Duality and interpolation for symmetric Banach spaces of noncommutative quasi-martingales

Abstract

Let E be a symmetric Banach space with the Fatou property and $1<p_E\le q_E<p$. We prove the duality for symmetric Banach space $_p\widehat{E}(\mathcal {M})$ which is a kind of noncommutative quasi-martingale space. As its applications, we discuss concrete description of the symmetric Banach space $_p\widehat{E}(\mathcal {M})$ as interpolations of quasi-martingale $L_p$-spaces.

1 Introduction

The theory of noncommutative symmetric spaces has been rapidly developed. Many of the noncommutative martingale results have been transferred to the noncommutative symmetric case. Especially, in [1], J. Yong proved Burkholder-Gundy inequalities for symmetric Banach spaces of noncommutative martingales. In [9], T. N. Bekjan proved the duality for conditional Hardy spaces of martingales in noncommutative symmetric Banach spaces.

The quasi-martingales are generalizations of martingales and play important roles in many different areas of mathematics. In [15], we studied duality theorems for $L_p$-spaces of noncommutative quasi-martingales. In this paper, we will extend the above results to the noncommutative symmetric case. Let E be a symmetric Banach space on [0, 1] with the Fatou property and $1<p_E\le q_E<p$. Then

$$\begin{aligned} (_p\widehat{E}(\mathcal {M}))^*= {_{p'}{\widehat{E}^\times }(\mathcal {M})}, \end{aligned}$$

where $_p\widehat{E}(\mathcal {M})$ and ${_{p'}{\widehat{E}^\times }(\mathcal {M})}$ denote the symmetric Banach spaces of noncommutative quasi-martingales which we refer to the next section for formal definitions. As applications of this result, we obtain the description of the symmetric space $_p\widehat{E}(\mathcal {M})$ as interpolations of noncommutative quasi-martingale $L_p$-spaces.

The organization of the paper is as follows. In Section 2, we give some preliminaries and notations on symmetric Banach spaces, quasi-martingale spaces and interpolations. We prove the main results in Section 3.

2 Preliminaries

Let E be a symmetric Banach space on [0, 1]. The K${\ddot{\text{ o }}}$the dual of E is the function space defined by setting:

$$\begin{aligned} E^{\times }=\{f\in L_0([0,1]) :\int ^1_0|f(t)g(t)|dt<\infty , \forall g\in E \}. \end{aligned}$$

When equipped with the norm $\Vert f\Vert _{E^\times }:=\sup \{\int ^1_0|f(t)g(t)|dt:\Vert g\Vert _E\le 1\},$ $E^{\times }$ is a symmetric Banach space.

A symmetric Banach space E on [0, 1] is said to have the Fatou property if for every sequence $(x_n)_{n}$ in E satisfying $0\le x_n \uparrow$ and $\sup _{n} \Vert x_n\Vert _E<\infty$, the supremum $x=\sup _{n}x_n$ belongs to E and $\Vert x_n\Vert _E\uparrow \Vert x\Vert _E$. Note that E has the Fatou property if and only if $E=E^{\times \times }$ isometrically. Examples of symmetric spaces with the Fatou property are separable symmetric spaces and duals of separable symmetric spaces.

For any $s > 0$ we define the dilation operator $D_s$ on $L_0[0,1]$ by

$$\begin{aligned} (D_sf)(t)= f(st)\chi _{[0,1]}(st),\ \ t\in [0,1]. \end{aligned}$$

If E is a symmetric Banach space on [0, 1], then $D_s$ is a bounded linear operator. Define the lower and upper Boyd indices of E by

$$\begin{aligned} p_E:=\lim \limits _{s\rightarrow \infty }\frac{\log s}{\log \Vert D_s\Vert } \ \ \ \ \ \text{ and } \ \ \ \ \ q_E:=\lim \limits _{s\rightarrow 0^{+}}\frac{\log s}{\log \Vert D_s\Vert }, \end{aligned}$$

respectively. It is well known that $1\le p_E\le q_E\le \infty$ and E has non-trivial Boyd indices, whenever $1<p_E\le q_E<\infty$. We shall need the following duality for Boyd indices:

$$\begin{aligned} \frac{1}{p_E}+\frac{1}{q_{E^{\times }}}=1,\ \ \frac{1}{q_{E}}+\frac{1}{p_{E^{\times }}}=1. \end{aligned}$$

Let E be a symmetric Banach space on [0, 1]. For $0<r<\infty$, we define $E^{(r)}$ and $E_{(r)}$ by

$$\begin{aligned}&E^{(r)}:=\{x: |x|^r\in E\}, \ \ \Vert x\Vert _{E^{(r)}}:=\big \Vert |x|^r\big \Vert ^{\frac{1}{r}}_E, \\&E_{(r)}:=\{x: |x|^{\frac{1}{r}}\in E\}, \ \ \Vert x\Vert _{E_{(r)}}:=\big \Vert |x|^{\frac{1}{r}}\big \Vert ^{r}_E, \end{aligned}$$

respectively. It is clear from the definitions that $E^{(r)}$, $E_{(r)}$ are symmetric and

$$\begin{aligned}&p_{E_{(r)}}={\frac{1}{r}}p_E,\ \ q_{E_{(r)}}={\frac{1}{r}}q_E,\ \ p_{E^{(r)}}=rp_E, \ \ q_{E^{(r)}}=rq_E. \end{aligned}$$

Let $E_i$ be a quasi Banach idea space on [0, 1], $i=1,2.$ The pointwise product space of $E_1$ and $E_2$ is defined as

$$\begin{aligned} E_1\odot E_2=\{x: x=x_1x_2, x_i\in E_i, i=1,2\} \end{aligned}$$

with a functional $\Vert x\Vert _{E_1\odot E_2}$ defined by

$$\begin{aligned} \Vert x\Vert _{E_1\odot E_2}=\inf \{\Vert x\Vert _{E_1}\Vert x\Vert _{E_2}: x=x_1x_2, x_i\in E_i, i=1,2\}. \end{aligned}$$

Note that if E and F are symmetric Banach spaces on [0, 1], then we have the following results (see Theorem 1 in [1]).

