TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 361, Number 5, May 2009, Pages 2689–2718 S 0002-9947(08)04631-X Article electronically published on November 20, 2008
ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS
YAOZHONG HU AND DAVID NUALART
Abstract. b Using fractional calculus we define integrals of the form af(xt)dyt, where x and y are vector-valued H¨older continuous functions of order β ∈ 1 1 ( 3 , 2 )andf is a continuously differentiable function such that f is λ-H¨older 1 − continuous for some λ> β 2. Under some further smooth conditions on f the integral is a continuous functional of x, y, and the tensor product x ⊗ y with respect to the H¨older norms. We derive some estimates for these integrals and we solve differential equations driven by the function y. We discuss some applications to stochastic integrals and stochastic differential equations.
1. Introduction The theory of rough path analysis has been developed from the seminal paper by Lyons [10]. The purpose of this theory is to analyze d-dimensional dynamical systems dxt = f(xt)dyt, where the control function y is not differentiable. The natural assumption is that the control function y has bounded p-variation for some p>1. If 1 Received by the editors October 2, 2006 and, in revised form, September 6, 2007. 2000 Mathematics Subject Classification. Primary 60H10, 60H05; Secondary 26A42, 26A33, 46E35. Key words and phrases. Rough path, fractional calculus, integral, integration by parts, differ- ential equation, stability, stochastic differential equation, Wong-Zakai approximation, convergence rate. The work of the first author was supported in part by the National Science Foundation under Grant No. DMS0204613 and DMS0504783. The work of the second author was partially supported by the MCyT Grant BFM2000-0598 and the NSF Grant No. DMS-0604207. c 2008 American Mathematical Society Reverts to public domain 28 years from publication 2689 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2690 YAOZHONG HU AND DAVID NUALART where π :0=t0 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS 2691 on any approximation argument and it leads to precise H¨older estimates (see Propositions 3.4, 6.4 and Corollary 6.5). ii) We transform the dynamical system dxt = f(xt)dyt into a closed system of equations involving only x, x ⊗ y and x ⊗ (y ⊗ y), which can be solved by a classical fixed point argument. These results can be applied to implement a path-wise approach to define sto- chastic integrals and to solve stochastic differential equations driven by a multidi- mensional Brownian motion. As an application of the deterministic results obtained for dynamical systems we derive a sharp rate of almost sure convergence of the Wong-Zakai approximation for multidimensional diffusion processes. We couldn’t find these kinds of estimates elsewhere. Similar results hold in the case of a frac- ∈ 1 1 tional Brownian motion with Hurst parameter H ( 3 , 2 ); however they are more involved and will be treated in a forthcoming paper. 2. Fractional integrals and derivatives Let a, b ∈ R with a License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2692 YAOZHONG HU AND DAVID NUALART • 1 If α> p ,then α− 1 α p ∪ α p ⊂ p Ia+ (L ) Ib− (L ) C (a, b) . We have the following composition formulas: α β α+β α β α+β (2.3) Da+Da+ = Da+ ,Db−Db− = Db− , and similar formulas hold for the fractional integral operator. The following inver- sion formulas hold: (2.4) Iα Dα f = f, ∀f ∈ Iα (Lp) , a+ a+ a+ α α ∀ ∈ α p (2.5) Ia− Da−f = f, f Ia− (L ) , and α α α α ∀ ∈ 1 (2.6) Da+ Ia+f = f, Da− Ia−f = f, f L (a, b) . On the other hand, for any f,g ∈ L1(a, b)wehave b b α − α α (2.7) Ia+f(t)g(t)dt =( 1) f(t)Ib−g(t)dt , a a ∈ α p ∈ α p and for f Ia+ (L )andg Ia− (L )wehave b b α − −α α (2.8) Da+f(t)g(t)dt =( 1) f(t)Db−g(t)dt. a a ∈ λ ∈ µ Suppose that f C (a, b)andg C (a, b)withλ + µ>1. Then, from the b classical paper by Young [17], the Riemann-Stieltjes integral a fdg exists. The following proposition can be regarded as a fractional integration by parts formula, b and provides an explicit expression for the integral a fdg in terms of fractional derivatives (see [18]). ∈ λ ∈ µ Proposition 2.1. Suppose that f C (a, b) and g C (a, b) with λ + µ>1. − b Let 1 µ<α<λ. Then the Riemann-Stieltjes integral a fdg exists and it can be expressed as b b − α α 1−α (2.9) fdg =( 1) Da+f (t) Db− gb− (t) dt, a a where gb− (t)=g (t) − g (b). 