On the Wiener integral with respect to the fractional Brownian motion

Constantin Tudor^{∗} University of Bucharest and CIMAT Department of Mathematics 70109 Bucharest, Romania

Abstract We give examples of inner product spaces of functions defined on a finite interval which are isometric with some subspaces of the Gaussian space generated by a fractional Brownian motion. Such pre-Hilbert spaces represent classes of integrands for the Wiener fractional integral. Key Words: Fractional Brownian motion, Wiener integral, Inner product spaces AMS Classification: 60H05

1 Introduction

Let {B_{t}^{H}}_{t∈[0,T]} be a fractional Brownian motion of Hurst parameter H ∈ (0, 1), defined on a probability space (Ω, F, P) . Recall that B^{H} is a continuous centered Gaussian process with B_{0}^{H} = 0 and covariance

R_{H}(t, s) = \frac{1}{2} (t^{2H} + s^{2H} − |t − s|^{2H}) , s, t ∈ [0, T] .

We denote by H_{H}(resp. H(R_{H})) the Gaussian space (resp. the Reproducing Kernel Hilbert Space) associated to B^{H}. Recall that if H = \frac{1}{2} then B^{\frac{1}{2}} is the standard Brownian motion.

∗E-mail addresses: ctudor@pro.math.unibuc.ro, tudor@cimat.mx

It is well known that H_{H} and H(R_{H}) are isometric and moreover the Wiener integral is a isometry between L^{2} ([0, T]) and H_{\frac{1}{2}} . Therefore if H = \frac{1}{2} then L^{2} ([0, T]) is the class of integrands for the Wiener integral. In the present paper we are interested in a similar problem for the fractional Brownian motion {B_{t}^{H})}_{t∈[0,T]}. An equivalent formulation of the problem is to characterize the largest inner product spaces of functions on [0, T] (integrands) Λ_{H} which contain the elementary functions and are isometric with subspaces of H_{H}. The situation is different in the cases 0 < H < \frac{1}{2}, \frac{1}{2} < H < 1. When 0 < H < \frac{1}{2} we see that we find a Hilbert space of integrands Λ_{H} and thus Λ_{H} is isometric with H(R_{H}), while if \frac{1}{2} < H < 1 the space Λ_{H} is not complete and consequently it is not isometric with H(R_{H}). For f ∈ Λ_{H} we can define the Wiener fractional integral ∫_{0}^{T} f(t)dB_{t}^{H} in the usual manner by setting

∫_{0}^{T} f(t)dB_{t}^{H} = ∑_{i=0}^{k−1} f_{i} (B_{t_{i+1}}^{H} − B_{t_{i}}^{H}) ,  (1.1)

if f has the form

f(t) = ∑_{i=0}^{k−1} f_{i}1_{[t_{i},t_{i+1})}(t), t ∈ [0, T] ,  (1.2)

where 0 = t_{0} < ... < t_{k} = T, f_{i} ∈ R. Then the definition is extended by continuity to Λ_{H}. This procedure is in fact equivalent with the definition of the fractional integral via a transfer from the standard Wiener integral. We also show that the operators used by Alòs, Mazet and Nualart [?] and by Decreusefond and A.S. Ūstunel [?] (see also Decreusefond [?]) for defining the anticipating fractional integral agree. The case of a fractional Brownian motion on an infinite interval is considered by Pipiras and Taqqu [?].

2 Preliminaries

We fix 0 < T < ∞ and 0 < α < 1. For f : [0, T] → R, g : R_{+} → R we define the Liouville fractional integrals by

(I_{T −}^{α}f) (s) = \frac{1}{Γ(α)} ∫_{s}^{T} \frac{f(u)}{(u − s)^{1−α}} du, s ∈ [0, T] ,  (2.1)

(I_{0+}^{α}f) (s) = \frac{1}{Γ(α)} ∫_{0}^{s} \frac{f(u)}{(s − u)^{1−α}} du, s ∈ [0, T] .

(I_{−}^{α}g) (s) = \frac{1}{Γ(α)} ∫_{s}^{∞} \frac{g(u)}{(u − s)^{1−α}} du, s ∈ R_{+}.