• (i) If $0<p<\infty$, then $(E\odot F)^{(p)}=E^{(p)}\odot F^{(p)}.$

• (ii) If $1<p<\infty$, then $(E^{(p)})^{\times }=(E^\times )^{(p)}\odot L_{p'}[0,1].$

Let $\mathcal {M}$ be a semi-finite von Neumann algebra with a faithful normal semi-finite trace $\tau$. The set of all $\tau$-measurable operators is denoted by $L_0(\mathcal {M})$. For $x\in L_0(\mathcal {M})$, define its generalized singular number by

$$\begin{aligned} \mu _t(x)=\inf \{\lambda>0:\tau (\chi _{(\lambda ,\infty )}(|x|))\le t\}, \ \ t>0. \end{aligned}$$

For a given symmetric Banach function space E on [0, 1], we define the corresponding noncommutative space by setting:

$$\begin{aligned} E(\mathcal {M},\tau )=\{x\in L_0(\mathcal {M}):\mu _t(x)\in E\}. \end{aligned}$$

Equipped with the norm $\Vert x\Vert _{E(\mathcal {M},\tau )}:=\Vert \mu _t(x)\Vert _E$, the space $E(\mathcal {M},\tau )$ is a Banach space and is referred to as the noncommutative symmetric Banach space associated with $(\mathcal {M},\tau )$ corresponding to the function space $(E, \Vert \cdot \Vert _E).$ Note that if $1\le p<\infty$ and $E=L_p([0,1])$, then $E(\mathcal {M},\tau )=L_p(\mathcal {M},\tau )$ is the usual noncommutative $L_p$-space associated with $(\mathcal {M},\tau )$.

2.1 Noncommutative quasi-martingales

We first recall the general setup for noncommutative martingales. Let $({\mathcal {M}}_n)_{n\ge 1}$ be an increasing sequence of von Neumann subalgebras of $\mathcal {M}$ such that the union of the ${\mathcal {M}}_n$’s is $\hbox {weak}^*$-dense in $\mathcal {M}$. For every $n\ge 1$, the restriction $\tau |_{\mathcal {M}_n}$ of $\tau$ to $\mathcal {M}_n$ remains semi-finite, still denoted by $\tau$, and we assume that there exists a trace preserving conditional expectation ${\mathcal {E}}_n$ from $\mathcal {M}$ onto $\mathcal {M}_n$. In this case, $(\mathcal {M}_n)_{n\ge 1}$ is called a filtration of $\mathcal {M}$. Note that ${\mathcal {E}}_n$ extends to a contractive projection from $L_p(\mathcal {M})$ onto $L_p(\mathcal {M}_n)$ for all $1\le p\le \infty$. A noncommutative $E(\mathcal {M})$-martingale with respect to $({\mathcal {M}}_n)_{n\ge 1}$ is a sequence $x=(x_n)_{n\ge 1}$ such that $x_n\in E(\mathcal {M}_n)$ and ${\mathcal {E}}_{n}(x_{n+1})=x_n\ \ \text{ for } \text{ any }\ \ n\ge 1.$ Let $\Vert x\Vert _{E(\mathcal {M})}=\sup _{n\ge 1}\Vert x_n\Vert _{E(\mathcal {M})}$. If $\Vert x\Vert _{E(\mathcal {M})}<\infty$, then x is called a bounded $E(\mathcal {M})$-martingale. The martingale difference sequence $dx=(dx_n)_{n\ge 1}$ of x is defined by $dx_n=x_n-x_{n-1}$ for $n\ge 1$. Here and in the following, we set $x_0=0$ and $\mathcal {E}_0=\mathcal {E}_1$ for the sake of convenience.

In this paper, we are concerned with the following quasi-martingales in noncommutative symmetric Banach spaces.

Definition 2.1

Let E be a symmetric Banach space on [0, 1] and $1\le p\le \infty$. A noncommutative ${_p}E(\mathcal {M})$-quasi-martingale with respect to $({\mathcal {M}}_n)_{n\ge 1}$ is a sequence $x=(x_n)_{n\ge 1}$ such that $x_n\in E(\mathcal {M}_n)$ for $n\ge 1$ and (with $\mathcal {E}_0 = 0, x_0=0$)

$$\begin{aligned} \sum \limits _{n=1}^\infty \Vert {\mathcal {E}}_{n-1}(dx_n)\Vert _{E(\mathcal {M})}^p<\infty . \end{aligned}$$

Let $y_n=\sum \limits _{k=1}^n(dx_k-{\mathcal {E}}_{k-1}(dx_k))$ for $n\ge 1$. We set

$$\begin{aligned} \Vert x\Vert _{{_p}\widehat{E}(\mathcal {M})}:=\sup _n\Vert y_n\Vert _{E(\mathcal {M})} +(\sum \limits _{n=1}^\infty \Vert {\mathcal {E}}_{n-1}(dx_n)\Vert _{E(\mathcal {M})}^p)^{\frac{1}{p}}. \end{aligned}$$

If $\Vert x\Vert _{{_p}\widehat{E}(\mathcal {M})}<\infty$, then x is called a bounded ${_p}E(\mathcal {M})$-quasi-martingale. The quasi-martingale space ${_p}\widehat{E}(\mathcal {M})$ is defined as the space of all bounded ${_p}E(\mathcal {M})$-quasi-martingales, equipped with the norm $\parallel \cdot \parallel _{{_p}\widehat{E}(\mathcal {M})}$. We remark that if $1\le q\le \infty$ and $E=L_q([0,1])$ then ${_p}\widehat{E}(\mathcal {M})={_p}\widehat{L_q}(\mathcal {M})$, where ${_p}\widehat{L_q}(\mathcal {M})$ consists of $x=(x_n)_{n\ge 1}\subset L_q(\mathcal {M})$ for which

$$\begin{aligned} \Vert x\Vert _{{_p}\widehat{L_q}(\mathcal {M})}=\sup _n\Vert y_n\Vert _{L_q(\mathcal {M})} +(\sum \limits _{n=1}^\infty \Vert {\mathcal {E}}_{n-1}(dx_n)\Vert ^p_{L_q(\mathcal {M})})^{\frac{1}{p}}. \end{aligned}$$

Now we define the noncommutative space ${_p}G_E(\mathcal {M})$ which is used in the proof of our main results.