3. Integration of rough functions 1 1 Fix 3 <β< 2 and a time interval [0,T]. Following [10] we introduce the following definition. Definition 3.1. We will say that (x, y, x ⊗ y)isan(m, d)-dimensional β-H¨older continuous multiplicative functional if x :[0,T] → Rm and y :[0,T] → Rd are β-H¨older continuous functions and x ⊗ y is a continuous function defined on ∆:={(s, t):0≤ s ≤ t ≤ T } with values on Rm ⊗ Rd and x ⊗ y satisfies the following properties: i) For all s ≤ u ≤ t we have (multiplicative property) ⊗ ⊗ − ⊗ − ⊗ (3.1) (x y)s,u +(x y)u,t +(xu xs) (yt yu)=(x y)s,t . License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS 2693 ii) For all (s, t) ∈ ∆, ⊗ ≤ | − |2β (3.2) (x y)s,t c t s . If x and y are continuously differentiable functions, then ⊗ i,j i j (3.3) (x y)s,t = dxξdyη, s<ξ<η License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2694 YAOZHONG HU AND DAVID NUALART compensated fractional derivative for r ∈ [0,a]: 1 f (x ) D α f (x)(r)= r a+ Γ(1− α) (r − a)α r − − m i − i f (xr) f (xθ) i=1 ∂if(xθ)(xr xθ) (3.9) +α α+1 dθ . a (r − θ) This derivative is well defined under our hypotheses, because there exists a constant K such that for all r, θ ∈ [0,a], θ License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS 2695 Proof. To simplify the proof we take m = d =1.Wehave b b b − − − f(xr)yrdr = [f(xr) f (xr)(xr xa)] yrdr + f (xr)(xr xa)yrdr a a a := A1 + A2. We can express the terms A1 and A2 using the fractional integration by parts formula (2.9). In this way we obtain b − α α − − 1−α (3.12) A1 =( 1) Da+ [f(x·) f (x·)(x· xa)] (r)Db− yb−(r)dr. a Let us compute the first fractional derivative in the expression (3.12): α D [f(x·) − f (x·)(x· − x )] (r) a+ a 1 f (x ) − f (x )(x − x ) = r r r a Γ(1− α) (r − a)α r f (xr) − f (xr)(xr − xa) − [f (xθ) − f (xθ)(xθ − xa)] +α α+1 dθ a (r − θ) (x − x ) f (x ) r f (x ) − f (x ) = D α f (x)(r) − r a r + α r θ dθ a+ − − α α+1 Γ(1 α) (r a) a (r − θ) α − − α = Da+f (x)(r) (xr xa)Da+f (x)(r). As a consequence, we obtain b − α α 1−α A1 =(1) Da+f (x)(r)Db− yb−(r)dr a b − − α − α 1−α (3.13) ( 1) (xr xa)Da+f (x)(r)Db− yb−(r)dr. a On the other hand, applying again (2.9) yields b A2 = f (xr)d(x ⊗ y)a,·(r) a b − α α 1−α ⊗ (3.14) =(1) Da+f (x)(r)Db− (x y)a,·,b−(r) dr. a We have, using (3.1), 1−α ⊗ Db− (x y)a,·,b−(r) 1−α b (−1) (x ⊗ y)a,r − (x ⊗ y)a,b (x ⊗ y)a,r − (x ⊗ y)a,θ = − +(1− α) − dθ Γ(α) (b − r)1 α (θ − r)2 α r 1−α (−1) −(x ⊗ y)r,b − (xr − xa)(yb − yr) = − Γ(α) (b − r)1 α b − ⊗ − − − − (x y)r,θ (xr xa)(yθ yr) +(1 α) 2−α dθ r (θ − r) −D1−α ⊗ − 1−α = b− (x y)(r)+(xr xa)Db− yb−(r). (3.15) License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2696 YAOZHONG HU AND DAVID NUALART Substituting (3.15) into (3.14) and using (3.13) we get b b − α α 1−α f(xr)dyr =(1) Da+f (x)(r)Db− yb−(r)dr a a b − − α α D1−α ⊗ ( 1) Da+f (x)(r) b− (x y)(r) dr. a Finally, using the fractional integration by parts formula (2.8) we obtain b α D1−α ⊗ Da+f (x)(r) b− (x y)(r) dr a b − α−1 2α−1 1−αD1−α ⊗ =(1) Da+ f (x)(r)Db− b− (x y)(r) dr, a which yields the desired result. In the sequel, k will denote a generic constant that may depend on the parameters β, α,andλ. We have the following estimate. ⊗ β Rm→ Rd Proposition 3.4. Let (x, y, x y) be in Mm,d(0,T). Assume that f : is a continuously differentiable function such that f is bounded and λ-H¨older contin- 1 − ≤ ≤ uous, where λ> β 2. Then for any 0 a where Φa,b,β(x, y) is defined in (3.6). Remark: In (3.16) we can replace f ∞ by f (x) a,b,∞ and f λ by the λ-H¨older norm of f on the interval [min x, max x]. Proof. First we have, for any r ∈ [a, b], α ≤ | | − −α 1+λ − (λ+1)β−α (3.17) Da+f(x)(r) k f(xr) (r a) + f λ x a,r,β (r a) , and 1−α ≤ − α+β−1 (3.18) Db− yb−(r) k y r,b,β (b r) . From (3.10) we get: D1−α ⊗ ≤ ⊗ − 2β+α−1 (3.19) b− (x y)(r) k x y r,b,2β (b r) . Using (3.10), (6.3) and (3.19) yields 1−αD1−α ⊗ ≤ − 2β+2α−2 (3.20) Db− b− (x y)(r) kΦr,b,β (x, y)(b r) . Moreover, (3.21) 2α−1 ≤ | | − 1−2α λ − λβ−2α+1 Da+ f (x)(r) k f (xr) (r a) + f λ x a,r,β (r a) . License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS 2697 Thus, using (3.11) and the inequalities (3.17), (3.18), (3.20), and (3.21) yields b b ≤ | | − −α − α+β−1 f(xr)dyr k y a,b,β f(xr) (r a) (b r) dr a a b 1+λ − (1+λ)β−α − α+β−1 + f λ x a,b,β (r a) (b r) dr a b |f (x )| − +k r + f x λ (r − a)λβ 2α+1 − 2α−1 λ a,b,β a (r a) 2β+2α−2 × Φa,b,β(x, y)(b − r) dr. Therefore, we obtain b ≤ | | − β f(xr)dyr k f(xa) y a,b,β (b a) a − 2β +k y a,b,β x a,b,β f ∞ (b a) 1+λ − (λ+2)β +k y a,b,β f λ x a,b,β (b a) 2β +kΦa,b,β (x, y) f ∞ (b − a) λ − (λ+2)β +k f λ x a,b,β Φa,b,β(x, y)(b a) , and this implies (3.16) easily. The estimate (3.16) implies that for a fixed x, the mapping (y, x⊗y)→ f(xr)dyr is continuous with respect to the β-norm. That is, if (x, yn,x⊗ yn) is a sequence of β-H¨older continuous multiplicative functionals such that − n → (3.22) y y β 0, ⊗ − ⊗ n → (3.23) x y x y 2β 0, as n tends to infinity, then − n → (3.24) f(xr)dyr f(xr)dyr 0. β ⊗ β Suppose that (x, y, x y) belongs to Mm,d(0,T). Then, the integral f(xr)dyr introduced in Definition 3.2 does not depend on the parameter α, and it coincides n n with the limit of the classical integrals f(xr)dyr when y is a sequence of piece- ⊗ i,j wise continuously differentiable functions such that (3.22) holds and (x y)s,t = t i − i j s xξ xs y ξ dξ satisfies (3.23). n iT Set ti = n for i =0, 1,...,n.Ify is β-H¨older continuous, then the sequence of functions n n n n y = y01{0}(t)+ 1(tn ,tn](t) ytn + t − t − (ytn − ytn ) t i−1 i i−1 T i 1 i i−1 i=1 converges to y