BMO-Lorentz martingale spaces

Abstract

In this paper BMO-Lorentz martingale spaces are investigated. We give the characterization of BMO-Lorentz martingale spaces. Moreover, we discuss the relationship between the Carleson measure and BMO-Lorentz martingales. As a consequence, we find a new way to characterize the geometrical properties of a Banach space.

1 Introduction and preliminaries

Since 1951 when they were first introduced by Lorentz in [1], Lorentz spaces have attracted more and more attention. A lot of results were obtained such as normability, duality, interpolation, and so on [2–7].

We know that martingale theory is intimately related to harmonic analysis. In martingale case, Weisz [8] and Long [9] considered the spaces $H_{p,q}$ and the interpolations between them, respectively. It is well known that the validity of a classical (scalar-valued) result in the vector-valued setting, i.e., for functions or martingales with values in a Banach space X, depends on the geometrical or topological properties of X.

It was exactly with this in mind that Xu [10] developed the vector-valued Littlewood-Paley theory, which was inspired by Pisier’s celebrated work [8] on martingale inequalities in uniformly convex spaces. Very recently, Ouyang and Xu [11] studied the endpoint case of the main results of [10] by means of the classical relationship between BMO functions and Carleson measures. Jiao [12] discussed the relationship between Carleson measures and vector-valued martingales.

Let $(\mathrm{\Omega },\mu )$ be a nonatomic σ-finite measure space. Suppose that f is a measurable function on a measure space $(\mathrm{\Omega },\mu )$. We define its distribution function

$${\lambda }_f(t)=\mu \{\omega :\parallel f(\omega )\parallel >t\},t\ge 0,$$

and its decreasing rearrangement function

$$f^{\ast }(t)=inf\{s>0:{\lambda }_f(s)\le t\}.$$

Given a measurable function f on a measure space $(\mathrm{\Omega },\mu )$ and $0<p,q\le \mathrm{\infty }$, define

$${\parallel f\parallel }_{L^{p,q}}=\{\begin{array}{cc}{({\int }_0^{\mathrm{\infty }}{(t^{\frac{1}{p}}{f}^{\ast }(t))}^q\frac{dt}{t})}^{\frac{1}{q}}\hfill & \text{if }q<\mathrm{\infty },\hfill \\ {sup}_{t>0}{t}^{\frac{1}{p}}{f}^{\ast }(t)\hfill & \text{if }q=\mathrm{\infty }.\hfill \end{array}$$

Remark 1.1 Observe that for all $0<p,r<\mathrm{\infty }$ and $0<q\le \mathrm{\infty }$ we have

$${\parallel {|g|}^r\parallel }_{L^{p,q}}={\parallel g\parallel }_{L^{pr,qr}}^r.$$

(1.1)

Unfortunately, the functions ${\parallel \cdot \parallel }_{L^{p,q}}$ do not satisfy the triangle inequality. However, since for all $t>0$, ${(f+g)}^{\ast }(t)\le f^{\ast }(t/2)+g^{\ast }(t/2)$, we have

$${\parallel f+g\parallel }_{L^{p,q}}\le c_{p,q}({\parallel f\parallel }_{L^{p,q}}+{\parallel g\parallel }_{L^{p,q}}),$$

(1.2)

where $c_{p,q}=2^{1/p}max\{1,2^{(1-q)/q}\}$.

The set of all f with ${\parallel f\parallel }_{L^{p,q}}<\mathrm{\infty }$ is denoted by $L^{p,q}(X,\mu )$ and is called the Lorentz space with indices p and q. Observe that the definition implies that $L^{\mathrm{\infty },\mathrm{\infty }}=L^{\mathrm{\infty }}$.

Let $(\mathrm{\Omega },\mathrm{\Sigma },P)$ be a complete probability space, and let ${({\mathrm{\Sigma }}_n)}_{n\ge 0}$ be a nondecreasing sequence of sub-σ-algebras of Σ with $\mathrm{\Sigma }=\sigma ({\bigcup }_{n\ge 0}{\mathrm{\Sigma }}_n)$. We denote by E and $E_n$ the expectation and conditional expectation with respect to Σ and ${\mathrm{\Sigma }}_n$, respectively. For a martingale $f={(f_n)}_{n\ge 0}$ with martingale difference $d{f}_n=f_n-f_{n-1}$, $n\ge 0$, $f_{-1}\equiv 0$, we define its maximal function, p-square function, respectively, as usual:

$$Mf=\underset{n}{sup}\parallel f_n\parallel ,S^{(p)}(f)={\left(\sum _{n=1}^{\mathrm{\infty }}{\parallel d{f}_n\parallel }^p\right)}^{1/p}.$$

We say that an X-valued martingale $f={(f_n)}_{n\ge 0}\in L^{p,q}(X)$ if ${sup}_n{\parallel f_n\parallel }_{L^{p,q}}<\mathrm{\infty }$.

