Fractional Fokker-Planck equation

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1 Fractional Fokker-Planck equation Isabelle Tristani To cite this version: Isabelle Tristani. Fractional Fokker-Planck equation. 17 pages <hal v1> HAL I: hal Submitte on 4 Dec 2013 (v1), last revise 23 Oct 2014 (v2) HAL is a multi-isciplinary open access archive for the eposit an issemination of scientific research ocuments, whether they are publishe or not. The ocuments may come from teaching an research institutions in France or abroa, or from public or private research centers. L archive ouverte pluriisciplinaire HAL, est estinée au épôt et à la iffusion e ocuments scientifiques e niveau recherche, publiés ou non, émanant es établissements enseignement et e recherche français ou étrangers, es laboratoires publics ou privés.

2 FRACTIONAL FOKKER-PLANCK EQUATION ISABELLE TRISTANI Abstract. This paper eals with the long time behavior of solutions to a fractional Fokker-Planck equation of the form tf = I[f]+iv(xf) where the operator I stans for a fractional Laplacian. We prove an exponential in time convergence towars equilibrium in new spaces. Inee, such a result was alreay obtaine in a L 2 space with a weight prescribe by the equilibrium in [6]. We improve this result obtaining the convergence in a L 1 space with a polynomial weight. To o that, we take avantage of the recent paper [7] in which an abstract theory of enlargement of the functional space of the semigroup ecay is evelope. Mathematics Subject Classication (2010): 47G20 Integro-ifferential operators; 35B40 Asymptotic behavior of solutions; 35Q84 Fokker-Planck equations. Keywors: Fractional Laplacian; Fokker-Planck equation; spectral gap; exponential rate of convergence; long-time asymptotic. Contents 1. Introuction Moel an main result Known results Metho of proof an outline of the paper 3 2. Preliminaries on the fractional Laplacian Definition on S(R ) Fractional Laplacian an Fourier transform Fractional Laplacian an fractional Sobolev spaces 5 3. Theorem of enlargement of the functional space of the semigroup ecay Notations The abstract theorem 6 4. Semigroup ecay in L 2 (µ 1/2 ) where µ is the steay state Preliminaries on steay states Decay properties in L 2 (µ 1/2 ) 8 5. Semigroup ecay in L 1 ( x k ) Splitting of the operator Regularization properties of ( Ae Bt) ( n) Proof of the main result 16 References 17 Date: December 4,

3 2 ISABELLE TRISTANI 1. Introuction 1.1. Moel an main result. For α (0, 2), we consier the following generalization of the Fokker-Planck equation: (1.1) t f = ( ) α/2 f +iv(xf), in R with an initial ata f 0. In the sequel, we will use the shorthan notations I[f] = ( ) α/2 f an Lf = I[f]+iv(xf). The operator ( ) α/2 is a fractional Laplacian, we first efine it on the space of Schwartz functionss(r )anwethenextentheefinitiontoothersfunctions. WerefertoSection2 for the exact efinition an for properties. We also efine here weighte L p spaces in the following way: for some given Borel weight function m 0 on R, let us efine L p (m), 1 p +, as the Lebesgue space associate to the norm h L p (m) = hm L p. We now state our main result concerning the equation (1.1). Theorem 1.1. Let us consier k (0,α). For any a ( min(λ,k),0) (where λ > 0 will be efine in Corollary 4.5) an for any initial ata f 0 L 1 ( x k ), the solution f(t) of the equation (1.1) satisfies the following ecay: f(t) µ f 0 L 1 ( x k ) C ae at f 0 µ f 0 L 1 ( x k ) where f 0 = R f 0 an for some constant C a > Known results. The main references to mention here are the papers [4] an [6]. In these two papers, Lévy-Fokker-Planck equations (the fractional Laplacian is replace by a Lévy operator) are stuie using the entropy prouction metho. There is a proof of existence an uniqueness of a nonnegative steay state of mass 1 of the associate stationary equation. Then, in a weighte L 2 space with a weight prescribe by the equilibrium, a convergence (with an exponential rate) of the solution of the full equation towars equilibrium is obtaine. Let us give more etails about these results. We first introuce the main tools use. Consierasmoothconvex functionφ : R + Ranµpositivesuchthat R µ(x)x = 1 an efine the Φ-entropy: for any nonnegative function f, ( ) Ent Φ µ R (f) := Φ(f)µx Φ f µx. R Jensen s inequality gives that Ent Φ µ(f) 0. Let f 0 be an initial conition of a Lévy- Fokker-Planck equation or of the classical Fokker-Planck equation: (1.2) t f = f +iv(xf), in R ( ) Then, let us introuce the quantity E Φ (f 0 )(t) := Ent Φ f(t) µ µ which is well-efine for any t > 0. In the case of the classical Fokker-Planck equation (1.2), by using functional inequalities as Poincaré, logarithmic Sobolev or Φ-entropy inequalities, one obtains exponential ecays to zero of E Φ (f 0 ). Then, the solution f of (1.2) converges towars the steay state of mass 1 in the sense of Φ-entropy. Methos to prove such results are usually base on entropy/entropy-prouction tools. See [3, 1, 2, 5] for ifferent methos an applications.