We refer to [?] for a complete survey of fractional integrals and derivatives. If f ∈ I_{T −}^{α} (L^{2} ([0, T])) (resp. f ∈ I_{0+}^{α} (L^{2} ([0, T])) then the function φ ∈ L^{2} ([0, T]) such that I_{T −}^{α}φ = f (resp. I_{0+}^{α}φ = f) is unique and coincides with the Marchaud derivative of order α defined by

D_{T −}^{α}f(s) = \frac{1}{Γ(1 − α)} [\frac{f(s)}{(T − s)^{α}} − α ∫_{s}^{T} \frac{f(u) − f(s)}{(u − s)^{α+1}}du] ,  (2.2)

resp.,

D_{0+}^{α}f(s) = \frac{1}{Γ(1 − α)} [\frac{f(s)}{s^{α}} + α ∫_{0}^{s} \frac{f(s) − f(u)}{(s − u)^{α+1}}du] ,  (2.3)

or with the Riemann-Liouville derivative

D_{T −}^{α}f(s) = −\frac{1}{Γ(1 − α)} \frac{d}{ds} ∫_{s}^{T} \frac{f(u)du}{(u − s)^{α}} = −\frac{d}{ds} (I_{T −}^{1−α}f) (s),  (2.4)

resp.,

D_{0+}^{α}f(s) = \frac{1}{Γ(1 − α)} \frac{d}{ds} ∫_{0}^{s} \frac{f(u)du}{(s − u)^{α}} = \frac{d}{ds} (I_{0+}^{1−α}f) (s).  (2.5)

The following lemma is very useful ( see Lemma 3.2-page 70 and Corollary 1-page 208 of [?]). Lemma 2.1. (a) If φ ∈ L^{p} ([0, T]) , 1 < p < ∞, μ > −1 + \frac{1}{p}, then we have the equalities

I_{0+}^{α}x^{μ}φ = x^{μ}I_{0+}^{α} (φ + A_{1}φ) ,  (2.6)
x^{μ}I_{0+}^{α}φ = I_{0+}^{α}x^{μ} (φ + A_{2}φ) ,  (2.7)

where A_{i} are bounded operators in L^{p} ([0, T]) .

(b) If 1 < p < \frac{1}{α} then I_{T −}^{α} (L^{p} ([0, T])) = I_{0+}^{α} (L^{p} ([0, T])) .

Lemma 2.2. For 1 ≤ p < ∞, the operator

I_{T −}^{α} (x^{−α}.) : L^{p} ([0, T]) → L^{p} ([0, T]) ,

is continuous. In particular the operator

x^{α}I_{T −}^{α} (x^{−α}.) : L^{p} ([0, T]) → L^{p} ([0, T]) ,

is continuous.

Proof. We use the following estimate (see [?]-(5.45′),

∫_{0}^{∞} x^{−(α+β+γ)p} |x^{β}I_{−}^{α} (x^{γ}f_{1}) (x)|^{p} dx ≤ k_{p} ∫_{0}^{∞} |f_{1}(x)|^{p} dx, f_{1} ∈ L^{p}(R_{+}),

for 1 ≤ p < ∞ and (α + γ) p < 1. Now for f ∈ L^{p} ([0, T]) define f_{1} = f on [0, T] and f_{1} = 0 on (T, ∞) . Applying the above inequality for β = 0, γ = −α we obtain

∫_{0}^{T} |I_{T −}^{α} (x^{−α}f) (x)|^{p} dx = ∫_{0}^{∞} |I_{−}^{α} (x^{−α}f_{1}) (x)|^{p} dx ≤ k_{p} ∫_{0}^{T} |f(x)|^{p} dx.

■

Remark 2.3. It is known that there is a standard Brownian motion \{Bt^{\frac{1}{2}} \}_{t∈[0,T]}

defined on the same probability space (Ω, F, P) such that:

(j) \{B_{t}^{H}\}_{t∈[0,T]} and \{Bt^{\frac{1}{2}} \}_{t∈[0,T]} generate the same filtration F.

(jj) (Representation formula),

B_{t}^{H} = ∫_{0}^{t} K_{H}(t, s)dBs^{\frac{1}{2}} ,  (2.8)

where

K_{H}(t, s) = c_{H} [(t − s)^{H− \frac{1}{2}} +

(\frac{1}{2} − H) ∫_{s}^{t} (u − s)^{H− \frac{3}{2}} (1 − (\frac{s}{u})^{\frac{1}{2} −H}) du] 1_{[0,t]} (s) =

c_{H} (t − s)^{H− \frac{1}{2}} F (H − \frac{1}{2}, \frac{1}{2} − H, H + \frac{1}{2}, 1 − \frac{t}{s})1_{[0,t]} (s) ,  (2.9)

where F is the Gauss hypergeometric function (see Theorems 4.3 and 4.5 of [?] or Theorem 4.8 of [?], or Theorems 3.2 and 5.2 and Remark 5.1 of [?]). If H > \frac{1}{2},

