The dual of the sequence spaces with mixed norms $l^{(p, \infty )}$

Abstract

We identify a norm-dense subspace of the dual of the sequence space $l^{(p,\infty )}$, thus closing the existing gap in the literature. We based our approach on the notion of James orthogonality, absolutely continuous norms and on the uniform convexity and the uniform smoothness of the underlying subspaces.

1 Introduction

The main focus of this work is to provide a full characterization of the dual of the space of double sequences endowed with mixed norms in the case when the outside norm is $l^{\infty }$, i.e. $l^{(p,\infty )}$. This instance is, as of yet, not covered in the literature.

Aiming at the precise formulation of the problem we recall some standard terminology and throughout this paper we will employ the notation introduced in [1].

Let $\textbf{P}=(p_1,p_2)$ with $p_i \in [1,\infty ]$, $i=1,2$. The adjoint of $\textbf{P}$ is defined as $\textbf{P}'=(p_1',p_2')$, where $1/p_i+ 1/p_i'=1$, with the usual convention when $p_i$ is 1 or $\infty$. As is customary, the characteristic function of a set S will be denoted by ${\textbf{1}}_S$. For a given Banach function space X, $X^{*}$ will stand for its dual, $X^{'}$ for its associate space, $B_X$ for the unit ball in X and $S_X$ for the corresponding unit sphere. The action of $\Lambda \in X^{*}$ on $y\in X$ will be written as $\Lambda (y)$. The Gateaux derivative of a norm $\Vert \cdot \Vert$ that is differentiable at the point $x\in X$ will be denoted by $\Vert \cdot \Vert ^{\prime }(x)$.

Definition 1.1

Consider the vector space consisting of all sequences $\textbf{f}=\{\textbf{f}^k\}_k=\{f_{n,k}\}_{n,k}$, where $\textbf{f}^k=\{ f_{n,k}\}_n \in l^{p_1}$. Endow this vector space with the mixed $l^{\textbf{P}}$-norm defined by

$$\begin{aligned} \Vert \textbf{f}\Vert _{\textbf{P}}:= \left( \sum _{k=1}^{\infty } \left( \sum _{n=1}^{\infty } |f_{n.k}|^{p_1}\right) ^{p_2/p_1} \right) ^{1/p_2}= \left\| \{ \Vert \textbf{f}^k\Vert _{l^{p_1} } \}_k \right\| _{l^{p_2}}, \end{aligned}$$

for $1\le p_1,p_2 <\infty$ or using the corresponding $l^{\infty }$ norm if $p_i=\infty$. Furnished with this norm, $l^{\textbf{P}}$ becomes a Banach space known as the mixed $l^{\textbf{P}}$-space [1].

The duality question for $l^{\textbf{P}}$ has been studied in [1] for $p_1,p_2<\infty .$ More precisely, [1, Theorem 1.(a) in Section 3], one has

$$\begin{aligned} \left( l^{\textbf{P}}\right) ^*= l^{\textbf{P}'} \end{aligned}$$

when $1\le p_1, p_2 <\infty .$ In particular, it is easy to verify that $l^{{\textbf{P}}}$ is a Banach function space [8] of functions over ${\mathbb N}\times {\mathbb N}$ whose associate space $\left( l^{{\textbf{P}}}\right) ^{\prime }$ is $l^{{\textbf{P}}^{\prime }}$ (see [2, 7, 8] for a detailed discussion).

This paper will focus on the duality question for the case $\textbf{P}=(p_1, \infty )$ when $1\le p_1 < \infty$. In this setting, the corresponding norm is:

$$\begin{aligned} \Vert \textbf{f}\Vert _{\textbf{P}}:= \left\| \{ \Vert \textbf{f}^k\Vert _{l^{p_1} } \}_k \right\| _{l^{\infty }}. \end{aligned}$$

Our main result is Theorem 2.10, in which we show that a set of relatively simple linear functionals is dense in $(l^{(p_1,\infty )})^{*}$.

We recall the notion of absolutely continuous norm in our case (see [2, 8] for the general setting):

Definition 1.2

Let $1\le p\le \infty$, $1\le q\le \infty$ and set ${\textbf{P}}=(p,q)$. An element ${\textbf{f}}\in l^{{\textbf{P}}}$ is said to have absolutely continuous norm if for any sequence of sets $(E_j)$ in $2^{{\mathbb N}\times {\mathbb N}}$ such that ${\textbf{1}}_{E_j}\rightarrow 0$ pointwise as $j\rightarrow \infty$, it holds that $\Vert {\textbf{f}}\,{\textbf{1}}_{E_j}\Vert _{{\textbf{P}}} \rightarrow 0$ as $j\rightarrow \infty$.

The set of all elements in $l^{{\textbf{P}}}$ with absolutely continuous norm will be denoted with $l^{{\textbf{P}}}_a$. Note that $l^{{\textbf{P}}}_a$ is a closed subspace of $l^{{\textbf{P}}}$ [2, Theorem I.3.8].

