Function Spaces with Bounded L p Means and Their Continuous Functionals

and Applied Analysis 3 Inequality (10) gives a bound for the norm of g: 󵄩󵄩󵄩󵄩g 󵄩󵄩󵄩󵄩Mp ⩽ 󵄩󵄩󵄩󵄩f 󵄩󵄩󵄩󵄩Mp 󵄩󵄩󵄩󵄩μ 󵄩󵄩󵄩󵄩M. (14) The convolution defined in this way is a linear continuous operator fromM toM, with the norm bounded by ‖μ‖ M . Since L(R) embeds isometrically into the space of finite Borel measures on R, we can convolve every ρ ∈ L with functions inM:


Introduction
Families of Banach spaces of locally   functions whose   means satisfy various boundedness conditions on finite intervals were introduced in [1][2][3] and references therein as a natural environment to extend the notion of almost periodic functions originally introduced in [4][5][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 R: Marcinkiewicz spaces M  , 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 S  , 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  ⩾ 1).
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 S  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 M (Section 7.2) are a greatly expanded and somewhat improved revision of the approach of [7], Inequality (10) The convolution defined in this way is a linear continuous operator from M  to M  , with the norm bounded by ‖‖  .
Since  1 (R) embeds isometrically into the space of finite Borel measures on R, we can convolve every  ∈  1 with functions in M  : In particular, As for the spaces introduced before, also   is complete.However, translations are not isometries here (see also [16, inequality (30)]).
Proof.Without loss of generality, let  > 0. Choose 0 <  <  and let  be the characteristic function of the interval [ − , +].Suppose that  ⩽ +.Then, for  ⩾ , the -mean (1/2) ∫  − |()|   is largest for  = +, and its maximum value ‖‖    is (/( + )) 1/ .On the other hand, the translate    is a characteristic function centered at 0, and its norm in    is 1 if  ⩾  and (/) The constant in this inequality is maximum for  =  and its largest value is 1 + /.

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, sup ∈R ∫ +1  ||  , defines a norm for the Stepanoff space S  , 1 ⩽  < ∞.The Stepanoff space S ∞ coincides with  ∞ and is not considered in this paper.It is immediate to prove (see Corollary 9 below) that the M  -norm is bounded by the S  -norm, and therefore the Stepanoff space embeds into the Marcinkiewicz space.
More generally, for every  > 0, one could introduce the following - Now let  ∈ Z be such that Since the Stepanoff norms are clearly invariant under translations, we can limit attention to positive  and .Now, Therefore the limit exists; it is infinite if and only if ‖‖ () S  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.

Completeness
It is easily seen that the spaces   and S  are complete.Instead, it is considerably more difficult to show that M  , hence M , are complete.The proof given here essentially reproduces the ingenious argument given by Marcinkiewicz in [9], except for correcting minor computational mistakes.
In particular, For the sake of simplicity, for every  ∈ M  and  ∈ R, we rewrite the norm in    , defined in (18), as follows: Then, for every  ∈ M  , ‖‖ M  = inf >0   (); therefore   (  ) is a Cauchy sequence.
Let us choose a sequence   such that We claim that the sequence   converges to the following function : Indeed, for  ⩾ 1 define and observe that and so 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 lim sup with respect to  satisfies the inequality and the claim is proved, hence the theorem.

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.

Inclusions.
First of all, it is obvious from the inclusions between   spaces over compact sets that   ⊂   if 1 ⩽  ⩽  ⩽ ∞, and the inclusions are continuous.The same is true for the spaces   and S  , and for the Banach quotient M .All these families of spaces coincide with  ∞ (R) 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 S  embeds continuously in   and M  , as follows.Proof.Recall that I  is the null space of the semi-norm; hence it is obviously closed in M  .If   ∈ I  converges to  in the norm of M  , then, by the first inequality of Lemma 8,   →  also in the seminorm of M  ; hence  ∈ I  since I  is closed in this seminorm.
Remark 11.The embedding of S  in   is proper; it is not an isomorphism, or, equivalently, the norm of S  cannot be bounded by a multiple of the   -norm.Indeed, we have already observed that translations are isometries on S  .Instead, ‖  ()‖   tends to 0 as  → ±∞ for every  with compact support.
As a consequence of Corollary 9, one has a continuous embedding   ⊂ M  .Therefore the embedding is projected onto the Banach quotient M ; one has   /I  ⊂ M .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. M = M  /I  is isometrically isomorphic to   /I  .Proof.We have already observed that the latter quotient is embedded in the former, and, for every  ∈   , one has ‖‖ M ⩽ ‖‖   /I  .Let  denote the coset of  mod I  .We only need to show that the coset  of every  ∈ M  contains a function in   and the norms are equal.
Let  ∈ M  and  > We also recall that the null space I  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.
(iii) If ,  are conjugate indices and 1 <  ⩽ ∞, then   is the dual space of   (it will follow from Theorem 60(iii) that  1 is not a dual space).
(iv) Moreover but   /  = 2 − , so Therefore the closed subspace of M  generated by the sequence {  } is isomorphic to ℓ ∞ , and the same argument also works for M .