The space ${\mathit{BMO}}_{L^{p,q}(X)}^a$ ($a\ge 1$, $1<p,q\le \mathrm{\infty }$) consists of all martingale $f\in L^{p,q}(X)$ such that

$${\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}^a}=\underset{n}{sup}{\parallel {\left(E({\parallel f-f_{n-1}\parallel }^a|{\mathrm{\Sigma }}_n)\right)}^{1/a}\parallel }_{L^{p,q}(X)}<\mathrm{\infty }.$$

Remark 1.2 The spaces ${\mathit{BMO}}_{L^{p,q}(X)}^a$ are independent of a and all corresponding norms are equivalent. This allows us to denote any of them by ${\mathit{BMO}}_{L^{p,q}(X)}$.

Proof If $a\ge 1$, $\phi (x)=x^a$ is a convex function, by Jensen’s inequality, we have $E(\parallel f-f_{n-1}\parallel |{\mathrm{\Sigma }}_n)\le {(E({\parallel f-f_{n-1}\parallel }^a|{\mathrm{\Sigma }}_n))}^{1/a}$, which implies $E{(\parallel f-f_{n-1}\parallel |{\mathrm{\Sigma }}_n)}^{\ast }(t)\le {({(E({\parallel f-f_{n-1}\parallel }^a|{\mathrm{\Sigma }}_n))}^{1/a})}^{\ast }(t)$, i.e., ${\mathit{BMO}}_{L^{p,q}(X)}^a\subset {\mathit{BMO}}_{L^{p,q}(X)}^1$.

On the contrary, let $g_n=E(\parallel f-f_{n-1}\parallel |{\mathrm{\Sigma }}_n)$, $Mg={sup}_n\parallel g_n\parallel$, $h_n=E{({\parallel f-f_{n-1}\parallel }^a|{\mathrm{\Sigma }}_n)}^{1/a}$, $Mh={sup}_n\parallel h_n\parallel$. Now we set $A_t=\{\omega :Mh>t\}$. Then $Mg>t$ a.e. on $A_t$. (Factually, if there is a $B\subset A_t$ with $\mu (B)>0$ such that $Mg\le t$ on B, then $E(\parallel f-f_{n-1}\parallel |{\mathrm{\Sigma }}_n)\le t=E(t|{\mathrm{\Sigma }}_n)$ a.e. on B, which implies $\parallel f-f_{n-1}\parallel \le t$ a.e. on B. This is a contradiction for $A_t$.) So, $P\{\omega :Mh>t\}\le cP\{\omega :Mg>t\}$. Then

$$\begin{array}{rcl}{\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}^a}& \le & {\left(q{\int }_0^{\mathrm{\infty }}{\left[tP{(Mh(\omega )>t)}^{1/p}\right]}^q\frac{dt}{t}\right)}^{1/q}\\ \le & c{\left(q{\int }_0^{\mathrm{\infty }}{\left[tP{(Mg(\omega )>t)}^{1/p}\right]}^q\frac{dt}{t}\right)}^{1/q}\le c{\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}^1},\end{array}$$

i.e., ${\mathit{BMO}}_{L^{p,q}(X)}^1\subset {\mathit{BMO}}_{L^{p,q}(X)}^a$. Thus we complete the proof. □

Remark 1.3 If $p=q=\mathrm{\infty }$, ${\mathit{BMO}}_{L^{p,q}(X)}$ is the classical BMO space.

Remark 1.4 For the classical BMO space, we have the following statement (see [12]):

$${\parallel f\parallel }_{\mathit{BMO}}\sim \underset{\tau }{sup}\mu {(\tau <\mathrm{\infty })}^{-1/p}{\parallel f-f_{\tau -1}\parallel }_{L_p},1\le p<\mathrm{\infty }.$$

(1.3)

It is well known that $L^{p,q}$ is a subspace of $L^{p,r}$ for $0<p\le \mathrm{\infty }$, $0<q<r\le \mathrm{\infty }$ and $L^{p_2,q_2}$ is a subspace of $L^{p_1,q_1}$ for $1<p_1\le p_2\le \mathrm{\infty }$, $1<q_1,q_2\le \mathrm{\infty }$. Thus we have the following proposition.