4 FRACTIONAL FOKKER-PLANCK EQUATION 3 In [4], Biler an Karch stuy Lévy-Fokker-Planck equations where the Lévy operators are Fourier multipliers associate to symbols a(ξ) satisfying for some real number β (0, 2] 0 < liminf ξ 0 a(ξ) ξ β limsup ξ 0 They prove that there exist C > 0 an ǫ > 0 such that a(ξ) a(ξ) < an 0 < inf ξ β ξ 2. E 2 /2(f 0 )(t) Ce ǫt, which means that the solution converges towars equilibrium at an exponential rate in L 2 (µ 1/2 ) where we enote µ the only steay state of mass 1. They euce a similar result in L 2 an finally, uner some more restrictive regularity an ecay assumptions on f 0, they prove that the exponential convergence hols in L 1. In [6], taking avantage of the paper [4], Gentil an Imbert prove an exponential ecay of the Φ-entropies for a class of convex functions Φ an for a larger class of operators which inclues the fractional Laplacian. In the present paper, we only consier the equation (1.1) but we are able to enlarge the space where we have a ecay towars equilibrium with minimal assumptions on f 0. If we compare our result to the one obtaine in [4] for others operators efine above, we have to unerline the fact that the result of convergence of the solution towars equilibrium in L 1 from [4] requires aitional assumptions on f 0 (f 0 must have finite moments of a large orer), it is not the case in our main result where f 0 is only suppose to belong to L 1 ( x k ) with k < α Metho of proof an outline of the paper. The main outcome of the present paper is a result of ecay towars equilibrium with an exponential rate of convergence in L 1 ( x k ) (with k < α) for solutions of our equation (1.1). To o that, we aopt the same strategy as the one aopte in [7] by Gualani, Mischler an Mouhot for the classical Fokker-Planck equation. Let us explain in more etails this strategy. It is base on the theory of enlargement of the functional space of the semigroup ecay evelope in [7]. It enables to get a spectral gap in a larger space when we alreay have one in a smaller space. It applies to operators L which can be splitte into two parts, L = A+B with A boune an B issipative. Moreover, if we enote e Bt the semigroup associate to the operator B, the semigroup ( Ae Bt) is require to have some regularization properties. The fact that we can use this theory for our operator is base on two facts: we know from [6] that our operator has a spectral gap in L 2 (µ 1/2 ) where µ is the only steay state of mass 1 of (1.1), we are able to get a splitting satisfying the previous properties using computations base on properties of the fractional Laplacian. In section 2, we recall some technical tools about the fractional Laplacian that are useful in orer to get a splitting of the operator. In section 4, we state results from [6] which are necessary to apply the abstract theorem of enlargement of spectral gap, which is remine in Section 3. Finally, in Section 5, we apply this theorem to obtain our main result on the convergence towars equilibrium of the solution of (1.1) in L 1 ( x k ) with k < α. Acknowlegements We woul like to thank Stéphane Mischler an Robert Strain for enlightene iscussions an their help.

