A non commutative sewing lemma

In a preceding paper [E.J.ofProb.34,860-892,(2006)], we proved a sewing lemma which was a key result for the study of Holder continuous functions. In this paper we give a non-commutative version of this lemma with some applications.


Introduction
In a preceding paper [1] we proved a sewing lemma which was a key result for the study of Hölder continuous functions.In this paper we give a non-commutative version of this lemma.
In the first section we recall the commutative version, and give some applications (Young integral and stochastic integral).
In the second section we prove the non-commutative version.This last result has interesting applications : an extension of the so-called integral product, a simple case of the semigroup Trotter type formula, and a sharpening of the Lyons theorem about multiplicative functionals [3,4,5].
Note that we replaced the Hölder modulus of continuity t α by a more general modulus V (t).This paper was elaborated with the regretted G. Mokobodzki.The writing has only been done after his death.
I. The additive sewing lemma 1 Definition : We say that a function V (t) defined on [0, T [ is a control function if it is non decreasing, V (0) = 0 and n≥1 V (1/n) < ∞.
As easily seen, this is equivalent to the property for every t ≥ 0. For example, t α and t/(log t −1 ) α with α > 1 are control functions.
Observe that we have from which follows that Lim t→0 V (t)/t = 0.
We say that u is midpoint-additive.Now, we prove that u is the unique midpoint-additive function hence we have w = u, that is u is in fact rationally-additive.As µ is continuous, then so also is u, as the defining series converges uniformly for 0 ≤ a ≤ b < T .Then u is additive, and it suffices to put ϕ(t) = u(0, t).
4 Remarks : a) In fact the result holds even in the case of discontinuous µ, as proved in the appendix.b) The result obviously extends to Banach spaces valued functions µ.
In the case V (t) = t α with α > 1, we get 5 Example : The Young integral Take V (t) = t 2α with α > 1/2.If x and y are two α-Hölder continuous functions on where x α is the norm in the space C α .Let ϕ be the function of theorem 2, put b a This is a Young integral (cf.also [7]).
7 Example : The stochastic integral Let X t be the standard IR m -valued Brownian motion.As is well known, t where the last integral is taken in the Ito or in the Stratonovitch sense.By straightforward computations, we get By the additive sewing lemma, there exists a unique in the Ito or in the Stratonovitch sense.
Observe that as the stochastic integral 8 Remark : For the FBM with α > 1/4, the reader is referred to our previous paper [1].

II. The multiplicative sewing lemma
Here we need a strong notion of control function 9 Definition : We say that a function V (t) defined on [0, T [ is a strong control function if it is a control function and there exists a θ > 2 such that for every t We consider an associative monoide M with a unit element I, and we assume that M is complete under a distance d satisfying for every x, y, z ∈ M, where z → |z| is a Lipschitz function on M with |I| = 1.
Let µ(a, b) be an M-valued function defined for 0 ≤ a ≤ b < T .We assume that µ is continuous, that µ(a, a) = I for every a, and that for every a ≤ c ≤ b we have Proof : Put µ 0 = µ and by induction The functions h n and U n continuous non decreasing with h n (0) = 1 and U n (0) = 0. Let κ be the Lipschitz constant of z → |z|.One has Let τ > 0 be such that h 0 (τ ) + κV (τ ) ≤ θ/2.Assume that U i (t) ≤ θ i V (t/2 i ) for t ≤ τ and i ≤ n.One has h n+1 (t) ≤ θ/2, then By the second step of the proof, we get w = u.The same proof extends to every m, so that u is in fact rationally multiplicative.As u is continuous, it is multiplicative.We then have Particularly we can take A t = tA and B t = tB with α = 1, this yields ab ∈ A, λ is a real parameter.We have Following [5], we suppose the algebraic hypothesis for k ≤ n (3) Moreover we have where the series is normally convergent for every λ.
Proof : The only problem is to prove formula (4), that is to prove that u is the sum of its Taylor expansion with respect to λ.In the case where A is a complex Banach algebra, the proof of the multiplicative sewing lemma yields a sequence of holomorphic functions which converges uniformly with respect to λ in every compact set of | C .Hence u(a, b) in holomorphic in λ ∈ | C .If A is only a real Banach algebra, we get a sequence of holomorphic functions with values in the complexified Banach space of A, and the result follows.It remains to observe that the n + 1 first terms of the Taylor expansion are the same for every function of the sequence converging to u.
Application to the Lyons theorem : Let E be a Banach space.Denote for every k and every a ≤ c ≤ b, and 16 Remark : This theorem sharpens the theorem 3.2.1 of [5].

X
t ⊗ dX t has C α -trajectories almost surely for 1/3 < α < 1/2, analoguous computations as above yield C α -trajectories for b a f (X t ) ⊗ dX t on the same set of paths as b a X t ⊗ dX t .