The Dual Spaces of 𝑀 𝑝 and M𝑝
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.

Functionals on Seminormed Spaces.
Let us consider the dual space of the Marcinkiewicz space M  .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 I  of the semi-norm, that is, the dual of the Banach quotient M = M  /I  .Indeed, all continuous linear functionals on a seminormed complete space  vanish on the null space  of the semi-norm, because if  ∈   does not vanish on , then () ̸ = 0 for some non-zero  ∈ , but ‖‖  = 0 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 I  , the dual of M  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 M  and thereby extended to continuous functionals on the whole of M  by the Hahn-Banach theorem: There are interesting instances of M  -discontinuous functionals defined on appropriate subspaces of M  .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 M  ; the proof of this elementary fact will be given in Corollary 50.
Here are some other interesting M  -discontinuous functionals.For 0 ⩽  ⩽ 1, let as  → +∞.It is clear that  1 = M  and, for  < 1,   is contained in the closed subspace of M  of the functions with semi-norm 0; in particular, these subspaces are not dense in Since   is a subspace of the null space of the semi-norm, the only linear functional that is continuous in the semi-norm of M  is the zero functional.We now exhibit some interesting nontrivial (hence discontinuous) linear functionals on   .For simplicity, we first describe them in the case  = 1: Clearly, Hölder's inequality shows that, for  > 1, the correct way to define  ±  and  ±  is by replacing   at the denominator with  1−(1−)/ .
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.

Uniformly Convex Normed Spaces
Definition 19 (see [18]).A normed (or semi-normed) space is uniformly convex if, for every  > 0 and all vectors ,  in the unit ball such that ‖ − ‖ ⩾ , there exists () > 0 such that (1/2)‖ + ‖ ⩽ 1 − ().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 ℓ  be a continuous linear functional on  of norm 1 that attains its norm at a vector  with ‖‖ = 1, in the sense that ℓ  () = 1.Then, for every  in the unit ball  of  with ‖ − ‖ ⩾ , one has |ℓ  ()| ⩽ 1 − 2().
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 {ℎ : ℓ  (ℎ) = 1} 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 {ℓ  = 1} and orthogonal to this hyperplane, whose length is ℓ  () − ℓ  () = 1 − ℓ  ().This segment is twice as long as its parallel segment  drawn from the mid-point (+)/2, 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 ‖‖ ⩽ ‖‖ < 1, the distance from /‖‖ and  is larger than ‖ − ‖ and to apply part (i).