5 4 ISABELLE TRISTANI 2. Preliminaries on the fractional Laplacian In this section, we recall some elementary properties of the fractional Laplacian that we will nee through this paper. The usual reference for this kin of operators is Lankof s book [9] Definition on S(R ). Let us consier α (0,2). The fractional Laplacian ( ) α/2 is an operator efine on S(R ) by: (2.1) f S(R ), ( ) α/2 f(x) f(y) f(x) = y. R +α x y This efinition has to be unerstoo in the sense of principal value: ( ) α/2 f(x) f(y) f(x) = lim y. ǫ 0 +α x y ǫ x y Due to the singularity of the kernel, the right han-sie of (2.1) is not well efine in general. However, when α (0,1), the integral is not really singular near x. Inee, since f S(R ), both f an f are boune. We hence euce the following inequality: +α y f(x) f(y) R x y f L B(x,1) y x y +α 1 + f L R \B(x,1) y x y +α. When α (0,2), we can also write the fractional Laplacian with a non principal value integral. For any f S(R ), we have (2.2) x R, ( ) α/2 f(x) = 1 f(x+y)+f(x y) 2f(x) 2 R y +α y an this integral is well efine. We can exten the integral efinition of the fractional Laplacian to the following set of functions: { f : R R, R } f(x) x < 1+ x +α In particular, we can efine ( ) α/2 x k when k < α Fractional Laplacian an Fourier transform. Let us remin a well-known fact about the Fourier transform of the fractional Laplacian of a Schwartz function. Lemma 2.1. There exists C > 0 such that for any f S ( R ), we have: ( ) F ( ) α/2 f (ξ) = C ξ α f(ξ). If f is a Schwartz function, there is a singularity at 0 in the Fourier transform of ( ) α/2 f. It implies a lack of ecay at infinity for ( ) α/2 f itself, ( ) α/2 f is not a Schwartz function. We can prove that ( ) α/2 f ecays at infinity as x α. We now mention a very useful property of the fractional Laplacian which can be seen as a sort of integration by parts. Lemma 2.2. Let us consier f an g two Schwartz functions. Then, we have ( ) α/2 f(x)g(x)x = f(x)( ) α/2 g(x)x. R R

6 FRACTIONAL FOKKER-PLANCK EQUATION 5 If k < α, we can also prove that ( ) α/2 f(x) x k x = f(x)( ) α/2 x k x. R R 2.3. Fractional Laplacian an fractional Sobolev spaces. Most of the time, fractional Sobolev spaces H s( R ) are efine in the following [ way: H s( R ) = W s,2( R ) for ( s 0 is the set of functions f L 2( R ) such that 1+ 2) s/2 f] is also in L 2( R ). We remin here an equivalent efinition which is going to be useful in what follows. Lemma 2.3. Let us consier s (0,1). We have: { H s( R ) = f L 2( R ) f(x) f(y) ( : L 2 R R )}. x y 2 +s We also have the following fact: ( ) α/2 f 2 L 2 (R ) = C for some C > 0. R f(x) f(y) R x y +α 2 yx 3. Theorem of enlargement of the functional space of the semigroup ecay 3.1. Notations. For a given real number a R, we efine the half complex plane a := {z C, Rez > a}. For some given Banach spaces (E, E ) an (E, E ) we enote by B(E,E) the space of boune linear operators from E to E an we enote by B(E,E) or E E the associate norm operator. We write B(E) = B(E,E) when E = E. We enote by C(E,E) the space of close unboune linear operators from E to E with ense omain, an C(E) = C(E,E) in the case E = E. For a Banach space X an Λ C(X) we enote by e Λt, t 0, its semigroup, by D(Λ) its omain, by N(Λ) its null space an by R(Λ) its range. We also enote by Σ(Λ) its spectrum, so that for any z belonging to the resolvent set ρ(λ) := C\Σ(Λ) the operator Λ z is invertible an the resolvent operator R Λ (z) := (Λ z) 1 is well-efine, belongs to B(X) an has range equal to D(Λ). We recall that ξ Σ(Λ) is sai to be an eigenvalue if N(Λ ξ) {0}. Moreover, an eigenvalue ξ Σ(Λ) is sai to be isolate if Σ(Λ) {z C, z ξ r} = {ξ} for some r > 0. In the case when ξ is an isolate eigenvalue, we may efine Π Λ,ξ B(X) the associate spectral projector by Π Λ,η := 1 (Λ z) 1 z 2iπ z ξ =r with0 < r < r. Notethatthisefinitionisinepenentofthevalueofr astheapplication C\Σ(Λ) B(X), z R Λ (z) is holomorphic. For any ξ Σ(Λ) isolate, it is well-known (see [8] paragraph III-6.19) that Π 2 Λ,ξ = Π Λ,ξ, so that Π Λ,ξ is inee a projector.

