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Abstract and Applied Analysis

Volume 2014 (2014), Article ID 609525, 26 pages

http://dx.doi.org/10.1155/2014/609525

## Function Spaces with Bounded Means and Their Continuous Functionals

^{1}Department of Mathematics, University of Rome “Tor Vergata”, Via Ricerca Scientifica, 00133 Rome, Italy^{2}Department of Mathematics, University of Maryland, College Park, MD 20742, USA

Received 7 October 2013; Accepted 14 November 2013; Published 19 February 2014

Academic Editor: Vakhtang M. Kokilashvili

Copyright © 2014 Massimo A. Picardello. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

This paper studies typical Banach and complete seminormed spaces of locally summable functions and their continuous functionals. Such spaces were introduced long ago as a natural environment to study almost periodic functions (Besicovitch, 1932; Bohr and Fölner, 1944) and are defined by boundedness of suitable means. The supremum of such means defines a norm (or a seminorm, in the case of the full Marcinkiewicz space) that makes the respective spaces complete. Part of this paper is a review of the topological vector space structure, inclusion relations, and convolution operators. Then we expand and improve the deep theory due to Lau of representation of continuous functional and extreme points of the unit balls, adapt these results to Stepanoff spaces, and present interesting examples of discontinuous functionals that depend only on asymptotic values.

#### 1. Introduction

Families of Banach spaces of locally functions whose means satisfy various boundedness conditions on finite intervals were introduced in [1–3] and references therein as a natural environment to extend the notion of almost periodic functions originally introduced in [4–6].

All the spaces of bounded -means contain , but usually they consist of functions that are not small at infinity and have norms defined by the asymptotic behaviour of their integral means. Therefore a relevant part of the information carried by these functions is at infinity, where they may become large. Which consequences does this fact yield for convolution operator, and for continuous functionals on these spaces? Should we expect the same behavior that is typical of spaces? This paper presents old and (some) new results and proofs for this question: it is aimed to show that bounded -mean spaces behave as spaces on the issue of completeness but (some or all of them) are completely different for what concerns isometric properties of translations, convolution operators, separability, representation theorems for continuous linear functionals, and extreme points of the unit balls.

For this goal, we focus our attention onto three significant families of locally spaces on : Marcinkiewicz spaces , consisting of functions whose values on finite intervals are irrelevant (they can be changed without changing the norm, which is indeed only a seminorm), Stepanoff spaces , whose norm depends only on the maximum content of functions on all finite intervals, and finite -mean spaces, where the -means with respect to intervals are bounded with respect to , for sufficiently large (say ).

We give an expanded and revised presentation of some known results, mostly taken from the fundamental paper [7], and prove some new ones. The results on Marcinkiewicz spaces and bounded -mean spaces are taken from [7]; the duality results for Stepanoff spaces in Section 6 are new and so are the examples of discontinuous asymptotic functionals on and . The analysis of the integral representation of continuous functionals on (Theorem 29) follows again [7] very closely but provides many more details, gives slightly more general statements, and improves several steps. Also the description of extreme points of the unit ball of (Section 7.1) is taken from [7]; the results on extreme points for the unit ball of (Section 7.2) are a greatly expanded and somewhat improved revision of the approach of [7], leading to a full characterization. In Section 7.3 we extend this approach to the extreme points of the unit ball of .

#### 2. Spaces with Bounded Means and Convolutions

##### 2.1. Marcinkiewicz Spaces

The Besicovitch-Marcinkiewicz spaces (that we briefly call Marcinkiewicz spaces) have been introduced by Besicovitch (see [8]), and their completeness was proved in [9] (we present this result in Section 3; a later but independent proof was given in [10]).

Here is their definition. Let be the space of functions on all compact sets in , , such that We equip , , with the following seminorm: The quotient of with respect to the null space of the semi-norm is therefore a Banach space, which we denote by .

Marcinkiewicz spaces have been studied or used in [3, 7, 9–14]. In particular, it has been observed in [11] that all regular bounded Borel measures give rise to bounded convolution operators on , with norm bounded by the norm of the measure, just as it happens for spaces. This follows by the fact that translation operators on have norm 1 (they are isometries!), and On the other hand, this is not true on other spaces defined by boundedness of -means, as we will now see.

