Ann. Acad. Rom. Sci. Ser. Math. Appl. Vol. 8, No. 1/2016. The space l (X) Namita Das

ISSN 2066-6594 Ann. Acad. Rom. Sci. Ser. Math. Appl. Vol. 8, No. 1/2016 The space l (X) Namita Das Astract In this paper we have shown that the space c n n 0 cannot e complemented in l n n and c 0 (H) cannot e complemented in l (H) where H is a Hilert space. Further, extending these results we show that if X is a Banach space then c 0 (X) cannot e complemented in l (X). MSC: 46B25, 46B45,46C07 keywords: Hilert space, Banach space, projection operator, contraction, Schauder asis. 1 Introduction A closed suspace M of a Banach space X is said to e complementd in X if and only if there exists a ounded linear projection from X onto M. It is not difficult to see that if M is complemented y the closed suspace N, then there exists a c > 0 such that m + n c m for all m M and n N. Murray [12] showed that l p, p > 1, p 2 has suspaces that cannot e complemented. Philips [14] and Lindenstrauss and Tzafriri [11] proved that c 0 cannot e complemented in l. In fact, Lindenstrauss and Tzafriri [11] estalished that every infinite dimensional Banach space which is not isomorphic to a Hilert space contains a closed suspace that cannot e complemented. Similar results were also proved in [2],[4] [6],[15],[17] and [20]. Pelczynski [13] showed that complemented suspaces of l 1 are isomorphic to Accepted for pulication on Octoer 22-nd 2015 namitadas440@yahoo.co.in, P.G. Department of Mathematics Utkal University Vanivihar, Bhuaneswar, 751004, Odisha, India 14

The space l (X) 15 l 1. Lindenstrauss [9] proved that every infinite dimensional complemented suspace of l is isomorphic to l. This also holds if l is replaced y l p, 1 p <, c 0 or c. It is also shown y Lindenstrauss [10] that if the Banach space X and its closed suspace Y are generated y weakly compact sets (in particular, if X is reflexive) then Y is complemented in X. Thorp [18] has shown that for X and Y, certain Banach spaces of sequences, the suspace LC(X, Y ) of compact linear operators from X to Y cannot e complemented in L(X, Y ), the space of ounded linear operators from X to Y. Arterurn and Whitley [1] proved similar results for either X an astract L space or Y a space of type C(S) and considered projections on the suspace W (X, Y ) of weakly compact linear operators mapping X to Y. Tong [19] studied the existence of ounded projections from the space L(X, Y ) onto the suspace LC(X, Y ), where X and Y are normed spaces. Johnson [5] showed that if X and Y are Banach spaces, the space X is infinite dimensional and if Y contains a complemented copy of c 0, then LC(X, Y ) cannot e complemented in L(X, Y ). Kuo [7] showed that if X and Z are Banach spaces and (a) if X contains an isomorph of c 0, then LC(X, l ) cannot e complemented in L(X, l ) () if S is a compact Hausdorff space which is not scattered, then LC(C(S), Z) cannot e complemented in W (C(S), Z) for Z = c 0 or l. In particular, LC(l, c 0 ) cannot e complemented in L(l, c 0 ). Let L(X) e the set of all ounded linear operators from the Banach space X into itself and LC(X) e the set of all compact operators in L(X). Conway [3] has estalished that LC(H) cannot e complemented in L(H) where H is an infinite dimensional separale Hilert space. Let l n n and = {f : N C n n : f(m) = A n n m and sup f(m) C n n < } m c n n 0 = {f : N C n n : f(m) = Am n n and f(m) C n n 0 as m }. It is not difficult to see that c0 n n l n n. If f l n n, we define f = sup m f(m) C n n. The space l n n with the sup norm is a Banach space and c n n 0 is a closed suspace of l n n. Let H e a Hilert space and and l (H) = {f : N H : f(n) = x n H sup f(n) H < } n