Proposition 1.5 (1) If $0<p\le \mathrm{\infty }$, $0<q<r\le \mathrm{\infty }$, ${\mathit{BMO}}_{L^{p,q}}\subseteq {\mathit{BMO}}_{L^{p,r}}$.

• (2) If $1<p_1\le p_2\le \mathrm{\infty }$, $1<q_1,q_2\le \mathrm{\infty }$, ${\mathit{BMO}}_{L^{p_2,q_2}}\subseteq {\mathit{BMO}}_{L^{p_1,q_1}}$.

Theorem 1.6 Let $f={(f_n)}_{n\ge 0}$ be an X-valued martingale in $L^{p,q}(X)$, $1<p,q\le \mathrm{\infty }$. Then $f\in {\mathit{BMO}}_{L^{p,q}(X)}$ if and only if there exists an adapted process $\theta ={(\theta )}_{n\ge 0}$ such that

$$C_{\theta }=\underset{n}{sup}{\parallel E(\parallel f-{\theta }_{n-1}\parallel |{\mathrm{\Sigma }}_n)\parallel }_{L^{p,q}(X)}<\mathrm{\infty }.$$

And, in any case, we have

$$⦀f{⦀}_{{\mathit{BMO}}_{L^{p,q}(X)}}:=\underset{\theta }{inf}{C}_{\theta }\le {\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}}\le c⦀f{⦀}_{{\mathit{BMO}}_{L^{p,q}(X)}}.$$

Proof Assume $f\in {\mathit{BMO}}_{L^{p,q}(X)}$. Then, obviously,

$$⦀f{⦀}_{{\mathit{BMO}}_{L^{p,q}(X)}}\le {\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}}.$$

Now let $⦀f{⦀}_{{\mathit{BMO}}_{L^{p,q}(X)}}<\mathrm{\infty }$ and $\theta ={(\theta )}_{n\ge 0}$ be any one such that $C_{\theta }<\mathrm{\infty }$. Then we have

$$\begin{array}{rcl}E(\parallel f-f_{n-1}\parallel |{\mathrm{\Sigma }}_n)& \le & E(\parallel f-{\theta }_{n-1}\parallel |{\mathrm{\Sigma }}_n)+\parallel {\theta }_{n-1}-f_{n-1}\parallel \\ =& E(\parallel f-{\theta }_{n-1}\parallel |{\mathrm{\Sigma }}_n)+\parallel E((f-{\theta }_{n-1})|{\mathrm{\Sigma }}_{n-1})\parallel \\ \le & E(\parallel f-{\theta }_{n-1}\parallel |{\mathrm{\Sigma }}_n)+E(\parallel f-{\theta }_{n-1}\parallel |{\mathrm{\Sigma }}_{n-1})\\ =& E(\parallel f-{\theta }_{n-1}\parallel |{\mathrm{\Sigma }}_n)+E(E(\parallel f-{\theta }_{n-1}\parallel |{\mathrm{\Sigma }}_n)|{\mathrm{\Sigma }}_{n-1}).\end{array}$$

Taking ‘inf’ over all possible θ and (1.2), we get the desired inequality. □

Theorem 1.7 Let $f={(f_n)}_{n\ge 0}$ be an X-valued martingale in ${\mathit{BMO}}_{L^{p,q}(X)}$, where $\frac{1}{p}+\frac{1}{s_1}=\frac{1}{r}$ and $\frac{1}{q}+\frac{1}{s_2}=\frac{1}{s}$, $1<p,q,r,s_1,s_2\le \mathrm{\infty }$. Then

$${\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}}\sim \underset{\tau }{sup}\mu {(\{\tau <\mathrm{\infty }\})}^{-\frac{1}{s_1}}{\parallel f-f_{\tau -1}\parallel }_{L^{r,s}(X)},$$

where ‘sup’ is taken over all stopping times τ.

Proof Assume that ${\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}}<\mathrm{\infty }$, τ is any stopping time. Then, by Hölder’s inequality, we have

