aa r X i v : . [ m a t h . P R ] N ov PATHWISE DEFINITION OF SECOND ORDER SDES
LLUÍS QUER-SARDANYONS AND SAMY TINDEL
Abstract.
In this article, a class of second order differential equations on [0 , , drivenby a γ -Hölder continuous function for any value of γ ∈ (0 , and with multiplicative noise,is considered. We first show how to solve this equation in a pathwise manner, thanksto Young integration techniques. We then study the differentiability of the solutionwith respect to the driving process and consider the case where the equation is drivenby a fractional Brownian motion, with two aims in mind: show that the solution wehave produced coincides with the one which would be obtained with Malliavin calculustools, and prove that the law of the solution is absolutely continuous with respect to theLebesgue measure. Introduction
During the last past years, a growing activity has emerged, aiming at solving stochasticPDEs beyond the Brownian case. In some special situations, namely in linear (additivenoise) or bilinear (noisy term of the form u ˙ B ) cases, stochastic analysis techniques canbe applied [14, 30]. When the driving process of the equation exhibits a Hölder continuityexponent greater than / , Young integration or fractional calculus tools also allow tosolve those equations in a satisfying way [10, 17, 25]. Eventually, when one wishes totackle non-linear problems in which the driving noise is only Hölder continuous withHölder regularity exponent ≤ / , rough paths analysis must come into the picture. Thissituation is addressed in [4, 11, 28].It should be mentioned however that all the articles mentioned above only handle thecase of parabolic or hyperbolic systems, letting apart the case of elliptic equations. This isof course due to the special physical relevance of heat and wave equations, but also stemsfrom a specific technical difficulty inherent to elliptic equations. Indeed, even in the usualBrownian case, the notion of filtration and adapted process is useless in order to solve non-linear elliptic systems, so that Itô’s integration theory is not sufficient in this situation.A natural idea in this context is then to use the power of anticipative calculus, based onMalliavin type techniques (see e.g. [18]). This method has however a serious drawbackin our context, mainly because the Picard type estimates involve Malliavin derivatives of Date : March 9, 2017.2010
Mathematics Subject Classification.
Key words and phrases.
Second order SDEs, Young integration, fractional Brownian motion, Malliavincalculus.L. Quer-Sardanyons is supported by the grant MCI-FEDER Ref. MTM2009-08869 and S. Tindel ispartially supported by the (French) ANR grant ECRU. any order, and cannot be closed. To the best of our knowledge, all the stochastic ellipticequations considered up to now involve thus a mere additive noise. Let us mention forinstance the pioneering works [3, 20] for the existence and uniqueness of solutions, thestudy of Markov’s property [5, 20], the numerical approximations of [15, 26, 29], as well asthe recent and deep contribution [22], which relates stochastic elliptic systems, anticipativeGirsanov’s transforms and deterministic methods.With these preliminary considerations in mind, the aim of the current paper is twofold: (i)
We wish to solve a nonlinear elliptic equation of the form ∂ tt z t = σ ( z t ) ˙ x t , t ∈ [0 , , y = y = 0 , (1)where σ is a smooth enough function from R to R , and x is a Hölder continuous noisyinput with any Hölder continuity exponent γ ∈ (0 , . To this purpose, we shall writeequation (1) in a variant of the so-called mild form, under which it becomes obviousthat the system can be solved in the space C κ of κ -Hölder continuous functions, for any − γ < κ < (see Section 2.1 for a precise definition of this space).Let us observe however that, when dealing with a non-linear multiplicative noise, oneis not allowed to use the monotonicity methods invoked in [3]. This forces us to usecontraction type arguments, which can be applied only provided the Hölder norm of x issmall enough. In order to overcome this restriction, we shall introduce a positive constant M , and replace the diffusion coefficient σ by a function σ M : R × C γ → R such that y σ M ( y, x ) is regular enough and σ M ( · , x ) ≡ whenever k x k γ ≥ M + 1 . We shall thusproduce a local solution to equation (1), in the sense given for instance in [18] concerningthe localization of the divergence operator on the Wiener space. Once this change is made,a proper definition of the solution plus a fixed point argument leads to the existence anduniqueness of solution for equation (1). (ii) Having produced a unique solution to our system in a reasonable class of functions,one may wonder if this solution could have been obtained thanks to Malliavin calculustechniques, in spite of the fact that a direct application of those techniques to our equationdo not yield a satisfying solution in terms of fixed point arguments. In order to answerthis question, we shall prove that, when x is a fractional Brownian motion (fBm in thesequel), the solution is differentiable enough in the Malliavin calculus sense, so that thestochastic integrals involved in the mild formulation of (1) can be interpreted as Skorohodintegrals plus a trace term, or better said as Stratonovich integrals. This will be achievedby differentiating the deterministic equation (1) with respect to the driving noise x andidentifying this derivative with the usual Malliavin derivative, as done in [2, 13, 21]. As aby-product, we will also be able to study the density of the random variable z t for a fixedtime t ∈ (0 , .We shall thus obtain the following result, which is stated here in a rather loose form(the reader is sent to the corresponding sections for detailed statements): Theorem 1.1.
Consider x ∈ C γ for a given γ > , a constant M > and a C ( R ) function σ , such that k σ ( j ) k ∞ ≤ c j M +1 for any j = 0 , , with some small enough constants ATHWISE SECOND ORDER SDES 3 c j . Let σ M be the localized diffusion coefficient alluded to above (see Definition 2.5 formore details). Then (1) The equation ∂ tt z t = σ M ( x, z t ) ˙ x t , t ∈ [0 , , z = z = 0 (2) admits a unique solution, lying in a space of the form C κ for any − γ < κ < . (2) Assume x to be the realization of a fractional Brownian motion with Hurst parameter H > / . Then for any t ∈ [0 , , z t is an element of the Malliavin-Sobolev space D , and the integral form of (2) can be interpreted by means of Skorohod integrals plus traceterms (see Section 4 for further definitions). (3) Still in the fBm context, with a slight modification of our cutoff coefficient σ M andunder the non-degeneracy condition | σ ( y ) | ≥ σ > for all y ∈ R , one gets the followingresult: for any t ∈ (0 , and a > , the restriction of L ( z t ) to R \ ( − a, a ) admits a densitywith respect to Lebesgue’s measure. The reader might wonder why we have made the assumption of a small coefficient σ here, through the assumption k σ ( j ) k ∞ ≤ c j M +1 . This is due to the fact that monotonicitymethods, which are essential in the deterministic literature (see e.g. [7]) as well as in thestochastic references quoted above, are ruled out here by the presence of the diffusioncoefficient in front of the noise ˙ x . We have thus focused on contraction type properties,which are also mentioned in [20]. Let us also say a word about possible generalizationsto elliptic equations in dimension d = 2 , : the main additional difficulty lies in the factthat the fundamental solution to the elliptic equation exhibits some singularities on thediagonal, which should be dealt with. In particular, if one wishes to handle the case of ageneral Hölder continuous signal x , rough paths arguments in higher dimensions shouldbe used. This possibility goes far beyond the current article.At a technical level, let us mention that the first part of Theorem 1.1 above relies on anappropriate formulation of the equation, which enables to quantify the increments of thecandidate solution in a reasonable way, plus some classical contraction arguments. As faras the Malliavin differentiability of the solution is concerned, it hinges on rather standardmethods (see [13, 21]). However, our density result for L ( y t ) is rather delicate, for twomain reasons: • The lack of a real time direction or filtration in equation (2) makes many usuallower bounds on the Malliavin derivatives rather clumsy. • One has to take care of the derivatives of our cutoff function σ M with respect tothe driving process, for which upper bounds are to be provided and compared tosome leading terms in the Malliavin derivatives.Solutions to these additional problems are given at Section 4, which can be seen as themost demanding part of our paper. It should also be pointed out that we are able to solveequation (2) for any Hölder regularity of the driving noise x , while our stochastic analysispart is devoted to fBm with Hurst parameter H > / . This is only due to the fact that LLUÍS QUER-SARDANYONS AND SAMY TINDEL
Malliavin calculus is much easier to handle in the latter situation, and we firmly believethat our results could be generalized to
H < / .Here is how our article is structured: our equation is defined and solved at Section 2.Differentiation properties of its solution with respect to the driving process are investigatedat Section 3. Finally, the Malliavin calculus aspects for fractional Brownian motion,including the existence of a density, are handled at Section 4.Unless otherwise stated, any constant c or C appearing in our computations below isunderstood as a generic constant which might change from line to line without furthermention. 2. Existence and uniqueness of solution
Recall that we wish to solve the one-dimensional second order differential equation (1).Towards this aim, we shall change a little its formulation thanks to some heuristic con-siderations, and introduce our localization coefficient σ M . We will then be able to solvethe equation thanks to a fixed point argument.2.1. Heuristic considerations.
Assume for the moment that x is a smooth functiondefined on [0 , . Hence, if σ is small and regular enough, it is easily shown (see [20] forsimilar arguments) that equation (1) can be solved thanks to contraction arguments.It is also well-known in this case that equation (1) can be understood in the mildsense. Specifically, let the kernel K : [0 , → [0 , be the fundamental solution of thelinear elliptic equation with Dirichlet boundary conditions, and notice that this kernel isexplicitly given by K ( t, ξ ) = t ∧ ξ − tξ, t, ξ ∈ [0 , . (3)Then { z t , t ∈ [0 , } solves (1) if it satisfies the integral equation z t = Z K ( t, ξ ) σ ( z ξ ) dx ξ , t ∈ [0 , , (4)where the integrals above are understood in the Riemann sense as soon as x is continuouslydifferentiable.Still assuming that x is continuously differentiable, let us retrieve some more informationabout the increments of the solution y to our elliptic equation. In order to do so, set first δf st = f t − f s , ≤ s ≤ t ≤ , for any continuous function f . Let us also give an expression for the increments of K , bynoticing that this kernel can be differentiated with respect to its first variable. Indeed,one has ∂ u K ( u, ξ ) = { u ≤ ξ } − ξ = ⇒ K ( t, ξ ) − K ( s, ξ ) = Z ts (cid:0) { u ≤ ξ } − ξ (cid:1) du. ATHWISE SECOND ORDER SDES 5
Then, thanks to an obvious application of Fubini’s theorem, the increments of z can bewritten as ( δz ) st = Z (cid:18)Z ts ( { u ≤ ξ } − ξ ) du (cid:19) σ ( z ξ ) dx ξ = Z ts du Z ( { u ≤ ξ } − ξ ) σ ( z ξ ) dx ξ = Z ts du (cid:18)Z u σ ( z ξ ) dx ξ (cid:19) − ( t − s ) Z ξσ ( z ξ ) dx ξ . The latter equation is the one which is amenable to generalization to a non-smoothsetting, and we will thus interpret our elliptic system in this way: we say that a continuousfunction z : [0 , → R is a solution to (1) if, for any ≤ s ≤ t ≤ , δz st = Z ts du (cid:18)Z u σ ( z ξ ) dx ξ (cid:19) − ( t − s ) Z ξσ ( z ξ ) dx ξ , (5)where the integrals with respect to the driving noise x are interpreted in the Young sense.2.2. Hölder spaces and cutoff.