Definition 2.2

Let E be a symmetric Banach space on [0, 1] and $1\le p\le \infty$. The noncommutative space ${_p}G_E(\mathcal {M})$ is defined as the subspace of $l_p(E(\mathcal {M}))$ consisting of all sequences $dx=(dx_n)_{n\ge 1}$ such that $x=(x_n)_{n\ge 1}$ is a predictable ${_p}E(\mathcal {M})$-quasi-martingale with $x_1=0$, and is equipped with the norm

$$\begin{aligned} \Vert x\Vert _{{_p}G_E(\mathcal {M})}=(\sum \limits _{n=1}^\infty \Vert dx_n\Vert ^p_{E(\mathcal {M})})^{\frac{1}{p}}. \end{aligned}$$

Note that if $1\le q\le \infty$ and $E=L_q([0,1])$ then ${_p}G_E(\mathcal {M})={_p}G_q(\mathcal {M})$, where ${_p}G_q(\mathcal {M})$ denotes the space of $x=(x_n)_{n\ge 1}\subset L_q(\mathcal {M})$ for which

$$\begin{aligned} \Vert dx\Vert _{{_p}G_q(\mathcal {M})}=(\sum \limits _{n=1}^\infty \Vert dx_n\Vert ^p_{L_q(\mathcal {M})})^{\frac{1}{p}}. \end{aligned}$$

The following theorem plays an important role in our paper which we call Doob’s decomposition.

Theorem 2.3

(Doob’s decomposition) Let E be a symmetric Banach space on [0, 1] and $1\le p\le \infty$. Then each bounded ${_p}E(\mathcal {M})$-quasi-martingale $x=(x_n)_{n\ge 1}$ can be uniquely decomposed as a sum of two sequences $y=(y_n)_{n\ge 1}$ and $z=(z_n)_{n\ge 1}$, where $y=(y_n)_{n\ge 1}$ is a bounded $E(\mathcal {M})$-martingale and $z=(z_n)_{n\ge 1}$ is a predicable ${_p}E(\mathcal {M})$-quasi-martingale with $z_1=0$.

Proof

The proof is similar with Lemma 2.2 in [15].

2.2 Interpolations

For a compatible Banach couple $(X_0,X_1)$, we define the K-functional by setting for any $x\in X_0+X_1$ and $t>0$,

$$\begin{aligned} K_t(x;X_0,X_1) = \inf \{\Vert x_0\Vert _{X_0}+t\Vert x_1\Vert _{ X_1}:x=x_0+x_1, x_0\in X_0,x_1\in X_1\}. \end{aligned}$$

The interpolation space $(X_0,X_1)_{E,K}$ is defined as the space of all elements $x\in X_0+X_1$ such that

$$\begin{aligned} \Vert x\Vert _{(X_0,X_1)_{E,K}}:=\left\| \frac{K_t(x;X_0,X_1)}{t}\right\| _E<\infty . \end{aligned}$$

We may state the following interpolation result which is needed in the sequel (see Theorem 2.2 in [19] and [20]).

Theorem 2.4

Let E be a symmetric Banach space on [0, 1] with the Fatou property and $1<p<p_E\le q_E<q<\infty .$ Then there exists a symmetric Banach space F with nontrivial Boyd indices such that

$$\begin{aligned} E(\mathcal {M})=(L_p(\mathcal {M}), L_q(\mathcal {M}))_{F,K}\ \ \text{(with } \text{ equivalent } \text{ norms) }. \end{aligned}$$

(2.1)

Proof

By Theorem 2.2 in [19], there is a symmetric Banach function space F on [0, 1] such that $f\in E$ if and only if $t\rightarrow K_t(f;L_p[0,1],L_q[0,1]) \in F$ and there exist a constant C such that

$$\begin{aligned} C^{-1}\Vert t\rightarrow K_t(f;L_p[0,1],L_q[0,1])\Vert _F\le \Vert f\Vert _E\le C\Vert t\rightarrow K_t(f;L_p[0,1],L_q[0,1]). \end{aligned}$$

(2.2)

For any $x\in E(\mathcal {M})$, using the results $K_t(\mu (x);L_p[0,1],L_q[0,1])\approx K_t(x;L_p(\mathcal {M}),L_p(\mathcal {M})$ and $\Vert \mu (x)\Vert _E=\Vert x\Vert _{E(\mathcal {M})}$, we can extend (2.2) to the noncommutative setting. The proof is completed.

Throughout the paper $p'$ will denote the conjugate index of p. $\square$

3 Main results

Our first result in this section is concerned with the dual space of $_p\widehat{E}(\mathcal {M})$ which is the symmetric Banach space of noncommutative quasi-martingales.

Theorem 3.1

Let E be a symmetric Banach space on [0, 1] with the Fatou property and $1<p_E\le q_E<p$. Then

$$\begin{aligned} \big ({_p}\widehat{E}(\mathcal {M})\big )^*={_{p'}}\widehat{E}^{\times }(\mathcal {M}) \end{aligned}$$

isometrically, with associated duality bracket given by

$$\begin{aligned} \forall x\in {_p}\widehat{E}(\mathcal {M}), \ \ \forall u\in {_{p'}}\widehat{E}^{\times }(\mathcal {M}),\ \ (x,u)=\tau (\nu y)+\sum \limits _{n=1}^\infty \tau (d\omega _ndz_n), \end{aligned}$$

where $\mu _n=\nu _n+\omega _n$ and $x_n=y_n+z_n(n\ge 1)$ are the Doob’s decomposition of u and x respectively.

For the proof we need the following Lemma.

Lemma 3.2

Let E be a symmetric Banach space on [0, 1] with the Fatou property and $1<p_E\le q_E<p$. Then

$$\begin{aligned} E=F\odot {E^{\times }}^{(\frac{p}{p'})}, \end{aligned}$$

where $F=(E^{\times (\frac{1}{p'})})^{\times }$ is separable.

