn-normed spaces are discussed. A sufficient condition for the space ofn-linear operators to be a Banach space is given.Further, we prove the equality of two different formulae of norms of ann-linear operator .Also we introduce ann-norm on the dual space of a real linear space.

Keywords:norm;n-norm; boundedn-linear operator; Banach space.

2010 AMS Subject Classification:46B20, 46B99, 46E15, 47A30, 47L05.

1. Introduction

LetX be a real linear space of dimension greater than 1 andk., .kbe a real valued function onX×X satisfying the following conditions:

(2N₁)kx,yk=0 if and only ifxandyare linearly dependent.

(2N₂)kx,yk=ky,xk.

(2N₃)kαx,yk=|α|kx,yk ∀x,y∈ X andα ∈_R.

(2N₄)kx+y,zk ≤ kx,zk+ky,zk ∀x,y,z∈ X. ∗_{Corresponding author}

E-mail address:[email protected] Received July 31, 2017

Then,k., .kis called a 2-norm onX and(X,k., .k)is called a linear 2-normed space.2-norms are non-negative andkx,y+αxk=kx,ykfor everyx,y∈X andα∈_R.

The concept of 2-normed spaces was initially investigated and developed by S.G ¨ahler in 1960s and has been extensively developed by C. Diminnie, S.G ¨ahler, A. White and many others [1, 2, 3, 4, 5, 10, 15].

Let X be a real vector space with dimX ≥n where n is a positive integer. A real valued functionk., ..., .k:Xn→_Ris called ann-norm onX if the following conditions hold:

(1) kx₁, . . . ,x_nk=0 iffx₁, . . . ,x_nare linearly dependent.

(2) kx₁, . . . ,x_nkremains invariant under permutations ofx₁, . . . ,x_n. (3) kαx1,x2, . . . ,xnk=|α|kx1, . . . ,xnk ∀x1, . . . ,xn∈X andα ∈R.

(4) kx₀+x₁,x₂, . . . ,x_nk ≤ kx₀, . . . ,x_nk+kx₁, . . . ,x_nkfor allx₀,x₁, . . . ,x_n∈X. The pair(X,k., . . . , .k)is called ann-normed space.

LetX be a real vector space with dimX≥n,nis a poitive integer and be equipped with an inner producth., .i.Then the standardn-norm onX is given by

kx₁, . . . ,x_nk_S:= q

det[hx_i,x_ji].

A standard example of ann-normed space isX =_Rnequipped with the Euclideann-norm:

kx1, . . . ,xnkE =abs



   

x₁₁ · · · x_1n

..

. . .. ...

x_n1 · · · x_nn



   

wherexi= (xi1, ...,xin)∈Rnfor eachi=1,2, ...,n.

Note that the value ofkx₁, ...,x_nks represents the volume of n-dimensional parallelepiped

s-panned byx₁, ...,x_n.

The theories of 2-normed spaces and n-normed spaces were initially developed by G¨ahler [2]-[5] in 1960s . The theory of n-normed space was much developed later by Misiak [10]. Notions of boundedness in 2-normed space was then introduced by White [15]. Related works

Gozali et al. also introduced the notion of boundedn-linear functionals inn-normed spaces in [6]. Zofia Lewandowska introduced notions of 2-linear operators on 2-normed sets in [9]. Agus

L. Soenjaya then introduced the notions of continuity and boundedness ofn-linear operators in [14]. Notions of different formulae ofn-norms can be seen in [6, 12, 13].

The above papers motivate us to write this paper. In this paper we introduce the notion of

boundedn-linear operators as further extensions of the corresponding notions in [14] and a new

n-norm as a further work of the notions in [12, 13].

2. Preliminaries

Let(X,k., . . . , .k)be ann-normed space and(Y,k.k)be a normed space.

Definition 2.1.

An operatorT :Xn→Y is ann-linear operator onX ifT is linear in each of the variables.

Definition 2.2.

Ann-linear operatorT is bounded if there is a constantK such that

kT(x₁, . . . ,x_n)k ≤Kkx₁, . . . ,x_nkfor all(x₁, . . . ,x_n)∈Xn.

