Banach Spaces II: Elementary Banach Space Theory

Transcription

1 BS II c Gabriel Nagy Banach Spaces II: Elementary Banach Space Theory Notes from the Functional Analysis Course (Fall 07 - Spring 08) In this section we introduce Banach spaces and examine some of their important features. Sub-section B collects the five so-called Principles of Banach space theory, which were already presented earlier. Convention. Throughout this note K will be one of the fields R or C, and all vector spaces are over K. A. Banach Spaces Definition. A normed vector space (X,. ), which is complete in the norm topology, is called a Banach space. When there is no danger of confusion, the norm will be omitted from the notation. Comment. Banach spaces are particular cases of Frechet spaces, which were defined as locally convex (F)-spaces. Therefore, the results from TVS IV and LCVS III will all be relevant here. In particular, when one checks completeness, the condition that a net (x λ ) λ Λ X is Cauchy, is equivalent to the metric condition: (mc) for every ε > 0, there exists λ ε Λ, such that x λ x µ < ε, λ, µ λ ε. As pointed out in TVS IV, this condition is actually independent of the norm, that is, if one norm. satisfies (mc), then so does any norm equivalent to.. The following criteria have already been proved in TVS IV. Proposition 1. For a normed vector space (X,. ), the following are equivalent. (i) X is a Banach space. (ii) Every Cauchy net in X is convergent. (iii) Every Cauchy sequence in X is convergent. (iv) Every Cauchy sequence in X has a convergent subsequence. (v) The norm. satisfies the Summability Test: (st) every sequence (x n ) n=1 X, with the property n=1 x n+1 x n <, is convergent. (v ) Any norm which is equivalent to. has property (st). 1

2 Proof. This is a special case of Proposition 3 from TVS IV. The next results (Remark 1 and Propositions 2-6 below) illustrate several standard methods for constructing Banach spaces. Remark 1. The completeness condition for a normed vector space (X,. ) is topological. In other words, if (Y,. ) is a Banach space, and there exists a topological linear isomorphism between X and Y (relative to the norm topologies), then X is also a Banach space. Proposition 2. If (X,. ) is a normed vector space, then its completion X, equipped with the canonical extension. of. is a Banach space. (See the proof for the construction of..) Proof. The norm. is the unique uniformly continuous map X [0, ), satisfying the condition Jx = x, x X, where J : X X denotes the natural inclusion. The fact that. is a norm on X, and the fact that it defines the completion topology, is well known, either from Exercises from LCVS III, or from the theory of metric spaces, since the natural metric d on X, implemented by the norm metric d(x, y) = x y on X, is obviously given by d(v, w) = v w. Convention. From now on, a normed vector space (X,. ) will always be regarded as a dense linear subspace in its completion X, whose norm will be denoted again by.. This way, the norm on X is the induced norm from X. Using this convention, we agree that, if X is already a Banach space, then X = X. (Strictly speaking, the map J : X X is an isometric linear isomorphism.) Exercise 1. Suppose X and Y are normed vector spaces, and T : X Y is linear and continuous. Let T : X Ỹ be the unique linear continuous map that extends T. (See TVS IV.) Prove that T = T. Proposition 3. Suppose X is a Banach space, and Y X is a linear subspace. If we equip Y with the norm induced from X, the following are equivalent: (i) Y is a Banach space; (ii) Y is closed in X. Proof. This is a special case of Remark 1 from TVS IV. Proposition 4. Suppose X is a Banach space, and Y X is a closed linear subspace. If we equip the quotient space X /Y with the quotient norm 1, then X /Y is a Banach space. Proof. This is a special case of Remark 3 from TVS IV. Proposition 5. If X 1, X 2,..., X n are Banach spaces, and we equip the product space X = X 1 X 2 X n with the product topology T prod, then X becomes a Banach space, relative to any norm. which has T prod as its associated norm topology. Proof. This is a special case of Remark 2 from TVS IV. 1 See Proposition-Definition 1 from BS I. 2

3 Corollary 1. Given any norm. on a finite dimensional vector space X, it follows that (X,. ) is a Banach space. Proof. Choose n N so that X K n (linear isomorphism). On K n all norms define the product topology, so K n is Banach relative to any norm. On X all norms are equivalent, so they define the same topology. Therefore the linear isomorphism X K n is topological, and everything follows from Remark 1. Comments. The product space construction is limited to the case of finitely many spaces. As it turns out, an infinite product X = i I X i of normed vector spaces is never normable, when equipped with the product topology T prod. (As seen in??, every neighborhood of 0 in (X, T prod ) contains an infinite dimensional linear subspace.) In the finite case, however, there are many norms on X 1 X 2 X n which define the product topology, for instance: (x 1, x 2,..., x n ) 1 = x 1 + x x n (x 1, x 2,..., x n ) p = [ x 1 p + x 2 p + + x n p] 1/p, 1 < p < (x 1, x 2,..., x n ) = max { x 1, x 2,..., x n } In the case of an infinite family (X i ) i I of Banach spaces, it is possible to construct various versions of the direct sum, as will be shown in sub-section E. Proposition 6. Suppose X is a normed vector space, and Y is a Banach space. When equipped with the operator norm, the space L(X, Y) is a Banach space. Proof. Let (T n ) n=1 be a Cauchy sequence in L(X, Y). Using the triangle inequality we have Tm T n Tm T n, m, n, so the sequence ( T n ) C > 0, such that n=1 is Cauchy (in R), thus bounded, so there exists some constant For every x X, using the operator norm inequality we have T n c, n. (1) T m x T n x = (T m T n )x T m T n x, m, n, (2) so the sequence (T n x) n=1 is Cauchy in Y. Since Y is a Banach space, it follows that the sequence (T n x) n=1 is convergent, for every x X. If we define T : X Y by T x = lim n T n x (in the norm topology in Y), then clearly T is linear. Since T x = lim n T n x, and by (1) we have T n x C x, n, it follows that T x C x, x X, so T is indeed continuous. Finally, to prove that (T n ) n=1 converges in norm to T, we use the Cauchy condition, which gives the existence, for any ε > 0, of an index n ε, such that T m T n ε, m, n n ε. 3