Though it could be intuited from the original equation,our formulation (5) of the elliptic system indicates clearly that the candidate solutionshould be κ -Hölder continuous for any κ < , independently of the smoothness of x .More precisely, let C γ be the space of continuous functions f ∈ C ([0 , such that k f k γ < + ∞ , where k f k γ = k f k ∞ + sup ≤ s Let f ∈ C γ , g ∈ C κ with γ + κ > , and ≤ s ≤ t ≤ . Thenthe integral R ts g ξ df ξ is well-defined as limit of Riemann sums along partitions of [ s, t ] .Moreover, the following estimation is fulfilled: (cid:12)(cid:12)(cid:12)(cid:12)Z ts g ξ df ξ (cid:12)(cid:12)(cid:12)(cid:12) ≤ c γ,κ k f k γ k g k κ | t − s | γ , (6) where the constant c γ,κ only depends on γ and κ . A sharper estimate is also available: (cid:12)(cid:12)(cid:12)(cid:12)Z ts g ξ df ξ (cid:12)(cid:12)(cid:12)(cid:12) ≤ | g s | k f k γ | t − s | γ + c γ,κ k f k γ k g k κ | t − s | γ + κ . (7)The following straightforward property will also be used in the sequel: if f, g ∈ C γ , thenthe product f g defines an element in C γ such that k f g k γ ≤ k f k γ k g k γ . Remark . It might be clear to the reader that the solution to our elliptic system willlive in fact in a space of Lipschitz functions. We have chosen here to work in the Youngsetting because this does not induce any additional difficulty, and is more likely to begeneralized to higher dimensions of the parameter t . LLUÍS QUER-SARDANYONS AND SAMY TINDEL The following Fubini type theorem for Young integrals is a slight modification of [13,Proposition 2.6] and shall be needed in the sequel: Proposition 2.3. Consider γ i , λ i ∈ (0 , , i = 1 , , such that γ i + λ j > for all i, j = 1 , .Let g ∈ C γ and f ∈ C γ , and h : { ( t, s ) ∈ [0 , ; 0 ≤ s ≤ t ≤ } → R a function such that h ( · , t ) (resp. h ( t, · ) ) belongs to C λ ([ t, (resp. C λ ([0 , t ]) ) uniformly in t ∈ [0 , , and k h ( r , · ) − h ( r , · ) k λ ≤ C | r − r | γ k h ( · , u ) − h ( · , u ) k λ ≤ C | u − u | γ . (8) Then Z ts Z rs h ( r, u ) dg u df r = Z ts Z tu h ( r, u ) df r dg u , ≤ s ≤ t ≤ T, (9) and Z ts Z u h ( r, u ) df r dg u = Z s Z t ∧ rs h ( r, u ) dg u df r , ≤ s ≤ t ≤ T. Let us describe now our cutoff procedure on the coefficient σ . Recall that we wishto produce a smooth function σ M : R × C γ → R such that σ M ( · , x ) ≡ whenever k x k γ ≥ M + 1 . This also means that the Hölder norm of x should enter into the picturein a smooth manner. To this purpose, let us consider the Sobolev type norm k f k γ,p := (cid:18)Z Z ( f ( ζ ) − f ( η )) p | ζ − η | pγ +2 dζ dη (cid:19) p , for p ≥ . It will be seen below that k f k pγ,p can be differentiated with respect to f in a suitablesense. Furthermore, Garsia’s lemma (see e.g. [9, Lemma 1]) assesses that, whenever pγ > , we have k f k γ ≤ C k f k γ,p . Otherwise stated, we have the following: Remark . Let γ ∈ (0 , . Assume that ε > and p ≥ satisfy ε > p . Then: f ∈ C γ + ε = ⇒ k f k γ,p < ∞ . This being said, our local coefficient is built in the following manner: let M > be anarbitrary strictly positive number. We introduce a smooth cutoff function ϕ M satisfying: Definition 2.5. We consider a function ϕ M ∈ C ∞ b ((0 , ∞ )) such that ϕ M ( r ) = 0 for all r > M + 1 , and ϕ M ( r ) = 1 for r < M . For any x : [0 , → R for which k x k γ,p < ∞ , forsome γ ∈ (0 , and p ≥ , set G M ( x ) := ϕ M ( k x k pγ,p ) . (10) Eventually, for such x and any y ∈ R , we define σ M ( x, y ) := G M ( x ) σ ( y ) . (11) Hence, in particular, σ M ( x, y ) = 0 whenever k x k pγ,p ≥ M + 1 . ATHWISE SECOND ORDER SDES 7 We shall consider now the modified elliptic integral equation: δz st = Z ts du (cid:18)Z u σ M ( x, z ξ ) dx ξ (cid:19) − ( t − s ) Z ξσ M ( x, z ξ ) dx ξ , ≤ s ≤ t ≤ . (12)That is, we will solve Equation (5) for any control x ∈ C γ such that k x k pγ,p < M . Noticein particular that the solution z to (12) depends on M , though we have avoided most ofthe explicit references to this fact for notational sake.2.3. Fixed point argument. After the preliminary considerations of Sections 2.1 and 2.2,we now consider a driving signal x in a Hölder space C γ , and we will seek for a uniquesolution to equation (12) in C κ with − γ < κ < .As it will be illustrated in the proof of Theorem 2.7, we will need some regularityproperties of σ when considered as a map defined on C κ with values into itself. Moreprecisely, we will make use of the following result: Lemma 2.6. Suppose that σ : R → R is a bounded function that belongs to C ( R ) andhas bounded derivatives. Then, for any κ ∈ (0 , , σ : C κ → C κ satisfies the followingproperties: for all y, z ∈ C κ , k σ ( y ) k κ ≤ k σ ′ k ∞ k y k κ + k σ k ∞ , k σ ( y ) − σ ( z ) k κ ≤ C k y − z k κ {k σ ′ k ∞ + k σ ′′ k ∞ ( k y k κ + k y − z k κ ) } . Proof. The first part in the statement is an immediate consequence of the fact that σ and σ ′ are bounded functions.For the second part, let us fix s, t ∈ [0 , and y, z ∈ C κ , so that we need to analyze theincrement δ ( σ ( y ) − σ ( z )) st = σ ( y t ) − σ ( z t ) − σ ( y s ) + σ ( z s ) . To this aim, let us consider the following path: for any λ, µ ∈ [0 , , set a ( λ, µ ) = y s + λ ( z s − y s ) + µ ( y t − y s ) + λµ ( y s − y t − z s + z t ) . Notice that, in particular, a (0 , 0) = y s , a (0 , 1) = y t , a (1 , 0) = z s and a (1 , 1) = z t . Then,we can write δ ( σ ( y ) − σ ( z )) st = Z dλ Z dµ ∂ λ ∂ µ σ ( a ( λ, µ ))= Z dλ Z dµ [ σ ′ ( a ( λ, µ )) ∂ λ ∂ µ a ( λ, µ ) + σ ′′ ( a ( λ, µ )) ∂ λ a ( λ, µ ) ∂ µ a ( λ, µ )] . (13)On the other hand, we have the following estimates: | ∂ λ ∂ µ a ( λ, µ ) | ≤ k y − z k κ | t − s | κ , | ∂ λ a ( λ, µ ) ∂ µ a ( λ, µ ) | ≤ C k y − z k κ ( k y k κ + k y − z k κ ) | t − s | κ . Using these bounds and expression (13), we end up with | δ ( σ ( y ) − σ ( z )) st | ≤ C k y − z k κ {k σ ′ k ∞ + k σ ′′ k ∞ ( k y k κ + k y − z k κ ) } | t − s | κ . LLUÍS QUER-SARDANYONS AND SAMY TINDEL Therefore, we conclude the proof. (cid:3) We are now in position to state the following existence and uniqueness result for Equa-tion (12): Theorem 2.7. Let γ, κ ∈ (0 , be such that γ + κ > . Assume that ε > and p ≥ satisfy ε > p and let x ∈ C γ + ε . Suppose that σ : R → R is bounded, belongs to C ( R ) and has bounded derivatives. Suppose also that the derivatives of σ satisfy the followingcondition: k σ ( j ) k ∞ ≤ c M + 1 , j = 0 , , , (14) for a small enough constant c < . Then, there exists a unique solution of Equation (12)in C κ . Moreover, it holds that k z k κ ≤ C ( M ) , (15) where C ( M ) is a positive constant depending on M .Proof. As mentioned above, we will apply a fixed-point argument. Let us thus considerthe following map on C κ : for any z ∈ C κ , Γ( z ) is the element of C ([0 , given by Γ( z ) t = Z t du (cid:18)Z u σ M ( x, z ξ ) dx ξ (cid:19) − t Z ξ σ M ( x, z ξ ) dx ξ , t ∈ [0 , . Owing to Lemma 2.6 and the definition of σ M , one easily proves that, for all z ∈ C κ , Γ( z ) is well-defined and belongs to C κ . We aim to prove that Γ : C κ → C κ has a unique fixedpoint. For this, we will find an invariant ball in C κ under Γ and check that Γ , restrictedto that ball, defines a contraction.To begin with, let us fix a real number K > and consider the following closed ball inthe Hölder space C κ : B K := { z ∈ C κ , k z k κ ≤ K } . Next, for z ∈ B K , we are going to analyze the norm k Γ( z ) k κ . Indeed, for any s, t ∈ [0 , , s < t , we have that | δ (Γ( z )) st | ≤ Z ts du (cid:12)(cid:12)(cid:12)(cid:12)Z u σ M ( x, z ξ ) dx ξ (cid:12)(cid:12)(cid:12)(cid:12) + | t − s | (cid:12)(cid:12)(cid:12)(cid:12)Z ξ σ M ( x, z ξ ) dx ξ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C G M ( x ) k x k γ k σ ( z ) k κ | t − s |≤ C G M ( x ) k x k γ ( k σ ′ k ∞ k z k κ + k σ k ∞ ) | t − s | , where in the last inequality we have applied Lemma 2.6 and C denotes a positive constant.Furthermore, the above estimate let us also infer that k Γ( z ) k ∞ ≤ C G M ( x ) k x k γ ( k σ ′ k ∞ k z k κ + k σ k ∞ ) . Hence, k Γ( z ) k κ ≤ C G M ( x ) k x k γ ( k σ ′ k ∞ k z k κ + k σ k ∞ ) . (16) ATHWISE SECOND ORDER SDES 9 Since z ∈ B K and G M ( x ) k x k γ < M , we get k Γ( z ) k κ ≤ C M ( K k σ ′ k ∞ + k σ k ∞ ) . ≤ C c ( K + 1) , thanks to (14). Moreover, recall that we have chosen a constant K > . Therefore, bythe hypothesis on σ , if we take for instance c < (2 C ) − , we obtain k Γ( z ) k κ ≤ ( K + 1) / ,and thus k Γ( z ) k κ ≤ K whenever k z k κ ≤ K. This implies that B K is invariant under Γ .Let us now prove that Γ | B K : B K → B K is a contraction. For this, it suffices to showthat Γ | B K is Lipschitz with a Lipschitz constant smaller than . Namely, we shall provethe existence of a constant L < such that, for all y, z ∈ B K , k Γ( y ) − Γ( z ) k κ ≤ L k y − z k κ . Let s, t ∈ [0 , , s < t , and y, z ∈ B K . Then, δ (Γ( y ) − Γ( z )) st = Z ts du (cid:18)Z u [ σ M ( x, y ξ ) − σ M ( x, z ξ )] dx ξ (cid:19) − ( t − s ) Z ξ [ σ M ( x, y ξ ) − σ M ( x, z ξ )] dx ξ . (17)By Lemma 2.6 and the properties of the Young integral, it turns out that the absolutevalue of both terms on the right-hand side of (17) can be bounded, up to some positiveconstant, by G M ( x ) k x k γ k y − z k κ {k σ ′ k ∞ + k σ ′′ k ∞ ( k y k κ + k y − z k κ ) } | t − s | . We have a similar bound for k Γ( y ) − Γ( z ) k ∞ as well. Thus, because y, z ∈ B K , weeventually end up with k Γ( y ) − Γ( z ) k κ ≤ C M K ( k σ ′ k ∞ + k σ ′′ k ∞ ) k y − z k κ . It suffices now to consider that k σ ′ k ∞ and k σ ′′ k ∞ are sufficiently small (that is we cantake c < ( C K + 1) − ∧ (2 C ) − , where C is the constant of the first part of the proof)so that the right-hand side above is bounded by L k y − z k κ , with L < . Therefore, Γ hasa unique fixed point in B K , which means that Equation (12) has a unique solution in C κ .Eventually, using (16) one proves that k z k κ ≤ C M ( k σ ′ k ∞ k z k κ + k σ k ∞ ) . In addition, invoking (14) and the fact that c ≤ (2 C ) − , we obtain k z k κ ≤ C M k σ k ∞ − C M k σ ′ k ∞ ≤ C ( M ) , which concludes the proof. (cid:3) Remark . Having been able to solve equation (12) in C κ for any κ < , one can nowapply the Fubini type Proposition 2.3 in order to assess that z is the unique solution tothe integral equation z t = Z K ( t, ξ ) σ M ( x, z ξ ) dx ξ , t ∈ [0 , , where we recall that the kernel K ( t, ξ ) is defined by K ( t, ξ ) = t ∧ ξ − tξ .3. Differentiability of the solution with respect to the control This section is devoted to show that the solution of Equation (12) is differentiable, inthe sense of Fréchet, when considered as a function of the control x driving the equation.For this, we need two auxiliary results.Let us remind that the diffusion coefficient under consideration (see Equation (12))is introduced in our Definition 2.5. Furthermore, the following differentiation rule holdstrue: Proposition 3.1. Let γ, κ ∈ (0 , be such that γ + κ > , p ≥ and ε > p . Assume that σ ∈ C ( R ) is bounded together with all its derivatives and let σ M be given by Definition 2.5.Consider x an element of C γ + ε and define the following map: F : C γ + ε × C κ −→ C κ , where, for all h ∈ C γ + ε and z ∈ C κ , F ( h, z ) t := z t − Z t du (cid:18)Z u σ M ( x + h, z ξ ) d ( x + h ) ξ (cid:19) + t Z ξ σ M ( x + h, z ξ ) d ( x + h ) ξ . Then, the map F is Fréchet differentiable with respect to the first and second variableand the Fréchet derivatives are given by, respectively: for all t ∈ [0 , , k ∈ C γ + ε and g ∈ C κ , ( D F ( h, z ) · k ) t (18) = − Z t du (cid:20)Z u σ M ( x + h, z ξ ) dk ξ + Z u ( DG M ( x + h ) · k ) σ ( z ξ ) d ( x + h ) ξ (cid:21) + t (cid:20)Z ξ σ M ( x + h, z ξ ) dk ξ + Z ξ ( DG M ( x + h ) · k ) σ ( z ξ ) d ( x + h ) ξ (cid:21) , and ( D F ( h, z ) · g ) t = g t − Z t du (cid:20)Z u G M ( x + h ) σ ′ ( z ξ ) g ξ d ( x + h ) ξ (cid:21) + t Z ξ G M ( x + h ) σ ′ ( z ξ ) g ξ d ( x + h ) ξ . (19) ATHWISE SECOND ORDER SDES 11 Remark . In the above formulae (18) and (19), the Fréchet derivative of G M ( · ) is well-defined and can be computed explicitly. Indeed, G M is defined on the Hölder space C γ + ε ,takes values in R and is defined by G M ( x ) = ϕ M ( k x k pγ,p ) , with some p ≥ . Moreover, ϕ M is a smooth function which fulfills Hypothesis 2.5. Hence, the Fréchet derivative DG M ( x ) at any point x ∈ C γ + ε defines a linear map on C γ + ε with values in R , and it isstraightforward to check that it is given by DG M ( x ) · k = 2 p ϕ ′ M ( k x k pγ,p ) Z Z ( x ζ − x η ) p − ( k ζ − k η ) | ζ − η | γp +2 dζ dη, k ∈ C γ + ε . Moreover, we have that k DG M ( x ) k := k DG M ( x ) k L ( C γ + ε ; R ) ≤ C p k x k γ + ε , (20)where the norm on the left-hand side denotes the corresponding operator norm. Remark . As in Remark 2.8, one can apply Fubini’s theorem for Young integrals inorder to obtain some more compact expressions for the derivatives of F . Indeed, it isreadily checked that ( D F ( h, z ) · k ) t = − G M ( x + h ) Z K ( t, ξ ) σ ( z ξ ) dk ξ − ( DG M ( x + h ) · k ) Z K ( t, ξ ) σ ( z ξ ) d ( x + h ) ξ (21)and ( D F ( h, z ) · g ) t = g t − G M ( x + h ) Z K ( t, ξ ) σ ′ ( z ξ ) g ξ d ( x + h ) ξ , where K is the kernel defined by (3). Remark . As it will be explained later on in the paper, we will apply the results ofthis section to the case where x is a fractional Brownian motion with Hurst parameter H > , defined on a complete probability space (Ω , F , P ) . In particular, the paths of x are almost surely γ -Hölder continuous for all γ < H , with γ -Hölder norm in L p (Ω) forany p ≥ . Thus, if we fix γ < H , we will be able to find ε > / (2 p ) satisfying γ + ε < H .This opens the possibility to apply the results of the current section to this particularcase. Proof of Proposition 3.1. Though the following considerations might be mostly standard(see [13, 21] for similar calculations), we include most of the details here for the sake ofclarity. We will develop the proof in several steps. Step 1. First of all, let us prove that F is continuous. For this, let h, ˜ h ∈ C γ + ε and z, ˜ z ∈ C κ ,so that we need to study the increment δ ( F ( h, z ) − F (˜ h, ˜ z )) st , for ≤ s < t ≤ . Indeed,we have that | δ ( F ( h, z ) − F (˜ h, ˜ z )) st | ≤ A + A + A , (22)where A = | δ ( z − ˜ z ) st | , A = Z ts du (cid:12)(cid:12)(cid:12)(cid:12)Z u σ M ( x + h, z ξ ) d ( x + h ) ξ − Z u σ M ( x + ˜ h, ˜ z ξ ) d ( x + ˜ h ) ξ (cid:12)(cid:12)(cid:12)(cid:12) ,A = ( t − s ) (cid:12)(cid:12)(cid:12)(cid:12)Z ξ σ M ( x + h, z ξ ) d ( x + h ) ξ − Z ξ σ M ( x + ˜ h, ˜ z ξ ) d ( x + ˜ h ) ξ (cid:12)(cid:12)(cid:12)(cid:12) . It is clear that A ≤ k z − ˜ z k κ ( t − s ) κ . On the other hand, the term A can be decomposed as A ≤ A + A , with: A = Z ts du (cid:12)(cid:12)(cid:12)(cid:12)Z u (cid:16) σ M ( x + h, z ξ ) − σ M ( x + ˜ h, ˜ z ξ ) (cid:17) d ( x + h ) ξ (cid:12)(cid:12)(cid:12)(cid:12) A = Z ts du (cid:12)(cid:12)(cid:12)(cid:12)Z u σ M ( x + ˜ h, ˜ z ξ ) d ( h − ˜ h ) ξ (cid:12)(cid:12)(cid:12)(cid:12) (23)In addition, our bound (6) on Young type integrals easily yields A ≤ G M ( x + ˜ h )( k σ ′ k ∞ k ˜ z k κ + k σ k ∞ ) k h − ˜ h k γ + ε ( t − s ) . (24)We still need to bound the term A by a sum B + B , where the latter terms are definedby: B = Z ts du (cid:12)(cid:12)(cid:12)(cid:12)Z u ( σ M ( x + h, z ξ ) − σ M ( x + h, ˜ z ξ )) d ( x + h ) ξ (cid:12)(cid:12)(cid:12)(cid:12) ,B = Z ts du (cid:12)(cid:12)(cid:12)(cid:12)Z u (cid:16) σ M ( x + h, ˜ z ξ ) − σ M ( x + ˜ h, ˜ z ξ ) (cid:17) d ( x + h ) ξ (cid:12)(cid:12)(cid:12)(cid:12) . Now, invoking Lemma 2.6, we get B ≤ CG M ( x + h ) k x + h k γ + ε k σ ( z ) − σ (˜ z ) k κ ( t − s ) ≤ CG M ( x + h ) k x + h k γ + ε ( k σ ′ k ∞ + k σ ′′ k ∞ ( k z k κ + k z − ˜ z k κ )) k z − ˜ z k κ ( t − s ) . (25)Concerning the term B , notice that we clearly have B ≤ (cid:12)(cid:12)(cid:12) G M ( x + h ) − G M ( x + ˜ h ) (cid:12)(cid:12)(cid:12) Z ts (cid:12)(cid:12)(cid:12)(cid:12)Z u σ (˜ z ξ ) d ( x + h ) ξ (cid:12)(cid:12)(cid:12)(cid:12) du ≤ (cid:12)(cid:12)(cid:12) G M ( x + h ) − G M ( x + ˜ h ) (cid:12)(cid:12)(cid:12) k x + h k γ + ε ( k σ ′ k ∞ k ˜ z k κ + k σ k ∞ )( t − s ) . Let us eventually analyse the difference | G M ( x + h ) − G M ( x + ˜ h ) | on the right hand-sideabove: by definition of G M and the properties of ϕ M summarized in Hypothesis 2.5, wecan argue as follows: | G M ( x + h ) − G M ( x + ˜ h ) | = | ϕ M ( k x + h k pγ,p ) − ϕ M ( k x + ˜ h k pγ,p ) |≤ C M,p |k x + h k γ,p − k x + ˜ h k γ,p |≤ C M,p k h − ˜ h k γ,p (26) ATHWISE SECOND ORDER SDES 13 and this last term may be bounded, up to some constant, by k h − ˜ h k γ + ε , because we havechosen ε to be small but verifying ε > p (see Remark 2.4) . This implies that B ≤ C k h − ˜ h k γ + ε k x + h k γ + ε ( k σ ′ k ∞ k ˜ z k κ + k σ k ∞ )( t − s ) . (27)Plugging the bounds (24), (25) and (27) in (23), we obtain that A ≤ C ( k σ ′ k ∞ k ˜ z k κ + k σ k ∞ )( G M ( x + ˜ h ) + k x + h k γ + ε ) k h − ˜ h k γ + ε ( t − s )+ C G M ( x + h ) k x + h k γ + ε ( k σ ′ k ∞ + k σ ′′ k ∞ ( k z k κ + k z − ˜ z k κ )) k z − ˜ z k κ ( t − s ) , (28)where C , C denote some positive constants.The analysis for the term A is very similar to that of A and, indeed, for the formerwe end up with a similar bound as in (28). Therefore, going back to expression (22), wehave proved that k F ( h, z ) − F (˜ h, ˜ z ) k κ ≤ C ( M, σ, x, z, ˜ z, h, ˜ h )( k z − ˜ z k κ + k h − ˜ h k γ + ε ) , which implies that F is continuous. Step 2. Let us prove now that the Fréchet derivative of F with respect to h is given by (18).First of all, let us check that D F ( h, z ) : C γ + ε → C κ , as defined by expression (18), is acontinuous