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 alreadypresented earlier.

Convention. Throughout this note K will be one of the fields R or C, and all vectorspaces 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 omittedfrom the notation.

Comment. Banach spaces are particular cases of Frechet spaces, which were definedas locally convex (F)-spaces. Therefore, the results from TVS IV and LCVS III will allbe relevant here. In particular, when one checks completeness, the condition that a net(xλ)λ∈Λ ⊂ X is Cauchy, is equivalent to the metric condition:

(mc) for every ε > 0, there exists λε ∈ Λ, such that ‖xλ − xµ‖ < ε, ∀λ, µ λε.

As pointed out in TVS IV, this condition is actually independent of the norm, that is, if onenorm ‖ . ‖ 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 (xn)∞n=1 ⊂ X , with the property∑∞

n=1 ‖xn+1 − xn‖ < ∞, is con-vergent.

(v’) Any norm which is equivalent to ‖ . ‖ has property (st).

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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 meth-ods for constructing Banach spaces.

Remark 1. The completeness condition for a normed vector space (X , ‖ . ‖) is topo-logical. In other words, if (Y , ‖ . ‖) is a Banach space, and there exists a topological linearisomorphism between X and Y (relative to the norm topologies), then X is also a Banachspace.

Proposition 2. If (X , ‖ . ‖) is a normed vector space, then its completion X , equippedwith the canonical extension ‖ . ‖ of ‖ . ‖ is a Banach space. (See the proof for the construc-tion of ‖ . ‖ .)

Proof. The norm ‖ . ‖ is the unique uniformly continuous map X → [0,∞), satisfying thecondition ‖Jx‖ = ‖x‖, ∀x ∈ X , where J : X → X denotes the natural inclusion. The factthat ‖ . ‖ is a norm on X , and the fact that it defines the completion topology, is well known,either from Exercises 10-12 from LCVS III, or from the theory of metric spaces, since thenatural metric d on X , implemented by the norm metric d(x, y) = ‖x−y‖ on X , is obviouslygiven by d(v, w) = ‖v − w‖ .

Convention. From now on, a normed vector space (X , ‖ . ‖) will always be regarded asa dense linear subspace in its completion X , whose norm will be denoted again by ‖ . ‖. Thisway, the norm on X is the induced norm from X . Using this convention, we agree that, ifX is already a Banach space, then X = X . (Strictly speaking, the map J : X → X is anisometric linear isomorphism.)

Exercise 1.♥ Suppose X and Y are normed vector spaces, and T : X → Y is linear andcontinuous. Let T : X → Y be the unique linear continuous map that extends T . (See TVSIV.) Prove that ‖T‖ = ‖T‖.

Proposition 3. Suppose X is a Banach space, and Y ⊂ X is a linear subspace. If weequip 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 norm1, then X/Y is a Banach space.

Proof. This is a special case of Remark 3’ from TVS IV.

Proposition 5. If X1,X2, . . . ,Xn are Banach spaces, and we equip the product spaceX = X1 × X2 × · · · × Xn with the product topology Tprod, then X becomes a Banach space,relative to any norm ‖ . ‖ which has Tprod 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

Corollary 1. Given any norm ‖ . ‖ on a finite dimensional vector space X , it followsthat (X , ‖ . ‖) is a Banach space.

Proof. Choose n ∈ N so that X ' Kn (linear isomorphism). On Kn all norms define theproduct topology, so Kn is Banach relative to any norm. On X all norms are equivalent, sothey define the same topology. Therefore the linear isomorphism X ' Kn is topological, andeverything follows from Remark 1.

Comments. The product space construction is limited to the case of finitely manyspaces. As it turns out, an infinite product X =

∏i∈I Xi of normed vector spaces is never

normable, when equipped with the product topology Tprod. (As seen in ??, every neighbor-hood of 0 in (X ,Tprod) contains an infinite dimensional linear subspace.)

In the finite case, however, there are many norms on X1×X2× · · ·×Xn which define theproduct topology, for instance:

‖(x1, x2, . . . , xn)‖1 = ‖x1‖+ ‖x2‖+ · · ·+ ‖xn‖

‖(x1, x2, . . . , xn)‖p =[‖x1‖p + ‖x2‖p + · · ·+ ‖xn‖p

]1/p, 1 < p <∞

‖(x1, x2, . . . , xn)‖∞ = max‖x1‖, ‖x2‖, . . . , ‖xn‖

In the case of an infinite family (Xi)i∈I of Banach spaces, it is possible to construct variousversions 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. Whenequipped with the operator norm, the space L(X ,Y) is a Banach space.

Proof. Let (Tn)∞n=1 be a Cauchy sequence in L(X ,Y). Using the triangle inequality we have∣∣‖Tm‖ − ‖Tn‖∣∣ ≤ ‖Tm − Tn‖, ∀m,n,

so the sequence(‖Tn‖

)∞n=1

is Cauchy (in R), thus bounded, so there exists some constantC > 0, such that

‖Tn‖ ≤ c, ∀n. (1)

For every x ∈ X , using the operator norm inequality we have

‖Tmx− Tnx‖ = ‖(Tm − Tn)x‖ ≤ ‖Tm − Tn‖ · ‖x‖, ∀m,n, (2)

so the sequence (Tnx)∞n=1 is Cauchy in Y . Since Y is a Banach space, it follows that the se-

quence (Tnx)∞n=1 is convergent, for every x ∈ X . If we define T : X → Y by Tx = limn→∞ Tnx

(in the norm topology in Y), then clearly T is linear. Since ‖Tx‖ = limn→∞ ‖Tnx‖, and by(1) we have ‖Tnx‖ ≤ C‖x‖, ∀n, it follows that

‖Tx‖ ≤ C‖x‖, ∀x ∈ X ,

so T is indeed continuous. Finally, to prove that (Tn)∞n=1 converges in norm to T , we use theCauchy condition, which gives the existence, for any ε > 0, of an index nε, such that

‖Tm − Tn‖ ≤ ε, ∀m,n ≥ nε.

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Using the operator norm inequality (2) we then have

‖Tmx− Tnx‖ ≤ ε‖x‖, ∀m,n ≥ nε, x ∈ X .

If we fix n and x and we take limm→∞ in the above inequality, we get

‖(T − Tn)x‖ = ‖Tx− Tnx‖ ≤ ε‖x‖, ∀n ≥ nε, x ∈ X ,

which yields ‖T − Tn‖ ≤ ε, ∀n ≥ nε, thus proving that ‖T − Tn‖ → 0.

Corollary 2. The topological dual X ∗ = (X , ‖ . ‖)∗ of a normed vector space (X , ‖ . ‖) isa 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 (`pK(S), ‖ . ‖p) is a Banach space. This isdue to the fact that `pK(S) is isometrically isomorphic to the topological dual a normedvector space X . Specifically: `1K(S) ' c0,K(S)∗, and `pK(S) ' `qK(S)∗, for p ∈ (1,∞],where q is the Holder conjugate of p. (See BS I, sub-section C.)

2. For p ∈ [1,∞), the Banach space `pK(S) is also naturally identified as the completionof (

⊕S K, ‖ . ‖p).

3. The normed vector space (c0,K(S), ‖ . ‖∞) is also a Banach space. This can be derived,for example, from the observation that c0,K(S) is norm closed in `∞K (S), being in fact thenorm closure of

⊕S K. Alternatively, c0,K(S) is naturally identified as the completion

of (⊕

S K, ‖ . ‖∞).

Comment. The fact that the `p-spaces, p ∈ [1,∞], are Banach spaces can also be proveddirectly, without their presentations as duals of normed vector spaces. This will be discussedin sub-section E (see Comment following Theorem 7.)

This sub-section concludes with a discussion on summability in Banach spaces. All resultsare left as exercises.

Definition. Given a Banach space2 X , and some (infinite) non-empty set S, a system(xs)s∈S ∈

∏S X is said to be summable in X , if the set of finite sums (yF )F∈Pfin(S), defined

by yF =∑

s∈F xs, is convergent to some x ∈ X . This means that: for every ε > 0, thereexists Fε ∈ Pfin(S), such that ‖yF − x‖ < ε, for all F ∈ Pfin(S) with F ⊃ Fε. In this case wewrite

∑s∈S xs = x.