( 1 )
d(µ(a, b), µ(a, c)µ(c, b)) ≤ V (b − a) We say that an M-valued u(a, b) is multiplicative if u(a, b) = u(a, c)u(c, b) for every a ≤ c ≤ b. 10 Theorem : There exists a unique multiplicative function u such that d(µ(a, b), u(a, b)) ≤ Cst V (b − a) for every a ≤ b.
(a, b) converges locally uniformly to a continuous function u(a, b) which is midpoint-multiplicative, that is u(a, b) = u(a, c)u(c, b) for c = (a + b)/2.One has d(u, µ) ≤ Cst V .Next we prove the unicity of u.Let v be a function with the same properties asu.Put K(t) = Sup b−a≤t [u(a, b), v(a, b)].Let τ 1 > 0 be such that K(τ 1 ) ≤ θ/2.One has d(u(a, b), v(a, b)) ≤ k V (b − a) with some constant k, then d(u(a, b), v(a, b)) ≤ 2K(t/2)kV (t/2) ≤ k θV (t/2) for b − a ≤ t ≤τ 1 , and by induction d(u(a, b), v(a, b)) ≤ k θ n V (t/2 n ).It follows that u(a, b) = v(a, b) for b − a ≤ τ 1 .This equality extends to every b − a by midpoint-multiplicativity. Finally we prove that u is multiplicative.We argue as in the additive case, and we put for an integer m w(a, b) = m−1 i=0 u(t i , t i+1 ) where t i = a + i.(b − a)/m.For simplicity we limit ourselves to the case m = 3, that is w(a, b) = u(a, c ′ )u(c ′ , c ′′ )u(c ′′ , b) with c ′ = a + (b − a)/3, c ′′ = a + 2(b − a)/3.Observe that w is obviously midpointmultiplicative.Take a ≤ b ≤ T 0 < T , we get successively with a constant k which can be changed from line to line d(w(a, b), µ(a, b)) ≤kV (b − a) + d(w(a, b), µ(a, c ′ )µ(c ′ , b))

11 0 H
Example : The integral product Let t → A t a C α function with values in a Banach algebra A with a unit I. Put A ab = A b − A a and µ(a, b) = I + A ab We get µ(a, b) − µ(a, c)µ(c, b) = −A ac A cb Suppose that α > 1/2, then the multiplicative sewing lemma applies with the obvious distance, and there exists a unique multiplicative function u(a, b) with values in A such that |u(a, b) − µ(a, b)| ≤ Cst |b − a| 2α We get the same u(a, b) by taking µ(a, b) = e A ab .A good notation for u(a, b) is u(a, b) = b a (I + dA t ) = b a e dA t 12 Theorem : Put H t = u(0, t).Then this is the solution of the EDO H t = I + t s dA s Proof : We have only to verify that |u(0, b) − u(0, a) − u(0, a)A ab | ≤ Cst |b − a| 2α .The first member is worth u(0, a)[u(a, b) − I − A ab ] = u(0, a)[u(a, b) − µ(a, b)] so that we are done.13 Example : A Trotter type formula Let t → A t and t → B t as in the previous paragraph, and put µ(a, b) = [I + A ab ][I + B ab ] It is straightforward to verify the good inequality |µ(a, b) − µ(a, c)µ(c, b)| ≤ Cst |b − a| 2α so that we get a multiplicative u(a, b) such that |u(a, b) − µ(a, b)| ≤ Cst |b − a| 2α or equivalently |u(a, b) − I − A ab − B ab | ≤ Cst |b − a| 2α u(a, b) = b a (I + dA t + dB t ) = b a e dA t e dB t 2 n e B/2 n 14 Example : Extending the Lyons theorem Let A be a Banach algebra with a unit I. Take µ(a, b) of the form µ(a, b) = c)µ(c, b) = µ(a, b) + 2n k=n+1 λ k C (k) acb 15 Theorem : Under the condition (3) and the inequality |A (k) ab | ≤ M |b − a| kα for every k ≤ n, where α > 1/(n + 1), there exists a unique multiplicative function u(a, b) such that |u(a, b) − µ(a, b)| ≤ Cst |b − a| (n+1)α

17
Corollary : Put c ′ = Max(c, 1), we have |u(a, b)| ≤ c ′ e β (|λ|x|b − a| α ) 2 n+1 ) for t ≤ τ and every n by induction.Hence for t ≤ τ the series U n (t) converges, so that the sequence h n (τ ) is bounded.By inequality (2) the series U n (2τ ) converges, and the sequence h n (2τ ) is bounded.From one step to the other we see that the sequence h n is locally bounded, and that the series U n converges locally uniformly on [0, T [.It follows that the sequence µ n This is a Banach algebra.The previous theorem applies, so that there exists a unique (a, b) → Y ab for every k such that Y