in the β-norm for any β <β.Fix1 <β <βand assume that 3 t − n ⊗ s (xr xs)dyr converges in the β -norm to (x y)s,t as n tends to infinity. Then License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2698 YAOZHONG HU AND DAVID NUALART (3.24) holds with β = β. In particular, this means that, under these assumptions, we have T n tn n i f(x )dy = lim f(x )ds (y n − y n ). r r s ti ti−1 n→∞ T n 0 i=1 ti−1 For any p ≥ 1, the p variation of a function x :[0,T] → R is defined as n 1/p n − n p Varp(x)=sup x(ti ) x(ti−1) , π i=1 where π = {0=t0 < ··· where π = {0=t0 < ··· License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS 2699 As a consequence, using (2.3) and (2.9) yields b b b − − 2α−1 2α−1 1−α 1−α f(xr)dyr f(xr)dyr =( 1) Da+ f (x)(r)Db− Db− ϕb−(r)dr a a a b − 2α−1 2α−1 2−2α =( 1) Da+ f (x)(r)Db− ϕb−(r)dr a b = f (xr)dϕ(r). a In order to handle differential equations we need to introduce the tensor product of two multiplicative functionals: ⊗ ⊗ β Definition 3.6. Suppose that (x, y, x y)and(y, z, y z)belongtoM1,1(0,T). Then, for all a ≤ b ≤ c, we define ⊗ ⊗ x (y z)·,c a,b (−1)α b (x − x )(z − z ) r (z − z )(x − x ) = r a c r + α r θ θ r dθ − − α α+1 Γ(1 α) a (r a) a (r − θ) 1−α ×D − yb−(r)dr b (−1)2α−1 b z − z r z − z − c r +(2α − 1) θ r dθ − − 2α−1 − 2α Γ(2 2α) a (r a) a (r θ) 1−α 1−α ×D − D − (x ⊗ y)(r)dr b b (−1)2α−1 b x − x r x − x + r a +(2α − 1) r θ dθ − − 2α−1 − 2α Γ(2 2α) a (r a) a (r θ) × 1−αD1−α ⊗ (3.26) Db− b− (y z)(r)dr. We have the following result. Proposition 3.7. If the function y is continuously differentiable and for all a ≤ b, b ⊗ − (y z)a,b = (zb zr) yrdr, a b ⊗ − (x y)a,b = (xr xa)yrdr, a then b ⊗ ⊗ − − x (y z)·,c = (xr xa)(zc zr)yrdr. a,b a Proof. It suffices to apply formula (3.11) with m =2,d =1,f(x, z)=xz and the functions xt − xa and zc − zt. Then, the following estimate follows from Definition 3.6 and the inequalities (3.18) and (3.20). ⊗ ⊗ β Proposition 3.8. Suppose that (x, y, x y) and (y, z, y z) belong to M1,1(0,T). Then, for any a ≤ b we have ⊗ ⊗ ≤ − 2β − β x (y z)·,c kΦa,b,β (x, y) z a,b,β (b a) (c a) a,b ⊗ − 3β +k x a,b,β y z a,b,2β (b a) . License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2700 YAOZHONG HU AND DAVID NUALART ⊗ ⊗ ⊗ ⊗ If b = c we write x (y z)·,b =(x y z)a,b. Notice that if the functions a,b x, y and z are continuously differentiable, then ⊗ ⊗ (x y z)a,b = xryθzσdrdθdσ. a where Φa,b,β(x, y, z)isdefinedin(3.7). ⊗ ⊗ ⊗ Proposition 3.8 also implies that x, (y z)·,c , x (y z)·,c is a β-H¨older continuous functional on the interval [0,c]. As a consequence, if f satisfies the b ⊗ assumptions of Definition 3.2, we can define the integral a f(xr)dr (y z)r,c,for all a ≤ b ≤ c. The following estimate for this integral will be needed to solve differential equations. ⊗ ⊗ β Proposition 3.9. Suppose that (x, y, x y) and (y, z, y z) belong to Mm,d(0,T). Let f : Rm→ Rd be a continuously differentiable function such that f is λ-H¨older 1 − − continuous and bounded, where λ> β 2.Fixα>0 such that 1 β<α<2β, λβ+1 α< 2 . Then the following estimate holds: b ⊗ ≤ | | − 2β f(xr)dr(y z)r,b k f(xa) Φa,b,β (y, z)(b a) a λ − λβ +k f ∞ + f λ x a,b,β (b a) 3β (3.28) ×Φa,b,β (x, y, z)(b − a) . Proof. To simplify the proof we will assume d = m = 1. From (3.1) it is easy to see that ⊗ ≤ − β (3.29) (x y)·,b Φa,b,β (x, y)(b a) , a,b,β and from Proposition 3.8 we have ⊗ ⊗ ≤ − β (3.30) x (y z)·,b a,b,2β kΦa,b,β (x, y, z)(b a) . From (3.16), (3.29), and (3.30) we obtain b ⊗ f(xr)dr(y z)r,b a β ≤ k |f(x )| (y ⊗ z)· (b − a) a ,b a,b,β ⊗ λ − λβ − 2β +k[Φa,b,β (x, (y z)·,b) f ∞ + f λ x a,b,β(b a) ](b a) ≤ k |f(x )| Φ (y, z)(b − a)2β a a,b,β λ − λβ − 3β +k[Φa,b,β (x, y, z) f ∞ + f λ x a,b,β (b a) ](b a) , which implies the desired result. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS 2701 4. Differential equations driven by rough paths ⊗ β Suppose that (y, y, y y) belongs to Md,d(0,T). We aim to solve the differential equation t (4.1) xt = x0 + f(xr)dyr, 0 where f : Rm → Rm ⊗ Rd. The main idea is to consider the enlarged unknown (x, x ⊗ y), where d t ⊗ k,l k ⊗ j,l (4.2) (x y)s,t = fj (xr)d(y y)·,t (r). j=1 s By formula (3.11), the right-hand side of Equation (4.1) is a function of (x, y, x⊗y), and the right-hand side of Equation (4.2) is a function of (x, y ⊗ y, x⊗ (y ⊗ y)). On the other hand, by Equation (3.26), x ⊗ (y ⊗ y) is a functional of (x, y, x ⊗ y, y ⊗ y). β By definition, a solution of Equation (4.1) is an element of Mm,d(0,T)such that (4.1) and (4.2) hold. This is equivalent to saying that the following system of equations holds d t − α α 1−α j xt = x0 +( 1) D0+fj (x)(s)Dt− yt−(s)ds j=1 0 m d t − − 2α−1 2α−1 ( 1) D0+ ∂ifj (x)(s) i=1 j=1 0 × 1−αD1−α ⊗ i,j (4.3) Dt− t− (x y) (s)ds, d t ⊗ k,l − α α k 1−α ⊗ i,l (x y)s,t =( 1) Ds+fi (x)(r)Dt− (y y)·,t− (r)dr i=1 s m d t − − 2α−1 2α−1 ( 1) Ds+ ∂ifj (x)(r) i=1 j=1 s × 1−αD1−α ⊗ ⊗ k,i,l (4.4) Dt− t− (x (y y)·,t) (r)dr, and i,j,k ⊗ ⊗ x (y y)·,t s,t (−1)α t (xi − xi )(yk − yk) r (yk − yk)(xi − xi ) = r s t r + α r θ θ r dθ − − α α+1 Γ(1 α) s (r s) s (r − θ) 1−α j ×D − y −(r)dr t t (−1)2α−1 t yk − yk r yk − yk − t r +(2α − 1) θ r dθ − − 2α−1 − 2α Γ(2 2α) s (r s) s (r θ) 1−α 1−α i j ×D − D − (x ⊗ y )(r)dr t t (−1)2α−1 t xi − xi r xi − xi + r a +(2α − 1) r θ dθ − − 2α−1 − 2α Γ(2 2α) s (r s) s (r θ) × 1−αD1−α ⊗ j,k (4.5) Dt− t− (y y) (r)dr. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2702 YAOZHONG HU AND DAVID NUALART First we establish the existence of a solution. ⊗ β Rm → Rmd Theorem 4.1. Let (y, y, y y) be in Md,d(0,T).Letf : be a continu- 1 − ously differentiable function such that f is λ-H¨older continuous, where λ> β 2, and f and f are bounded. Set ρf = f ∞ + f ∞ + f λ. ⊗ β Then there is a solution (x, y, x y) in Mm,d(0,T) to Equations (4.1)–(4.2).More- over, the function x satisfies the estimate 1/β y ⊗ y 2β (4.6) sup |xt|≤|x0| +1+ T 2kρf y β + , 0≤t≤T y β where k is a universal constant depending only on α, β and λ, with the convention y⊗y2β =0if y β =0. yβ Proof. To simplify the proof we will assume d = m = 1. The proof will be done in several steps. − λβ+1 Step 1. Fix α>0 such that 1 β<α<2β, α< 2 . β → β ⊗ Consider the mapping J : M1,1(0,T) M1,1(0,T)givenbyJ(x, y, x y)= (J1,y,J2), where J1 and J2 are the right-hand sides of equations (4.1) and (4.2), ⊗ β respectively. This mapping is well defined because for each (x, y, x y)inM1,1(0,T), (J1,y,J2) is a real-valued β-H¨older continuous multiplicative functional. We need some a priori estimates of the H¨older norms of J1 and J2 in terms of the H¨older norms of x, y and x ⊗ y. From (3.16) and (3.28) it follows that J ≤ k[|f(x )| y +Φ (x, y) 1 s,t,β s s,t,β s,t,β × λ − λβ − β (4.7) f (x) s,t,∞ + f λ x s,t,β (t s) (t s) ] and J ≤ k |f(x )| Φ (y, y) 2 s,t,2β s s,t,β λ − λβ +k f ∞ + f λ x s,t,β (t s) β (4.8) ×Φs,t,β (x, y, y)(t − s) . Step 2. Set 1/β y ⊗ y 2β α(y):= 2kρf y β + , y β where k is the constant appearing in formulas (4.7) and (4.8). Fix s, t such that 1 (4.9) 0 (4.11) x s,t,β ≤ 2kρf y β, − β ⊗ ≤ (4.12) (t s) x y s,t,2β y β. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS 2703 y ⊂ y ⊗ ∈ y We claim that J(Cs,t) Cs,t. In fact, suppose that (x, y, x y) Cs,t. Then, (4.10) and (4.11) imply β (4.13) (t − s) x s,t,β ≤ 1. The inequalities (4.12) and (4.13) imply − β − β ⊗ ≤ (4.14) (t s) Φs,t,β (x, y)=(t s) x y s,t,2β + x s,t,β y s,t,β 2 y β and − β − β 2 (t s) Φs,t,β(x, y, y)=(t s) x s,t,β y s,t,β ⊗ ⊗ + x s,t,β y y s,t,2β + y s,t,β x y s,t,2β ≤ 2 ⊗ (4.15) 2 y s,t,β + y y s,t,2β . From (4.7), (4.13) and (4.14) we obtain ≤ (4.16) J1 s,t,β 2kρf y β , and from (4.8), (4.13), (4.15) and (4.9) we get ≤ 2 ⊗ J2 k f ∞ Φs,t,β(y, y)+( f ∞ + f ) 2 y β + y y s,t,2β λ 2β ≤ 2 ⊗ β 2kρf y β + y y 2β = α(y) y β −β (4.17) ≤ (t − s) y β. ∈ y Hence, (J1,y,J2) Cs,t and the claim is proved. Step 3. We can now proceed with the proof of the existence. Let N be a natural T ≤ 1 number such that N = δ α(y) . We partition the interval [0,T]intoN subintervals iT − of the same length and set ti = N , i =0, 1,...,N 1. We will make use of the ⊗ ⊗ − notation x i = x ti−1,ti,β and x y i = x y ti−1,ti,2β,fori =1,...,N 1. From Step 2 we know that if that x and x ⊗ y satisfy x i ≤ 2kρf y β, ⊗ ≤ −β x y i y βδ , for any i =1,...,N − 1, then the same inequalities hold for J1 and J2,thatis, J1 i ≤ 2kρf y β, ≤ −β J2 i y βδ . Consequently, there is a constant C1 such that n n ≤ J1 β + J2 2β C1 . n This implies that the sequence of functions J1 is equicontinuous and bounded in Cβ(0,T). Therefore, there exists a subsequence which converges in the β-H¨older n norm if β <β. In the same way, there is a subsequence of J2 which converges in the β-H¨older norm. The limit defines a β-H¨older continuous multiplicative functional (x, y, x ⊗ y). Using the continuity of the solution in this norm it is not difficult to show that the limit is a solution. This implies the existence of a solution, which satisfies (4.11) and (4.12) if (4.9) holds. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2704 YAOZHONG HU AND DAVID NUALART Step 4. Let us now prove the estimate (4.6). By step 2, the solution we have constructed satisfies the estimates (4.11) and (4.12) if (4.9) holds. Then it follows that for any r ∈ [s, t], β sup |xr|≤|xs| +(t − s) x s,t,β ≤|xs| +1. r∈[s,t] 1 Since the interval [0,T] can be divided into [Tα(y)] + 1 intervals of length α(y) ,the inequality (4.6) follows. The following result provides the uniqueness of a solution under stronger hy- potheses; moreover it provides the continuity of the solution in the input parame- 2N ters x0, y and y ⊗ y. For any vector y in R ,wesety =(y, y ), where y (resp. y ) is the vector formed with the first (resp. last) N components of y. ⊗ β Rm → Rmd Theorem 4.2. Let (y, y, y y) be an element of Md,d(0,T).Letf : be a twice continuously differentiable function such that f is λ-H¨older continuous, 1 − where λ> β 2,andf, f and f are bounded. Then there is a unique solution ⊗ ∈ β (x, y, x y) Mm,d(0,T) to Equations (4.1)–(4.2). β Suppose that y belongs to