The following statements will be of importance in the sequel.

Lemma 1.3

The following are equivalent and true for ${\textbf{P}}=(p,q)$ with $1<p<\infty$ and $1<q<\infty$:

(i):

$l^{{\textbf{P}}}_a$=$l^{{\textbf{P}}}$,

(ii):

$\left( l^{{\textbf{P}}}\right) ^{*}=\left( l^{{\textbf{P}}}\right) ^{\prime }=l^{{\textbf{P}^{\prime }}}$.

Proof

See [2, 8] for a detailed proof of this Lemma. $\square$

The next theorem has been proved in [1].

Theorem 1.4

(i):

$\left( l_a^{(p,\infty )}\right) ^*=l^{(p',1)}$ when $1\le p < \infty$.

(ii):

$\left( l_a^{(\infty ,\infty )}\right) ^*=l^{(1,1)}.$

A bounded linear functional $\Lambda \in X^{*}$ is said to be norm-attaining if there exists $x_0 \in B_{X}$ such that $\Vert \Lambda \Vert _{X^{*}}=\Lambda (x_0)$. The following theorem is well known [3]:

Theorem 1.5

A Banach space X is reflexive if and only if every functional $\Lambda \in X^{*}$ is norm-attaining. The subspace of all norm-attaining functionals on a given Banach space X is norm-dense in $B_{X^{*}}$.

We next recall the notions of uniform convexity and uniform smoothness in Banach spaces, these concepts will be at the heart of our approach. The interested reader will find a detailed survey on the geometry of Banach spaces, as well as further references, in [5]. To measure the degree of convexity of a Banach space X, its modulus of convexity $\delta _X:[0,2] \rightarrow [0,1]$ is defined by the quantity:

$$\begin{aligned} \delta _X(\varepsilon )=\inf \left\{ 1-\frac{1}{2}\Vert x+y\Vert : x,y \in X, \Vert x\Vert \le 1, \Vert y\Vert \le 1, \Vert x-y\Vert \ge \varepsilon \right\} . \end{aligned}$$

(1.1)

Note that the same function is obtained if the infimum is taken over all $x,y \in X$ with $\Vert x\Vert =\Vert y\Vert =1$ and $\Vert x-y\Vert =\varepsilon$. We say that the Banach space X is uniformly convex if for all $\varepsilon \in (0,2], \delta _X(\varepsilon )>0$.

The modulus of smoothness of X is the function $\rho _X:(0,\infty ) \rightarrow [0,\infty )$ defined by:

$$\begin{aligned}\rho _X(\varepsilon ):=\sup \left\{ {\Vert x+y\Vert +\Vert x-y\Vert \over 2 }-1: \Vert x\Vert =1, \Vert y\Vert =\varepsilon \right\} . \end{aligned}$$

The space X is called uniformly smooth if

$$\begin{aligned}\lim _{\varepsilon \rightarrow 0} \rho _X(\varepsilon ) / \varepsilon =0. \end{aligned}$$

We recall that X is uniformly smooth if and only if

$$\begin{aligned} \lim _{t\rightarrow 0} {\Vert x+ty\Vert - \Vert x \Vert \over t} \text{ exists, } \text{ uniformly } \text{ on } \{(x,y): \Vert x\Vert =\Vert y\Vert =1 \}. \end{aligned}$$

(1.2)

We underline the fact that X is uniformly convex (respectively, uniformly smooth) if and only if $X^*$ is uniformly smooth (respectively, uniformly convex), [5].

It is well known that the space $l^p$ is uniformly convex and uniformly smooth for $p\in (1, \infty ).$

2 The dual space of $\ell ^{(p,\infty )}$

It is easy to see that any linear functional $\Lambda \in (l^{(p,\infty )})^{*}$ can be written as

$$\begin{aligned}\Lambda =\Lambda _a+\Lambda _d,\end{aligned}$$

where $\Lambda _a\in \left( l^{(p,\infty )}\right) ^{\prime }$ (i.e, $\Lambda _a$ can be identified with an element of the associate space of $l^{(p,\infty )}$) and $\Lambda _d$ vanishes on the subspace of $l^{(p,\infty )}$ that consists of elements that have absolutely continuous norm, henceforth denoted by $l^{(p,\infty )}_a$.

Since, for $1<p<\infty$, the associate space of $l^{(p,\infty )}$ is $l^{(\frac{p}{p-1},1)}$, a full characterization of those linear functionals on $l^{(p,\infty )}$ which vanish on $l_a^{(p,\infty )}$ will yield a complete description of $\left( l^{(p,\infty )}\right) ^{*}.$

Along this line of thought, we will focus on a linear functional

$$\begin{aligned} {L:l^{(p,\infty )}\rightarrow {\mathbb {R}}} \end{aligned}$$

such that $L(f)=0$ for each $f\in l_a^{(p,\infty )}$ (i.e. such that $\ker (L) \supset l_a^{(p,\infty )}$) and that $\Vert L\Vert _{l^{(p,\infty )} \rightarrow {\mathbb {R}}} > 0$ (see Definition 2.2).

