# 2-NSR lemma and compact operators on 2-normed space.

1. Introduction

The concept of a linear 2-normed space was introduced as a natural 2-metric analogue of that of a normed space. In 1963, Gahler introduced the notion of a 2-metric space, a real valued function of point-triples on a set X, whose abstract properties were suggested by the area function for a triangle determined by a triple in Euclidean space.

We begin with a lemma which was given by Fatemeh Lael and Kourosh Nouruzi[2], by using the norm [parallel]x + <e>[parallel] = [parallel]x,e[parallel]/[parallel]e,e'[parallel] in the quotient space X/<e>. Even though this norm seems to be many valued, they have reached the conclusion on the basis of the same norm. Inspired by their finding we have proved the same lemma by using the norm

(1) [[parallel]x + <e>[parallel].sub.Q] = [parallel]x,e[parallel] + sup{[parallel]x,e[parallel]/[parallel]e,e'[parallel]: e' [not member of] <e> and [parallel]e,e'[parallel] > 1}. Thus all the results in paper [2] can be proved by using the specific norm(1) in the quotient space X/<e>. Here we prove a lemma which is similar to Riesz lemma in normed space and (e) using this we list the properties of compact operator.

In [4] S. Gahler introduced the following definition of a 2-normed space.

2. Preliminaries

Definition 2.1[4]: Let X be a real linear space of dimension greater than 1. Suppose [parallel].,.[parallel] is a real valued function on X x X satisfying the following conditions:

A1: [parallel]x,y[parallel] = 0 [??] x and y are linearly dependent.

A2: [parallel]x,y[parallel] = [parallel]y,x[parallel]

A3: [parallel][alpha]x,y[parallel] = [absolute value of [alpha]][parallel]x,y[parallel]

A4: [parallel]x,y+z[parallel] [less than or equal to] [parallel]x,y[parallel] + [parallel]x,z[parallel].

Then [parallel].,.[parallel] is called a 2-norm on X and the pair (X,[parallel].,.[parallel]) is called a 2-normed space. Some of the basic properties of 2-norms, that they are non-negative and [parallel]x,y + [alpha]x[parallel] = [parallel]x,y[parallel] for all x,y [member of] X and for all [alpha] [member of] [Real part].

Definition 2.2[2]: Let X and Y be two 2-normed spaces and T: X [right arrow] Y be a linear operator. For any e [member of] X, we say that the operator T is e-bounded if there exist [M.sub.e] >0 such that [parallel]T(x), T(e)[parallel] [less than or equal to] [M.sub.e][parallel]x,e[parallel] for all x [member of] X. An e-bounded operator T for every e will be called bounded.

Definition 2.3[4]: A sequence {[x.sub.n]} in a 2-normed space X is said to be convergent if there exists an element x [member of] X such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] for all y [member of] X.

Definition 2.4 [4]: Let X and Y be two 2-normed spaces and T: X [right arrow] Y be a linear operator. The operator T is said to be sequentially continuous at x [member of] X if for any sequence {[x.sub.n]} of X converging to x we have T([x.sub.n]) [right arrow] T(x).

Definition 2.5 [2]: The closure of a subset E of a 2-normed space X is denoted by [bar.E] and defined by the set of all x [member of] X such that there is a sequence {[x.sub.n]} of E converging to x. We say that E is closed if E = [bar.E].

For a 2-normed space, we consider the subsets

[B.sub.e](a,r) = {x: [parallel]x-a,e[parallel] < r}

[B.sub.e][a,r] = {x: [parallel]x-a,e[parallel] [less than or equal to] r} of X.

Definition 2.6 [2]: A subset A of a 2-normed space X is said to be locally bounded if there exist e [member of] X--{0} and r > 0 such that A [subset or equal to] [B.sub.e](0,r).

Definition 2.7 [2]: A subset B of a 2-normed space X is said to be compact if every {[x.sub.n]} of B has a convergent subsequence in B.

Definition 2.8 [2]: Let X and Y be two 2-normed spaces. A linear operator T: X [right arrow] Y is called a compact operator if it maps every locally bounded sequence {[x.sub.n]} of X onto a sequence {T([x.sub.n])} in Y which has a convergent subsequence.

Lemma 2.9[2]: Let X and Y be two 2-normed spaces. If T: X [right arrow] Y is a surjective bounded linear operator then T is sequentially continuous.

Corollary 2.10[2]: Let X and Y be two 2-normed spaces. Then every compact operator T: X [right arrow] Y is bounded.