K_{H}(t, s) = [c_{H}Γ (H + \frac{1}{2}) s^{\frac{1}{2} −H}I_{T −}^{H− \frac{1}{2}} (x^{H− \frac{1}{2}} 1_{[0,t]}) (s)] 1_{[0,t]} (s) , (2.10)

and if H < \frac{1}{2},

K_{H}(t, s) = [c_{H} (t − s)^{H− \frac{1}{2}} + s^{H− \frac{1}{2}} F_{1}(\frac{t}{s})] 1_{[0,t]} (s) ,  (2.11)

where

F_{1}(z) = c_{H} (\frac{1}{2} − H) ∫_{0}^{z−1} u^{H− \frac{3}{2}} (1 − (u + 1)^{H− \frac{1}{2}} ) du  (2.12)

is a continuous nonnegative and bounded function on [1, ∞), which vanishes at 1. Also, by Theorem 3.2 of [?] the following inequality hold:

|K_{H}(t, s)| ≤ d_{H}s^{−|H− \frac{1}{2}|} (t − s)^{−(\frac{1}{2} −H)_{+}} 1_{[0,t]}(s), ∀t, s ∈ [0, T] .  (2.13)

Remark 2.4. According to [?]-page 187, the operator

K_{H} : L^{2} ([0, T]) → I_{0+}^{H+ \frac{1}{2}} (L^{2} ([0, T])),

defined by the kernel K_{H}(t, s), is an isomorphism and the following equalities hold:

K_{H}f = c_{H}I_{0+}^{1}x^{H− \frac{1}{2}} I_{0+}^{H− \frac{1}{2}} x^{\frac{1}{2} −H}f ifH ≥ \frac{1}{2},  (2.14)
K_{H}f = c_{H}I_{0+}^{2H}x^{\frac{1}{2} −H}I_{0+}^{\frac{1}{2} −H}x^{H− \frac{1}{2}} f ifH ≤ \frac{1}{2}.  (2.15)

We denote by E the family of elementary functions on [0, T] , i.e., f ∈ E if has the form (??).

3 Class of integrands for 0 < H < \frac{1}{2}

We assume that 0 < H < \frac{1}{2}. Define

Λ_{H} = I_{T −}^{\frac{1}{2} −H} (L^{2} ([0, T])) , Γ_{H} : Λ_{H} ⟶ L^{2} ([0, T]) ,
(Γ_{H}f) (s) = c_{H}Γ (H + \frac{1}{2}) s^{\frac{1}{2} −H}D_{T −}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} f) (s).  (3.16)

Remark 3.1. (a) Of course Λ_{H} is dense in L^{2} ([0, T]) and by Lemma 2.1 it follows that if f ∈ Λ_{H} then s^{H− \frac{1}{2}} f ∈ Λ_{H} and thus D_{T −}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} f) ∈ L^{2} ([0, T]) and Γ_{H}f ∈ L^{2} ([0, T]) .

(b) Also from (??) and (??) it follows that

(Γ_{H}1_{[0,t]}) (s) = K_{H}(t, s).  (3.17)

Remark 3.2. The operator Γ_{H} is used in [?] for defining the Skorohod integral in the particular case of the fractional Brownian motion.

Lemma 3.3. (a) The operator Γ_{H} : Λ_{H} ⟶ L^{2} ([0, T]) is one to one and onto.

(b) If f ∈ C_{c}^{1} ([0, T)) then

Γ_{H}f(s) = − ∫_{s}^{T} K_{H}(t, s)f′(t)dt, s ∈ (0, T) .  (3.18)

(c) For f ∈ Λ_{H} there exist f_{n} ∈ C_{c}^{∞} ([0, T)) such that

f_{n} ⟶^{L^{2}} f, Γ_{H}f_{n} ⟶^{L^{2}} Γ_{H}f.

Proof. (a) The one to one property is obvious. Now if f ∈ L^{2} ([0, T]) then φ = [c_{H}Γ (H + \frac{1}{2})]^{−1} s^{\frac{1}{2} −H}I_{T −}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} f) ∈ Λ_{H} and Γ_{H}φ = f.

(b) By Proposition 6 of [?], integration by parts and (??), we have

Γ_{H}f(s) = f(s)K_{H}(T, s) + ∫_{s}^{T} [f(t) − f(s)] \frac{∂K_{H}}{∂t} (t, s)dt =

− lim_{t→s} [f(t) − f(s)] K_{H}(t, s) − ∫_{s}^{T} K_{H}(t, s)f′(t)dt = − ∫_{s}^{T} K_{H}(t, s)f′(t)dt.