4 Using the operator norm inequality (2) we then have T m x T n x ε x, m, n n ε, x X. If we fix n and x and we take lim m in the above inequality, we get (T T n )x = T x T n x ε x, n n ε, x X, which yields T T n ε, n n ε, thus proving that T T n 0. Corollary 2. The topological dual X = (X,. ) of a normed vector space (X,. ) is a Banach space, when equipped with the dual norm. Proof. Immediate from Proposition 6, applied to the Banach space Y = (K,. ). Examples 1-3. Fix some non-empty set S. 1. For all p [1, ], the normed vector space (l p K (S),. p) is a Banach space. This is due to the fact that l p K (S) is isometrically isomorphic to the topological dual a normed vector space X. Specifically: l 1 K (S) c 0,K(S), and l p K (S) lq K (S), for p (1, ], where q is the Hölder conjugate of p. (See BS I, sub-section C.) 2. For p [1, ), the Banach space l p K (S) is also naturally identified as the completion of ( S K,. p). 3. The normed vector space (c 0,K (S),. ) is also a Banach space. This can be derived, for example, from the observation that c 0,K (S) is norm closed in l K (S), being in fact the norm closure of S K. Alternatively, c 0,K(S) is naturally identified as the completion of ( S K,. ). Comment. The fact that the l p -spaces, p [1, ], are Banach spaces can also be proved directly, without their presentations as duals of normed vector spaces. This will be discussed in sub-section E (see Comment following Theorem 7.) This sub-section concludes with a discussion on summability in Banach spaces. All results are left as exercises. Definition. Given a Banach space 2 X, and some (infinite) non-empty set S, a system (x s ) s S S X is said to be summable in X, if the set of finite sums (y F ) F Pfin (S), defined by y F = s F x s, is convergent to some x X. This means that: for every ε > 0, there exists F ε P fin (S), such that y F x < ε, for all F P fin (S) with F F ε. In this case we write s S x s = x. Exercises 2-5. Fix a Banach space X. 2. Prove that, if (x x ) s S is summable in X, if and only if, for every ε > 0, there exists F ε P fin (S), such that s G x s < ε, for all G Pfin (S) with G F ε =. 2 Summability makes sense in an arbitrary topological vector space. We restrict to Banach spaces simply because they offer the only (reasonable) framework in which the properties listed here are true. 4

5 3. Prove that, if (x x ) s S is summable in X, then so is any sub-system (x s ) s S0, for any S 0 S. 4. Prove that, if (x s ) s S is such that s S x s <, then (x s ) s S is summable. 5. Give an example of a Banach space X, in which there exists a summable system (x s ) s S with s S x s =. 6. Prove that if (x n ) n N is summable, then (i) the sequence (z n ) n=1, given by z n = n k=1 x k is convergent (to n N x n); (ii) lim n x n = 0. 7*. Prove that (x n ) n=1 is summable, if and only if, for every permutation σ : N N, the sequence (zn) σ n=1, given by zn σ = n k=1 x σ(k) is convergent. Moreover, in this case it follows that lim n zn σ = n N x n. (Hint: Argue by contradiction, using Exercise 2, by showing that if (x N ) n N is not summable, there exist ε > 0, and a sequence of finite sets F 1 F 2 F 3 N, such that F n n and n F n+1 F n x n ε, n.) 8*. Prove that, if (x s ) s S is summable, then (i) the S-tuple ( x s ) s S belongs to c 0,R(S); (ii) the set {s S : x s 0} is at most countable. (Hint: Use Exercises 3 and 6 to show that, for every ε > 0, the set F ε = {s S : x s ε} is finite.) Comment. As the above exercises show, summability reduces to the case S = N. In this case, for a sequence (x n ) n=1 X we are dealing with three different conditions, each of which will be expressed using series terminology. (a) The sequence of partial sums (z n ) n=1, given by z n = n k=1 x n is convergent, in which case we say that the series n=1 x n is convergent, (b) As an element in N X, the N-tuple (x n) n N is summable. Equivalently (using the terminology above), for every permutation σ, the series n=1 x σ(n) is convergent. For this reason, this condition is sometimes stated by saying that the series n=1 x n is unconditionally convergent, (c) n=1 x n <, in which case we say that the series n=1 x n is absolutely convergent, In a Banach space one has the implications (c) (b) (a). According to the Summability Test, we also know that, if the implication (c) (a) is true, then X is forced to be a Banach space. B. The Five Principles of Banach Space Theory 5

6 Since Banach spaces are a special case of (F)-spaces, the Principles discussed in TVS IV apply to them. In particular, Theorems 1-2 below are direct results of this observation, so they do not require a proof. Theorem 1 (Open Mapping Principle). If X and Y are Banach spaces, and T : X Y is linear, continuous, and surjective, then T is an open mapping, i.e. whenever A is open in X, it follows that T (A) is open in Y. Theorem 2 (Inverse Mapping Principle). If X and Y are Banach spaces, and T : X Y is linear, continuous, and bijective, then the inverse mapping T 1 : Y X is also continuous. Theorem 3 (Closed Graph Principle). If X and Y are Banach spaces, and T : X Y is linear, then the following are equivalent: (i) T is continuous; (ii) the graph of T, i.e. the space G T product topology; = {(x, T x) : x X } is closed in X Y, in the It should be noted here that X Y is Banach space (relative to any norm that has the product topology as the norm topology), so condition (ii) is also equivalent that G T is a Banach space. Theorem 4 (Alaoglu Compactness Theorem). If X is a normed vector space, then for any r > 0, the ball (X ) r = {φ X : φ r} is compact in the w -topology. Proof. This is a special case of Alaoglu-Bourbaki Theorem from LCVS V, sub-section F. Indeed, if we consider the unit ball (X ) 1 = {x X : x 1}, then (X ) 1 is a neighborhood of 0 in X, so the set K r = {φ X : sup φ(x) r} x (X ) 1 is compact in (X, w ), by the Alaoglu-Bourbaki Theorem. Now everything is trivial, since by the definition of the dual norm, we have K r = (X ) r. Theorem 5 (Uniform Boundedness Principle). Suppose X is a Banach space and Y is an arbitrary normed vector space. If Θ L(X, Y) is point-wise bounded, i.e. sup T x <, x X, T Θ then Θ is bounded in the operator norm, i.e. sup T Θ T <. Proof. Of course, the point-wise boundedness condition stated here is the same as the condition: for every x X, the set Θ x = {T x : T Θ} Y is bounded in the norm topology. In particular, we can apply the Equi-continuity Principle (Theorem2 from TVS IV) to conclude that Θ is equi-continuous, which in this case means: 6