Proof

From the proof of Lemma 2.1 in [1], we know that $E^{\times (\frac{1}{p'})}$ is reflexive and F is separable. By (ii) of the properties of pointwise product spaces, we have

$$\begin{aligned} E=(E^\times )^\times =([E^{\times (\frac{1}{p'})}]^{(p')})^\times = ([E^{\times (\frac{1}{p'})}]^\times )^{(p')}\odot L_{p}[0,1]=F^{(p')}\odot L_{p}[0,1]. \end{aligned}$$

Using the equality $L_{1}[0,1]=E\odot E^\times$ (see Theorem 1.2 in [1]) and (i) of the properties of pointwise product spaces, we obtain that

$$\begin{aligned} E=F^{(p')}\odot (F\odot E^{\times (\frac{1}{p'})})^{(p)} =F^{(p')}\odot (F^{(p)}\odot E^{\times (\frac{1}{p'})(p)}) =F\odot E^{\times (\frac{p}{p'})} . \end{aligned}$$

The proof is completed. $\square$

We also require the following duality result (see Theorem 5.6 in [6]).

Lemma 3.3

Let E be a symmetric Banach space on [0, 1] with the Fatou property, then $\big (E(\mathcal {M})\big )^*=E^\times (\mathcal {M})$ isometrically, with associated duality bracket given by

$$\begin{aligned} (x,y)=\tau (xy), \ \ x\in E(\mathcal {M}), y\in E^\times (\mathcal {M}). \end{aligned}$$

Now, we concern the dual space of $l_p(E(\mathcal {M}))$ which is the main ingredient in the proof of Theorem 3.1.

Lemma 3.4

Let E be a symmetric Banach space on [0, 1] with the Fatou property and $1<p_E\le q_E<p$. Then

$$\begin{aligned} \big (l_p({E}(\mathcal {M}) )\big )^*=l_{p'}({E}^{\times }(\mathcal {M})) \end{aligned}$$

with equivalent norms.

Proof

Let $x=(x_n)_{n\ge 1}\in l_p({E}(\mathcal {M}))$ and $y=(y_n)_{n\ge 1}\in l_{p'}({E}^{\times }(\mathcal {M}))$. Now, we define a linear functional on $l_p({E}(\mathcal {M}))$ by

$$\begin{aligned} l_y(x)=\sum \limits _{n=1}^\infty \tau (x_ny_n). \end{aligned}$$

Then by Lemma 3.3 and Hölder’s inequality,

$$\begin{aligned} \left| \sum \limits _{n=1}^\infty \tau (x_ny_n)\right|\le & {} \sum \limits _{n=1}^\infty |\tau (x_ny_n)|\\\le & {} \sum \limits _{n=1}^\infty \Vert x_n\Vert _{{E}(\mathcal {M})}\Vert y_n\Vert _{{E}^{\times }(\mathcal {M})}\\\le & {} (\sum \limits _{n=1}^\infty \Vert x_n\Vert ^p_{{E}(\mathcal {M})})^{\frac{1}{p}} (\sum \limits _{n=1}^\infty \Vert y_n\Vert ^{p'}_{{E}^{\times }(\mathcal {M})})^{\frac{1}{{p'}}}\\= & {} \Vert x\Vert _{ l_p({E}(\mathcal {M}))}\Vert y\Vert _{l_{p'}({E}^{\times }(\mathcal {M}))}. \end{aligned}$$

Thus the series $\sum \limits _{n=1}^\infty \tau (x_ny_n)$ converges absolutely. Therefore, $l_y(x)$ is continuous on $l_p({E}(\mathcal {M}))$ and $\Vert l_y\Vert \le \Vert y\Vert _{l_{p'}({E}^{\times }(\mathcal {M}))}.$

We pass to the converse inclusion. Let $l \in \big (l_p({E}(\mathcal {M}) )\big )^*$ of norm one. For every $n\ge 1$, set

$$\begin{aligned} l_n(x_n)=l(\theta ),\ \ x_n\in E(\mathcal {M}), \end{aligned}$$

where $\theta =(\mathop {\underbrace{{0,\ldots ,0,x_n,}}}\limits _n0,\ldots )$.Then

$$\begin{aligned} |l_n(x_n)|=|l(\theta )|\le \Vert l\Vert \Vert \theta \Vert _{l_p({E}(\mathcal {M}))}=\Vert x_n\Vert _{E(\mathcal {M})}. \end{aligned}$$

This implies that $l_n\in (E(\mathcal {M}))^*$. Since $\big ({E}(\mathcal {M})\big )^*={E}^{\times }(\mathcal {M})$, the representation theorem allows us to find an element $y_n\in {E}^{\times }(\mathcal {M})$ such that

$$\begin{aligned} l_n(x_n)=\tau (x_ny_n),\ \ x_n\in E(\mathcal {M}). \end{aligned}$$

Thus we have that

$$\begin{aligned} l(x)=\sum \limits _{n=1}^\infty l_n(x_n)=\sum \limits _{n=1}^\infty \tau (x_ny_n) \end{aligned}$$

(3.1)

for any finite sequence $x=(x_n)_{n\ge 1}\in l_p({E}(\mathcal {M}))$. We must show that $y=(y_n)_{n\ge 1}\in l_{p'}(E^{\times }(\mathcal {M}))$ and is of norm $\le 1$. Now, fix an n. Note that for any $k\le n$

$$\begin{aligned} \Vert y_k\Vert ^{p'}_{{E}^{\times }(\mathcal {M})} =\Vert (|y_k|^{p'})^{\frac{1}{{p'}}}\Vert ^{p'}_{E^{\times }(\mathcal {M})} =\Vert |y_k|^{p'}\Vert _{E^{\times (\frac{1}{{p'}})}(\mathcal {M})} =\sup \{\tau (a_k|y_k|^{p'}):a_k\in F(\mathcal {M}),\Vert a_k\Vert _{F(\mathcal {M})}\le 1\}, \end{aligned}$$

where $\big (F(\mathcal {M})\big )^*={E}^{\times (\frac{1}{{p'}})}(\mathcal {M}).$ Thus for an arbitrarily given $\varepsilon >0$, there exists $a_k^\varepsilon \in F(\mathcal {M})$ and $\Vert a_k^\varepsilon \Vert _{F(\mathcal {M})}\le 1$ such that

$$\begin{aligned} \Vert y_k\Vert ^{p'}_{{E}^{\times }(\mathcal {M})}\le \tau (a_k^\varepsilon |y_k|^{p'})+\frac{\varepsilon }{2^k}. \end{aligned}$$

(3.2)