Exercises 2-5. Fix a Banach space X .

2. Prove that, if (xx)s∈S is summable in X , if and only if, for every ε > 0, there existsFε ∈ Pfin(S), such that

∥∥∑s∈G xs

∥∥ < ε, for all G ∈ Pfin(S) with G ∩ Fε = ∅.

2 Summability makes sense in an arbitrary topological vector space. We restrict to Banach spaces simplybecause they offer the only (reasonable) framework in which the properties listed here are true.

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3. Prove that, if (xx)s∈S is summable in X , then so is any sub-system (xs)s∈S0 , for anyS0 ⊂ S.

4. Prove that, if (xs)s∈S is such that∑

s∈S ‖xs‖ <∞, then (xs)s∈S is summable.

5. Give an example of a Banach space X , in which there exists a summable system (xs)s∈S

with∑

s∈S ‖xs‖ = ∞.

6. Prove that if (xn)n∈N is summable, then

(i) the sequence (zn)∞n=1, given by zn =∑n

k=1 xk is convergent (to∑

n∈N xn);

(ii) limn→∞ ‖xn‖ = 0.

7*. Prove that (xn)∞n=1 is summable, if and only if, for every permutation σ : N → N, thesequence (zσ

n)∞n=1, given by zσn =

∑nk=1 xσ(k) is convergent. Moreover, in this case it

follows that limn→∞ zσn =

∑n∈N xn.

(Hint: Argue by contradiction, using Exercise 2, by showing that if (xN)n∈N is notsummable, there exist ε > 0, and a sequence of finite sets F1 ( F2 ( F3 ( · · · ( N,such that Fn 3 n and

∥∥∑n∈Fn+1rFn

xn

∥∥ ≥ ε, ∀n.)

8*. Prove that, if (xs)s∈S is summable, then

(i) the S-tuple(‖xs‖

)s∈S

belongs to c0,R(S);

(ii) the set s ∈ S : xs 6= 0 is at most countable.

(Hint: Use Exercises 3 and 6 to show that, for every ε > 0, the set Fε = s ∈ S :‖xs‖ ≥ ε is finite.)

Comment. As the above exercises show, summability reduces to the case S = N. Inthis case, for a sequence (xn)∞n=1 ⊂ X we are dealing with three different conditions, each ofwhich will be expressed using series terminology.

(a) The sequence of partial sums (zn)∞n=1, given by zn =∑n

k=1 xn is convergent, in whichcase we say that the series

∑∞n=1 xn is convergent,

(b) As an element in∏

NX , the N-tuple (xn)n∈N is summable. Equivalently (using theterminology above), for every permutation σ, the series

∑∞n=1 xσ(n) is convergent. For

this reason, this condition is sometimes stated by saying that the series∑∞

n=1 xn isunconditionally convergent,

(c)∑∞

n=1 ‖xn‖ <∞, in which case we say that the series∑∞

n=1 xn is absolutely convergent,

In a Banach space one has the implications (c) ⇒ (b) ⇒ (a). According to the SummabilityTest, we also know that, if the implication (c) ⇒ (a) is true, then X is forced to be a Banachspace.

B. The Five Principles of Banach Space Theory

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Since Banach spaces are a special case of (F)-spaces, the Principles discussed in TVS IVapply to them. In particular, Theorems 1-2 below are direct results of this observation, sothey do not require a proof.

Theorem 1 (Open Mapping Principle). If X and Y are Banach spaces, and T : X → Yis linear, continuous, and surjective, then T is an open mapping, i.e. whenever A is open inX , 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 alsocontinuous.

Theorem 3 (Closed Graph Principle). If X and Y are Banach spaces, and T : X → Yis linear, then the following are equivalent:

(i) T is continuous;

(ii) the graph of T , i.e. the space GT = (x, Tx) : x ∈ X is closed in X × Y, in theproduct topology;

It should be noted here that X × Y is Banach space (relative to any norm that has theproduct topology as the norm topology), so condition (ii) is also equivalent that GT is aBanach space.

Theorem 4 (Alaoglu Compactness Theorem). If X is a normed vector space, then forany 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 neighborhoodof 0 in X , so the set

Kr = φ ∈ X ∗ : supx∈(X )1

|φ(x)| ≤ r

is compact in (X ∗,w∗), by the Alaoglu-Bourbaki Theorem. Now everything is trivial, sinceby the definition of the dual norm, we have Kr = (X ∗)r.

Theorem 5 (Uniform Boundedness Principle). Suppose X is a Banach space and Y isan arbitrary normed vector space. If Θ ⊂ L(X ,Y) is point-wise bounded, i.e.

supT∈Θ

‖Tx‖ <∞, ∀x ∈ X ,

then Θ is bounded in the operator norm, i.e. supT∈Θ ‖T‖ <∞.

Proof. Of course, the point-wise boundedness condition stated here is the same as the con-dition:

• for every x ∈ X , the set Θx = Tx : T ∈ Θ ⊂ Y is bounded in the norm topology.

In particular, we can apply the Equi-continuity Principle (Theorem2 from TVS IV) to con-clude that Θ is equi-continuous, which in this case means:

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(∗) for every ε > 0, there exists δε > 0, such that

‖x‖ ≤ δε ⇒ ‖Tx‖ < ε, ∀T ∈ Θ.

Apply the above condition to ε = 1, and let δ1 > 0 be the corresponding constant. Noticethat if we start with some arbitrary x ∈ (X )1, then ‖δ1x‖ = δ1‖x‖ ≤ δ1, so using (∗) andlinearity, we have δ1‖Tx‖ = ‖T (δ1x)‖ < 1, ∀T ∈ Θ, which shows that

‖Tx‖ < 1/δ1, ∀x ∈ (X )1, T ∈ Θ,

thus proving that ‖T‖ ≤ 1/δt, ∀T ∈ Θ.

C. Miscellaneous Results

This sub-section contains various application of the five Principles discussed in the previ-ous sub-section. Some of them are formulated as exercises. Another set of applications willbe given BS VI.

Exercise 9.♥ Suppose X is a Banach space. Prove that for a set A ⊂ X ∗, the followingare 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 (φn)∞n=1 ⊂ X ∗, such that φnw∗−→ 0, but such that ‖φn‖ = n, ∀n.

On the one hand, the set A = φn : n ∈ N ⊂ X ∗ is w∗-compact (see TVS I), hencew∗-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 everyT-Cauchy sequence is T-convergent.

Using this terminology, Exercise 11 states that the topological dual X ∗ of a Banach spaceX 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

complete, when equipped with the strong operator topology Tso. Recall that TΛTso−−→ T ⇔

Tλx→ Tx, ∀x ∈ X .

Exercise 13. Prove that, if X is an infinite dimensional Banach space, then the locallyconvex space (X ∗,w∗) is not metrizable. (Hint: Use Exercise 11.)

Exercise 14.♥. Prove that, if X is a separable normed vector space, then the compactspace ((X ∗)1,w

∗) is metrizable. (Hint: Let xn∞n=1 be a dense subset in the unit ball (X )1.

Prove that if (φλ)λ∈Λ and φ are in (X ∗)1, then φλw∗−→ φ⇔ φλ(xn) → φ(xn), ∀n. Therefore,

if we consider Ω =∏

N K, equipped with the product topology, which is metrizable, the map

7

F : (X ∗)1 3 φ 7−→(φ(xn)

)∞n=1

∈ Ω is w∗-continuous and injective, thus a homeomorphismonto its range.)

Exercises 15-17 below make use of the weak topology, introduced3 in LCVS V. In ourparticular setting, on a normed vector space X , this topology is denoted simply by w, andby definition we have xλ

w−→ x⇔ φ(xλ) → 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 3 x 7−→ 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. xλw−→ x⇒ Txλ

w−→ Tx.

(Hint: For (ii) ⇒ (i) use the Closed Graph Principle.)

We conclude this section with a useful technical result that characterizes operators withclosed range.

