Behavior with Respect to the Hurst Index of the Wiener Hermite Integrals and Application to Spdes

Home , Hermite distribution

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H .., , d Abstract ) ∈ 1 1 2 , 1 00,6H5 60G22. 60H15, 60H05, d n eaayetelmtbhvo in behavior limit the analyze we and H 1 2 edt and/or 1 to tend , 1 gas alai acls Fourth calculus; Malliavin egrals; d i qain rvnb Hermite by driven equations tic hrceie h anproper- main the characterizes , prmtrHrieprocess Hermite -parameter lta h tnadspace-time standard the than al mmr n ttoayincre- stationary and -memory oa rwinmto.Their motion. Brownian ional oe ob h emt noise. Hermite the be to hosen yrsacesepoe the explored researchers ny 0 r[6 o complete a for [26] or 20] ain.TeHermite The cations. SDs rvnby driven (SPDEs) s emt noise Hermite e 2 1 sexamples, As . noises have been considered by many authors. We refer, among others, to [3], [10], [11], [12], [17], [25], [8], [13], [21], [22]. Our purpose is to analyze the asymptotic behavior in distribution of the solution to the stochastic heat equation with additive Hermite noise, when the Hurst parameter (which is also the self-similarity index of the Hermite process) converges to the extreme values of its interval of definition, i.e when it tends to one and to one half. Our work continues a recent line of research that concerns the limit behavior in distribution with respect to the Hurst parameter of Hermite and related fractional- type stochastic processes. In particular, the papers [5] and [2] deal with the asymptotic behavior of the generalized Rosenblatt process, the work [1] studies the multiparamter Hermite processes while the paper [22] investigates the Ornstein-Uhlenbeck process with Hermite noise of order q = 2. The solution to the heat equation with Hermite noise in Rd is a (d + 1)- parameter d+1 random field depending on a Hurst index H 1 , 1 . We prove that the solution ∈ 2 converges in distribution to a Gaussian limit when at least one of the components of H 1 converges to 2 and to a random variable in a Wiener chaos of higher order when at least one 1 of the components of H tends to 1 (and none of them converges to 2 ). Moreover, the limit always coincides in distribution with the solution to the stochastic heat equation driven by the limit of the Hermite noise. The results show that these models offer a large flexibilitily, covering a large class of probability distributions, from Gaussian laws to distribution of random variables in Wiener chaos of higher order. For the proofs we use various techniques, such as the Malliavin calculus and the Fourth Moment Theorem for the normal convergence, the properties of the Wiener integrals with respect to the Hermite process and the so-called power counting theorem. Since the solution to the Hermite-driven heat equation can be expressed as a Wiener integral with respect to a Hermite sheet, we start our analysis by some more general results, i.e by studying the behavior with respect to the Hurst index of such Wiener integrals. This allows to consider other examples, in particular the Hermite Ornstein-Uhlenbeck process. We organized our paper as follows. Section 2 contains some preliminaries. We introduce the multidimensional Hermite processes and the Wiener integral with respect to them. We also recall some known results concerning the asymptotic behavior of the Hermite sheet. In Section 3, we state general results on the asymptotic behavior of the Wiener- Hermite integrals with respect to the Hurst parameter. We will give two applications of the main results obtained. In Section 4 we analyse the asymptotic behavior of the mild solution of the stochastic heat equation with Hermite noise and finally Section 5 contains the case of the Hermite Ornstein -Uhlenbeck process. The Appendix (Section 6) contains the basic elements of the stochastic analysis on Wiener spaces needed in the paper.

2 Preliminaries

In this preliminary section we will introduce the Hermite sheet and the Wiener integral with respect to this multiparameter process. We also recall the main ﬁndings from [1] concerning

2 the behavior of the Hermite sheet with respect to its Hurst multi-index. We start with some multidimensional notation, that we will use throughout our work.

2.1 Notation For d N 0 we will work with multi-parametric processes indexed by elements of d ∈ \ { } R . We shall use bold notation for multi-indexed quantities, i.e., a = (a1, a2, . . . , ad), b = (b , b , .., b ), α = (α , .., α ), ab = d a b , a b α = d a b αi , a/b = 1 2 d 1 d i=1 i i | − | i=1 | 1 − 1| d d N N1 N2 Nd Q Q (a1/b1, a2/b2, . . . , ad/bd), [a, b]= [ai, bi], (a, b)= (ai, bi), ai = . . . ai1,i2,...,id i=1 i=1 i=0 i1=0 i2=0 id=0 d Y Y X X X X b bi if N = (N1, .., Nd), a = ai , and a < b iﬀ a1 < b1, a2 < b2, . . . , ad < bd (analogously for i=1 the other inequalities). Y We write a 1 to indicate the product d (a 1). By β we denote the Beta − i=1 i − function β(p,q)= 1 zp 1(1 z)q 1dz,p,q > 0 and we use the notation 0 − − − Q R d β(a, b)= β a(i), b(i) Yi=1 if a = (a(1), .., a(d)) and b = (b(1), .., b(d)). Let us recall that the increment of a d-parameter process X on a rectangle [s, t] Rd, ⊂ s = (s ,...,s ), t = (t ,...,t ), with s t (denoted by ∆X([s, t])) is given by 1 d 1 d ≤ d d i ri ∆X([s, t]) = ( 1) −P =1 Xs+r (t s). (1) − · − r 0,1 d ∈{X} When d = 1 one obtains ∆X([s, t]) = X X while for d = 2 one gets ∆X([s, t]) = t − s X X X + X . t1,t2 − t1,s2 − s1,t2 s1,s2 2.2 Hermite processes and Wiener-Hermite integrals We recall the deﬁnition and the basic properties of multiparameter Hermite processes. For a more complete presentation, we refer to [9], [20] or [26]. Let q 1 integer and the ≥ Hurst multi-index H = (H ,H ,...,H ) ( 1 , 1)d. The Hermite sheet of order q and with 1 2 d ∈ 2 self-similarity index H , denoted (Zq,d(t), t Rd ) in the sequel, is given by H ∈ +

(1) (d) q 1 1 H 1 1 H t t + − 1 + − d q,d − 2 q − 2 q ZH (t) = c(H,q) . . . (s1 y1,j)+ . . . (sd yd,j)+ Rd q · 0 0  − −  Z Z Z jY=1 dsd ...ds1 dW (y1,1,...,yd,1) ...dW (y1,q,...,yd,q) 

3 t q 1 1 H + − − 2 q = c(H,q) (s yj)+ ds dW (y1) ...dW (yq) (2) Rd q · 0 − Z Z jY=1 for every t = (t1, ..., td) Rd , where x = max(x, 0). The above stochastic integral ∈ + + is a multiple stochastic integral with respect to the Wiener sheet (W (y), y Rd), see ∈ Section 6.1. The constant c(H,q) ensures that E Zq (t) 2 = t2H for every t Rd . As H ∈ + pointed out before, when q = 1, (2) is the fractional Brownian sheet with Hurst multi-index H = (H ,H ,...,H ) ( 1 , 1)d. For q 2 the process Zq,d is not Gaussian and for q = 2 1 2 d ∈ 2 ≥ H we denominate it as the Rosenblatt sheet. The Hermite sheet is a H-self-similar stochastic process and it has stationary increments. Its paths are H¨older continuous of order δ < H, see [20] or [26]. Its covariance is the same for every q 1 and it coincides with the covariance of the d-parameter fractional ≥ Brownian motion, i.e.

d q,d q,d 1 2H 2H 2H EZ (t)Z (s)= t i + s i t s i =: RH(t, s), t ,s 0. (3) H H 2 i i − | i − i| i i ≥ jY=1

We will denote by H the space of measurable functions f : Rd R such that |H | → f 2 < H k k|H | ∞ where

2 H H 1 u v u v u v 2H 2 f H := (2 ) d d f( ) f( ) − (4) k k − Rd Rd | | · | || − | |H | Z Z = H(2H 1) du(1)...du(d)dv(1)...dv(d) − Rd Rd Z Z d (1) (d) (1) (d) (j) (j) 2Hj 2 f(u , .., u )f(v , .., v ) u v − × | − | jY=1 where u = (u(1), .., u(d)), v = (v(1), .., v(d)) Rd. ∈ Notice that the space H satisfies the following inclusion (see Remark 3 in [9]) |H | 1 d 2 d 1 d L (R ) L (R ) L H (R ) H . (5) ∩ ⊂ ⊂ |H | q,d The Wiener integral with respect to the Hermite sheet ZH has been defined in [9] (following the idea of [15] in the one-parmeter case). In particular, it is well-defined for measurable integrands f H via the formula ∈ |H | q,d f(s)dZH (s)= (Jf)(y1, ..., yq)dW (y1)...dW (yq) (6) Rd Rd.q Z Z

4 where W (y), y Rd is a d-parameter Wiener process and ∈

1 1 H 1 1 H + − + − − 2 q − 2 q (Jf)(y1, ..., yq)= c(H,q) duf(u)(u y1)+ . . . (u yq)+ (7) Rd − − Z with c(H,q) from (2). The stochastic integral Rd.q (Jf)(y1, ..., yq)dW (y1)...dW (yq) is a multiple Wiener-Itˆointegral with respect to the Wiener sheet W . R We have the isometry formula, for f, g H ∈ |H |

q,d q,d 2H 2 E f(s)dZH (s) g(s)dZH (s) = H(2H 1) dudvf(u)g(v) u v − Rd Rd − Rd Rd | − | Z Z Z Z := f, g H . (8) h iH By f 2 we denote f,f . H H k kH h iH 2.3 Behavior of the Hermite sheet with respect to the Hurst parameter q,d In a ﬁrst step, we analyze the convergence of the integral Rd f(s)dZH (s) when the Hurst indices H goes to 1 and/or 1 . i 2 R Let us introduce the following notation: if j , .., j 1, .., d with 1 k d we { 1 k} ⊂ { } ≤ ≤ will denote

1 k A = j , .., j , H = (H , ..., H ) , 1 , t = t(j1)....t(jk) if t = (t(1), .., t(d)). k { 1 k} Ak j1 jk ∈ 2 h iAk (9) We will separate our study into following two situations:

1 1. At least one parameter converges to 1 and none to 2 . Then the limit will be a non- Gaussian random variable related to the Hermite distribution.