(c) Choose φ_{n} ∈ C_{c}^{∞} (0, T) such that φ_{n} ⟶^{L^{2}} s^{\frac{1}{2} −H}D_{T −}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} f) and define

f_{n} = s^{\frac{1}{2} −H}I_{T −}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} φ_{n}) ∈ C_{c}^{∞} ([0, T)) .

It is clear that f_{n} ∈ Λ_{H} (by Lemma 2.1 or Lemma 2.2 of [?]),

Γ_{H}f_{n} = c_{H}Γ (H + \frac{1}{2}) φ_{n} ⟶^{L^{2}} Γ_{H}f,

and f_{n} ⟶^{L^{2}} f by Lemma 2.2. ■ Proposition 3.4. The operator Γ_{H} : Λ_{H} ⟶ L^{2} ([0, T]) is closed and agrees with the adjoint of the closable operator K_{H}^{(1)} defined by

K_{H}^{(1)} = K−1_{\frac{1}{2}} ∘ K_{H} : C_{c}^{∞} ((0, T]) → L^{2} ([0, T]) ,

Proof. Assume that f_{n} ⟶^{L^{2}} f, φ_{n} = Γ_{H}f_{n} ⟶^{L^{2}} g. Then by Lemma 2.2

f_{n} = [c_{H}Γ (H + \frac{1}{2})]^{−1} s^{\frac{1}{2} −H}I_{T −}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} φ_{n}) ⟶^{L^{2}}

[c_{H}Γ (H + \frac{1}{2})]^{−1} s^{\frac{1}{2} −H}I_{T −}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} g) ,

so that

f = [c_{H}Γ (H + \frac{1}{2})]^{−1} s^{\frac{1}{2} −H}I_{T −}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} g) ∈ Λ_{H}, Γ_{H}f = g,

and therefore Γ_{H} is closed. Next if f ∈ C_{b}^{1} ([0, T)) , g ∈ C_{c}^{∞} ((0, T]) we have (K−1_{\frac{1}{2}} is the derivative operator),

⟨K_{H}^{(1)}g, f⟩_{L^{2}([0,T])} = ∫_{0}^{T} (K_{H}g)′ (t)f(t)dt =

− ∫_{0}^{T} (K_{H}g) (t)f′(t)dt = − ∫_{0}^{T} (K_{H}^{∗}f′) (t)g(t)dt.

Therefore (K_{H}^{(1)})^{∗} f(t) = − (K_{H}^{∗}f′) (t) = − ∫_{0}^{T} K_{H}(s, t)f′(s)ds = − ∫_{t}^{T} K_{H}(s, t)f′(s)ds, and by Lemma 3.3 we deduce that

(K_{H}^{(1)})^{∗} = Γ_{H} on C_{c}^{∞} ([0, T)) .  (3.19)

Again by Lemma 3.3 and the fact that (K_{H}^{(1)})^{∗} is closed (since K_{H}^{(1)} is densely defined) it follows that Γ_{H} ⊂ (K_{H}^{(1)})^{∗} . Now if g ∈ C_{c}^{∞} ((0, T]) then the function

f = s^{\frac{1}{2} −H}D_{0+}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} D_{0+}^{2H}g) ∈ C_{c}^{∞} ((0, T]) ,

and from (??), it follows that K_{H}f = g. Taking g(t) = ∫_{0}^{t} g_{1}(s)ds, g_{1} ∈ C_{c}^{∞} ((0, T]) , we obtain (K_{H}f)′ = g_{1} which implies the inclusion C_{c}^{∞} ((0, T]) ⊂ R (K_{H}^{(1)}) (in fact is equality) and consequently R (K_{H}^{(1)}) is dense in L^{2} ([0, T]) . Now if f ∈ Dom (K_{H}^{(1)})^{∗} then

g = [c_{H}Γ (H + \frac{1}{2})]^{−1} s^{\frac{1}{2} −H}I_{T −}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} (K_{H}^{(1)})^{∗} (f)) ∈ Λ_{H},

and

Γ_{H}g = (K_{H}^{(1)})^{∗} (g) = (K_{H}^{(1)})^{∗} (f).

Then f − g ∈ N ((K_{H}^{(1)})^{∗}) = R ((K_{H}^{(1)}))^{⊥} = {0} and hence f = g and Γ_{H}f = (K_{H}^{(1)})^{∗} (f). Since (K_{H}^{(1)})^{∗} is densely defined it follows that K_{H}^{(1)} is closable. ■

Remark 3.5. The operator (K_{H}^{(1)})^{∗} is used in [?] (see also [?]) for defining the anticipating stochastic integral with respect to the fractional Brownian motion.