IfT is bounded,

kTk:= sup

kx1,...,xnk6=0

kT(x₁, . . . ,x_n)k kx1, . . . ,xnk

.

Proposition 2.1.

[14, Proposition 2.3] LetT :Xn→Y be ann-linear operator onX, where

(X,k., . . . , .k)is ann-normed space and(Y,k.k)is a normed space. T is bounded if and only if for all(x₁, . . . ,x_n),(y₁, . . . ,y_n)∈Xn,

kT(x1, . . . ,xn)−T(y1, . . . ,yn)k ≤K(kx1−y1,x2, . . . ,xnk+ky1,x2−y2, . . . ,xnk+

Proposition 2.2.

[14, Proposition 2.4] LetT be a boundedn-linear operator. Then

kTk=inf{K:kT(x₁, . . . ,x_n)k ≤Kkx₁, . . . ,x_nk}

=inf{K:kT(x₁, . . . ,x_n)−T(y₁, . . . ,y_n)k ≤K(kx₁−y₁,x₂, . . . ,x_nk

+ky₁,x₂−y₂, . . . ,x_nk+· · ·+ky₁, . . . ,y_n−1,x_n−y_nk)} (2.2.1)

= sup

kx1,...,xnk≤1

kT(x1, . . . ,xn)k.

IfX is ann-normed space with dualX0, the following formula-formulated by G ¨ahler [3]

kx1, ...,xnkG= sup fj∈X0,kfjk≤1

f₁(x₁) · · · f_n(x₁)

..

. . .. ...

f₁(x_n) · · · f_n(x_n)

defines ann-norm onX.

Proposition 2.3.

[12, Proposition 2.1] Let X be a normed space of dimX ≥n with dual

X0.Then the function

kx1, ...,xnk=Abs

    

f₁(x₁) · · · f_n(x₁)

..

. . .. ...

f₁(x_n) · · · f_n(x_n)     

defines an n-norm onX for fixed linearly independent n funtionals

f₁,f₂, ...,f_n∈X0

The above concepts motivate us to have the following results.

3. Main results

n-normed space and(Y,k.k)is a normed space. Then

kTk=sup

Σ6=0

kT(x₁, . . . ,x_n)−T(y₁, . . . ,y_n)k

Σ ,

whereΣ=kx1−y1,x2, . . . ,xnk+ky1,x2−y2, . . . ,xnk+· · ·+ky1, . . . ,yn−1,xn−ynk.

Proof.Let

kTkn=sup

Σ6=0

kT(x₁, . . . ,x_n)−T(y₁, . . . ,y_n)k

Σ .

SinceT is bounded, by definition

kTk= sup

kx1,...,xnk6=0

kT(x₁, . . . ,x_n)k kx1, . . . ,xnk

.

It is enough to prove thatkTkn=kTk.

Let

P =

K:K= kT(x1, . . . ,xn)−T(y1, . . . ,yn)k

Σ andΣ6=0

,

Q=

K:K= kT(x1, . . . ,xn)k kx1, . . . ,xnk

andkx₁, . . . ,x_nk 6=0

.

Clearly,

Q⊆P

⇒supQ≤supP

⇒ kTk ≤ kTkn. (3.1.1)

Let

For eachK∈W ,

kT(x1, . . . ,xn)−T(y1, . . . ,yn)k ≤K(Σ)

⇒ kT(x1, . . . ,xn)−T(y1, . . . ,yn)k

Σ ≤K,which is true for allK&Σ6=0 ⇒ kT(x1, . . . ,xn)−T(y1, . . . ,yn)k

Σ ≤infW for allΣ6=0

⇒ kT(x1, . . . ,xn)−T(y1, . . . ,yn)k

Σ ≤ k

Tkfor allΣ6=0 ,using(2.2.1)

⇒sup

Σ6=0

kT(x₁, . . . ,x_n)−T(y₁, . . . ,y_n)k

Σ ≤ kTk

⇒supP≤ kTk

⇒ kTkn≤ kTk. (3.1.2)

Conclusion follows(3.1.1)and(3.1.2).

