ON RELAXING THE ASSUMPTION OF DIFFERENTIAL SUBORDINATION IN SOME MARTINGALE INEQUALITIES

Let X , Y be continuous-time martingales taking values in a separable Hilbert spaceH . (i) Assume that X , Y satisfy the condition [X , X ]t ≥ [Y, Y ]t for all t ≥ 0. We prove the sharp inequalities sup t ||Yt ||p ≤ (p− 1)−1 sup t ||X t ||p, 1< p ≤ 2, P(sup t |Yt | ≥ 1)≤ 2 Γ(p+ 1) sup t ||X t ||p, 1≤ p ≤ 2, and for any K > 0 we determine the optimal constant L = L(K) depending only on K such that sup t ||Yt ||1 ≤ K sup t E|X t | log |X t |+ L(K). (ii) Assume that X , Y satisfy the condition [X , X ]∞ − [X , X ]t− ≥ [Y, Y ]∞ − [Y, Y ]t− for all t ≥ 0. We establish the sharp bounds sup t ||Yt ||p ≤ (p− 1) sup t ||X t ||p, 2≤ p <∞ and P(sup t |Yt | ≥ 1)≤ pp−1 2 sup t ||X t ||p, 2≤ p <∞. This generalizes the previous results of Burkholder, Suh and the author, who showed the above estimates under the more restrictive assumption of differential subordination. The proof is based on Burkholder’s technique and integration method. 1RESEARCH PARTIALLY SUPPORTED BY MNISW GRANT N N201 397437

and for any K > 0 we determine the optimal constant L = L(K) depending only on K such that sup

Introduction
Let (Ω, , ) be a probability space, filtered by a nondecreasing family ( n ) n≥0 of sub-σ-fields of .Let f = ( f n ), g = (g n ) be adapted martingales taking values in a separable Hilbert space (which may and will be assumed to be equal to 2 ), with a norm | • | and a scalar product denoted by •.The difference sequences d f = (d f n ), d g = (d g n ) of f and g are given by the equalities The following notion of differential subordination is due to Burkholder: we say that g is differentially subordinate to f , if for any n we have |d g n | ≤ |d f n |.This condition implies many interesting estimates which have numerous applications in many areas of mathematics, see the surveys [4], [6] by Burkholder and references therein.Consult also Bañuelos and Wang [1], Wang [13], Bañuelos and Méndez-Hernández [2], Geiss, Montgomery-Smith and Saksman [8], Suh [12] and the papers [10], [11] by the author for some more recent results in this direction.To begin, let us recall the following classical moment inequality, due to Burkholder [3].We use the notation Here p * = max{p − 1, (p − 1) −1 } and the constant is the best possible.
Theorem 1.2.Assume that g is differentially subordinate to f .Then and, if f and g are real-valued, Both inequalities are sharp.
In the case p = 1 the moment inequality does not hold with any finite constant.The author established in [10] the following substitute.
Theorem 1.3.Assume that g is differentially subordinate to f .Then for K > 1, where The constant is the best possible.Furthermore, for K ≤ 1 the inequality does not hold in general with any universal L(K) < ∞.
Let us now turn to the continuous-time setting.Assume that the probability space is complete and is equipped with a filtration ( t ) t∈[0,∞) such that 0 contains all the events of probability 0. Let X = (X t ) t≥0 , Y = (Y t ) t≥0 be -valued martingales, which have right-continuous trajectories with limits from the left.The generalization of the differential subordination is as follows (see [1] and t≥0 is nonnegative and nondecreasing as a function of t.Here [X , X ] denotes the square bracket (quadratic variation) of X : that is, if where [X k , X k ] is the usual square bracket of a real-valued martingale X k , see e.g.Dellacherie and Meyer [7] for details.We use the notation ||X || p = sup t ||X t || p and X * = sup t |X t |, analogous to the one in the discrete-time case.Furthermore, throughout the paper, we set The inequalities (1.1), (1.2), (1.3) and (1.4) can be successfully extended to the continuous-time setting (this will be clear from our results below, see also the papers by Wang [13] and Suh [12] for the proofs of (1.1) and (1.3)).The motivation in the present paper comes from the interesting and challenging question raised in [4]: Burkholder asked whether the moment inequality remains valid under a weaker assumption that the process ([X , X ] t − [Y, Y ] t ) t≥0 is nonnegative (and not necessarily nondecreasing).We will prove that this is true for p ≤ 2, and introduce a dual condition, weaker than the differential subordination, which implies the validity of (1.6) for p ≥ 2. Furthermore, we will show that under these relaxed conditions the corresponding weak-type and logarithmic bounds hold.
The main results of the paper are stated in the two theorems below. and (1.9) where L(K) is given by (1.5).For K ≤ 1 the inequality does not hold in general with any universal L(K) < ∞.All the inequalities above are sharp.
Theorem 1.5.Suppose that X , Y are -valued martingales such that and The inequalities are sharp.
Obviously, the condition (1.11) is weaker than the differential subordination.It concerns the quadratic variations of X and Y on the intervals [t, ∞), t ≥ 0, and hence it can be seen as a dual to (1.7), which compares the square brackets on the intervals [0, t], t ≥ 0. It should also be stressed that Suh's result (the weak-type inequality for p ≥ 2) concerned only real-valued martingales.Our approach is not only much simpler, but it also enables us to obtain the bound for processes taking values in a Hilbert space.
A few words about our approach and the organization of the paper.The original proofs of (1.1), (1.2), (1.3) and (1.4) are based on Burkholder's method, which reduces the problem of showing a given martingale inequality to the problem of finding a certain biconcave function.This approach has also been been successful in a number of other estimates, see [3], [5] for the detailed description of the method and related remarks, and [13] for the extension of the technique to the continuous-time setting.Our approach is slightly different and exploits an integration argument developed by the author in [9].In Section 2 we introduce two "basis" functions which are used in two simple inequalities (2.7) and (2.8) for martingales satisfying (1.7) or (1.11).Then we complicate these inequalities by integrating the basis functions against various kernels; this yields the desired estimates.Section 3 contains the description of the integration argument, and in Section 4 we present the detailed calculations leading to (1.8), (1.9), (1.10), (1.12) and (1.13).