3. Main Results

Lemma 3.1: Let X be a 2-normed space. If [B.sub.e][a,r] is compact in X for some a,e [member of] X and r > 0 then X is of finite dimension.

Proof: Suppose that [B.sub.e][a,r] is compact. The quotient space X/<e> is a normed space equipped with the norm;

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Define [A.sub.e']= {x + <e>: [parallel]x - a + <e>[parallel] [less than or equal to] r/[parallel]e,e'[parallel]}

and let A = [intersection]{[A.sub.e']: e and e' are linearly independent]. Then A is a closed ball in the normed space X/<e>. We aim to show that A is a compact set in the normed space X/<e>. For that let {[x.sub.n] + <e>] be any sequence in A.

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

In particular, [parallel][x.sub.n]-a,e[parallel] [less than or equal to] r; [for all]n

[??] [x.sub.n] [member of] [B.sub.e][a,r].

Hence {[x.sub.n]} has a convergent subsequence [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] converges to a point [x.sub.0].

We have [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Hence [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] is a convergent subsequence of {[x.sub.n] + <e>}. This implies that A is compact and so X/<e> is of finite dimension.

[??] X is of finite dimension.

Remark: Here we introduce a result which is similar to Riesz Lemma.

Lemma 3.2(2-NSR LEMMA): Let X be a 2-normed space and let 0 [not equal to] e [member of] X. Let r be any number such that 0 < r < 1. Then there exist some [x.sub.r] [member of] X such that [parallel][x.sub.r],e[parallel] = 1 and r < [parallel][x.sub.r],e[parallel] [less than or equal to] 2.

Proof: Since [parallel]x,e[parallel] > 0, [for all]x [not member of] <e>,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Also, as r < 1 [parallel]x,e[parallel] [less than or equal to] [[parallel]x + <e>[parallel].sub.Q] < [[parallel]x + <e>[parallel].sub.Q]/r, [for all]x [not member of] <e>.

Put [x.sub.r] = x/[parallel]x,e[parallel]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[less than or equal to] 2.

Thus there exist [x.sub.r] [member of] X such that [parallel][x.sub.r],e[parallel] = 1 and r < [parallel][x.sub.r],e[parallel] [less than or equal to] 2.

Remark: Let X be a 2-normed space ad let Y be a finite dimensional subspace of X generated by {[e.sub.1],[e.sub.2],...[e.sub.n]]. Then X/Y is a normed space equipped with the norm

[[parallel]x + Y[parallel].sub.Q] = [n.summation over (k=1)][[parallel]x + <[e.sub.k]>[parallel].sub.Q].

Corollary 3.3: Let X be a 2-normed space and let Y be a finite dimensional subspace of X. Let r be any number such that 0 < r < 1.Then there exist some [x.sub.r] [member of] X such that r < [parallel][x.sub.r] + Y[parallel] [less than or equal to] 2.

Proof: Let {[e.sub.1],[e.sub.2],...[e.sub.n]] be a basis for Y. Then for any x [not member of] Y,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Put [x.sub.r] = x/[n.summation over (k=1)][parallel]x,[e.sub.k][parallel] so that r < [parallel][x.sub.r] + Y[parallel] [less than or equal to] 2.

Lemma 3.4: Every finite dimensional subspace Y of a 2-normed space X is complete.

Proof: To prove the completeness of Y, we use mathematical induction on the dimension m of Y. Let m = 1. Then Y = {ke: k [member of] R} with e [not equal to] 0. If {[x.sub.n]} is a Cauchy sequence in Y with [x.sub.n] = [k.sub.n]e, then for every x [member of] X, [parallel][x.sub.n] - [x.sub.m],x[parallel][right arrow]0

[parallel][x.sub.n] - [x.sub.m],x[parallel] = [parallel]([k.sub.n] - [k.sub.m])e,x[parallel] = [absolute value of [k.sub.n] - [k.sub.m]][parallel]e,x[parallel]; [for all]x [member of] X.

[??] [absolute value of [k.sub.n] - [k.sub.m]] = [parallel][x.sub.n] - [x.sub.m],x[parallel]/[parallel]x,e[parallel], [for all]x [not member of] <e>.

[??] {[k.sub.n]} is a Cauchy sequence in R which is complete.