Theorem 3.6. The space Λ_{H} is a Hilbert space with the inner product

⟨f, g⟩_{H} = ⟨Γ_{H} (f) , Γ_{H} (g)⟩_{L^{2}([0,T])} .  (3.20)

Moreover E is dense in Λ_{H} and consequently Λ_{H} is isometric with H(R_{H}).

Proof. It is easy to see that Λ_{H} is a scalar product. To show the completeness, let (f_{n})_{n} be a Cauchy sequence in Λ_{H} and define the finite measure on [0, T] by μ_{H}(ds) = s^{1−2H}λ(ds), λ is the Lebesgue measure. Then the sequence

\{D_{T −}^{\frac{1}{2} −H} (s^{H− \frac{1}{2}} f_{n})\}_{n} is Cauchy in L^{2}(μ_{H}) and therefore D_{T −}^{\frac{1}{2} −H} (s^{H− \frac{1}{2}} f_{n}) ⟶^{L^{2}(μ_{H})}

ψ. Set φ = s^{\frac{1}{2} −H}ψ ∈ L^{2} ([0, T]) and f = s^{\frac{1}{2} −H}I_{T −}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} φ) . Then by Lemma 2.1 f ∈ Λ_{H} and

|f_{n} − f|_{H} = ∥s^{\frac{1}{2} −H}D_{T −}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} f_{n}) − s^{\frac{1}{2} −H}D_{T −}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} f)∥_{L^{2}([0,T])}

= ∥D_{T −}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} f_{n}) − ψ∥_{L^{2}(μ_{H})} → 0.

Next, since it easily seen that for every 0 ≤ a < b < T ,

s → ∫_{s}^{T} \frac{1_{(a,b)}(t)dt}{(t − s)^{\frac{1}{2} −H}}

is absolutely continuous and vanishes in T, we have E ⊂ Λ_{H} (see Theorem 2.1 of [?]). By Lemma 3.3-(c) it is sufficient to show that for every f ∈ C_{c}^{∞} ([0, T)) there

is f_{n} ∈ E such f_{n} ⟶^{Λ_{H}} f.

Let Δ : 0 = t_{0} < ... < t_{k} = T by such that ∥Δ∥ → 0 and define

f_{Δ}(t) = ∑_{i} (\frac{1}{t_{i+1} − t_{i}} ∫_{t_{i}}^{t_{i+1}} f(s)ds) 1_{[t_{i},t_{i+1})}(t) ∈ E.  (3.21)

Next it is easily seen that

((K_{H}^{(1)})^{∗} 1_{[0,t]}) (s) = K_{H}(t, s),  (3.22)

and thus by (??) Γ_{H} = (K_{H}^{(1)})^{∗} on E. Therefore Γ_{H} = (K_{H}^{(1)})^{∗} on E ∪

C_{c}^{∞} ([0, T)) ⊂ Λ_{H} ( see (??)) and by Lemma 6.3 of [?] we obtain that f_{Δ} ⟶^{Λ_{H}} f. ■

Remark 3.7. From the above theorem one defines the fractional Wiener integral by

∫_{0}^{T} f(t)dB_{t}^{H} = ∫_{0}^{T} (Γ_{H}f) (t)dBt^{\frac{1}{2}} , f ∈ Λ_{H}.  (3.23)

We have H_{H} = {∫_{0}^{T} f(t)dB_{t}^{H}}_{f∈Λ_{H}} , the map

f → ∫_{0}^{T} f(t)dB_{t}^{H} : Λ_{H} → H_{H},

is an unitary operator and by (??) and (??) the relation (??) holds.

4 Class of integrands for \frac{1}{2} < H < 1

We assume that \frac{1}{2} < H < 1. We define

Λ_{H} = \{f : I_{T −}^{H− \frac{1}{2}} (x^{H− \frac{1}{2}} |f|) (s) < ∞fora.e.s, s^{\frac{1}{2} −H}I_{T −}^{H− \frac{1}{2}} (x^{H− \frac{1}{2}} f) ∈ L^{2} ([0, T])\} , (4.24)

Γ_{H} : Λ_{H} ⟶ L^{2} ([0, T]) , (Γ_{H}f) (s) = c_{H}Γ (H + \frac{1}{2}) s^{\frac{1}{2} −H}I_{T −}^{H− \frac{1}{2}} (x^{H− \frac{1}{2}} f) (s).

(4.25)

Lemma 4.1. For a, b ∈ R, a < b, define

ψ_{a,b}(s) = \frac{1}{Γ (\frac{3}{2} − H)} [(s − a)_{+}^{\frac{1}{2} −H} − (s − b)_{+}^{\frac{1}{2} −H}] , s ∈ R.