##### 2.2. Convolution on Marcinkiewicz Spaces

It is obvious that functions can be convolved with functions with compact support, and the convolution integral converges. At first glance, it might look obvious that functions, and more generally finite Borel measures, are bounded convolution operators on , by the following argument. Let be a normed or semi-normed space of functions on , where translations are isometries and, for every , , the map is measurable. Then, for every finite Borel measure on , one has Unfortunately, here one needs that the vector valued integral be defined, and this condition is not obvious in our case, since the integrand is not continuous (see, e.g., [14, 15]). The lack of continuity is usually expressed by saying that is a semi-homogeneous but not homogeneous Banach space.

However, let us show that the vector valued integral in (4) exists and therefore that convolution with finite Borel measures makes sense on . The proof is taken from [11, Chapter I, Section 4].

Let be the normed linear space of all finite Borel measures with the norm given by the total variation of ; that is, Let and be a Radon measure. If is a finite interval in , set By Hölder’s inequality By Fubini’s theorem, the function is in on compact sets. Since is uniformly bounded on for , Lebesgue’s dominated convergence theorem yields Hence Let us consider as a function of the variable with values in , that is, the function Since the measure is finite, it follows from inequality (10) that this function verifies the Cauchy condition with respect to the Marcinkiewicz semi-norm as tends to .

Therefore there exists a unique function of such that where the limit is taken in the -norm.

The function in (12) is the convolution of and , and we write

Inequality (10) gives a bound for the norm of : The convolution defined in this way is a linear continuous operator from to , with the norm bounded by .

Since embeds isometrically into the space of finite Borel measures on , we can convolve every with functions in : In particular,

##### 2.3. Spaces with Upper Bounded -Means

A related family of spaces are the bounded -mean spaces introduced again in [3] and studied in [7, 16]. Their norm is defined as Obviously, for every this norm is equivalent to the norm of the dilated space defined by

Since , the space embeds continuously in , but the embedding is not an isometric isomorphism, as the next lemma will show.

But first let us clarify the reason for which the length of the interval in the definition of the bounded -mean space is bounded away from 0.

*Remark 1. *The space of all functions on such that on compact sets in and
is just and the supremum above is .

Indeed, it is obvious that if , then . For the converse inequality, let us show that if , then is bounded on a subset whose complement has measure zero, for instance, the set of Lebesgue points of (as is integrable on compacts). Let us choose a representative of in its Lebesgue class. If is a Lebesgue point, then the norm verifies for small enough. By the triangle inequality for this norm, if is small enough. Therefore for almost every Lebesgue point : since almost every point is a Lebesgue point, .

As for the spaces introduced before, also is complete. However, translations are not isometries here (see also [16, inequality (30)]).

Lemma 2. *The translation operator is bounded on , with norm .*

*Proof. *Without loss of generality, let . Choose and let be the characteristic function of the interval . Suppose that . Then, for , the -mean is largest for , and its maximum value is . On the other hand, the translate is a characteristic function centered at 0, and its norm in is 1 if and otherwise. Therefore
The constant in this inequality is maximum for and its largest value is .

##### 2.4. Stepanoff Spaces

Stepanoff functions, introduced in [1], are those measurable functions whose -norm on intervals of length, say, 1, is bounded. The supremum of these norms, , defines a norm for the Stepanoff space , . The Stepanoff space coincides with and is not considered in this paper. It is immediate to prove (see Corollary 9 below) that the -norm is bounded by the -norm, and therefore the Stepanoff space embeds into the Marcinkiewicz space.

More generally, for every , one could introduce the following -Stepanoff norm: However, we have the following.

Proposition 3. *The norms are equivalent for all . Moreover, if , then
*

*Proof. *For every ,
Now let be such that
Since the Stepanoff norms are clearly invariant under translations, we can limit attention to positive and . Now,
That is, the mean over is the average of the means over . Hence
Then, by the two previous inequalities,
Now, by (26), . Therefore (28) yields

On the other hand, let . Then the following inequality is similar to (27):
Hence .

Corollary 4. *If , then its Weyl norm
**
is finite.*

*Proof. *It follows from Proposition 3 that
for every . Therefore
Therefore the limit exists; it is infinite if and only if is infinite for some, hence for all, .

The Weyl norm defines a normed space called the Weyl space . This was one of the first bounded mean spaces introduced in order to extend the definition of almost periodic functions [2]. However, it was proved in [10] that the Weyl space is not complete; therefore it will not be considered in this paper.

#### 3. Completeness

It is easily seen that the spaces and are complete. Instead, it is considerably more difficult to show that , hence , are complete. The proof given here essentially reproduces the ingenious argument given by Marcinkiewicz in [9], except for correcting minor computational mistakes.