map. Indeed, owing to inequality (6) and Remark 3.2, one can easily checkfrom expressions (18) and (20) that k D F ( h, z ) · k k κ ≤ C ( k σ ′ k ∞ k z k κ + k σ k ∞ )( G M ( x + h ) + k x + h k γ + ε ) k k k γ + ε , which implies that D F ( h, z ) is continuous.In order to prove that (18) also represents the Fréchet derivative of F with respect tothe first variable, we fix h ∈ C γ + ε and z ∈ C κ , so that we need to prove that lim k k k γ + ε → k F ( h + k, z ) − F ( h, z ) − D F ( h, z ) · k k κ k k k γ + ε = 0 . (29)For this, let ≤ s < t ≤ , h, k ∈ C γ + ε and z ∈ C κ , and we proceed to analyze theincrement | δ ( F ( h + k, z ) − F ( h, z ) − D F ( h, z ) · k ) st | . (30)According to (18), the above increment can be split into a sum of four terms, which wedenote by E i , i = 1 , . . . , , and are defined as follows: E = Z ts du Z u [ G M ( x + h ) − G M ( x + h + k ) − DG M ( x + h ) · k ] σ ( z ξ ) d ( x + h ) ξ ,E = Z ts du Z u [ G M ( x + h ) − G M ( x + h + k )] σ ( z ξ ) dk ξ ,E = ( t − s ) Z [ G M ( x + h ) − G M ( x + h + k ) − DG M ( x + h ) · k ] ξ σ ( z ξ ) d ( x + h ) ξ ,E = ( t − s ) Z [ G M ( x + h ) − G M ( x + h + k )] ξ σ ( z ξ ) dk ξ . We will only deal with the study of the terms E and E , since the remaining ones involveanalogous arguments. First, note that we have the following estimates: | E | ≤ | G M ( x + h ) − G M ( x + h + k ) − DG M ( x + h ) · k | (cid:12)(cid:12)(cid:12)(cid:12)Z ts du Z u σ ( z ξ ) d ( x + h ) ξ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C | G M ( x + h ) − G M ( x + h + k ) − DG M ( x + h ) · k |× k x + h k γ + ε ( k σ ′ k ∞ k z k κ + k σ k ∞ )( t − s ) . (31)By Remark 3.2, the map G M : C γ + ε → R is Fréchet differentiable and its derivative canbe computed explicitly. Hence, lim k k k γ + ε → | G M ( x + h ) − G M ( x + h + k ) − DG M ( x + h ) · k |k k k γ + ε = 0 , and this implies that the contribution of | E | is of order o ( k k k γ + ε ) .On the other hand, using the same arguments as in (26), we have: | E | ≤ C | G M ( x + h ) − G M ( x + h + k ) | k k k γ + ε ( k σ ′ k ∞ k z k κ + k σ k ∞ )( t − s ) ≤ C k k k γ + ε ( k σ ′ k ∞ k z k κ + k σ k ∞ )( t − s ) , (32)which is obviously also of order o ( k k k γ + ε ) .For the terms | E | and | E | we obtain, respectively, the same bounds as in (31) and (32).Eventually, plugging all these estimates in (30), we end up with the limit (29). Step 3. In this part, we prove that the Fréchet derivative of F with respect to the secondvariable is given by (19). The continuity of D F ( h, z ) in (19) can be proved as we havedone in Step 2 for D F ( h, z ) . Hence, we will check that, for all h ∈ C γ + ε and z ∈ C κ , itholds: lim k g k κ → k F ( h, z + g ) − F ( h, z ) − D F ( h, z ) · g k κ k g k κ = 0 . (33)Throughout this step we will use the fact that σ , considered as a map defined on andtaking values into C κ , is Fréchet differentiable and its derivative is given by (see Lemma3.5 below): ( Dσ ( z ) · g ) t = σ ′ ( z t ) g t , z, g ∈ C κ . This means that, for all z ∈ C κ , lim k g k κ → k σ ( z + g ) − σ ( z ) − Dσ ( z ) · g k κ k g k κ = 0 . In order to prove (33), let us fix ≤ s < t ≤ and observe that | δ ( F ( h, z + g ) − F ( h, z ) − D F ( h, z ) · g ) st | ≤ F + F , where F := (cid:12)(cid:12)(cid:12)(cid:12)Z ts Z u G M ( x + h )[ σ ( z ξ + g ξ ) − σ ( z ξ ) − σ ′ ( z ξ ) g ξ ] d ( x + h ) ξ du (cid:12)(cid:12)(cid:12)(cid:12) ≤ CG M ( x + h ) k x + h k γ + ε k σ ( z + g ) − σ ( z ) − σ ′ ( z ) g k κ ( t − s ) ATHWISE SECOND ORDER SDES 15 and F := ( t − s ) (cid:12)(cid:12)(cid:12)(cid:12)Z G M ( x + h ) ξ [ σ ( z ξ + g ξ ) − σ ( z ξ ) − σ ′ ( z ξ ) g ξ ] d ( x + h ) ξ (cid:12)(cid:12)(cid:12)(cid:12) , for which the same inequality as for F is available. Therefore, we obtain that k F ( h, z + g ) − F ( h, z ) − D F ( h, z ) · g k κ ≤ CG M ( x + h ) k x + h k γ + ε k σ ( z + g ) − σ ( z ) − σ ′ ( z ) g k κ , and the latter κ -norm, as we have mentioned above, is of order o ( k g k κ ) whenever k g k κ tends to zero. This implies that (33) holds, and ends the proof. (cid:3) Let us quote now the relation needed in the previous proof in order to compute theFréchet derivative of the process σ ( z ) : Lemma 3.5. Let σ ∈ C ( R ) be a bounded function with bounded derivatives. Then σ ,understood as a map σ : C κ → C κ , is Fréchet differentiable and its derivative is given by: ( Dσ ( z ) · g ) t = σ ′ ( z t ) g t , z, g ∈ C κ . Proof. We refer to [13, Proposition 3.5] for the proof of this fact, and in particular for theidentification of ( Dσ ( z ) · g ) t with the quantity σ ′ ( z t ) · g t . (cid:3) As in [21], a crucial step in order to differentiate z with respect to the driving noise x is to solve the following class of linear elliptic PDEs: Proposition 3.6. Let γ, κ ∈ (0 , be such that γ + κ > . Assume that we are given x ∈ C γ and w, R ∈ C κ such that the κ -norm of R verifies k R k κ < c M + 1 , (34) for some small enough constant c < . Then, there exists a unique solution { y t , t ∈ [0 , } in C κ of the following linear integral equation: y t = w t − G M ( x ) Z K ( t, ξ ) R ξ y ξ dx ξ , t ∈ [0 , . Moreover, there exists a positive constant c ( M ) that only depends on M such that k y k κ ≤ c ( M ) k w k κ . (35) Proof. As for Theorem 2.7, we will use a fixed point argument, and solve our equationunder the form ( δy ) st = ( δw ) st − G M ( x ) Z ts du (cid:18)Z u R ξ y ξ dx ξ (cid:19) + ( t − s ) G M ( x ) Z ξ R ξ y ξ dx ξ , (36) for any ≤ s < t ≤ . More precisely, let us define the map Θ : C κ → C κ by Θ( y ) t := w t − G M ( x ) Z t du (cid:18)Z u R ξ y ξ dx ξ (cid:19) + t G M ( x ) Z ξ R ξ y ξ dx ξ , for any y ∈ C κ and t ∈ [0 , . Using elementary properties of Young integrals, one easilychecks that the map Θ is well-defined, that is Θ( y ) belongs to C κ whenever y ∈ C κ .On the other hand, in order to prove that Θ defines a contraction, we will show that itexhibits a Lipschitz property with Lipschitz constant L < . Indeed, let us fix y, ˜ y ∈ C κ and proceed to study the increment δ (Θ( y ) − Θ(˜ y )) st for any ≤ s < t ≤ : | δ (Θ( y ) − Θ(˜ y )) st |≤ G M ( x ) Z ts du (cid:12)(cid:12)(cid:12)(cid:12)Z u R ξ ( y ξ − ˜ y ξ ) dx ξ (cid:12)(cid:12)(cid:12)(cid:12) + ( t − s ) G M ( x ) (cid:12)(cid:12)(cid:12)(cid:12)Z ξ R ξ ( y ξ − ˜ y ξ ) dx ξ (cid:12)(cid:12)(cid:12)(cid:12) ≤ CG M ( x ) k x k γ k R k κ k y − ˜ y k κ ( t − s ) . Hence, taking into account that G M ( x ) k x k γ < M and the assumptions on R , we concludethat k Θ( y ) − Θ(˜ y ) k κ ≤ c CMM + 1 k y − ˜ y k κ . Choosing the constant c conveniently, we get that Θ is Lipschitz with Lipschitz constant L < . Therefore, Θ defines a contraction and it has a unique fixed point, which solvesequation (36).Eventually, the bound (35) can be easily obtained using similar arguments as the onesdeveloped so far. Indeed, observe that we have | ( δy ) st | ≤ CM k R k κ k y k κ ( t − s ) + k w k κ ( t − s ) κ and also k y k ∞ ≤ CM k R k κ k y k κ + k w k ∞ . Thus k y k κ ≤ CM k R k κ k y k κ + k w k κ , from which one deduces (35), provided our constant c is chosen small enough. (cid:3) At this point, we can proceed to state and prove the main result of the section. Theorem 3.7. Let γ, κ ∈ (0 , be such that γ + κ > . Let ε > and a sufficientlylarge p ≥ so that ε > p . Assume that σ ∈ C ( R ) is a bounded function with boundedderivatives such that: k σ ( j ) k ∞ ≤ c M + 1 , j = 0 , , , (37) for some constant c < c C ( M ) ∧ c , where c and C ( M ) are the constants in the statementof Theorem 2.7, and c the one of Proposition 3.6.Let z ( x ) = { z t , t ∈ [0 , } be the solution of Equation (12) with control x ∈ C γ + ε and diffusion coefficient σ M (see (10) and (11)). Then, the map x z ( x ) , defined in ATHWISE SECOND ORDER SDES 17 C γ + ε with values in C κ is Fréchet differentiable. Moreover, for all h ∈ C γ + ε , the Fréchetderivative of z ( x ) is given by: ( Dz ( x ) · h ) t = Z Φ s ( t ) dh s , (38) where the kernels Φ s ( t ) satisfy the following equation: Φ s ( t ) = Ψ s ( t ) + G M ( x ) Z K ( t, ξ ) σ ′ ( z ξ )Φ s ( ξ ) dx ξ , (39) with Ψ s ( t ) = G M ( x ) σ ( z s ) K ( t, s ) + 2 ϕ ′ M ( k x k pγ,p ) µ s z t , (40) and µ s := Z s Z s ρ ζη dζ dη, where ρ ζη = 2 p ( x ζ − x η ) p − | ζ − η | γp +2 . (41) Proof. We will adapt the arguments used in the proof of Proposition 4 in [21]. That is,we will apply the Implicit Function Theorem to the functional F defined in the statementof Proposition 3.1. For this, notice first that we have proved there that, for any h ∈ C γ + ε and z ∈ C κ , F ( h, z ) belongs to C κ and F is Fréchet differentiable with partial derivativeswith respect to h and z given by (18) and (19), respectively. Moreover, since z is thesolution of (12), we have that F (0 , z ) = 0 .We need to check now that D F (0 , z ) defines a