The Dual of 𝑀 𝑝 : 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 R 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 R, regarded as a continuous functional on (R), 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 onedimensional subspace Ṽ generated by , for some linear functional ã on Ṽ; the requested element  ∈   is the normpreserving Hahn-Banach extension of ã to a functional on the whole of .
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  ∞ (R) attains its norm as a functional on  ∞ .Indeed, if () = ∫ R   ℎ  for every measurable set  and for some ℎ ∈  1 , then In general, though, discrete non-positive measures on R do not attain their norm; for instance, let {  } be an enumeration of the rationals and let  = ∑  (−1)  2 −    ; since Q is everywhere dense in R, there is no continuous function  such that (  ) = (−1)  , and so  cannot attain its norm as a functional on (R).
To finish the proof, let us provide examples of continuous functionals on  ∩  ∞ (R) that are not represented by countably additive measures.Consider the closed subspace of  ∩  ∞ of functions that have a finite limit for, say,  → +∞.The limit lim  → +∞ () 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 (R) = 1; therefore  is finitely but not countably additive.
Definition 23.For every locally   function  on R, let ♮  be the operator on   defined by if () ̸ = 0, and  ♮ = 0 otherwise (here as usual, for  ∈ C,  =   , we write  = phase ()).
Proof.If  ∈   , one has  ♮ ∈   , because 1/ + 1/ = 1 implies that ( − 1) = , and so  can be uniquely written as 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 1 < ,  < ∞, and ℓ 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 [1, ∞], one has for every  ∈   .Moreover,  ⩾ 0, ‖‖ = ‖ℓ‖, (, ||  ) = 1 on the support of  and ℓ attains its norm on  ♮ .Conversely, let ℓ be a functional as in (90), with respect to a finitely additive measure .Then this integral representation of ℓ is unique (except for the identification mentioned in the proof of Corollary 28), and ℓ is continuous on   .Moreover, ℓ attains its norm if and only if the measure  is positive, and (, ||  ) = 1 on the support of .
Proof.Without loss of generality, assume ‖ℓ‖ = 1.By Lemma 27 the dual of   is isometric to the space of countably additive measures on (  ); therefore, for some ] ∈ ((  )) with ‖]‖ = 1 and for all  ∈   , one has Let  ∈   be a function on which ℓ attains its norm: As ‖ ♮ ‖   = 1, the integrand has modulus less than or equal to 1 for every  in the unit ball of   .On the other hand, As ‖‖ = 1, the measure  must be positive.Moreover, (, ||  ) = 1 on the support of .
Conversely, let us write where ] is the finitely additive Borel measure on   given by ] =  ×   .This integral representation is uniquely associated to an integral representation over (  ), of the form ℓ() = ∫ (  )  † ]. is also necessary.But the necessity holds in general also for complex spaces, as we have shown in the first half of the proof.That proof also shows that (, ||  ) = 1 on the support of  and ℓ( ♮ ) = 1.
It is important to remember that, in general, the Stone-Čech compactification of a cartesian product does not coincide with the product of the compactification (for an example, see [20,Chapter 6, problem 6N2]), and so the points of the compactification have not be written as pairs, as is done, for the aim of brevity, in the original reference [7,Lemma 4.3].
We now apply the Yosida-Hewitt decomposition theorem for finitely additive measures [21].We need the following definition.
Definition 31.A Borel measure  on a topological space  is purely finitely additive (p.f.a.) if whenever ] is a nonnegative countably additive Borel measure on  bounded by , in the sense that 0 ⩽ ] ⩽ ||, then ] = 0.