Proposition 7. Suppose X and Y are Banach spaces. For an operator T ∈ L(X ,Y),the following are equivalent:

(i) RangeT is closed;

(ii) there exists C > 0 such that:

(∗) for every y ∈ RangeT , there exists x ∈ X with Tx = 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 simplyreplace Y with RangeT , which is of course a Banach space, when equipped with the normcoming from Y .) Consider the quotient space Z = X/KerT , equipped with the quotientnorm. We know that Z is a Banach space, and furthermore, if we consider the quotient mapπ : X → Z, there exists a (unique) injective linear continuous map T ∈ L(Z,Y), such thatT π = T . Of course, since T is surjective, it follows that T is a linear isomorphism. Bythe Inverse mapping Principle, the inverse operator S = T−1 : Y → Z is also linear andcontinuous. 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 wT on X , imple-mented by T is the smallest topology that makes all φ ∈ (X ,T)∗ continuous. In other words, xλ

wT−−→ x meansφ(xλ) → φ(x), ∀φ ∈ (X ,T)∗.

8

so that T z = y and ‖z‖ ≤ ‖S‖ · ‖y‖. If y 6= 0, then C‖y‖ = ‖S‖ · ‖y‖+ ‖y‖ > ‖z‖, so by thedefinition of the quotient norm (‖z‖ = infπ(x)=z ‖x‖), there exists x ∈ X with ‖x‖ ≤ C‖y‖and π(x) = z, which clearly yields Tx = y. The case y = 0 is trivial, because we can choosex = 0.

(ii) ⇒ (i). Assume there exists C > 0, so that T has property (∗), and let us provethat RangeT is closed in Y . It suffices to prove that RangeT is a Banach space, whenequipped with the norm coming from Y , so it suffices to prove that RangeT satisfies theSummability Test. Start with a sequence (yn)∞n=1 ⊂ RangeT , with

∑∞n=1 ‖yn‖ <∞, and let

us prove that the series∑∞

n=1 yn is convergent in RangeT . For each n, use condition (∗)to choose some xn ∈ X , such that Txn = yn and ‖xn‖ ≤ C‖yn‖. Obviously we now have∑∞

n=1 ‖xn‖ ≤∑∞

n=1C‖yn‖ < ∞, so using the fact that X is a Banach space it follows that∑∞n=1 xn is convergent in X to some x. By continuity, it follows that

∑∞n=1 yn =

∑∞n=1 Txn

is convergent in Y to y = Tx, and we are done.

Corollary 3. If X and Y are Banach spaces, and T ∈ L(X ,Y) is injective, then thefollowing are equivalent:

(i) RangeT is closed;

(ii) T is bounded from below, i.e. there exists D > 0, such that

‖Tx‖ ≥ D‖x‖, ∀x ∈ X .

Proof. We know that condition (i) is equivalent to the existence of some C > 0, satisfyingcondition (∗). Notice, however, that since T is injective, any y ∈ RangeT can be uniquelyrepresented as y = Tx with x ∈ X , so condition (∗) is equivalent to

‖x‖ ≤ C‖Tx‖, ∀x ∈ X , (3)

which clearly is equivalent to the boundedness from below condition.

D. The Dual Banach Spaces `∞K (S)∗ and Related Issues

In this sub-section we take a closer look at the topological duals of the Banach spaces ofthe form `∞K (S). Most results concerning these spaces are left as Exercises. The only resultspresented with proofs are those concerning the inclusion c0,K(S) ⊂ `∞K (S) an the ProjectionProblem mentioned in LCVS V.

Notations. Throughout this sub-section S will denote an infinite set. We denote byP(S) the set of all subsets of, and by Pfin(S) the set of all finite subsets of S. For everyA ∈ P(S), we denote by χA the indicator function of A, defined as χA(s) = 1, if s ∈ A, andχA(s) = 0, if s 6∈ A. Clearly, χA ∈ `∞K (A). Furthermore, if A ∈ Pfin(S), then χA ∈

⊕S K ⊂

c0,K(S). We denote the linear span spanχA : A ∈ P(S) by elemK(S), and we declare thefunctions that belong to this space elementary.

Exercise 18.♥ Prove that elemK(S) is dense in `∞K (S) in the norm topology.

Exercises 19-24. Declare a map ν : P(S) → K finitely additive, if wheneverA1, . . . , An ∈P(S) are disjoint, it follows that4 ν(A1 ∪ · · · ∪ An) = ν(A1) + · · ·+ ν(An).

4 It is pretty obvious that, in order to check this, one only needs to consider the case n = 2.

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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∑

j=1

|ν(Aj)| : n ∈ N, A1, . . . , An disjoint,n⋃

j=1

Aj = A∈ [0,∞]

is finite. (Hint: One can assume K = R. In this case, show that, if one defines thequantity M = supA∈P(S) |ν(A)|, then |ν|(A) ≤ 2M , ∀A.)

20.♥ Denote by FK(S) the space of all bounded finitely additive maps P(S) → K. Provethat, if ν ∈ FK(S), then |ν| ∈ FR(S).

21.♥ Prove that FK(S) is a vector space, when equipped with

• point-wise addition: (ν1 + ν2)(A) = ν1(A) + ν2(A);

• point-wise scalar multiplication: (αν)(A) = αν(A).

Prove that the map FK(S) 3 ν 7−→ |ν|(S) ∈ [0,∞) defines a norm on FK(S), denotedby ‖ . ‖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 : elemK(S) → K, such that

s(χA) = ν(A), ∀A ∈ P(S); (4)

moreover the map s is unique.

23.♥ Prove that, if ν ∈ FK(S), then there exists a unique linear continuous map sν : `∞K (S) →K, satisfying (4), and furthermore, one has: ‖sν‖ = ‖ν‖var. (Hint: Use Exercise 20combined with Exercise 16.)

24.♥ Using the notations from the preceding Exercise, prove that the correspondence

FK(S) 3 ν 7−→ sν ∈ `∞K (S)∗

is an isometric linear isomorphism.

Notation. Using the notation from Exercise 24, given ν ∈ FK(S), and x ∈ `∞K (S),the quantity sν(x) ∈ K is denoted by SS x dν. One thinks the operation S as a version ofintegration.

Comments. Among the maps in FK(S) one finds the σ-additive ones. These are the(finite) K-valued measures on P(S), which are identified with the vectors in `1K(S), as follows:for x = (xs)s∈S ∈ `1K(S), one defines νx : P(S) → K by νx(A) =

∑s∈A xs. It is not hard to

see that,

SS y dµx =

∫S

y dµx = x#(y) =∑s∈S

xsys, ∀ y = (ys)s∈S ∈ `∞K (S),

10

where the map x# : `∞K (S) → K comes from the dual pairing between `1K(S) and `∞K (S), andthe “honest” integral in the middle is in the sense of the classical theory of measure andintegration. This gives rise to a linear isometric inclusion

J : `1K(S) 3 x 7−→ x# ∈ `∞K (S)∗. (5)

There exist, however, maps ν ∈ FK(S) which are not σ-additive (see Exercise 25 and theComment that precedes it), therefore the map (5) is not surjective.

Composing the restriction map

`∞K (S)∗ 3 φ 7−→ φ∣∣c0,K(S)

∈ c0,K(S) (6)

with the isometric linear isomorphism c0,K(S)∗∼−→ `1K(S) gives rise to a linear continuous

map E : `∞K (S) → `1K(S). If we consider the linear isometric inclusion (5), then we clearlyhave EJ = Id`1K(S). So, when we view – using J – the space `1K(S) as a closed linear subspace

in `∞K (S)∗, the map E is a projection of `∞K (S)∗ onto `1K(S), which satisfies the equality

‖Eφ‖ =∥∥φ∣∣

c0,K(S)

∥∥, (7)

where the norm on the right-hand side is computed in c0,K(S)∗.Let us agree with the following convention, which refers to a functional φ ∈ `∞K (S)∗:

We say that φ originates in `1K(S), if φ ∈ Range J , i.e. there exists x ∈ `1K(S) such thatφ(y) =

∑s∈S xsys, ∀ y = (ys)s∈S ∈ `∞K (S). Remark that, a necessary condition5 for a

functional φ ∈ `∞K (S)∗ to originate in `1K(S) is the equality

‖φ‖ = ‖Eφ‖ (8)

Exercise 25*. Let U be a free ultrafilter on S, i.e. an ultrafilter that contains the filterF = S r A : A ∈ Pfin. Consider the limit operation limU : `∞K (S) → K and the mapµU : P(S) → K defined by

µU(A) =

1 if A ∈ U0 if A 6∈ U

Prove that

(i) µU ∈ FK(S);

(ii) limU(y) = SS y dµU , ∀ y ∈ `∞K (S).