linear homeomorphism from C κ intoitself for which, by the Open Map Theorem, it suffices to prove that it is bijective (wealready know that it is continuous). For this, we apply Proposition 3.6 to the case where R ξ = σ ′ ( z ξ ) , so that ( D F (0 , z ) · g ) t = g t − G M ( x ) Z K ( t, ξ ) σ ′ ( z ξ ) g ξ dx ξ (42)defines a one-to-one mapping. Indeed, observe that condition (37) guarantees that (34)in Proposition 3.6 is satisfied. On the other hand, if we fix w ∈ C κ , applying againProposition 3.6 we deduce that there exists g ∈ C κ such that w = D F (0 , z ) · g , whichimplies that D F (0 , z ) is onto and therefore a bijection.Hence, by the Implicit Function Theorem, the map x z ( x ) is continuously Fréchetdifferentiable and Dz ( x ) = − D F (0 , z ) − ◦ D F (0 , z ) . (43)Moreover, by (42), for any h ∈ C γ + ε , Dz ( x ) · h is the unique solution to the differentialequation ( Dz ( x ) · h ) t = w t + G M ( x ) Z K ( t, ξ ) σ ′ ( z ξ )( Dz ( x ) · h ) ξ dx ξ , with w t = − ( D F (0 , z ) · h ) t . Let us proceed to prove (38). Consider Equation (39) and integrate both sides withrespect to some h ∈ C γ + ε : Z Φ s ( t ) dh s = Z Ψ s ( t ) dh s + G M ( x ) Z (cid:20)Z K ( t, ξ ) σ ′ ( z ξ )Φ s ( ξ ) dx ξ (cid:21) dh s . (44)At this point, we can use the same arguments as in the proof of Proposition 4 in [21]:apply our Fubini type Proposition 2.3 to the last term in the right-hand side of (44),which yields Z Φ s ( t ) dh s = Z Ψ s ( t ) dh s + G M ( x ) Z K ( t, ξ ) σ ′ ( z ξ ) (cid:20)Z Φ s ( ξ ) dh s (cid:21) dx ξ . (45)In order to conclude the proof, thanks to uniqueness part of Proposition 3.6, it is nowsufficient to show that w = − D F (0 , z ) · h can be represented in the form w t = Z Ψ s ( t ) dh s . (46)For this, let us observe that, by (21), it holds: ( D F (0 , z ) · h ) t = − G M ( x ) Z K ( t, ξ ) σ ( z ξ ) dh ξ − ( DG M ( x ) · h ) z t . Hence, owing to Lemma 3.8 below, we obtain the representation (46) with Φ s ( t ) givenby (40), which concludes the proof. (cid:3) We close this section by giving an expression for DG M ( x ) , which has already been usedin the proof above. Lemma 3.8. For all h ∈ C γ + ε , it holds that DG M ( x ) · h = 2 ϕ ′ M ( k x k pγ,p ) Z µ s dh s , (47) where the function µ is defined at equation (41).Proof. As we have mentioned in Remark 3.2, the map G M : C γ + ε → R is Fréchet differ-entiable at any point x ∈ C γ + ε , and its Fréchet derivative is given by: DG M ( x ) · k = 2 p ϕ ′ M ( k x k pγ,p ) Z Z ( x ζ − x η ) p − ( k ζ − k η ) | ζ − η | γp +2 dζ dη, k ∈ C γ + ε . According to the definition of ρ ζη , this derivative can be written in the form: DG M ( x ) · k = ϕ ′ M ( k x k pγ,p ) Z Z ρ ζη ( k ζ − k η ) dζ dη. (48) ATHWISE SECOND ORDER SDES 19 Then, applying Fubini Theorem, one can argue as follows: Z Z ρ ζη ( k ζ − k η ) dζ dη = Z Z ρ ζη (cid:18)Z ζη dk r (cid:19) { η ≤ ζ } dζ dη + Z Z ρ ζη (cid:18)Z ζη dk r (cid:19) { ζ ≤ η } dζ dη = Z (cid:20)Z r dζ Z r dη ρ ζη (cid:21) dk r + Z (cid:20)Z r dη Z r dζ ρ ζη (cid:21) dk r = 2 Z (cid:20)Z r dζ Z r dη ρ ζη (cid:21) dk r . Plugging this expression in (48) we obtain (47) and we conclude the proof. (cid:3) Stochastic elliptic equations driven by a fractional Brownian motion Let us first describe the probabilistic setting in which we will apply the results obtainedin the previous section. For some fixed H ∈ (0 , , we consider (Ω , F , P ) the canonicalprobability space associated with the fractional Brownian motion with Hurst parameter H . That is, Ω = C ([0 , is the Banach space of continuous functions vanishing at equipped with the supremum norm, F is the Borel sigma-algebra and P is the uniqueprobability measure on Ω such that the canonical process B = { B t , t ∈ [0 , } is afractional Brownian motion with Hurst parameter H . Remind that this means that B isa centered Gaussian process with covariance R H ( t, s ) = 12 ( s H + t H − | t − s | H ) . In particular, the paths of B are γ -Hölder continuous for all γ ∈ (0 , H ) . Then, we considerEquation (12) where the driving trajectory is a path of B . Namely: δz st = Z ts du (cid:18)Z u σ M ( B, z ξ ) dB ξ (cid:19) − ( t − s ) Z ξσ M ( B, z ξ ) dB ξ , ≤ s < t ≤ , which can be written in the reduced form z t = G M ( B ) Z K ( t, ξ ) σ ( z ξ ) dB ξ , t ∈ [0 , . (49)Assuming that σ ∈ C ( R ) is bounded, has bounded derivatives and satisfies (14), Theo-rem 2.7 implies that Equation (49) has a unique solution z = { z t , t ∈ [0 , } such that z ∈ C κ for any κ ∈ (1 − γ, , and almost surely in ω ∈ Ω .4.1. Malliavin differentiability of the solution. This subsection is devoted to presentthe Malliavin calculus setting which we shall work in, so that we will be able to obtainthat the solution of (49) belongs to the domain of the Malliavin derivative. Notice that,in spite of the fact that we can solve Equation (49) driven by a fBm with arbitrary Hurstparameter, our Malliavin calculus section will be restricted to the range H ∈ (1 / , .This is due to the fact that stochastic analysis of fractional Brownian motion becomescumbersome for H < / , and we have thus imposed this restriction for sake of conciseness. Consider then a fixed parameter H > / , and let us start by briefly describing theabstract Wiener space introduced for Malliavin calculus purposes (for a more general andcomplete description, we refer the reader to [21, Section 3]).Let E be the set of R -valued step functions on [0 , and H the completion of E withrespect to the semi-inner product h [0 ,t ] , [0 ,s ] i H := R H ( s, t ) , s, t ∈ [0 , . Then, one constructs an isometry K ∗ H : H → L ([0 , such that K ∗ H ( [0 ,t ] ) = [0 ,t ] K H ( t, · ) ,where the kernel K H is given by K H ( t, s ) = c H s − H Z ts ( u − s ) H − u H − du and verifies that R H ( t, s ) = R s ∧ t K H ( t, r ) K H ( s, r ) dr , for some constant c H . Moreover, letus observe that K ∗ H can be represented in the following form: [ K ∗ H ϕ ] t = Z t ϕ r ∂ r K H ( r, t ) dr. The fractional Cameron-Martin space can be introduced in the following way: let K H : L ([0 , → H H := K H ( L ([0 , be the operator defined by [ K H h ]( t ) := Z t K H ( t, s ) h ( s ) ds, h ∈ L ([0 , . Then, H H is the Reproducing Kernel Hilbert space associated to the fractional Brownianmotion B . Observe that, in the case of the classical Brownian motion, one has that K H ( t, s ) = [0 ,t ] ( s ) , K ∗ H is the identity operator in L ([0 , and H H is the usual Cameron-Martin space.In order to deduce that (Ω , H , P ) defines an abstract Wiener space, we remark that H is continuously and densely embedded in Ω . In fact, one proves that the operator R H : H → H H given by R H ψ := Z · K H ( · , s )[ K ∗ H ψ ]( s ) ds defines a dense and continuous embedding from H into Ω ; this is due to the fact that R H ψ is H -Hölder continuous (for details, see [21, p. 9]).At this point, we can introduce the Malliavin derivative operator on the Wiener space (Ω , H , P ) . Namely, we first let S be the family of smooth functionals F of the form F = f ( B ( h ) , . . . , B ( h n )) , where h , . . . , h n ∈ H , n ≥ , and f is a smooth function having polynomial growthtogether with all its partial derivatives. Then, the Malliavin derivative of such a functional ATHWISE SECOND ORDER SDES 21 F is the H -valued random variable defined by D F = n X i =1 ∂f∂x i ( B ( h ) , . . . , B ( h n )) h i . For all p > , it is known that the operator D is closable from L p (Ω) into L p (Ω; H ) (seee.g. [18, Section 1]). We will still denote by D the closure of this operator, whose domainis usually denoted by D ,p and is defined as the completion of S with respect to the norm k F k ,p := ( E ( | F | p ) + E ( kD F k p H )) p . The local property of the operator D allows to define the localized version of D ,p , asfollows. By definition, F ∈ D ,ploc if there is a sequence { (Ω n , F n ) , n ≥} in F × D ,p suchthat Ω n increases to Ω with probability one and F = F n on Ω n . In this case, one sets D F := D F n on Ω n .We will first prove now that the solution of (49) at any t ∈ [0 , belongs to D ,ploc .For this, we need to introduce the notion of differentiability of a random variable F inthe directions of H , and we shall apply a classical result of Kusuoka (see [12] or [18,Proposition 4.1.3]). Indeed, a random variable F is H -differentiable if, by definition, foralmost all ω ∈ Ω and for any h ∈ H , the map ν F ( ω + ν R H h ) is differentiable. Then,the above-mentioned result of Kusuoka states that any H -differentiable random variable F belongs to the space D ,ploc , for any p > . We have the following result: Proposition 4.1. Let γ, κ ∈ (0 , be such that γ + κ > . Let ε > and a sufficientlylarge p ≥ so that ε > p and γ + ε < H (this latter condition guarantees that B ∈ C γ + ε ).Assume that σ satisfies the