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On a representation of the inverse F -transform q Sabir Umarov1, Constantino Tsallis2,3 8 0 1 Department of Mathematics, Tufts University, Medford, MA 02155, USA 0 2 Centro Brasileiro de Pesquisas Fisicas, Xavier Sigaud 150, 22290-180 Rio de Janeiro-RJ, Brazil 2 3 Santa Fe Institute, 1399 Hyde Park Road, Santa Fe, NM 87501, USA n a J Abstract 8 A representation formula for the inverse q-Fourier transform is obtained in the class of functions = , where = f =ae−βx2, a>0, β >0 . ] G 1≤q<3Gq Gq { q } h c S e m InthispaperwefindarepresentationformulafortheinverseF -transforminaclassoffunctions - q t a defined below. Fq-transform is a useful tool in the study of limit processes in nonextensive G st statistical mechanics (see [1] and references therein). Note that, in this theory, random states are t. correlated in a special manner, and a knowledge on Fq-inverse of data is helpful in understanding a the nature of such correlations. m Throughout the paper we assume that 1 q < 3. The F -transform, called also q-Fourier - ≤ q d transform, of a nonnegative f(x) L (R1) is defined by the formula (see [2, 3]) 1 n ∈ o c F [f](ξ) = eixξ f(x)dx, (1) [ q Zsuppf q ⊗q 1 where isthesymbolof theq-product,andex isaq-exponential. Thereaderisreferredfordetails v ⊗q q 1 of q-algebra and q-functions to [4, 5, 6]. Fq coincides with the classic Fourier transform if q = 1. If 1 1 < q < 3 then F is a nonlinear mapping in L . The representation formula for the inverse, F 1, q 1 q− 3 1 is defined in the class of function of the form Ae−q1Bξ2, since it uses a specific operator I defined in . this class. The question on extension of this operator to wider classes is remaining a challenging 1 0 question. 8 The obvious equality eixξ f(x) = f(x)eixξ[f(x)]q−1, which holds for all x suppf, implies 0 q ⊗q q ∈ : the following lemma, which gives an expression for the q-Fourier transform without usage of the v q-product. i X r Proposition 0.1 The q-Fourier transform can be written in the form a F [f](ξ) = f(x)eixξ[f(x)]q−1dx. (2) q q Zsuppf Introduce the operator F [f](ξ) = f(x)e ixξ[f(x)]q−1dx. (3) q∗ −q Zsuppf 1 Forarbitrarynonnegativef L (R)bothoperators,F andF arecorrectlydefined. Moreover, ∈ 1 q q∗ sup F [f](ξ) f and sup F [f](ξ) f . (4) | q | ≤ k kL1 | q∗ | ≤ k kL1 ξ R1 ξ R1 ∈ ∈ Introduce the set of functions = f : f(x)= ae βx2, a> 0, β > 0 . (5) Gq { q− } Obviously, L for all q < 3. The set is fully identified by the triplet (q,a,β). We denote q 1 q G ⊂ G = (q,a,β) :q < 3,a > 0,β > 0 . R { } For any function f we have f( x) = f(x), so f is symmetric about the origin. Moreover, q ∈ G − √β it follows from the symmetry that F [f](ξ) = F [f](ξ). Further, a function f with a = , q q∗ ∈ Gq Cq where 2 π/2(cos t)13−−qqdt = 2√πΓ(1−1q) , < q < 1, √1 q 0 (3 q)√1 qΓ( 3−q ) −∞  − − − 2(1−q) Cq =  √√qπ2,1R0∞(1+y2)q−−11dy = √√qπΓ1(Γ2((3q−−q11))), q1=<1q,< 3. (6) is called a q-Gaussian, and−isRdenoted by Gq(β;x). −Thusq,−t1he set of all q-Gaussians forms a subset of . q G The following statement was proved in [2]. Proposition 0.2 Let 1 q < 3. For the q-Fourier transform of a q-Gaussian, the following ≤ formula holds: ξ2 F [G (β;x)](ξ) = e−4β2−qCq2(q−1) 3−2q. (7) q q q (cid:16) (cid:17) Assume a sequence q is given by k 2q k(q 1) 2 q = − − , < k < 1, (8) k 2 k(q 1) −∞ q 1 − − − − for q > 1, and q = 1 for all k = 0, 1,..., if q = 1. k ± It follows from this proposition the following result. Corollary 0.3 Let 1 q < 3 and k < 2 1. Then ≤ q 1 − − Fqk[Gqk(β;x)](ξ) = eq−kβ+k1+1ξ2, (9) where qk+1 = 31+−qqkk and βk+1 = 8β2−q3k−Cqq2kk(qk−1). Remark 0.4 It follows from (9) that aC F [ae βx2](ξ) = qk e Bξ2, qk q−k √β −qk+1 where B = a2(qk−1)(3−qk). 