(iii) limU∣∣c0,K(S)

= 0, but limU 1 = 1.

Conclude that so µU is not σ-additive, because limU does not satisfy (8).

In what follows we are going to discuss several unusual features of the dual spaces `∞K (S)∗,the most intriguing being the one stated in Theorem 7 below. In preparation for it weintroduce the following.

5 As we shall see in BS III, this condition is also sufficient.

11

Notations and Conventions. Given some A ∈ P(S), we identify, for each p ∈ [1,∞]the space `pK(A) as a closed linear subspace in `pK(S) as follows. Given an A-tuple x =(xs)s∈A ∈ `pK(A), we identify it with the S-tuple x = (xs)s∈S ∈ `pK(S) given by

xs =

xs if s ∈ A0 if s 6∈ A

Similarly we can view c0,K(A) as a closed linear subspace in c0,K(S).Another way to make these identifications is to consider the multiplication operators

QpA : `pK(S) 3 y 7−→ χA · y ∈ `pK(S),

which are linear, continuous (in fact one has ‖QpA‖ = 1, if A 6= ∅), idempotent, i.e. (Qp

A)2 =Qp

A. With the help of this operator `pK(A) is identified with6 RangeQpA. Likewise, if one

restricts Q∞A to c0,K(S) one obtains an linear continuous idempotent operator Q0A : c0,K(S) →

c0,K(S), whose range is identified with c0,K(A). When there is no danger of confusion, thesuperscript will be omitted, so all these idempotents will simply be denoted by QA.

Thinking `pK(A) as a closed linear subspace in `pK(S), and c0,K(A) as a closed linearsubspace of c0,K(S), allows us to define the dual restriction operations

`pK(S)∗ 3 φ 7−→ φA ∈ `∞K (A)∗ (9)

c0,K(S)∗ 3 ψ 7−→ ψA ∈ c0,K(A)∗ (10)

as φA = φ∣∣`pK(A)

and ψA = ψ∣∣c0,K(A)

. For p = ∞, another way of defining (9) is to use bounded

finitely additive maps: if φ is represented as φ(y) = SS y dν, for some ν ∈ FK(S), then wecan restrict ν to P(A), thus producing a νA ∈ FK(A), and then φA(y) = SA y dνA.

For the alternative equivalent definitions of (9) with 1 ≤ p < ∞ and of (10), one canemploy the ordinary restriction map

`pK(S) 3 y = (ys)s∈S 7−→ yA = (ys)s∈A ∈ `pK(A)

and the linear isometric identifications of the dual spaces (`p)∗ ' `q (p, q > 1 and 1p+ 1

q= 1),

(`1)∗ ' `∞, and (c0)∗ ' `1, so we have

(y#)A = (yA)#, ∀ y ∈ `∞ ' (`1)∗

(y#)A = (yA)#, ∀ y ∈ `q ' (`p)∗

(y#)A = (yA)#, ∀ y ∈ `1 ' (c0)∗

Remark 2. The map E : `∞K (S)∗ → `1K(S) is compatible with (dual) restrictions tosubsets of S. Namely, if we start with some (non-empty) A ∈ P(S), then (use subscripts toindicate the ambient set):

EAφA = (ESφ)A, ∀φ ∈ `∞K (S)∗. (11)

Furthermore, if φ ∈ `∞K (S)∗ has the property that φA originates in `1K(A) for some A ∈ P(S),then φB originates in `1K(B), for every B ∈ P(A).

Exercise 26.♥ Fix some φ ∈ `∞K (S)∗.

6 Since QpA is idempotent, Range Qp

A is a closed linear subspace.

12

(i) Prove that the map P(S) 3 A 7−→ ‖φA‖ ∈ [0,∞) is bounded and finitely additive.

(ii) Prove that if (Ai)i∈I ⊂ P(S) is a system of disjoint subsets, then∑

i∈I ‖φAi‖ ≤ ‖φ‖.

(iii) Prove that, if A1, . . . , An ∈ P(S) are such that φAjoriginates in `1K(Aj), ∀ j = 1, . . . , n,

then if we consider the union A = A1 ∪ · · · ∪An, it follows that φA originates in `1K(A).

Exercise 27.♥ For every A ∈ P(S), consider the idempotent QA : `∞K (S) → `∞K (S)introduced previously.

(i) Prove that the correspondence

P(S) 3 A 7−→ QA ∈ L(`∞K (S))

is finitely additive, in the sense that, whenever A1, . . . , an ∈ P(S) are disjoint, itfollows that QA1∪···∪An = QA1 + · · · + QAn . Prove also that, if A,B ∈ P(S), thenQAQB = QBQA = QA∩B.

(ii) Given φ ∈ `∞K (S)∗, prove that the functional7 φ QA ∈ `∞K (S)∗ has the followingproperties:

(a) ‖φ QA‖ = ‖φA‖,(b) φ QA originates in `1K(S), if and only if φA originates in `1K(A).

Exercise 28.♥ Let (φλ)λ∈Λ ⊂ `∞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 ((φλ)A)λ∈Λ ⊂ `∞K (A)∗ also converges to 0 inthe w∗-topology.

(ii) Prove that, if A is finite, then ((φλ)A)λ∈Λ also converges to 0 in the norm topology.

Lemma 2. Let I be an infinite set, and let (Ai)i∈I ⊂ P(S) be a collection of disjointsubsets. Suppose (φn)∞n=1 is a sequence in `∞K (S)∗, such that (φn)Ai

originates in `1K(Ai),∀n ∈ N, i ∈ I. Then there exists an infinite subset M ⊂ I, so that when we consider theunion B =

⋃i∈M Ai, it follows that (φn)B originates in `1K(B), ∀n ∈ N.

Proof. For the sake of avoiding an unnecessary repetition, we will assume that φ1 = 0.During the proof, let us agree to use the following notation. For any subset M ⊂ I, we

denote the union⋃

i∈M Ai by A(M).

Claim 1: There exists a sequence I1 ) I2 ) I3 ) . . . of infinite subsets of I, such that,for every k ∈ N we have

‖(φj)A(Ik)‖ ≤ 2−k, ∀ j ∈ 1, . . . , k. (12)

The desired sequence is constructed recursively as follows (since φ1 = 0, we can start withI1 = I). Assume we have constructed the infinite sets I1 ) · · · ⊃ In, so that (12) holds

7 The functional φ QA is of course the same as Q∗Aφ.

13

for k = 1, . . . , n, and let us indicate how the next set In+1 ( In is constructed. Since In isinfinite, we can write it as a disjoin union In =

⋃∞m=1 Tm of a sequence of infinite subsets of

In. Since the sets (A(Tm))∞m=1 are disjoint, by Exercise 27 we know that, for every φ ∈ `∞K (S)∗

we have∑∞

m=1 ‖φA(Tm)‖ <∞, which implies that

limm→∞

‖φA(Tm)‖ = 0 ∀φ ∈ `∞K (S)∗

In particular, we can find m, such that

‖(φj)A(Tm)‖ ≤ 2−(n+1), ∀ j ∈ 1, . . . , n+ 1,

and we will set In+1 to be this Tm.Having proved Claim 1, let us now choose a sequence (in)∞n=1 ⊂ I of indices, such that

in ∈ In r In+1, ∀n, and let us from the sets M = in : n ∈ N and B = A(M). For eachn, let Mn = i1, . . . , in, Bn = A(Mn), Rn = M r Mn, and Dn = A(Rn), so that we haveB = Bn ∪Dn (disjoint union).

Claim 2: For every j ∈ N one has limn→∞ ‖φj QDn‖ = 0.