The following is an extension of the notions of boundedn-linear functionals, operators found in [6, 9, 10].

Let(X,k.k)and(Y,k.k)be normed spaces.

Definition 3.1. An operator T : Xn →Y is an n-linear operator on X if T is linear in each

variable.

Definition 3.2.Ann-linear operatorT is bounded if there exists a positive constantKsuch that

kT(x₁, ...,x_n)k ≤Kkx₁k...kx_nkfor all(x₁, ...,x_n)∈Xn.

Whenn=1, it is reduced to the usual notion of bounded operator in a normed space. IfT is bounded, norm ofT is defined to be

kTk= sup

xi∈X

kx_ik 6=0

kT(x₁, ...,x_n)k kx1k...kxnk

Proposition 3.2. LetT be a boundedn-linear operator. Then,

kTk = sup

xi∈X,kxik=1

kT(x₁, . . . ,x_n)k

= sup

xi∈X,kxik≤1

kT(x₁, . . . ,x_n)k

= inf{K:kT(x1, . . . ,xn)k ≤Kkx1k...kxnk}

Proof.Let

kTka= sup xi∈X,kxik=1

kT(x₁, . . . ,x_n)k,

kTkb= sup xi∈X,kxik≤1

kT(x₁, . . . ,x_n)k,

kTkc=inf{K:kT(x1, . . . ,xn)k ≤Kkx1k...kxnk}.

SinceT is boundedn-linear operator,

kTk= sup

xi∈X

kx_ik 6=0

kT(x₁, . . . ,x_n)k kx1k · · · kxnk

Clearly,kTka≤ kTk.

Conversely, definey_i= xi

kxik;i=1,2, ...,n.

Now,

kT(x₁, . . . ,x_n)k

kx₁k...kx_nk =kT(y1, . . . ,yn)k

≤ sup

kyik=1

kT(y₁, . . . ,y_n)k

≤ kTka∀xi∈X &xi6=0

⇒ kTk ≤ kTka

Again,kTk_a≤ kTk_bis obvious. Conversely, definey_i= xi

kxik ; 0<kxik ≤1.

Then,

kT(x1, . . . ,xn)k=kx1k...kxnkkT(y1, . . . ,yn)k

≤ kT(y₁, . . . ,y_n)k, becausekx_ik ≤1∀i

≤ sup

kyik=1

kT(y₁, . . . ,y_n)k

=kTka∀xi∈X & 0<kxik ≤1

⇒ kTkb≤ kTka

Therefore,kTkb=kTka. (3.2.2)

Again, letL ={K:kT(x1, ...,xn)k ≤Kkx1k...kxnk}.

For eachK∈L,

kT(x1, . . . ,xn)k

kx₁k...kx_nk ≤K∀xi&kxik 6=0

⇒ sup

kxik6=0

kT(x₁, . . . ,x_n)k kx₁k...kx_nk ≤K ⇒ kTk ≤K,which is true for allk∈L.

⇒ kTk ≤infL

⇒ kTk ≤ kTkc.

Conversely,

kTk= sup

kxik6=0

kT(x₁, . . . ,x_n)k kx₁k...kx_nk

⇒kTk ∈L

⇒kTk ≥infL

⇒kTk ≥ kTkc

Thus, kTk=kTkc. (3.2.3)

Hence,kTk=kTka=kTkb=kTkcusing(3.2.1),(3.2.2)and(3.2.3).

Proposition 3.3. LetT :Xn→Y be ann-linear operator wherenis a positive integer andX,Y

are normed spaces. Then, T is bounded iff there exists a positive constant K such that for all

(x1, ...,xn),(y1, ...,yn)∈Xn,

kT(x1, . . . ,xn)−T(y1, . . . ,yn)k ≤ K(kx1−y1kkx2k...kxnk+ky1kkx2−y2k...kxnk+

· · ·+ky₁k...ky_n−1kkx_n−y_nk).

Proof.Suppose the inequality holds. We take(y1, . . . ,yn) = (0, . . . ,0).