$$\begin{array}{rcl}{\parallel f-f_{\tau -1}\parallel }_{L^{r,s}(X)}& =& {\parallel (f-f_{\tau -1}){\chi }_{\{\tau <\mathrm{\infty }\}}\parallel }_{L^{r,s}(X)}\\ \le & c{\parallel f-f_{\tau -1}\parallel }_{L^{p,q}(X)}{\parallel {\chi }_{\{\tau <\mathrm{\infty }\}}\parallel }_{L^{s_1,s_2}(X)}\\ \le & c\underset{{\parallel g\parallel }_{{(L^{p,q})}^{\ast }}\le 1}{sup}\left|{\int }_{\{\tau <\mathrm{\infty }\}}(\parallel f-f_{\tau -1}\parallel )gd\mu \right|\cdot {\parallel {\chi }_{\{\tau <\mathrm{\infty }\}}\parallel }_{L^{s_1,s_2}(X)}\\ =& c{\left(\frac{q}{p}\right)}^{1/q}\underset{{\parallel g\parallel }_{{(L^{p,q})}^{\ast }}\le 1}{sup}\left|{\int }_{\{\tau <\mathrm{\infty }\}}E(\parallel f-f_{\tau -1}\parallel \cdot g|{\mathrm{\Sigma }}_{\tau })d\mu \right|\cdot \mu {(\tau <\mathrm{\infty })}^{\frac{1}{s_1}}\\ \le & c{\left(\frac{q}{p}\right)}^{1/q}{\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}}}\cdot \mu {(\tau <\mathrm{\infty })}^{\frac{1}{s_1}}.\end{array}$$

This proves one half of the assertion. Conversely, assume that $\beta ={sup}_{\tau }\mu {(\{\tau <\mathrm{\infty }\})}^{-\frac{1}{s_1}}{\parallel f-f_{\tau -1}\parallel }_{L^{r,s}(X)}<\mathrm{\infty }$, and τ is any stopping time, $F\in {\mathrm{\Sigma }}_{\tau }$, $F\subset \{\tau <\mathrm{\infty }\}$. Define

$${\tau }_F=\{\begin{array}{cc}\tau \hfill & \text{if }\omega \in F,\hfill \\ \mathrm{\infty }\hfill & \text{if }\omega \notin F.\hfill \end{array}$$

Thus we get

$$\begin{array}{rcl}\beta & \ge & \mu {(\{\tau <\mathrm{\infty }\})}^{-\frac{1}{s_1}}{\parallel f-f_{\tau -1}\parallel }_{L^{r,s}(X)}\\ =& \frac{1}{\mu {(F)}^{1/s_1}}{\parallel f-f_{{\tau }_F-1}\parallel }_{L^{r,s}(X)}\\ \ge & \frac{1}{\mu {(F)}^{1/s_1}}{\parallel f-f_{{\tau }_F-1}\parallel }_{L^{r\wedge 1,r\wedge 1}(X)}\\ \ge & \frac{1}{\mu (F)}{\parallel f-f_{{\tau }_F-1}\parallel }_{L^{1,1}(X)}\\ =& \frac{1}{\mu (F)}{\int }_F\parallel f-f_{{\tau }_F-1}\parallel du.\end{array}$$

That is, $E(\parallel f-f_{{\tau }_F-1}\parallel |{\mathrm{\Sigma }}_{{\tau }_F})\le \beta$. By Remark 1.2 we have

$${\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}}\le c\beta .$$

Thus we complete the proof of the theorem. □

Proposition 1.8 Particularly, if $s=r$ and $p=q=\mathrm{\infty }$, we get Remark 1.4.

By Theorem 1.6 and Theorem 1.7, we have the following proposition.

Proposition 1.9 Let $f={(f_n)}_{n\ge 0}$ be an X-valued martingale in ${\mathit{BMO}}_{L^{p,q}(X)}$, where $\frac{1}{p}+\frac{1}{s_1}=\frac{1}{r}$ and $\frac{1}{q}+\frac{1}{s_2}=\frac{1}{s}$, $1<p,q,r,s_1,s_2\le \mathrm{\infty }$. Then

$${\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}}\sim \underset{\theta }{inf}\underset{\tau }{sup}\mu {(\{\tau <\mathrm{\infty }\})}^{-\frac{1}{s_1}}{\parallel f-{\theta }_{\tau -1}\parallel }_{L^{r,s}(X)},$$

where ‘sup’ is taken over all stopping times τ and ‘inf’ is taken over all adapted process $\theta ={(\theta )}_{n\ge 0}$.

2 Carleson measure and BMO-Lorentz martingale spaces

Definition 2.1 Let ν be a nonnegative measure on $\mathrm{\Omega }\times N$, where N is equipped with the counting measure dm. Let $\hat{\tau }$ denote the tent over τ:

$$\hat{\tau }=\{(\omega ,k):k\ge \tau (\omega ),\tau (\omega )<\mathrm{\infty }\}.$$

ν is said to be an s-Carleson measure if

$$⦀\nu ⦀=\underset{\tau }{sup}\frac{\nu (\hat{\tau })}{\mu {(\{\tau <\mathrm{\infty }\})}^s}<\mathrm{\infty },$$

where τ runs through all stopping times.