We next introduce the following definitions.

Definition 2.1

Let $\textbf{f}\in l^{(p,\infty )}$. The seminorm

$$\begin{aligned} \Vert \textbf{f}\Vert _{l^{(p,\infty )}\setminus l^{(p,\infty )}_a}:= \lim _{n\rightarrow \infty } \sup _{i\ge n} \Vert \textbf{f}^i \Vert _{l^p({\mathbb {N}})} \end{aligned}$$

is referred to as the norm of discontinuity of $\textbf{f}$.

The following properties are obvious by definition:

(i):

$\Vert \textbf{f}\Vert _{l^{(p,\infty )}} \ge \Vert \textbf{f}\Vert _{l^{(p,\infty )}\setminus l^{(p,\infty )}_a}$

(ii):

$\textbf{f}\in l^{(p,\infty )}_a \iff \Vert \textbf{f}\Vert _{l^{(p,\infty )}\setminus l^{(p,\infty )}_a} =0.$

Definition 2.2

$(l^{(p,\infty )}\setminus l^{(p,\infty )}_a)^*$ stands for the subset of $(l^{(p,\infty )})^*$ consisting of linear functionals L such that $L(\textbf{f})=0$ whenever $\Vert \textbf{f}\Vert _{l^{(p,\infty )}{\setminus } l^{(p,\infty )}_a}=0$.

Definition 2.3

For $L\in (l^{(p,\infty )})^*$ we define:

$$\begin{aligned} \Vert L\Vert _{l^{(p,\infty )} \rightarrow {\mathbb {R}}}= & {} \sup _{0\ne f\in l^{(p,\infty )}} \frac{|L(f)|}{\Vert f\Vert _{l^{(p,\infty )}}}, \\ \Vert L\Vert _{l^{(p,\infty )}\setminus l^{(p,\infty )}_a \rightarrow {\mathbb {R}}}= & {} \sup _{0\ne f\in l^{(p,\infty )}\setminus l^{(p,\infty )}_a} \frac{|L(f)|}{\Vert f\Vert _{l^{(p,\infty )}\setminus l^{(p,\infty )}_a}} \end{aligned}$$

and

$$\begin{aligned} \Vert L\Vert _{l^{(p,\infty )}_a \rightarrow {\mathbb {R}}}=\sup _{0\ne f\in l^{(p,\infty )}_a} \frac{|L(f)|}{\Vert f\Vert _{l^{(p,\infty )}}}. \end{aligned}$$

Next the notion of James orthogonality will be recalled.

Definition 2.4

Let $(X,\Vert \cdot \Vert )$ be a Banach space, $x,y\in X$. An element $0\ne x \in X$ is said to be j-orthogonal to y, denoted by $x\perp ^X y$ (or $x\perp ^j y$) iff the inequality

$$\begin{aligned} \Vert x\Vert \le \Vert x+\alpha y\Vert \end{aligned}$$

holds for any $\alpha \in {\mathbb R}$.

Note that James’ orthogonality on a Banach space is not symmetric. An observation is in order at this point: if the norm $\Vert \cdot \Vert _X$ is Gateaux differentiable at x, say with derivative $\Vert \cdot \Vert ^{\prime }(x)$, then $x\perp ^X y$ if and only if $\langle \Vert \cdot \Vert ^{\prime }(x),y\rangle =0$ [5] (the ‘only if’   part is obvious).

We also emphasize the fact that if the norm of X is uniformly smooth, then, for $0\ne x \in X$, the set

$$\begin{aligned}x^{\perp ^j}:= \left\{ y: y\in X, x \perp ^j y \right\} \end{aligned}$$

is a vector subspace of X and $\textrm{codim}(x^{\perp ^j}) =1.$

In particular, for $1<p<\infty$ and ${\textbf{a}}, {\textbf{b}}\in l^p$, it follows that for ${\textbf{a}}=(a_k)$ and ${\textbf{b}}=(b_k)$, then ${\textbf{a}}\perp ^j{\textbf{b}}$ iff

$$\begin{aligned} \sum _{k=1}^{\infty }b_k| a_{k}|^{p-2}a_{k}=0. \end{aligned}$$

(2.1)

Indeed, (2.1) can be obtained by observing that the derivative of the real-valued function

$$\begin{aligned}t\rightarrow \sum _{k=0}^{\infty }|a_k+tb_k|^p \end{aligned}$$

is given by

$$\begin{aligned} \sum _{k=0}^{\infty }b_k|a_k + tb_k|^{p-2}a_k\end{aligned}$$

and that in view of the orthogonality condition, it should vanish at $t=0$. Next we adapt the definition of James orthogonality to the case under consideration.