Indeed, if we fix j, and we start with some n > j, then by construction Rn ⊂ In, soDn = A(Rn) ⊂ A(In), and then by Claim 1 (use Exercises 26 and 27, plus the fact that‖QDn‖ ≤ 1) we have

‖φj QDn‖ = ‖(φj QA(In)) QDn‖ ≤ ‖φj QA(In)‖ = ‖(φj)A(In)‖ ≤ 2−n, ∀n ≥ j.

To finish the proof, we now notice that, since Bn = Ai1 ∪ · · · ∪ Ain , by Exercises 26 and27 it follows that φj QBn originates in `1K(S), for every n, j ∈ N. Then the obvious equality

φj QB = φj QBn + φj QDn ,

combined with Claim 2, yields limn→∞ φj QBn = φj QB, in norm in `∞K (S)∗, therefore8

φj QB originates in `1K(S), ∀ j.

Theorem 6. (R.S. Phillips). Suppose (xn)∞n=1 is a sequence in `∞K (S)∗, which convergesto 0 in the w∗-topology. Then the sequence (Eφn)∞n=1 converges to 0 in `1K(S) in the normtopology.

Proof. We argue by contradiction, assuming that the sequence (Eφn)n does not converge innorm to 0. Passing to a subsequence, and multiplying by a bounded sequence of scalars, wecan assume that

‖Eφn‖1 = 1, ∀n. (13)

Claim 1: There exists a sequence of integers 1 = r1 < r2 < . . . , and a sequence(Bn)∞n=0 ⊂ Pfin(S), consisting of disjoint sets, such that

(∗) ‖(φrn)Bn‖ ≥ 1− 2−n, ∀n ∈ N;

(∗∗) if we take the union B =⋃∞

n=1Bn, then (φk)B originates in `1K(B), ∀ k ∈ N.

8 Of course, the set of all functionals in `∞ that originate in `1 is norm-closed, being the range of theisometric linear map (5).

14

To prove the above statement, we start off by constructing a sequence of integers 1 =k1 < k2 < . . . and a sequence of finite sets ∅ = F0 ⊂ F1 ⊂ F2 ⊂ · · · ⊂ S, such that, forevery n ∈ N one has:

(a) ‖(Eφkn)Fn‖1 ≥ 1− 2−(n+1);

(b) ‖(φm)Fn−1‖ ≤ 2−(n+1), ∀m ≥ kn.

The two sequences are constructed recursively as follows. Let us assume the integers 1 =k1 < k2 < · · · < kN and the sets F0, . . . , FN−1 have been constructed, satisfying (a) for1 ≤ n ≤ N − 1, and (b) for 1 ≤ n ≤ N , and let us indicate how FN and kN+1 areconstructed, so that (a) holds for n = N and (b) holds for n = N + 1. (We start withk1 = 1 and F0 = ∅, with nothing else to verify.) To construct the set FN , we use the factthat for any x ∈ `1K(S), one has9 ‖x‖1 = limG∈Pfin(S) ‖xG‖1. If we apply this to x = EφkN

,then there exists G ∈ Pfin(S), such that

‖(EφkN)H‖1 ≥ ‖EφkN

‖ − 2−(N+1) = 1− 2−(N+1), ∀H ⊃ G

thus (a) holds for n = N , if we put FN = FN−1 ∪ G. To construct kN+1 so that (b) holdsfor n = N + 1, all we need to do is to use Exercise 28, which gives limm→∞ ‖(φm)FN

‖ = 0.Having constructed our two sequences (kn)∞n=1 and (Fn)∞n=0 satisfying (a) and (b) for

all n, we now consider the finite sets An = Fn r Fn−1. Let us notice that, by Exercise 26we have ‖(φkn)An‖ = ‖(φkn)Fn‖ − ‖(φkn)Fn−1‖, so using (a), (b), and the obvious inequality‖(φkn)Fn‖ ≥ ‖(Eφkn)Fn‖1, we have

‖(φkn)An‖ ≥ ‖(Eφkn)Fn‖1 − 2−(n+1) ≥ 1− 2−n. (14)

Since all the A’s are finite, it is trivial that (φk)An originates in `1K(An), for all k, n. UsingLemma 2, there exists a sequence of integers 1 ≤ p1 < p2 < . . . , such that if we take theunion B =

⋃∞n=1Apn , then (φk)B originates in `1K(B), for every k. If we put Bn = Apn and

rn = kpn then the Claim follows, since by (14) we have10

‖(φrn)Bn‖ = ‖(φkpn)Apn

‖ ≥ 1− 2−pn ≥ 1− 2−n.

Having proved Claim 1, let us now restrict everything to B, and let us denote, forsimplicity, the functionals (φrn)B ∈ `∞K (B)∗ by ψn. To summarize what we have constructedso far, we have B =

⋃∞n=1Bn – disjoint union of finite sets – and a sequence (ψn)∞n=1 ⊂

`∞K (B)∗, with all ψn’s originating in `1K(B), such that ψnw∗−→ 0, and

(i) ‖ψn‖ ≤ 1,

(ii) ‖(ψn)Bn‖ ≥ 1− 2−n,

for all n ∈ N. (To check condition (i) we notice that, since ψn = (φrn)B originates in `1,using the assumption (13) we have ‖ψn‖ = ‖(Eφrn)B‖1 ≤ ‖Eφrn‖ = 1.)

9 Here we view Pfin(S) as a directed set, with the direct inclusion.10 Of course, pn ≥ n, ∀n, which yields 1− 2−pn ≥ 1− 2−n.

15

For every n ∈ N let xn = (xns )s∈B ∈ `1K(B) be B-tuple that represents ψn, i.e.

ψn(y) = x#n (y) =

∑s∈B

xns ys, ∀ y = (ys)s∈B ∈ `∞K (B),

so thatlim

n→∞

[∑s∈B

xns ys

]= 0, ∀ y = (ys)s∈B ∈ `∞K (B), (15)

and the above two properties read:

(i’)∑

s∈B |xns | ≤ 1,

(ii’)∑

s∈Bn|xn

s | ≥ 1− 2−n,

for all n.Let us consider the B-tuple y = (ys)s∈B ∈ `∞K (B) defined by

ys =

|xn

s |xn

s

if s ∈ Bn and xns 6= 0

0 if s ∈ Bn and xns = 0

Claim 2: limn→∞[∑

s∈B xns ys

]= 1.

Using (i’) and (ii’) we clearly have ∑s∈BrBn

|xns | ≤ 2−n, ∀n.

So if we fix for the moment n, using the inequalities |ys| ≤ 1, ∀ s ∈ B, we have∣∣ ∑s∈B

xns ys −

∑s∈Bn

xns ys

∣∣ =∣∣ ∑

s∈BrBn

xns ys

∣∣ ≤ ∑s∈BrBn

|xns ys| ≤

∑s∈BrBn

|xns | ≤ 2−n.

This means that in order to prove the Claim, it suffices to prove that the sequence (θn)n,given by θn =

∑s∈Bn

xns ys, converges to 1. But this is pretty clear, since θn =

∑s∈Bn

|xns |,

so by properties (i’) and (ii’) above we have the inequalities

1 ≥ θn ≥ 1− 2−n.

Having proved Claim 2, it is now obvious that we reached a contradiction, becausecondition (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 (xn)∞n=1 is a sequence in `1K(S), such that xnw−→ 0, i.e. φ(xn) → 0,

∀φ ∈ `1K(S)∗, then ‖xn‖1 → 0. (Using the dual pairing between `1 and `∞, the condition

xnw−→ 0 is equivalent to the condition x#

nw∗−→ 0 in `∞K (S)∗).

Of course, one cannot expect the implication xλw−→ 0 ⇒ ‖xλ‖1 → 0 to be true for

arbitrary nets (xλ)λ in `1K(S), because this would force the weak topology to coincide with

16

the norm topology. This is an indication that the sequential nature of Theorem 6 is a deepphenomenon. More on this will be discussed in BS ??

We now conclude with two interesting applications of Theorem 6. The first one – for-mulated as an Exercise – is an application to topology, which shows that sequences andsubsequences are inadequate for compactness.