16 Namita Das and and c 0 (H) = {f : N H : f(n) = x n H, f l (H) f(n) H 0 as n }. It is also not difficult to verify that c 0 (H) l (H). The space l (H) is a Banach space [8] with the norm f = sup n f(n) H and c 0 (H) is a closed suspace of l (H). Now let (X, X ) e a Banach space. Define l (X) = {f : N X : f(n) = x n X and sup f(n) X < }. n The space l (X) is a Banach space with the norm f = sup n f(n) X and define c 0 (X) = {f : N X : f(n) = x n X, f l (X) and f(n) X 0 as n }. The space c 0 (X) is a closed suspace of l (X). In this paper we have shown that the space c n n 0 cannot e complemented in l n n and c 0 (H) cannot e complemented in l (H) where H is a separale Hilert space. Further, extending these results we show that if X is a Banach space then c 0 (X) cannot e complemented in l (X). For this we need to introduce the Hardy space. Let T denote the unit circle in the complex plane C. Let dθ e the arc-length measure on T. For 1 p +, L p (T) will denote the Leesgue space of T induced y dθ 2π. Given f L 1 (T), the Fourier coefficients of f are a n (f) = 1 2π f(θ)e inθ dθ, n Z (1) 2π 0 where Z is the set of all integers. Let Z + denote the set of nonnegative integers. For 1 p +, the Hardy space of T denoted y H p (T), is the suspace of L p (T) consisting of functions f with a n (f) = 0 for all negative integers n. We shall let H p (D) denote the space of analytic functions on D which are harmonic extensions of functions in H p (T). The Hardy space H 2 (D) is a reproducing kernel Hilert space and the reproducing kernel 1 (called the Cauchy or Szego kernel) K w (z) =, for z, w D. It is not 1 wz

The space l (X) 17 so important to distinguish H p (D) from H p (T). The sequence of functions {e inθ } n Z+ forms an orthonormal asis for H 2 (T). Let L 2 C n(t) denote the Hilert space of Cn -valued, norm square integrale, measurale functions on T. When endowed with the inner product defined y the equality f, g = f(z), g(z) C ndm, f, g L 2 C n(t), T the space L 2 C n (T) ecomes a separale Hilert space. Here the measure m denotes the normalized Leesgue measure on T. For a function F L 2 C n(t), we define the nth Fourier coefficient of F as c n (F ) = 1 2π 2π 0 e int F (e it )dt, n Z. The integral is understood in the strong sense. Let H 2 C n (T) e the Hardy space of functions in L 2 C n (T) with vanishing negative Fourier coefficients. Notice that L 2 C n (T) = L 2 (T) C n and H 2 C n (T) = H 2 (T) C n where the Hilert space tensor product is used. 2 The space l n n In this section we show that c0 n n cannot e complemented in l n n. For this purpose, we introduce the operators U a and V a acting on a direct sum H, with each H the same Hilert space H. Define the ounded linear B operators U a : H H, V a : H H, for each a in B, as follows. When x H and u = {x } H, V a u = x a and U a x is the family {z } in which z a = x and all other z are 0; H a is the range of U a, and so consists of all elements {z } of H in which z = 0 when a. The space H a is a closed suspace of H, and oserve that V a U a is the identity operator on H and U a V a is the projection E a from H onto H a. Since the suspaces H a(a B) are pairwise orthogonal, and H a = H, it follows that the sum a B E a is strong-operator convergent to I. Note that U a = V a, since U a x, {x } = x, x a = x, V a {x }

18 Namita Das whenever x H and {x } H. With each ounded linear operator T acting on H, we associate a matrix [T a ] a, B, with entries T a in L(H) defined y T a = V a T U. (2) If u = {x } H, then T u is an element {y } of H, and ( ) y a = V a T u = V a T E u = V a T U V u = T a x. Thus ( ) T x = y where y a = T a x (a B). (3) B The usual rules of matrix algera have natural analogues in this situation. From (2.1), the matrix elements T a depend linearly on T. Since V a T U = U a T V = (V T U a ) = (T a ), the matrix of T has (T a ) in the (a, ) position. If S and T are ounded linear operators acting on H, and R = ST, then R a = V a RU = V a ST U = c B V a SE c T U = V a SU c V c T U = S ac T c, c B c B the sum converging in the strong-operator topology if the index set B is infinite. In this way, we estalish a one-to-one correspondence etween elements of L( H ) and certain matrices [T a ] a, B with entries T a in B L(H). When the index set B is finite, each such matrix corresponds to some ounded operator T acting on H ; indeed, T is defined y (2.2), and its oundedness follows at once from the relations ( ) 2 {y } 2 = a y a 2 = a T a x 2 a T a x a ( ) ( T a 2 ) ( ) x 2 = T a 2 {x } 2. a

