MATEMATIQKI VESNIK
originalni nauqni rad
research paper
63, 1 (2011), 1–6
March 2011
2-NSR LEMMA AND QUOTIENT SPACE IN 2-NORMED SPACE
P. Riyas and K. T. Ravindran
Abstract. In this paper we discuss properties of compactness and compact operator on
2-normed space. Also we consider a result which is similar to Riesz Lemma and its applications in
2-normed space. We introduce quotient space from the finite dimensional subspace of a 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, Gähler 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. In [4] S. Gähler introduced the following definition of a 2normed space.
2. Preliminaries
Definition 2.1. [4] Let X be a real linear space of dimension greater than 1.
Suppose k , k is a real valued function on X ×X satisfying the following conditions:
1. kx, yk = 0 if and only if x and y are linearly independent,
2. kx, yk = ky, xk,
3. kαx, yk = |α|kx, yk,
4. kx + y, zk ≤ kx, zk + ky, zk.
Then k , k is called a 2-norm on X and the pair (X,k , k) is called a 2-normed
space. Some of the basic properties of 2-norms, are that they are non-negative and
kx, y + xk = kx, yk, ∀x, y ∈ X and ∀α ∈ R.
Definition 2.2. [2] Let X and Y be two 2-normed spaces and T : X → Y be
a linear operator. For any e ∈ X, we say that the operator T is e-bounded if there
2010 AMS Subject Classification: 41A65, 41A15.
Keywords and phrases: 2-normed space, locally bounded, compact operator, Riesz Lemma,
e-bounded operator.
1
2
P. Riyas, K. T. Ravindran
exists Me > 0 such that kT (x), T (e)k ≤ Me kx, ek for all x ∈ X. An e-bounded
operator T for every e will be called bounded.
Definition 2.3. [4] A sequence {xn } in a 2-normed space X is said to be
convergent if there exists an element x ∈ X such that
lim k{xn − x}, yk = 0
n→∞
for all x ∈ X.
Definition 2.4. [4] Let X and Y be two 2-normed spaces and T : X → Y be
a linear operator. The operator T is said to be sequentially continuous at x ∈ X if
for any sequence {xn } of X converging to x we have T ({xn }) → T (x).
Definition 2.5. [2] The closure of a subset E of a 2-normed space X is
denoted by E and defined as the set of all x ∈ X such that there is a sequence {xn }
of E converging to x. We say that E is closed if E = E.
For a 2-normed space we consider the subsets
Be (a, r) = {x : kx − a, ek < r},
Be [a, r] = {x : kx − a, ek ≤ r}.
Definition 2.6. [2] A subset A of a 2-normed space X is said to be locally
bounded if there exist e ∈ X − {0} and r > 0 such that A ⊆ Be (0, r).
Definition 2.7. [2] A subset B of a 2-normed space X is said to be compact
if every sequence {xn } in 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 → Y is called a compact operator if it maps every locally bounded sequence
{xn } of X onto a sequence {T (xn )} in Y which has a convergent subsequence.
Lemma 2.9. [2] Let X and Y be two 2-normed spaces. If T : X → 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 → Y is bounded.
3. Main results
Lemma 3.1. Let X be a 2-normed space. If Be [a, r] is compact in X for some
a, e ∈ X and r > 0 then X is of finite dimension.
Proof. Suppose that Be [a, r] is compact. The quotient space X/hei is a normed
space equipped with the norm
½
¾
kx, ek
0
0
kx + heikQ = inf
:
e
∈
/
hei
&
ke,
e
k
≤
1
ke, e0 k
½
¾
kx, ek
0
0
+ sup
:
e
∈
/
hei
&
ke,
e
k
>
1
ke, e0 k
½
¾
kx, ek
0
0
= kx, ek + sup
:
e
∈
/
hei
&
ke,
e
k
>
1
.
ke, e0 k
2-nsr lemma and quotient space in 2-normed space
3
T
r
{Ae0 :
Define A0e = {x + hei : kx − a + heikQ ≤ ke,e
0 k } and let A =
e and e0 are linearly independent}. Then A is a closed ball in the normed space
X/hei. We aim to show that A is a compact set in the normed space X/hei. For
that let {xn + hei} be any sequence in A. Then
kxn + hei − (a + hei)kQ = kxn − a + heikQ ≤
⇒ kxn − a, ek ≤
r
; ∀ e0 ∈
/ hei and ∀ n
ke, e0 k
r
; ∀ e0 ∈
/ hei and ∀ n.
ke, e0 k
In particular,
kxn − a, ek ≤ r ; ∀ n ⇒ xn ∈ Be [a, r].