The Yosida-Hewitt decomposition theorem [21, Theorem 1.24] states that, for every finitely additive Borel measure , there exists a unique pair of Borel measure  1 ,  2 with  1 countably additive and  2 purely finitely additive, such that  =  1 +  2 .If  is nonnegative, then so are  1 and  2 .As a consequence one has the following.Lemma 32.If  is a purely finitely additive positive measure vanishing on sets of Lebesgue measure zero, then () = 0 for every measurable function  that vanishes at infinity.Proof.Since the finitely additive measures vanishing on sets of Lebesgue measure zero are the Banach dual of  ∞ , then the restriction of such a measure  to the space  0 of continuous functions vanishing at infinity yields a continuous functional on  0 , hence a countably additive measure.This restriction is dominated by , and so it must be zero if  is purely finitely additive.
Proof.Let l be a norm-preserving Hahn-Banach extension of ℓ to   .By Theorem 29, we know that for some finitely additive measure  and  ∈ (  ) defined as in Proposition 25.Now the previous Lemma states that the purely finitely additive measure  2 in the Yosida-Hewitt decomposition  =  1 +  2 satisfies the identity for every  ∈ I  , because lim  → ∞  † (, ) = 0 by definition of null space.
Since ‖ 1 ‖ ⩽ 1, the measure  1 must be positive and of norm 1.
We now extend this result by proving that, on I  , the condition that the functional attains its norm is not needed.
Theorem 34.For 1 <  < ∞, all continuous linear functionals on I  (attaining their norm or not) can be represented as in (79).
Then, by Lemma 22(i) (with  = I  and   =   = (  )  ) every element of   = (I  )  attains its norm as a functional on (  )  =   .Now the statement follows from Corollary 33.
Our next goal is to provide a similar integral representation for the dual of the Marcinkiewicz Banach quotient M = M  /I  , 1 <  < ∞.For this we need to remind and extend some previous results.
(ii) Regard the predual (I  )  of   as a subspace of (  )  ; then (iv) For every coset  ∈   /I  there exists  ∈ belongs to I  and so it is in the kernel of ℓ 2 ∈ I  ⊥ .By all this and (111) we have It now follows from (114) and (116) that This proves part (iii).Part (iv) is a bit more technical.Indeed, in the terminology of [22], the statement of part (ii) says that for all  ∈   (with this realization, all functions belong to   or   and we do not need to pay attention to equivalence classes modI  , in accordance with the remark at the end of the statement).
We only need to prove that the representing measure  is supported at infinity.Recall from the proof of Theorem 29 that, in this integral representation, the function  ≡  ♮ ∈   is built as in Definition 23 in terms of the function  ∈   where ℓ attains its norm and has the property that  = ||  ⩾ 0; of course  is not the zero function except in the trivial case ℓ = 0.
Suppose that, for some  > 0, the interval [−, ] has positive -measure, and consider the truncation   =  [−,] .Take  large enough so that   is not identically zero (this is possible since  is not identically zero; we are disregarding the trivial case ℓ = 0).Then   ∈ I  and (,   ) = (, |  |  ) is a positive function on − ⩽  ⩽ .Therefore ℓ(  ) = ∫ ∞ 1 (,   ) () > 0. This contradicts the assumption that ℓ vanishes on I  .These theorems on the integral representation of almost every continuous functionals shed light upon the examples of functionals on   and M  supported at infinity, which we built in Section 5.1.Those examples are all functionals of the type ℓ() = ∫ ∞ 1 (, ) () where the representing measure is supported at infinity.In other words, here  is a finitely additive positive finite measure given by the restriction of a countably additive positive finite Borel measure supported in (  ) \   , where   is the product space introduced in Definition 26 and (  ) is its Stone-Čech compactification.As observed at the beginning of Section 5.1, all continuous functionals on M  vanish on the null space of the seminorm, hence they must depend only on asymptotic values, and so, if they have an integral representation of the type ℓ() = ∫ ∞ 1 (, ) (), the measure  must be supported at infinity.Instead, functionals on   can be represented by measures that have a part at finite (necessarily countably additive, by the Yosida-Hewitt representation theorem and Remark 38).In the next section we extend this analysis to Stepanoff spaces.