M2d,2d(0,T),andsety =(y, y),wherey and y are d-dimensional vectors. Let (x, y, x ⊗ y) ∈ M β (0,T) be the unique solution to 2m,2d t R2m → R2d the Equation xt = x0 + 0 f(xr)dyr,wheref : is defined by f(x)= (f(x),f(x )).Then (4.18) sup |xt − x t|≤C {|x0 − x 0| + y − y β + (y − y ) ⊗ y 2β + y ⊗ (y − y ) 2β} , 0≤t≤T where C depends on y β, y β, y ⊗ y 2β, β, λ, T ,andρf ,andwhere ρf = f ∞ + f ∞ + f λ + f ∞ + f λ . Proof. To simplify the proof we will assume d = m = 1. Notice that uniqueness follows from the estimate (4.18). So it suffices to show this inequality. To simplify, − ≤ 1 ∧ assume y β > 0and y β > 0. We fix s License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS 2705 The term I1,s,t can be estimated using (6.4) and we obtain ≤ 1 − 2 − 3 − ⊗ (4.24) I1,s,t k[Hs,t(y) x x s,t,∞ +Hs,t(y) x x s,t,β +Hs,t (x x) y s,t,2β], 1 2 3 where Hs,t(y), Hs,t(y), and Hs,t are defined in (6.5), (6.6), and (6.7), respectively. Then, using the inequalities (4.20), (4.21), and (4.22) we get the following estimates: 1 ≤ (4.25) Hs,t(y) k y β ρf , 2 ≤ − β (4.26) Hs,t(y) k y β ρf (t s) , 3 ≤ − β (4.27) Hs,t ρf (t s) . It remains to handle the term (x − x ) ⊗ y s,t,2β in (4.24). From (4.2) for the solution (x, y, x ⊗ y)weobtain t t |((x − x ) ⊗ y)s,t| = f(xr)dr(y ⊗ y)·,t − f(x r)dr(y ⊗ y)·,t s s t = [f(xr) − f(x r)] dr(y ⊗ y)·,t s t + f(x r)dr((y − y ) ⊗ y)·,t s (4.28) := B1 + B2. For the term B1 we have, using the inequality (6.4), β 1 2 B ≤ k(t − s) H ((y ⊗ y) ) x − x ∞ + H ((y ⊗ y) ) x − x 1 s,t .,t s,t, s,t .,t s,t,β 3 − ⊗ ⊗ (4.29) +Hs,t (x x) (y y)·,t s,t,2β . 1 ⊗ The definition of Hs,t((y y).,t) together with the estimates (3.29) and (3.30) leads to 1 2β H ((y ⊗ y) ) ≤ Φ (y, y) f ∞(t − s) s,t .,t s,t,β λ λ − βλ + f ∞ + f λ x s,t,β + x s,t,β (t s) 2β × (Φs,t,β(x, y, y)+( x s,t,β + x s,t,β)Φs,t,β (y, y)) (t − s) , and the estimates (4.20), (4.21) and (4.22) imply 1 ⊗ ≤ − β (4.30) Hs,t((y y).,t) kρf Φβ(y, y)(t s) , ⊗ 2 2 ⊗ where Φβ(y, y)= y y 2β+ y β. In a similar way, the definition of Hs,t((y y).,t) together with the estimates (3.29) and (3.30) leads to 2 ⊗ ≤ Hs,t((y y).,t) k f ∞ (Φs,t,β(x, y, y) 3β +( x s,t,β + x s,t,β )Φs,t,β(y, y)) (t − s) 2β + f ∞Φs,t,β(y, y)(t − s) 2β (4.31) ≤ kρf Φβ(y, y)(t − s) . On the other hand, from (3.30) we get β (4.32) (x − x˜) ⊗ (y ⊗ y)·,t s,t,2β ≤ kΦs,t(x − x, y, y)(t − s) . License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2706 YAOZHONG HU AND DAVID NUALART Thus, substituting (4.30), (4.31), (4.27) and (4.32) into (4.29) yields 2β B1 ≤ kρf (t − s) Φβ(y, y) x − x s,t,∞ 3β +kρf (t − s) Φβ(y, y) x − x s,t,β +kρ (t − s)3βΦ (x − x, y, y) f s,t,β 2β β ≤ k(t − s) ρf Φβ(y, y) x − x s,t,∞ + x − x s,t,β(t − s) − 3β − ⊗ (4.33) +k(t s) ρf y β (x x) y s,t,β . For the term B2 we obtain, using Proposition 3.9, B ≤ k |f(x )| Φ (y − y,y )(t − s)2β 2 s s,t,β λ − λβ − − 3β +k f ∞ + f λ x s,t,β (t s) Φs,t,β(x, y y, y)(t s) . (4.34) From (4.2) when x and y are replaced by x and y − y we get t ⊗ − ⊗ − (x (y y))s,t = f(xr)dr(y (y y)·,t. s ⊗ − As a consequence, we can estimate x (y y) s,t,2β using Proposition 3.9 and we obtain x ⊗ (y − y ) ≤ k |f(x )| Φ (y, y − y ) s,t,2β s s,t,β λ − λβ +k f ∞ + f λ x s,t,β (t s) β (4.35) ×Φs,t,β (x, y, y − y )(t − s) , and using (4.21) and (4.23), − β − ≤ − ⊗ − (t s) Φs,t,β (x, y, y y) y β y y β + y (y y) 2β (4.36) ≤ Φβ(y, y − y ). Substituting (4.36) into (4.35), and using (4.21) yields ⊗ − ≤ − (4.37) x (y y) s,t,2β kρf Φβ(y, y y). Then, substituting (4.37) into (4.34) we obtain 2β β (4.38) B2 ≤ kρf (t − s) Φβ(y − y, y)+ρf y β(t − s) Φβ(y, y − y ) . From (4.33), (4.38) and (4.28) we get − ⊗ ≤ − − − β (x x) y s,t,2β kρf Φβ(y, y) x x s,t,∞ + x x s,t,β(t s) − ⊗ − β +kρf y β (x x) y s,t,2β (t s) +kρf Φβ(y − y, y) 2 β +k (ρf ) y βΦβ(y, y − y )(t − s) . The condition t−s ≤ 1/β(y), if the constant k in β(y) is chosen in an appropriate way, implies that 1 k(t − s) βρ y ≤ . f β 2 Hence, assuming k ≥ 2, we have − ⊗ ≤ − − − β (x x) y s,t,2β kρf Φβ(y, y) x x s,t,∞ + x x s,t,β(t s) (4.39) +kρf [Φβ(y − y,y )+ρf Φβ(y, y − y )] . License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS 2707 Substituting (4.25), (4.26), (4.27) and (4.39) into (4.24) yields β I1,s,t ≤ kρf y x − x s,t,∞ + x − x s,t,β (t − s) β 2 β β +k (ρf ) Φβ(y, y) x − x s,t,∞ + x − x s,t,β(t − s) (t − s) 2 β +k (ρf ) (Φβ(y − y, y)+ρf Φβ(y, y − y )) (t − s) . Again, condition t−s ≤ 1/β(y), if the constant k in β(y) is chosen in an appropriate way, implies that 1 I ≤ kρ y x − x ∞ + x − x 1,s,t f β s,t, 2 s,t,β 2 β (4.40) +k (ρf ) (Φβ(y − y, y)+ρf Φβ(y, y − y )) (t − s) . For the term I2,s,t we have the following estimates, using (3.16), (4.21) and (4.37): I ≤ k f y − y + kΦ (x, y − y ) 2,s,t ∞ s,t,β s,t,β λ λβ β × f ∞ + f x (t − s) (t − s) λ s,t,β ≤ − ⊗ − − β kρf y y β + x (y y˜) s,t,2β (t s) ≤ − − − β (4.41) kρf y y β + ρf Φβ(y, y y)(t s) . Now from (4.40) and (4.41) we get 1 x − x ≤ kρ y x − x ∞ + x − x +Ψ(y, y ), s,t,β f β s,t, 2 s,t,β where 2 − − − β − Ψ(y, y)=k (ρf ) (Φβ(y y, y)+ρf Φβ(y, y y)) (t s) + kρf y y β ≤ K( y − y β + (y − y ) ⊗ y 2β + y ⊗ (y − y ) 2β), for some constant K, depending on y β, y β, y ⊗ y 2β, β, λ,andρf . Hence, − ≤ − (4.42) x x s,t,β kρf y β x x s,t,∞ +2Ψ(y, y) . Notice that − ≤| − | − β − (4.43) x x s,t,∞ xs xs +(t s) x x s,t,β . Hence, − ≤ | − | − β − x x s,t,β kρf y β [ xs xs +(t s) x x s,t,β]+2Ψ(y, y). Consequently, − ≤ | − | (4.44) x x s,t,β kρf y β xs xs +4Ψ(y, y). Substituting (4.44) into (4.43) yields − ≤| − | − β | − | − β (4.45) x x s,t,∞ xs xs +(t s) kρf y β xs x˜s +4(t s) Ψ(y, y). 