Then for every 0 < a < b ≤ T we have

(I_{0+}^{H− \frac{1}{2}} ψ_{a,b}) (s) = 1_{(a,b]}(s) foreverys ∈ [0, T] .

Proof. If s ≤ a then ψ_{a,b}(s) = 0. If s > b then

(I_{0+}^{H− \frac{1}{2}} ψ_{a,b}) (s) = \frac{1}{Γ (\frac{3}{2} − H) Γ (H − \frac{1}{2})}×

× \{∫_{a}^{s} (u − a)^{\frac{1}{2} −H} (s − u)^{H− \frac{3}{2}} du − ∫_{b}^{s} (u − b)^{\frac{1}{2} −H} (s − u)^{H− \frac{3}{2}} du\} = 0,

since

∫_{a}^{s} (u − a)^{\frac{1}{2} −H} (s − u)^{H− \frac{3}{2}} du = ∫_{b}^{s} (u − b)^{\frac{1}{2} −H} (s − u)^{H− \frac{3}{2}} du

= Γ (\frac{3}{2} − H) Γ (H − \frac{1}{2}) .

If s ∈ (a, b] then

(I_{0+}^{H− \frac{1}{2}} ψ_{a,b}) (s) = \frac{1}{Γ (\frac{3}{2} − H) Γ (H − \frac{1}{2})} ∫_{a}^{s} (u − a)^{\frac{1}{2} −H} (s − u)^{H− \frac{3}{2}} du = 1.■

Proposition 4.2. (a) If |f| ∈ Λ_{H} then

∥f∥_{H} = H(2H − 1) ∫_{0}^{T} ∫_{0}^{T} f(t)f(s) |t − s|^{2H−2} dtds.  (4.26)

(b) We have the inclusion E ⊂ Γ_{H}(Λ_{H}). Moreover if \frac{1}{2} < H < \frac{3}{4} then E ⊂ Γ_{H}(L^{2} ([0, T])).

(c) We have the strict and dense inclusions

Γ_{H}(L^{2} ([0, T])) ⊂ Γ_{H}(Λ_{H}) ⊂ L^{2} ([0, T]) .

Proof. (a) It is proved in [?].

(b) We have 1_{(0,a)} ∈ Γ_{H}(Λ_{H}) if and only if 1_{(0,a)}x^{H− \frac{1}{2}} = I_{T −}^{H− \frac{1}{2}} (x^{H− \frac{1}{2}} φ) for some φ ∈ Λ_{H} and this follows easily from [?]-page 33. Assume now that \frac{1}{2} < H < \frac{3}{4}. By Lemma 2.2 of [?] we have

x^{H− \frac{1}{2}} = I_{T −}^{H− \frac{1}{2}} (ψ) , ψ ∈ L^{r} ([0, a]) , ∀1 ≤ r < (H − \frac{1}{2})^{−1} .

Next by [?]- page 237 we obtain that

1_{(0,a)}x^{H− \frac{1}{2}} = I_{T −}^{H− \frac{1}{2}} (ψ_{1}) , ψ_{1} ∈ L^{r} ([0, T]) .

Clearly that φ = x^{\frac{1}{2} −H}ψ_{1} ∈ L^{2} ([0, T]) .

(c) Clearly we have L^{2} ([0, T]) ⊂ Λ_{H}. Define

f(s) = s^{\frac{1}{2} −H}(T − s)^{β}, β = −\frac{1}{2} − ε, 0 < ε < H − \frac{1}{2}.

Then

∫_{0}^{T} |f(s)|^{2} ds ≥ T^{1−2H} ∫_{0}^{T} (T − s)^{−1−2ε} = ∞,

I_{T −}^{H− \frac{1}{2}} (x^{H− \frac{1}{2}} f) (s) = C (T − s)^{H− \frac{1}{2} +β} ,

∫_{0}^{T} |(Γ_{H}f) (s)|^{2} ds = C_{1} ∫_{0}^{T} s^{1−2H}(T − s)^{2(H− \frac{1}{2} +β)}ds = C_{2} < ∞,

so that f ∈ Λ_{H} and f ∉ L^{2} ([0, T]) . The strict inclusion Γ_{H}(L^{2} ([0, T])) ⊂ Γ_{H}(Λ_{H}) follows from the strict inclusion L^{2} ([0, T]) ⊂ Λ_{H} and the fact that Γ_{H} is one to one. We show now the strict inclusion Γ_{H}(Λ_{H}) ⊂ L^{2} ([0, T]) . For simplicity assume T = 1. Let b > 1, 0 < p < H − \frac{1}{2}, φ : [0, 1] → R,

φ(s) = c_{H}Γ(H + \frac{1}{2})Re [∑_{n=1}^{∞} b^{−pn}e^{ib^{n}s}] , ψ(s) = c_{H}Γ(H + \frac{1}{2})s^{\frac{1}{2} −H}φ(s).