Definition 2.5

The sequence $\textbf{f}\in l^{(p,\infty )}$ is said to be j-orthogonal to $\textbf{g}\in l^{(p,\infty )}$ ($\textbf{f}\perp ^{l^{(p,\infty )}} \textbf{g}$, or simply $\textbf{f}\perp ^j \textbf{g}$) iff for every $i \in {\mathbb {N}}$ we have $\textbf{f}^i \perp ^{l^p} \textbf{g}^i$ (i.e. $\Vert \textbf{f}^i\Vert _{l^p}\le \Vert \textbf{f}^i+ \alpha \textbf{g}^i\Vert _{l^p}$ for each $\alpha \in {\mathbb {R}}$).

For given $\textbf{f}\in l^{(p,\infty )}$ we define the James orthogonal complement of $\textbf{f}$ as the vector subspace

$$\begin{aligned}\textbf{f}^{\perp ^j}:= \left\{ \textbf{g}: \textbf{g}\in l^{(p,\infty )}, \textbf{f}\perp ^j \textbf{g}\right\} . \end{aligned}$$

It is easily seen that $\textrm{codim}(\textbf{f}^{\perp ^j}) =1.$

It is important to mention that the validity of $\textrm{codim}(\textbf{f}^{\perp ^j}) =1$ results from each space in the sequence (i.e $l^p$) having uniformly smooth norm (1.2), and that the uniqueness of the subspace $\textbf{f}^{\perp ^j}$ is guaranteed by the uniform convexity (1.1) of the norm of each space in the given sequence.

Note that $\textbf{f}\perp ^j \textbf{g}$ iff for each $i\in {\mathbb {N}}$ we have:

$$\begin{aligned}\left[ \textbf{f}^i,\textbf{g}^i\right] :=\sum _{k=1}^{\infty } g_{k,i} |f_{k,i}|^{p-2} f_{k,i}=0 \end{aligned}$$

where $\textbf{f}=\{\textbf{f}^i\}_i=\{ \{f_{k,i}\}_k \}_i$ and $\textbf{g}=\{\textbf{g}^i\}_i=\{ \{g_{k,i}\}_k \}_i$.

Definition 2.6

For $\textbf{g}\in l^{(p,\infty )}$, $\textbf{f}\in l^{(p,\infty )}$, define

$$\begin{aligned}\left[ \textbf{f}, \textbf{g}\right] : = \left\{ \sum _{k=1}^{\infty } g_{k,i} |f_{k,i}|^{p-2} f_{k,i} \right\} _{i}=\left\{ \left[ \textbf{f}^i,\textbf{g}^i\right] \right\} _i; \end{aligned}$$

set $(\textbf{f}^i)_p:=\{ |\textbf{f}_{k,i}|^{p-2} \textbf{f}_{k,i}\}^{\infty }_{k=1}$. Note that if $\textbf{f}^i \in l^p$, then $(\textbf{f}^i)_p \in l^{p'}$.

For the sake of clarity, we recall that a set function

$$\begin{aligned} \mu : 2^{{\mathbb N}}\rightarrow {\mathbb R} \end{aligned}$$

is a finitely additive measure if $-\infty<\mu ({\mathbb N})<\infty$, $\mu$ is finitely additive and

$$\begin{aligned} \Vert \mu \Vert _{ba({\mathbb N})}:=\sup \left\{ |\mu (A)|,\,\,A\in 2^{{\mathbb N}} \right\} <\infty . \end{aligned}$$

It is well known that $\left( l^{\infty }\right) ^{*}$ can be naturally identified with the space ba(${\mathbb N}$) of all finitely additive measures of finite total variation, via the one-to-one assignment

$$\begin{aligned} \left( l^{\infty }\right) ^{*} \ni \lambda \leftrightarrow \mu (A)=\lambda ({\textbf{1}}_A)\,\,\,\text {for any}\,\,\,A\subset {\mathbb N}, \end{aligned}$$

where ${\textbf{1}}_A$ stands for the indicator function of the set A [4]. Given a finitely additive measure $\mu \in$ ba(${\mathbb N}$) it is customary to denote the linear functional $\lambda \in \left( l^{\infty }\right) ^{*}$) associated to $\mu$ in the above sense, by $\int \nolimits _{\mathbb N}\cdot \mathrm{{d}}\mu$ and its action on ${\textbf{a}}=({\textbf{a}}_i)\in l^{\infty }$ with $\int \nolimits _{{\mathbb N}}\{{\textbf{a}}_i\}_i \mathrm{{d}}\mu .$

Example 2.7

Let $1<p<\infty$, $\frac{1}{p}+\frac{1}{p^{\prime }}=1$. Consider a double sequence

$$\begin{aligned} {\textbf{h}}=\{{\textbf{h}}^i\}_i=\{ \{h_{k,i}\}_k \}_i\in l^{(p^{\prime },\infty )}, \end{aligned}$$

that is

$$\begin{aligned} \Vert \textbf{h}\Vert _{l^{(p^{\prime },\infty )}}=\sup _{i\in {\mathbb N}}\Vert \textbf{h}^i\Vert _{ l^{p'}}<\infty . \end{aligned}$$