Exercise 29.♥. Suppose S is infinite, and (sn)∞n=1 is a sequence of distinct points in S, i.e.sm 6= sn, ∀m 6= n. Define, for every n, the functional φn : `∞K (S) 3 y = (ys)s∈S 7−→ ysn ∈ K.We know that the φn’s are linear and continuous. In fact, they also originate in `1K(S), soone has ‖φn‖ = ‖Eφn‖1 = 1, ∀n. Furthermore, by Exercise 7 from BS I, it follows that

φn

∣∣c0,K(S)

w∗−→ 0, in c0,K(S)∗. (16)

Prove that the sequence (φn)∞n=1 does not converge in the w∗-topology to any φ ∈ `∞K (S)∗.Conclude that in the compact Hausdorff space

((`∞K (S)∗)1,w

∗) there exists sequences which

do not have convergent subsequences. (Hint: Argue by contradiction, assuming φnw∗−→ φ

for some φ ∈ `∞K (S)∗. Using Theorem 6, this forces ‖Eφn − Eφ‖1 → 0. By (16), however,one has Eφ = 0, so we will also have ‖Eφn‖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 `∞K (S) ontoc0,K(S).

Proof. Argue by contradiction. Assume there exists a linear continuous operator P : `∞K (S) →c0,K(S), such that Px = x, ∀x ∈ c0,K(S). Taking the transpose, we obtain a linear continuousoperator P ∗ : (c0,K(S)∗,w∗) → (`∞K (S)∗,w∗), such that

(P ∗γ)∣∣c0,K(S)

= γ, ∀ γ ∈ c0,K(S)∗. (17)

The big point here is the fact that P ∗ is continuous, with respect to the w∗-topologies .Fix now a sequence (sn)∞n=1 of distinct points in S. As remarked in Exercise 29, we know

that the functionalsγn : c0,K(S) 3 y = (ys)s∈S 7−→ ysn ∈ K

are linear, continuous, of norm 1, and they form a sequence which is w∗-convergent to 0 inc0,K(S)∗. By the w∗-continuity of P ∗, it follows that the sequence (ψn)∞n=1 ⊂ `∞K (S)∗, given byψn = P ∗γn, is w∗-convergent in `∞K (S)∗ to 0, so by Theorem 6, it follows that ‖Eψn‖1 → 0.Of course, using the isometric isomorphism Θ : `1K(S)

∼−→ c0,K(S)∗, and (17), which impliesΘ(Eψn) = γn, we now get ‖γn‖ → 0, which is impossible.

E. Vector valued `p-spaces

In this sub-section we introduce several versions of Banach direct sums and products, forinfinite systems of Banach spaces. Such constructions are heavily dependent on the normson the Banach spaces involved, so they are not topological.

Definitions. Assume S is some non-empty set and (Xs, ‖ . ‖s)s∈S is an S-tuple of normedvector spaces.

17

A. The space of all S-tuples x = (xs)s∈S ∈∏

s∈S Xs, with sups∈S ‖xs‖Xs < ∞, is denotedby `∞(Xs, ‖ . ‖s)s∈S. For x ∈ `∞(Xs, ‖ . ‖s)s∈S, the above supremum is denoted by ‖x‖∞.It is pretty clear that `∞(Xs, ‖ . ‖s)s∈S is a linear subspace in

∏s∈S Xs, and that ‖ . ‖∞

is a norm on `∞(Xs, ‖ . ‖s)s∈S.

B. The space of all x = (xs)s∈S ∈∏

s∈S Xs, for which limF∈Pfin(S)

[sups∈SrF ‖xs‖s

]= 0, is

denoted by c0(Xs, ‖ . ‖s)s∈S. It is pretty clear that c0(Xs, ‖ . ‖s)s∈S is a linear subspacein `∞(Xs, ‖ . ‖s)s∈S.

C. Given p ∈ [1,∞), the space of all x = (xs)s∈S ∈∏

s∈S Xs, for which∑

s∈S ‖xs‖ps < ∞

is denoted by `p(Xs, ‖ . ‖s)s∈S. It is pretty clear that `p(Xs, ‖ . ‖s)s∈S is contained in

c0(Xs, ‖ . ‖s)s∈S. For x = (xs)s∈S ∈ `p(Xs, ‖ . ‖s)s∈S, we define ‖x‖p =[∑

s∈S ‖xs‖ps

]1/p.

When there is no confusion about the norms on the spaces Xs, s ∈ S, they will be omittedfrom the notation. It is recommended, however, that we do not do that, since a change innorms will produce different spaces!

Theorem 7. Assume all spaces (Xs, ‖ . ‖s), s ∈ S are Banach spaces.

(i) The normed vector space(`∞(Xs, ‖ . ‖s)s∈S, ‖ . ‖∞

)is a Banach space.

(ii) The space c0(Xs, ‖ . ‖s)s∈S is norm-closed in(`∞(Xs, ‖ . ‖s)s∈S, ‖ . ‖∞

). Specifically,

c0(Xs, ‖ . ‖s)s∈S is the norm-closure of the direct sum11⊕

s∈S Xs. In particular, thenormed vector space

(c0(Xs, ‖ . ‖s)s∈S, ‖ . ‖∞

)is also a Banach space.

(iii) For every p ∈ [1,∞):

• the space `p(Xs, ‖ . ‖s)s∈S is a linear subspace in c0(Xs, ‖ . ‖s)s∈S;

• the map ‖ . ‖p : `p(Xs, ‖ . ‖s)s∈S → [0,∞) is a norm;

• the normed vector space(`p(Xs, ‖ . ‖s)s∈S, ‖ . ‖p

)is a Banach space.

Proof. (i). Assume (xn)∞n=1 is a Cauchy sequence in(`∞(Xs, ‖ . ‖s)s∈S, ‖ . ‖∞

), and let us

prove it is convergent. Using the triangle inequality, we have∣∣‖xm‖∞ − ‖xn‖∞∣∣ ≤ ‖xm − xn‖∞, ∀m,n ≥ 1,

so the sequence(‖xn‖∞)∞n=1 is Cauchy, thus convergent in [0,∞), so in particular the quantity

M = supn∈N ‖xn‖∞ is finite.For each n ∈ N, write xn = (xns)s∈S, with xns ∈ Xs. By the construction of the norm

‖ . ‖∞, we have the inequalities

‖xms − xns‖s ≤ ‖xm − xn‖∞, ∀m,n ∈ N, ∀ s ∈ S, (18)

so in particular it follows that, for each s ∈ S, the sequence (xns)∞n=1 is Cauchy in Xs, thus

convergent to some xs ∈ Xs. Consider then the vector x = (xs)s∈S ∈∏

s∈S Xs, and letus prove that: (a) x ∈ `∞(Xs, ‖ . ‖s)s∈S, and (b) limn→∞ ‖x − xn‖∞ = 0. To prove these

11 It is trivial that⊕

s∈S Xs is contained in `∞(Xs, ‖ . ‖s)s∈S .

18

features, we fix for the moment some ε > 0, and we use the Cauchy condition, to producenε ∈ N, such that ‖xm − xn‖∞ < ε, ∀m,n ≥ nε, so by (18) we have

‖xms − xns‖s ≤ ε, ∀m,n ≥ nε, s ∈ S.