7 6 ISABELLE TRISTANI When moreover the so-calle algebraic eigenspace R(Π Λ,ξ ) is finite imensional we say that ξ is aiscreteeigenvalue, written as ξ Σ (Λ). Inthat case, R Λ is ameromorphic function on a neighborhoo of ξ, with non-removable finite-orer pole ξ. Finally for any a R such that Σ(Λ) a = {ξ 1,...,ξ k } where ξ 1,...,ξ k are istinct iscrete eigenvalues, we efine without any risk of ambiguity Π Λ,a := Π Λ,ξ1 +...Π Λ,ξk. We shall also nee the following efinition on the convolution of semigroups. Consier some Banach spaces X 1, X 2 an X 3. For two given functions S 1 L 1( R + ;B(X 1,X 2 ) ) an S 2 L 1( R + ;B(X 2,X 3 ) ), the convolution S 2 S 1 L 1 (R + ;B(X 1,X 3 )) is efine by t 0, S 2 S 1 (t) = t 0 S 2 (s)s 1 (t s)s. When S 1 = S 2 an X 1 = X 2 = X 3, S ( l) is efine recursively by S ( 1) = S an S ( l) = S ( (l 1)) for any l The abstract theorem. Let us now present an enlargement of the functional space of a quantitative spectral mapping theorem (in the sense of semigroup ecay estimate). The aim is to enlarge the space where the ecay estimate on the semigroup hols. The version state here comes from [7, Theorem 2.13] an [7, Lemma 2.17]. Theorem 3.1. Let E, E be two Banach spaces such that E E with ense an continuous embeing, an consier L C(E), L C(E) with L E = L an a R. We assume: (1) L generates a semigroup e tl an Σ(L) a = {ξ 1,...,ξ k } Σ (L) (with ξ k ξ k if k k an {ξ 1,...,ξ k } = if k = 0) an L a is issipative on R(I Π L,a ). (2) There exist A, B C(E) such that L = A+B (with corresponing restrictions A an B on E) an some constants l 0 N, C 1, b R an γ [0,1) so that (i) B a an B a are issipative respectively on E an E, (ii) A B(E) an A B(E), (iii) T l0 := ( Ae Bt) ( l 0 ) satisfies t 0, T l0 (t) B(E,E) C ebt t γ. Then the following estimate on the semigroup hols: a k > a, t 0, elt e Lt Π L,ξj C a e a t. B(E) j=1 Remark 3.2. The assumption (2)-(iii) implies that for any a > a, there exist some constructive constants n N, C a 1 such that t 0, T n (t) B(E,E) C a e a t.

8 FRACTIONAL FOKKER-PLANCK EQUATION 7 4. Semigroup ecay in L 2 (µ 1/2 ) where µ is the steay state 4.1. Preliminaries on steay states. We recall results obtaine in [6] about existence of steay states. They prove such a theorem for a more general equation than ours: t f = I[f]+iv(f V) x R, t > 0 f(0,x) = f 0 (x) x R where f 0 L 1( R ). The operator I is a Lévy operator efine as: I[f](x) = iv(σ f)(x) b f(x)+ (f(x+z) f(x) f(x) zh(z)) ν(z) R where σ is a symmetric semi-efinite matrix, b R an ν enotes a nonnegative singular measure on R that satisfies ν({0}) = 0 an ( R min 1, z 2) ν(z) <. The fractional Laplacian correspons to a particular Lévy operator. Inee, with σ = 0, b = 0 an ν(z) = z α z, we obtain the fractional Laplacian. In this particular case, the proof of existence of steay states of (1.1) is easier, we hence give a sketch of a proof of it (it is aapte from the proof of [6, Theorem 1]). We suppose that µ is an equilibrium of the equation (1.1). At least formally, we have: (4.1) I[µ]+iv(xµ) = 0. We o the following computation in orer to take the Fourier transform of (4.1) F (iv(xµ))(ξ) = F ( j (x j µ))(ξ) = j=1 = i ξ j F (x j µ)(ξ) j=1 ξ j j µ(ξ) = ξ µ(ξ). j=1 We euce that an equilibrium µ satisfies ξ α µ(ξ)+ξ µ(ξ) = 0, which implies that µ(ξ) = Ce ξ α /α for a constant C. In the remaining part of the paper, we enote µ the only steay state of (1.1) of mass 1: µ F 1( e α /α ). Remark 4.1. In the case of the classical Fokker-Planck equation (1.2), the steay state is a Maxwellian, it is hence a Schwartz function. In our case, the steay state is not anymore a Schwartz function because its Fourier transform has a singularity at 0. If we enote χ 1 a smooth function which is nonnegative, supporte on x 2 an such that χ 1 (x) = 1 for x 1, we can write the following ecomposition of µ: µ(ξ) = χ 1 (ξ)(1+a 1 ξ α +a 2 ξ 2α +...)+(1 χ 1 (ξ))e ξ α /α. We see that the secon part of the right-han sie is a Scwhartz function an the first one inuces a singularity at 0. We can hence prove that µ(x) x α when x.