Hence {xn } has a convergent subsequence {xnk } converges to a point x0 . We have
kxnk + hei − (x0 + hei)kQ = kxnk − x0 + heikQ
½
¾
kxnk − x0 , ek 0
0
= kxnk − x0 , ek + sup
:e ∈
/ hei and ke, e k > 1
ke, e0 k
·
½
¾¸
1
0
0
:e ∈
/ hei and ke, e k > 1
= kxnk − x0 , ek 1 + sup
ke, e0 k
⇒ lim kxnk + hei − (x0 + hei)kQ = 0.
k→∞
Hence {xnk + hei} is a convergent subsequence of {xn + hei}. This implies that A
is compact and so X/hei is of finite dimension. Hence X is of finite dimension.
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 6= e ∈ X.
Let r be any number such that 0 < r < 1. Then there exists some xr ∈ X such that
kxr , ek = 1 and r < kxr , ek.
Proof. Since kx, ek > 0, ∀ x ∈
/ hei, we have
½
¾
kx, ek
0
0
kx + heikQ = kx, ek + sup
:
e
∈
/
hei
and
ke,
e
k
>
1
ke, e0 k
> 0; ∀ x ∈
/ hei.
Also, as r < 1, kx, ek ≤ kx + heikQ <
that kxr , ek = 1 and
kx+heikQ
,
r
kxr + heikQ = k
∀x∈
/ hei. Put xr = x/kx, ek, so
x
+ heikQ > r.
kx, ek
Thus there exist xr ∈ X such that kxr , ek = 1 and r < kxr , ek.
Remark. Let X be a 2-normed space ad let Y be a finite dimensional subspace
of X generated by {e1 , e2 , . . . , en }. Then X/Y is a normed space equipped with
the norm
n
X
kx + Y kQ =
kx + hek ikQ .
k=1
4
P. Riyas, K. T. Ravindran
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 exists some
xr ∈ X such that r < kxr + Y kQ ≤ 2.
Proof. Let {e1 , e2 , . . . , en } be a basis for Y . Then for any x ∈
/ Y,
x
r<k
+ hek ikQ ≤ 2; for k = 1, 2, . . . , n
kx, ek k
⇒ rkx, ek k < kx + hek ikQ
≤ 2kx, ek k; for k = 1, 2, . . . , n and for every x ∈
/ Y.
⇒ r
n
X
kx, ek k < kkx + Y kQ ≤ 2
k=1
n
X
kx, ek k.
k=1
Pn
Put xr = x/ k=1 kx, ek k so that r < kxr + Y k ≤ 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 ∈ R} with e 6= 0. If {xn } is a
Cauchy sequence in Y with xn = kn e then for every x ∈ X, kxn − xm , xk → 0.
kxn − xm , xk = k(kn − km )e, xk = |kn − km |ke, xk; ∀ x ∈ X.
⇒ |kn − km | =
kxn − xm , xk
, ∀x ∈
/ hei.
kx, ek
It follows that {kn } is a Cauchy sequence in R which is complete.
If kn → k in R then xn → 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
{xn } be a Cauchy sequence in Y . Let {e1 , e2 , . . . , en } be a basis for Y and let
Z = span{e2 , e3 , . . . , en }. Now for each n = 1, 2, 3, . . . , xn = kn e1 + zn for some
kn ∈ R and zn ∈ Z.
kxn − xm , xk = k(kn − km )e1 + (zn − zm ), xk
= |kn − km |ke1 +
(zn − zm )
, xk; ∀ x ∈ X.
kn − km )
In particular,
kxn − xm , e2 k = |kn − km |ke1 +
(zn − zm )
|kn − km |
, e2 k >
ke1 + ZkQ .
kn − km )
2
It follows that {kn } a Cauchy sequence in R which is complete. As zn = xn − kn e1 ,
it follows that {zn } is a Cauchy sequence in Z which is complete. If kn → k in R
and zn → z in Z, then xn → ke1 + 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 6= k ∈ R. If {xn } is a locally bounded sequence in X such
2-nsr lemma and quotient space in 2-normed space
5
that T (xn ) − kxn → y in X then there is a subsequence {xnk } of {xn } such that
{xnk } converges to x in X and T (x) − kx = y.
