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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. 6 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: [email protected] K. T. Ravindran, P. G. Department and Research Centre, Payyannur College, Payyannur E-mail: [email protected]