Clearly ψ ∈ L^{2} ([0, T]) (φ is bounded) and we show that ψ ∉ Γ_{H}(Λ_{H}). Assume the contrary, i.e. there is f ∈ Λ_{H} such that Γ_{H}(f) = ψ, or equivalently,

φ = I_{T −}^{H− \frac{1}{2}} (x^{H− \frac{1}{2}} f) = I_{T −}^{H− \frac{1}{2}} (f̃).

Because I_{T −}^{H− \frac{1}{2}} (|f̃|)(s) < ∞ for a.e. s, we have that

∫_{b}^{1} |f̃(s)| ds < ∞foreachb ∈ (0, 1) ,

and then by Lemma 4.1,

∫_{0}^{1} |f̃(s)| (∫_{0}^{1} ψ_{b,1}(u) (s − u)_{+}^{H− \frac{3}{2}} du) ds =

∫_{0}^{1} |f̃(s)| I_{0+}^{H− \frac{1}{2}} (ψ_{b,1}) (s)ds = ∫_{b}^{1} |f̃(s)| ds < ∞.

Next by Lemma 4.1 and Fubini’s theorem

∫_{0}^{1} ψ_{b,1}(s)φ(s)ds = ∫_{0}^{1} ψ_{b,1}(s)I_{T −}^{H− \frac{1}{2}} \binom{}{f̃} (s)ds =

∫_{0}^{1} f̃(s)I_{0+}^{H− \frac{1}{2}} (ψ_{b,1}) (s)ds = ∫_{b}^{1} f̃(s)ds,

which is a.e. differentiable in b. But by Lemma 5.7 of [?]

∫_{0}^{1} ψ_{b,1}(s)φ(s)ds = \frac{1}{Γ(\frac{3}{2} − H)} ∫_{0}^{1} φ(s) (s − b)_{+}^{\frac{1}{2} −H} ds

is not differentiable on a set of positive probability. To show the dense inclusion Γ_{H} (L^{2} ([0, T])) ⊂ Γ_{H} (Λ_{H}), or equivalently L^{2} ([0, T]) ⊂ Λ_{H} in the norm |.|_{H}, for f ∈ Λ_{H} choose φ_{n} ∈ C_{c}^{1} ((0, T))

such that φ_{n} ⟶^{L^{2}} Γf. Then

f_{n} = [c_{H}Γ (H + \frac{1}{2})]^{−1} s^{\frac{1}{2} −H}D_{T −}^{\frac{1}{2} −H} (x^{H− \frac{1}{2}} φ_{n}) ∈ L^{2} ([0, T])

and Γ_{H}f_{n} ⟶^{L^{2}} Γf.

For f ∈ L^{2} ([0, T]) choose f_{n} ∈ E such that f_{n} ⟶^{L^{2}} f. Then the dense inclusion Γ_{H}(Λ_{H}) ⊂ L^{2} ([0, T]) follows from (b). ■

¿From Lemma 2.2 (see also Theorem 2.2 and Remark 6.2 of [?]) it follows: Proposition 4.3. The operator Γ_{H} : L^{2} ([0, T]) ⟶ L^{2} ([0, T]) is continuous and agrees with the adjoint of the continuous operator K_{H}^{(2)} defined by

K_{H}^{(2)} = K−1_{\frac{1}{2}} ∘ K_{H} : L^{2} ([0, T]) → L^{2} ([0, T]) .

Remark 4.4. Again the operators Γ_{H} and (K_{H}^{(2)})^{∗} are used in [?] and [?] for defining the stochastic integral with respect to the fractional Brownian motion.

Theorem 4.5. The space Λ_{H} is an incomplete inner product space with respect to

⟨f, g⟩_{H} = ⟨Γ_{H} (f) , Γ_{H} (g)⟩_{L^{2}([0,T])} .  (4.27)

Moreover E is dense in Λ_{H}.

Proof. It is easily seen that (Λ_{H}, |.|_{H}) is a pre-Hilbert space (Γ_{H} is one to one). For the density of E in Λ_{H}, it is enough to show that Γ_{H}(E) is dense in Γ_{H}(L^{2} ([0, T])) (by Proposition 4.2-(c)).Or for f ∈ L^{2} ([0, T]) the functions f_{Δ} defined in (??) are in E, f_{Δ} ⟶^{L^{2}} f and thus by Proposition 4.3,

Γ_{H}f_{Δ} ⟶^{L^{2}} Γ_{H}f.