Correlation Functionals.
By the representation theorems proved in this section we know that M is not the dual space of M .However, the following construction of functionals, called correlation functionals in [11], shows that at least it is possible to embed M as a quotient space of the dual of M .
Let  be a separable linear subspace of M  (one such subspace is I  , by Remark 17).Let  ∈ M  and consider a sequence {  } dense in .By Hölder's inequality in   ([−, ], /2), one has lim sup Therefore there is an increasing unbounded sequence { 1, } such that the limit Clearly this functional vanishes on the null space I  of the semi-norm; hence it defines a continuous functional on /I  ⊂ M .By Hahn-Banach extension, we produce in this way a continuous functional of M that depends only on limits of means, even when these limits are not defined directly by integration.

Summary and Open Problems.
We have proved representation theorems for continuous functionals on   and M  (1 <  < ∞) as integrals with respect to measures ] over (  ).For the norm-attaining functionals, the representing measure splits as the product of a finitely additive positive measure  on [1, ∞) times a delta measure   on the unit ball (  ), and so the representation becomes more specific: a average over  ∈ [1, ∞) of -weighted means over intervals of length 2.
In particular, the convex hull of these functionals contains those whose measure ], regarded as a finitely additive measure on   , splits as a product.
Is there a characterization of those functionals arising from measures  that are not countably additive and are supported at infinity, as for instance the Banach limits?
We have seen that the same problem is not interesting in the case  = ∞, since the representation of continuous functionals on M ∞ =  ∞ as finitely additive measures is already known.But can we prove interesting representation theorems for continuous functionals on M 1 ?Theorem 41. S  is the dual space of the Banach space