1 1 2. At least one parameter Hi converges to 2 and the other indices are ﬁxed in ( 2 , 1) or converges to 1, i.e. if Ak is as above, Bp = l1, .., lp 1, .., d with p + k d and 1 1 R{k } ⊂ { } Rp ≤ Ak Bp = , we assume HAk ( 2 , ..., 2 ) and HBp (1, .., 1) . In this case ∩ ∅ → q,d∈ → ∈ we will see that the limit of Rd f(s)dZH (s) is a centered Gaussian random variable with an explicit variance. R We start by recalling the main result in [1] concerning the asymptotic behavior of the Hermite sheet.

q,d Theorem 1 Let ZH (t) be given by (2) and let Ak,Bp be as in (9). Fix T > 0. t 0 ≥

5 Rk 1. Assume HAk (1, .., 1) . Assume that the parameters Hj, j Ak are fixed. → q,d ∈ d ∈ Then the process ZH converges weakly in C([0, T ] ) to the d-parameter stochastic process (Xt)t 0 defined by ≥ q,d k Xt = t Z − (t ) (10) Ak H Ak h i Ak q,d k where Z − (t ) is a (d k)-parameter Hermite process of order q with H Ak Ak d k t R − − Ak ∈ + d k Hurst index H 1 , 1 − . Ak ∈ 2 d q,d 2. Assume (H1, .., Hd) (1, .., 1) R . Then the process ZH converges weakly in d → ∈ C([0, T ] ) to the d-parameter stochastic process (Xt)t 0 defined by ≥ 1 Xt = t Hq(Z) (11) h id √q! where Z N(0, 1) and H is the qth Hermite polynomial (see (64)). ∼ q 1 1 Rk 3. Assume HAk 2 , ..., 2 . Assume that the parameters Hj, j Ak are fixed. → q,d ∈ d ∈ Then the process ZH converges weakly in C([0, T ] ) to a d-parameter centered Gaus- sian process (X(t))t 0 with covariance ≥

(a) (a) (b) (b) EXtXs = t s R (t ,s ) (12)  ∧   Hb  a Ak b A Y∈ Y∈ k     with RHb deﬁned in (3). 1 1 Rk Rp 4. Assume HAk 2 , .., 2 and HBp (1, .., 1) . Assume that the Hj → ∈ → ∈ q,d with j 1, 2, .., d (Ak Bp) are ﬁxed. Then the process ZH converges weakly in ∈d { } \ ∪ C([0, T ] ) to a d-parameter Gaussian process (X(t))t 0 with covariance ≥

(a) (a) (b) (b) (c) (c) EXtXs = (t s ) t s R (t ,s ) . (13)  ∧     Hc  a Ak b Bp c A Bp Y∈ Y∈ ∈ Yk∪       We will use the above result in order to get the limit behavior with respect to the Hurst parameter of the Hermite Wiener integral.

3 Convergence of the Wiener-Hermite integrals with respect to the Hurst parameter

Let us start the analysis of the behavior of the Wiener-Hermite integral (6) when the components of the self-similarity index H tends to their extreme values. As mentioned above, we will separate our study into two cases: at least one component of H converges to 1 1 (and no component tends to 2 ) and at least one component of H converges to one-half.

6 3.1 Convergence around 1 We need to introduce new spaces for the deterministic integrand in (6). Working on these spaces will ensure the convergence of the Hermite-Wiener integral. Let Ak be as in (9) and assume 1 k < d. We introduce the space of ≤ HAk measurable functions f : Rd R such that → f := (14) k kHAk k 1 2H 2 2 Aj − duAj dvA dwA f(uAj , vA ) f(uAj , wA ) vA wA Rj Rd j j Rd j j | j | · | j || j − j | j=1 Z Z − Z − X k

= duAj f(uAj , ) H < (15) Rj k · kH Aj ∞ Xj=1 Z with the norm deﬁned in (4). Notice that for f , the integral H Ak k · kH Aj ∈H

q,d duAk dZH (uA )f(u) (16) Rk Rd k k Z Z − is well-deﬁned in L1(Ω). Indeed,

q,d E duAk dZH (uA )f(u) Rk Rd k k Z Z − 1 2 2 q,d q,d duAk E dZH (uA )f(u) duAk E dZH (uA )f(u) ≤ Rk Rd k k ≤ Rk Rd k k Z Z − Z Z − ! 1 2 = HA (2HA 1) duAk k k − Rk Z 1 2H 2 2 Aj − dvA dwA f(uAk , vA ) f(uAk , wA ) vA wA Rd k k Rd k k | k | · | k || j − j | Z − Z − k 1 2 HA (2HA 1) duAj ≤ k k − Rj Xj=1 Z 1 2H 2 2 Aj − dvA dwA f(uAj , vA ) f(uAj , wA ) vA wA Rd j j Rd j j | j | · | j || j − j | Z − Z − 1 2 = HA (2HA 1) f < . k k − k kHAk ∞ If k = d, we deﬁne = to be the set of measurable functions f : Rd R such that HAk HAd →

f := f L1(Rd) k kHAk k k

7 d 1 1 − 2H 2 2 Aj − + duAj dvA dwA f(uAj , vA ) f(uAj , wA ) vA wA Rj Rd j j Rd j j | j | · | j || j − j | j=1 Z Z − Z − X := f L1(Rd) + f < . (17) k k k k Ad 1 ∞ H − Remark 1 Notice that the order of integration in (16) is important. That is, the integral

q,d dZH (uA ) duAk f(u) Rd k k Rk Z − Z is not necesarily well-deﬁned for f . ∈HAk We have the following non-central limit theorem.

Proposition 1 Let Ak be as in (9) and assume f H . ∈HAk ∩ |H | Assume 1 k < d and • ≤

d k k 1 − HA (1, .., 1) R and H , 1 is ﬁxed. k → ∈ Ak ∈ 2 d Then the family of random variables XH, H 1 , 1 ∈ 2 H q,d X := f(u)dZH (u) (18) Rd Z converges in distribution to the random variable

(1) (d) q,d k q,d k X := f(u , .., u )dZ − (uA )duAk = f(uAk , uA )dZ − (uA ) duAk . Rd Ak k Rk Rd k Ak Ak k − Z Z Z (19)

Assume k = d and • H (1, .., 1) Rd. → ∈ d Then the limit in distribution of the family XH, H 1 , 1 given by (18) is ∈ 2 (1) (d) 1 f(u , .., u )du Hq(Z) Rd √q! Z with Z N(0, 1) and H the Hermite polinomial of degree q (64). ∼ q Proof: We will check the convergence of the characteristic function of XH. That is, we will show that for every α R, ∈ iαXH iαX E k E e HA (1,..,1) R e . → k → ∈

8 The idea is to approximate ﬁrst X by a sequence of random variables that can be q,d written in terms of the linear combinaisons of ZH and to use the result in Theorem 1. Consider a sequence of step functions

n n (1) (d) d d fn(u)= al1(tl,tl+1](u)= al1 (1) (1) (u )....1 ( ) ( ) (u ) (tl ,tl+1] (tl ,tl+1] Xl=1 Xl=1 (1) (d) (1) (d) (where we used again the notation u = (u , .., u ) and tl = (tl , .., tl ) for l = 1, .., n) such that

fn f n 0 and fn f H n 0. (20) k − kHAk → →∞ k − k|H | → →∞ The choice of such a sequence (fn)n 1 is possible because for any positive function f ≥ ∈ H , there exists an increasing sequence of step functions in fn H HAk ∩ |H | ∈ HAk ∩ |H | which converges poinwise to f and satisﬁes f f f , and by dominated convergence | n − | ≤ | | theorem, it converges in and in H . Then, we use the fact that a general function HAk |H | can be decomposed into its positive and negative parts. Consider the Hermite Wiener integral of fn with respect to the Hermite sheet

n n,H q,d q,d X = fn(u)dZH (u)= al(∆ZH )((tl, tl+1]) Rd Z Xj=1 q,d n,H 2 with ∆ZH given by (1). Then we know from [9], Section 3 that X converges in L (Ω) H to X if f converges to f in H due to the isometry of the Hermite Wiener integral (8). n |H | So we have n,H H q,d 2 X n X := f(s)dZH (s) in L (Ω). → →∞ Rd Z Consequently, we can write

H n,H lim EeiαX = lim lim EeiαX . (21) k k HA (1,..,1) R HA (1,..,1) R n k → ∈ k → ∈ →∞ Now, we aim at exchanging the two limits above. Recall that if f , j 1 is a j ≥ sequence of functions on D R converging uniformly to f on D and if a is a limit point for ⊂ D, then limj limx a fj(x) = limx a f(x) provided that limx a f(x), limx a fj(x) exist. →∞ → →iαXn,H → → Therefore it suﬃces to show that Ee converges uniformly with respect to HAk to H EeiαX . By the mean value theorem

1 n,H H H H H H 2 2 EeiαX EeiαX α E Xn, X α E Xn, X . − ≤ | | − ≤ | | −

Thus, in order to invert the limits in (21), it suﬃces to show that for some ε> 0

9 n,H H 2 sup E X X n 0 1 k − → →∞ HA [ +ε,1] k ∈ 2 that is proved in Lemma 1 below. The relation (21) becomes

H n,H lim EeiαX = lim lim EeiαX . (22) k k HA (1,..,1) R n HA (1,..,1) R k → ∈ →∞ k → ∈ q,d Assume k < d. Since, from Theorem 1 ZH converges weakly to the process (Ut)t 0 given by ≥ q,d k Ut = t Z − (t ) Ak H Ak h i Ak it follows from (22) that

H n q,d iαX iα al(∆Z )((tl,tl+1]) lim Ee = lim lim Ee Pl=1 H k k HA (1,..,1) R n HA (1,..,1) R k → ∈ →∞ k → ∈ n iα al(∆U)((tl,tl+1]) = lim Ee Pl=1 . (23) n →∞ At this point we need to study the convergence as n of the sequence →∞ n n X := al(∆U)((tl, tl+1]) (24) Xl=1 as n . If A = j , .., j , let us use the notation →∞ k { 1 k} (t , t ] = (t(j1),t(j1)] .... (t(jk),t(jk)]. l l+1 Ak l l+1 × × l l+1 Then it is not diﬃcult to see that

q,d k (∆U)((tl, tl+1]) = (∆ t Ak )(tl, tl+1]Ak (∆Z − )(tl, tl+1]A . h i Ak k and therefore the sequence (24) can be expressed as follows

n n n q,d k X = a (∆U)((t , t ]] = a (∆ t )(t , t ] (∆Z − )(t , t ] l l l+1 l Ak l l+1 Ak H l l+1 Ak h i Ak Xl=1 Xl=1 (1) (d) q,d k = f (u , .., u )du dZ − (u ). n Ak H Ak Rd Ak Z Now, we show that

n 1 X n X in L (Ω) (25) → →∞ where the random variable X is given by (19). We have

n q,d E X X = E duAk dZH (uA )(fn(uAk , uA ) f(uAk , uA )) | − | Rk Rd k k k − k Z Z −