7 ( ) for every ε > 0, there exists δ ε > 0, such that x δ ε T x < ε, T Θ. Apply the above condition to ε = 1, and let δ 1 > 0 be the corresponding constant. Notice that if we start with some arbitrary x (X ) 1, then δ 1 x = δ 1 x δ 1, so using ( ) and linearity, we have δ 1 T x = T (δ 1 x) < 1, T Θ, which shows that thus proving that T 1/δ t, T Θ. T x < 1/δ 1, x (X ) 1, T Θ, C. Miscellaneous Results This sub-section contains various application of the five Principles discussed in the previous sub-section. Some of them are formulated as exercises. Another set of applications will be given BS VI. Exercise 9. Suppose X is a Banach space. Prove that for a set A X, the following are equivalent: (i) A is bounded in the w -topology; (ii) A is bounded in the norm topology. Exercise 10*. Show that the condition that X is complete is essential. Specifically, consider X = N K, equipped with the norm., and construct a sequence of linear continuous functionals (φ n ) n=1 X, such that φ n 0, but such that φ n = n, n. On the one hand, the set A = {φ n : n N} X is w -compact (see TVS I), hence w -bounded. On the other hand, however, A is clearly unbounded in the norm topology. Exercise 11. Prove that if X is a Banach space, then every Cauchy sequence in (X, w ) is w -convergent. Definition. A topological vector space (X, T) is said to be sequentially complete, if every T-Cauchy sequence is T-convergent. Using this terminology, Exercise 11 states that the topological dual X of a Banach space X is sequentially complete, when equipped with the w -topology. Exercise 12. Prove that, if X and Y are Banach spaces, then L(X, Y) is sequentially T complete, when equipped with the strong operator topology T so. Recall that T so Λ T T λ x T x, x X. Exercise 13. Prove that, if X is an infinite dimensional Banach space, then the locally convex space (X, w ) is not metrizable. (Hint: Use Exercise 11.) Exercise 14.. Prove that, if X is a separable normed vector space, then the compact space ((X ) 1, w ) is metrizable. (Hint: Let {x n } n=1 be a dense subset in the unit ball (X ) 1. Prove that if (φ λ ) λ Λ and φ are in (X ) 1, then φ λ φ φ λ (x n ) φ(x n ), n. Therefore, if we consider Ω = N K, equipped with the product topology, which is metrizable, the map 7 w w

8 F : (X ) 1 φ ( φ(x n ) ) n=1 Ω is w -continuous and injective, thus a homeomorphism onto its range.) Exercises below make use of the weak topology, introduced 3 in LCVS V. In our particular setting, on a normed vector space X, this topology is denoted simply by w, and by definition we have x λ w x φ(x λ ) φ(x), φ X. Exercise 15. Suppose X is a normed vector space. Prove that, for a set A X, following are equivalent: (i) A is bounded in the w-topology; (ii) A is bounded in the norm topology. (Hint: Use the isometric linear map Θ : X x x # X.) Exercise 16. Let X be a normed vector space. Prove that, for any r > 0, the ball (X ) r is w-closed. Exercise 17. Let X and Y be Banach spaces. Prove that, for a linear map T : X Y, the following are equivalent: (i) T is norm-continuous; (ii) T is weakly continuous, i.e. x λ w x T x λ w T x. (Hint: For (ii) (i) use the Closed Graph Principle.) We conclude this section with a useful technical result that characterizes operators with closed range. Proposition 7. Suppose X and Y are Banach spaces. For an operator T L(X, Y), the following are equivalent: (i) Range T is closed; (ii) there exists C > 0 such that: ( ) for every y Range T, there exists x X with T x = y and x C y. Proof. (i) (ii). Assume T has closed range, and let us show that T has property ( ). Without any loss of generality, we can assume that T is surjective. (Otherwise, we simply replace Y with Range T, which is of course a Banach space, when equipped with the norm coming from Y.) Consider the quotient space Z = X /Ker T, equipped with the quotient norm. We know that Z is a Banach space, and furthermore, if we consider the quotient map π : X Z, there exists a (unique) injective linear continuous map ˆT L(Z, Y), such that ˆT π = T. Of course, since T is surjective, it follows that ˆT is a linear isomorphism. By the Inverse mapping Principle, the inverse operator S = ˆT 1 : Y Z is also linear and continuous. We now check ( ) with C = 1 + S. Start with some y Y, and let z = Sy, 3 Recall that, given a (Hausdorff) locally convex topology T on X, the weak topology w T on X, implemented by T is the smallest topology that makes all φ (X, T) w continuous. In other words, x T λ x means φ(x λ ) φ(x), φ (X, T). 8

9 so that ˆT z = y and z S y. If y 0, then C y = S y + y > z, so by the definition of the quotient norm ( z = inf π(x)=z x ), there exists x X with x C y and π(x) = z, which clearly yields T x = y. The case y = 0 is trivial, because we can choose x = 0. (ii) (i). Assume there exists C > 0, so that T has property ( ), and let us prove that Range T is closed in Y. It suffices to prove that Range T is a Banach space, when equipped with the norm coming from Y, so it suffices to prove that Range T satisfies the Summability Test. Start with a sequence (y n ) n=1 Range T, with n=1 y n <, and let us prove that the series n=1 y n is convergent in Range T. For each n, use condition ( ) to choose some x n X, such that T x n = y n and x n C y n. Obviously we now have n=1 x n n=1 C y n <, so using the fact that X is a Banach space it follows that n=1 x n is convergent in X to some x. By continuity, it follows that n=1 y n = n=1 T x n is convergent in Y to y = T x, and we are done. Corollary 3. If X and Y are Banach spaces, and T L(X, Y) is injective, then the following are equivalent: (i) Range T is closed; (ii) T is bounded from below, i.e. there exists D > 0, such that T x D x, x X. Proof. We know that condition (i) is equivalent to the existence of some C > 0, satisfying condition ( ). Notice, however, that since T is injective, any y Range T can be uniquely represented as y = T x with x X, so condition ( ) is equivalent to x C T x, x X, (3) which clearly is equivalent to the boundedness from below condition. D. The Dual Banach Spaces l K (S) and Related Issues In this sub-section we take a closer look at the topological duals of the Banach spaces of the form l K (S). Most results concerning these spaces are left as Exercises. The only results presented with proofs are those concerning the inclusion c 0,K (S) l K (S) an the Projection Problem mentioned in LCVS V. Notations. Throughout this sub-section S will denote an infinite set. We denote by P(S) the set of all subsets of, and by P fin (S) the set of all finite subsets of S. For every A P(S), we denote by χ A the indicator function of A, defined as χ A (s) = 1, if s A, and χ A (s) = 0, if s A. Clearly, χ A l K (A). Furthermore, if A P fin(s), then χ A S K c 0,K (S). We denote the linear span span{χ A : A P(S)} by elem K (S), and we declare the functions that belong to this space elementary. Exercise 18. Prove that elem K (S) is dense in l K (S) in the norm topology. Exercises Declare a map ν : P(S) K finitely additive, if whenever A 1,..., A n P(S) are disjoint, it follows that 4 ν(a 1 A n ) = ν(a 1 ) + + ν(a n ). 4 It is pretty obvious that, in order to check this, one only needs to consider the case n = 2. 9