Duality for Stepanoff Spaces
A continuous functional  on S  is represented by a function  in the sense that if and only if  ∈ T  .In this case, the following quasi-isometry holds up to a factor 2: In other words, the subspace of the dual of S  of functionals that can be represented by a function is the bidual of T  .
Proof.Let  be a continuous functional on T  .On all functions  ∈ T  with support in [,  + 1],  is represented by a function ℎ  in   with support in [, +1]: () = ∫ ℎ  .Therefore, for all functions  ∈ T  with support in [−, ], () = ∫    where   = ∑ −1 =− ℎ  .Now observe that, by the way the norm in T  is defined, compactly supported   functions are dense in T  .Therefore every continuous functional on T  is represented by a locally   function .Again by the way the norm is defined, it is clear that the norm of this functional   is given by ‖  ‖ = sup  ‖  ‖  ; in other words, by Lemma 8, the norm of   is quasi-isometric with the norm of  in S  .Thus the dual of T  is S  .
On the other hand, we have already observed that if a functional on S  is represented by a function, then this function must belong to T  and the correspondence is quasiisometric.
Since S  embeds continuously in   , we know by Proposition 15 that the predual   of   embeds continuously in T  .The following is an independent simple proof of the embedding of   into T  .Lemma 42.One has ‖‖ T  ⩽ ‖‖   .Conversely, there is no  > 0 such that ‖‖   ⩽ ‖‖ T  .
Proof.Let  ∈   and, as before,   =  [,+1] .Let   be the characteristic functions of the dyadic intervals introduced in the definition of   .Hölder's inequality for ℓ 1 yields Therefore In these inequalities we have made use of Hölder's inequality, which is not an equality if  ∈ T  because then the sequence {‖  ‖  } cannot be constant.This yields the fact that the converse inequality does not hold.

S 𝑝 as a Bidual.
In analogy with the null space I  of the Marcinkiewicz semi-norm, we now introduce a similar subspace in S  .
The next lemma is proved by the same argument of Theorem 41.As a consequence, S  is the second dual of J  .
Lemma 46.J  ⊂ I  , and the inclusion is proper.

Proof. The inclusion means that lim
For the sake of simplicity, we first show that this is true for  = 1.Indeed, as  ranges from − to  − 1; then, as the sequence   goes to 0, so do its averages.In general, for ] .If we extract the th root of both sides this equality becomes an inequality: ] (because the ℓ  norm is dominated by the ℓ 1 norm).Therefore the previous argument still applies.
It is easy to show that the inclusion is proper: a function that belongs to 6.3.The Dual of S  .We now turn our attention to the dual of S  .As we did with M  , we first exhibit some examples of linear functionals that depend only on asymptotic values, and then we prove a representation theorem.For this goal, we introduce some interesting subspaces of S  , as follows.
Definition 47. (i)  ±  is the subspace of S  of all functions that have limits at ±∞; (ii)  ±  is the subspace of S  of all functions  such that the sequence ∫ The same argument yields the following.
Corollary 50.The spaces  ±  of functions vanishing at infinity are not closed in M  , and the functionals are not continuous in the semi-norm of M  .
The previous lemma shows that the spaces  ±  do not yield natural linear functionals that extend continuously to S  .On the other hand, the spaces  ±  allow to construct continuous functionals on S  which do not depend on values over finite intervals, that is, that do not belong to ⊗ ℓ 1   [, +1].This can be done as follows.Given a subspace  of a Banach space  and a continuous functional  on  (continuous in the -norm), denote by F its (many) Hahn-Banach extensions to .For instance, the Hahn-Banach extensions to ℓ ∞ of the continuous functional ({  }) = lim    , defined on the subspace of convergent sequences, are usually called Banach limits.In the same way we can now define on S  some Banach limits induced by the subspace   .The following result is now clear.Proof.The only thing left to prove is the continuity of  ± in the S  -norm, which is obvious since It is obvious that  ± () does not depend on values of  on compact sets; hence it cannot be expressed as an integral of the type ∫ ∞ −∞ ()() .
Remark 52.Since S  embeds continuously in   and   embeds continuously in M  , the limit functionals on M  described in (70) are also continuous functionals on S  and   .