Then,T(y₁, . . . ,y_n) =0. Using the inequality, we have

kT(x1, . . . ,xn)−0k ≤ K(kx1k...kxnk+0+...+0)

⇒ kT(x₁, . . . ,x_n)k ≤ Kkx₁k...kx_nk.

Therefore,T is bounded.

Conversely, supposeT is bounded. Then byn-linearity, we have

kT(x₁, . . . ,x_n)−T(y₁, . . . ,y_n)k

=kT(x1−y1,x2, . . . ,xn) +T(y1,x2−y2, . . . ,xn) +

· · ·+T(y1,y2, . . . ,yn−1,xn−yn)k

≤ kT(x₁−y₁,x₂, . . . ,x_n)k+kT(y₁,x₂−y₂, . . . ,x_n)k+

(by triangle inequality)

≤K(kx₁−y₁kkx₂k...kx_nk+ky₁kkx₂−y₂k...kx_nk+

· · ·+ky1k...kyn−1kkxn−ynk).

It completes the proof.

Proposition 3.4. LetT be a boundedn-linear operator from a normed spaceX into a normed

spaceY. Then,

kTk = inf{K:kT(x₁, . . . ,x_n)k ≤Kkx₁k...kx_nk}

= inf{K:kT(x₁, . . . ,x_n)−T(y₁, . . . ,y_n)k ≤K(kx₁−y₁kkx₂k...kx_nk +ky1kkx2−y2k...kxnk+...+ky1k...kyn−1kkxn−ynk)}

= sup

Σ6=0

kT(x₁, . . . ,x_n)−T(y₁, . . . ,y_n)k Σ

whereΣ=kx1−y1kkx2k...kxnk+ky1kkx2−y2k...kxnk+· · ·

+ky₁k...ky_n−1kkx_n−y_nk

Proof.By definition,

kTk= sup

xi∈X

kx_ik 6=0

kT(x₁, . . . ,x_n)k kx1k...kxnk

Let

kTkc = inf{K:kT(x1, . . . ,xn)k ≤Kkx1k...kxnk},

kTk_d = inf{K:kT(x₁, . . . ,x_n)−T(y₁, . . . ,y_n)k ≤K(kx₁−y₁kkx₂k...kx_nk +ky₁kkx₂−y₂k...kx_nk+· · ·+ky₁k...ky_n−1kkx_n−y_nk)}

andkTke = sup_Σ6=0

kT(x1,...,xn)−T(y1,...,yn)k

Also, let

K = {K:kT(x1, . . . ,xn)−T(y1, . . . ,yn)k ≤K(kx1−y1kkx2k...kxnk

+ky₁kkx₂−y₂k...kx_nk+· · ·+ky₁k...ky_n−1kkx_n−y_nk)}

andC = {K:kT(x1, . . . ,xn)k ≤Kkx1k...kxnk}.

Clearly,C ⊆K .

Therefore, infK ≤ infC.

⇒ kTkd ≤ kTkc

⇒ kTk_d ≤ kTk, becausekTk=kTk_cby proposition(3.2).

(3.4.1)

Let

A =

K:K= kT(x1, . . . ,xn)k

kx₁k...kx_nk ,kxik 6=0

,

B =

K:K= kT(x1, . . . ,xn)−T(y1, . . . ,yn)k

Σ ,Σ6=0

.

Obviuosly,

A ⊆ B

⇒supA ≤ supB

⇒ kTk ≤ kTke.

(3.4.2)

For eachK∈K ,

kT(x1,...,xn)−T(y1,...,yn)k

Σ ≤K ∀ Σ6=0

⇒sup

Σ6=0

kT(x₁, . . . ,x_n)−T(y₁, . . . ,y_n)k

Σ ≤

K

⇒ kTk_e≤K , which is true for allK.

⇒ kTke≤infK

⇒ kTk_e≤ kTk_d.

(3.4.3)

From(3.4.1),(3.4.2),(3.4.3)and proposition 3.2, it follows that

kTk ≤ kTke≤ kTkd≤ kTk and kTk=kTkc.

It completes the proof.