Assume $\textbf{f}=(\textbf{f}^i)_i \in l^{(p,\infty )}$, set

$$\begin{aligned} \textbf{h}^i(\textbf{f}^i)=\sum ^{\infty }_{k=1} h_{k,i}f_{k,i}. \end{aligned}$$

(2.2)

Notice that for each $i\in {\mathbb N}$,

$$\begin{aligned} |\textbf{h}^i(\textbf{f}^i)|=\left| \sum ^{\infty }_{k=1} h_{k,i}f_{k,i}\right| \le \Vert \textbf{h}^i\Vert _{l^{p^{\prime }}} \Vert \textbf{f}^i\Vert _{l^{p}}\le \Vert \textbf{h}\Vert _{l^{(p^{\prime },\infty )}}\Vert \textbf{f}\Vert _{{l^{(p,\infty )}}}. \end{aligned}$$

Let $\mu \in \textrm{ba}({\mathbb {N}})$ and define $L\in (l^{(p,\infty )})^{*}$ by

$$\begin{aligned} l^{(p,\infty )}\ni \textbf{g}\longrightarrow L(\textbf{g})= \int \limits _{{\mathbb {N}}} \{\textbf{h}^i(\textbf{g}^i)\}_i \mathrm{{d}}\mu . \end{aligned}$$

(2.3)

It is clear that L can be regarded as the composition of two linear maps, namely

$$\begin{aligned} A:&l^{(p,\infty )} \rightarrow l^{\infty }, \\ B:&l^{\infty } \rightarrow {\mathbb {R}}, \end{aligned}$$

with $A(\textbf{g})= \{\textbf{h}^i(\textbf{g}^i)\}_{i\in {\mathbb {N}}}$ and $B(\{k_i\}_{i\in {\mathbb {N}}})=\int _{{\mathbb {N}}} \{k_i\}_{i\in {\mathbb {N}}} \, \mathrm{{d}}\mu$.

Obviously L is an element of $(l^{(p,\infty )})^*$ and

$$\begin{aligned}\Vert L\Vert _{l^{(p,\infty )}\rightarrow {\mathbb {R}}} \le \Vert \{ \Vert \textbf{h}^i \Vert _{l^{p'}} \}_i \Vert _{l^{\infty }({\mathbb {N}})} \cdot \Vert \mu \Vert _{\textrm{ba}} = \Vert \textbf{h}\Vert _{l^{(p',\infty )}} \cdot \Vert \mu \Vert _{\textrm{ba}}. \end{aligned}$$

We underline the fact that for $\textbf{h}^i \in l^{p'}$ we have (Definition 2.6 and (2.2))

$$\begin{aligned} \textbf{h}^i((\textbf{h}^i)_{p'})= \Vert \textbf{h}^i \Vert ^{p^{\prime }}_{l^{p^{\prime }}}. \end{aligned}$$

Moreover, $\textbf{h}^i\in \left( l^p\right) ^{*}$ (see (2.2)) and if $\textbf{h}^i\ne 0$, the norm of the bounded linear functional on $l^p$ associated to $\textbf{h}^i$ (2.2) is reached at

$$\begin{aligned} \frac{(\textbf{h}^i)_{p^{\prime }}}{\Vert (\textbf{h}^i)_{p^{\prime }}\Vert _{l^p}}. \end{aligned}$$

Observe that for a given $\textbf{f}\in l^{(p,\infty )}$, any $\textbf{g}\in l^{(p,\infty )}$ can be decomposed as

$$\begin{aligned} \textbf{g}= \{\alpha _i\}_i \textbf{f}+\textbf{h}\end{aligned}$$

(2.4)

with $\textbf{f}{\perp ^j}\textbf{h}$. This can be achieved by setting, for each $i\in {\mathbb N}$,

$$\begin{aligned} \alpha _i:= \left( \sum _{k=1}^{\infty } g_{k,i} |f_{k,i}|^{p-2} f_{k,i}\right) / \Vert \textbf{f}^i\Vert _{l^p}^p \end{aligned}$$

if $\Vert \textbf{f}^i\Vert _{l^p}^p\ne 0$ or $\alpha _i=0$ if $\Vert \textbf{f}^i\Vert _{l^p}^p=0$. We will use the notation

$$\begin{aligned} \{\alpha _i\}_i =\left[ \textbf{f},\textbf{g}\right] /\left[ \textbf{f},\textbf{f}\right] . \end{aligned}$$