A Summary of Duality and Inclusions.
Let us summarize the inclusions between these families of spaces and their duals.We have shown that These embeddings are proper.By the usual embedding of topological vector spaces into their biduals: For the same reason, since (T  )  = S  , The last embedding yields the part of the dual of S  consisting of functionals represented by functions.The embedding (  )  → (S  )  encompasses the previous construction of Banach limit functionals depending only on asymptotic values.It is intriguing to exhibit explicit examples of continuous functionals on S  that are not continuous on   .For instance, an interesting subspace of (  )  is the bidual of its predual I  ; not all these functionals are represented by functions (as functionals on   ), because most functions in I  are not small at infinity and do not belong to   .For a similar reason, they are not represented by functions when they act on S  .So here we have other exotic functionals on S  ; we leave to the reader to verify that they are different from the Banach limits considered before.

Integral Representation of Continuous Functionals on
S  .We now extend to the Stepanoff spaces the integral representation theorem for continuous functionals attaining their norms that we proved in Theorem 29 for   and in Theorem 40 for M .The proof is similar; we resume it skipping many details.
Definition 53.For  ∈ S  (R) and  ∈ R, put The next statement follows immediately from Lemma 24.Proposition 55.Let ,  be conjugate indices, with 1 < ,  < ∞.Let  be a -additive finite Borel measure on [1, ∞] and  ∈ S  (R).If  is the functional on S  given by then Definition 56.Denote by (S  ) the unit sphere, that is, the subset of S  of all functions of norm 1, by   the cartesian product [1, ∞] × (S  ), and by (  ) its Stone-Čech compactification.
Lemma 57.For  ∈ S  , let us define a function  ‡ on   by Then  ‡ is an isometric isomorphism from S  to  ∩  ∞ (  ) and therefore also from S  to ((  )).
Proof.By Lemma 24 the function for every  ∈ S  .
Proof.We follow the guidelines of the proof of the same result for   in Theorem 29.Again, we can assume ‖ℓ‖ = 1, and, by Lemma 57, we know that, for some ] ∈ ((  )) with ‖]‖ = 1 and for all  ∈ S  , Let  ∈ S  be a function on which ℓ attains its norm: ℓ() = 1.We have seen in ( 92) that ‖ ♮ ‖   = 1.
Denote by  the subset of   where | † (, )| attains its maximum value 1.Consider the family Φ of all nets (  ,   ) in   that converge to points of .
As (155) The rest of the proof is as in Theorem 29.

Extreme Points in the Unit Balls
Compact convex sets  in many Banach spaces (and more generally, in topological vector spaces) have plenty of extreme points.Indeed, the celebrated Krein-Milman theorem states that if the dual space separates points, then  is the closed convex hull of its extreme points.In particular, this is what happens for the unit ball of   spaces when  > 1 (including the case  = ∞, which is compact in the weak * topology by another well-known fact, the Banach-Alaoglu theorem).Therefore the Krein-Milman theorem applies to the unit ball of a normed (or semi-normed) space if the linear functionals that are weak * continuous separate points.On the other hand, the Hahn-Banach theorem shows that the dual of a locally convex space  separates points.So, if  is a normed space that is the dual of another normed space , then  separates points on ; hence , regarded as a subspace of   , separates points of  and of course the functionals in this subspace are weak * continuous.Therefore the unit ball of a Banach space  that is the dual of a normed space is the closed convex hull of its extreme points.This property generally fails if  is not a dual space.For instance, if ] is a finite measure on a measure space  which has no atoms, that is, such that every set  with ]() > 0 splits as the disjoint union  =  1 ∪  2 with 0 < ](  ) < ](), then the unit ball of  1 (, ]) has no extreme points, because every  of  1 -norm 1 is a proper convex combination of its (renormalized) truncations to two disjoint subsets of positive finite measure.Instead, the characteristic function of an atom is clearly an extreme point.
In this section we study the extreme points of the unit balls of the other spaces considered in this paper.We follow again [7] for the spaces   and M .Then we handle the easier case of S  , never considered before.
Remark 59.To simplify the presentation, we shall check extremality in the following form.A vector  in the unit ball  of a normed space  is an extreme point of  if and only if there does not exists  ̸ = 0 in  such that  ±  ∈ .Indeed, if such  exists then  is the mid-point of the chord connecting  +  and  − ; hence it is not extreme in .Conversely, if  is not extreme in  then it is an interior point of some chord in , hence it is the mid-point of some other chord.
As a consequence, the unit ball of a semi-normed but not normed space has no extreme points, since every  of seminorm less than or equal to 1 is the average of  ± , and ‖ ± ‖ ⩽ ‖‖ + ‖‖ = ‖‖ ⩽ 1 whenever ‖‖ = 0.This makes extremality a trivial issue on M  .for some  > 0 and some unbounded sequence   , then  is an extreme point of the unit ball (  ).
Therefore  is not an extreme point, once again by Remark 59.We are now ready to characterize the extreme points of the unit ball of M .Part (i) of the next theorem is a slightly more detailed proof of [7,Theorem 3.11]; parts (ii) extends results in [7, Theorems 3.8 and 3.10], where a clever argument is aimed to show that extremality in the unit sphere of M is critically related to the rate of speed of those subsequences (  , ||  ) that converge to their maximum limit 1.Our proof of part (ii) is a considerable revision of the argument in [7].
Theorem 65. (i) The unit ball ( M1 ) does not contain any extreme point.
Proof of Theorem 65.We can as well restrict attention to the unit sphere, that is, to functions  of norm Letting  → +∞ we see that ‖ + ‖ ⩽ 1.By Remark 63, the same is true for the norm in M1 ; thus  is not an extreme point of M1 .This proves (i).