10 q,d duAk E dZH (uA )(fn(uAk , uA ) f(uAk , uA )) ≤ Rk Rd k k k − k Z Z − 1 2 2 q,d duAk E dZH (uA )(fn(uAk , uA ) f(uAk , uA )) ≤ Rk Rd k k k − k Z Z − ! 1 2 2H 2 Ak − = HA (2HA 1) duAk dvA dwA vA wA k k − Rk Rd k Rd k k k | k − k | Z Z − Z − 1 2 fn(uAk , v ) f(uAk , v ) fn(uAk , w ) f(uAk , w ) × Ak − Ak Ak − Ak 1 2 HA (2HA 1) fn f n 0 ≤ k k − k − kHAk → →∞ where the last convergence comes from (20). We obtain from (23) and (25)

H n lim EeiαX = lim EeiαX = EeiαX k HA (1,..,1) R n k → ∈ →∞ and the proof is complete for 1 k < d. ≤ q,d If k = d, the proof is similar. We know that the process ZH converges weakly in C[0, T ] to the process 1 t Hq(Z). h id √q! Using the same lines as above, we get

H n lim EeiαX = lim EeiαX H (1,..,1) Rd n → ∈ →∞ and in this case the sequence (24) becomes

n n 1 1 X = (∆ t d)[tl, tl+1] Hq(Z)= fn(u)du Hq(Z) h i √q! R √q! Xi=1 Z Clearly, by (20)

n 1 1 E X f(u)du Hq(Z) fn(u) f(u) du Hq(Z) n 0. | − Rd √q! |≤ R | − | √q! → →∞ Z Z using the deﬁnition of the norm in for k = d. Then HAk

n u u 1 iαX iα( Rd f( )d ) Hq(Z) lim Ee = Ee R √q! . n →∞

The below lemma has been needed in the proof of Proposition 1.

11 Lemma 1 Let A be as in (9) with 1 k d. Assume f H and consider a k Ak ≤d ≤ ∈ H ∩ |H | sequence (fn)n 1 of step functions on R such that (20) holds true. Let ≥ n n,H q,d X = al(∆ZH )((tl, tl+1]). Xl=1 Then for every ε> 0 small enough

n,H H 2 sup E X X n 0. 1 k − → →∞ HA [ +ε,1] k ∈ 2

Proof: From the isometry property (8) and from (20) we have for every H ( 1 , 1)d, ∈ 2 H H 2 E Xn, X 0. (26) − → Let us show that the above convergence is uniform with respect to H [ 1 + ε, 1]k. By Ak ∈ 2 (8),

n,H H 2 2H 2 E X X = H(2H 1) fn(u)fn(v) u v − dudv − − Rd Rd | − | Z Z 2H 2 2H(2H 1) fn(u)f(v) u v − dudv − − Rd Rd | − | Z Z 2H 2 +H(2H 1) f(u)f(v) u v − dudv − Rd Rd | − | Z Z := G(HAk ) (27)

1 k with the function G considered on the interval [ 2 + ε, 1] . Assume k < d. Let 1(Ak) = (1, .., 1) Rk. Then from (27) ∈

G(1(Ak)) 2H 2 Ak − = HA (2HA 1) duAk dvAk duA dvA fn(uAk , uA )fn(vAk , vA ) uA vA k k − Rk Rk Rd k k Rd k k k k | k − k | Z Z Z − Z − 2H 2 Ak − 2HA (2HA 1) duAk dvAk duA dvA fn(uAk , uA )f(vHA , vA ) uA vA − k k − Rk Rk Rd k k Rd k k k k k | k − k | Z Z Z − Z − 2H 2 Ak − +HA (2HA 1) duAk dvAk duA dvA f(uAk , uA )f(vAk , vA ) uA vA k k − Rk Rk Rd k k Rd k k k k | k − k | Z Z Z − Z − and this can be written

G(1(Ak))

= duAk dvAk (fn f)(uAk , ), (fn f)(vAk , ) H Rk Rk h − · − · iH Ak Z Z

duAk dvAk (fn f)(uAk , ) H (fn f)(vAk , ) H ≤ Rk Rk k − · kH Ak k − · kH Ak Z Z 12 2

= duAk (fn f)(uAk , ) H Rk k − · kH Ak Z 1 2 2H 2 2 Ak − = HA (2HA 1) duAk dvA dwA f(uAk , vA ) f(uAk , wA ) vA wA k k − Rk Rd k k Rd k k | k | · | k || k − k | "Z Z − Z − # 2 HA (2HA 1) fn f (28) ≤ k k − k − k H Ak where we used the deﬁnition (15). 1 k Now, the function G is continuous on [ 2 +ε, 1] so there exists H0 = (H0,1, .., H0,k) 1 k ∈ [ 2 + ε, 1] such that

sup G(HAk )= G(H0). 1 k HA [ +ε,1] k ∈ 2

If H0 = 1(Ak), then the conclusion follows from (28) and the assumption (20). If H0 has the form H0 = (1, .., 1,H0,j+1, ..., H0,k) with j < k then a similar calculation to (28) shows that

G(H0) 1 2 2H 2 2 H (2H 1) du dv dw f(u , v ) f(u , w ) v w Aj − Aj Aj Aj Aj Aj Aj Aj Aj Aj Aj Aj ≤ − Rj Rd j Rd j | | · | || − | "Z Z − Z − # 2 HA (2HA 1) fn f (29) ≤ j j − k − k H Ak Rk and again G(H0) 0 as HAk (1, .., 1) from (20). → → ∈ 1 Otherwise, if all H0,i, i = 1, .., k are in [ 2 + ε, 1), then the conclusion follows from (26). If k = d, the conclusion follows in the same way. Let G be given by (27) and let 1 H0 = (H , .., H ) [ + ε, 1]d such that 0,1 0,k ∈ 2

sup G(H)= G(H0). H [ 1 +ε,1]g ∈ 2 d If H0 = (1, .., 1) R , notice that in this case G(1d) = G(1, ..., 1) = fn 2 ∈ k − f 1 Rd n 0. If H0 has the form kL ( ) → →∞

H0 = (1, .., 1,H0,j+1, ..., H0,d) with j < d then G(H0) satisﬁes (29) and consequently it converges to zero from the as- 1 sumption (20). Il all components of H0 are strictly contained in the interval 2 , 1 , then we conclude by (26).

13 1 3.2 Convergence around 2 In this section, we will study the convergence in distribution of the Hermite Wiener integral (18) when at least one Hurst index converges to one half. Actually, we will assume (recall notation (9) from the previous section) 1 1 H , ..., Rk Ak → 2 2 ∈ and H (1, .., 1) Rp Bp → ∈ with 1 k d, 0 p d and p + k d. Note that k 1 means that at least one Hurst ≤ ≤ ≤ ≤ ≤ ≥ parameter converges to 1 while p 0 means that some Hurst parameters (possibly zero) 2 ≥ converges to 1. We have the following result.

Proposition 2 Assume A is as in (9) and B = l , .., l 1, .., d with 0 p d, 1 k p { 1 p} ⊂ { } ≤ ≤ ≤ k d, p + k d and A B = (if p = 0 then B = .). Let f H . Assume that the ≤ ≤ k ∩ p ∅ p ∅ ∈ |H | following limit exists

2H 2 2 lim H(2H 1) f(u)f(v) u v − dudv := σf,H (30) 1 1 k A HA ( ,..., ) R − Rd Rd | − | k k → 2 2 ∈ Z Z and that

sup dudvdu′dv′f(u)f(u′)f(v)f(v′) 1 k Rd Rd Rd Rd HA [ ,1] k ∈ 2 Z Z Z Z 2(H 1)r 2(H 1)r 2(H 1)(q r) 2(H 1)(q r) − − − − − − u v q u′ v′ q u u′ q v v′ q < . (31) ×| − | | − | | − | | − | ∞ If

d k p 1 1 k p 1 − − HA , ..., R , HB (1, .., 1) R and H , 1 is ﬁxed k → 2 2 ∈ p → ∈ Ak Bp ∈ 2 ∪ q,d then the Hermite Wiener integral Rd f(u)dZH (u) converges in distribution to the Gaussian 2 law N(0,σf,H ). Ak R

q,d Proof: Recall that by (6), Rd f(u)dZH (u)= Iq(Jf) with the operator J deﬁned in (7). We can apply the Fourth Moment Theorem to study the normal convergence of (18). R First notice that by assumption (30), we have

2 q,d 2H 2 E f(u)dZH (u) = H(2H 1) f(u)f(v) u v − dudv Rd − Rd Rd | − | Z Z Z 14 2 converges to σf,H . Therefore, in order to apply the Fourth Moment Theorem (see Theo- Ak rem 4 in the Appendix), it suﬃces to show that

Jf r Jf 2 Rd(2q 2r) 0 k ⊗ kL ( − ) → for every r = 1, ..., q 1. − Now, as in the proof of Theorem 3 in [1] (based on relation (13) in this reference)

(Jf r Jf)(y1, .., y2q 2r) = Jf(u1, .., ur, y1, .., yq r)Jf(u1, .., ur, yq r 1, .., y2q 2r)du1...dur ⊗ − Rd r − − − − Z( ) 2 = c(H,q) du1...dur Rd r Z( ) q r 1 1 H r 1 1 H − + − + − − 2 q − 2 q f(u) (u yj)+ (u uj)+ du Rd  −   −  Z jY=1 jY=1  2q 2r 11H r 1 1 H − + − + − − 2 q − 2 q f(v) (v yj)+ (v uj)+ dv × Rd  −   −  j=q r+1 j=1 Z Y− Y r 1  1 H 2 2H   2(H 1)r = c(H,q)2β − , − dudvf(u)f(v) u v −q 2 − q q Rd Rd | − | Z Z q r 1 1 H 2q 2r 1 1 H − + − − + − (u y )− 2 q (v y )− 2 q  − j +   − j +  j=1 j=q r+1 Y Y−     by using the Fubini theorem and again relation (13) in [1], this leads to

2 Jf r Jf 2 Rd(2q 2r) k ⊗ kL ( − ) 2r 2q 2r 1 1 H 2 2H 1 1 H 2 2H − = c(H,q)4β − , − β − , − 2 − q q 2 − q q dudvdu′dv′(u)f(u′)f(v)f(v′) Rd Rd Rd Rd Z Z Z2(HZ1)r 2(H 1)r 2(H 1)(q r) 2(H 1)(q r) − − − − − − u v q u′ v′ q u u′ q v v′ q ×| − | | − | | − | | − | 1 H H 1 2 u v u v u u v v = 2 ( (2 )) d d d ′d ′f( )f( ′)f( )f( ′) q! − Rd Rd Rd Rd 2(H 1)r Z Z 2(HZ 1)Zr 2(H 1)(q r) 2(H 1)(q r) − − − − − − u v q u′ v′ q u u′ q v v′ q . ×| − | | − | | − | | − | The last quantity converges to zero under assumption (31). Notice that q = 2 and d = 1 we retrieve the results in [22]. For f = 1, the results in this section reduces to those in Theorem 1 from [1].