8β 2 Theorem 0.5 The operator F : is invertible. qk Gqk → Gqk+1 Proof. With the operator F : we associate the mapping defined as qk Gqk → Gqk+1 R → R (qk,a,β) → (qk+1,A,B), where A = a√Cqβk and B = a2(qk−81β)(3−qk). Consider the system of equations 1+q k = Q, 3 q k − aC qk = A, √β a2(qk−1)(3 qk) − = B 8β with respect to q ,a,β assuming that Q,A and B are given. The first equation is autonomous and k has a unique solution q = (3Q 1)/(Q+1). If the condition k < 2 1 is fulfilled then the other k − q 1− two equations have a unique solution as well, namely − 1 1 A√3 qk 2−qk A2(qk−1)(3 qk) 2−qk a = − , β = − . (10) 2Cqk√2B! 8Cq2k(qk−1)B ! It follows from (8) that Q and q are related as Q = q . Hence, the inverse mapping (F ) 1 : k k+1 q − exists and maps each element Ae Bξ2 to the element ae βx2 with a and β Gk+1 → Gk −qk+1 ∈ Gk+1 q−k ∈ Gk defined in (10). Now we find a representation formula for the inverse operator F 1. Denote by T the mapping q− T : (a,β) (A,B), where A = aCqk and B = a2(qk−1)(3−qk), as indicated above. We have seen → √β 8β that T is invertible and T 1 : (A,B) (a,β) with a and b in (10). Assume (a¯,β¯) = T 2(A,B) = − − → T 1(T 1(A,B)) = T 1(a,β).Further,weintroducetheoperatorI : defined − − − (qk+1,qk−1) Gqk+1 → Gqk−1 by the formula I [Ae Bξ2]= a¯e β¯ξ2. (11) (qk+1,qk−1) −qk+1 −qk−1 Consider the composition H = F I . By definition, it is clear that I : qk q∗k−1 ◦ (qk+1,qk−1) (qk+1,qk−1) . Since F : , we have H : . Let fˆ , that is Gqk+1 → Gqk−1 q∗k−1 Gqk−1 → Gqk qk Gqk+1 → Gqk ∈ Gk+1 fˆ(ξ) = Ae Bξ2. Then, taking into account the fact that supp fˆ = R1 if q 1, one obtains an q−k+1 ≥ explicit form of the operator H : qk H [fˆ(ξ)](x) = ∞ a¯e β¯ξ2 e ixξdξ = ∞ I [fˆ(ξ)] e ixξdξ. (12) qk −qk−1 ⊗qk−1 −qk−1 (qk+1,qk−1) ⊗qk−1 −qk−1 Z−∞(cid:16) (cid:17) Z−∞ Theorem 0.6 1. Let f . Then H F [f]= f; ∈ Gqk qk ◦ qk 2. Let f . Then F H [f]= f. ∈ Gqk+1 qk ◦ qk Proof. 1. We need to show validity of the equation F I F = , where is the qk−1◦ (qk+1,qk−1)◦ qk J J identity operator in . This equation is equivalent to T T 2 T = J (J is the identity operator Gqk ◦ − ◦ in with fixed q), which is correct by construction. R ′ ′ 2. Now the equation F F I = ( is the identity operator in ) is qk ◦ qk−1 ◦ (qk+1,qk−1) J J Gqk+1 equivalent to the identity T2 T 2 = J. − ◦ 3 Corollary 0.7 The operator H :G G is the inverse to the q Fourier transform: H = qk qk+1 → qk k− qk F 1. q−k Corollary 0.8 For q = 1 the inverse F 1 coincides with the classical inverse Fourier transform. q−k Proof. If q = 1, then by definition one has q = q = q = 1. We find a¯ and β¯ taking k k 1 k+1 (A,B) = (1,1). It follows from relationships (10) that (a−,β) = T 1(1,1) = ( 1 , 1). Again using − 2√π 4 (10) we obtain (a¯,β¯) = T 2(1,1) = T 1( 1 ,1) = ( 1 ,1). This means that I fˆ(ξ) = 1 fˆ(ξ). − − 2√π 4 2π (1,1) 2π Hence, the formula (12) takes the form 1 F 1[fˆ(ξ)](x) = ∞ fˆ(ξ)e ixξdx, 1− 2π − Z−∞ recovering the classic formula for the inverse Fourier transform. Summarizing, we have proved that, if fˆ(ξ) is a function in , where q with k < 2 1 is Gqk+1 k q 1 − defined in Eq. (8) for q [1,3), then − ∈ F 1[fˆ(ξ)](x) = ∞ I [fˆ(ξ)] e ixξdξ, (13) q−k (qk+1,qk−1) ⊗qk−1 −qk−1 Z−∞ with the operator I given in (11). This might constitute a first step for finding a repre- (qk+1,qk−1) sentation of F 1[fˆ(ξ)](x) for generic fˆ(ξ) L (R), which would be of great usefulness. q− ∈ 1 References [1] M. Gell-Mann and C. Tsallis, eds., Nonextensive Entropy - Interdisciplinary Applications (Oxford University Press, New York, 2004). [2] S. Umarov, C. Tsallis, S. Steinberg. A generalization of the central limit theorem consistent with nonextensive statistical mechanics. Arxiv: cond-mat/0603593 (version 14 Nov 2007). [3] S. Umarov and C. Tsallis, On multivariate generalizations of the q-central limit theorem consistent with nonextensive statistical mechanics, in Complexity, Metastability and Nonex- tensivity, eds. S. Abe, H.J. Herrmann, P. Quarati, A. Rapisarda and C. Tsallis, American Institute of Physics Conference Proceedings 965, 34 (New York, 2007). [4] L. Nivanen, A. Le Mehaute and Q.A. Wang, Generalized algebra within a nonextensive statistics, Rep. Math. Phys. 52, 437 (2003). [5] E.P. Borges, A possible deformed algebra and calculus inspired in nonextensive thermostatis- tics, Physica A 340, 95 (2004). [6] C. Tsallis, Nonextensive statistical mechanics, anomalous diffusion and central limit theo- rems, Milan Journal of Mathematics 73, 145 (2005). 4