Theorem 5 .
The Marcinkiewicz spaces M  are complete.Proof.Let {  } be a Cauchy sequence in the Marcinkiewicz semi-norm.Choose a subsequence {  } such that        −        M  ⩽ 2 − for every  >   .
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          if  is small enough.Therefore |‖‖| > |()| for almost every Lebesgue point : since almost every point is a Lebesgue point, |‖‖| > ‖‖ ∞ .
−       = inf {      − ℎ      : ℎ ∈ I  } =      + I      /I  .(118) This proves that any coset  in the quotient   /I  has a representative  such that ‖‖   /I  = ‖‖   , hence (iv).(observe that, although the integrand involves functions instead of I  cosets, the statement is well posed because ℓ vanishes on I  by Corollary 36 and for large  the integrand does not depend on the choice of coset representatives by Corollary 39).Proof.By the previous Corollary, ℓ is identified with a continuous functional on   vanishing on I  and attaining its norm.Then Theorem 29 yields the following integral representation: |⟨,  1 ⟩ { 1, } | ⩽ ‖ 1 ‖ M  ‖‖ M  .Let us now extract from { 1, } a subsequence { 2, } such that the limit⟨,  2 ⟩ { 2, } := lim Here again, one has |⟨,  2 ⟩ { 2, } | ⩽ ‖ 2 ‖ M  ‖‖ M  .We iterate this process to build a family of nested subsequences { , } that define limits ⟨,   ⟩ { , } satisfying the Hölder inequality above for the Marcinkiewicz seminorms.Then the diagonal sequence {  :=  , } gives rise to a limit ⟨,   ⟩ := lim that exists for every  and satisfies |⟨,   ⟩| ⩽ ‖  ‖ M  ‖‖ M  .Therefore ⟨, ⋅⟩ is a continuous functional on the subspace of M  generated by the sequence {  }, hence on  by density.
Observe that   (R) embeds continuously in S  .Indeed, if  =  [,+1] ,         S  = sup  ≡ ⊗ ℓ 1   [,  + 1], that is, the space of all  = ∑      , where {  } ∈ ℓ 1 and   ∈   [,  + 1] with 1/ + 1/ = 1 and ‖  ‖  = 1.T  is a Banach space with respect to the norm given by the infimum of ∑  |  | over all such representations.By the same argument, T  embeds continuously in   (R).Observe that both inequalities, hence both embeddings, are proper except for the cases S ∞ =  ∞ (R) andT 1 =  1 (R).It is clear that the dual Banach space of S  contains T  .Indeed, the functions  = ∑      , with {  } ∈ ℓ 1 and   ∈   [,  + 1], define continuous functionals on S  with norm ‖‖ (S  )  = ∑  |  |, since, for every  ∈ S  , 6.1.The Predual of S  .Before considering (S  )  , we need to examine the Banach space structure of S  .byLemma8.More generally, every  ∈   (R) with compact support defines a continuous functional on S  by the rule  () = ∫ ∞ −∞ () () .Indeed, if supp  ⊂ [−, ], then        ()      ⩽(127)It is clear that the restriction to [,  + 1] of every function  such that the functional   is defined on S  must belong to   [,  + 1].In general, however,   functions whose support is not compact do not yield continuous functionals on S  , unless they belong to T  (we have already observed that T  is properly contained in   (R), except for  = 1).For instance, choose a non-negative real sequence { ±  is the subspace of S  of all functions  such that the sequence ‖‖   [,+1] has limits at ±∞. Remark 48.It is clear that   is contained in   and   is contained in   .The other inclusions are false.The function ∑ >0  1/  [,+1/] belongs to   but not to   .The function ∑ ∞ =−∞ (−1)   [,+1] belongs to   but not to   .The function with values  in [,  + 1/2] and − in [ + 1/2,  + 1] belongs to   but not   .A variant of   and of   is obtained by integrating with respect to any sequence of finite Borel measure over [, +1] instead of Lebesgue measure; the same inclusion properties hold for this variant.defined on  ±  , are not continuous in the norm of S  .Proof.Let   = [,  + 2 −|| ],   the characteristic function of   and   = ∑  =−   .Each   is in S  and has compact support, hence it belongs to  ±  .It is immediately verified that the sequence   converges in S  to  = ∑ ∞ =−∞   .This  does not have limits at infinity; hence it does not belong to  ±  .For the same reason, the functional  ± and  have norm 1, as in the proof of Theorem 29 it follows by Hölder's inequality (145) that if {(  ,   )} ∈ Φ, then the interval [−  ,   ] must verify the condition For every  ∈ S  ,  ∈  and for every net {(  ,   )} in Φ that converges to , one has  ‡ () = lim   ‡ (  ,  ♮ ), and so       ‡     ∞ ⩽ sup The functional l attains its norm at  ♮ /‖ ♮ ‖   [  ,  +1] .(iv)Also lim  |(  ,   )| = 1 = lim  |(  ,  ♮ )| = lim  ‖‖    [  ,  +1] .(v)Moreover lim sup  (  ,   −  ♮ ) = 0.This follows as in the proof of Claim 3 in Theorem 29, by using now the uniform convexity of the spaces   [  ,   + 1].By applying again Hölder's inequality (145) to the identity that we have just proved in point (V) above, we finally obtain lim  (  ,   ) = lim  (  ,  ♮ ) for every  ∈ S  .Hence, for every  ∈  and every net converging to , f () = lim   (  ,  ♮ ) .
[7]ed in[7, Theorem 3.8]) is well defined, but (, ||  ) is not, because it depends on the coset representative.Instead, we need to project it to the quotient.In the next proofs, we shall consider representatives in the equivalence classes modulo I  and make use of the following simple remark.Remark 63.For every 1 ≤  < ∞, the M  (semi-)norm of a function.The M  (semi-)norm of a function  is equal to the M norm of its equivalence class modulo I  .Indeed, for  ∈ I  , one has ‖ + ‖ M  ⩽ ‖‖ M  + ‖‖ M  = ‖‖ M  , and‖ + ‖ M  ⩾ ‖‖ M  − ‖‖ M  = ‖‖ M  .We need a preliminary lemma that clarifies some comments in our reference ([7], Remark at page 161).Let ‖‖ M  = 1 and for 0 <  < 1 write There exist two unbounded sequences of positive numbers   ,   , with   <  +1 for every , such that the connected components of   are the open intervals (  ,  +1 ).By passing to suitable increasing subsequences, which we still denote by {  } and {  }, we have   <   <  +1 for every , and the disjoint union ⋃ ∞ =1 [  ,   ] is contained in the complement   of   : Proof.Since   → (, ||  ) is continuous,   is open, hence a countable union of open intervals.Because ‖‖ M  = 1,   is not all of the real line.As the union of overlapping open intervals is again an open interval,   is a disjoint countable union of intervals that we write as   = ⋃  (  ,  +1 ) for suitable   <  +1 .Then the complement   is given by   = ⋃  [  ,   ] (here   <   ).Again as ‖‖ M  = 1,   is not compact; therefore, by passing to a subsequence, we may assume that   ⊃ ⋃  [  ,   ] with   ,   monotonically increasing and unbounded.