Set $z_k=\frac{1}{\gamma _n}a_k^\varepsilon |y_k|^{{p'}-2}y^*,$ where $\gamma _n=(\sum \limits _{k=1}^n\Vert y_k\Vert _{E^{\times }(\mathcal {M})}^{p'})^{\frac{1}{{p}}}.$ Then noting that $|y_k|^{{p'}-2}y^*\in E^{\times (\frac{p}{p'})}$ and by Lemma 3.2, we get $z_k\in E(\mathcal {M})$ and

$$\begin{aligned} \Vert z_k\Vert _{E(\mathcal {M})}\le \frac{1}{\gamma _n}\Vert a_k^\varepsilon \Vert _{F(\mathcal {M})}\Vert |y_k|^{{p'}-2}y^*\Vert _{E^{\times (\frac{p}{p'})}}\le \frac{1}{\gamma _n}\Vert y_k\Vert ^{{p'}-1}_{E^{\times }(\mathcal {M})}. \end{aligned}$$

(3.3)

Thus we have that

$$\begin{aligned} \left( \sum \limits _{k=1}^n\Vert z_k\Vert _{E(\mathcal {M})}^p\right) ^{\frac{1}{p}} \le \frac{1}{\gamma _n}(\sum \limits _{k=1}^n\Vert y_k\Vert _{E^\times (\mathcal {M})}^{((p'-1)p)})^{\frac{1}{p}} =\frac{1}{\gamma _n}\left( \sum \limits _{k=1}^n\Vert y_k\Vert _{E^{\times }(\mathcal {M})}^{p'}\right) ^{\frac{1}{{p}}}=1. \end{aligned}$$

Let $z^{(n)}=(z_1,\ldots ,z_n,0,\ldots ).$ Then $z^{(n)}\in l_p({E}(\mathcal {M}))$ and $\Vert z^{(n)}\Vert _{l_p ({E}(\mathcal {M}))}\le 1.$ Using (3.1) and (3.2), we obtain that

$$\begin{aligned} l(z^{(n)})= & {} \sum \limits _{k=1}^n\tau (z_ky_k)\\= & {} \frac{1}{\gamma _n}\sum \limits _{k=1}^n\tau (a_k^\varepsilon |y_k|^{p'})\\\ge & {} \frac{1}{\gamma _n}\sum \limits _{k=1}^n(\Vert y_k\Vert ^{p'}_{{E}^{\times }(\mathcal {M})}-\frac{\varepsilon }{2^k})\\\ge & {} (\sum \limits _{k=1}^n\Vert y_k\Vert ^{p'}_{{E}^{\times }(\mathcal {M})})^{\frac{1}{{p'}}}- \frac{1}{\gamma _n}\varepsilon . \end{aligned}$$

Thus we have that

$$\begin{aligned} \left( \sum \limits _{k=1}^n\Vert y_k\Vert _{E^{\times }(\mathcal {M})}^{p'})^{\frac{1}{{p'}}}\le \frac{1}{\gamma _n}\varepsilon +l(z^{(n)}\right) \le 1. \end{aligned}$$

It follows that

$$\begin{aligned} \left( \sum \limits _{n=1}^\infty \Vert y_n\Vert _{E^{\times }(\mathcal {M})}^{p'}\right) ^{\frac{1}{{p'}}}\le 1 \end{aligned}$$

as $n\rightarrow \infty$ which implies

$$\begin{aligned} y\in l_{p'}(E^{\times }(\mathcal {M}))\ \ \text{ and } \ \ \ \Vert y\Vert _{l_{p'}(E^{\times }(\mathcal {M}))}\le 1. \end{aligned}$$

For any $x=(x_n)_{n\ge 1}\in l_p({E}(\mathcal {M}))$, let $x^{(n)}=(x_1,\ldots ,x_n,0,\ldots )$ $(n\ge 1)$. Then

$$\begin{aligned} \Vert x-x^{(n)}\Vert _{l_p({E}(\mathcal {M}))}\rightarrow 0\ \ (n\rightarrow \infty ). \end{aligned}$$

Using (3.1), we have

$$\begin{aligned} l(x)=\lim \limits _{n\rightarrow \infty }l(x^{(n)})=\lim \limits _{n\rightarrow \infty }\sum \limits _{i=1}^n\tau (x_iy_i) =\sum \limits _{i=1}^\infty \tau (x_iy_i). \end{aligned}$$

The proof is completed. $\square$

The proof of Theorem 3.1

Let $\mu =(\mu _n)_{n\ge 1}\in {_{p'}}\widehat{E}^{\times }(\mathcal {M})$ and $x=(x_n)_{n\ge 1}\in {_p}\widehat{E}(\mathcal {M})$. Let $\mu _n=\nu _n+\omega _n$ and $x_n=y_n+z_n(n\ge 1)$ be the Doob’s decomposition of $\mu$ and x respectively. Then $y=(y_n)_{n\ge 1}$ is a bounded $E(\mathcal {M})$-martingale and $\nu = (\nu _n)_{n\ge 1}$ is a bounded $E^{\times }(\mathcal {M})$-martingale. Thus there exist $y_\infty \in E(\mathcal {M})$ and $\nu _\infty \in E^{\times }(\mathcal {M})$ such that $y_n\overset{E(\mathcal {M})}{\longrightarrow }y_\infty ,\ \ \nu _n\overset{E^{\times }(\mathcal {M})}{\longrightarrow }\nu _\infty .$

Now we define a linear functional on $_{p}\widehat{{E}}(\mathcal {M})$ by

$$\begin{aligned} l_\mu (x)=\tau (\nu _\infty y_\infty )+\sum \limits _{n=1}^\infty \tau (d\omega _ndz_n). \end{aligned}$$