9 8 ISABELLE TRISTANI 4.2. Decay properties in L 2 (µ 1/2 ). We again use results obtaine in [6]. We just use them in our particular case, the fractional Laplacian. For Φ a convex function, we introuce D Φ on (R + ) 2 as: D Φ (a,b) = Φ(a) Φ(b) Φ (b)(a b), which is nonnegative on (R + ) 2. We will not prove the next two lemmas which are going to enable us to prove the ecay towars equilibrium in L 2 (µ 1/2 ). The first one is [6, Proposition 1] an the secon one is [6, Theorem 2] an comes from [5]. Lemma ( 4.2. ) Consier f 0 a nonnegative initial ata for the equation (1.1) which satisfies Ent Φ f0µ µ <. Then, for any smooth convex function Φ an for any t 0, the solution f(t) satisfies t E Φ(f 0 )(t) = D Φ (u(t,x),u(t,x z)) R R where u(t,x) = f(t,x)/µ(x). z µ(x)x +α z Lemma 4.3. Let us suppose that Φ is a smooth convex function such that (a,b) D Φ (a+b,b) is convex on {a+b 0,b 0}. Then, for any smooth function v, we have: Ent Φ z µ (v(t, )) K D Φ (v(t,x),v(t,x +z)) µ(x)x R R +α z for some K > 0. We can now state the main theorem ([6, Theorem 1]) of this section, its proof is a irect consequence of the two previous lemmas an the Gronwall lemma. ( ) Theorem 4.4. Consier a nonnegative initial ata f 0 such that Ent Φ f0µ µ <. We then have: ( ) ( ) f(t) t 0, Ent Φ µ e t/k Ent Φ f0 µ. µ µ In what follows, we enote µ(x) := x α. We now give a corollary of this theorem which gives the ecay property in the space L 2 ( µ 1/2 ) i.e L 2 ( x (+α)/2 ). Corollary 4.5. Consier a nonnegative initial ata f 0 such that f 0 µ f 0 L2 ( µ 1/2 ) is finite. Then, there exist λ > 0 an C > 0 such that: f(t) µ f 0 L2 ( µ 1/2 ) Ce λt f 0 µ f 0 L2 ( µ 1/2 ). Proof. Theorem 4.4 applie with Φ(s) = (s f 0 ) 2 gives the result in L 2 (µ 1/2 ) an the conclusion of the Corollary is a irect consequence of it because of Remark 4.1.

10 FRACTIONAL FOKKER-PLANCK EQUATION 9 5. Semigroup ecay in L 1 ( x k ) 5.1. Splitting of the operator. We woul like to get a splitting of our operator L into two operators which satisfies hypothesis of Theorem 3.1 with E = L 2 ( µ 1/2 ) an E = L 1 ( x k ) with k < α. In what follows, we enote m(x) := x k, k < α. Lemma 5.1. Consier a ( min(k,λ),0) where λ > 0 is efine in Corollary 4.5. There exist two operators A an B which satisfy the following conitions: (i) L = A+B, (ii) A B(L 2 ( µ 1/2 )) an A B(L 1 (m)), (iii) B a is issipative on L 2 ( µ 1/2 ) an L 1 (m). Proof. We are going to estimate the integral Lf sign(f)m with f a Schwartz function. The inequality obtaine will also hol for any f L 1 (m) because of the ensity of S(R ) in L 1 (m). We split the integral into two parts: Lf sign(f)m = I[f]sign(f)m+ iv(xf)sign(f)m R R R := T 1 +T 2. As far as T 1 is concerne, we introuce the function Φ(s) := s on R which is convex an its erivative is Φ (s) = sign(s). We also introuce the notation K(x) := x α. Let us o the following computation: (f(y) f(x))k(x y)y sign(f(x)) R ( = (f(y) f(x))φ (f(x))+φ(f(x)) φ(f(y)) ) K(x y)y R + (Φ(f(y)) Φ(f(x))) K(x y)y R ( f (y) f (x)) K(x y)y = I[ f ](x), R where the last inequality comes from the convexity of Φ. We hence euce that T 1 I[ f ]m = f I[m] = f m I[m] R R R m, because of Lemma 2.2. Let us now eal with T 2. Performing integrations by parts, we obtain: T 2 = iv(xf)sign(f)m R = f m+ x f signf m R R = f m+ x f m R R = f m f m f x m R R R = f m x m R m