For $F \in (l^{(p,\infty )})^*$ one obviously has $F(\textbf{g})=F(\{\alpha _i \textbf{f}^i\}_i)+F(\textbf{h})$; this can also be written as

$$\begin{aligned} F(\textbf{g})=F( \textbf{f}\cdot \left[ \textbf{f}, \textbf{g}\right] / \left[ \textbf{f}, \textbf{f}\right] )+F(\textbf{h}). \end{aligned}$$

Lemma 2.8

Let $L\in (l^{(p,\infty )}\setminus l^{(p,\infty )}_a)^*$. Assume that L attains its norm at $\textbf{f}\in l^{(p,\infty )}$, say $L({\textbf{f}})= \Vert L\Vert _{l^{(p,\infty )}\rightarrow {\mathbb {R}}}$. Furthermore, let $\Vert \textbf{f}^i \Vert _{l^p}=1$ or 0 for each i and assume $\Vert \textbf{f}\Vert _{l^{(p,\infty )}}= \Vert \textbf{f}\Vert _{l^{(p,\infty )}{\setminus } l^{(p,\infty )}_a}=1$. Then for each $\textbf{h}\in \textbf{f}^{\perp ^j}$ one must have $L(\textbf{h})=0$.

Proof

The claim is obvious if $\textbf{h}\in l^{(p,\infty )}_a$. Let $\textbf{h}\notin l^{(p,\infty )}_a, \textbf{h}=\{\textbf{h}^i\}_i\in \textbf{f}^{\perp ^j}$. Then $\Vert \textbf{h}\Vert _{l^{(p,\infty )}{\setminus } l^{(p,\infty )}_a}>0$. On account of the uniform smoothness of $l^p$ and of the fact that for each $i\in {\mathbb N}$, $\textbf{f}^i \perp ^j \textbf{h}^i$ (see (2.1)), for each i it must hold that:

$$\begin{aligned} \lim _{t\rightarrow 0} \frac{\Vert \textbf{f}^i+t\textbf{h}^i\Vert _{l^p}-\Vert \textbf{f}^i \Vert _{l^p} }{ |t|}=0, \end{aligned}$$

(2.5)

uniformly for $\textbf{f}^i$, $\textbf{h}^i$. Thus, for each i it holds

$$\begin{aligned} \Vert \textbf{f}^i+t \textbf{h}^i\Vert _{l^p}\le \Vert \textbf{f}^i \Vert _{l^p} +c(t)|t|, \end{aligned}$$

(2.6)

where $0\le c(t)\searrow 0$ as $t\rightarrow 0$ and c(t) depends only on the uniform smoothness of $l^p$ (see connection with (1.2)). It follows that

$$\begin{aligned} \Vert \textbf{f}+t \textbf{h}\Vert _{l^{(p,\infty })}\le \Vert \textbf{f}\Vert _{l^{(p,\infty })} +c(t)|t|, \end{aligned}$$

whence

$$\begin{aligned} \Vert L\Vert \ge \frac{L(\textbf{f}+t\textbf{h})}{\Vert \textbf{f}+ t\textbf{h}\Vert _{l^{(p,\infty )}} } = \frac{L(\textbf{f})+tL(\textbf{h})}{ \Vert \textbf{f}+ t\textbf{h}\Vert _{l^{(p,\infty )}} }. \end{aligned}$$

In all,

$$\begin{aligned} \Vert L\Vert \Vert \textbf{f}+ t\textbf{h}\Vert _{l^{(p,\infty )}} \ge L(\textbf{f})+tL(\textbf{h}), \end{aligned}$$

which by virtue of (2.6) leads to:

$$\begin{aligned}{} & {} \Vert L\Vert (1+c(t)|t|) \ge \Vert L\Vert + t L(\textbf{h})\\{} & {} \Vert L\Vert c(t) |t| \ge tL(\textbf{h}),\\{} & {} \Vert L\Vert c(t) \ge L(\textbf{h})\, \textrm{sgn}(t). \end{aligned}$$

Letting $t\rightarrow 0^+$, $c(t)\searrow 0$ and it readily follows that $L(\textbf{h}) \le 0.$ A similar argument (letting $t\rightarrow 0^+$, $c(t)\searrow 0$) yields $L(\textbf{h}) \ge 0.$ In all, $L(\textbf{h})=0$ as claimed.

$\square$

Theorem 2.9

Let $L\in (l^{(p,\infty )}{\setminus } l^{(p,\infty )}_a)^*$ and $\textbf{f}\in l^{(p,\infty )}$ satisfy $\Vert \textbf{f}\Vert _{l^{(p,\infty )}}=1$ and

$$\begin{aligned} L(\textbf{f})= \Vert L\Vert _{l^{(p,\infty )}\rightarrow {\mathbb {R}}}. \end{aligned}$$

Then:

(i):

It can be assumed without loss of generality that $\left\{ j:0<\Vert {\textbf{f}}^j\Vert _{l^p}<1\right\} =\emptyset$,

(ii):

$\Vert \textbf{f}\Vert _{l^{(p,\infty )}\setminus l^{(p,\infty )}_a}=1$,

(iii):

$L(\textbf{f})=\Vert L\Vert _{l^{(p,\infty )}\setminus l^{(p,\infty )}_a \rightarrow {\mathbb {R}}}$,

(iv):

$L(\textbf{g})=0$ for each $\textbf{g}$ for which $\textbf{f}\perp ^j \textbf{g}$,

(v):

L is of the form (2.3).