1 Denote δ = β(y) and tn = nδ.Set Zn =sup|xs − x s|. 0≤s≤tn Then inequality (4.45) states that β β Zn+1 ≤ (1 + kρf δ )Zn +4δ Ψ(y, y ). License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2708 YAOZHONG HU AND DAVID NUALART Therefore β [ T ] ZT ≤ (1 + kρf δ ) δ |x0 − x 0| T [ δ ] β l β +k (1 + kρf δ ) 4δ Ψ(y, y ). l=0 This implies the desired estimate. The following corollary is a direct consequence of (4.42) and (4.18). Corollary 4.3. Under the assumptions of Theorem 4.2,ifx and x˜ satisfy t t xt = x0 + f(xs)dys andx ˜t =˜x0 + f(˜xs)dy˜s , 0 0 then (4.46) xt − x˜t β ≤ C {|x0 − x˜0| + y − y˜ β +[ (y − y˜) ⊗ y) 2β + y˜ ⊗ (y − y˜) 2β]} . 5. Stochastic differential equations { 1 2 d ≥ } Suppose that B = Bt =(Bt ,Bt ,...,Bt ),t 0 is a d-dimensional Brownian motion. Fix a time interval [0,T]. Define t ⊗ − ◦ (B B)s,t = (Br Bs) d Br, s where the stochastic integral is a Stratonovich integral. That is, 1 (Bi − Bi )2 if i = j, (B ⊗ B)i,j = 2 t s s,t t i − i j s Br Bs dBr if i = j, where the stochastic integral is an Itˆo integral. It is not difficult to show that we ⊗ ⊗ can choose a version of (B B)s,t in such a way that (B,B,B B)constitutesa β-H¨older continuous multiplicative functional, for a fixed β ∈ (1/3, 1/2). As a first application of Theorem 3.3 and (3.11) we deduce that the Stratonovich T ◦ stochastic integral 0 f(Br)d Br has the following path-wise expression: T d T ◦ − α α 1−α j f(Br)d Br =(1) D0+fj (B)r (DT − BT −)rdr 0 j=1 0 d T − − 2α−1 2α−1 1−αD1−α ⊗ i,j ( 1) D0+ ∂ifj (B)r DT − T − (B B) dr . i,j=1 0 i,j In the same way, if we set B⊗B =(B ⊗ B)i,j − 1 (t − s)δ , then the path-wise s,t s,t 2 i,j T ⊗ integral 0 f(Br)dBr corresponding to the multiplicative functional (B,B B)is the Itˆointegral: T d T − α α 1−α j f(Br)dBr =(1) D0+fj (B)r (DT − BT −)rdr 0 j=1 0 d T − − 2α−1 2α−1 1−αD1−α ⊗ i,j ( 1) D0+ ∂ifj (B)r DT − T − (B B) dr . i,j=1 0 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS 2709 Notice that Proposition 3.5 yields the well-known relation between both integrals: T T 1 d T f(B )d◦B − f(B )dB = ∂ f (B )dr. r r r r 2 i i r 0 0 i=1 0 We can apply Theorem 4.1 and deduce the existence of a solution for the sto- chastic differential equation in Rm: t ◦ (5.1) Xt = X0 + f(Xs)d Bs, 0 where the initial condition X0 is an arbitrary random variable, and the function f : Rm → Rmd is a continuously differentiable function such that f is λ-H¨older 1 − continuous, where λ> β 2, and f and f are bounded. By Theorem 4.2 the solution is unique if f is twice continuously differentiable with bounded derivatives 1 − and f is λ-H¨older continuous, where λ> β 2. The stochastic integral here is a path-wise integral which depends on B and B ⊗ B. We have also the stability-type results (4.18) and (4.46). In particular, assume that Bε is a piece-wise smooth approximation of B such that ε ε ε ε B − B β, B ⊗ (B − B ) 2β, and (B − B ) ⊗ B 2β converge to zero with a certain rate. Let X =(X, Xε) be the solution of t Xt = X0 + f(Xs)dBs, 0 ε where X0 =(X0,X0), B =(B,B ), and we use the notation of Theorem 4.2. Then ε according to Corollary 4.3, X − X β will also converge to 0 with the same rate. In particular, this implies that the stochastic process X coincides with the solution of the Stratonovich stochastic differential equation t ◦ H (5.2) Xt = X0 + f(Xs)d Bs . 0 In this section we will apply these results in order to obtain the almost sure rate of convergence of the Wong-Zakai approximation to the stochastic differential equation (5.1). That is, we will consider the rate of convergence in the H¨older norm when we approximate the Brownian motion by a polygonal line. In order to get a precise rate for these approximations we will make use of the following exact modulus of continuity of the Brownian motion. There exists a random variable G such that almost surely for any s, t ∈ [0,T]wehave 1/2 −1 (5.3) |Bt − Bs|≤G|t − s| log (|t − s| ). Let π = {0=t0 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2710 YAOZHONG HU AND DAVID NUALART Lemma 5.1. There exists a random variable CT,β such that (5.4) B − Bπ ≤ C nβ−1/2 log n, β T,β (5.5) B ⊗ (B − Bπ) ≤ C nβ−1/2 log n, 2β T,β (5.6) Bπ ⊗ (B − Bπ) ≤ C nβ−1/2 log n, 2β T,β π β−1/2 (5.7) (B − B ) ⊗ B 2β ≤ CT,βn log n. Proof. Fix 0 On the other hand, if s ∈ [tk−1,tk]andt ∈ [tk,tk+1] we have, again if n is large enough, 1 n |h (s, t)|≤ B − B + (t − t ) B − B 1 − β tk t k tk+1 tk (t s) T n − − − − − Btk Bs (tk s) Btk Btk−1 T 1 n ≤ |B − B | + (t − s) |B − B | + |B − B | (t − s)β t s T tk tk−1 tk+1 tk G n 1/2 ≤ |t − s|1/2 log |t − s|−1 +2(t − s) log (n/T ) (t − s)β T ≤ 3GT −β+1/2n−1/2+β log (n/T ). This proves (5.4). Now we turn to the estimate of the term 1 t t h (s, t)= (Bi − Bi )dBj,π − (Bi − Bi )dBj 2 − 2β u s u u s u (t s) s s for i = j (the case i = j is obvious from (5.4)). We claim that there exists a random variable Z such that, almost surely, for all s, t ∈ [0,T]wehave t i − i j ≤ | − | | − |−1 (5.8) (Bu Bs)dBu Z t s log t s . s License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS 2711 In fact, it suffices to show this inequality almost surely for all s and t rational numbers. If we fix s, the process {Mt,t∈ [s, T ]}, t i − i j Mt = (Bu Bs)dBu, s is a continuous martingale and it can be represented as a time-changed Brownian motion: Mt = W t i − i 2 . s (Bu Bs) du As a consequence, applying (5.3) there exists a random variable G1 such that 1/2 −1 t t i i 2 i i 2 |Mt| = W t i − i 2 ≤ G (B − B ) du log (B − B ) du (Bu Bs) du u s u s s s s and again (5.3), applied to Bi − Bi , yields u s t 1/2 t −1 | |≤ − | − |−1 2 − | − |−1 Mt G1G2 (u s)log u s du log G2 (u s)log u s du , s s for some random variable G2.Wehavefor|t − s|≤1, t 1 1 (u − s)log|u − s|−1du = |t − s|2 + log |t − s|−1 , s 4 2 and this implies easily the estimate (5.8). − ≥ T Suppose first that t s n .Then t t 1 i i j,π i i j h2(t, s)= (B − B )dB − (B − B )dB (t − s)2β u s u u s u s s t 1 i i j,π j,π j,π j,π i = (B − B )(B − B ) − (B − B )dB (t − s)2β t s t s u s u s t − i − i j − j j − j i (Bt Bs)(Bt Bs )+ (Bu Bs )dBu s 1 ≤ (Bi − Bi )(Bj,π − Bj,π − Bj − Bj) (t − s)2β t s t s t s 1 t + Bj,π − Bj dBi − 2β u u u (t s) s = A1 + A2. Using (5.3) and (5.4) the term A1 can be estimated as follows: A ≤ G|t − s|1/2−β log |t − s|−1 Bj,π − Bj 1 β β−1/2 (5.9) ≤ CT,βn log n. For the term A2 we proceed as in the proof of the estimate (5.8). We have t j,π − j i Bu Bu dBu = W t j,π j 2 , (Bu −Bu) du s s where W is a Brownian motion. As a consequence, using that π −1/2 B − B ∞ ≤ CT,βn log (n/T ) License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2712 YAOZHONG HU AND DAVID NUALART (this estimate is proved as (5.4)) we get 1/2 −1 G t t A ≤ (Bj,π − Bj )2du log (Bj,π − Bj )2du 2 (t − s)2β u u u u s s − 1/2−2β −1/2 −1 1 ≤ CT,β(t − s) n log n log (t − s) n (log n) β−1/2 (5.10) ≤ CT,βn log n. − T Suppose now that t s< n . We make the decomposition 1 t t h (s, t) ≤ (Bi − Bi)dBj,π + (Bi − Bi )dBj 2 − 2β u s u u s u (t s) s s = B1 + B2. Then (5.8) yields 1−2β −1 2β−1 (5.11) B2 ≤ Z|t − s| log |t − s| ≤ CT,βn log n. In order to handle the term B1, assume first that s, t ∈ [tk,tk+1]. Then t n t (Bi − Bi )dBj,π = Bj − Bj (Bi − Bi )du, u s u tk+1 tk u s s T s and we obtain 1−2β 1−2β B1 ≤ CT,β(t − s) log n ≤ CT,βn log n. Finally, if s ∈ [tk−1,tk]andt ∈ [tk,tk+1]wehave tk −2β n j j i i B1 ≤ (t − s) B − B (B − B )du T tk tk−1 u s s t + Bj − Bj (Bi − Bi )du tk+1 tk u s tk 1−2β ≤ CT,βn log n. The inequality (5.6) can be proved similarly, and (5.7) is a direct consequence of (5.10). The proof is now complete. As a consequence, we can establish the following result. Theorem 5.2. Let f : Rm → Rmd be continuously differentiable with bounded derivative up to fourth order and let X satisfy t ◦ Xt = X0 + f(Xs)d Bs. 0 π If Xt satisfies the following ordinary differential equation, t π π π Xt = X0 + f(Xs )dBs , 0 then for any β ∈ (1/3, 1/2), there is a random constant CT,β ∈ (0, ∞) such that π β−1/2 (5.12) X − X β ≤ CT,βn log n. Proof. The result is a straightforward consequence of Lemma 5.1 and Theorem 4.2. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS 2713 6. Appendix Lemma 6.1. We have for any x Lemma 6.2. For any z ∈ [0, 1] we have 1 − α−2 2β α−2 − α−2 ≤ 2 2 2β+α−1 (6.2) x (x (x + z) )dx − z . 0 2β + α 1 Proof. We can decompose the integral as follows: 1 z x2β(xα−2 − (x + z)α−2)dx = x2β(xα−2 − (x + z)α−2)dx 0 0 1 + x2β(xα−2 − (x + z)α−2)dx z z 1 ≤ x2β+α−2 dx + x2β(xα−2 − (2x)α−2)dx 0 z 2 − 2α−2 = z2β+α−1. 2β + α − 1 Lemma 6.3. For any a ≤ b ≤ T we have (6.3) D1−α(x ⊗ y) ≤ kΦ (x, y)(b − a)β+α−1, b− a,b,β a,b,β where the constant k depends on α and β. D1−α ⊗ Proof. Let us first consider the first term in the definition of b− (x y). Fix a ≤ s 1−α 1−α (x ⊗ y)t,b (x ⊗ y)s,b (x ⊗ y)t,b (b − s) − (x ⊗ y)s,b (b − t) − − − = − − (b − t)1 α (b − s)1 α (b − t)1 α (b − s)1 α 1−α 1−α |(x ⊗ y)t,b| (b − t) − (b − s) ≤ − − (b − t)1 α (b − s)1 α |(x ⊗ y)t,b − (x ⊗ y)s,b| + − (b − s)1 α = A1 + A2. Using (3.2) yields ≤ ⊗ − 2β+a−1 − α−1 − 1−α − − 1−α A1 x y t,b,2β (b t) (b s) (b t) (b s) , License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2714 YAOZHONG HU AND DAVID NUALART and (3.1) together with (3.2) implies α−1 A2 ≤ (|(x ⊗ y)s,t| + |(xt − xs)(yb − yt)|)(b − s) ≤ ⊗ − 2β − α−1 − β − α−1 − β x y s,t,2β (t s) (b s) + x s,t,β y t,b,β (t s) (b s) (b t) β α+β−1 ≤ Φa,b,β(x, y)(t − s) (b − s) . Then, (6.1) implies ≤ ⊗ − α+β−1 − β A1 x y a,b,2β (b s) (t s) . D1−α ⊗ ≤ ≤ Consider now the second term in the definition of b− (x y). Fix a s = B1 + B2. For the term B1 we have using (3.2), x ⊗ y B ≤ s,t,2β (t − s)2β+α−1. 