Theorem 2.2 Let $1<p,q\le \mathrm{\infty }$, and $g={(g_n)}_{n\ge 0}$ be a real-valued martingale and $d\nu ={|{\mathrm{\Delta }}_{k}g|}^2{\delta }_{k}d\mu$, where ${\delta }_k$ is the Dirac measure centered at k. So, if $g\in {\mathit{BMO}}_{L^{2p,2q}}^2$, ν is a $1/p^{\prime }$-Carleson measure. Moreover, if $1<p,q<\mathrm{\infty }$, the converse is also true, where $\frac{1}{p^{\prime }}+\frac{1}{p}=1$, $\frac{1}{q^{\prime }}+\frac{1}{q}=1$.

Proof Let $g={(g_n)}_{n\ge 0}$ be a real-valued martingale, let ν be generated by g as above, and let τ be any stopping time. Then, for $1<q<\mathrm{\infty }$,

$$\begin{array}{rcl}\nu (\hat{\tau })& =& E\left(\sum _{k=0}^{\mathrm{\infty }}{|{\mathrm{\Delta }}_{k}g|}^2{\chi }_{\{\tau (\omega )\le k\}}\right)\\ =& E\left(E\left(\sum _{k=\tau (\omega )}^{\mathrm{\infty }}{|{\mathrm{\Delta }}_{k}g|}^2\right|{\mathrm{\Sigma }}_{\tau }){\chi }_{\{\tau (\omega )\le k\}}\right)\\ =& E\left(E({|g-g_{\tau -1}|}^2|{\mathrm{\Sigma }}_{\tau }){\chi }_{\{\tau (\omega )\le k\}}\right)\\ \le & c{\parallel E({|g-g_{\tau -1}|}^2|{\mathrm{\Sigma }}_{\tau })\parallel }_{L^{p,q}}\cdot {\parallel {\chi }_{\{\tau (\omega )\le k\}}\parallel }_{L^{p^{\prime },q^{\prime }}}\\ =& c{\parallel E{({|g-g_{\tau -1}|}^2|{\mathrm{\Sigma }}_{\tau })}^{\frac{1}{2}\cdot 2}\parallel }_{L^{p,q}}\cdot {\parallel {\chi }_{\{\tau (\omega )\le k\}}\parallel }_{L^{p^{\prime },q^{\prime }}}\\ \le & c{\parallel E{({|g-g_{\tau -1}|}^2|{\mathrm{\Sigma }}_{\tau })}^{1/2}\parallel }_{L^{2p,2q}}^2\cdot \mu {\left(\{\tau (\omega )\le k\}\right)}^{1/p^{\prime }}\\ \le & c{\parallel g\parallel }_{{\mathit{BMO}}_{L^{2p,2q}}^2}^2\cdot \mu {\left(\{\tau (\omega )\le k\}\right)}^{1/p^{\prime }}.\end{array}$$

So, $g\in {\mathit{BMO}}_{L^{2p,2q}}^2$ implies that ν is a $1/p^{\prime }$-Carleson measure and $⦀\nu ⦀\le {\parallel g\parallel }_{{\mathit{BMO}}_{L^{2p,2q}}^2}^2$.

Conversely, for any n and any $F\in {\mathrm{\Sigma }}_n$, we define

$$\tau =\{\begin{array}{cc}n\hfill & \text{if }\omega \in F,\hfill \\ \mathrm{\infty }\hfill & \text{if }\omega \notin F.\hfill \end{array}$$

Since ν is a $1/p^{\prime }$-Carleson measure, we have

$$\begin{array}{rcl}⦀\nu ⦀& \ge & \frac{1}{\mu {(\{\tau <\mathrm{\infty }\})}^{1/p^{\prime }}}\nu \left(\{(\omega ,k):k\ge \tau (\omega ),\tau (\omega )<\mathrm{\infty }\}\right)\\ =& \frac{1}{\mu {(F)}^{1/p^{\prime }}}{\int }_F\sum _{k=n}^{\mathrm{\infty }}{|{\mathrm{\Delta }}_{k}g|}^{2}d\mu \\ \ge & \frac{1}{\mu (F)}{\int }_F\sum _{k=n}^{\mathrm{\infty }}{|{\mathrm{\Delta }}_{k}g|}^{2}d\mu \\ =& \frac{1}{\mu (F)}{\int }_F{|g-g_{n-1}|}^{2}d\mu .\end{array}$$