15 4 Applications to the stochastic heat equation with Hermite noise

We will apply the main results in the previous section to some particular cases. First, we look to the solution to the heat equation driven by an Hermite noise. That is, we consider the following linear stochastic heat equation driven by an additive Hermite sheet with d + 1 parameters

∂u q,d+1 d (t, x) = ∆u(t, x)+ Z˙ H (t, x), t 0, x R ∂t H0, ≥ ∈ (32) u(0, x) = 0, x Rd ( ∈ d q,d q,d+1 d We denoted by ∆ the Laplacian on R and Z H = Z H (t, x); t 0, x R H0, { H0, ≥ ∈ } denotes the (d + 1)-parameter Hermite sheet whose covariance is given by

q,d+1 q,d+1 E Z (s, x)Z (t, y) = R (t,s)RH(x, y) H0,H H0,H H0 d+1 if (H , H) = (H ,H ,...,H ) 1 , 1 . We denoted by H = (H ,...,H ) and 0 0 1 d ∈ 2 1 d d 1 2H 2H 2H R (t,s)= ( t + s t s ),RH(x, y)= R (x ,y ) H 2 | | | | − | − | Hj j j jY=1 if s,t R and x = (x , .., x ), y = (y , .., y ) Rd. ∈ 1 d 1 d ∈ The solution to (32) is understood in the mild sense. That is, the mild solution to (32) is a square-integrable process u = u(t, x); t 0, x Rd defined by: { ≥ ∈ } t q,d+1 d H R uH0, (t, x)= G(t s, x y)ZH ,H (ds, dy), t 0, x (33) Rd − − 0 ≥ ∈ Z0 Z living in the space of jointly measurables random fields X(t, x),t 0, x Rd such that 2 ≥ ∈ for every T > 0, supt [0,T ],x Rd E X(t, x) < . ∈ ∈ | | ∞ The above integral is a Wiener integral with respect to the Hermite sheet, as introduced in Section 2 and G(t, x) is the Green function (or the fundamental solution) that satisfies ∂u ∆u = 0, i.e. ∂t − d/2 x 2 d (2πt) exp | | if t> 0, x R , G(t, x)= − − 2t ∈ (34) ( 0 if t 0,x Rd. ≤ ∈

The stochastic heat equation (32) admits a unique mild solution (uH0,H(t, x))t 0,x Rd if and only if (see [21]) ≥ ∈

d d< 4H + (2H 1) := γ. (35) 0 i − Xi=1 16 In this case, for every T > 0, sup E u(t, x)2 < . t [0,T ],x Rd ∞ ∈ ∈ We will use the following Parseval-type formula (see Lemma A1 in [4]): for every f, g L2(a, b) and for every 0 <α< 1 ∈ b b (1 α) α dudvf(u)g(v) u v − − = qα τ − a,bf(τ) a,bg(τ) (36) | − | R | | F F Za Za Z b iξy where ( a,bf)(ξ)= f(y)e− dy (we use the notation f = , f) and F a F F−∞ ∞ R 1 α 1/2 1 Γ(α/2) q = (2 − π )− . (37) α Γ((1 α)/2) − We recall that the Fourier transform of the function y Rd G(u, y) is G(u, )(ξ) = 1 u ξ 2 ∈ → F · e− 2 | | .

4.1 Limit behavior of the solution when the Hurst index tends to 1 The expression ”Hurst index tends to 1”means that at least one component of the Hurst multi-index tends to 1. We will apply Proposition 1 to obtain the asymptotic behavior of the solution (33) when at least one of the Hurst parameters H0,H1, .., Hd converges to 1 and the other parameters are ﬁxed.

Theorem 2 Assume (35) and let A be as in (9). Fix T > 0 and x Rd. Then k ∈ 1. If

(H , H ) (1, .., 1) Rk+1 and H , j A are fixed 0 Ak → ∈ j ∈ k then the stochastic process (u H(t, x),t [0, T ]) converges weakly in C[0, T ] to the H0, ∈ process (u(t, x),t [0, T ]) defined by ∈ t q,d k u(t, x)= du dy dZ − (y )G(t u, x y). (38) Ak H Ak Rk Rd k Ak − − Z0 Z Z − 2. If H (1, .., 1) Rk and H ,H , j A are fixed, then (u H(t, x),t [0, T ]) Ak → ∈ 0 j ∈ k H0, ∈ converges weakly in C[0, T ] to the stochastic process (u(t, x),t [0, T ]) ∈ t q,d+1 k u(t, x)= dy dZ − (u, y )G(t u, x y). Ak H0,H A Rk Rd k Ak k − − Z Z0 Z − 3. If (H , H) (1, ..., 1) Rd+1, then the weak limit of (u H(t, x),t [0, T ]) in 0 → ∈ H0, ∈ C[0, T ] is (u(t, x),t [0, T ]) with ∈ t 1 u(t, x)= G(t u, x y)dydu Hq(Z). Rd − − √q! Z0 Z 17 Remark 2 As usual, by the weak convergence of the family (u H(t, x),t [0, T ]) to H0, ∈ (u(t, x),t [0, T ]) in C[0, T ] for fixed x Rd we mean the weak convergence of the family ∈ ∈ of distributions of u H( , x) to the law of u( , x) in (C[0, T ], (C[0, T ])). H0, · · B Proof: Consider the function F defined on R R given by + × x y 2 d | − | 2(t u) F : (u, y) 1 (u)(2π(t u))− 2 e− − . (39) → (0,t) − We first show the convergence of finite dimensional distributions Consider the case 1. Let us show that this function belongs to H ,H , with these two spaces defined by (4) |H 0 |∩HAk and (15) respectively. We know from [4] that, under (35), the function F (39) belongs to the space H . |HH0, | Let us check that this function belongs to the space . Writting HAk 2 x y 2 x y d | − Ak | | − Ak | F (u, y)= F (u, y , y ) = (2πu)− 2 e− 2u e− 2u Ak Ak we have by the definition of the norm in (see (15)), HA

k t y y z F A = du d Aj d A d A k kH k Rj Rd j Rd j j j j=1 0 − − X Z Z Z Z 1 2 2 y 2 y y 2 z 2 d | Aj | | Aj | d | Aj | | Aj | 2HA 2 2 2u 2u 2 2u 2u j − (2πu)− e− e− (2πu)− e− e− yA zA × | j − j |

k t y 2 j | Aj | 2 2u = du dyAj (2πu)− e− 0 Rj Xj=1 Z Z 1 y 2 z 2 2 d j | Aj | d j | Aj | 2HA 2 −2 2u −2 2u j − dyA dzA (2πu)− e− (2πu)− e− yA zA . × Rd j Rd j j j | j − j | Z − Z −

By using Parseval’s identity (36)

y 2 z 2 d j | Aj | d j | Aj | H 2HA 2 u ξ 2 1 2 A dy dz (2πu) −2 e 2u (2πu) −2 e 2u y z j − = C dξe ξ − j Aj Aj − − − − Aj Aj j − | | Rd j Rd j | − | Rd j | | Z − Z − Z − so with Cj,C > 0

k t 1 2 1 2H 2 u ξ − Aj F A = Cj du dξe− | | ξ k kH k Rd j | | j=1 Z0 Z − X t d j 1 −4 + 4 a A (2Ha 1) = C u− P ∈ j − du Z0 18 and the last integral is ﬁnite if for every j = 1, .., k d j 1 1 − + (2H 1) > 0 or d< 4+ j + (2H 1). (40) − 4 4 a − a − a Aj a Aj X∈ X∈ The last bound is true due to (35), so the function F given by (39) belongs to H ,H . |H 0 |∩HAk Take λ R,t 0 for j = 1, .., N and denote by j ∈ j ≥

N N ∞ q,d+1 Y (x)= λ u H(t , x)= λ 1 (u)G(t u, x y) dZ (u, y). N j H0, j j (0,tj ) j H0,H 0 Rd  − −  Xj=1 Z Z Xj=1   (41) From the above computations, the integrand N λ 1 (u)G(t u, x y) in (41) belongs j=1 j (0,tj ) j − − to H . Therefore, by Proposition 1, the sequence Y (x) (41) converges, as |HH0, |∩HA P N (H , H ) (1, .., 1) Rk+1 to 0 Ak → ∈ N tj N q,d k y − y x y x λj du d Ak dZH ( A )G(tj u, )= λju(tj, ) Rk Rd k Ak k 0 − − − Xj=1 Z Z Z Xj=1 with u defined in (38). This gives the convergence of the finite dimensional distribution of (u H(t, x),t [0, T ]) to the finite dimensional distributions of (u(t, x),t [0, T ]). H0, ∈ ∈ For the case 2., we have similarly

1 k t t 1 H 2 2H 2 (u+v) ξ 2 1 2 A F = C dudv u v 0 dξe 2 ξ − j A j − − | | k kH k | − | Rd j | | j=1 0 0 − X Z Z Z 1 t t d j 1 2 − + (2H 1) 2H0 2 2 2 a A a = C dudv u v − (u + v)− P ∈ j − | − | Z0 Z0 and the above integral is ﬁnite under (35). For the case 3., we notice in addition that the function F given by (39) belongs to L1(Rd+1). Concerning the tightness, we recall that (see [26]), for every s,t [0, T ], x Rd, ∈ ∈ 2 γ E u H(t, x) u H(s, x) C t s | H0, − H0, | ≤ | − | with γ > 0 from (35) and C is a constant not depending on s,t, x. Since uH0,H(t, x) is an element of the (q + 1)th Wiener chaos, we use the hypercontractivity property for multiple stochastic integrals to get for every p 2 ≥ 2p γp E u H(t, x) u H(s, x) C t s (42) | H0, − H0, | ≤ | − | and the tightness follows from (42) and the Billingsley criterion (see [6, Theorem 12.3] or [7]).