If we keep n and s fixed, and take limm→∞ this yields

‖xs − xns‖s ≤ ε, ∀n ≥ nε. (19)

On the one hand, using the triangle inequality, we have

‖xs‖s ≤ ‖xs − xns‖s + ‖xns‖s ≤ ε+ ‖xn‖∞ ≤ ε+M, ∀ s ∈ S,

thus proving that x = (xs)s∈S indeed belongs to `∞(Xs, ‖ . ‖s)s∈S. On the other hand, takingsups∈S in (19) clearly yields

‖x− xn‖∞ ≤ ε, ∀n ≥ nε,

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 ∈ `p(Xs, ‖ . ‖s)s∈S, then αx ∈

`p(Xs, ‖ . ‖s)s∈S, and furthermore one has ‖αx‖p = |α| · ‖x‖p, ∀α ∈ K. To prove the first twostatements in (iii), it suffices them to show that

(∗) whenever x, y ∈ `p(Xs, ‖ . ‖s)s∈S, it follows that x+y ∈ `p(Xs, ‖ . ‖s)s∈S, and furthermoreone has the inequality

‖x+ y‖p ≤ ‖x‖p + ‖y‖p. (20)

This above statement is, however, pretty obvious, by considering the S-tuples u = (us)s∈S, v =(vs)s∈S, w = (ws)s∈S ⊂

∏s∈S R, given by us = ‖xs‖s and vs = ‖ys‖s, ws = ‖xs + ys‖s, which

will satisfy the following conditions

• u and v both belong to `pR(S),

• 0 ≤ ws ≤ us + vs,

which imply that w also belongs to `pR(S) – thus showing that x + y indeed belongs to`p(Xs, ‖ . ‖s)s∈S – as well as the inequality ‖w‖p ≤ ‖u‖p + ‖v‖p, which is precisely (20)

Assume now (xn)∞n=1 is a Cauchy sequence in(`p(Xs, ‖ . ‖s)s∈S, ‖ . ‖p

), and let us prove it

is convergent. Arguing exactly as in (i), it follows that the quantity M = supn∈N ‖xn‖p isfinite.

For each n ∈ N, write xn = (xns)s∈S, with xns ∈ Xs. By the construction of the norm‖ . ‖p, we have the inequalities∑

s∈F

‖xms − xns‖ps ≤ ‖xm − xn‖p

p, ∀m,n ∈ N, ∀F ∈ Pfin(S), (21)

so in particular – using singeltons F = s, s ∈ S – it follows that, for each s ∈ S, thesequence (xns)

∞n=1 is Cauchy in Xs, thus convergent to some xs ∈ Xs. Consider then the

vector x = (xs)s∈S ∈∏

s∈S Xs, and let us prove that: (a) x ∈ `p(Xs, ‖ . ‖s)s∈S, and (b)

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limn→∞ ‖x− xn‖p = 0. To prove these features, we fix for the moment some ε > 0, and weuse the Cauchy condition, to produce nε ∈ N, such that ‖xm − xn‖p < ε, ∀m,n ≥ nε, so by(21) we have ∑

s∈F

‖xms − xns‖ps ≤ εp, ∀m,n ≥ nε, F ∈ Pfin(S).

If we keep n and F fixed, and take limm→∞ this yields∑s∈F

‖xs − xns‖ps ≤ εp, ∀n ≥ nε. (22)

On the one hand, using the triangle inequality in Xs, we have ‖xs‖s ≤ ‖xs − xns‖s + ‖xns‖,so using the triangle inequality in `pR(F ), we get have[ ∑

s∈F

‖xs‖ps

]1/p

≤[∑

s∈F

[‖xs − xns‖s + ‖xns‖s

]p]1/p

≤

≤[∑

s∈F

‖xs − xns‖ps

]1/p

+

[∑s∈F

‖xns‖ps

]/1p

≤

≤ ε+ ‖xn‖p ≤ ε+M, ∀F ∈ Pfin(S),

so taking supF∈Pfin(S) we obtain∑

s∈S ‖xs‖ps ≤ (ε+M)p, thus proving that x = (xs)s∈S indeed

belongs to `p(Xs, ‖ . ‖s)s∈S. On the other hand, taking supF∈Pfin(S) in (22) clearly yields

‖x− xn‖p ≤ ε, ∀n ≥ nε,

thus proving (b).

Comment. The spaces introduced above are to be understood as vector-valued gen-eralizations of the `p-spaces. In the case when (Xs, ‖ . ‖s) = (K, | . |), ∀ s ∈ S, the spaces`p(K, | . |)s∈S, p ∈ [1,∞], and c0(K, | . |)s∈S, coincide with `pK(S), and c0,K(S), respectively.

Remarks 3-5. Let (Xs, ‖ . ‖s)s∈S be an S-tuple of normed vector spaces.

3. Exactly as in the scalar case, the direct sum⊕

s∈S Xs is a dense linear subspace inc0(Xs, ‖ . ‖s)s∈S and in all `p(Xs, ‖ . ‖s)s∈S, p ∈ [1,∞).

4. The norm topology on each one of the spaces c0(Xs, ‖ . ‖s)s∈S and `p(Xs, ‖ . ‖s)s∈S,p ∈ [1,∞] is stronger than the induced product topology from

∏s∈S Xs. This follows

from the observation that the coordinate maps

Et : `p(Xs, ‖ . ‖s)s∈S 3 x = (xs)s∈S 7−→ xt ∈ (Xt, ‖ . ‖t)

are contractive, for all t ∈ S and every p ∈ [1,∞].

5. For every t ∈ S, the inclusion map Jt : Xt →⊕

s∈S Xs is isometric, when the targetspace is equipped with any one of the norms ‖ . ‖p, p ∈ [1,∞]. This way we see that(Xt, ‖ . ‖t) is isometrically represented as a closed linear subspace in either one of thespaces c0(Xs, ‖ . ‖s)s∈S or `p(Xs, ‖ . ‖s)s∈S, p ∈ [1,∞].

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6. If (X , ‖ . ‖) denotes either one of the spaces c0(Xs, ‖ . ‖s)s∈S or `p(Xs, ‖ . ‖s)s∈S, p ∈[1,∞], one can use the isometric linear maps Jt : (Xt, ‖ . ‖t) → (X , ‖ . ‖) to define amap Ψ : X ∗ →

∏s∈S X ∗

s by Ψ(φ) = (J∗sφ)s∈S = (φ Js)s∈S, φ ∈ X ∗. Theorem 8 belowgives more information on the map Ψ in two special cases of interest.

Theorem 8. Let (Xs, ‖ . ‖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 Holder conjugate of p, then the map

Ψ :[`p(Xs, ‖ . ‖s)s∈S

]∗ 3 φ 7−→ (φ Js)s∈S ∈∏s∈S

X ∗s

has the following properties

(a) Range Ψ = `q(X ∗s , ‖ . ‖∗s)s∈S;

(b) ‖Ψ(φ)‖q = ‖φ‖, ∀φ ∈ `p(Xs, ‖ . ‖s)∗s∈S.

Therefore, the map Ψ establishes an isometric linear isomorphism

Ψ :[`p(Xs, ‖ . ‖s)s∈S

]∗ ∼−→ `q[(X ∗

s , ‖ . ‖∗s)s∈S

].

(ii) The map

Ψ0 :[c0(Xs, ‖ . ‖s)s∈S

]∗ 3 φ 7−→ (φ Js)s∈S ∈∏s∈S

X ∗s

has the following properties

(a) Range Ψ0 = `1(X ∗s , ‖ . ‖∗s)s∈S;

(b) ‖Ψ0(φ)‖1 = ‖φ‖, ∀φ ∈[c0(Xs, ‖ . ‖s)s∈S

]∗.

Therefore, the map Ψ0 establishes an isometric linear isomorphism

Ψ0 :[c0(Xs, ‖ . ‖s)s∈S

]∗ ∼−→ `1(X ∗s , ‖ . ‖∗s)s∈S.

Proof. The proof will be divided into several steps, which are ordered somehow differentlyfrom the statement.

Claim 1: If p ∈ (1,∞), and φ ∈[`p(Xs, ‖ . ‖s)s∈S

]∗, then:∑

s∈S

(‖φ Js‖∗s)q ≤ ‖φ‖q. (23)

In particular, it follows that Range Ψ ⊂ `q(X ∗s , ‖ . ‖∗s)s∈S, and

‖Ψ(φ)‖q ≤ ‖φ‖, ∀φ ∈[`p(Xs, ‖ . ‖s)s∈S

]∗. (24)

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The desired inequality (23) is equivalent to∑s∈S0

(‖φ Js‖∗s)q ≤ ‖φ‖q, (25)

where S0 = s ∈ S : φ Js 6= 0. Fix some set F ∈ Pfin(S0) and some ε > 0, small enoughso that ‖φ Js‖∗s > ε, for all s ∈ F . Consider the vector u = (us)s∈F ∈ `qR(F ), given byus = ‖φ Js‖∗s − ε, s ∈ S Using the isometric isomorphism `qR(F ) ' `pR(F )∗, there exists avector v = (vs)s∈F ∈ `pR(F ), such that ‖v‖p ≤ 1, and such that

∣∣ ∑s∈F usvs

∣∣ ≥ ‖u‖q − ε.Since us > 0, and

∑s∈F |us| · |vs| ≥

∣∣ ∑s∈F usvs

∣∣, we can in fact assume that all vs, s ∈ Fare non-negative, so in fact we have the inequality∑

s∈F

usvs ≥ ‖u‖q − ε.