10 19. Prove that, if ν : P(S) K is finitely additive and bounded, then for every A P(S), the quantity ν (A) = sup { n ν(a j ) : n N, A 1,..., A n disjoint, j=1 n A j = A } [0, ] j=1 is finite. (Hint: One can assume K = R. In this case, show that, if one defines the quantity M = sup A P(S) ν(a), then ν (A) 2M, A.) 20. Denote by F K (S) the space of all bounded finitely additive maps P(S) K. Prove that, if ν F K (S), then ν F R (S). 21. Prove that F K (S) is a vector space, when equipped with point-wise addition: (ν 1 + ν 2 )(A) = ν 1 (A) + ν 2 (A); point-wise scalar multiplication: (αν)(a) = αν(a). Prove that the map F K (S) ν ν (S) [0, ) defines a norm on F K (S), denoted by. var. This norm is referred to as the variation norm. 22. Prove that, for a map ν : P(S) K, the following are equivalent: (i) ν is finitely additive; (ii) there exists a linear functional s : elem K (S) K, such that moreover the map s is unique. s(χ A ) = ν(a), A P(S); (4) 23. Prove that, if ν F K (S), then there exists a unique linear continuous map s ν : l K (S) K, satisfying (4), and furthermore, one has: s ν = ν var. (Hint: Use Exercise 20 combined with Exercise 16.) 24. Using the notations from the preceding Exercise, prove that the correspondence is an isometric linear isomorphism. F K (S) ν s ν l K (S) Notation. Using the notation from Exercise 24, given ν F K (S), and x l K (S), the quantity s ν (x) K is denoted by SS x dν. One thinks the operation S as a version of integration. Comments. Among the maps in F K (S) one finds the σ-additive ones. These are the (finite) K-valued measures on P(S), which are identified with the vectors in l 1 K (S), as follows: for x = (x s ) s S l 1 K (S), one defines νx : P(S) K by ν x (A) = s A x s. It is not hard to see that, SS y dµ x = x s y s, y = (y s ) s S l K (S), S y dµ x = x # (y) = s S 10

11 where the map x # : l K (S) K comes from the dual pairing between l1 K (S) and l K (S), and the honest integral in the middle is in the sense of the classical theory of measure and integration. This gives rise to a linear isometric inclusion J : l 1 K(S) x x # l K (S). (5) There exist, however, maps ν F K (S) which are not σ-additive (see Exercise 25 and the Comment that precedes it), therefore the map (5) is not surjective. Composing the restriction map l K (S) φ φ c0,k (S) c 0,K(S) (6) with the isometric linear isomorphism c 0,K (S) l 1 K (S) gives rise to a linear continuous map E : l K (S) l1 K (S). If we consider the linear isometric inclusion (5), then we clearly have EJ = Id l 1 K (S). So, when we view using J the space l 1 K (S) as a closed linear subspace in l K (S), the map E is a projection of l K (S) onto l 1 K (S), which satisfies the equality Eφ = φ c0,k (S), (7) where the norm on the right-hand side is computed in c 0,K (S). Let us agree with the following convention, which refers to a functional φ l K (S) : We say that φ originates in l 1 K (S), if φ Range J, i.e. there exists x l1 K (S) such that φ(y) = s S x sy s, y = (y s ) s S l K (S). Remark that, a necessary condition5 for a functional φ l K (S) to originate in l 1 K (S) is the equality φ = Eφ (8) Exercise 25*. Let U be a free ultrafilter on S, i.e. an ultrafilter that contains the filter F = {S A : A P fin }. Consider the limit operation lim U : l K (S) K and the map µ U : P(S) K defined by { 1 if A U µ U (A) = 0 if A U Prove that (i) µ U F K (S); (ii) lim U (y) = SS y dµ U, y l K (S). (iii) lim U c0,k (S) = 0, but lim U 1 = 1. Conclude that so µ U is not σ-additive, because lim U does not satisfy (8). In what follows we are going to discuss several unusual features of the dual spaces l K (S), the most intriguing being the one stated in Theorem 7 below. In preparation for it we introduce the following. 5 As we shall see in BS III, this condition is also sufficient. 11

12 Notations and Conventions. Given some A P(S), we identify, for each p [1, ] the space l p K (A) as a closed linear subspace in lp K (S) as follows. Given an A-tuple x = (x s ) s A l p K (A), we identify it with the S-tuple x = ( x s) s S l p K (S) given by { xs if s A x s = 0 if s A Similarly we can view c 0,K (A) as a closed linear subspace in c 0,K (S). Another way to make these identifications is to consider the multiplication operators Q p A : lp K (S) y χ A y l p K (S), which are linear, continuous (in fact one has Q p A = 1, if A ), idempotent, i.e. (Qp A )2 = Q p A. With the help of this operator lp K (A) is identified with6 Range Q p A. Likewise, if one restricts Q A to c 0,K(S) one obtains an linear continuous idempotent operator Q 0 A : c 0,K(S) c 0,K (S), whose range is identified with c 0,K (A). When there is no danger of confusion, the superscript will be omitted, so all these idempotents will simply be denoted by Q A. Thinking l p K (A) as a closed linear subspace in lp K (S), and c 0,K(A) as a closed linear subspace of c 0,K (S), allows us to define the dual restriction operations l p K (S) φ φ A l K (A) (9) c 0,K (S) ψ ψ A c 0,K (A) (10) as φ A = φ and ψ l p K (A) A = ψ c0,k. For p =, another way of defining (9) is to use bounded (A) finitely additive maps: if φ is represented as φ(y) = SS y dν, for some ν F K (S), then we can restrict ν to P(A), thus producing a ν A F K (A), and then φ A (y) = SA y dν A. For the alternative equivalent definitions of (9) with 1 p < and of (10), one can employ the ordinary restriction map l p K (S) y = (y s) s S y A = (y s ) s A l p K (A) and the linear isometric identifications of the dual spaces (l p ) l q (p, q > 1 and 1 p + 1 q = 1), (l 1 ) l, and (c 0 ) l 1, so we have (y # ) A = (y A ) #, y l (l 1 ) (y # ) A = (y A ) #, y l q (l p ) (y # ) A = (y A ) #, y l 1 (c 0 ) Remark 2. The map E : l K (S) l 1 K (S) is compatible with (dual) restrictions to subsets of S. Namely, if we start with some (non-empty) A P(S), then (use subscripts to indicate the ambient set): E A φ A = (E S φ) A, φ l K (S). (11) Furthermore, if φ l K (S) has the property that φ A originates in l 1 K (A) for some A P(S), then φ B originates in l 1 K (B), for every B P(A). Exercise 26. Fix some φ l K (S). 6 Since Q p A is idempotent, Range Qp A is a closed linear subspace. 12