Then by Lemma 3.3 and Hölder’s inequality,

$$\begin{aligned} |l_\mu (x)|\le & {} \Vert \nu _{\infty }\Vert _{E^{\times }(\mathcal {M})}\Vert y_{\infty }\Vert _{E(\mathcal {M})}+ \sum \limits _{n=1}^\infty \Vert d\omega _n\Vert _{E^{\times }(\mathcal {M})}\Vert dz_n\Vert _{E(\mathcal {M})}\\\le & {} \Vert \nu _{\infty }\Vert _{E^{\times }(\mathcal {M})}\Vert y_{\infty }\Vert _{E(\mathcal {M})}+ \left( \sum \limits _{n=1}^\infty \Vert d\omega _n\Vert ^{p'}_{E^{\times }(\mathcal {M})}\right) ^{\frac{1}{{p'}}} \left( \sum \limits _{n=1}^\infty \Vert dz_n\Vert ^{p}_{E^(\mathcal {M})}\right) ^{\frac{1}{p}}\\\le & {} \Bigg (\Vert \nu _\infty \Vert _{E^{\times }(\mathcal {M})}+\left( \sum \limits _{n=1}^\infty \Vert d\omega _n\Vert ^{p'}_{E^{\times }(\mathcal {M})}\right) ^{\frac{1}{{p'}}}\Bigg ) \big (\Vert y_\infty \Vert _{E(\mathcal {M})}+\left( \sum \limits _{n=1}^\infty \Vert dz_n\Vert ^p_{E^(\mathcal {M})}\right) ^{\frac{1}{p}}\big )\\= & {} \Vert \mu \Vert _{{_{p'}\widehat{E}^{\times }(\mathcal {M})}}\Vert x\Vert _{_{p}\widehat{E}(\mathcal {M})} . \end{aligned}$$

Thus $l_\mu (x)$ is continuous on $_{p}\widehat{E}(\mathcal {M})$ and $\Vert l_\mu \Vert \le \Vert \mu \Vert _{{_{p'}\widehat{E}^{\times }(\mathcal {M})}}.$

We pass to the converse inclusion. Let $l\in \big (_{p}\widehat{E}(\mathcal {M})\big )^{*}$ of norm one. Let $l_1$ be the restriction of l on $E(\mathcal {M})$. Noting that $\big ({E}(\mathcal {M})\big )^*={E}^{\times }(\mathcal {M})$, there exists $\nu \in E^{\times }(\mathcal {M})$ and $\Vert \nu \Vert _{E^{\times }(\mathcal {M})}\le 1$ such that

$$\begin{aligned} l_1(a)=\tau (av),\ \ a\in E(\mathcal {M}). \end{aligned}$$

(3.4)

On the other hand, define a functional on $_{p}G_E(\mathcal {M})$ by

$$\begin{aligned} l_2(db)=l(b),\ \ db=(db_n)_{n\ge 1}\in {_{p}}G_E(\mathcal {M}). \end{aligned}$$

Then $|l_2(db)|\le \Vert l\Vert \Vert b\Vert _{_{p}\widehat{E}(\mathcal {M})}=\Vert db\Vert _{_{p}G_E(\mathcal {M})}.$ Thus we have that $l_2$ is a continuous linear functional on $_{p}G_E(\mathcal {M})$ and $\Vert l_2\Vert \le 1$. Recall that $_{p}G_E(\mathcal {M})$ is the closed subspace of $l_p(E(\mathcal {M}))$. By the Hahn-Banach theorem, $l_2$ extends to a norm one functional $\widetilde{l}_2$ on $l_p(E(\mathcal {M}))$. Consequently, by Lemma 3.4, $\widetilde{l}_2$ is given by a norm one element $\omega '=(\omega _n')_{n\ge 1}$ of $l_{p'}(E^\times (\mathcal {M}))$. Thus

$$\begin{aligned} l_2(db)=\sum \limits _{n=1}^\infty \tau (d\omega _n'db_n)\ \ db=(db_n)_{n\ge 1}\in {_{p}G_E}(\mathcal {M}). \end{aligned}$$

(3.5)

Set $\omega _1=0$ and $\omega _n=\sum \limits _{k=1}^{n}{\mathcal {E}}_{k-1}(\omega _k') (n\ge 2)$. For any $db=(db_n)_{n\ge 1}\in {_p}G_E(\mathcal {M})$, noting that $db=(db_n)_{n\ge 1}$ is predicable, it follows from (3.5) that

$$\begin{aligned} l_2(db)=\sum \limits _{n=1}^\infty \tau ({\mathcal {E}}_{n-1}(\omega _n'db_n)) =\sum \limits _{n=1}^\infty \tau (db_n{\mathcal {E}}_{n-1}(\omega _n')) =\sum \limits _{n=1}^\infty \tau (d\omega _ndb_n). \end{aligned}$$

(3.6)

It is easy to see that $\omega =(\omega _n)_{n\ge 1}$ is predicable with $\omega _1=0$ and

$$\begin{aligned} \left( \sum \limits _{n=1}^\infty \Vert d\omega _n\Vert ^{p'}_{E^{\times }(\mathcal {M})}\right) ^{\frac{1}{{p'}}} =\left( \sum \limits _{n=1}^\infty \Vert \mathcal {E}_{n-1}(\omega _n')\Vert ^{p'}_{E^{\times }(\mathcal {M})}\right) ^{\frac{1}{{p'}}}\le \left( \sum \limits _ {n=1}^\infty \Vert \omega _n'\Vert ^{p'}_{E^{\times }(\mathcal {M})}\right) ^{\frac{1}{{p'}}}=1. \end{aligned}$$

Set $\mu _n=\nu _n+\omega _n(n\ge 1)$, where $\nu _n=\mathcal {E}_n(\nu )(n\ge 1)$. Then $\mu =(\mu _n)_{n\ge 1}\in {_{p'}}\widehat{E}^{\times }(\mathcal {M})$ and

$$\begin{aligned} \Vert \mu \Vert _{_{p'}\widehat{E}^{\times }(\mathcal {M})} =\Vert \nu \Vert _{E^{\times }(\mathcal {M})}+ \left( \sum \limits _{n=1}^\infty \Vert d\omega _n\Vert ^{p'}_{E^{\times }(\mathcal {M})}\right) ^{\frac{1}{{p'}}}\le 2. \end{aligned}$$

For any $x=(x_n)_{n\ge 1}\in {_p}\widehat{{E}}(\mathcal {M})$, let $x_n=y_n+z_n(n\ge 1)$ be its Doob’s decomposition. Noting that $y=(y_n)_{n\ge 1}$ is a bounded $E(\mathcal {M})$-martingale and $dz=(dz_n)_{n\ge 1}\in {_p}G_E(\mathcal {M})$, it follows from (3.4) and (3.6) that

$$\begin{aligned} l(x)=l_1(y)+l_2(dz)=\tau (y_{\infty }\upsilon _{\infty })+\sum \limits _{n=1}^\infty \tau (d\omega _ndz_n), \end{aligned}$$

where $y_\infty$ is the limit of $(y_n)_{n\ge 1}$ in $E(\mathcal {M})$. The proof is completed. $\square$

As applications of Theorem 3.1, we shall consider the symmetric space ${_p}\widehat{E}(\mathcal {M})$ as interpolations of noncommutative quasi-martingale $L_p$-spaces, which is a generalization of Theorem 2.4.