Theorem 5. *The Marcinkiewicz spaces are complete.*

*Proof. *Let be a Cauchy sequence in the Marcinkiewicz semi-norm. Choose a subsequence such that
In particular,
For the sake of simplicity, for every and , we rewrite the norm in , defined in (18), as follows:
Then, for every , ; therefore is a Cauchy sequence.

Let us choose a sequence such that

We claim that the sequence converges to the following function :
Indeed, for define
and observe that
and so
Now let . Then
Let us estimate the four integrals on the right-hand side. Remember that is a Cauchy sequence, hence uniformly bounded with respect to . Therefore there exists a constant such that, for every ,
On the other hand, on . So, for , from inequalities (36), (42) and the fact that the measure of is less than it follows that
for some constant depending only on . Therefore
by the first inequality in (38). The same argument for yields
Hence, again by the first inequality (38),
Finally, by the same argument,
and so
Now, by (44), (46), (48), and (50); one has
Therefore, the with respect to satisfies the inequality
Now, finally, if , this and inequality (35) yield
and the claim is proved, hence the theorem.

#### 4. Inclusions and Banach Space Structure

For the goal of understanding duality, it is appropriate to discuss first the inclusions between all these spaces, and their structure.

##### 4.1. Inclusions

First of all, it is obvious from the inclusions between spaces over compact sets that if , and the inclusions are continuous. The same is true for the spaces and , and for the Banach quotient . All these families of spaces coincide with when , and obviously embeds continuously in all of them, but in this paper we do not consider the special case .

Let us come to more interesting inclusions. It is easy to see that embeds continuously in and , as follows.

*Definition 6. *For every locally function , denote by
its -averages and by
the usual average.

Lemma 7. *.*

*Proof. *If , it is clear that
Since this yields .

Lemma 8. *.*

*Proof. *In computing the power of the norm, that is, , just split the domain of integration into the intervals with integer endpoints that overlap it.

Corollary 9. *.*

*Proof. *The first inequality is obvious. For the second, by splitting the interval into subintervals of length 1, we see that ; therefore, by Lemma 7, .

Corollary 10. * is closed in .*

*Proof. *Recall that is the null space of the semi-norm; hence it is obviously closed in . If converges to in the norm of , then, by the first inequality of Lemma 8, also in the seminorm of ; hence since is closed in this seminorm.

*Remark 11. *The embedding of in is proper; it is not an isomorphism, or, equivalently, the norm of cannot be bounded by a multiple of the -norm. Indeed, we have already observed that translations are isometries on . Instead, tends to as for every with compact support.

As a consequence of Corollary 9, one has a continuous embedding . Therefore the embedding is projected onto the Banach quotient ; one has . It turns out that these two quotients coincide. This has been proved in [17, Proposition 2.2(ii)]; here we give a slightly different proof.

Proposition 12. * is isometrically isomorphic to .*

*Proof. *We have already observed that the latter quotient is embedded in the former, and, for every , one has . Let denote the coset of . We only need to show that the coset of every contains a function in and the norms are equal.

Let and , and denote by the function that coincides with outside the interval and is zero inside: . Observe that since it is compactly supported. Moreover, for every , because the semi-norm of a function in does not change by adding another function with semi-norm zero. Now, if ,
Therefore
As belongs to the coset of modulo and , this shows that, for large enough, this coset contains functions with finite -norm, and

##### 4.2. Tensor Products

*Definition 13. *From every sequence of Banach spaces of functions on , one obtains a product Banach space by taking the completion of all finite linear combinations of functions in the norm
where the infimum is taken over all possible decompositions of of this type. Similarly, for every sequence , one introduces a weighted product where the norm is defined by
It is clear how to extend these definitions to the case , or to products over instead of .

When we consider spaces of functions over disjoint intervals, for instance if , where or , then the above representations are unique (up to normalization), and we can choose to be the truncation and .

*Remark 14. *It follows from Lemma 8 and Definition 13 that

##### 4.3. The Predual of and Tensor Products

We now consider another Banach space of functions with appropriate averages, the space , introduced in [16], that is defined as follows. Let , and and . Then

We also recall that the null space of the Marcinkiewicz semi-norm, endowed with the norm of , is a Banach space.

Now the following result, proved in [16, Theorems 2 and 3], is easy.