Proof. Since T is compact, there exist a subsequence {xnk } of {xn } such
that T (xnk ) converges to some z ∈ X. Then kxnk = [kxnk − T (xnk )] + T (xnk )
converges to −y + z and so {xnk } → z−y
k = x. Since T is sequentially continuous
[2.9], T (xnk ) → T (x). It follows that
T (x) − kx = lim [T (xnk ) − kxnk ] = z − (−y + z) = y.
k→∞
Theorem 2.6. Let X be a 2-normed space and T : X → X. Let 0 6= k ∈ R
and 0 6= e ∈ X such that (T − kI)X ⊆ hei. Then there is some x0 ∈ X such that
kx0 , ek = 1 and for every y ∈ hei, kT (x0 ) − T (y), ek > |k|
4 .
Proof. Let Y = hei. Then T (y) = (T (y) − ky) + ky ⊆ Y + Y = Y, ∀ y ∈ Y .
It follows that T (Y ) ⊆ Y . Choose some x0 ∈ X such that kx0 , ek = 1 and
1
2 < kx0 + heikQ . For any y ∈ Y ,
kT (x0 ) − T (y), ek = kkx0 − [kx0 − T (x0 ) + T (y)] , ek
1
= |k|kx0 − [kx0 − T (x0 ) + T (y)] , ek
k
1
|k|
kx0 − [kx0 − T (x0 ) + T (y)] + heik
≥
2
k
Q
|k|
|k|
=
kx0 + heikQ >
.
2
4
Corollary 2.7. Let X be a 2-normed space and T : X → X. Let 0 6= k ∈ R
and let Y be a finite dimensional proper subspace of X such that (T − kI)X ⊆ Y .
Then there is some x0 ∈ X such that for every x0 , y0 ∈ Y , kT (x0 ) − T (y), xk > |k|
4 .
Proof. Let {e1 , e2 , . . . , em } be a basis for Y . Then T (y) = (T (y) − ky) +
ky ⊆ Y + Y = Y, ∀ y ∈ Y and so T (Y ) ⊆ Y . Choose some x0 ∈ X such that
1
2 < kx0 + Y kQ . For any x, y ∈ Y ,
kT (x0 ) − T (y), xk = kkx0 − [kx0 − T (x0 ) + T (y)] , xk
1
= |k|kx0 − [kx0 − T (x0 ) + T (y)] , xk
k
|k|
1
≥
kx0 − [kx0 − T (x0 ) + T (y)] + Y k
2
k
Q
|k|
|k|
=
kx0 + Y kQ >
.
2
4
Acknowledgement. The authors wish to thank the referee for many helpful
comments and suggestions.
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P. Riyas, K. T. Ravindran
REFERENCES
[1] Y.J. Cho, P.C.S. Lin, S.S. Kim, A. Misiak, Theory of 2-Inner Product Spaces, Nova Science,
New York, 2001.
[2] F. Lael, K. Nouruzi, Compact operators defined on 2-normed and 2-probablistic normed spaces,
Mathematical Problems in Engineering , Vol. 2009, Article ID 950234.
[3] S. Gähler, Siegfrid 2-metrische Räume und ihre topologische Struktur, Math. Nachr. 26 (1963),
115–148.
[4] S. Gähler, Lineare 2-normierte Räume, Math. Nachr. 28 (1964), 1–43.
[5] F. Raymond W, C. Yeol Je, Geometry of Linear 2-Normed Spaces, Nova science publishers,
Inc., Hauppauge, Ny, 2001.
(received 23.07.2009; in revised form 12.06.2010)
P. Riyas, P. G. Department and Research Centre, Payyannur College, Payyannur
E-mail: riyasmankadavu@gmail.com
K. T. Ravindran, P. G. Department and Research Centre, Payyannur College, Payyannur
E-mail: drravindran@gmail.com