If [k.sub.n] [right arrow] k in R then [x.sub.n] [right arrow] ke in Y. Thus Y is complete. Now assume that every m - 1 dimensional subspace of X is complete. Let dim Y = m and let {[x.sub.n]} be a Cauchy sequence in Y. Let {[e.sub.1],[e.sub.2],...[e.sub.m]} be a basis for Y and let Z = span {[e.sub.2],...[e.sub.m]}. Now for each n = 1,2,3,...[x.sub.n] = [k.sub.n][e.sub.1] + [z.sub.n] for some [k.sub.n] [member of] R and [z.sub.n] [member of] Z.

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

In particular, [parallel][x.sub.n]-[x.sub.m][parallel],[e.sub.2][parallel] = [absolute value of [k.sub.n]-[k.sub.m]][parallel][e.sub.1]+[([z.sub.n]- [z.sub.m])/([k.sub.n]-[k.sub.m])],[e.sub.2][parallel];[for all]x[member of] X

> [absolute value of [k.sub.n] - [k.sub.m]]/2[parallel][e.sub.1] + Z[parallel].

[??] {[k.sub.n]} is a Cauchy sequence in R which is complete. As [z.sub.n] = [x.sub.n] - [k.sub.n][e.sub.1], it follows that {[z.sub.n]} is a Cauchy sequence in Z which is complete. If [k.sub.n] [right arrow] k in R and [z.sub.n] [right arrow] z in Z, then [x.sub.n] [right arrow] k[e.sub.1] + z in Y. Hence Y is complete.

Theorem 3.5: Let X be a 2-normed space and let T be a surjective compact operator on X and 0 [not equal to] k [member of] R. If {[x.sub.n]} is a locally bounded sequence in X such that

T([x.sub.n]) - k[x.sub.n] [right arrow] y in X then there is a subsequence [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] of {[x.sub.n]} such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] converges to x in X and T(x) - kx = y.

Proof: Since T is compact, there exist a subsequence [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] of {[x.sub.n]} such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] converges to some z [member of] X. Then [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] converges to -y + z and so [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]. Since T is sequentially continuous [2.9], [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII].

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Theorem 3.6: Let X be a 2-normed space and T: X [right arrow] X. Let 0 [not equal to] k [member of] R and 0 [not equal to] e [member of] X such that (T - kI)X [subset or equal to] <e>. Then there is some [x.sub.0] [member of] X such that [parallel][x.sub.0],e[parallel] = 1 and for every y [member of] <e>, [parallel]T([x.sub.0]) - T(y),e[parallel] > [absolute value of k]/4.

Proof: Let Y = <e>. Then T(y) = (T(y) - ky) + ky [subset or equal to] Y + Y = Y, [for all]y [member of] Y.

[??] T(Y) [subset or equal to] Y. Choose some [x.sub.0] [member of] X such that [parallel][x.sub.0],e[parallel] = 1 and 1/2 < [parallel][x.sub.0] + <e>[parallel].

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Corollary 3.7: Let X be a 2-normed space and T: X [right arrow] X. Let 0 [not equal to] k [member of] R and let Y be a finite dimensional proper subspace of X such that (T - kI)X [subset or equal to] Y. Then there is some [x.sub.0] [member of] X such that for every x,y [member of] Y, [parallel]T([x.sub.0]) - T(y),x[parallel] > [absolute value of k]/4.

Proof: Let {[e.sub.1],[e.sub.2],...[e.sub.m]} be a basis for Y. Then T(y) = (T(y) - ky) + ky [subset or equal to] Y + Y = Y, [for all] y [member of] Y and so T(Y) [subset or equal to] Y. Choose some [x.sub.0] [member of] X such that 1/2 < [parallel][x.sub.0] + Y[parallel].

For any x,y [member of] Y,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

References

[1] Y.J. Cho, P.C.S. Lin, S.S. Kim and A. Misiak, Theory of 2-Inner product spaces, Nova Science, New York, 2001.

[2] Fatemeh Lael and Kourosh Nouruzi, compact operators defined on 2-normed and 2-Probablistic normed spaces, Mathematical Problems in Engineering Volume 2009 (2009), Article ID 950234,

[3] S. Gahler, Siegfrid 2-metrische Raume und ihre topologische struktur, Math.Nachr.26(1963)115-148.

[4] S. Gahler, Lineare 2-normierte Raume, Math.Nachr.28(1964)1-43.

[5] F. Raymond W and C. Yeol Je, Geometry of Linear 2-normed spaces, Nova science publishers, Inc., Hauppauge, Ny, 2001.

(1) P. Riyas and (2) K T. Ravindran

P.G Department and Research Centre in Mathematics, Payyanur College, Payyanur, Kerala, India.