Theorem 3.5

Let E be a symmetric Banach space on [0, 1] with the Fatou property and $1<p<p_E\le q_E<q<\infty .$ Then there exists a symmetric Banach space F with nontrivial Boyd indices such that

$$\begin{aligned} _s\widehat{E}(\mathcal {M})=(_s\widehat{L}_p(\mathcal {M}), {_s\widehat{L}}_q(\mathcal {M}))_{F,K}, \end{aligned}$$

where $1<s<p_E.$

For the proof of Theorem 3.5, we need the following lemmas (see [2]).

Lemma 3.6

Let E be a symmetric Banach space on [0, 1] with the Fatou property, and let $(X_1, X_2)$ be a compatible Banach couple. Then

$$\begin{aligned} (X_1, X_2)^*_{F,K}=(X_2^*, X_1^*)_{F^\times ,K}. \end{aligned}$$

Lemma 3.7

Let E be a symmetric Banach space on [0, 1] with the Fatou property and $1<p<p_E\le q_E<q<\infty .$ Then there exists a symmetric Banach space F with nontrivial Boyd indices such that

$$\begin{aligned} l_s({E}(\mathcal {M}))=\big (l_s({L}_p(\mathcal {M})), l_s({L}_q(\mathcal {M}))\big )_{F,K}, \end{aligned}$$

where $1<s<p_E.$

Proof

Let $x=(x_n)_{n\ge 1}\in l_s(E(\mathcal {M}))$. Then by Theorem 2.4, there exists a symmetric Banach space F with nontrivial Boyd indices such that $x_n\in ({L}_p(\mathcal {M}), {L}_q(\mathcal {M}))_{F,K}$ for any $n\ge 1$. For any $a,b>0$, $\alpha _s(a^s+b^s)\le (a+b)^s\le \beta _s (a^s+b^s)$ for some constants $\alpha _s,\beta _s$ depending only on s. Using this fact, it is easy to show that

$$\begin{aligned}&\big ( K_t(x;l_s(L_{p}(\mathcal {M})),l_s(L_{q}(\mathcal {M})))\big )^s\\&\quad \le \beta _s\sum \limits _{n=1}^\infty \inf \limits _{x_n=x_n^0+x_n^1}(\Vert x_n^0\Vert ^s_{L_{p}(\mathcal {M})}+t^s\Vert x_n^1\Vert ^s_{L_{q}(\mathcal {M})})\\&\quad \le \frac{\beta _s}{\alpha _s}\sum \limits _{n=1}^\infty \big ( K_t(x_n;L_{p}(\mathcal {M}),L_{q}(\mathcal {M}))\big )^s. \end{aligned}$$

Noting that$F_{(s)}$ is a quasi-Banach space, we have that

$$\begin{aligned}&\Vert x\Vert ^s_{(l_s(L_{p}(\mathcal {M})),l_s(L_{q}(\mathcal {M})))_{F,K}}\\&\quad = \left\| \frac{ K_t(x;l_s(L_{p}(\mathcal {M})),l_s(L_{q}(\mathcal {M})))^s}{t^s}\right\| _{F_{(s)}}\\&\quad \le C_s \sum \limits _{n=1}^\infty \left\| \frac{ K_t(x_n;L_{p}(\mathcal {M}),L_{q}(\mathcal {M}))^s}{t^s}\right\| _{F_{(s)}}\\&\quad = C_s \sum \limits _{n=1}^\infty \left\| \frac{ K_t(x_n;L_{p}(\mathcal {M}),L_{q}(\mathcal {M}))}{t}\right\| ^s_{F}, \end{aligned}$$

where $C_s$ is a constant depending on s. This means that

$$\begin{aligned} \Vert x\Vert _{\big (l_s({L}_p(\mathcal {M})), l_s({L}_q(\mathcal {M}))\big )_{F,K}}\le C_s\Vert x\Vert _{l_s({E}(\mathcal {M}))} \end{aligned}$$

and

$$\begin{aligned} l_s({E}(\mathcal {M}))\subset \big (l_s({L}_p(\mathcal {M})), l_s({L}_q(\mathcal {M}))\big )_{F,K}. \end{aligned}$$

(3.7)

Similarly, we have that $l_{s'}({E^\times }(\mathcal {M}))\subset \big (l_{s'}({L}_{q'}(\mathcal {M})), l_{s'}({L}_{p'}(\mathcal {M}))\big )_{F^\times ,K}.$ It follows that

$$\begin{aligned} (l_{s'}({E^\times }(\mathcal {M})))^*\supset \big ((l_{s'}({L}_{q'}(\mathcal {M})), l_{s'}({L}_{p'}(\mathcal {M})))\big )^*_{F^\times ,K}. \end{aligned}$$

Observe that $p_{E^\times }\le q_{E^\times }\le s'$. Thus by Lemma  3.4 and Lemma 3.6, we have that

$$\begin{aligned} l_s({E}(\mathcal {M}))\supset \big (l_s({L}_p(\mathcal {M})), l_s({L}_q(\mathcal {M}))\big )_{F,K}. \end{aligned}$$

(3.8)

Putting (3.7) and (3.8) together, we obtain that

$$\begin{aligned} l_s({E}(\mathcal {M}))=\big (l_s({L}_p(\mathcal {M})), l_s({L}_q(\mathcal {M}))\big )_{F,K}. \end{aligned}$$

The proof is completed.

The following is an interpolation result on the space $_pG_E(\mathcal {M})$.