13 (i) Prove that the map P(S) A φ A [0, ) is bounded and finitely additive. (ii) Prove that if (A i ) i I P(S) is a system of disjoint subsets, then i I φ A i φ. (iii) Prove that, if A 1,..., A n P(S) are such that φ Aj originates in l 1 K (A j), j = 1,..., n, then if we consider the union A = A 1 A n, it follows that φ A originates in l 1 K (A). Exercise 27. For every A P(S), consider the idempotent Q A introduced previously. : l K (S) l K (S) (i) Prove that the correspondence P(S) A Q A L(l K (S)) is finitely additive, in the sense that, whenever A 1,..., a n P(S) are disjoint, it follows that Q A1 A n = Q A1 + + Q An. Prove also that, if A, B P(S), then Q A Q B = Q B Q A = Q A B. (ii) Given φ l K (S), prove that the functional 7 φ Q A l K (S) has the following properties: (a) φ Q A = φ A, (b) φ Q A originates in l 1 K (S), if and only if φ A originates in l 1 K (A). Exercise 28. Let (φ λ ) λ Λ l K (S) be a net which converges to 0 in the w -topology, and let A P(S). (i) Prove that the net of (dual) restrictions ((φ λ ) A ) λ Λ l K (A) also converges to 0 in the w -topology. (ii) Prove that, if A is finite, then ((φ λ ) A ) λ Λ also converges to 0 in the norm topology. Lemma 2. Let I be an infinite set, and let (A i ) i I P(S) be a collection of disjoint subsets. Suppose (φ n ) n=1 is a sequence in l K (S), such that (φ n ) Ai originates in l 1 K (A i), n N, i I. Then there exists an infinite subset M I, so that when we consider the union B = i M A i, it follows that (φ n ) B originates in l 1 K (B), n N. Proof. For the sake of avoiding an unnecessary repetition, we will assume that φ 1 = 0. During the proof, let us agree to use the following notation. For any subset M I, we denote the union i M A i by A(M). Claim 1: There exists a sequence I 1 I 2 I 3... of infinite subsets of I, such that, for every k N we have (φ j ) A(Ik ) 2 k, j {1,..., k}. (12) The desired sequence is constructed recursively as follows (since φ 1 = 0, we can start with I 1 = I). Assume we have constructed the infinite sets I 1 I n, so that (12) holds 7 The functional φ Q A is of course the same as Q A φ. 13

14 for k = 1,..., n, and let us indicate how the next set I n+1 I n is constructed. Since I n is infinite, we can write it as a disjoin union I n = m=1 T m of a sequence of infinite subsets of I n. Since the sets (A(T m )) m=1 are disjoint, by Exercise 27 we know that, for every φ l K (S) we have m=1 φ A(T m) <, which implies that In particular, we can find m, such that lim φ A(T m m) = 0 φ l K (S) (φ j ) A(Tm) 2 (n+1), j {1,..., n + 1}, and we will set I n+1 to be this T m. Having proved Claim 1, let us now choose a sequence (i n ) n=1 I of indices, such that i n I n I n+1, n, and let us from the sets M = {i n : n N} and B = A(M). For each n, let M n = {i 1,..., i n }, B n = A(M n ), R n = M M n, and D n = A(R n ), so that we have B = B n D n (disjoint union). Claim 2: For every j N one has lim n φ j Q Dn = 0. Indeed, if we fix j, and we start with some n > j, then by construction R n I n, so D n = A(R n ) A(I n ), and then by Claim 1 (use Exercises 26 and 27, plus the fact that Q Dn 1) we have φ j Q Dn = (φ j Q A(In)) Q Dn φ j Q A(In) = (φ j ) A(In) 2 n, n j. To finish the proof, we now notice that, since B n = A i1 A in, by Exercises 26 and 27 it follows that φ j Q Bn originates in l 1 K (S), for every n, j N. Then the obvious equality φ j Q B = φ j Q Bn + φ j Q Dn, combined with Claim 2, yields lim n φ j Q Bn = φ j Q B, in norm in l K (S), therefore 8 φ j Q B originates in l 1 K (S), j. Theorem 6. (R.S. Phillips). Suppose (x n ) n=1 is a sequence in l K (S), which converges to 0 in the w -topology. Then the sequence (Eφ n ) n=1 converges to 0 in l 1 K (S) in the norm topology. Proof. We argue by contradiction, assuming that the sequence (Eφ n ) n does not converge in norm to 0. Passing to a subsequence, and multiplying by a bounded sequence of scalars, we can assume that Eφ n 1 = 1, n. (13) Claim 1: There exists a sequence of integers 1 = r 1 < r 2 <..., and a sequence (B n ) n=0 P fin (S), consisting of disjoint sets, such that ( ) (φ rn ) Bn 1 2 n, n N; ( ) if we take the union B = n=1 B n, then (φ k ) B originates in l 1 K (B), k N. 8 Of course, the set of all functionals in l that originate in l 1 is norm-closed, being the range of the isometric linear map (5). 14

15 To prove the above statement, we start off by constructing a sequence of integers 1 = k 1 < k 2 <... and a sequence of finite sets = F 0 F 1 F 2 S, such that, for every n N one has: (a) (Eφ kn ) Fn (n+1) ; (b) (φ m ) Fn 1 2 (n+1), m k n. The two sequences are constructed recursively as follows. Let us assume the integers 1 = k 1 < k 2 < < k N and the sets F 0,..., F N 1 have been constructed, satisfying (a) for 1 n N 1, and (b) for 1 n N, and let us indicate how F N and k N+1 are constructed, so that (a) holds for n = N and (b) holds for n = N + 1. (We start with k 1 = 1 and F 0 =, with nothing else to verify.) To construct the set F N, we use the fact that for any x l 1 K (S), one has9 x 1 = lim G Pfin (S) x G 1. If we apply this to x = Eφ kn, then there exists G P fin (S), such that (Eφ kn ) H 1 Eφ kn 2 (N+1) = 1 2 (N+1), H G thus (a) holds for n = N, if we put F N = F N 1 G. To construct k N+1 so that (b) holds for n = N + 1, all we need to do is to use Exercise 28, which gives lim m (φ m ) FN = 0. Having constructed our two sequences (k n ) n=1 and (F n ) n=0 satisfying (a) and (b) for all n, we now consider the finite sets A n = F n F n 1. Let us notice that, by Exercise 26 we have (φ kn ) An = (φ kn ) Fn (φ kn ) Fn 1, so using (a), (b), and the obvious inequality (φ kn ) Fn (Eφ kn ) Fn 1, we have (φ kn ) An (Eφ kn ) Fn 1 2 (n+1) 1 2 n. (14) Since all the A s are finite, it is trivial that (φ k ) An originates in l 1 K (A n), for all k, n. Using Lemma 2, there exists a sequence of integers 1 p 1 < p 2 <..., such that if we take the union B = n=1 A p n, then (φ k ) B originates in l 1 K (B), for every k. If we put B n = A pn and r n = k pn then the Claim follows, since by (14) we have 10 (φ rn ) Bn = (φ kpn ) Apn 1 2 pn 1 2 n. Having proved Claim 1, let us now restrict everything to B, and let us denote, for simplicity, the functionals (φ rn ) B l K (B) by ψ n. To summarize what we have constructed so far, we have B = n=1 B n disjoint union of finite sets and a sequence (ψ n ) n=1 l K (B), with all ψ n s originating in l 1 K (B), such that ψ w n 0, and (i) ψ n 1, (ii) (ψ n ) Bn 1 2 n, for all n N. (To check condition (i) we notice that, since ψ n = (φ rn ) B originates in l 1, using the assumption (13) we have ψ n = (Eφ rn ) B 1 Eφ rn = 1.) 9 Here we view P fin (S) as a directed set, with the direct inclusion. 10 Of course, p n n, n, which yields 1 2 pn 1 2 n. 15