Definition 4.1. An n-linear operator T :Xn→Y where X andY are normed linear spaces is continuous at(x₁, ...,x_n)∈Xnif forε>0, there isδ >0 such that

kT(x₁, ...,x_n)−T(y₁, ...,y_n)k<ε

whenever

kx₁−y₁kkx₂k...kx_nk<δ, ky1kkx2−y2k...kxnk<δ, . . . and

ky1kky2k...kyn−1kkxn−ynk<δ

or

kx₁−y₁kky₂k...ky_nk<δ, kx1kkx2−y2k...kynk<δ, . . . and

kx₁kkx₂k...kx_n−1kkxn−ynk<δ,

where(y₁, . . . ,y_n)∈Xn.

Definition 4.2. T is continuous onXnif it is continuous at every(x1, . . . ,xn)∈Xn.

Whenn=1, it reduces to the usual notion of continuity in the normed space.

The above definition is similar to the definition given by A. L. Soenjaya in [14] as an

ex-tension of that given by White in [15]. In Soenjaya’s paper, the n-linear operator is from an

n-normed space to a normed space. But, in this paper then-linear operator is from a normed space to a normed space.

Theorem 4.1. LetT :Xn→Y be ann-linear operator where(X,k.k)and(Y,k.k)are normed

spaces. Then the followings are equivalent:

(a) T is continuous.

(b) T is continuous at(0, . . . ,0)∈Xn. (c) T is bounded.

Proof.(a)⇒(b)is obvious.

(b)⇒(c): IfT=0, the proof is through. SupposeT 6=0.Let(x1, ...,xn)∈Xnwithkxik 6=0∀i.

Set

ui=

δ

2kx₁k · · · kx_nk 1_n

xi.

Then,

ku1k...kunk=

δ

Now,

kT(u₁, . . . ,u_n)k = δ

2kx1k...kxnkkT(x1, . . . ,xn)k ⇒ kT(x1, . . . ,xn)k = 2_δkx1k...kxnkkT(u1, ...,un)k.

(4.1.2)

SinceT is continuous at(0, . . . ,0)∈Xn, forε =1 there existsδ >0 such that

kT(u₁, . . . ,u_n)k<1 wheneverku₁k...ku_nk<δ.

Therefore,

kT(x₁, . . . ,x_n)k< 2 δk

x₁k...kx_nk.1, using(4.1.1)and(4.1.2)

= 2

δkx1k · · · kxnk

⇒ kT(x1, . . . ,xn)k<

2

δkx1k · · · kxnk

⇒T is bounded.

(c)⇒(a): SupposeT is bounded. IfT =0, the statement is trivial. LetT 6=0. Consider any point(x0₁, . . . ,x0_n)∈Xn.

Letε >0 be given. For every(x1, . . . ,xn)∈Xnsuch that

kx1−x 0

1kkx2k...kxnk<δ, kx 0

1kkx2−x 0

2k...kxnk<δ , . . . and

kx0₁kkx0₂k...kx_n0₋₁kkx_n−x0_nk<δ,

(4.1.3)

whereδ = _nkε_Tk,

kT(x₁, . . . ,x_n)−T(x0₁, . . . ,x0_n)k ≤ kTk(kx₁−x₁0kkx₂k...kx_nk+kx₁0kkx₂−x0₂k...kx_nk+ · · ·+kx0₁k...kx0_n−1kkxn−x

0

nk)

< kTk(δ+δ+· · ·+δ

| {z }

n terms

), using(4.1.3)

= nkTkδ

= ε

LetB(Xn,Y)denote the space of all boundedn-linear operators fromXnintoY, whereX andY

are normed spaces. Then we have the following theorems.

Theorem 4.2. (B(Xn,Y),k.k)is a normed space with respect to the norm given by

kTk= sup

xi∈X

kx_ik 6=0

kT(x₁, . . . ,x_n)k kx1k...kxnk

.

Proof.(i)

kαTk= sup kxik6=0

kαT(x1, . . . ,xn)k

kx₁k...kx_nk

=|α| sup

kxik6=0

kT(x₁, . . . ,x_n)k kx1k...kxnk

=|α|kTk.