Lemma 3.8

Let E be a symmetric Banach space on [0, 1] with the Fatou property and $1<p<p_E\le q_E<q<\infty .$ Then there exists a symmetric Banach space F with nontrivial Boyd indices such that

$$\begin{aligned} _sG_E(\mathcal {M})=\big ({_sG_p}(\mathcal {M}), {_sG_q}(\mathcal {M})\big )_{F,K}, \end{aligned}$$

where $1<s<p_E.$

Proof

Note that ${_sG_p}(\mathcal {M})$ consists of quasi-martingale difference sequences in $l_s(L_p(\mathcal {M}))$. So ${_sG_p}(\mathcal {M})$ is 1-complemented in $l_s(L_p(\mathcal {M}))$ via the projection

$$\begin{aligned} P: {\left\{ \begin{array}{ll} \ l_s(L_p(\mathcal {M}))\rightarrow {_sG_p}(\mathcal {M});\\ \ (a_n)_{n\ge 1} \rightarrow (\mathcal {E}_{n-1}(a_n))_{n\ge 1}. \end{array}\right. } \end{aligned}$$

It follows that for any $x\in \big ({_sG_p}(\mathcal {M}), {_sG_q}(\mathcal {M})\big )_{F,K}$,

$$\begin{aligned} K_t(x;{_sG_p}(\mathcal {M}),{_sG_q}(\mathcal {M}))= K_t(x;l_s(L_{p}(\mathcal {M})),l_s(L_{q}(\mathcal {M})))\ \ t>0. \end{aligned}$$

Thus

$$\begin{aligned} \Vert x\Vert _{\big ({_sG_p}(\mathcal {M}), {_sG_q}(\mathcal {M})\big )_{F,K}}=\Vert x\Vert _{\big (l_s(L_{p}(\mathcal {M})),l_s(L_{q}(\mathcal {M}))\big )_{F,K}}. \end{aligned}$$

Therefore, using Lemma 3.6, we have finished the proof of the theorem.

Proof of Theorem 3.5

Let $x\in ({_s\widehat{L}_p}(\mathcal {M}), {_s\widehat{L}}_q(\mathcal {M}))_{F,K}$ and $x=x^0+x^1$ be a decomposition of x where $x^0\in {_s\widehat{L}_p}(\mathcal {M})$ and $x^1\in {_s\widehat{L}}_q(\mathcal {M}).$ Let $x_n^k=y_n^k+z_n^k \ (n\ge 1)$ be the Doob’s decomposition of $x^k \ (k=0,1)$. Then we have that $y^0\in L_p(\mathcal {M})$, $y^1\in L_q(\mathcal {M})$ and $dz^0\in {_sG_p}(\mathcal {M})$, $dz^1\in {_sG_q}(\mathcal {M})$. Set $y=y^0+y^1$ and $z=z^0+z^1$. Then

$$\begin{aligned}&K_t(y;L_{p}(\mathcal {M}),L_{q}(\mathcal {M}))+K_t(dz; {_sG_p}(\mathcal {M}), {_sG_q}(\mathcal {M}))\\&\quad \le \Vert y^0\Vert _{L_{p}(\mathcal {M})}+t\Vert y^1\Vert _{L_{q}(\mathcal {M})} +\Vert dz^0\Vert _{{_sG_p}(\mathcal {M}))}+t\Vert dz^1\Vert _{{_sG_q}(\mathcal {M})}\\&\quad = \Vert x^0\Vert _{_s\widehat{L}_{p}(\mathcal {M})}+t\Vert x^1\Vert _{_s\widehat{L}_{q}(\mathcal {M})}. \end{aligned}$$

Thus we get that

$$\begin{aligned} K_t(y;L_{p}(\mathcal {M}),L_{q}(\mathcal {M}))+K_t(dz; {_sG_p}(\mathcal {M}), {_sG_q}(\mathcal {M})) \le K_t(x;{_s\widehat{L}}_p(\mathcal {M}),{_s\widehat{L}}_q(\mathcal {M})), \end{aligned}$$

where the infimum runs over all decomposition of x. Using the equality $\Vert x\Vert _{(X_0,X_1)_{F,K}}=\Vert \frac{K_t(x;X_0,X_1)}{t}\Vert _F,$ we have that

$$\begin{aligned} \Vert y\Vert _{(L_{p}(\mathcal {M}),L_{q}(\mathcal {M}))_{F,K}}+\Vert dz\Vert _{({_sG_p}(\mathcal {M}), {_sG_q}(\mathcal {M}))_{F,K}} \le 2\Vert x\Vert _{({_s\widehat{L}}_{p}(\mathcal {M}),{_s\widehat{L}}_{q}(\mathcal {M}))_{F,K}}. \end{aligned}$$

By Lemma 2.4 and Lemma 3.8, we get that

$$\begin{aligned} \Vert y\Vert _{E(\mathcal {M})}+\Vert dz\Vert _{{_sG_E}(\mathcal {M})} \le 2\Vert x\Vert _{({_s\widehat{L}}_{p}(\mathcal {M}),{_s\widehat{L}}_{q}(\mathcal {M}))_{F,K}} \end{aligned}$$

which implies that $\Vert x\Vert _{_s\widehat{E}(\mathcal {M})} \le 2\Vert x\Vert _{({_s\widehat{L}}_{p}(\mathcal {M}),{_s\widehat{L}}_{q}(\mathcal {M}))_{F,K}}$ and

$$\begin{aligned} _s\widehat{E}(\mathcal {M})\supset ({_s\widehat{L}}_{p}(\mathcal {M}),{_s\widehat{L}}_{q}(\mathcal {M}))_{F,K}. \end{aligned}$$

By Theorem 3.1 and Lemma 3.6, we have that

$$\begin{aligned} _s\widehat{E}(\mathcal {M})= (_{s'}\widehat{E^\times }(\mathcal {M}))^*\subset \big (({_{s'}\widehat{L}}_{q'}(\mathcal {M}),{_{s'}\widehat{L}}_{p'}(\mathcal {M}))_{F^\times ,K}\big )^* =({_s\widehat{L}}_{p}(\mathcal {M}),{_s\widehat{L}}_{q}(\mathcal {M}))_{F,K}. \end{aligned}$$

Therefore,

$$\begin{aligned} _s\widehat{E}(\mathcal {M})= ({_s\widehat{L}}_{p}(\mathcal {M}),{_s\widehat{L}}_{q}(\mathcal {M}))_{F,K}. \end{aligned}$$

The proof is completed.