The space l (X) 19 Theorem 1. The space c n n 0 cannot e complemented in l n n. Proof. The set of vectors {e n } n=0 form an orthonormal asis for H2 (T) where e n (z) = z n, z T. It is easy to verify that HC 2 n (T) = H 2 (T) H 2 (T) H 2 (T) (direct sum of n-copies of H 2 (T)) and if T L(HC 2 n (T)) then T = T 11 T 12 T 1n T 21 T 22 T 2n... T n1 T n2 T nn, T ij L(H 2 (T)), 1 i, j n. Define the maps V i and U i, 1 i n, as we have discussed earlier in this section on the space HC 2 n (T) = H 2 (T) H 2 (T) H 2 (T). It is not difficult to see that T ij = V i T U j, 1 i, j n. Further, define the map σ : L(HC 2 n (T)) into l n n as σ(t )(m) = 1 n T 11 e m, e m T 12 e m, e m T 1n e m, e m T 21 e m, e m T 22 e m, e m T 2n e m, e m... T n1 e m, e m T n2 e m, e m T nn e m, e m if T = (T ij ) 1 i,j n. Notice that T ij = V i T U j T. Then σ(t ) = sup m σ(t )(m) C n n 1 = sup m max n 1 j n n i=1 T ije m, e m 1 sup m max n 1 j n n i=1 T ij 1 sup m max 1 j n nn T = T., Thus σ is a contraction. Now if T LC(HC 2 n (T)), then T ij LC(H 2 (T)) for all i, j {1, 2, n} and T ij e m, e m 0 as m. Hence σ(t )(m) 0 as m. Thus σ(lc(hc 2 n (T))) c n n 0. Define ρ : l n n L(HC 2 n (T)) as ρ(f ) = T where T = T 11 T 12 T 1n T 21 T 22 T 2n... T n1 T n2 T nn, and T ij = 1 n m=1 F ij(m)psp{e m} if F 11 F 12 F 1n F 21 F 22 F 2n F =... ln n F n1 F n2 F nn.

20 Namita Das Now F l n n = sup m F (m) C n n = sup m max 1 j n n i=1 F ij(m). Hence ρ(f ) = T ( n i,j=1 T ij 2 ) 1 2 ( n i,j=1 ( 1 n sup m F (m) ) 2 ) 1 2 ( = 1 (sup n 2 m F (m) ) 2)] 1 2 [ n i,j=1 = sup m F (m) C n n = F l n n. This proves that ρ is a contraction and ρ(c0 n n ) LC(HC 2 n (T)). Now if there is a projection P from l n n onto c n n 0, then Q = ρ P σ is a projection from L(HC 2 n (T)) onto LC(HC 2 n (T)) since Q 2 = Q. This is a contradiction since LC(H) cannot e complemented [3] in L(H) if H is an infinite dimensional separale Hilert space. 3 Projection onto c 0 (H) In this section we shall estalish that c 0 (H) cannot e complemented in l (H) if H is an infinite dimensional separale Hilert space. Theorem 2. If H is an infinite dimensional separale Hilert space, then c 0 (H) cannot e complemented in l (H). Proof. Let {e n } n=0 e an orthonormal asis for H. Then it is well known [8] that e n 0 weakly. Define σ : L(H) l (H) as σ(t )(n) = T e n for all n N. The map σ is well defined and is a contraction since sup n σ(t )(n) = sup T e n T <. n If T LC(H), then T e n 0 strongly and σ(t )(n) = T e n 0 as n. Thus σ(t ) c 0 (H) and σ(lc(h)) c 0 (H). Now define ρ : l (H) L(H) as ρ(f ) = T where T = F (n), e n Psp{e n} n=0 if F = (F (n)) n=0 l (H). Notice that if F l (H), then F l (H) = sup F (n) H < n

The space l (X) 21 and T = sup n F (n), e n sup n F (n) H well-defined, ρ(c 0 (H)) LC(H) and <. Hence the map ρ is ρ(f ) = T = sup n F (n), e n sup n F (n) H = F l (H). Thus the map ρ is a contraction. Now if there exists a projection P from l (H) onto c 0 (H) then Q = ρ P σ is a projection from L(H) onto LC(H). Since there exists no projection [3] from L(H) onto LC(H), hence there exists no projection from l (H) onto c 0 (H). 4 Projection onto c 0 (X) In this section we shall estalish that c 0 (X) cannot e complemented in l (X) where X is an infinite dimensional separale Banach space. Definition 1. Let (X, ) e a normed linear space over C. A countale suset {x 1, x 2, } of X is called a Schauder asis for X if x n = 1 for each n and if for every x X, there are unique α 1, α 2, in C such that x = n=1 α nx n with the series converging in the norm of X. If {x 1, x 2, } is a Schauder asis for X, then for n = 1, 2,, define f n : X C y f n (x) = α n, for x = n=1 α nx n X. The uniqueness condition in the definition of a Schauder asis shows that each f n is well-defined and linear on X. It is called the nth coefficient functional on X corresponding to the Schauder asis {x 1, x 2, } for X. If X is a Banach space then it is not difficult to verify [8] that each f n is a continuous linear functional and f n α for all n = 1, 2, and some α > 0. If x = n=1 α nx n, define x = sup n n k=1 α kx k. Then is a norm and (X, ) is complete. The norm and are equivalent. Let X e a Banach space, U = {f : f X and f 1} e the unit all of X and let E e the set of extreme points of U. Theorem 3. Let X e a separale Banach space and {x n } n=1 e a Schauder asis for X. Suppose lim n f(x n ) = f(0) for each f in E. Then c 0 (X) cannot e complemented in l (X). Proof. Let X e a separale Banach space with a Schauder asis {x n } n=1 such that x n = 1 for all n N. From [16], it follows that lim n f(x n ) = f(0) for each f in E if and only if x n 0 weakly. Define σ : L(X) l (X) such that σ(t )(n) = T x n. Since σ(t ) = sup n σ(t )(n) X = sup n T x n T x n = T. The map σ is well-defined and σ is a