Proof

(i) If $0<\Vert \textbf{f}^j\Vert _{l^p}<1$ for some $j\in {\mathbb N}$, then one could set ${\textbf{h}^j}$=$\left( \Vert \textbf{f}^j\Vert _{l^p}\right) ^{-1}{\textbf{f}^j}$ for those j’s, and $\textbf{h}^j=\textbf{f}^j$ when $\Vert \textbf{h}^j\Vert _{l^{p}}=\Vert \textbf{f}^j\Vert _{l^{p}}=1$. Obviously $1=\Vert \textbf{h}\Vert _{l^{(p,\infty )}}=\Vert \textbf{f}\Vert _{l^{(p,\infty )}}$ and due the maximality of $\textbf{f}$ we get $L({\textbf{h}})= L({\textbf{f}})$.

(ii) If ${\textbf{f}}$ is as given and

$$\begin{aligned} \Vert \textbf{f}\Vert _{l^{(p,\infty )}\setminus l^{(p,\infty )}_a}<1, \end{aligned}$$

then, for some $N>1$,

$$\begin{aligned} \sup \limits _{i\ge N}\Vert \textbf{f}^i\Vert _{l^p}<1. \end{aligned}$$

Since L vanishes on $l^{(p,\infty )}_a$, one easily concludes

$$\begin{aligned} L({\textbf{f}})=L({\textbf{g}}), \end{aligned}$$

where ${\textbf{g}}=({\textbf{f}}^N,{\textbf{f}}^{N+1},...).$ In this case, it would follow

$$\begin{aligned} \Vert L\Vert&=L({\textbf{f}})=\Vert {\textbf{g}}\Vert _{l^{(p,\infty )}\setminus l^{(p,\infty )}}L\left( {\textbf{g}}/\Vert {\textbf{g}}\Vert _{l^{(p,\infty )}\setminus l^{(p,\infty )}}\right) \\&\le \Vert {\textbf{g}}\Vert _{l^{(p,\infty )}\setminus l^{(p,\infty )}}\Vert L\Vert <\Vert L\Vert , \end{aligned}$$

which is a contradiction.

(iii) The proof is obvious, since L vanishes on $l_a^{(p,\infty )}$.

(iv) Already proved (Lemma 2.8).

(v) Set $I_1=\{i: \Vert \textbf{f}^{i}\Vert _{l^p}=1\}$ (From (i) and since $\Vert \textbf{f}\Vert _{l^{(p,\infty )}}=1$, $I_1\ne \emptyset$). We set $\textbf{g}^i:=(\textbf{f}^i\chi _{I_1})_{p}$. Define the map,

$$\begin{aligned} l^{\infty }\ni \{\alpha _i\}_i\longrightarrow \eta (\{\alpha _i\}_i ):=L(\{ \alpha _i \textbf{f}^i \}_i). \end{aligned}$$

It is clear that $\eta \in (l^{\infty })^*$.

Next, for arbitrary $\textbf{g}\in l^{(p,\infty )}$ the decomposition given in (2.4) holds, namely

$$\begin{aligned} \textbf{g}=\{\left[ \textbf{f}^i,\textbf{g}^i\right] \textbf{f}^i\}_i+\textbf{h}, \end{aligned}$$

with $\textbf{f}\perp ^j \textbf{h}$. By virtue of Lemma 2.8 and the linearity of L, one readily obtains

$$\begin{aligned} L({\textbf{g}})=L(\{\left[ \textbf{f}^i,\textbf{g}^i\right] \textbf{f}^i\}_i)=\eta (\{\textbf{g}^i(\textbf{f}^i)\}_i). \end{aligned}$$

$\square$

Theorem 2.10

Let $1<p<\infty$. The subset of $\left( l^{p,\infty }\right) ^*$ consisting of functionals of the form (2.3) is norm-dense in $\left( l^{p,\infty }\right) ^*$.

Proof

The theorem is an immediate consequence of Theorem 2.9 in concert with the theorem of Bishop-Phelps [3]. $\square$

3 Open problems and further comments

We conclude the paper with a generalization of the previous section results and some open problems related to $l^{\{p_n\}}$.

In the proof of Theorem 2.9 the essential requirement is that all $l^p$ spaces are uniformly smooth and uniformly convex Banach spaces. Replacing spaces $l^p$ by a sequence of uniformly convex spaces for which exist a suitable uniform upper bound for the quantity c(t) in inequality (2.6) one can easily generalize the above arguments and derive to the next theorem.