Proposition 15. *(i) One has with and .**(ii) Also .**(iii) If , are conjugate indices and , then is the dual space of (it will follow from Theorem 60(iii) that is not a dual space).**(iv) Moreover .**(v) If , then is the dual space of .*

*Proof. *We give a sketch of the proof. Part (i) follows directly from the definition of . Let (for this is just ). Then part (ii) is equivalent to the statement that the norm is equivalent to ; we outline the argument of [16, Theorem 2]. Let and observe that . Therefore . For the opposite inequality, let and choose such that . Let on and zero elsewhere. Then
This proves part (ii). The rest of the proof follows easily from this.

*Remark 16. *For , the duality property of Proposition 15(iii) does not hold; is strictly smaller than the dual of , because the restrictions of the spaces and to functions supported, say, in a dyadic interval are, respectively, and . Similarly, for the duality of part (v) does not hold, because the restriction of to the dyadic interval is .

*Remark 17. *It is easy to see, as in [15, Theorem 3.1.C], that compactly supported functions and Schwartz functions are dense in . It has been observed in [16, Theorem 2.E] that the same is true for the null space of the semi-norm. Therefore and are separable, and translation is strongly continuous on them. Instead, , , and are not separable, as we show in the next theorem.

##### 4.4. Separability

It follows from the tensor product structure of (Remark 14) that contains a closed subspace isometric to and therefore is not separable. For the same reason, is not separable, by part (ii) of Proposition 15. Following [7, Proposition 2.5], we now show that and are not separable.

Theorem 18. *For all , the Marcinkiewicz spaces and contain a closed subspace isomorphic to ; hence they are not separable.*

*Proof. *We start by building a sequence of intervals with larger and larger distance and length. Start with and and let and . Then , so
but , so
In particular, . Hence, if we denote by a partition of into a sequence of infinite subsets and let , then , because there are infinitely many intervals with , all disjoint, and . Then, for every sequence with , the function satisfies
(the last statement holds because, for every , there is such that , and because all the are nonnegative).

Let now be the integer such that . Then, by (67), the fact that has support in , the disjointness of the intervals, and (68), one has
So . Therefore the closed subspace of generated by the sequence is isomorphic to , and the same argument also works for .

#### 5. The Dual Spaces of and

The Riesz representation theorem shows that all continuous linear functionals on spaces can be represented as (integrals versus) functions in , and so they depend mostly on the values of the functions on compact sets. Our aim here is to show that, on spaces of locally summable functions, that can be large at infinity, some continuous functionals depend on asymptotic values and cannot be represented by functions in the usual integral sense (we shall see that most of them can be represented by integrals of means). Continuous linear functionals on Marcinkiewicz spaces have been studied in [7] on bounded -mean spaces, in [16]. We present these results here and construct interesting examples of functionals that are not represented by functions; in the next Section, we extend these results to Stepanoff spaces.

##### 5.1. Functionals on Seminormed Spaces

Let us consider the dual space of the Marcinkiewicz space . This is a complete semi-normed space but not a Banach space. It is clear that its continuous linear functionals are precisely those that factor through the null space of the semi-norm, that is, the dual of the Banach quotient .

Indeed, all continuous linear functionals on a semi-normed complete space vanish on the null space of the semi-norm, because if does not vanish on , then for some non-zero , but because is the null space of the semi-norm; hence there is no constant such that . The converse is obvious.

Since every compactly supported function is in , the dual of does not contain non-zero functionals that can be represented as functions; that is, it consists of linear functionals that depend only on the asymptotic behaviour of Marcinkiewicz functions. Here are the two most natural ones, defined and continuous on a closed subspace of and thereby extended to continuous functionals on the whole of by the Hahn-Banach theorem:

There are interesting instances of -discontinuous functionals defined on appropriate subspaces of . For instance, the functionals defined on the subspaces of functions vanishing at infinity, are discontinuous. The lack of continuity is equivalent to the fact that the subspaces are not closed in ; the proof of this elementary fact will be given in Corollary 50.

Here are some other interesting -discontinuous functionals. For , let as . It is clear that and, for , is contained in the closed subspace of of the functions with semi-norm 0; in particular, these subspaces are not dense in . Moreover, if .

Since is a subspace of the null space of the semi-norm, the only linear functional that is continuous in the semi-norm of is the zero functional. We now exhibit some interesting nontrivial (hence discontinuous) linear functionals on . For simplicity, we first describe them in the case : Clearly, if .

Hölder’s inequality shows that, for , the correct way to define and is by replacing at the denominator with .

In the next sections we expand these ideas to achieve a more complete representation, developed in [7], where the above Hahn-Banach extensions are reinterpreted as Dirac measures on the points at infinity of a suitable Stone-Čech compactification.