22 Namita Das contraction. Further, if T LC(X), then T x n 0 strongly. Hence σ(t )(n) = T x n 0 strongly and in this case σ(t ) c 0 (X). Thus σ(lc(x)) c 0 (X). Now define a map ρ : l (X) L(X) as ρ(f ) = T where T = n=1 f n(f (n))psp{x n} where f n s are the coordinate functions. If F l (X) then F is a function from N to X such that sup n F (n) X < and F (n) X. If F (n) = j=1 α jx j, α j C then f j (F (n)) = α j for all j N. The map ρ is well-defined since ρ(f ) = T = sup n f n (F (n)) α sup n F (n) X = α F for some constant α > 0 and ρ(c 0 (X)) LC(X). Now if there exists a projection P from l (X) onto c 0 (X) then Q = ρ P σ is a projection from L(X) onto LC(X). But from [19] it follows that there exists no ounded projection from L(X) onto LC(X). Hence there exists no ounded projection P from l (X) onto c 0 (X). This proves the claim. References [1] D. Arterurn, and R. Whitley: Projections in the space of ounded linear operators, Pacific J. of Maths. 15(3)(1965), 739 746. [2] J, Bourgain, H.P. Rosenthal, G. Schechtman: An ordinal L p -index for Banach spaces, with application to complemented suspaces of L p, Ann. of Math.(2) 114(2) (1981), 193 228. [3] J. B. Conway :The compact operators are not complemented in B(H), Proc. Amer. Math. Soc. 32(2), (1972), 549 550. [4] W.T. Gowers, B. Maurey: The unconditional asic sequence prolem, J. Amer. Math. Soc. 6(4), (1993), 851 874. [5] J. Johnson: Remarks on Banach spaces of compact operators, Journal of Funct. Anal. 32, (1979), 304 311. [6] W.B. Johnson, J. Lindenstrauss: Examples of L 1 spaces, Ark. Mat., 18(1), (1980), 101 106. [7] T.H. Kuo: Projections in the spaces of ounded linear operators, Pacific J. of Maths. 52(2), (1974), 475 480.

The space l (X) 23 [8] B.V. Limaye: Functional Analysis, New Age International Pvt. Ltd., 2nd Edition, 2004. [9] J. Lindenstrauss: On complemented suspaces of m, Israel J. Math. 5, (1967), 153 156. [10] J. Lindenstrauss: On a theorem of Murray and Mackey, An. Acad. Brasil. Ci. 39, (1967), 1 6. [11] J. Lindenstrauss, L. Tzafriri: Classical Banach spaces, I. Sequence spaces, Ergenisse der Mathematik und iher Grenzgeiete, Vol. 92, Springer-Verlag, Berlin-New York, 1977. [12] F.J. Murray: On complementary manifolds and projections in spaces L p and l p,, Trans. Amer. Math. Soc. 41(1), (1937), 138 152. [13] A. Pelczynski: Projections in certain Banach spaces, Studia Math. 19, (1960), 209 228. [14] R.S. Phillips: On linear transformations, Trans. Amer. Math. Soc. 48, (1940), 516 541. [15] G. Pisier: Remarks on complemented suspaces of von Neumann algeras, Proc. Roy. Soc. Edinurgh Sect. A, 121, no. 1 2, (1992), 1 4. [16] J. Rainwater: Weak convergence of ounded sequences, Proc. Amer. Math. Soc. 14, (1963), 999. [17] B. Randrianantoanina: On isometric staility of complemented suspaces of L p,, Israel J. Math. 113, (1999), 45 60. [18] E. Thorp: Projections onto the suspace of compact operators, Pacific J. of Maths. 10, (1960), 693 696. [19] A. Tong: Projecting the space of ounded operators onto the space of compact operators, Proc. Amer. Math. Soc. 24, (1970), 362 365. [20] R. Whitley: Projecting m onto c 0, Amer. Math. Monthly, 73, (1966), 285 286.