On Doob’s inequality and Burkholder’s inequality in weak spaces

Abstract

Let X be a Banach function space over a nonatomic probability space. We associate with X the weak space ${\mathrm {w}}\text {-}{X}$, which is a quasi-Banach space defined in a natural way. We give necessary and sufficient conditions for Doob’s inequality in ${\mathrm {w}}\text {-}{X}$ to be valid, and necessary and sufficient conditions for Burkholder’s inequality in ${\mathrm {w}}\text {-}{X}$ to be valid.

Appendix

As the following example shows, there is a quasi-concave function $\varphi :[0,1]\rightarrow [0,\infty )$ such that both (3.1) and (3.6) hold for some $A>1$ but do not hold for all $A>1$.

Example

Define a function $\varphi :[0,1]\rightarrow [0,\infty )$ by letting

$$\begin{aligned} \varphi (t)=\left\{ \begin{array}{ll} 2^{n-1}t &{}\quad \text {if }n \in \mathbb {N}\hbox { and }2\cdot 10^{-n}<t\le 10^{-n+1},\\ 5^{-n} &{}\quad \text {if }n \in \mathbb {N}\hbox { and }10^{-n}<t\le 2\cdot 10^{-n},\\ 0 &{}\quad \text {if }t=0. \end{array} \right. \end{aligned}$$

Then $\varphi$ is quasi-concave. It is easily seen that

$$\begin{aligned} \varliminf _{t\rightarrow 0+}\frac{\varphi (2t)}{\varphi (t)}=1 \quad \text {and} \quad \varlimsup _{t\rightarrow 0+}\frac{\varphi (2t)}{\varphi (t)}=2, \end{aligned}$$

while

$$\begin{aligned} 1<5=\varliminf _{t\rightarrow 0+}\frac{\varphi (10\,t)}{\varphi (t)} \quad \text {and} \quad \varlimsup _{t\rightarrow 0+}\frac{\varphi (10\,t)}{\varphi (t)}=5<10. \end{aligned}$$

On the other hand, we have the following:

Proposition

Suppose that a quasi-concave function $\varphi :[0,1]\rightarrow [0,\infty )$ is concave. Then:

• (i) If (3.1) holds for some $A>1$, then (3.1) holds for all $A>1$.

• (ii) If (3.6) holds for some $A>1$, then (3.6) holds for all $A>1$.

Proof

• (i) Suppose that $A>B>1$. It suffices to show that if (3.1) holds for A, then it holds with A replaced by B (cf. Remark 3.3). Since (3.3) holds for all $c>1$, it suffices to show that if

  $$\begin{aligned} \varlimsup _{t \rightarrow 0+}\frac{\varphi (Bt)}{\varphi (t)}=B, \end{aligned}$$

  (6.1)

  then

  $$\begin{aligned} \varlimsup _{t \rightarrow 0+}\frac{\varphi (At)}{\varphi (t)}=A. \end{aligned}$$

  (6.2)

  Assume that (6.1) holds. Then there exists a decreasing sequence $\{t_n\}_{n\in \mathbb {N}}$ of numbers in (0, 1 / B] such that $t_n \rightarrow 0$ and $\varphi (Bt_n)/\varphi (t_n)\rightarrow B$ as $n \rightarrow \infty$. For each $n\in \mathbb {N}$, let $\psi _n$ be the affine function whose graph passes through the points $(t_n,\varphi (t_n))$ and $(Bt_n,\varphi (Bt_n))$; that is,

  $$\begin{aligned} \psi _n(t)=\frac{\varphi (Bt_n)-\varphi (t_n)}{(B-1)t_n}\,t +\frac{B\varphi (t_n)-\varphi (Bt_n)}{B-1}. \end{aligned}$$

  Note that since $\varphi$ is a concave function, we have $\psi _n(t)\ge \varphi (t)$ for $t\in (0,t_n]$. In particular, $\psi _n(t_n/A)\ge \varphi (t_n/A)$, and therefore

  $$\begin{aligned} \sup _{0<t\le t_n/A}\frac{\varphi (At)}{\varphi (t)} \ge \frac{\varphi (t_n)}{\varphi (t_n/A)} \ge \frac{\varphi (t_n)}{\psi _n(t_n/A)} =\frac{A(B-1)\varphi (t_n)}{(1-A)\varphi (Bt_n)+(AB-1)\varphi (t_n)}. \end{aligned}$$

  Since $\varphi (Bt_n)/\varphi (t_n)\rightarrow B$ as $n \rightarrow \infty$, the right-hand side converges to A as $n\rightarrow \infty$. Hence

  $$\begin{aligned} \varlimsup _{t\rightarrow 0+}\frac{\varphi (At)}{\varphi (t)}\ge A. \end{aligned}$$

  Since the reverse inequality always holds, this implies (6.2).

• (ii) Suppose that $A>B>1$. It suffices to show that if

  $$\begin{aligned} \varliminf _{t \rightarrow 0+}\frac{\varphi (Bt)}{\varphi (t)}=1, \end{aligned}$$

  (6.3)

  then

  $$\begin{aligned} \varliminf _{t \rightarrow 0+}\frac{\varphi (At)}{\varphi (t)}=1. \end{aligned}$$

  (6.4)

  Assume that (6.3) holds. Then there exists a decreasing sequence $\{t_n\}_{n\in \mathbb {N}}$ of numbers in (0, 1 / A] such that $t_n \rightarrow 0$ and $\varphi (Bt_n)/\varphi (t_n)\rightarrow 1$ as $n\rightarrow \infty$. Again, let $\psi _n$ be the affine function whose graph passes through the points $(t_n,\varphi (t_n))$ and $(Bt_n,\varphi (Bt_n))$. Note that $\varphi (t)\le \psi _n(t)$ for $t\in [Bt_n,1]$. Then we have $\varphi (At_n)\le \psi _n(At_n)$, and hence

  $$\begin{aligned} \inf _{0<t\le t_n}\frac{\varphi (At)}{\varphi (t)} \le \frac{\varphi (At_n)}{\varphi (t_n)} \le \frac{\psi _n(At_n)}{\varphi (t_n)} =\frac{A-1}{B-1}\cdot \frac{\varphi (Bt_n)}{\varphi (t_n)}+\frac{B-A}{B-1}. \end{aligned}$$

  Since $\varphi (Bt_n)/\varphi (t_n)\rightarrow 1$ as $n\rightarrow \infty$, the right-hand side converges to 1 as $n \rightarrow \infty$. Thus

  $$\begin{aligned} \varliminf _{t\rightarrow 0+}\frac{\varphi (At)}{\varphi (t)}\le 1, \end{aligned}$$

  which implies (6.4). $\square$