11 10 ISABELLE TRISTANI We now introuce ψ m,1 := I[m]/m x m/m. Let us stuy the behavior of ψ m,1 at infinity. First, x m(x)/m(x) tens to k as x tens to infinity. Then, we prove that I[m](x)/m(x) tens to 0 as x tens to infinity. We use both representations (2.1) an (2.2) to split I[m](x) into two parts: I[m](x) = 1 (m(x+z)+m(x z) 2m(x)) K(z)z 2 z 1 + (m(y) m(x)) K(x y)y x y 1 := I 1 [m](x)+i 2 [m](x). Concerning I 1 [m], using a Taylor expansion, we obtain: m(x+z)+m(x z) 2m(x) sup D 2 m(x+z) z 2 z 1 from which we euce that (5.1) I 1 [m](x) C x k 2 C x k 2 z 2, z 1 z z +α 2. Concerning I 2 [m], let us introuce the function ψ(s) := s k/2 on R +. Using the fact that ψ is k/2-höler continuous on R + because k/2 1, we obtain for any x, y R : ψ(1+ x 2 ) ψ(1+ y 2 ) C x 2 y 2 k/2 for some C > 0. We euce the following inequalities: m(x) m(y) C x y k/2 ( x + y ) k/2 Finally, we obtain the following estimate on I 2 [m]: ( x k/2 (5.2) I 2 [m](x) C C x y k/2 ( x + x y + x ) k/2 ( 2C x y k/2 x k/2 + x y k). z 1 z z +α k/2 + z 1 ) z z +α k, where we notice that the integrals are convergent because k < α. Gathering (5.1) an (5.2), we euce that I[m]/m tens to 0 at infinity. Finally, we obtain: Lf signf m f mψ m,1 with lim ψ m,1(x) = k < 0. R R x We introuce the smooth function χ R (R > 0) which is nonnegative, supporte on x 2R an such that χ R (x) = 1 for x R. For any a > min(λ,k), we may fin M an R large enough so that (5.3) x R, ψ m,1 (x) Mχ R (x) a. Inee, if we choose R large enough such that for any x R, ψ m,1 (x) a an M := max x R ψ m,1 (x) a, we have (5.3).

12 FRACTIONAL FOKKER-PLANCK EQUATION 11 We then introuce A := Mχ R an B := L Mχ R. We finally obtain: (B a)f signf m = (L Mχ R a)f signf m R R (ψ m,1 Mχ R a) f m R 0, which implies that B a is issipative on L 1 (m). Let us now check that B a is issipative on L 2 ( µ 1/2 ): Bf f µ 1 = Lf f µ 1 M χ R f 2 µ 1 R R R Lf f µ 1 R = L(f µ f )f µ 1 + f Lµµ 1 R R = L(f µ f )(f µ f )µ 1 R λ f 2 L 2 (µ 1/2 ), where the last inequality comes from Corollary 4.5. We thus euce that B a is issipative on L 2 (µ 1/2 ) using that λ a. To conclue that we also have the issipativity of B a on L 2 ( µ 1/2 ), we use the fact that there exists C > 0 such that µ C µ on R. We can now conclue. This splitting L = A+B fulfills conitions (i), (ii) an (iii) of Lemma 5.1. Inee, it is immeiate to check assumption (ii) because A is a truncation operator Regularization properties of ( Ae Bt) ( n). We are now going to show that there exists n N such that ( Ae Bt) ( n) has a regularizing effect. In orer to get such a result, we are going to use the negative term in the computations one to get the issipativity of B. Let us state a result which is going to be useful to get an estimate on this negative term. Lemma 5.2 (Fractional Nash inequality). Consier α (0, 2). There exists a constant C > 0 such that for any g L 1 (R ) H α/2 (R ), we have: ( ) 2α ( g(x) 2 +α x C g(x) x R R R g(x) g(y) R x y +α yx 2 ) +α Proof. We use the Plancherel formula to get the following equality for any R > 0: ( ) g(x) 2 x = C ĝ(ξ) 2 ξ + ĝ(ξ) 2 ξ. R ξ R ξ R The first part of the integral can be boune as follows: ĝ(ξ) 2 ξ R ĝ 2 L R g 2 L 1. ξ R.

13 12 ISABELLE TRISTANI As far as the secon part is concerne, we use Lemma 2.3 ξ R ĝ(ξ) 2 ξ 1 R α ξ α ĝ(ξ) 2 ξ = C (g(x) g(y)) R R α R R x y +α 2 yx. We enote a = g 2 L 1 et b = C R (g(x) g(y)) R x y +α. 2 an the aim is to minimize the function φ(r) := ar +br α to get an optimal inequality. We compute φ (R) = 0 ar 1 1 αb = 0 R = Rα+1 ( ) 1 αb +α a an φ ( (αb a ) 1 ) +α = Ca α +α b +α, which conclues the proof. Let us now prove the following lemma which is the cornerstone of the proof of the regularizing effect of ( Ae Bt) ( n). We introuce the following measure: m 0 (x) := x k 0 with k 0 < min(k,α/2). Let us notice that this assumption on k 0 allows us to efine I[m p 0 ] for any p [1,2] an that m 0 satisfies L 2 ( µ 1/2 ) L q (m 0 ) for any q [1,2]. Lemma 5.3. There are b,c > 0 such that for any p an q, 1 p q 2, we have: t 0, Proof. For p [1,2], we enote e Bt f L q (m 0 ) Cebt ( 1 f L p q) p (m 0 ). 1 ψ m0,p := I[mp 0 ] pm p 0 t α + p 1 p an we introuce b R such that sup q [1,2] ψ m0,q b. Let us prove that for any p [1,2], we have: x (mp 0 ) pm p 0 (5.4) t 0, e Bt f L p (m 0 ) e bt f L p (m 0 ). We now o same kin of computations as in the proof of Lemma 5.1: Lf f p 1 signf m p 0 = I[f] f p 1 signf m p 0 + iv(xf) f p 1 signf m p 0 R R R := T 1 + T 2.