Two basis functions
Let D 1 , D ∞ be subsets of × × [0, ∞), defined as follows: Furthermore, introduce the functions φ 1 , Here be the functions given by The key property of the functions u 1 and u ∞ is described in the lemma below.
Proof.(i) We start from the observation that Indeed, both sides are equal on D c 1 , while on D 1 we have Suppose then, that (x, y, s) lies in and we are done.(ii) This follows immediately from (i) and the identities valid for all x, y ∈ and t ≥ 0. The lemma above leads to the following martingale inequalities.Let u 0 1 (x, y) = u 1 (x, y, 0) and u 0 ∞ (x, y) = u ∞ (x, y, 0).Lemma 2.2.Suppose that X , Y are -valued martingales.(i) Suppose that (1.7) holds.Then for all t ≥ 0, Proof.Using standard approximation (cf.[13]), it suffices to show the inequalities for finite dimensional case: = d for some positive integer d. (i) Let t ≥ 0 be fixed.We will prove that (2.9) Since u 1 is of class C 2 on D 1 and the range of the process (Z τ∧s− ) s≤t is contained in D 1 , we may apply Itô's formula to obtain where As one easily verifies, I 2 /2 + I 3 = 0; this is due to the fact that for all i, j; and u 1s (Z s− ) = 1.Furthermore, I 4 + I 5 ≤ 0, in view of (1.7) and (2.3).By the properties of stochastic integrals, I 1 has mean 0. Combine the above facts about the terms I k with (2.9) and (2.10) to get u 1 (Z t ) ≤ u 1 (Z 0 ).However, the latter expression is nonpositive; this is due to are finite almost surely.We will show a stronger statement: for any stopping time η, and τ = inf{s : Z s / ∈ D ∞ }.By Doob's optional sampling theorem, . Applying Itô's formula for u ∞ (Z τ∧η ), we get the analogue of (2.10), with similar terms I 0 -I 5 (simply replace u 1 by u ∞ ).We have that I 1 = I 2 = I 3 = 0 and I 4 + I 5 ≤ 0 due to (1.11) and (2.4); this implies u ∞ (Z η ) ≤ u ∞ (Z 0 ).The proof is completed by the observation that u(Z 0 ) ≤ 0 almost surely, which can be verified readily.