16 For every n N let x n = (x n s ) s B l 1 K (B) be B-tuple that represents ψ n, i.e. ψ n (y) = x # n (y) = s B x n s y s, y = (y s ) s B l K (B), so that lim n [ ] x n s y s = 0, y = (ys ) s B l K (B), (15) s B and the above two properties read: (i ) s B xn s 1, (ii ) s B n x n s 1 2 n, for all n. Let us consider the B-tuple y = (y s ) s B l K (B) defined by x n s if s B x y s = n n and x n s 0 s 0 if s B n and x n s = 0 Claim 2: lim n [ s B xn s y s ] = 1. Using (i ) and (ii ) we clearly have s B B n x n s 2 n, n. So if we fix for the moment n, using the inequalities y s 1, s B, we have x n s y s x n s y s = x n s y s x n s y s x n s 2 n. s B s B n s B B n s B B n s B B n This means that in order to prove the Claim, it suffices to prove that the sequence (θ n ) n, given by θ n = s B n x n s y s, converges to 1. But this is pretty clear, since θ n = s B n x n s, so by properties (i ) and (ii ) above we have the inequalities 1 θ n 1 2 n. Having proved Claim 2, it is now obvious that we reached a contradiction, because condition (15) is clearly violated. Comment. Theorem 6 is fascinating even in one of it special cases (which in fact was employed in its proof): if (x n ) n=1 is a sequence in l 1 K (S), such that x w n 0, i.e. φ(x n ) 0, φ l 1 K (S), then x n 1 0. (Using the dual pairing between l 1 and l, the condition x w n 0 is equivalent to the condition x # w n 0 in l K (S) ). Of course, one cannot expect the implication x w λ 0 x λ 1 0 to be true for arbitrary nets (x λ ) λ in l 1 K (S), because this would force the weak topology to coincide with 16

17 the norm topology. This is an indication that the sequential nature of Theorem 6 is a deep phenomenon. More on this will be discussed in BS?? We now conclude with two interesting applications of Theorem 6. The first one formulated as an Exercise is an application to topology, which shows that sequences and subsequences are inadequate for compactness. Exercise 29.. Suppose S is infinite, and (s n ) n=1 is a sequence of distinct points in S, i.e. s m s n, m n. Define, for every n, the functional φ n : l K (S) y = (y s) s S y sn K. We know that the φ n s are linear and continuous. In fact, they also originate in l 1 K (S), so one has φ n = Eφ n 1 = 1, n. Furthermore, by Exercise 7 from BS I, it follows that φ n c0,k (S) w 0, in c 0,K (S). (16) Prove that the sequence (φ n ) n=1 does not converge in the w -topology to any φ l K (S). Conclude that in the compact Hausdorff space ( (l K (S) ) 1, w ) there exists sequences which w do not have convergent subsequences. (Hint: Argue by contradiction, assuming φ n φ for some φ l K (S). Using Theorem 6, this forces Eφ n Eφ 1 0. By (16), however, one has Eφ = 0, so we will also have Eφ n 1 0, which is impossible.) The next application concerns the Projection Problem (see LCVS V). Proposition 8. If S is infinite, there is no linear continuous projection of l K (S) onto c 0,K (S). Proof. Argue by contradiction. Assume there exists a linear continuous operator P : l K (S) c 0,K (S), such that P x = x, x c 0,K (S). Taking the transpose, we obtain a linear continuous operator P : (c 0,K (S), w ) (l K (S), w ), such that (P γ) c0,k (S) = γ, γ c 0,K(S). (17) The big point here is the fact that P is continuous, with respect to the w -topologies. Fix now a sequence (s n ) n=1 of distinct points in S. As remarked in Exercise 29, we know that the functionals γ n : c 0,K (S) y = (y s ) s S y sn K are linear, continuous, of norm 1, and they form a sequence which is w -convergent to 0 in c 0,K (S). By the w -continuity of P, it follows that the sequence (ψ n ) n=1 l K (S), given by ψ n = P γ n, is w -convergent in l K (S) to 0, so by Theorem 6, it follows that Eψ n 1 0. Of course, using the isometric isomorphism Θ : l 1 K (S) c 0,K (S), and (17), which implies Θ(Eψ n ) = γ n, we now get γ n 0, which is impossible. E. Vector valued l p -spaces In this sub-section we introduce several versions of Banach direct sums and products, for infinite systems of Banach spaces. Such constructions are heavily dependent on the norms on the Banach spaces involved, so they are not topological. Definitions. Assume S is some non-empty set and (X s,. s ) s S is an S-tuple of normed vector spaces. 17

18 A. The space of all S-tuples x = (x s ) s S s S X s, with sup s S x s Xs <, is denoted by l (X s,. s ) s S. For x l (X s,. s ) s S, the above supremum is denoted by x. It is pretty clear that l (X s,. s ) s S is a linear subspace in s S X s, and that. is a norm on l (X s,. s ) s S. B. The space of all x = (x s ) s S s S X s, for which lim F Pfin (S) [ sups S F x s s ] = 0, is denoted by c 0 (X s,. s ) s S. It is pretty clear that c 0 (X s,. s ) s S is a linear subspace in l (X s,. s ) s S. C. Given p [1, ), the space of all x = (x s ) s S s S X s, for which s S x s p s < is denoted by l p (X s,. s ) s S. It is pretty clear that l p (X s,. s ) s S is contained in c 0 (X s,. s ) s S. For x = (x s ) s S l p (X s,. s ) s S, we define x p = [ s S x s p s] 1/p. When there is no confusion about the norms on the spaces X s, s S, they will be omitted from the notation. It is recommended, however, that we do not do that, since a change in norms will produce different spaces! Theorem 7. Assume all spaces (X s,. s ), s S are Banach spaces. (i) The normed vector space ( l (X s,. s ) s S,. ) is a Banach space. (ii) The space c 0 (X s,. s ) s S is norm-closed in ( l (X s,. s ) s S,. ). Specifically, c 0 (X s,. s ) s S is the norm-closure of the direct sum 11 s S X s. In particular, the normed vector space ( c 0 (X s,. s ) s S,. ) is also a Banach space. (iii) For every p [1, ): the space l p (X s,. s ) s S is a linear subspace in c 0 (X s,. s ) s S ; the map. p : l p (X s,. s ) s S [0, ) is a norm; the normed vector space ( l p (X s,. s ) s S,. p ) is a Banach space. Proof. (i). Assume (x n ) n=1 is a Cauchy sequence in ( ) l (X s,. s ) s S,., and let us prove it is convergent. Using the triangle inequality, we have xm x n xm x n, m, n 1, so the sequence ( x n ) n=1 is Cauchy, thus convergent in [0, ), so in particular the quantity M = sup n N x n is finite. For each n N, write x n = (x ns ) s S, with x ns X s. By the construction of the norm., we have the inequalities x ms x ns s x m x n, m, n N, s S, (18) so in particular it follows that, for each s S, the sequence (x ns ) n=1 is Cauchy in X s, thus convergent to some x s X s. Consider then the vector x = (x s ) s S s S X s, and let us prove that: (a) x l (X s,. s ) s S, and (b) lim n x x n = 0. To prove these 11 It is trivial that s S X s is contained in l (X s,. s ) s S. 18