Now assume that (Λ_{H}, |.|_{H}) is complete. We show in this situation that Γ_{H} (Λ_{H}) = L^{2} ([0, T]) which is a contradiction accordind to Proposition 4.2- (c). If f ∈ L^{2} ([0, T]) choose f_{n} ∈ E such that f_{n} ⟶^{L^{2}} f. We know that f_{n} = Γ_{H}φ_{n}. Therefore (φ_{n})_{n} is a Cauchy sequence in (Λ_{H}, |.|_{H}) and thus φ_{n} ⟶^{Λ_{H}} φ or

equivalently, f_{n} ⟶^{L^{2}} Γ_{H}φ = f.

Remark 4.6. One defines the fractional Wiener integral as in (??). This time {∫_{0}^{T} f(t)dB_{t}^{H}}_{f∈Λ_{H}} is a strict subspace of H_{H}, the map

f → ∫_{0}^{T} f(t)dB_{t}^{H} : Λ_{H} → H_{H},

is an isometry and again (??) holds.

Remark 4.7. For \frac{1}{2} < H < 1, as in the case of an infinite time interval (see [?]), it is still an open problem if there exists a Hilbert space of real functions defined [0, T] such that the fractional Wiener integral is well defined for elements of this space. Therefore we do not know if the completion of (Λ_{H}, |.|_{H}) is a space of functions, or equivalently, if Γ_{H} : Λ_{H} ⟶ L^{2} ([0, T]) can be extended to an isometry between a Hilbert space of functions onto L^{2} ([0, T]) .

Remark 4.8. We could also consider the class of integrands

|Λ_{H}| = {f : [0, T] → R : |f| ∈ Λ_{H}} ,

with the scalar product (??) , or alternatively

≪ f, g ≫_{H}= H(2H − 1) ∫_{0}^{T} ∫_{0}^{T} f(t)g(s) |t − s|^{2H−2} dtds,  (4.28)

(by Proposition 4.2 -(a)). The space |Λ_{H}| is still incomplete.

Remark 4.9. If we define for f, g ∈ |Λ_{H}| ,

≺ f, g ≻_{H}= ⟨Γ_{H} (|f|) , Γ_{H} (|g|)⟩_{L^{2}([0,T])} ,

then as in Theorem 4.1 of [?] it follows that (|Λ_{H}| , ≺ ., . ≻_{H}) is a Hilbert space.

Comment. Taking the multidimensional forms Λ_{H}^{(p)}, Γ_{H}^{(p)} of the Λ_{H}, Γ_{H} (for any 0 < H < 1) we can define the multiple fractional integrals I_{p}^{H} by

I_{p}^{H}(f) = Ip^{\frac{1}{2}} (Γ_{H}^{(p)} (f)), f ∈ Λ_{H}^{(p)},

and then we can consider the fractional chaos decomposition of square integrable functionals. The previous results show a rather surprising fact. For H > \frac{1}{2} not all square integrable functionals has a fractional chaos decom-

position( take I1^{\frac{1}{2}} (f) with f ∈ L^{2} ([0, T]) and f ∉ Γ_{H} (Λ_{H}) ) and if H < \frac{1}{2} every square integrable functional has a chaos decomposition.

References

[1] E. Alós, O. Mazet and D. Nualart. Stochastic calculus with respect to Gaussian processes. Preprint.Universitat de Barcelona, 1999.

[2] R.J. Barton and H. Vincent Poor. Signal detection in fractional Gaussian noise. IEEE Transactions on Information Theory, Vol. 34, Nr. 5 (1988), 943-959.

[3] L. Decreusefond. A Skorohod-Stratonovitch integral for the fractional Brownian motion. Progress in Probability, pp. 177-1999. Birkhauser, 2000.

[4] L. Decreusefond and A.S. Ūstunel. Stochastic Analysis with respect to fractional Brownian motion. Potential Analysis 10(2) (1999), 177-214.

[5] I. Norros, E. Valkeila and J. Virtamo. An elementary approach to a Girsanov formula and other analytical results on fractional Brownian motions. Bernoulli 5(4) (1999), 571-587.

[6] V. Pérez Abreu and C. Tudor. A transfer principle for multiple stochastic fractional integrals. Preprint. CIMAT, 2001.

[7] V. Pipiras and M.S. Taqqu. Integration questions related to fractional Brownian motion. Probability Theory and Related Fields 118(2) (2000), 251-291.

[8] S.G. Samko, A.A. Kilbas and O.I. Marichev. Fractional Integrals and Derivatives. Gordon and Breach Science, 1993.