##### 5.2. Uniformly Convex Normed Spaces

*Definition 19 (see [18]). *A normed (or semi-normed) space is uniformly convex if, for every and all vectors , in the unit ball such that , there exists such that . The function is called the modulus of convexity; its geometrical meaning is the infimum of the distance from the midpoint of and to the unit sphere (the boundary of the ball). Observe that is a nondecreasing function of .

The following results are stated without proof in [7].

Lemma 20. *(i) Let be a uniformly convex space and be a continuous linear functional on of norm 1 that attains its norm at a vector with , in the sense that . Then, for every in the unit ball of with , one has .**(ii) The same statement holds if , and attains its norm at .**(iii) If, more generally, and are any vectors in the unit ball such that and and attains its norm at , then .*

*Proof. *We can restrict attention to the bidimensional subspace of generated by and . The proof is illustrated in Figure 1. For simplicity, we have drawn the figure under the implicit assumption that the restriction of the -norm to this bidimensional space is the Euclidean norm, and indeed the spotted line that represents the hyperplane is drawn as perpendicular to the radius of the unit ball, but the only property that we are using is that all of the ball is on one side of this line, that is, we only use the fact that the ball is convex: that is, the triangular inequality.

Part (i) follows by considering the segment in Figure 1, drawn from to the hyperplane and orthogonal to this hyperplane, whose length is . This segment is twice as long as its parallel segment drawn from the mid-point , and in turn is longer than the distance between the mid-point and the unit sphere, hence longer than .

For part (ii), it is enough to observe that, whenever , the distance from and is larger than and to apply part (i).

To prove (iii) consider the triangle whose vertices are , , and . We may as well consider the worst possible case where the segment from to has maximal length . The segment from to has length larger than or equal to . Hence the third side, from to , has length not less than . Now part (i) and the monotonicity of yield .

Proposition 21 (see [18, 19]). *For , is uniformly convex. Its modulus of convexity satisfies
**
if , and
**
if and is the conjugate index.*

##### 5.3. The Dual of : Integral Representation of Norm-Attaining Continuous Functionals

Now we describe the dual of the spaces of bounded -means, studied in [7, 16]; in particular, we describe an integral representation, obtained in [7], for those continuous linear functions that attain their norm. All the forthcoming results on integral representation are taken from [7]; our proofs are more detailed and expanded than those in the original paper.

We start with some easy comments on functionals that attain their norm.

Lemma 22. *(i) Let be a Banach space and its dual space. Then every element of , regarded as a functional on , attains its norm on some element . In particular, all continuous functionals on reflexive spaces attain their norms.**(ii) Every real finitely additive finite Borel measure on a Borel space , regarded as a continuous functional on , attains its norm. The norm is attained on a function that has modulus 1 on the support of the measure. The same is true for complex-valued finitely additive measures on provided that they are absolutely continuous with respect to Lebesgue measure.**(iii) If is not compact, finitely additive measures are continuous functionals on (by restriction from functionals on : so, not all continuous functionals on this space are given by countably additive measures. A real finitely additive measure attains its norm on if and only if it is positive.*

*(iv) Not every (countably additive) finite (real or complex) Borel measure on , regarded as a continuous functional on , attains its norm, but it attains its norm if it is positive (up to multiplication by a constant).*

*Proof. *We first observe that, for every , there is such that . This is trivially true in the one-dimensional subspace generated by , for some linear functional on ; the requested element is the norm-preserving Hahn-Banach extension of to a functional on the whole of .

Then, for every , ; therefore , as a functional on , attains its norm. This proves part (i).

The real finitely additive finite Borel measures on a Borel space are the continuous dual of . Let and be the characteristic functions of the supports of the positive and negative parts of , respectively. Then . This proves the first half of part (ii), and it also proves parts (iii) and (iv); a real countably additive measure attains its norm if and only if it is positive (because it attains its norm only on the function , which is discontinuous unless one of its two terms vanishes), or, slightly more generally, if it is a constant multiple of a positive measure.

Let us show that an absolutely continuous measure on attains its norm as a functional on . Indeed, if for every measurable set and for some , then
In general, though, discrete non-positive measures on do not attain their norm; for instance, let be an enumeration of the rationals and let ; since is everywhere dense in , there is no *continuous* function such that , and so cannot attain its norm as a functional on .

To finish the proof, let us provide examples of continuous functionals on that are not represented by countably additive measures. Consider the closed subspace of of functions that have a finite limit for, say, . The limit is continuous on this subspace, and, by Hahn-Banach theorem, extends to a continuous functional on all of that vanishes on all compactly supported functions. Represented as a measure , this functional vanishes on all bounded sets but ; therefore is finitely but not countably additive.