That is, $⦀\nu ⦀\ge E({|g-g_{n-1}|}^2|{\mathrm{\Sigma }}_n)$. Then we have ${\parallel g\parallel }_{{\mathit{BMO}}_{L^{2p,2q}}^2}^2\le c⦀\nu ⦀$. Thus, we complete the proof of the theorem. □

3 The characterization of Banach space’s geometrical properties

Let $1<q<\mathrm{\infty }$. Then a Banach space X has an equivalent q-uniformly convex norm if and only if for one $1<p<\mathrm{\infty }$ (or equivalently, for every $1<p<\mathrm{\infty }$) there exists a positive constant c such that

$${\parallel S^{(q)}(f)\parallel }_p\le c\underset{n}{sup}{\parallel f_n\parallel }_p$$

for all finite $L_p$-martingales f with values in X. Again, the validity of the converse inequality amounts to saying that X has an equivalent q-uniformly smooth norm.

Definition 3.1 Let $X_1$ and $X_2$ be two Banach spaces. Let $L(X_1,X_2)$ denote the space of all bounded linear operators from $X_1$ to $X_2$. Let $v={(v_n)}_{n\ge 1}$ be an adapted sequence such that $v_n\in L_{\mathrm{\infty }}(L(X_1,X_2))$ and ${sup}_{n\ge 1}{\parallel v_n\parallel }_{L_{\mathrm{\infty }}(L(X_1,X_2))}\le 1$. Then the martingale transform T associated to v is defined as follows. For any $X_1$-valued martingale $f={(f_n)}_{n\ge 1}$,

$${(Tf)}_n=\sum _{k=1}^n{v}_{k}d{f}_k.$$

We get the following results from [13, 14].

Lemma 3.2 With the assumptions above, the following statements are equivalent:

• (1) There exists a positive constant c such that

  $${\parallel Tf\parallel }_{\mathit{BMO}(X_2)}\le c{\parallel f\parallel }_{\mathit{BMO}(X_1)},\mathrm{\forall }f\in \mathit{BMO}(X_1).$$

• (2) There exists a positive constant c such that

  $${\parallel {(Tf)}^{\ast }\parallel }_{\mathit{BMO}(X_2)}\le c{\parallel f\parallel }_{\mathit{BMO}(X_1)},\mathrm{\forall }f\in \mathit{BMO}(X_1).$$

• (3) For some $1\le p<\mathrm{\infty }$ (or equivalently, for every $1\le p<\mathrm{\infty }$), there exists a positive constant c such that

  $${\parallel Tf\parallel }_{L^p(X)}\le c{\parallel f\parallel }_{L^p(X)},\mathrm{\forall }f\in L^p(X).$$

Theorem 3.3 Let X be a Banach space, $2\le q<\mathrm{\infty }$, $1<p<\mathrm{\infty }$. Then the following statements are equivalent:

• (1) There exists a positive constant c such that for any finite X-valued martingale,

  $${\parallel S^{(q)}(f)\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}}\le c{\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}}.$$

• (2) X has an equivalent norm which is q-uniformly convex.

Proof $\text{(1)}\Rightarrow \text{(2)}$ Let $\frac{1}{p}+\frac{1}{s_1}=\frac{1}{r}$, $\frac{1}{q}+\frac{1}{s_2}=\frac{1}{r}$, $1<p,q,r,s_1,s_2\le \mathrm{\infty }$ By Theorem 1.7 we have

$${\parallel S^{(q)}(f)\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}}\sim \underset{\tau }{sup}\mu {(\{\tau <\mathrm{\infty }\})}^{-\frac{1}{s_1}}{\parallel S^{(q)}(f)-S_{\tau -1}^{(q)}(f)\parallel }_{L^{r,r}(X)},$$

(3.1)

$${\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}^a}\sim \underset{\tau }{sup}\mu {(\{\tau <\mathrm{\infty }\})}^{-\frac{1}{s_1}}{\parallel f-f_{\tau -1}\parallel }_{L^{r,r}(X)}.$$

(3.2)

So, if (1) holds, we have

$${\parallel S^{(q)}(f)-S_{\tau -1}^{(q)}(f)\parallel }_{L^{r,r}(X)}\le c{\parallel f-f_{\tau -1}\parallel }_{L^{r,r}(X)}.$$

By Remark 1.4 we have

$${\parallel S^{(q)}(f)\parallel }_{\mathit{BMO}}\le c{\parallel f\parallel }_{\mathit{BMO}}.$$

(3.3)