19 features, we fix for the moment some ε > 0, and we use the Cauchy condition, to produce n ε N, such that x m x n < ε, m, n n ε, so by (18) we have x ms x ns s ε, m, n n ε, s S. If we keep n and s fixed, and take lim m this yields On the one hand, using the triangle inequality, we have x s x ns s ε, n n ε. (19) x s s x s x ns s + x ns s ε + x n ε + M, s S, thus proving that x = (x s ) s S indeed belongs to l (X s,. s ) s S. On the other hand, taking sup s S in (19) clearly yields x x n ε, n n ε, thus proving (b). (ii). This statement is pretty clear, the details being left to the reader. (iii). Fix p [1, ). It is pretty obvious that, if x l p (X s,. s ) s S, then αx l p (X s,. s ) s S, and furthermore one has αx p = α x p, α K. To prove the first two statements in (iii), it suffices them to show that ( ) whenever x, y l p (X s,. s ) s S, it follows that x+y l p (X s,. s ) s S, and furthermore one has the inequality x + y p x p + y p. (20) This above statement is, however, pretty obvious, by considering the S-tuples u = (u s ) s S, v = (v s ) s S, w = (w s ) s S s S R, given by u s = x s s and v s = y s s, w s = x s + y s s, which will satisfy the following conditions u and v both belong to l p R (S), 0 w s u s + v s, which imply that w also belongs to l p R (S) thus showing that x + y indeed belongs to l p (X s,. s ) s S as well as the inequality w p u p + v p, which is precisely (20) Assume now (x n ) n=1 is a Cauchy sequence in ( ) l p (X s,. s ) s S,. p, and let us prove it is convergent. Arguing exactly as in (i), it follows that the quantity M = sup n N x n p is finite. For each n N, write x n = (x ns ) s S, with x ns X s. By the construction of the norm. p, we have the inequalities x ms x ns p s x m x n p p, m, n N, F P fin (S), (21) s F so in particular using singeltons F = {s}, s S it follows that, for each s S, the sequence (x ns ) n=1 is Cauchy in X s, thus convergent to some x s X s. Consider then the vector x = (x s ) s S s S X s, and let us prove that: (a) x l p (X s,. s ) s S, and (b) 19

20 lim n x x n p = 0. To prove these features, we fix for the moment some ε > 0, and we use the Cauchy condition, to produce n ε N, such that x m x n p < ε, m, n n ε, so by (21) we have x ms x ns p s ε p, m, n n ε, F P fin (S). s F If we keep n and F fixed, and take lim m this yields x s x ns p s ε p, n n ε. (22) s F On the one hand, using the triangle inequality in X s, we have x s s x s x ns s + x ns, so using the triangle inequality in l p R (F ), we get have [ 1/p [ ] [ ] 1/p x s s] p p xs x ns s + x ns s s F s F [ 1/p [ /1p x s x ns s] p + x ns s] p s F s F ε + x n p ε + M, F P fin (S), so taking sup F Pfin (S) we obtain s S x s p s (ε+m) p, thus proving that x = (x s ) s S indeed belongs to l p (X s,. s ) s S. On the other hand, taking sup F Pfin (S) in (22) clearly yields thus proving (b). x x n p ε, n n ε, Comment. The spaces introduced above are to be understood as vector-valued generalizations of the l p -spaces. In the case when (X s,. s ) = (K,. ), s S, the spaces l p (K,. ) s S, p [1, ], and c 0 (K,. ) s S, coincide with l p K (S), and c 0,K(S), respectively. Remarks 3-5. Let (X s,. s ) s S be an S-tuple of normed vector spaces. 3. Exactly as in the scalar case, the direct sum s S X s is a dense linear subspace in c 0 (X s,. s ) s S and in all l p (X s,. s ) s S, p [1, ). 4. The norm topology on each one of the spaces c 0 (X s,. s ) s S and l p (X s,. s ) s S, p [1, ] is stronger than the induced product topology from s S X s. This follows from the observation that the coordinate maps E t : l p (X s,. s ) s S x = (x s ) s S x t (X t,. t ) are contractive, for all t S and every p [1, ]. 5. For every t S, the inclusion map J t : X t s S X s is isometric, when the target space is equipped with any one of the norms. p, p [1, ]. This way we see that (X t,. t ) is isometrically represented as a closed linear subspace in either one of the spaces c 0 (X s,. s ) s S or l p (X s,. s ) s S, p [1, ]. 20

21 6. If (X,. ) denotes either one of the spaces c 0 (X s,. s ) s S or l p (X s,. s ) s S, p [1, ], one can use the isometric linear maps J t : (X t,. t ) (X,. ) to define a map Ψ : X s S X s by Ψ(φ) = (Js φ) s S = (φ J s ) s S, φ X. Theorem 8 below gives more information on the map Ψ in two special cases of interest. Theorem 8. Let (X s,. s ) s S be an S-tuple of normed vector spaces. For every s S, let (X s,. s) denote the topological dual, where. s is the functional norm. (i) If p [1, ) and q (1, ] is the Hölder conjugate of p, then the map Ψ : [ l p (X s,. s ) s S ] φ (φ Js ) s S s S X s has the following properties (a) Range Ψ = l q (X s,. s) s S ; (b) Ψ(φ) q = φ, φ l p (X s,. s ) s S. Therefore, the map Ψ establishes an isometric linear isomorphism (ii) The map has the following properties Ψ : [ l p (X s,. s ) s S ] l q[ (X s,. s) s S ]. Ψ 0 : [ c 0 (X s,. s ) s S ] φ (φ Js ) s S s S X s (a) Range Ψ 0 = l 1 (X s,. s) s S ; (b) Ψ 0 (φ) 1 = φ, φ [ c 0 (X s,. s ) s S ]. Therefore, the map Ψ 0 establishes an isometric linear isomorphism Ψ 0 : [ c 0 (X s,. s ) s S ] l 1 (X s,. s) s S. Proof. The proof will be divided into several steps, which are ordered somehow differently from the statement. Claim 1: If p (1, ), and φ [ l p (X s,. s ) s S ], then: In particular, it follows that Range Ψ l q (X s,. s) s S, and ( φ J s s) q φ q. (23) s S Ψ(φ) q φ, φ [ l p (X s,. s ) s S ]. (24) 21