Remark 3 Notice that when (H , H ) (1, .., 1) Rk+1, the condition (35) ”converges” 0 Ak → ∈ to (40).

19 1 4.2 Limit behavior when the Hurst index tends to 2 Fix T > 0. When at least one of the components of Hurst multi-index goes to one-half, we have a central limit theorem.

Theorem 3 1. Assume 1 1 (H , H ) , ..., Rk+1 (43) 0 Ak → 2 2 ∈ and k d< 1+ + H . (44) 2 a a A X∈ k Then the process (u H(t, x),t [0, T ]) given by (33) converges weakly in C[0, T ] to H0, ∈ the process (u(t, x),t [0, T ]) where u is the mild solution to the heat equation ∈ ∂u q,d+1 d (t, x) = ∆u(t, x)+ W˙ H (t, x), t> 0, x R ∂t H0, ∈ (45) u(0, x) = 0, x Rd ( ∈

where W H(t, A A ),t [0, T ], A (Rk), A (Rd k) is a Gaussian ﬁeld H0, 1 × 2 ∈ 1 ∈Bb 2 ∈Bb − with covariance E [W H(t, A A )W H(s,B B )] H0, 1 × 2 H0, 1 × 2 2H 2 = (t s)λ (A B ) H (2H 1) y z Ak − dy dz . k 1 1 Ak Ak Ak Ak Ak Ak ∧ ∩ A2 B2 − | − | Z ∩ k We denoted by λk the Lebesque measure on R . 2. If H 1 , ..., 1 Rk , H (1, .., 1) Rp and Ak → 2 2 ∈ Bp → ∈ k d< 2H + + H . (46) 2 a a A X∈ k then the process (u H(t, x),t [0, T ]) given by (33) converges weakly in C[0, T ] to H0, ∈ the process (u(t, x),t [0, T ]) where u is the mild solution to the heat equation (45) ∈ where the Gaussian noise has the following covariance

E [W H(t, A A )W H(s,B B )] H0, 1 × 2 H0, 1 × 2 2H 2 = R (t,s)λ (A B ) H (2H 1) y z Ak − dy dz . H0 k 1 1 Ak Ak Ak Ak Ak Ak ∩ A2 B2 − | − | Z ∩

1 1 3. If (H , H) , ..., Rd+1 and d = 1, then the weak limit of (u H,t [0, T ]) in 0 → 2 2 ∈ H0, ∈ C[0, T ] is the solution to the heat equation (45) driven by a space-time white noise.

20 Remark 4 The conditions (44), (46) and d = 1 are the ”limits” of (35) in the cases 1., 2. and 3. respectively.

Proof: We will prove that the ﬁnite dimensional distributions of (u H(t, x),t [0, T ]) H0, ∈ converge to those of (u(t, x),t [0, T ]) which satisﬁes (45). In order to apply Proposition ∈ 2, we need to check conditions (30) and (31). Checking condition (30). Consider the case 1., i.e. assume (43) and (44). Take λ R,t 0 for j = 1, .., N and denote by j ∈ j ≥

N N ∞ q,d+1 Y (x)= λ u H(t , x)= λ 1 (u)G(t u, x y) dZ (u, y). N j H0, j j (0,tj ) j H0,H 0 Rd  − −  Xj=1 Z Z Xj=1   2 We ﬁrst check condition (30) for YN (x). Let us calculate E YN (x) . By using the isometry (8),

N E (Y (x))2 = λ λ H (2H 1)H(2H 1) N j k 0 0 − − j,kX=1 tj tk 2H0 2 2H 2 du dv u v − dy dzG(tj u, x y)G(tk v, x z) y z − . × | − | Rd Rd − − − − | − | Z0 Z0 Z Z Notice that, if x = (x(1), .., x(d)), y = (y(1), .., y(d)), z = (z(1), .., z(d)) we have

d x(a) y(a) 2 d | − | 2(t u) G(t u, x y) = 1 (u) (2π(t u))− 2 e− − − − (0,t) − aY=1 and so

2H 2 dy dzG(tj u, x y)G(tk v, x z) y z − Rd Rd − − − − | − | Z Z (a) (a) 2 d x y x(a) z(a) 2 1 1 | − | | − | (a) (a) 2(tj u) 2(t v) (a) (a) 2Ha 2 = dy dz (2π(tj u))− 2 (2π(tk v))− 2 e− − e− k− y z − . R R − − | − | a=1 Y Z Z We will apply the Parseval identity (36) with

α = 2H 1 for every a = 1, .., d. a − We get, for every a = 1, .., d,

(a) (a) 2 x y x(a) z(a) 2 1 1 | − | | − | (a) (a) 2(tj u) 2(t v) (a) (a) 2Ha 2 dy dz (2π(tj u))− 2 (2π(tk v))− 2 e− − e− k− y z − R R − − | − | Z Z 21 1 2 1 2 1 2Ha 2 (tj u) τ 2 (tk v) τ = q2Ha 1 dτ τ − e− − | | e− − | | . − R | | Z 1 Now, by the change of variablesτ ˜ = (t + t 2u) 2 τ, j k −

1 2 1 2 1 2Ha (tj u) τ (t v) τ dτ τ − e− 2 − | | e− 2 k− | | R | | Z 1 2Ha 1 1 2 1 2 + − 1 2Ha τ Ha 1 1 2Ha τ = (tj + tk u v)− 2 2 dτ τ − e− 2 | | = (tj + tk u v) − dτ τ − e− 2 | | . − − R | | − − R | | Z Z Thus

2 E (YN (x)) N = λjλkH0(2H0 1)H(2H 1)q2H 1 − − − j,kX=1 tj tk d 1 2 2H0 2 H1+...+Hd d 1 2Ha τ du dv u v − (tj + tk u v) − dτ τ − e− 2 | (47)| × 0 0 | − | − − R | | Z Z aY=1 Z where q2Ha 1 is deﬁned in (37) and − d

q2H 1 = q2Ha 1. − − a=1 Y Notice that for every H ( 1 , 1), we have ∈ 2 1 1 Γ(H + 2 ) H(2H 1)Γ(H )= H(2H 1) 1 H 1 2Γ(1) = 2 − − 2 − H → → 2 − 2 and then 1 H(2H 1)q2H 1 H 1 (2π)− . (48) − − → → 2 Relation (48) implies

k H(2H 1)q H 1 1 1 k+1 (2π) q H 1. (49) 2 (H0,HA ) ( ,.., ) R − 2 A − − → k → 2 2 ∈ k − Let γ := H + ... + H d. (50) 1 d − We have, by integrating by parts

t s 2H0 2 γ H (2H 1) dudv u v − (t + s u v)− 0 0 − | − | − − Z0 Z0 22 s s 2H0 2 γ = H0(2H0 1) dudv u v − (t + s u v)− − 0 0 | − | − − Z Zt s 2H0 2 γ +H0(2H0 1) dudv u v − (t + s u v)− − s 0 | − | − − Z s Z u 2H0 2 γ = H0(2H0 1)2 dudv u v − (t + s u v)− − 0 0 | − | − − Z tZ s 2H0 2 γ +H0(2H0 1) dudv u v − (t + s u v)− − s 0 | − | − − s Z Z 2H0 1 γ = 2H0 duu − (t + s u)− (51) 0 − Z t 2H0 1 γ 2H0 1 γ H duu − (t + s u)− (u s) − (t u)− − 0 − − − − Zs s u 2H0 1 γ 1 +2H0γ du dv(u v) − (t + s u v)− − 0 0 − − − Zt Zu 2H0 1 γ 1 +H0γ du dv(u v) − (t + s u v)− − s 0 − − − Zs Z 2H0 1 γ = 2H0 duu − (t + s u)− 0 − Z t 2H0 1 γ 2H0 1 γ +H duu − (t + s u)− (u s) − (t u)− 0 − − − − Zs t u 2H0 1 γ 1 +H γ du dv u v − (t + s u v)− − (52) 0 | − | − − Z0 Z0 Assuming (43), from (50)

k γ d H := γ →−  − 2 − a 0 a A X∈ k   and, by taking the limit as γ γ and H 1 in (52), we get → 0 0 → 2

t s 2H0 2 γ H0(2H0 1) dudv u v − (t + s u v)− − 0 0 | − | − − s Z Z γ0 du(t + s u)− → 0 − Z t 1 γ0 γ0 + du (t + s u)− (t u)− 2 − − − Zs t u 1 γ0 1 + γ du dv(t + s u v)− − 2 0 − − Z0 Z0 1 1 γ0+1 γ0+1 = (t + s)− t s − . (53) 2 ( γ + 1) − | − | − 0 23 Consequently, as the limit (43) holds true, by plugging (49) and (53) into (47), we obtain

N 2 1 1 k γ +1 γ +1 E x 0 0 H 1 YN ( ) (2π)− λjλk (tj + tk)− tj tk − q2 A → 2 γ0 + 1 − | − | k − − j,k=1 X 1 2 1 2 τ 1 2Ha τ dτe− 2 | | dτ τ − e− 2 | | × R R | | a Ak Z a A Z Y∈ Y∈ k N 1 1 k γ0+1 γ0+1 = (2π)− λjλk (tj + tk)− tj tk − 2 γ0 + 1 − | − | − j,k=1 X 1 2 k 1 2Ha τ q2H 1(√2π) dτ τ − e− 2 | | × Ak − R | | a A Z Y∈ k N k 1 1 γ0+1 γ0+1 = (2π)− 2 λjλk tj + tk)− tj tk − 2 γ0 + 1 − | − | − j,k=1 X 1 2 1 2Ha τ q2H 1 dτ τ − e− 2 | | . × Ak − R | | a A Z Y∈ k On the other hand, if u is the solution to (45), then

