Nonextensive Statistical Mechanics

136 4 Stochastic Dynamical Foundations of Nonextensive Statistical MechanicsF q [ f ](ξ) =∫ ∞−∞i ξ x[ f (x)]q−1dx eq f (x) (q ≥ 1) . (4.71)It is transparent that this transformation is, for q ≠ 1, nonlinear. Indeed, if wedo f (x) → λf (x), λ being any constant, we verify that F q [λf ](ξ) ≠ λF q [ f ](ξ)(q ≠ 1).This generalization of the standard Fourier transform (F 1 [ f ](ξ)) has a remarkableproperty: it transforms q-Gaussians into q-Gaussians. Indeed, we verifywhereandF q[ √ βA qβ 1 =e−β x2q](ξ) = e −β 1 ξ 2q 1, (4.72)q 1 = 1 + q3 − q , (4.73)3 − q8 β 2−q A 2(1−q)q, (4.74)with⎧2 √ ( )1π 1−q⎪⎨ (3−q) √ 1−q (3−q) if q < 1 ,2(q−1)√A q = π if q = 1 ,√(π 3−q)2(q−1)⎪⎩ √( ) if 1 < q < 3 .q−1 1q−1(4.75)It follows that the q-Fourier transform has the inverse transform in the set ofq-Gaussians (see [256]) (and for the (q,α)-stable distributions to be soon defined,as well). 77 Hilhorst [257] has recently produced an interesting example which is noninvertible. Considerf (x) = (λ/x) 1q−1if a < x < b, and zero otherwise; q > 1, 0 < a < b, andλ>0. Impositionof normalization straightforwardly yields λ =)] (b q−2 q−2 −(q−1).q−1 − aq−1 It immediately[q−1q−2follows that F q [ f ](ξ) = [1 + (1 − q)iξλ] 11−q . Therefore this solution is, for fixed q, one and thesame for a one-parameter family of normalized functions f (x). Indeed, for all (a, b) havingthesame λ, theq-Fourier transform is the same. Therefore, for this example, the inverse q-Fouriertransform does not exist in the sense that it does not yield a single function, but rather a familyof them. In other words, the q-Fourier transform is not invertible in the image of all probabilitydensity functions. To further understand the domain of impact of this example, let us consider amore general situation, namely F q ′[ f ](ξ) = ∫ [] 1ba dx 1 + (1 − q ′ )i ξ x[ f (x)] q′ −1 1−q ′ f (x)(q ′ > 1),i.e., F q ′[ f ](ξ) = ∫ [ba dx 1 + (1 − q ′ )i ξλ q′ −1 q−qq−1 x ′q−1] 11−q ′f (x). This integral can be expressed as a