Theorem 3.1

Given a sequence $\{X_i\}$ of uniformly convex and uniformly smooth Banach spaces with norms $\Vert \cdot \Vert _{X_i}$ which have uniform upper bound for modulus of smoothness (i.e. uniform c in (2.6)). Let $l^{\infty }(\{X_i\})$ denote the Banach space of all sequences $\{x_n\}$, $x_n\in X_n$ with the norm $\Vert \{x_n\}\Vert _{l^{\infty }(\{X_i\})}=\sup \limits _n\Vert x_n\Vert _{X_n}.$ Then, the subspace of functionals $\Lambda \in \left( l^{\infty }(\{X_i\})\right) ^{*}$ of the form

$$\begin{aligned} l^{\infty }(\{X_i\})\ni \{x_n\} \longrightarrow \int \limits _{\mathbb N}\lambda _i(x_i)\mathrm{{d}}\mu (i) \end{aligned}$$

(3.1)

is norm dense in $\left( l^{\infty }(\{X_i\}\right) )^{*}$. In (3.1), for each i, $\lambda _i\in X_i^{*}$, $\mu \in ba({\mathbb N})$ and the notation in the paragraph preceding Example 2.7 has been used.

In connection with our work, we highlight Theorem 3.2.4 in [6], which deals with a similar result for an order continuous Koethe function space E over a probability space and a Banach space X.“

Though the space $l^{(p,\infty )}$ we consider in our work can be described in the language of Koethe function spaces as E(X), where $E=l^{\infty }$ and $X=l^p$, Theorem 3.2.4 above does not apply in this case, since $l^{\infty }$ is not order continuous, which is easily seen by considering the sequence $x_1=(1,1,...1,...), x_2=(1/2,1/2,1,1....),... x_n=(1/n, 1/n,.....1/n, 1,1...),...$. The essential novelty of our paper is the consideration of duality in the absence of order continuity. The space $l^{p,\infty }$ whose dual we compute in our work is not covered by Theorem 3.2.4 above and our result does not follow from the work by Pei Kee Lin.

We conclude this section by considering an example of a sequence space in which modulus of smoothness can naturally "deteriorated" as $n\rightarrow \infty$.

Let $(p_n)\subset [1,\infty )$. The variable exponent sequence space $l^{\{p_n\}}$ is defined as the real; vector space

$$\begin{aligned} l^{\{p_n\}}:=\left\{ \{a_n\}\subset {\mathbb R}:\sum _{{\mathbb N}}|\lambda a_n|^{p_n}<\infty \,\,\text {for some}\,\lambda >0\right\} \end{aligned}$$

endowed with the norm

$$\begin{aligned} \Vert \{a_n\}\Vert _{\{p_n\}}:=\inf \left\{ \lambda >0: \sum _{{\mathbb N}}\left| \frac{a_n}{\lambda }\right| ^{p_n}\le 1\right\} . \end{aligned}$$

It is well known that furnished with this norm $l^{\{p_n\}}$ is a Banach space, separable if the sequence $\{p_n\}$ is bounded and reflexive if $\inf \limits _np_n>1$ (see [8])

In the sequel we will set $p^{\prime }_n=\frac{p_n}{p_n-1}$ if $p_n>1$, $p'_n=\infty$ if $p_n=1$.

Theorem 3.2

Let $\{p_n\}\subset (1,\infty )$. Then the following are equivalent:

• (i) $\sup \nolimits _{n}p_n<\infty$,

• (ii) $\left( l^{\{p_n\}}\right) ^{*}=l^{\{p^{\prime }_n\}},$

• (iii) $l^{\{p_n\}}=l_a^{\{p_n\}}.$

Proof

See [8]. $\square$

On another note, it was realized by Orlicz in 1931 [9] that for an unbounded sequence $\{p_n\}$, if $p_n$ tends to infinity fast enough, then $l^{\{p_n\}}$ is isomorphic to $l^{\infty }$.

More precisely:

Theorem 3.3

If $\{p_n\}\subset [1,\infty )$, then $l^{\{p_n\}}$ is isomorphic to $l^{\infty }$ if and only if ${\textbf{1}}_{{\mathbb N}}\in l^{\{p_n\}}$.

Theorems 3.2 and 3.3 provide a full characterizations of the dual space of $l^{\{p_n\}}$ except in the case when $\sup \limits _np_n=\infty$ and ${\textbf{1}}_{{\mathbb N}}\notin l^{\{p_n\}}$. In this case the uniform smoothness and convexity deteriorated as $p_n \rightarrow \infty$.

The characterization of the dual in the latter case remains open: a result in this direction would shed light on the description of the dual space of $L^{p(\cdot )}(\Omega )$ for a domain $\Omega \subset {\mathbb R}^n$ and a measurable $p:\Omega \rightarrow (1,\infty )$ with $\sup \limits _{\Omega }p(x)=\infty$, which is hitherto unknown.