hypotheses of Theorem 3.7.Let z = { z t , t ∈ [0 , } ∈ C κ be the unique solution of equation (49). Then, for any t ∈ [0 , , z t ∈ D , loc and we have: hD z t , h i H = [ Dz ( B )( R H h )] t , h ∈ H . (50) Proof. Recall that the process B is γ -Hölder continuous for any γ ∈ (0 , H ) . Hence, in thestatement of Theorem 3.7, we will be able to find ε (choosing p therein sufficiently large)such that γ + ε < H and k B k γ,p is finite almost surely.On the other hand, note that for all h ∈ H , we have: | ( R H h )( t ) − ( R H h )( s ) | = (cid:0) E ( | B t − B s | ) (cid:1) k h k H ≤ | t − s | H k h k H . Consequently, by Theorem 3.7 and Lemma 4.2 below, we can infer that z t is H -differen-tiable. Therefore, Kusuoka’s result implies that z t ∈ D , loc and we have hD z t , h i H = ddν z t ( ω + ν R H h ) (cid:12)(cid:12) ν =0 a.s., (51)which, together with Lemma 4.2, allows us to conclude that hD z t , h i H = Dz t ( B )( R H h ) = [ Dz ( B )( R H h )] ( t ) . (cid:3) Lemma 4.2. Let γ < H and ε > such that γ + ε < H , as in the statement ofTheorem 3.7. Let z be the solution of (49) and t ∈ [0 , . Then x z t ( x ) is Fréchetdifferentiable from C γ + ε into R . Furthermore, for x ∈ C γ + ε , it holds: Dz t ( x )( k ) = [ Dz ( x )( k )] t , k ∈ C γ + ε . Proof. It is very similar to that of [13, Lemma 4.2]. Indeed, the following estimates arereadily checked: | z ( x + k ) t − z ( x ) t − [ Dz ( x ) k ] t | ≤ k z ( x + k ) − z ( x ) − [ Dz ( x ) k ] k ∞ ≤ k z ( x + k ) − z ( x ) − [ Dz ( x ) k ] k γ + ε . In addition, Theorem 3.7 ensures that the latter term is of order o ( k k k γ + ε ) , from whichour claim is easily deduced. (cid:3) At this point, let us go a step further and prove that the solution z t of Equation (49),indeed, belongs to D , . Proposition 4.3. Let γ, κ ∈ (0 , be such that γ + κ > . Let ε > and a sufficientlylarge p ≥ so that ε > p and γ + ε < H . Assume that σ satisfies the hypotheses ofTheorem 3.7.Let z = { z t , t ∈ [0 , } be the unique solution of equation (49). Then, for any t ∈ [0 , , z t belongs to D , .Proof. By (50), formula (38) and the definition and properties of R H , we have the followingequalities: for any h ∈ H , hD z t , h i H = [ Dz ( B )( R H h )] t = Z Φ s ( t ) d ( R H h ) s = Z Φ s ( t ) (cid:18)Z s ∂K H ∂s ( s, r )( K ∗ H h )( r ) dr (cid:19) ds = Z ( K ∗ H Φ · ( t ))( s )( K ∗ H h )( s ) ds = h Φ · ( t ) , h i H . This implies that, as elements of H , D z t = Φ · ( t ) .On the other hand, let us observe that L H ([0 , ⊂ H continuously (see e.g. [18, Lemma5.1.1]), and clearly any Hölder space C κ is continuously embedded in L H ([0 , . Therefore,if we aim to prove that E ( kD z t k H ) < + ∞ , it suffices to verify that E ( k Φ · ( t ) k κ ) < ∞ , forany κ ∈ (0 , .Taking into account that Φ s ( t ) satisfies the linear equation (39), we are in position toapply Proposition 3.6 so that we end up with k Φ · ( t ) k κ ≤ C ( M ) k Ψ · ( t ) k κ , (52) ATHWISE SECOND ORDER SDES 23 where we remind that Ψ s ( t ) has been defined in (40). By the boundedness of G M and ϕ ′ M ,the fact that K ( t, · ) is Lipschitz with Lipschitz constant bounded by − t , Lemma 2.6and estimate (15), we can infer that k Ψ · ( t ) k κ ≤ C (1 + k µ k κ ) , (53)for some constant C depending on M and σ . Hence, it remains to study the κ -Hölderregularity of µ (recall that this process is defined by (41)). Namely, for any ≤ s < s ≤ , one easily verifies that µ s − µ s = Z s s Z s ρ ζη dζ dη − Z s Z s s ρ ζη dζ dη. At this point, let us observe that, in the statement, the condition relating p and ε isslightly stronger than the one considered in Proposition 4.1. In fact, the former allows usto infer that ρ ζη ≤ C k B k p − γ + ε , almost surely, which guarantees that µ ∈ C κ and k µ k κ ≤ C k B k p − γ + ε . (54)Plugging this bound in (53) and using (52), we end up with E ( kD z t k H ) ≤ E ( k Φ · ( t ) k κ ) ≤ CE ( k B k p − γ + ε ) , and the latter is a finite quantity since γ + ε < H and k B k γ + ε has moments of any orderby Fernique’s lemma [8, Theorem 1.2.3]. This concludes the proof. (cid:3) Stratonovich interpretation of the fractional elliptic equation. Up to now,we have succeeded in solving equation (49) by interpreting any integral with respect to B in the Young (pathwise) sense. In this particular situation, it is a well known fact[23] that our approach is equivalent to Russo-Vallois kind of techniques. Namely, if fora process V the integral R T V s dB s can be defined in the Young sense, then one also hasalmost surely Z T V s dB s = lim ε → ε Z T V s ( B s + ε − B s − ε ) ds. The latter limit is usually called Stratonovich integral with respect to B (see [18, Definition5.2.2]), and is denoted by R T V s ◦ dB s .Our point of view in this section is slightly different: we wish to show that the integralswith respect to B in equation (49) can also be interpreted as the sum of a Skorohodintegral plus a trace term. As we shall see below (see Proposition 4.4), this gives anotherdefinition of Russo-Vallois symmetric integral in the particular case of smooth integrandsin the Malliavin calculus sense. In particular we shall see that, at least a posteriori,Malliavin calculus might have been applied in order to solve our original elliptic equation,though a direct application of these techniques lead to non closed estimations. Let us thus introduce the space |H| , which is composed of measurable functions ϕ :[0 , → R such that k ϕ k |H| := α H Z Z | ϕ r || ϕ u || r − u | H − drdu < + ∞ , where α H = H (2 H − , and we denote by h· , ·i |H| the associated inner product. We defineStratonovich integrals thanks to the following result, borrowed from [1, Proposition 3]: Proposition 4.4. Let { u t , t ∈ [0 , } be a stochastic process in D , ( |H| ) such that Z Z |D s u t || t − s | H − dsdt < + ∞ a.s. (55) Then, the Stratonovich integral R u t ◦ dB t exists and can be written as Z u t ◦ dB t = δ ( u ) + Z Z D s u t | t − s | H − dsdt, (56) where δ ( u ) stands for the Skorohod integral of u . We are now in a position to apply this result to our elliptic equation: Proposition 4.5. Let z = { z t , t ∈ [0 , } be the solution to equation (49). Under thesame hypothesis as in Proposition 4.3, the process z belongs to D , ( |H| ) and also satisfiesthe equation z t = G M ( B ) Z K ( t, ξ ) σ ( z ξ ) ◦ dB ξ , t ∈ [0 , , (57) where the Stratonovich stochastic integral with respect to B is interpreted as in (56).Proof. Note first that the norm of z in D , ( |H| ) is given by k z k D , ( |H| ) = E ( k z k |H| ) + E ( kD z k |H|⊗|H| ) . By Theorem 2.7 (see (15) therein), we have E ( k z k |H| ) = Z Z E ( | z r z u | ) | r − u | H − drdu ≤ CE ( k z k ∞ ) Z Z | r − u | H − drdu ≤ C. (58)On the other hand, owing to (52)-(54) we can infer that, for any r, u ∈ [0 , : E ( |D u z r | ) ≤ C, ATHWISE SECOND ORDER SDES 25 for some positive constant C . Thus E ( kD z k |H|⊗|H| ) (59) = Z Z dr dr | r − r | H − Z Z du du E ( |D r z u | |D r z u | ) | u − u | H − ≤ C Z Z dr dr | r − r | H − Z Z du du | u − u | H − < + ∞ . (60)Putting together (58) and (60), we have seen that z ∈ D , ( |H| ) , and one also de-duces that (55) holds. By Proposition 4.4, this implies that z belongs to the domainof the Stratonovich integral. Therefore, thanks to the regularity properties of σ andthe fact that K ( t, · ) is a deterministic function, we obtain that the Stratonovich integral R K ( t, ξ ) σ ( z ξ ) ◦ dB ξ is well-defined. By [24, Section 2.2, Proposition 3], this Stratonovichintegral coincides with the pathwise Young integral on the right-hand side of (49), forwhich we can conclude that z solves (57). (cid:3) A modified elliptic equation. One of the major obstacles on our way to get theabsolute continuity of L ( z t ) is the following: associated to equation (49) is the process µ defined by (41), appearing in the expression for D z t . This process happens to have somefluctuations around s = 0 which are too high to guarantee the strict positivity of D z t atleast in a small interval. This is why we consider in this section a slight modification ofour elliptic equation (49) and we will prove that its solution, at any instant t , has a lawwhich is absolutely continuous with respect to the Lebesgue measure. Specifically, thecutoff term G M ( B ) in equation (49) will be replaced by a new ˜ G M ( B ) , whose motivationrelies on a variation of Garsia’s lemma given below: Proposition 4.6. Let f be a continuous function defined on [0 , . Set, for p ≥ , U γ,p ( f ) := (cid:18)Z dv Z v ∧ v | δf uv | p | v − u | γp +2 du (cid:19) / p , (61) and assume U γ,p ( f ) < ∞ . Then f ∈ C γ ([0 , ; more precisely, k f k γ ≤ c U γ,p ( f ) , (62) for a universal constant c > .Proof. Let ≤ s < t ≤ . We wish to show that | δf st | ≤ c U γ,p ( f ) | t − s | γ . (63)To this end, let us