*Definition 23. *For every locally function on , let be the operator on defined by
if , and otherwise (here as usual, for , , we write ).

Lemma 24. *If are conjugate indices, and on an interval (or a measurable subset of ). Then and . Moreover, the operator is the inverse of .*

*Proof. *If , one has , because implies that , and so .

Finally, if , then
again because .

Proposition 25. *Let be conjugate indices, let be a -additive finite Borel measure on and . If is the average operator introduced in Definition 6 and is the functional defined on by
**
then is continuous on , and
*

*Proof. *The first inequality follows from Hölder’s inequality, because

If , by Lemma 24 one has , , and . Therefore
This is the second inequality of the statement.

*Definition 26. *Denote by the subset of of all functions of norm 1, by the Cartesian product , and by its Stone-Čech compactification.

Lemma 27. *For and define the weighted mean operator on as
**
Then is an isometric isomorphism from to and therefore also from to . In particular, the -subspace of that consists of real valued functions is isometrically isomorphic to the spaces of real valued functions in and .*

*Proof. *Observe again that, since , one has hence ; that is, . Therefore Lemma 24 shows that if (notation as in Definition 23), then
This yields an inequality between the two norms:
Let us prove the converse inequality. Observe that, again by Hölder’s inequality (86) for every one has
This and (84) imply that
Hence
This proves the converse inequality. The last statement follows from the fact that every continuous and bounded function on has a unique continuous extension on the Stone-Čech compactification, and the extension is an isometry (see [20, Chapter 6]).

Corollary 28 (the dual of ). *The space of continuous functionals on , , is isometrically isomorphic to the space of countably additive regular finite Borel measure on and to the space of finitely additive regular finite Borel measures on , in the sense that every functional can be uniquely written as
**
with and such that , and conversely.**The extreme points in the unit ball of are the Dirac measures on , or the extreme points in the unit ball of ; these correspond to the functionals where is an extreme point in the unit ball of and , plus the purely finitely additive measures in the sense of the forthcoming Definition 31. (Necessary and sufficient conditions for extremality in the unit ball of , for , will be given later in Theorem 60.)*

*Proof. *Recall that, for every Borel space , the dual space of is . By restriction, the dual space of is again . More precisely, as is a norm-closed subspace of , its dual space is the quotient of obtained by identifying two finitely additive measures that give rise to the same functional when restricted to ; apart from this equivalence, the dual of is isometrically isomorphic to . Then the isometric isomorphism between and induces an isometry between the respective dual spaces and .

The characterization of extreme points, whose details are left to the reader, follows from this.

On the basis of the isomorphism between and , from now on, with abuse of notation, we shall write the measure in corresponding to again as . The representing measure can be described more precisely for functionals attaining their norm, as follows.

Theorem 29 (integral representation of functionals on attaining their norm). *Let , be conjugate indices, with , , and a continuous functional on that attains its norm. Then, for some (notation as in Definition 26) and for some finite finitely additive positive measure on , one has
**
for every . Moreover, , , on the support of and attains its norm on .**Conversely, let be a functional as in (90), with respect to a finitely additive measure . Then this integral representation of is unique (except for the identification mentioned in the proof of Corollary 28), and is continuous on . Moreover, attains its norm if and only if the measure is positive, and on the support of .*

*Proof. *Without loss of generality, assume . By Lemma 27 the dual of is isometric to the space of countably additive measures on ; therefore, for some with and for all , one has
Let be a function on which attains its norm: . Since , one has , and by Lemma 24
Denote by the subset of where attains its maximum value (i.e., 1, by Proposition 25). Consider the family of all nets (i.e., ultrafilters) in that converge to points of .

As and have norm 1, it follows by (86) that, for every ,
Therefore, if , then the interval must verify the condition
Since the measure in (91) has mass 1, by (93) must be supported in . We make the following *Claim **1*: for every , and for every net in that converges to , one has (notation as in Definition 23), and so

Indeed, remember that by Definition 23, and choose any point in and let be a net in that converges to it. Then by (94). This proves Claim 1.

In the remainder of this proof, we keep notation more compact by writing . Denote by the continuous functional on given by
Then, by (94),

We make the following *Claim **2*: the functional attains its norm at .

Indeed, by Lemma 24, and because and are conjugate indices. Therefore
This proves Claim 2.

Now observe that
Indeed, the last identity for