An integration method
In the proof of the announced inequalities we will use the following procedure.Suppose that V : × → is a given Borel function and assume that we are interested in proving the estimate for a pair (X , Y ) of martingales satisfying the condition (1.7 and take If the kernel k is chosen in such a way that U(x, y) ≥ V (x, y) for all x, y ∈ , (3.3) then (3.1) holds.Indeed, for any r > 0 and t ≥ 0 we have 3) and Fubini's theorem, permitted due to (3.2).Similarly, suppose we are interested in the bound for a pair (X , Y ) satisfying (1.11) and ||X || 2 < ∞.Arguing as previously, we see that it suffices to find a function k : [0, ∞) → [0, ∞) which enjoys the condition and the majorization property (3.3),where U : × → is given by This approach will be successful in proving (1.8), (1.9), (1.10) and (1.12).In the case of (1.13) we will need a slight modification of this method.The details are presented in the next section.

The proof of (1.8)
Clearly, we may assume that ||X || p < ∞, since otherwise there is nothing to prove.We will be done if we show that |Y t | p ≤ (p − 1) −p |X t | p for any t ≥ 0. Let V p , U p : × → be given by Now we show (3.2) and (3.3).The second estimate was shown by Burkholder, see page 17 in [5].To establish (3.2), note that by Burkholder-Davis-Gundy inequality, for some constant c p > 0, we obtain where It is straightforward to verify that J 1 , J 2 , J 3 are finite, and (3.2) follows.This completes the proof of (1.8).

The proof of (1.9)
This is a bit more technical.We start with the following auxiliary fact.Let Proof.(i) Denote the left-hand side by F (r).We have We turn to the proof of (1.9).It suffices to show that and letting → 0 yields (1.9).To prove (4.5), note that for p = 1 it follows directly from (2.7), since (see e.g.pages 20-21 in [5]).For 1 where where A is given by (4.2).Now the condition (3.2) can be easily verified using (4.1).To prove while for |x| + | y| > 1, the majorization reduces to (4.4).This completes the proof.

4.3
The proof of (1.10) where α > 0 will be chosen later.The condition (3.2) is verified easily using (4.1) and the observation that for some positive constants c, d, The inequality (3.3), for a proper choice of α, was shown in Lemma 3.3 in [10].

The proof of (1.12)
As previously, we restrict ourselves to the case ||X || p < ∞ and the inequality takes the form where α p = p 3−p (p − 1) p (p − 2)/2.It can be verified that The inequality (3.2) can be proved in the same manner as in the case 1 < p ≤ 2; the majorization (3.3) was established by Burkholder on page 17 in [5].

The proof of (1.13)
We assume that ||X || p < ∞.We cannot proceed as in the proof of (1.9): the inequality |X ∞ | p , the analogue of (4.5), is of no value to us.The problem is that if τ is a stopping time, then (X τ∧t , Y τ∧t ) may no longer satisfy (1.11).To overcome this difficulty, we consider a stopping time τ = inf{t : |Y t | ≥ 1} and show that This yields the claim: see the argumentation leading to (4.6) above.Introduce V p,∞ : It is easy to check the analogue of (3.2), that is, Use (2.11) with η = τ to get U p,∞ (Z τ ) ≤ 0, where Z is given by (2.12).We have Combining this with U p,∞ (Z τ )I {τ=∞} = U p,∞ (X ∞ , Y ∞ , 0)I {τ=∞} , we obtain and the proof will be complete if we show that U p,∞ (x, y, 0) ≥ V p,∞ (x, y).To this end, note that the function F given by But this bound is valid for all x, y ∈ .Indeed, observe that as a function of | y|, the left-hand side attains its minimum for | y| = 1 − 2/p, and one easily checks that the inequality again reduces to (4.7).This completes the proof.