15.2 Notations and Preliminaries 327The Davis’ inequality forH^1 -martingales ([RY 91, Theorem IV.4.1], see
also [M 76]) states that there are universal constants,c 1 andc 2 (only depend-
ing on the dimensiond), such that for eachH^1 -martingaleXwe have:
c 1 ‖X∞∗‖L 1 ≤‖X‖H 1 ≤c 2 ‖X∞∗‖L 1 ,whereXu∗=supt≤u|Xt|denotes the maximal function.
We denote byH^1 =H^1 (Ω,F,(Ft)t∈R+,P)andL^1 =L^1 (Ω,F,(Ft)t∈R+,P)
the Banach spaces of real-valued uniformly integrable martingales with fi-
niteH^1 -orL^1 -norm respectively. Note that the spaceL^1 (Ω,F,(Ft)t∈R+,P)
may be isometrically identified with the space of integrable random variables
L^1 (Ω,F,P) by associating to a uniformly integrable martingaleXthe random
variableX∞.
Also note that for a local martingale of the formH·Mwe have the formula
‖H·M‖H^1 =∥
∥
∥[H·M, H·M]
(^12)
∞
∥
∥
∥
L^1 (Ω,F,P)=E[(∫
∞0Ht′d[M, M]tHt) 12 ]
,
whereH′denotes the transpose ofH.
We now state and prove the result of Kadeˇc-Pelczy ́nski [KP 65] in a form
that will be useful in the rest of our paper.
Theorem 15.2.1.(Kadeˇc-Pelczy ́nski).If(fn)n≥ 1 is anL^1 -bounded sequence
in the positive coneL^1 +(Ω,F,P),andgis a non-negative integrable function,
then there is a subsequence(nk)k≥ 1 as well as an increasing sequence of strictly
positive numbers(βk)k≥ 1 such thatβktends to∞and(fnk∧(βk(g+ 1)))k≥ 1
is uniformly integrable.
The sequence(fnk∧(βk(g+ 1)))k≥ 1 is then relatively weakly compact by
the Dunford-Pettis theorem.
Proof.We adapt the proof of [KP 65]. Without loss of generality we may sup-
pose that the sequence (fn)n≥ 1 is bounded by 1 inL^1 -norm but not uniformly
integrable, i.e.,
E[fn]≤1; δ(β)=sup
nE[fn−fn∧β(g+1)]<∞;0<δ(∞) = inf
β> 0
δ(β)(it is an easy exercise to show thatδ(∞) = 0 implies uniform integrability).
Fork=1andβ 1 = 1 we selectn 1 so thatE[fn 1 −fn 1 ∧β 1 (g+1)]>δ(∞ 2 ).
Having chosenn 1 ,n 2 ,...,nk− 1 as well asβ 1 ,β 2 ,...,βk− 1 we putβk=2βk− 1
and we selectnk>nk− 1 so thatE[fnk−fnk∧βk(g+1)]>(1− 2 −k)δ(∞).
The sequence (fnk∧βk(g+1))k≥ 1 is now uniformly integrable. To see this,
let us fixKand letk(K) be defined as the smallest numberksuch that
βk>K. Clearlyk(K)→∞asKtends to∞.Forl<k(K)wethenhave
thatfnl∧βl(g+1) =fnl∧βl(g+1)∧K(g+ 1), whereas forl≥k(K)wehave