(ii)

kT₁+T₂k= sup

kxik6=0

k(T1+T2)(x1, . . . ,xn)k

kx₁k...kx_nk

= sup

kxik6=0

k(T₁(x₁, . . . ,x_n) +T₂(x₁, . . . ,x_n)k kx1k...kxnk

≤ sup

kxik6=0

k(T1(x1, . . . ,xn)k

kx₁k...kx_nk +_kxsup_ik6=0

k(T2(x1, . . . ,xn)k

kx₁k...kx_nk =kT₁k+kT₂k.

(iii) Clearly,kTk ≥0. Lastly, it is to show thatkTk=0⇐⇒T =0.

kTk=0⇔ sup

kxik6=0

kT(x₁, . . . ,x_n)k kx1k...kxnk

=0

⇔ kT(x₁, . . . ,x_n)k=0 forx_i6=0 for each i

Ifkx₁k...kx_nk=0, at least one ofx_isis 0. SinceT is boundedn-linear operator,T(x₁, . . . ,x_n) =0 ifkx₁k...kx_nk=0. So,

kTk=0⇔T(x₁, . . . ,x_n) =0 for all(x₁, . . . ,x_n)∈Xn

⇔T ≡0.

Therefore,k.kis a norm and it completes the proof.

Theorem 4.3. If(Y,k.k)is a Banach space,(B(Xn,Y),k.k)is a Banach space.

Proof.Let{T_k}be a cauchy sequence inB(Xn,Y). Letε>0 be given. Then there existsN>0

such thatkT_k−T_mk< ε

2 ∀ k,m>N.

Now,

kT_k(x₁, . . . ,x_n)−T_m(x₁, . . . ,x_n)k=k(T_K−T_m)(x₁, . . . ,x_n)k ≤ kT_k−T_mkkx₁k...kx_nk.

This shows that{T_k(x₁, . . . ,x_n)}is a cauchy sequence inY. SinceY is complete, there exists a vector inY, say T(x₁, . . . ,x_n)such thatT_k(x₁, . . . ,x_n)→T(x₁, . . . ,x_n). Clearly,T is ann-linear operator from Xn intoY. To conclude the proof, we have to show that T is bounded and that

T_k→T.

Now,

kT(x₁, . . . ,x_n)k=limkT_k(x₁, . . . ,x_n)k ≤sup(kT_kkkx₁k...kx_nk)

= (supkT_kk)(kx1k...kxnk). (4.3.1)

Fork,m>N,

kT_k(x₁, . . . ,xn)−Tm(x1, . . . ,xn)k ≤ kTk−Tmkkx1k...kxnk

< ε

Now, holdingkfixed and allowingm→∞, we have

kT_k(x₁, . . . ,x_n)−T(x₁, . . . ,x_n)k<ε ∀k>Nand∀(x1, . . . ,xn)∈Xn:kx1k...kxnk ≤1

⇒ k(T_k−T)(x₁, . . . ,x_n)k<ε∀k>Nand∀(x1, . . . ,xn)∈Xn:kx1k...kxnk ≤1

⇒ sup

kxik≤1

k(T_k−T)(x₁, . . . ,x_n)k ≤ ε

2

⇒ kT_k−Tk<ε ∀k>N

⇒T_k−→T as k−→∞and by (4.3.1),T ∈B(Xn,Y).

This completes the proof.

Theorem 4.4. Let X be a real vector space with dimX =d whered ≥n andn is a positive

integer. Let(Y,k.k)be a normed space andT:Xn−→Y be ann-linear operator. IfXis equipped with a normk.kand ann-normk., ..., .k, define

kTk₁=sup_{xi∈X,kx1,...,xnk6=0}kT(x1, ...,xn)k kx₁, ...,x_nk

and kTk2=supxi∈X,kxik6=0

kT(x₁, ...,x_n)k kx₁k...kx_nk .

Then,k.k1andk.k2are identical.