14 FRACTIONAL FOKKER-PLANCK EQUATION 13 As far as T 1 is concerne, we introuce the function Φ(x) := x p /p on R which is convex an its erivative is Φ (x) = x p 1 sign(x). Let us o the following computation: (f(y) f(x))k(x y)y f p 1 (x)sign(f(x)) R ( = (f(y) f(x))φ (f(x))+φ(f(x)) φ(f(y)) ) K(x y)y R + (Φ(f(y)) Φ(f(x))) K(x y)y R 1 R p ( f p (y) f p (x)) K(x y)y = 1 p I[ f p ](x), where the last inequality comes from the convexity of Φ. We hence euce that T 1 I[ f p ]m 0 = 1 f p I[m p 0 R p ] = f p m p 0 R R I[m p 0 ] pm p. 0 Concerning T 2, using an integration by part, we obtain: T 2 = Finally, the previous estimates imply that R [ p 1 x (mp 0 ) ] p pm p f p m p 0. 0 Bf f p 1 signf m p 0 (ψ m0,p Mχ R ) f p m p 0 b f p m p 0 R R R using the efinition of b. This implies the estimate (5.4). In orer to establish the gain of integrability estimate, we have to use the non positive term in a sharper way, i.e. not merely the fact that it is non-positive. It is enough to o that in the simplest case when p = 2. Let us consier a solution f t of the equation t f t = Bf t, f 0 = f L 2 (m 0 ). The previous computation involving the function Φ(x) is simpler in the case p = 2 an becomes: Bf f m 2 0 = 1 R R R f(x) f(y) R x y +α f(x) f(y) 2 x y 1 2 x y +α ym 2 0(x)x+ ym 2 0 (x)x+b f 2 m 2 0(ψ m0,2 Mχ R ) f 2 m 2 0

15 14 ISABELLE TRISTANI Let us eal with the negative part of the last inequality. f t (x) f t (y) 2 R x y 1 x y +α ym 2 0(x)x f t (x)m 0 (x) f t (y)m 0 (y)+f t (y)(m 0 (y) m 0 (x)) 2 = R x y 1 x y +α ym 2 0(x)x 1 f t (x)m 0 (x) f t (y)m 0 (y) 2 2 R x y 1 x y +α ym 2 0 (x)x m 0 (x) m 0 (y) 2 R x y 1 x y +α xft 2 (y)y 1 2 f t (x)m 0 (x) f t (y)m 0 (y) 2 R R x y +α ym 2 0(x)x 1 f t (x)m 0 (x) f t (y)m 0 (y) 2 2 R x y 1 x y +α ym 2 0(x)x m 0 (x) m 0 (y) 2 R x y 1 x y +α xft 2 (y)y We treat the first term using Lemma 5.2 with g = f t m 0 : (5.5) 2 ( f t (x)m 0 (x) f t (y)m 0 (y) R R x y +α yx C f t 2 m 2 0 R )+α ( R f t m 0 ) 2α. We cruely boun the secon term from above: (5.6) f t (x)m 0 (x) f t (y)m 0 (y) 2 R x y 1 x y +α ym 2 0(x)x ( f t (x)m 0 (x) 2 ) f t (y)m 0 (y) 2 C R x y 1 x y +α yx+ xy R +α x y 1 x y C f t 2 m 2 0. R Finally, the thir term is boune using the fact that sup B(y,1) m 0 2 Cm 2 0 (y): (5.7) m 0 (x) m 0 (y) 2 R x y 1 x y +α xft 2 (y)y x y 2 sup B(y,1) m 0 2 C R x y 1 x y +α xft 2 (y)y 1 C R z 1 z +α 2 zf2 t (y)m 2 0(y)y C ft 2 m 2 0. R