We now consider a martingale transform operator Q from the family of X-valued martingales to that of $l_q(X)$-valued martingales. Let $v\in L(X,l_q(X))$ be the operator defined by $v_k(x)={\{x_j\}}_{j=1}^{\mathrm{\infty }}$ for $x\in X$, where $x_j=x$ if $j=k$ and $x_j=0$ otherwise. Q is the martingale transform associated to the sequence $(v_k)$:

$${(Qf)}_n=\sum _{k=1}^n{v}_{k}d{f}_k=(d{f}_1,d{f}_2,\dots ,d{f}_n,0,\dots ).$$

Then

$${(Qf)}^{\ast }=S^{(q)}(f).$$

(3.4)

By (3.3) and (3.4) we have

$${\parallel {(Qf)}^{\ast }\parallel }_{\mathit{BMO}}\le c{\parallel f\parallel }_{\mathit{BMO}}.$$

By Lemma 3.2 we have

$${\parallel S^{(q)}(f)\parallel }_q={\parallel {(Qf)}^{\ast }\parallel }_q\le c{\parallel f\parallel }_q.$$

Thus, by Pisier’s theorem, X has an equivalent q-uniformly convex norm.

$\text{(2)}\Rightarrow \text{(1)}$ Suppose that X has an equivalent q-uniformly convex norm. By Pisier’s theorem [15], we find, for any $1\le n<\mathrm{\infty }$,

$$E\left(\sum _{i=n}^{\mathrm{\infty }}{\parallel d{f}_i\parallel }^q\right|{\mathrm{\Sigma }}_n)\le cE({\parallel f-f_{n-1}\parallel }^q|{\mathrm{\Sigma }}_n).$$

Since $E({|S^{(q)}(f)-S_{n-1}^{(q)}(f)|}^q|{\mathrm{\Sigma }}_n)\le cE(|{(S^{(q)}(f))}^q-{(S_{n-1}^{(q)}(f))}^q||{\mathrm{\Sigma }}_n)$, we have

$$E({|S^{(q)}(f)-S_{n-1}^{(q)}(f)|}^q|{\mathrm{\Sigma }}_n)\le cE({\parallel f-f_{n-1}\parallel }^q|{\mathrm{\Sigma }}_n).$$

Thus

$${\parallel S^{(q)}(f)\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}}\le c{\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}}.$$

We complete the proof. □

Theorem 3.4 Let X be a Banach space and $1<p\le 2$, $1<q<\mathrm{\infty }$. If there exists a positive constant c such that for any finite X-valued martingale

$${\parallel f\parallel }_{{\mathit{BMO}}_{L^p(X)}}\le c{\parallel S^{(p)}(f)\parallel }_{{\mathit{BMO}}_{L^p(X)}},$$

(3.5)

then X has an equivalent norm which is p-uniformly smooth.

On the contrary, if X has an equivalent norm which is p-uniformly smooth, then

$${\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}^a}\le c{\parallel S^{(p)}(f)\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}^a}$$

for every martingale f.

Proof Let $X^{\ast }$ be the dual space of X. It suffices to prove that $X^{\ast }$ has an equivalent $p^{\prime }$-uniformly convex norm, where $p^{\prime }$ is the conjugate index of p. By Pisier’s theorem, this is equivalent to showing that

$${\parallel S^{(p^{\prime })}(g)\parallel }_{L_{p^{\prime }}}\le c{\parallel g^{\ast }\parallel }_{L_{p^{\prime }}}=c{\parallel g\parallel }_{H_{p^{\prime }}^{\ast }(X^{\ast })}.$$

(3.6)

Recall that $H_{p^{\prime }}^{\ast }(X^{\ast })$ is defined by

$$H_{p^{\prime }}^{\ast }\left(X^{\ast }\right)=\{X^{\ast }\text{-valued martingale }g=(g_n):g^{\ast }\in L_{p^{\prime }}\}.$$

It is well known that ${\mathit{BMO}}_{L^p(X)}$ can be identified as a subspace of $H_{p^{\prime }}^{\ast }(X^{\ast })$. Thus, for any finite martingale, $f={(f_n)}_{n\ge 0}\in {\mathit{BMO}}_{L^p(X)}$ and $g={(g_n)}_{n\ge 0}\in H_{p^{\prime }}^{\ast }(X^{\ast })$.

$$|〈g,f〉|={\int }_{\mathrm{\Omega }}〈g(\omega ),f(\omega )〉dP\le c{\parallel f\parallel }_{{\mathit{BMO}}_{L^p(X)}}\cdot {\parallel g\parallel }_{H_{p^{\prime }}^{\ast }(X^{\ast })}.$$

On the other hand, ${\parallel S^{(p^{\prime })}(g)\parallel }_{L_{p^{\prime }}}$ is the norm of the difference sequence $(d{g}_n)$ in $L_{p^{\prime }}(l_{p^{\prime }}(X^{\ast }))$.