Proof.Define

y_i= kyikxi

n

p

kx₁, ...,xnk

= kyikxi

γ ;γ = n

p

Now,

T(x₁, ...,x_n) =T

γy1

ky₁k, ...,

γyn

ky_nk

= γ

n_T(y

1, ...,yn)

ky1k, ...,kynk

⇒ T(x1, ...,xn)

γn =

T(y₁, ...,y_n) ky₁k, ...,ky_nk ⇒ T(x1, ...,xn)

kx₁, ...,x_nk =

T(y1, ...,yn)

ky₁k, ...,ky_nk (4.4.1)

Taking supremum on the right side of(4.4.1)overyis∈X withkyik 6=0, we have

T(x1, ...,xn)

kx₁, ...,x_nk ≤_yi∈X,sup_kyik6=0

T(y1, ...,yn)

ky₁k, ...,ky_nk

=kTk₂

It is true for allxi∈X withkx1, ...,xnk 6=0.

Therefore

sup

xi∈X,kx1,...,xnk6=0

T(x₁, ...,x_n) kx1, ...,xnk

≤ kTk2

⇒ kTk1≤ kTk2 (4.4.2)

Again,Taking supremum on the left side of(4.4.1)over{(x₁, ...,x_n)∈Xn:kx₁, ...,x_nk 6=0}, we have

sup

xi∈X,kx1,...,xnk6=0

T(x1, ...,xn)

kx₁, ...,x_nk ≥

T(y1, ...,yn)

ky₁k, ...,ky_nk ⇒ kTk1≥

It is true for ally_i∈X withky_ik 6=0. Therefore,

kTk₁≥ sup

yi∈X,kyik6=0

T(y₁, ...,y_n) ky₁k, ...,ky_nk

⇒ kTk1≥ kTk2 (4.4.3)

Conclusion follows from(4.4.2)and(4.4.3).

5. NEW

n

-NORM

The idea of the following formula of n-norm is derived from [12, 13]. Its similar formula in Proposition 2.3 is defined on a real vector space with dimension ≥n but the forthcoming formula is defined on a dual space.

Theorem 5.1. LetX be a real vector space with dim(X)≥nwherenis a positive integer and

X0be the dual ofX. Then, the functionk., ..., .k:(X0)n−→_Rgiven by

kf₁, ...,f_nk=abs 

   

f₁(x₁) · · · f_n(x₁)

..

. . .. ...

f₁(x_n) · · · f_n(x_n)



   

for fixed linearly independent n

elements x₁, ...,x_n∈X,

defines ann-norm onX0.

Proof.(i)

f1,f2, ...,fnare linearly dependent.

⇔columns of the matrix[f_i(x_j)]are linearly dependent.

⇔abs     

f₁(x₁) · · · f_n(x₁)

..

. . .. ...

f₁(x_n) · · · f_n(x_n)      =0

⇔ kf₁, ...,f_nk=0.

(ii) By the properties of determinant and definition of absolute (abs), kf₁, ...,f_nk remains in-variant under the permutations of f₁,f₂, ...,f_n.

(iii)Forα ∈_R,

kαf1, ...,fnk=abs

    

αf1(x1) · · · fn(x1)

..

. . .. ...

αf1(xn) · · · fn(xn)

    

=|α|abs

    

f1(x1) · · · fn(x1)

..

. . .. ...

αf1(xn) · · · fn(xn)

    

=|α|kf1, ...,fnk

(iv)

kf₀+f₁, ...,f_nk=abs     

(f0+f1)(x1) · · · fn(x1)

..

. . .. ...

(f0+f1)(xn) · · · fn(xn)

     =abs     

f₀(x₁) · · · f_n(x₁)

..

. . .. ...

f₀(x_n) · · · f_n(x_n) +

f₁(x₁) · · · f_n(x₁)

..

. . .. ...

f₁(x_n) · · · f_n(x_n)      ≤abs     

f₀(x₁) · · · f_n(x₁)

..

. . .. ...

f₀(x_n) · · · f_n(x_n)      +abs     

f₁(x₁) · · · f_n(x₁)

..

. . .. ...

f₁(x_n) · · · f_n(x_n)     

It completes the proof.

Conflict of Interests

The authors declare that there is no conflict of interests.

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