2 N N tj t ∧ k E λju(tj, x) = λjλk du dyAk dyA dzA Rk Rd k Rd k k k   0 − − Xj=1 j,kX=1 Z Z Z Z 2  y 2 y 2 z 2 A | A | yA A d | k | k d | k | | k | 2(tj u) 2(tj u) 2(t u) 2(t u) (2π(t u)− 2 e− − e− − (2π(t u)− 2 e− k− e− k− × j − k − N tj tk 2 2 1 ∧ k (tj +tk 2u) ξ (tj +tk 2u) τ 2 a A (2Ha 1) = λ λ du(2π) dξe q H 1 dτe τ P ∈ k − j k − − − | | 2 A − − | | Rk k − Rd k 0 − | | j,kX=1 Z Z Z N tj tk ∧ γ k ξ 2 τ 2 1 2H 0 H 1 a = λjλk du(tj + tk 2u)− (2π)− dξe−| | q2 A dτe−| | τ − 0 − Rk k − R | | j,k=1 Z Z a A Z X Y∈ k N 1 1 k γ +1 γ +1 τ 2 1 2H = (2π) 2 λ λ (t + t ) 0 t t 0 q H 1 dτe τ a . − j k j k − j k − 2 A −| | − 2 γ0 + 1 − | − | k − R | | − j,k=1 a Ak Z X Y∈

The point 2. follows similarly. Let us discuss point 3. Assume H0,H1, .., Hd converge all to 1 2 . Notice that in this case condition (35) implies d< 2 so d = 1! Then, from (50) d 1 γ = . → 2 2

24 Therefore, from (52), as H ,H , .., H 1 0 1 d → 2 t s 2H0 2 γ H0(2H0 1) dudv u v − (t + s u v)− − 0 0 | − | − − t Z Z t s 1 1 1 1 1 3 du (t u)− 2 (t + s u)− 2 + du dv(t + s u v)− 2 → 2 0 − − − 2 × 2 0 0 − − Z 1 1 Z Z = (t + s) 2 t s 2 . (54) − | − | and we obtain, by combining (54) and (47), by taking the limit (43)

N 2 1 1 1 1 τ 2 EYN (x) (2π)− λjλk (tj + tk) 2 tj tk 2 dτe− 2 | | → − | − | R j,kX=1 Z N 1 1 = λ λ (t + t ) 2 t t 2 √2π j k j k − | j − k| j,kX=1 N 1 1 1 = (2π)− 2 λ λ (t + t ) 2 t t 2 j k j k − | j − k| j,kX=1 2 N which coincides with the E j=1 λju(tj, x) where u is the solution of the heat equation (45) driven by a space-time whiteP noise (see [23] or [26]). Checking condition (31). In order to check condition (31), we need to show in the case 1. (the other situations are similar) that for every t ,t ,t ,t [0, T ], 1 2 3 4 ∈ t1 t4 α0 α0 β0 β0 I := sup du1... du4 u1 u2 − u2 u3 − u3 u4 − u4 u1 − 1 k+1 | − | | − | | − | | − | (H0,HA ) [ ,1] 0 0 k ∈ 2 Z Z x y 2 x y 2 1 | − 1| 1 | − 2| 2(t1 u1) 2(t2 u2) dy1... dy4 d e− − d e− − × Rd Rd (2π(t u )) 2 (2π(t u )) 2 Z Z 1 − 1 2 − 2 x y 2 x y 2 1 | − 3| 1 | − 4| 2(t3 u3) 2(t4 u4) d e− − d e− − ×(2π(t3 u3)) 2 (2π(t4 u4)) 2 −α α −β β y y − y y − y y − y y − < | 1 − 2| | 2 − 3| | 3 − 4| | 4 − 1| ∞ with 2(1 H)r 2(1 H)(q r) 2(1 H )r 2(1 H )(q r) α = − , β = − − , α = − 0 , β = − 0 − q q 0 q 0 q for every r = 1, .., q 1. After the change of variables t u =u ˜ , y˜ = x y, we will have − i − i i − to show that

t1 t4 I = sup du1... du4 1 k+1 (H0,HA ) [ ,1] 0 0 k ∈ 2 Z Z

25 α0 α0 β0 β0 u u (t t ) − u u (t t ) − u u (t t ) − u u (t t ) − | 1 − 2 − 1 − 2 | | 2 − 3 − 2 − 3 | | 3 − 4 − 3 − 4 | | 4 − 1 − 4 − 1 | y 2 y 2 y 2 y 2 1 −| 1| 1 −| 2| 1 −| 3| 1 −| 4| 2u1 2u2 2u3 2u4 dy1... dy4 d e− d e− d e− d e− Rd Rd Z Z (2πu1) 2 (2πu2) 2 (2πu3) 2 (2πu4) 2 α α β β y y − y y − y y − y y − < . | 1 − 2| | 2 − 3| | 3 − 4| | 4 − 1| ∞

Next, we write for the integrals dyi

dy1... dy4... Rd Rd Z Z (j) (j) y 2 y 2 1 | 1 | 1 | 4 | (j) (j) 2u 2u = dy1 ... dy4 e− 1 ... e− 4 R R √2πu1 √2πu4 j Ak Z Z Y∈ (j) (j) y 2 y 2 1 | 1 | 1 | 4 | (j) (j) 2u 2u dy1 ... dy4 e− 1 ... e− 4 . × R R √2πu1 √2πu4 j A Z Z Y∈ k (j) We will separate the integral dy1 , for every j = 1, .., d, as follows

(j) (j) (j) dy1 = dy1 + dy1 (j) (j) R y1 >√2T y1 6√2T Z Z| | Z| | (j) (j) (j) and similarly for the integrals dy2 , dy3 , dy4 . We use the fact that on the set

y2 > 2T > 2u the function 1 y2 u e− 2u is increasing → √u and we majorize 1 y2 1 y2 e− 2u by e− 2T √u √T On the other hand, on the set y262T we majorize 1 y2 e− 2u by a constant. √u In this way, the quantity I can be bounded by

t1 t4 I C sup du1... du4 ≤ 1 k+1 (H0,HA ) [ ,1] 0 0 k ∈ 2 Z Z α0 α0 β0 β0 u u (t t ) − u u (t t ) − u u (t t ) − u u (t t ) − | 1 − 2 − 1 − 2 | | 2 − 3 − 2 − 3 | | 3 − 4 − 3 − 4 | | 4 − 1 − 4 − 1 |

26 (j) y 2 (j) (j) 1 | 1 | T dy ... dy e− 2 1 (j) + 1 (j) 1 4 y >√2T y1 6√2T R R √2πT | 1 | | | ! j Ak Z Z Y∈ (j) y 2 1 | 4 | ... e− 2T 1 (j) + 1 (j) y >√2T y4 6√2T × √2πT | 4 | | | !

(j) (j) αj (j) (j) αj (j) (j) βj (j) (j) βj y y − y y − y y − y y − ×| 1 − 2 | | 2 − 3 | | 3 − 4 | | 4 − 1 | R × 2(1 Hj )r 2(1 Hj )(q r) with αj = −q , βj = − q − for every j = 1, .., d and

(j) y 2 (j) (j) 1 | 1 | T R = dy ... dy e− 2 1 (j) + 1 (j) 1 4 y >√2T y1 6√2T R R √2πT | 1 | | | j A Z Z ! Y∈ k (j) y 2 1 | 4 | ... e− 2T 1 (j) + 1 (j) y >√2T y4 6√2T × √2πT | 4 | | | !

(j) (j) αj (j) (j) αj (j) (j) βj (j) (j) βj y y − y y − y y − y y − . ×| 1 − 2 | | 2 − 3 | | 3 − 4 | | 4 − 1 | Consequently, we can write

t1 t4 I C sup du1... du4 ≤ 1 H0 [ ,1] 0 0 ∈ 2 Z Z α0 α0 β0 β0 u u (t t ) − u u (t t ) − u u (t t ) − u u (t t ) − | 1 − 2 − 1 − 2 | | 2 − 3 − 2 − 3 | | 3 − 4 − 3 − 4 | | 4 − 1 − 4 − 1 | (j) y 2 (j) (j) 1 | 1 | sup dy ... dy e 2T 1 (j) + 1 (j) 1 4 − y >√2T y 6√2T H 1 k R R √2πT | 1 | | 1 | Ak [ 2 ,1] j A ! ∈ Y∈ k Z Z (j) y 2 1 | 4 | ..... e 2T 1 (j) + 1 (j) − y >√2T y 6√2T √2πT | 4 | | 4 | !

(j) (j) αj (j) (j) αj (j) (j) βj (j) (j) βj y y − y y − y y − y y − | 1 − 2 | | 2 − 3 | | 3 − 4 | | 4 − 1 | R. ×

Note that R does not depend on H0, HAk and

t1 t4 sup du1... du4 1 H0 [ ,1] 0 0 ∈ 2 Z Z α0 α0 β0 β0 u1 u2 (t1 t2) − u2 u3 (t2 t3) − u3 u4 (t3 t4) − u4 u1 (t4 t1) − | T− − T − | | − − − | | − − − | | − − − | α0 α0 β0 β0 du ... du u u − u u − u u − u u − ≤ 1 4| 1 − 2| | 2 − 3| | 3 − 4| | 4 − 1| Z0 Z0 which is ﬁnite by Lemma 3.3 in [2] since 2α + 2β + 4 = 2(2H 2) + 4 = 4H > 1. −

27 Therefore, in order to conclude, it remains to show that

(j) (j) sup dy1 ... dy4 H 1 k R R Ak [ 2 ,1] j A ∈ Y∈ k Z Z (j) (j) y 2 y 2 1 | 1 | 1 | 4 | e 2T 1 (j) + 1 (j) ... e 2T 1 (j) + 1 (j) − y >√2T y 6√2T − y >√2T y 6√2T √2πT | 1 | | 1 | ! √2πT | 4 | | 4 | !