Using the definition of the functional norm, for every s ∈ F , for which ‖φ Js‖ 6= 0, thereexists xs ∈ Xs, such that

|(φ Js)(xs)| > ‖φ Js‖∗s − ε‖xs‖s = us‖xs‖s.

Replacing xs with a suitable scalar multiple αsxs, we can assume that (φ Js)(xs) ≥ 0,and furthermore, that ‖xs‖s = vs. Consider not the vector x =

∑s∈F Jsxs ∈

⊕s∈S Xs ⊂

`p(Xs, ‖ . ‖s)s∈S. By construction (use the fact that (φ Js)(xs) = |(φ Js)(xs)|), we have

φ(x) =∑s∈F

(φ Js)(xs) ≥∑s∈S

usvs ≥ ‖u‖q − ε. (26)

Of course, we have ‖x‖p =[∑

s∈F ‖xs‖ps

]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:∑s∈F

(‖φ Js‖∗s − ε)q ≤ (‖φ‖+ ε)q.

Taking limε→0 this will imply ∑s∈F

(‖φ Js‖∗s)q ≤ ‖φ‖q.

Taking supF∈Pfin(S0) in the above inequality yields (25), and the Claim follows.

Claim 2: If φ ∈[c0(Xs, ‖ . ‖s)s∈S

]∗, then:∑

s∈S

‖φ Js‖∗s ≤ ‖φ‖. (27)

In particular, one has the inclusion Range Ψ0 ⊂ `1(X ∗s , ‖ . ‖∗s)s∈S, and

‖Ψ0(φ)‖1 ≤ ‖φ‖, ∀φ ∈[c0(Xs, ‖ . ‖s)s∈S

]∗. (28)

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The proof is almost identical to the proof of Claim 1. Fix some set F ∈ Pfin(S) and someε > 0, small enough so that ‖φJs‖∗s > ε, ∀ s ∈ F . Consider the vector u = (us)s∈F ∈ `1R(F ),given by us = ‖φ Js‖∗s − ε, s ∈ F . Using the definition of the functional norm, for everys ∈ F , there exists xs ∈ Xs, such that

|(φ Js)(xs) > (‖φ Js‖∗s − ε)‖xs‖s = us‖xs‖s.

Replacing xs with a suitable scalar multiple αsxs, we can assume that (φ Js)(xs) ≥ 0,and furthermore, that ‖xs‖s = 1. Consider not the vector x =

∑s∈F Jsxs ∈

⊕s∈S Xs ⊂

c0(Xs, ‖ . ‖s)s∈S. By construction (use the fact that (φ Js)(xs) = |(φ Js)(xs)|), we have

φ(x) =∑s∈F

(φ Js)(xs) ≥∑s∈S

us ≥ ‖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: ∑

s∈F

(‖φ Js‖∗s − ε) ≤ ‖φ‖.

Taking limε→0 this will imply ∑s∈F

‖φ Js‖∗s ≤ ‖φ‖,

and then taking supF∈Pfin(S) in the above inequality we get∑

s∈F ‖φ Js‖∗s ≤ ‖φ‖, and theClaim follows.

Claim 3: If φ ∈[`1(Xs, ‖ . ‖s)s∈S

]∗, then:

maxs∈S

‖φ Js‖∗s ≤ ‖φ‖. (30)

In particular, for p = 1, one has the inclusion Range Ψ ⊂ `∞(X ∗s , ‖ . ‖∗s)s∈S, and

‖Ψ(φ)‖∞ ≤ ‖φ‖, ∀φ ∈[`1(Xs, ‖ . ‖s)s∈S

]∗. (31)

This statement is trivial, since the inclusion Js : Xs → `1(Xs, ‖ . ‖s)s∈S is isometric, so‖φ Js‖∗s ≤ ‖φ‖ · ‖Js‖ = ‖φ‖, ∀ s ∈ S.

Having proved the above three Claims, we see that the remaining issues that need to beaddressed are the surjectivity of Ψ and Ψ0, as well as the inequalities

‖φ‖ ≤ ‖Ψ(φ)‖q, ∀φ ∈[`p(Xs, ‖ . ‖s)s∈S

]∗; (32)

‖φ‖ ≤ ‖Ψ0(φ)‖1, ∀φ ∈[c0(Xs, ‖ . ‖s)s∈S

]∗. (33)

As it turns out, one can use a unified approach for treating these issues. Assume p, q ∈ [1,∞]are Holder conjugate, so now we also include the case (p, q) = (∞, 1), and ψ = (ψs)s∈S

belongs to `q(X ∗s , ‖ . ‖∗s)s∈S. Define the vector u = (us)s∈S ∈ `qK(S), by us = ‖ψs‖∗s. Notice

now that, for every x = (xs)s∈S ∈ `p(Xs, ‖ . ‖s)s∈S, the vector v = (xs)s∈S, defined by

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vs = ‖xs‖s, s ∈ S, belongs to `pR(S), therefore the vector (usvs)s∈S belongs to `1R(S), andsatisfies (by Holder) the inequality∑

s∈S

usvs ≤ ‖u‖q · ‖v‖p = ‖ψ‖q · ‖x‖p.

Notice now that by the operator norm inequality, the vector w = (ws)s∈S, defined by ws =ψs(xs) satisfies

|ws| = |ψs(xs)| ≤ ‖ψs‖∗s · ‖xs‖s = usvs,

so w belongs to `1K(S), and satisfies the inequality∣∣ ∑

s∈S ws| ≤ ‖u‖q · ‖v‖p = ‖ψ‖q · ‖x‖p.This means that, for every x = (xs)s∈S ∈ `p(Xs, ‖ . ‖s)s∈S, one can define the sum

φ(x) =∑s∈S

ψs(xs) ∈ K, (34)

which satisfies the inequality|φ(x)| ≤ ‖ψ‖q · ‖x‖p. (35)

Of course, this yields a linear continuous functional φ : `p(Xs, ‖ . ‖s)s∈S → K, which byconstruction clearly satisfies φ Js = ψs, ∀ s ∈ S, so the surjectiviy of Ψ is clear. For thesurjectivity of Ψ0 one needs φ0 ∈

[c0(Xs, ‖ . ‖s)s∈S

]∗, such that φ0 Js = ψs. In this case we

use (34) with (p, q) = (∞, 1), and we take φ0 to be the restriction of φ ∈[`∞Xs, ‖ . ‖s)s∈S

]∗to the subspace c0(Xs, ‖ . ‖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[`∞(Xs, ‖ . ‖s)s∈S]∗,

which is a very complicated matter, even in the scalar case. The arguments in the proof ofTheorem 8, however, show that, applying the construction (34), to vectors ψ = (ψs)s∈S ∈`1(X ∗

s , ‖ . ‖∗s)s∈S, one obtains an isometric linear embedding

`1(X ∗s , ‖ . ‖∗s)s∈S →

[`∞(Xs, ‖ . ‖s)s∈S]∗.

Remark 7. In the case when S is finite, one has

c0(Xs, ‖ . ‖s)s∈S = `p(Xs, ‖ . ‖s)s∈S =⊕s∈S

Xs =∏s∈S

Xs, ∀ p ∈ [1,∞],

and the norms ‖ . ‖p, p ∈ [1,∞], on∏

s∈S Xs are all equivalent (they all define the product

topology). In this case, Ψ :[∏

s∈S Xs

]∗ → ∏s∈S X ∗

s is a linear isomorphism. Theorem 8 issignificant even in this case, because it provides an explicit calculation of the functional norm,namely, if p, q ∈ [1,∞] are Holder conjugate, then Ψ :

[∏s∈S Xs, ‖ . ‖p

]∗ ∼−→[∏

s∈S X ∗s , ‖ . ‖q

]is an isometric linear isomorphism.

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