construct a sequence of points ( s k ) k ≥ , s k ∈ [0 , , converging to t in thefollowing way: set s = s , suppose by induction that s , . . . , s k ≤ t have been constructed,and let V k := [ a k , b k ] , with a k = 2 s k ∧ (cid:18) s k + t (cid:19) , b k = 3 s k ∧ t. (64) Notice that the main differences between our proof an the original one by Garsia (orbetter said the one given by Stroock in [27]) stems from this definition of a k , b k . Indeed,in the classical proof, a k = s k + t and b k = t . Define then A k := (cid:26) v ∈ V k | I ( v ) > U pγ,p ( f ) | v − s k | (cid:27) (65)and B k := (cid:26) v ∈ V k | | δf s k v | p | v − s k | γp +2 > I ( s k ) | v − s k | (cid:27) (66)where we have set I ( v ) := Z t ∧ vv | δf uv | p | v − u | γp +2 du. Let us prove now that V k \ ( A k ∪ B k ) is not empty: observe that, for t ∈ [0 , , Z v ∧ v | δf uv | p | v − u | γp +2 du ≥ Z v ∧ tv | δf uv | p | v − u | γp +2 du = I ( v ) , and thus U pγ,p ( f ) ≥ Z A k I ( v ) dv > U pγ,p ( f ) | b k − s k | µ ( A k ) . Moreover, I ( s k ) = Z s k ∧ ts k | δf s k u | p | u − s k | γp +2 du ≥ Z B k | δf s k u | p | u − s k | γp +2 du> Z B k I ( s k ) | u − s k | du ≥ I ( s k ) | b k − s k | µ ( B k ) . All together one has obtained µ ( A k ) , µ ( B k ) < | b k − s k | , so that µ ( A k ) + µ ( B k ) < | b k − s k | .Next we show that | b k − s k | = 2 µ ( V k ) = 2 | b k − a k | . This study can be separated in twocases: (i) If s k ≤ t/ , then a k = 2 s k and b k = 3 s k . Thus b k − a k = s k and b k − s k = 2 s k . Thisobviously yields | b k − s k | = 2 µ ( V k ) . (ii) If s k > t/ , then a k = s k + t and b k = t . Thus b k − a k = t − s k and b k − s k = t − s k . Hereagain, we get | b k − s k | = 2 µ ( V k ) .We have thus proved that µ ( A k ) + µ ( B k ) < µ ( V k )3 , which means that V k \ ( A k ∪ B k ) is not empty. Let us thus choose s k +1 arbitrarily in this set. Note that, by construction, s k → t while staying inside [ s, t ] .Now, for an arbitrary n ≥ , decompose δf st into δf st = δf s n +1 t + n X k =0 δf s k s k +1 . (67) ATHWISE SECOND ORDER SDES 27 Applying (66) k and (65) k − , one gets | δf s k s k +1 | p | s k +1 − s k | γp +2 ≤ c I ( s k ) | s k +1 − s k | ≤ c U pγ,p ( f ) | s k +1 − s k || s k − s k − | , and hence | δf s k s k +1 | p ≤ c Q k U pγ,p ( f ) | s k +1 − s k | γp , where Q k := | s k +1 − s k || s k − s k − | . (68)Notice that in our definition (64), we have a k = 2 s k instead of ( s k + t ) / iff s k < t/ .Therefore, we can distinguish three cases in order to bound the quantity Q k above: (i) If s k − > t/ , then s k +1 − s k ≤ t − s k and s k − s k − ≥ ( t − s k ) / . Thus Q k ≤ . (ii) If s k ≤ t/ , then s k +1 − s k ≤ s k − s k = 2 s k and s k − s k − ≥ s k − s k / s k / . Thus Q k ≤ again. (iii) If s k − ≤ t/ and s k > t/ , then s k +1 − s k ≤ t − s k ≤ s k − s k = 2 s k and s k − s k − ≥ s k / . Thus Q k ≤ .Putting those estimates together, we end up with Q k ≤ in all cases, and plugging thisinequality into (68), we obtain | δf s k s k +1 | p ≤ c U pγ,p ( f ) | s k +1 − s k | γp . Now (67) reads | δf st | ≤ (cid:12)(cid:12) δf s n +1 t (cid:12)(cid:12) + n X k =0 (cid:12)(cid:12)(cid:12) δf s k s k +1 (cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12) δf s n +1 t (cid:12)(cid:12) + U γ,p ( f ) n X k =0 | s k +1 − s k | γ . (69)It remains to bound P nk =0 | s k +1 − s k | γ for an arbitrary n . This is achieved by separatingcases again: (i) If s > t/ , then it is easily shown that a k = s k + t and b k = t , for all k , for which we have s k +1 ∈ [( s k + t ) / , t ] . This implies that t − s k +1 ≤ ( t − s k ) / and hence t − s k ≤ − k ( t − s ) ,for any k ≥ . Therefore s k +1 − s k ≤ t − t + s k − t − s k − ≤ t − s k . Plugging this into (69): | δf st | ≤ | δf s n +1 t | + U γ,p ( f ) n X k =0 γk | t − s | γ . Let n → ∞ and use the continuity of f and the fact that s n +1 → t . Then | δf st | ≤ CU γ,p ( f ) | t − s | γ , where C denotes a positive constant which may depend on γ . This concludes the proofin the case s > t/ . (ii) If s ≤ t/ , then by definition of s k +1 we will have, for small enough k , that s k +1 = β k +1 s k for some β k +1 ∈ [2 , . Thus s k = k Y j =1 β j ! s. Set M := inf { k ∈ N ; Q kj =1 β j ≥ t/ (3 s ) } , so that we wish to evaluate P M − k =0 | s k +1 − s k | γ : M − X k =0 | s k +1 − s k | γ = M − X k =0 k Y j =1 β j s ! γ ( β k +1 − γ ≤ γ s γ M − X k =0 k Y j =1 β j ! γ . Notice that b M := Q M − j =0 β j ≤ t/ (3 s ) , by definition of M , and M − l Y j =0 β j ≤ b M l − , for l = 1 , . . . , M, since β j ≥ . Hence M − X k =0 k Y j =0 β j ! γ ≤ M X l =1 b γM ( l − γ < Cb γM ≤ C ( t/s ) γ and M − X k =0 | s k +1 − s k | γ ≤ γ s γ C ( t/s ) γ ≤ Ct γ . Observe that t − s > t/ whenever s < t/ . Therefore t < ( t − s ) and M − X k =0 | s k +1 − s k | γ ≤ C ( t − s ) γ . Let us go back now to (69) and write | δf st | ≤ (cid:12)(cid:12) δf s n +1 t (cid:12)(cid:12) + U γ,p ( f ) M − X k =0 | s k +1 − s k | γ + n X k = M | s k +1 − s k | γ ! := (cid:12)(cid:12) δf s n +1 t (cid:12)(cid:12) + U γ,p ( f ) ( A M + B M ) . (70)We have just seen that A M ≤ C ( t − s ) γ , and one can also prove that B M ≤ C ( t − s ) γ uniformly in n by means of the same kind of argument as for step (i). This ends the proofby taking limits in (70). (cid:3) We will now take advantage of the previous proposition in order to build a slightmodification of our elliptic equation which is amenable to density results. Namely, asbefore, let M > be any real number and ϕ M ∈ C ∞ b ((0 , ∞ )) such that ϕ M ( r ) = 0 for all ATHWISE SECOND ORDER SDES 29 r > M + 1 , and ϕ M ( r ) = 1 for r < M . For any x : [0 , → R for which U γ,p ( x ) < ∞ , forsome γ ∈ (0 , and p ≥ , set ˜ G M ( x ) := ϕ M ( U γ,p ( x ) p ) , and, for such x and any z ∈ R , we define ˜ σ M ( x, z ) := ˜ G M ( x ) σ ( z ) . (71)We shall thus consider another kind of modified elliptic integral equation driven by thefractional Brownian motion B : δz st = Z ts du (cid:18)Z u ˜ σ M ( B, z ξ ) dB ξ (cid:19) − ( t − s ) Z ξ ˜ σ M ( B, z ξ ) dB ξ , ≤ s ≤ t ≤ . (72)This equation can be equivalently formulated in its compact form: z t = ˜ G M ( B ) Z t K ( t, ξ ) σ ( z ξ ) dB ξ , t ∈ [0 , . (73)We will prove that the probability law of the solution to (73) taken at t ∈ (0 , isabsolutely continuous with respect to the Lebesgue measure.First of all, let us point out that the results of Sections 2.3 and 3 remain valid for thesolution of Equation (73). Moreover, using exactly the same arguments as in the proof ofProposition 4.3, one obtains that, for all t ∈ [0 , , the solution z t belongs to the domainof the Malliavin derivative. Altogether, we can state the following result: Theorem 4.7. Let γ, κ ∈ (0 , be such that γ + κ > . Let ε > and a sufficientlylarge p ≥ so that ε > p and γ + ε < H . Assume that σ satisfies the hypotheses ofTheorem 3.7.Then, there exists a unique solution z = { z t , t ∈ [0 , } of (73), which is an elementof C κ . For any t ∈ [0 , , z t belongs to D , and the Malliavin derivative D z t satisfies thefollowing linear integral equation: D s z t = Ψ s ( t ) + ˜ G M ( B ) Z K ( t, ξ ) σ ′ ( z ξ ) D s z ξ dB ξ , s ∈ [0 , , (74) with Ψ s ( t ) = ˜ G M ( B ) σ ( z s ) K ( t, s ) + 2 ϕ ′ M ( U γ,p ( B ) p ) ˜ µ s z t , (75) and ˜ µ s := Z s s Z η ∧ s ρ ζη dζ dη, where ρ ζη = (2 p − 1) ( B ζ − B η ) p − | ζ − η | γp +2 . (76) Remark . The term ˜ µ s in (75) comes from the fact that, as one can easily verify, theFréchet derivative of ˜ G M at x ∈ C γ + ε is given by D ˜ G M ( x ) · h = 2 ϕ ′ M ( U γ,p ( x ) p ) Z ˜ µ s dh s , h ∈ C γ + ε . We can now give the technical justification for our change in the elliptic equation weconsider: the lemma below (whose proof can be immediately deduced from (76)) showsthat ˜ µ can be made of order s q for an arbitrary large q and s in a neighborhood of 0.This simple fact will enable us to upper bound |D s z t | for s → in a satisfying way. Thefollowing result will thus be important in the sequel: Lemma 4.9. Assume that the hypothesis of Theorem 4.7 are satisfied. Then, for all s ∈ (0 , ) and p ≥ : ˜ µ s ≤ k B k p − γ + ε s β a.s., (77) where β = (2 p − ε − γ . Fix t ∈ (0 , , and consider z t solution to (73). Observe that the random variable z t cannot have a density p t ( y ) at y = 0 , since P ( z t = 0) > due to our cutoff procedure.Hence we will prove the existence of density for the law of the random variable z t on thesubset of Ω defined by Ω a := {| z t | ≥ a } , for all a > . The fact that we are restrictingour analysis to Ω a implies the following simple but useful properties: Lemma 4.10. On Ω a , we have k B k γ ≤ C and ˜ G M ( B ) ≥ C , a.s. where C , C denote some positive constants depending on a and M .Proof. Note that on Ω a we must clearly have that ˜ G M ( B ) = ϕ M ( U γ,p ( B ) p ) > a.s.Thus, by definition of ϕ M , we get U γ,p ( B ) p < M + 1 a.s. in Ω a , and the first part of thestatement follows after applying Proposition 4.6.Let us also estimate the integral appearing in equation (73): by (6), Lemma 2.6, The-orem 2.7, and the first part of the lemma, on Ω a we have (cid:12)(cid:12)(cid:12)(cid:12)Z K ( t, ξ ) σ ( z ξ ) dB ξ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C k K ( t, · ) k κ ( k σ ′ k ∞ k z k κ + k σ k ∞ ) k B k γ ≤ C, a.s. (78)where the constant C is positive, depends on M, σ, κ, γ and indeed can be small enoughwhenever k σ k ∞ and k σ ′ k ∞ are small. Note that here we have used the fact that k K ( t, · ) k κ ≤ Ct − κ , which can be easily deduced from the explicit expression of the kernel K .On the other hand, still playing with equation (73), ˜ G M ( B ) (cid:12)(cid:12)(cid:12)(cid:12)Z K ( t, ξ ) σ ( z ξ ) dB ξ (cid:12)(cid:12)(cid:12)(cid:12) ≥ a a.s. on Ω a . Hence, (78) yields ˜ G M ( B ) ≥ aC almost surely on Ω a , which concludes the proof. (cid:3) ATHWISE SECOND ORDER SDES 31 Absolute continuity of the law. With the previous changes in the equation weare considering, we are now ready to state and prove our result concerning the density ofthe law for z t : Theorem 4.11. Assume that σ satisfies the hypothesis of Theorem 3.7 and that | σ ( y ) | ≥ σ > for all y ∈ R , for some constant σ . For any t ∈ (0 , , we consider the randomvariable z t ∈ D , and a > . Then, we have that kD z t k H > a.s. on Ω a .As a consequence, the law of z t restricted to R \ ( − a, a ) is absolutely continuous withrespect to the Lebesgue measure. Let us say a few words about the methodology we have followed in order to prove theresult above: as in many instances, our density result will be obtained by bounding theMalliavin derivatives from below. Let us go back thus to equation (74), which is the onesatisfied by the Malliavin derivative D z t . We wish to prove that a density exists for therandom variable z t under a non-degeneracy condition of the form σ ( y ) > σ for any y ∈ R ;we can assume, without loosing generality, that σ is positive. Our strategy will be basedon the fact that D z t is a continuous function, and we will prove that, almost surely on Ω a ,the Malliavin derivative is negative on some non-trivial interval. This necessarily impliesthat the norm kD z t k H cannot vanish. Let us however make the following observations: (i) We will take advantage of the leading term Ψ s ( t ) in equation (74) and we will analyzeits increments. According to expression (75), these can only be assumed to be strictlynegative when s is small enough: we have not imposed any condition on ˜ µ s , and thus wecan only rely on the upper bound (77), which is valid for s close enough to 0. Let usinsist again here on the fact that our change of cutoff in the elliptic equation we consideris meant to have ˜ µ s very small in a neighborhood of 0. (ii) The estimation of the integral part in equation (74) involves some Hölder norms ofthe function ξ 7→ D s z ξ . It is thus natural to think that the same should occur on the lefthand side of this equation. Therefore, we are induced to consider increments of the form D s z t − D s z t and perform our estimations on these quantities. (iii) We shall tackle those increment estimates in a slightly more abstract setting, similarto Proposition 3.6: consider a function ( t, η ) w ηt , depending on two parameters t, η ∈ [0 , . For η ∈ [0 , , let z η be the solution to z ηt = w ηt − ˜ G M ( B ) Z K ( t, ξ ) R ξ z ηξ dB ξ . (79)In the equation above, w and R satisfy some suitable Hölder continuity assumption, andwe assume the increments of w to be also bounded from below. Notice that, for η ≤ t ,the function t 7→ D η z t satisfies an equation of the form (79). Our aim is then to getan appropriate lower bound on the increments of z η . This will be a consequence of thefollowing lemma: Lemma 4.12. Let γ < H and κ ∈ (0 , be such that γ + κ > . For any η ∈ [0 , , let w η be a function in C κ satisfying the relation | δw ηt t | ≤ c | t − t | η for any η ≤ t ≤ t ≤ and c < small enough. Moreover, let R ∈ C κ such that k R k κ ≤ c M + 1 , (80) for a small enough constant c < (see Proposition 3.6). Then the solution z η to equa-tion (79) is such that for all η ≤ t ≤ t ≤ , | δz ηt ,t | ≤ | t − t | η. (81) If we further suppose that δw ηt t ≤ − c | t − t | η for any η ≤ t ≤ t ≤ and c largeenough, then we also get the bound δz ηt ,t ≤ − c | t − t | η, (82) for all η ≤ t ≤ t ≤ and a small positive constant c .Proof. Let us start by proving (81): the solution z η to equation (79) is obtained as thefixed point of an application Θ constructed as in the proof of Proposition 3.6. Namely,let us define the map Θ : C κ → C κ by Θ( y ) t := w ηt − ˜ G M ( B ) Z t du (cid:18)Z u R ξ y ξ dB ξ (cid:19) + t ˜ G M ( B ) Z ξ R ξ y ξ dB ξ . Then under our standing assumptions, z η can be seen as the fixed point of the map Θ . Itis thus enough to check that, if y verifies | δy t t | ≤ | t − t | η for all η ≤ t ≤ t ≤ , then ˆ y := Θ( y ) fulfills the same condition.Let us write then δ ˆ y t t = A t t − C t t + D t t , with A t t = δw ηt t and C t t = ˜ G M ( B ) Z t t du (cid:18)Z u R ξ y ξ dB ξ (cid:19) , D t t = ( t − t ) ˜ G M ( B ) Z ξ R ξ y ξ dB ξ . We shall bound those 3 terms separately for η ≤ t ≤ t ≤ . | A t t | is bounded by assumption by c | t − t | η . Furthermore, | C t t | is easily estimatedas follows: | C t t | ≤ k R k κ k y k C κ ([ t , M | t − t | ≤ k R k κ M | t − t | η, thanks to our induction hypothesis. Hence, by (80), we have | C t ,t | ≤ c | t − t | η. Some similar considerations also yield | D t ,t | ≤ c | t − t | η for a small enough constant c . In order to complete the proof of (81), it suffices thus to consider that c , c smallenough so that c + c + c < .Let us turn now to the proof of (82): it is sufficient to go through the same computationsas for (81) and take into account the lower bound on δw ηt ,t . Details are left to the reader.We only notice that the constant c has to be taken such that c > c + c , where the c , c are the same constants of the proof of (81). (cid:3) ATHWISE SECOND ORDER SDES 33 At this point, we already have the main tools in order to prove the main result of thesection. Proof of Theorem 4.11. Taking into account that the Malliavin derivative D z t satisfiesequation (74), we will apply (82) to the following situation: z st = D s z t , R ξ = σ ( z ξ ) and w st = Ψ s ( t ) , where we recall that Ψ s ( t ) = ˜ G M ( B ) σ ( z s ) K ( t, s ) + 2 ϕ ′ M ( U γ,p ( B ) p ) ˜ µ s z t and ˜ µ s is defined by (76). We also remind that, throughout the proof, we have implicitlyfixed ω belonging to Ω a .First, note that the hypotheses on σ guarantee that (80) is satisfied. Secondly, weobserve that (82) is still true is we replace η ≤ t ≤ t ≤ by η ≤ t ≤ t ≤ T , for any T ∈ (0 , . In fact, we are going to apply that result for some small enough T .Let us prove that there exists T such that δw st ,t ≤ − c | t − t | s , for all s ≤ t ≤ t ≤ T .We clearly have that δw st ,t = Ψ s ( t ) − Ψ s ( t ) = − ˜ G M ( B ) σ ( z s )( t − t ) s + 2 ϕ ′ M ( U γ,p ( B ) p )˜ µ s ( δz t ,t ) . (83)By Lemma 4.10 and the non-degeneracy condition on σ , the first term on the right-handside of (83) can be bounded by − c ( t − t ) η , where c is some large enough constant (seethe proof of Lemma 4.10). We will check now that, for some small enough T , then ϕ ′ M ( U γ,p ( B ) p )˜ µ s ( δz t ,t ) ≤ c ( t − t ) s, (84)for some (small) constant c (which may depend on ω ). For this, we use the boundednessof ϕ ′ M , apply Lemma 4.9 (thus take T small enough) and take into account the fact that,as it can be deduced from the existence result Theorem 2.7, the solution z is indeedLipschitz continuous (with Lipschitz constant depending on M ). Altogether this yields ϕ ′ M ( U γ,p ( B ) p )˜ µ s ( δz t ,t ) ≤ C k B k p − γ + ε s β ( t − t ) ≤ c ( t − t ) s, with c = C k B k p − γ + ε T β − , where we recall that β = (2 p − ε − γ .Therefore, taking p large enough such that c < c and plugging (84) into (83), weobtain δw st ,t = Ψ s ( t ) − Ψ s ( t ) ≤ − c ( t − t ) s, for all s ≤ t ≤ t ≤ T, where c can be large enough (since c can be as well).Then, we are in position to apply (82) and we obtain that δ ( D z t ) t ,t ≤ −| t − t | s, for all s ≤ t ≤ t ≤ T. This implies that we will be able to find T < T such that δ ( D z t ) t ,t ≤ − T ( t − t ) < , for all T ≤ t ≤ t ≤ T (85)and this holds almost surely in Ω a . At this point, we have two possible situations:(i) If D T z t = 0 , the continuity of the Malliavin derivative implies that it does notvanish in an interval around T . Thus the norm kD z t k H must be strictly positivea.s. on Ω a . 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