(j) (j) αj (j) (j) αj (j) (j) βj (j) (j) βj y y − y y − y y − y y − < . ×| 1 − 2 | | 2 − 3 | | 3 − 4 | | 4 − 1 | ∞ Assume for simplicity A = 1, 2, .., k . To check that the above quantity is ﬁnite, k { } it suﬃces to prove that

y 2 y 2 | 1| | 4| sup dy .... dy e− 2T 1 + 1 .... e− 2T 1 + 1 1 4 y1 >√2T y1 6√2T y4 >√2T y4 6√2T H [ 1 ,1] R R | | | | | | | | ∈ 2 Z Z α α β β y y − y y − y y − y y − < . ×| 1 − 2| | 2 − 3| | 3 − 4| | 4 − 1| ∞ 4 Using i=1(Ai +Bi)= A1A2A3A4 +A1B2B3B4 +....+B1B2B3B4, the last integrals can be expressed as a sum of several terms, involving integrals on the sets y > √2T and Q | i| y 6 √2T . | i| Let us start with the ﬁrst summand, namely

2 2 2 2 y1 y2 y3 y4 T := sup dy .... dy e− 2T 1 e− 2T 1 e− 2T 1 e− 2T 1 1 1 4 y1 >√2T y2 >√2T y3 >√2T y4 >√2T H [ 1 ,1] R R | | | | | | | | ∈ 2 Z Z α α β β y y − y y − y y − y y − . ×| 1 − 2| | 2 − 3| | 3 − 4| | 4 − 1| Since y y 2 2(y2 + y2) we have | 1 − 2| ≤ 1 2

y y 2 + y y 2 + y y 2 + y y 2 4(y2 + y2 + y2 + y2) (55) | 1 − 2| | 2 − 3| | 3 − 4| | 4 − 1| ≤ 1 2 3 4 so

2 2 2 2 y1+y2+y3+y4 1 2 2 2 2 ( y1 y2 + y2 y3 + y3 y4 + y4 y1 ) e− 2T e− 8T | − | | − | | − | | − | . ≤ Hence, T1 can be bounded as follows

1 2 2 2 2 ( y1 y2 + y2 y3 + y3 y4 + y4 y1 ) T1 sup dy1.... dy4e− 8T | − | | − | | − | | − | ≤ H [ 1 ,1] R R ∈ 2 Z Z α α β β y y − y y − y y − y y − . | 1 − 2| | 2 − 3| | 3 − 4| | 4 − 1|

28 We apply the power counting theorem, see the Appendix. Consider the set of aﬃne functionals T ′ = y y ,y y ,y y ,y y . { 1 − 2 2 − 3 3 − 4 4 − 1} The only padded subset of T ′ is T ′ itself. We apply the power counting theorem with 2(1 H)r 2(1 H)r 2(1 H)(q r) 2(1 H)(q r) (α , α , α , α )= − , − , − − , − − 1 2 3 4 − q − q − q − q and (β , β , β , β ) = ( γ, γ, γ, γ) 1 2 3 4 − − − − with γ > 0 arbitrarly large. We have (d0 and d are given by (69) and (70) respectively) ∞ 4 1 d (T ′)= r(T ′)+ α = 3 + 2(2H 2) = 4H 1 > 0 for H > 0 i − − 4 Xi=1 and 3 d ( ) = 4 1 4γ < 0 if γ > . ∞ ∅ − − 4 Therefore T1 is ﬁnite. Let us regard the last summand, i.e.

T := sup dy .... dy 1 ...1 2 1 4 y1 6√2T y4 6√2T H [ 1 ,1] R R | | | | ∈ 2 Z Z α α β β y y − y y − y y − y y − < . ×| 1 − 2| | 2 − 3| | 3 − 4| | 4 − 1| ∞ This is clearly ﬁnite by Lemma 3.3 in [2] since

2α + 2β +4=4H 4+4=4H > 1 − 1 when H > 4 . The other summands can be handled by combining the arguments used for the two terms above. For instance, consider

2 y1 T := sup dy .... dy e− 2T 1 1 1 1 3 1 4 y1 >√2T y2 6√2T y3 6√2T y4 6√2T H [ 1 ,1] R R | | | | | | | | ∈ 2 Z Z α α β β y y − y y − y y − y y − . ×| 1 − 2| | 2 − 3| | 3 − 4| | 4 − 1| We use the bound (which follows from (55)

1 y2 ( y y 2 + y y 2 + y y 2 + y y 2) (y2 + y2 + y2) 1 ≥ 4 | 1 − 2| | 2 − 3| | 3 − 4| | 4 − 1| − 2 3 4

29 and then

2 2 2 2 y1 1 2 2 2 2 y2+y3+y4 ( y1 y2 + y2 y3 + y3 y4 + y4 y1 ) e− 2T e− 8T | − | | − | | − | | − | e 2T 1≤ 2 2 2 2 ( y1 y2 + y2 y3 + y3 y4 + y4 y1 ) Ce− 8T | − | | − | | − | | − | . ≤

The term T3 is thus bounded by

1 2 2 2 2 ( y1 y2 + y2 y3 + y3 y4 + y4 y1 ) T3 C sup dy1.... dy4e− 8T | − | | − | | − | | − | ≤ H [ 1 ,1] R R ∈ 2 Z Z α α β β y y − y y − y y − y y − ×| 1 − 2| | 2 − 3| | 3 − 4| | 4 − 1| and we follow the proof for the ﬁrst term.

Remark 5 Notice that the limit process in Theorem coincides in distribution with a bifractional Brownian motion with Hurst parameters H = 1 , K = γ +1= d k H 2 0 2 a Ak a 1 k − − − ∈ 1 (in the case i. ), H = , K = d Ha +(2H 1) (in the case ii.) and H = K = 2 − 2 − a Ak − P 2 (in the case iii.) We refer to [14], [26],∈ [27] for the deﬁnition of the bifractional Brownian P motion and for the link between this process and the solution to the heat equation.

5 Applications to Hermite Ornstein-Uhlenbeck process

Let Zq,1 := Zq be a (one-parameter) Hermite process deﬁned by (2). The Hermite Ornstein Uhlenbeck process has been introduced in [15]. It is deﬁned as the solution of Langevin equation driven by Hermite noise.

t X = ξ λ X ds + σZq (t),t 1 (56) t − s H ≥ Z0 where λ, σ > 0 and the initial condition ξ is a random variable in L2(Ω). The unique solution of (56) is given by

t H λt λu q Y (t)= e− ξ + σ e dZ (u) , t 0 (57) H ≥ Z0 t λu q where the integral 0 e dZ (u) exists in the Riemann-Stieljes sense. 0 λu H In particular,R by taking the initial condition ξ = σ e dZ (u) in (57). The H −∞ unique solution to (56), denoted in the sequel by (X (t))t 0, can be expressed as ≥ R t H λ(t u) q X (t)= σ e− − dZ (u), t 0 (58) H ≥ Z−∞ 30 and the stochastic integral in (58) can be also understood in the Wiener sense. The process H X (t) t 0 is a stationary process, H-self similar process with stationary increments. In≥ [22] the authors have established the asymptotic behavior with respect to H of the Rosenblatt Ornstein Uhlenbeck process which is the solution of (56) driven by the Rosenblatt process, i.e. q = 2. The proof was based on the analysis of the cumulants, but it is well-known that this method does not work for a Wiener chaos of order q 3. In this ≥ section, we will study the behavior as H 1 and as H 1 of the processes XH (t) → → 2 t [0,T ] H ∈ and Y (t) t [0,T ] when q > 2 . The results obtained give a complete picture for the asymptotic behavior∈ of the Hermite Ornstein Uhlenbeck of any order q 1. ≥ 5.1 Asymptotic behavior of the non stationary Hermite Ornstein-Uhlenbeck Assume that the initial condition ξ does not depend on H. Proposition 3 H 1 Assume H 1. Then the process Y (t) t [0,T ] converges weakly, → ∈ in the space of the continuous functions C[0, T ] to the process (Y (t))t [0,T ] given by ∈

λt λt Hq(Z) Y (t)= e− ξ + σ 1 e− (59) − √q! with Z (0, 1) ∼N 1 H 2 Assume H 2 , the process Y (t) t [0,T ] converges weakly, in the space of the → ∈ continuous functions C[0, T ] as H 1 to the standard Ornstein Uhlenbeck process → 2 (Y0(t))t [0,T ] given by ∈ t λ λu Y0(t)= e− ξ + σ e dW (u) (60) Z0 that is a Gaussian process with mean EY (t)= e λtEξ for any t 0 and covariance 0 − ≥ function 2 σ λ t s λ(t+s) Cov(Y (t),Y (s)) = e− | − | e− 0 0 2λ − for every s,t 0. ≥ Proof: Consider α ,...,α R and t ,...,t [0, T ]. We will study the convergence of 1 N ∈ 1 N ∈ the ﬁnite dimensional distributions of Y H .

N N N H λti λ(ti u) q YN = αiY (ti)= e− ξ + αi1[0,ti](u)e− − dZH (u) R Xi=1 Xi=1 Z Xi=1 N λti q = e− ξ + f(u)dZH (u) R Xi=1 Z 31 N λ(ti u) with f(u)= i=1 αi1[0,ti](u)e− − . Notice that in this case the space given by (15) coincides with L1(R). Since it P HAk is clear that f belongs to L1 (R) (see [22]), we get immediatly by Proposition 1 the |HH |∩ convergence as H 1 of f(u)dZq (u) to f(u)du Hq(Z) . → R H R √q! In order to prove the convergence when H 1 , we will apply Proposition 2. Using R R → 2 the same arguments as for the proof of Proposition 5 in [22], we get

2H 2 2 lim H(2H 1) f(u)f(v) u v − dudv = (f(u)) du H 1 − R R | − | R → 2 Z Z Z N N ti tj N N 2 ∧ λ(ti+tj 2u) σ λ ti tj λ(ti+tj ) = αiαj e− − du = αiαj e− | − | e− 0 2λ − Xi=1 Xj=1 Z Xi=1 Xj=1 N which coincides with the variance of j=1 αjY0(tj). The proof is completed by showing that (31) is satisﬁed. We have P H 1 H 1 H 1 H 1 du1...du4f(u1)...f(u4) u1 u2 − u2 u3 − u3 u4 − u4 u1 − R4 | − | | − | | − | | − | Z d T T

αj1 ....αj4 ... du1..du4 ≤ | | 0 0 j1,..,jX4=1 Z Z 2(H 1)r 2(H 1)r 2(H 1)(q r) 2(H 1)(q r) u u q− u u q− u u − q − u u − q − ×| 1 − 2| | 2 − 3| | 3 − 4| | 4 − 1| 1 is ﬁnite and continuous in H on the set ( 4 , 1]. This follows from Lemma 3.3 in [2] or by apply- 2(H 1)r 2(H 1)r 2(H 1)(q r) 2(H 1)(q r) ing the power counting theorem with (α1, α2, α3, α4)= q− , q− , − q − , − q − . We recall (see [22]) that for p 1, ≥ E Y H (t) Y H (s) 2p C (E Y H (t) Y H (s) 2)p c t s p. (61) | − | ≤ p | − | ≤ | − | The tighness follows from (61) and Bilingsley criterium (see [7]).