1 2β + α − 1 For the term B2 we write − − b (x ⊗ y) (θ − t)2 α − (x ⊗ y) (θ − s)2 α B = s,θ t,θ dθ 2 2−α 2−α t (θ − s) (θ − t) b α−2 α−2 ≤ |(x ⊗ y)t,θ| (θ − t) − (θ − s) dθ t b |(x ⊗ y)t,θ − (x ⊗ y)s,θ| + 2−α dθ t (θ − s) = B21 + B22. The term B21 can be estimated using (3.2): b ≤ ⊗ − 2β − α−2 − − α−2 B21 x y t,b,2β (θ t) (θ t) (θ s) dθ. t θ−t s−t Making the change of variables b−t = x and b−t = z, and using the estimate (6.2) yields b − − (θ − t)2β (θ − t)α 2 − (θ − s)α 2 dθ t 1 =(b − t)2β+α−1 x2β(xα−2 − (x + z)α−2)dx 0 2 − 2α−2 ≤ (t − s)2β+α−1. 2β + α − 1 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS 2715 Finally, using (3.1) and (3.2) we obtain b |(x ⊗ y)s,t| + |xt − xs||yθ − yt| B22 ≤ − dθ (θ − s)2 α t x ⊗ y x y ≤ s,b,2β + s,t,β t,b,β (t − s)2β+α−1. 1 − α α + β − 1 The proof of the result is now complete. The next two propositions provide stability results for the integral f(x)dy on the variables f and x. Proposition 6.4. Suppose that (x, y, x ⊗ y) and (˜x, y, x˜ ⊗ y) are elements of β Rm→ Rd Mm,d(0,T).Letf : be a twice continuously differentiable function such that the second derivative f is λ-H¨older continuous and f and f are bounded, 1 − where λ> β 2.Then f(xr)dyr − f(x r)dyr a,b,β ≤ − 2 − kHa,b(y) x x a,b,∞ + kHa,b(y) x x a,b,β 3 − ⊗ (6.4) +kHa,b (x x) y a,b,2β , where 1 β H (y)= y f ∞(b − a) a,b a,b,β λ λ βλ + f ∞ + f x + x (b − a) λ a,b,β a,b,β β (6.5) × x ⊗ y a,b,2β +( x a,b,β + x a,b,β ) y (b − a) , a,b,β 2 ⊗ − 2β Ha,b(y)= f ∞ x y a,b,2β +( x a,b,β + x a,b,β ) y a,b,β (b a) β (6.6) + f ∞ y (b − a) , a,b,β 3 − β − β (6.7) Ha,b = f ∞ + f ∞ x a,b,β (b a) (b a) . Proof. Note that for any a ≤ θ ≤ r ≤ b we can write f(xr) − f(xθ) − f (xθ)(xr − xθ) − [f(x r) − f(x θ) − f (x θ)(x r − x θ)] 1 = [f (xθ + z(xr − xθ)) − f (xθ)] (xr − xθ − x r + x θ)dz 0 1 + [f (xθ +z(xr −xθ))−f (x θ +z(x r − x θ)) − f (xθ)+f (x θ)] (x r − x θ)dz 0 = a1 − a2. We have 2β |a1|≤ f ∞ x a,b,β x − x a,b,β (r − θ) . License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 2716 YAOZHONG HU AND DAVID NUALART For the term a2 we make the decomposition 1 1 a2 =(x r − x θ) [(x θ − xθ)+z (x r − x θ − xr + xθ)] 0 0 ×f (x + z(x − x )+t(x − x )+tz(x − x − x + x ))dtdz θ r θ θ θ r θ r θ 1 −(x r − x θ)(x θ − xθ) f (xθ + t(x θ − xθ))dt 0 =(x − x − x + x )(x − x ) r θ r θ r θ 1 1 × zf (xθ + z(xr − xθ)+t(x θ − xθ)+tz(x r − x θ − xr + xθ))dtdz 0 0 1 1 −(x r − x θ)(x θ − xθ) [f (xθ + z(xr − xθ)+t(x θ − xθ) 0 0 + tz(x r − x θ − xr + xθ)) −f (xθ + t(x θ − xθ))] dtdz. Thus, |a |≤ f x x − x (r − θ)2β 2 ∞ a,b,β a,b,β − λ − λ − β(1+λ) + f λ x a,b,β x x˜ a,b,∞ x a,b,β + x x a,b,β (r θ) . As a consequence, |f(xr) − f(xθ) − f (xθ)(xr − xθ) − [f(x r) − f(x θ) − f (x θ)(x r − x θ)]| β(1+λ) (6.8) ≤ kI1(r − θ) , where I = f ∞ { x + x } x − x 1 a,b,β a,b,β a,b,β λ λ − + f λ x a,b,β x a,b,β + x a,b,β x x a,b,∞ . On the other hand, we have D2α−1f (x)(r) − D2α−1f (x )(r) s+ s+ 1 f (x ) − f (x ) = r r Γ(2 − 2α) (r − s)2α−1 r [f (x ) − f (x ) − f (x )+f (x )] +(2α − 1) r r θ θ dθ . − 2α s (r θ) Using the decomposition f (x ) − f (x ) − f (x )+f (x ) r r θ θ 1 1 = f (xr + t(x r − xr))(x r − xr)dt − f (xθ + t(x θ − xθ))(x θ − xθ)dt, 0 0 we obtain 2α−1 − 2α−1 Ds+ f (x)(r) Ds+ f (x)(r) 1−2α β−2α+1 ≤ k(r − s) f ∞ x − x ∞ + k(r − s) f ∞ x − x s,r, s,r,β − βλ−2α+1 λ λ − +k(r s) f λ x s,r,β + x s,r,β x x s,r,∞ 1−2α (6.9) = kI2(r − s) , License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ROUGH PATH ANALYSIS VIA FRACTIONAL CALCULUS 2717 where λ λ − βλ − I2 = f ∞ + f λ x a,b,β + x a,b,β (b a) x x a,b,∞ β + f ∞ x − x a,b,β(b − a) . Now using (3.18), (6.8), (6.9) we obtain b − ≤ [f(xr) f(xr)] dyr k y a,b,β a b −α α+β−1 × |f(xr) − f(x r)| (r − a) (b − r) dr a b β(1+λ)−α α+β−1 +I1 (r − a) (b − r) dr a b − 1−2α 1−αD1−α ⊗ +kI2 (b r) Db− b− (x y)(r) dr a b 2α−1 1−αD1−α − ⊗ + Da+ f (x)(r) Db− b− ((x x) y)(r) dr. a Finally, using (3.20) and (3.21), we get b − ≤ [f(xr) f(xr)] dyr k y a,b,β a 2β β(2+λ) × f ∞(b − a) x − x a,b,∞ + I1(b − a) +kI ( x ⊗ y + x y )(b − a)2β 2 a,b,2β a,b,β a,b,β β (6.10) + f ∞ + f ∞ x a,b,β (b − a) 2β ×{ (x − x ) ⊗ y a,b,2β + x − x a,b,β y a,b,β } (b − a) . This implies (6.4). The following corollary is a consequence of Propositions 3.4 and 3.5. Corollary 6.5. Under the hypotheses of Proposition 3.5, we have f(xr)dyr − f(x r)dyr a,b,β 1,f 2,f 3,f ≤ kH (y) x − x ∞ + kH x − x + kH (x − x ) ⊗ y a,b a,b, a,b a,b,β a,b a,b,2β − ⊗ +k f f y a,b,β + k x y a,b,2β + x a,b,β y a,b,β ∞ × − − λ − λβ − β (6.11) f f + f f x a,b,β (b a) (b a) . ∞ λ References 1. Coutin, L. and Lejay, A. 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MR1640795 (99j:60073) Department of Mathematics, University of Kansas, Lawrence, Kansas 66045-2142 E-mail address: [email protected] Department of Mathematics, University of Kansas, Lawrence, Kansas 66045-2142 E-mail address: [email protected] License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use