16 FRACTIONAL FOKKER-PLANCK EQUATION 15 Gathering (5.5), (5.6) et (5.7), we obtain: f t (x) f t (y) 2 R (5.8) x y 1 x y +α ym 2 0(x)x ( )+α ( ) C f t 2 m 2 2α ( ) 0 f t m 0 C ft 2 m 2 0, R R R for some constants C, C > 0. We introuce the following notations: X(t) := f t 2 L 2 (m 0 ) et Y(t) := f t L 1 (m 0 ). On the one han, if X 0 (2C /C) /α Y0 2, because of estimate (5.4), we have: t 0, X(t) 1/2 Ce bt X 1/2 0. We hence obtain t 0, X(t) 1/2 Ce bt Y 0. On the other han, we treat the case X 0 > (2C /C) /α Y0 2. By the previous step (5.8), we en up with the ifferential inequality (5.9) X(t) CY(t) 2α X(t) 1+ α +C X(t). t We also have from estimate (5.4): Y(t) Ce bt Y(0) for any t 0. So, we obtain for any t [0,1], Y(t) CY(0) changing the value of C. Putting this together with (5.9), we obtain: 2α (5.10) t [0,1], X(t) CY 0 X(t) 1+α +C X(t). t { Let us introuce τ := sup t [0,1] : X(s) (2C /C) /α Y0 }. 2, s [0,t] For any t ]0,τ[, we have 1/2CX(t) 1+α/ Y 2α/ 0 C X(t). Then, using (5.10), we obtain: t (0,τ), t X(t) 1 2α 2 CY 0 X(t) 1+α, which finally implies ( ) α C (5.11) t (0,τ), X(t) 2 Y 2α α 0 t. Moreover, because of estimate (5.4), we get: (5.12) t [τ,+ ), X(t) 1/2 Ce b(t τ) X(τ) 1/2 Ce b(t τ) ( 2C Therefore, gathering inequalities (5.11) an (5.12), we obtain: As a conclusion, we have t > 0, t > 0, X(t) 1 2 Ct 2α e bt Y 0. e Bt f L 2 (m 0 ) Cebt t 2α f L 1 (m 0 ). which means that the operator e Bt is continuous from L 1 (m 0 ) into L 2 (m 0 ). C ) 2α Y0. Let us now consier p an q, 1 p q, e Bt is continuous from L p (m 0 ) into L q (m 0 ) using Riesz-Thorin interpolation Theorem. Moreover, if we enote C ab the norm of e Bt : L a (m 0 ) L b (m 0 ), we get the following estimate: C pq C 2 2/p 22 C 2/q 1 11 C 2/p 2/q 12

17 16 ISABELLE TRISTANI an C 2 2/p 22 C 2/q 1 11 C 2/p 2/q 12 = Ce bt(2 2/p) e bt(2/q 1) e bt(2/p 2/q) t /(2α)(2/p 2/q) which yiels the result. = Cebt ( t 1 α p 1 q Using the same metho as in [7], we can euce the following corollary: ), Corollary 5.4. There exists a constant C such that for any p an q, 1 p q 2, we have: t 0, T l0 (t)f L q (m 0 ) C tl0 1 e bt ( )] where l 0 = E[ 1 α p 1 q +1. t α ( 1 p 1 q ) f L p (m 0 ) 5.3. Proof of the main result. As a conclusion, we can now apply Theorem 3.1 with E = L 2 ( µ 1/2 ) an E = L 1 (m). Hypothesis (1) comes from Corollary 4.5. Hypothesis (2)-(i) an (2)-(ii) come from Lemma 5.1. We can also prove that assumption (2)-(iii) is satisfie. Inee, we can check by an immeiate computation that we have the following estimate for any function f: Af L q (m) C Af L q (m 0 ). Moreover, we have that L p (m) L p (m 0 ) with continuous embeing(because k 0 < k). Usingthese two last facts an Corollary 5.4, we can euce that for any p an q, 1 p q 2, we have: (5.13) t 0, T l0 (t)f L q (m) C tl0 1 e bt ( 1 L p q) f p (m). 1 Moreover, we can show that Af L 2 ( µ 1/2 ) C Af L 2 (m). Finally, usingthe last estimate combine with (5.13) with p = 1, q = 2 an enoting γ := 2α E( 2α), we obtain: T l0 (t) L 1 (m) L 2 ( µ 1/2 ) Cebt t γ, with γ [0,1), which implies that (2)-(iii) is fulfille. We can conclue that Theorem 1.1 hols. Remark 5.5. To obtain a similar result as Theorem 1.1 in L p ( x k ) with p (1,2], we nee a very restrictive assumption: (1 1/p) < k < α. Inee, it implies that the limit at infinity of ψ m,p is negative, which allows us to get the issipativity of B a in L p ( x k ) for any a > (1 1/p) k. The rest of the proof can be one in the same way. t α

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