5.2 Asymptotic behavior of the stationary Hermite Ornstein-Uhlenbeck Now we will study the asymptotic behavior of (58). The diffrence to the non-stationary case is that the function f from the last proof has support of infinite Lebesque measure an we need to use an argument based on the power counting theorem when H tends to one half. The proof of this results is similar in spirit to the proofs of Proposition 6 and Proposition 7 in [22]. Proposition 4 H 1 Assume H 1. Then the process X (t) t [0,T ] converges weakly, → ∈ in the space of the continuous functions C[0, T ] to the process (X(t))t [0,T ] defined by ∈ σ H (Z) X(t)= q (62) λ √q! with Z (0, 1) ∼N

32 1 H 2 Assume H 2 , the process X (t) t [0,T ] converges weakly, in the space of the continuous → ∈ functions C[0, T ] as H 1 to the stationary Ornstein Uhlenbeck process (X (t)) given → 2 0 t [0,T ] by ∈ t λ(t u) X0(t)= σ e− − dW (u) (63) Z−∞ which is a stationary centered Gaussian process with covariance function

2 σ λ t s Cov(X (t), X (s)) = e− | − | 0 0 2λ for every s,t 0. ≥ Proof: Consider α ,...,α R and t ,...,t [0, T ]. We will study the convergence of 1 N ∈ 1 N ∈ the ﬁnite dimensional distributions of Y H .

N N H λ(ti u) q αiX (ti) = σαi1[ ,ti](u)e− − dZH (u) R −∞ Xi=1 Z Xi=1 q = g(u)dZH (u) R Z

N λ(ti u) with g(u)= i=1 αi1[ ,ti](u)e− − . The computations−∞ in proofs of Proposition 6 and Proposition 7 in [22] show that P g belongs to L1 (R), we get immediatly by Proposition 1 that the random variable |HH |∩ N α XH (t ) converges to N α X(t ) as H 1. i=1 i i i=1 i i → When H 1 , the proof with slight changes, follows along the same lines as the P → 2 P proof of Proposition 7 in [22]. We have

2 2 d d H E αjX (tj) E αjX0(tj) .   −−−→1   H 2 Xj=1 → Xj=1     It remains to prove that the condition (31) holds true. We have

2(H 1)r 2(H 1)r 2(H 1)(q r) 2(H 1)(q r) − − − − − − du1...du4g(u1)...g(u4) u1 u2 q u2 u3 q u3 u4 q u4 u1 q R4 | − | | − | | − | | − | Z d tj1 tj4 λ(tj u1) λ(tj u4) α ...α du .... du e− 1 − ....e− 4 − ≤ | j1 j4 | 1 m j1,j2X,..,j4=1 Z−∞ Z−∞ 2(H 1)r 2(H 1)r 2(H 1)(q r) 2(H 1)(q r) u u q− u u q− u u − q − u u − q − | 1 − 2| | 2 − 3| | 3 − 4| | 4 − 1| d ∞ ∞ λ(u1+..+u4) = αj1 ...αj4 du1.... du4e− | | 0 0 j1,j2X,..,j4=1 Z Z 33 2(H 1)r 2(H 1)r q− q− u1 u2 (tj1 tj2 ) u2 u3 (tj1 tj2 ) ×| − − − 2(|H 1)(q r|) − − − | 2(H 1)(q r) u u (t t ) − q − u u (t t ) − q − | 3 − 4 − j3 − j4 | | 4 − 1 − j4 − j1 | d λ ∞ ∞ 2 ( tj1 tj2 +...+ tj4 tj1 ) e | − | | − | αj1 ...αj4 du1... du4 ≤ | | 0 0 j1,j2X,..,j4=1 Z Z λ ( u1 u2 (tj tj ) +...+ u4 u1 (tj tj ) ) e− 2 | − − 1 − 2 | | − − 4 − 1 | 2(H 1)r 2(H 1)r 1 u u (t t ) q− 1 u u (t t ) q− × ∨ | 1 − 2 − j1 − j2 | ∨ | 2 − 3 − j1 − j2 | 2(H 1)(q r) 2(H 1)(q r) 1 u u (t t ) − q − 1 u u (t t ) − q − ∨ | 3 − 4 − j3 − j4 | ∨ | 4 − 1 − j4 − j1 |

We apply the power counting theorem on the set T ′ deﬁned by

T ′ = u u (t t ), ..., u u (t t ) { 1 − 2 − j1 − j2 4 − 1 − j4 − j1 } with 2(H 1)r 2(H 1)r 2(H 1)(q r) 2(H 1)(q r) (α , .., α )= − , − , − − , − − and (β , .., β ) = ( γ, ..., γ) 1 4 q q q q 1 4 − − with γ ( 3 , 1]. Since T is the only paddet subset of T , we have ∈ 4 ′ ′ 4(H 1)(q r) 4(H 1)(q r) 1 d (T ′) = 4 1+ − − + − − = 4H 1 > 0 if H > 0 − q q − 4 and 1 3 d ( ) = 4 1 4γ < 0 if γ > 1 = . ∞ ∅ − − − 4 4 Therefore, the function

2(H 1)r 2(H 1)r 2(H 1)(q r) 2(H 1)(q r) − − − − − − H ... du1...du4 g(u1)...g(um) u1 u2 q u2 u3 q u3 u4 q u4 u1 q → R R | || − | | − | | − | | − | Z Z is ﬁnite and continuous on the set D = H (0, 1],H > 1 . The conclusion follows from { ∈ 4 } Proposition 2. Again the tighness is obtained by (61).

6 Appendix

The basic tools from the analysis on Wiener space and the power counting theorem proven in [24] are presented in this appendix.

34 6.1 Multiple stochastic integrals and the Fourth Moment Theorem Here, we shall only recall some elementary facts; our main reference is [18]. Consider a H real separable infinite-dimensional Hilbert space with its associated inner product ., . , h i and (B(ϕ), ϕ ) an isonormal Gaussian process on a probability space (Ω, F, P), whichH ∈ H is a centered Gaussian family of random variables such that E (B(ϕ)B(ψ)) = ϕ, ψ , for h iH every ϕ, ψ . Denote by Iq the qth multiple stochastic integral with respect to B. This Iq ∈H q is actually an isometry between the Hilbert space ⊙ (symmetric tensor product) equipped 1 H with the scaled norm q and the Wiener chaos of order q, which is defined as the √q! k · kH⊗ closed linear span of the random variables Hq(B(ϕ)) where ϕ , ϕ = 1 and Hq is ∈ H k kH the Hermite polynomial of degree q 1 defined by: ≥ x2 dq x2 H (x) = ( 1)q exp exp , x R. (64) q − 2 dxq − 2 ∈ The isometry of multiple integrals can be written as: for p, q 1, f p and g q, ≥ ∈H⊗ ∈H⊗

˜ q q! f, g˜ ⊗ if p = q E Ip(f)Iq(g) = h iH (65) ( 0 otherwise. It also holds that: Iq(f)= Iq f˜ , where f˜ denotes the canonical symmetrization of fand it is deﬁned by: 1 f˜(x ,...,x )= f(x ,...,x ), 1 q q! σ(1) σ(q) σ q X∈S in which the sum runs over all permutations σ of 1,...,q . { } In the particular case when = L2(T, (T ),µ) , the rth contraction f g is the H B ⊗r element of (p+q 2r), which is deﬁned by: H⊗ −

(f r g)(s1,...,sp r,t1,...,tq r) ⊗ − − = r du1 . . . durf(s1,...,sp r, u1, . . . , ur)g(t1,...,tq r, u1, . . . , ur), (66) T − − for every f LR2([0, T ]p), g L2([0, T ]q) and r = 1,...,p q. ∈ ∈ ∧ An important property of ﬁnite sums of multiple integrals is the hypercontractivity. Namely, if F = n I (f ) with f k then k=0 k k k ∈H⊗ P p E F p C EF 2 2 . (67) | | ≤ p for every p 2. ≥ We will use the following famous result initially proven in [19] that characterizes the convergence in distribution of a sequence of multiple integrals torward the Gaussian law.

35 Theorem 4 Fix n 2 and let (F , k 1) , F = I (f ) ( with f n for every k 1 ≥ k ≥ k n k k ∈H⊙ ≥ ), be a sequence of square-integrable random variables in the nth Wiener chaos such that E F 2 1 as k . The following are equivalent: k → →∞ 1. the sequence (Fk)k 0 converges in distribution to the normal law (0, 1); ≥ N 2. E F 4 = 3 as k ; k →∞ 3. for all1 l n 1, it holds that lim fk l fk 2(n l) = 0; ≤ ≤ − k k ⊗ kH⊗ − →∞ Another equivalent condition can be stated in term of the Malliavin derivatives of Fk, see [16].

6.2 Power counting theorem We need to recall some notation and results from [24] which are needed in order to check the integrability assumption from Proposition 2. Consider a set T = M , .., M of linear functions on Rm. The power counting { 1 m} theorem (see Theorem 1.1 and Corollary 1.1 in [24]) gives suﬃcient conditions for the integral

I = ... du1...dumf1(M1(u1, .., um))....fm(Mm(u1, .., um)) (68) R R Z Z to be ﬁnite, where f : R R, i = 1, .., m are such that f is bounded above on (a , b ) i → | i| i i (0 < a < b < ) and i i ∞ f (y) c y αi if y < a and f (y) c y βi if y > b . | i |≤ i| | | i| i | i |≤ i| | | | i For a subset W T we denote by s (W )= span(W ) T . A subset W of T is said ⊂ T ∩ to be padded if s (W ) = W and any functional M W also belongs to s (W M ). T ∈ T \ { } Denote by span (W ) the linear span generated by W and by r(W ) the number of linearly independent elements of W . Then Theorem 1.1 in [24] says that the integral I (68) is ﬁnite if

d0(W )= r(W )+ αi > 0 (69) sTX(W ) for any subset W of T with sT (W )= W and

d (W )= r(T ) r(W )+ βi < 0 (70) ∞ − T sT (W ) \X for any proper subset W of T with s (W )= W , including the empty set. If α > 1 then T i − it suffices to check (69) for any padded subset W T . Also, it suffices to verify (70) only ⊂ for padded subsets of T if β 1. i ≥− The condition (69) implies the integrability at the origin while (70) gives the integrability of I at infinity. There is a similar result if one starts with a set T of affine functionals instead of linear functionals.

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