Thus

$$\begin{array}{rcl}{\parallel S^{(p^{\prime })}(g)\parallel }_{L_{p^{\prime }}}& =& \underset{(a_k)}{sup}\{|\sum 〈d{g}_k,a_k〉|:{\parallel (a_k)\parallel }_{L_p(l_p(X))}\le 1\}\\ =& \underset{(a_k)}{sup}\{|\sum 〈d{g}_k,E_k(a_k)-E_{k-1}(a_k)〉|:{\parallel (a_k)\parallel }_{L_p(l_p(X))}\le 1\}.\end{array}$$

Set $d{f}_k=E_k(a_k)-E_{k-1}(a_k)$ and $f=\sum d{f}_k$. Then f is an X-valued martingale. We have

$${\parallel S^{(p^{\prime })}(g)\parallel }_{L_{p^{\prime }}}=\underset{(a_k)}{sup}\left\{|\sum 〈d{g}_k,d{f}_k〉|\right\}\le c{\parallel f\parallel }_{{\mathit{BMO}}_{L^p(X)}}\cdot {\parallel g\parallel }_{H_{p^{\prime }}^{\ast }(X^{\ast })}.$$

(3.7)

It remains to estimate ${\parallel f\parallel }_{{\mathit{BMO}}_{L^p(X)}}$. Since ${\parallel (a_k)\parallel }_{L_p(l_p(X))}\le 1$, we can also get the conditional case $E(({\sum }_{k=n}^{\mathrm{\infty }}{\parallel a_k\parallel }^p)|{\mathrm{\Sigma }}_n)\le 1$. Then by (3.5)

$$\begin{array}{rcl}{\parallel f\parallel }_{{\mathit{BMO}}_{L^p(X)}}& \le & c{\parallel S^{(p)}(f)\parallel }_{{\mathit{BMO}}_{L^p(X)}}\\ \le & c\underset{n}{sup}{\parallel E({\parallel S^{(p)}(f)-S_{n-1}^{(p)}(f)\parallel }^p|{\mathrm{\Sigma }}_n)\parallel }_{L_p}\\ \le & c\underset{n}{sup}{\parallel E(\parallel S^{(p)}{(f)}^p-S_{n-1}^{(p)}{(f)}^p\parallel |{\mathrm{\Sigma }}_n)\parallel }_{L_p}\\ \le & c\underset{n}{sup}{\parallel E\left(\sum _{k=n}^{\mathrm{\infty }}{\parallel E_k(a_k)-E_{k-1}(a_k)\parallel }^p\right|{\mathrm{\Sigma }}_n)\parallel }_{L_p}\\ \le & c{\parallel E\left(\left(\sum _{k=n}^{\mathrm{\infty }}{\parallel a_k\parallel }^p\right)\right|{\mathrm{\Sigma }}_n)\parallel }_{L^p}\le c.\end{array}$$

On the contrary, we define ${\hat{f}}^{\tau }=({\hat{f}}_i^{\tau },)$, where ${\hat{\mathrm{\Sigma }}}_i={\mathrm{\Sigma }}_{\tau +i}$, ${\hat{f}}_i^{\tau }=f_{\tau +i}-f_{\tau }$, $i\ge 0$. So, we have

$${\left(S^{(p)}\left({\hat{f}}^{\tau }\right)\right)}^p=S^{(p)}{(f)}^p-S_{\tau }^{(p)}{(f)}^p.$$

Suppose that X has an equivalent p-uniformly smooth norm. Then by Pisier’s theorem, we have

$${\parallel {\hat{f}}^{\tau }\parallel }_p\le c{\parallel (S^{(p)}\left({\hat{f}}^{\tau }\right)\parallel }_p,$$

i.e.,

$$E\left({\parallel f-f\tau \parallel }^p\right)\le E(S^{(p)}{(f)}^p-S_{\tau }^{(p)}{(f)}^p).$$

Moreover, we conditionalize it, we will get

$$E({\parallel f-f\tau \parallel }^p|{\mathrm{\Sigma }}_{\tau +1})\le E(S^{(p)}{(f)}^p-S_{\tau }^{(p)}{(f)}^p|{\mathrm{\Sigma }}_{\tau +1}).$$

By the definition and Theorem 1.7, we get

$${\parallel f\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}^a}\le c{\parallel S^{(p)}(f)\parallel }_{{\mathit{BMO}}_{L^{p,q}(X)}^a}.$$

Thus we complete the proof. □