22 The desired inequality (23) is equivalent to s S 0 ( φ J s s) q φ q, (25) where S 0 = {s S : φ J s 0}. Fix some set F P fin (S 0 ) and some ε > 0, small enough so that φ J s s > ε, for all s F. Consider the vector u = (u s ) s F l q R (F ), given by u s = φ J s s ε, s S Using the isometric isomorphism l q R (F ) lp R (F ), there exists a vector v = (v s ) s F l p R (F ), such that v p 1, and such that s F u sv s u q ε. Since u s > 0, and s F u s v s s F u sv s, we can in fact assume that all vs, s F are non-negative, so in fact we have the inequality u s v s u q ε. s F Using the definition of the functional norm, for every s F, for which φ J s = 0, there exists x s X s, such that (φ J s )(x s ) > φ J s s ε x s s = u s x s s. Replacing x s with a suitable scalar multiple α s x s, we can assume that (φ J s )(x s ) 0, and furthermore, that x s s = v s. Consider not the vector x = s F J sx s s S X s l p (X s,. s ) s S. By construction (use the fact that (φ J s )(x s ) = (φ J s )(x s ) ), we have φ(x) = s F (φ J s )(x s ) s S u s v s u q ε. (26) Of course, we have x p = [ s F x s p s] 1/p = v q 1, which implies φ(x) φ x p φ, so by (26) we now get u q φ + ε, which yields: Taking lim ε 0 this will imply ( φ J s s ε) q ( φ + ε) q. s F ( φ J s s) q φ q. s F Taking sup F Pfin (S 0 ) in the above inequality yields (25), and the Claim follows. Claim 2: If φ [ c 0 (X s,. s ) s S ], then: φ J s s φ. (27) In particular, one has the inclusion Range Ψ 0 l 1 (X s,. s) s S, and s S Ψ 0 (φ) 1 φ, φ [ c 0 (X s,. s ) s S ]. (28) 22

23 The proof is almost identical to the proof of Claim 1. Fix some set F P fin (S) and some ε > 0, small enough so that φ J s s > ε, s F. Consider the vector u = (u s ) s F l 1 R (F ), given by u s = φ J s s ε, s F. Using the definition of the functional norm, for every s F, there exists x s X s, such that (φ J s )(x s ) > ( φ J s s ε) x s s = u s x s s. Replacing x s with a suitable scalar multiple α s x s, we can assume that (φ J s )(x s ) 0, and furthermore, that x s s = 1. Consider not the vector x = s F J sx s s S X s c 0 (X s,. s ) s S. By construction (use the fact that (φ J s )(x s ) = (φ J s )(x s ) ), we have φ(x) = s F (φ J s )(x s ) s S u s u 1. (29) Of course, we have x = 1, which implies φ(x) φ x 1 φ, so by (29) we now get u 1 φ, which yields: ( φ J s s ε) φ. Taking lim ε 0 this will imply s F φ J s s φ, s F and then taking sup F Pfin (S) in the above inequality we get s F φ J s s φ, and the Claim follows. Claim 3: If φ [ l 1 (X s,. s ) s S ], then: max s S φ J s s φ. (30) In particular, for p = 1, one has the inclusion Range Ψ l (X s,. s) s S, and Ψ(φ) φ, φ [ l 1 (X s,. s ) s S ]. (31) This statement is trivial, since the inclusion J s : X s l 1 (X s,. s ) s S is isometric, so φ J s s φ J s = φ, s S. Having proved the above three Claims, we see that the remaining issues that need to be addressed are the surjectivity of Ψ and Ψ 0, as well as the inequalities φ Ψ(φ) q, φ [ l p (X s,. s ) s S ] ; (32) φ Ψ 0 (φ) 1, φ [ c 0 (X s,. s ) s S ]. (33) As it turns out, one can use a unified approach for treating these issues. Assume p, q [1, ] are Hölder conjugate, so now we also include the case (p, q) = (, 1), and ψ = (ψ s ) s S belongs to l q (Xs,. s) s S. Define the vector u = (u s ) s S l q K (S), by u s = ψ s s. Notice now that, for every x = (x s ) s S l p (X s,. s ) s S, the vector v = (x s ) s S, defined by 23

24 v s = x s s, s S, belongs to l p R (S), therefore the vector (u sv s ) s S belongs to l 1 R (S), and satisfies (by Hölder) the inequality u s v s u q v p = ψ q x p. s S Notice now that by the operator norm inequality, the vector w = (w s ) s S, defined by w s = ψ s (x s ) satisfies w s = ψ s (x s ) ψ s s x s s = u s v s, so w belongs to l 1 K (S), and satisfies the inequality s S w s u q v p = ψ q x p. This means that, for every x = (x s ) s S l p (X s,. s ) s S, one can define the sum φ(x) = s S ψ s (x s ) K, (34) which satisfies the inequality φ(x) ψ q x p. (35) Of course, this yields a linear continuous functional φ : l p (X s,. s ) s S K, which by construction clearly satisfies φ J s = ψ s, s S, so the surjectiviy of Ψ is clear. For the surjectivity of Ψ 0 one needs φ 0 [ ], c 0 (X s,. s ) s S such that φ0 J s = ψ s. In this case we use (34) with (p, q) = (, 1), and we take φ 0 to be the restriction of φ [ ] l X s,. s ) s S to the subspace c 0 (X s,. s ) s S. The desired inequalities (32) and (33) are immediate from (35). Comment. Theorem 8 does not include the calculation of the dual space [ l (X s,. s ) s S ], which is a very complicated matter, even in the scalar case. The arguments in the proof of Theorem 8, however, show that, applying the construction (34), to vectors ψ = (ψ s ) s S l 1 (X s,. s) s S, one obtains an isometric linear embedding l 1 (X s,. s) s S [ l (X s,. s ) s S ]. Remark 7. In the case when S is finite, one has c 0 (X s,. s ) s S = l p (X s,. s ) s S = s S X s = s S X s, p [1, ], and the norms. p, p [1, ], on s S X s are all equivalent (they all define the product topology). In this case, Ψ : [ s S X ] s s S X s is a linear isomorphism. Theorem 8 is significant even in this case, because it provides an explicit calculation of the functional norm, namely, if p, q [1, ] are Hölder conjugate, then Ψ : [ s S X ] s,. p [ s S X ] s,. q is an isometric linear isomorphism. 24