Volume 9 (2008), Issue 3, Article 88, 23 pp. HEISENBERG-PAULI-WEYL UNCERTAINTY PRINCIPLE FOR THE RIEMANN-LIOUVILLE OPERATOR S. OMRI AND L.T. RACHDI D EPARTMENT OF M ATHEMATICS FACULTY OF S CIENCES OF T UNIS 2092 M ANAR 2 T UNIS T UNISIA . lakhdartannech.rachdi@fst.rnu.tn Received 29 March, 2008; accepted 08 August, 2008 Communicated by J.M. Rassias A BSTRACT. The Heisenberg-Pauli-Weyl inequality is established for the Fourier transform connected with the Riemann-Liouville operator.Also, a generalization of this inequality is proved. Lastly, a local uncertainty principle is studied. Key words and phrases: Heisenberg-Pauli-Weyl Inequality, Riemann-Liouville operator, Fourier transform, local uncertainty principle. 2000 Mathematics Subject Classification. 42B10, 33C45. 1. I NTRODUCTION Uncertainty principles play an important role in harmonic analysis and have been studied by many authors and from many points of view [8].These principles state that a function f and its Fourier transform fb cannot be simultaneously sharply localized. The theorems of Hardy, Morgan, Beurling, ... are established for several Fourier transforms in [4], [9], [13] and [14]. In this context, a remarkable Heisenberg uncertainty principle [10] states, according to Weyl [25] who assigned the result to Pauli, that for all square integrable functions f on Rn with respect to the Lebesgue measure, we have 2 Z Z Z 1 2 2 2 b 2 2 ξj |f (ξ)| dξ ≥ |f (x)| dx , j ∈ {1, . . . , n}. xj |f (x)| dx 4 Rn Rn Rn This inequality is called the Heisenberg-Pauli-Weyl inequality for the classical Fourier transform. Recently, many works have been devoted to establishing the Heisenberg-Pauli-Weyl inequality for various Fourier transforms, Rösler [21] and Shimeno [22] have proved this inequality for the Dunkl transform, in [20] Rösler and Voit have established an analogue of the HeisenbergPauli-Weyl inequality for the generalized Hankel transform. In the same context, Battle [3] has proved this inequality for wavelet states, and Wolf [26], has studied this uncertainty principle 100-08 2 S. O MRI AND L.T. R ACHDI for Gelfand pairs. We cite also De Bruijn [5] who has established the same result for the classical Fourier transform by using Hermite Polynomials, and Rassias [17, 18, 19] who gave several generalized forms for the Heisenberg-Pauli-Weyl inequality. In [2], the second author with others considered the singular partial differential operators defined by ∂ ∆1 = ∂x , ∆ = 2 ∂2 ∂r2 2α+1 ∂ r ∂r + − ∂2 ; ∂x2 (r, x) ∈]0, +∞[×R; α ≥ 0. and they associated to ∆1 and ∆2 the following integral transform, called the Riemann-Liouville operator, defined on C∗ (R2 ) (the space of continuous functions on R2 , even with respect to the first variable) by α R1 R1 √ 1 2 , x + rt (1 − t2 )α− 2 (1 − s2 )α−1 dtds; if α > 0, f rs 1 − t π −1 −1 Rα (f )(r, x) = √ R1 π1 −1 f r 1 − t2 , x + rt √ dt 2 ; if α = 0. (1−t ) In addition, a convolution product and a Fourier transform Fα connected with the mapping Rα have been studied and many harmonic analysis results have been established for the Fourier transform Fα (Inversion formula, Plancherel formula, Paley-Winer and Plancherel theorems, ...). Our purpose in this work is to study the Heisenberg-Pauli-Weyl uncertainty principle for the Fourier transform Fα connected with Rα .More precisely, using Laguerre and Hermite polynomials we establish firstly the Heisenberg-Pauli-Weyl inequality for the Fourier transform Fα , that is • For all f ∈ L2 (dνα ), we have Z 0 +∞ Z 12 (r + x )|f (r, x)| dνα (r, x) 2 2 2 R Z Z × 12 (µ + 2λ )|Fα (f )(µ, λ)| dγα (µ, λ) 2 2 2 Γ+ 2α + 3 ≥ 2 Z +∞ 0 Z |f (r, x)| dνα (r, x) , 2 R with equality if and only if f (r, x) = Ce −r 2 +x2 2t2 0 ; C ∈ C, t0 > 0, where • dνα (r, x) is the measure defined on R+ × R by dνα (r, x) = r2α+1 √ dr ⊗ dx. 2α Γ(α + 1) 2π • dγα (µ, λ) is the measure defined on the set Γ+ = R+ × R ∪ (it, x); (t, x) ∈ R+ × R; t ≤ |x| , J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ H EISENBERG -PAULI -W EYL U NCERTAINTY P RINCIPLE 3 by Z Z 1 √ g(µ, λ)dγα (µ, λ) = α 2 Γ(α + 1) 2π Γ+ +∞ Z Z 0 g(µ, λ)(µ2 + λ2 )α µdµdλ R Z Z ! |λ| g(iµ, λ)(λ2 − µ2 )α µdµdλ . + 0 R Next, we give a generalization of the Heisenberg-Pauli-Weyl inequality, that is • For all f ∈ L2 (dνα ), a, b ∈ R; a, b ≥ 1 and η ∈ R such that ηa = (1 − η)b, we have +∞ Z 0 Z η2 (r + x ) |f (r, x)| dνα (r, x) 2 2 a 2 R Z Z × 1−η 2 (µ + 2λ ) |Fα (f )(µ, λ)| dγα (µ, λ) 2 2 b 2 Γ+ ≥ 2α + 3 2 aη Z +∞ Z 0 12 |f (r, x)| dνα (r, x) , 2 R with equality if and only if a = b = 1 and f (r, x) = Ce −r 2 +x2 2t2 0 C ∈ C; ; t0 > 0. In the last section of this paper, building on the ideas of Faris [7], and Price [15, 16], we develop a family of inequalities in their sharpest forms, which constitute the principle of local uncertainty. Namely, we have established the following main results • For all real numbers ξ; 0 < ξ < 2α+3 , there exists a positive constant Kα,ξ such that for 2 2 all f ∈ L (dνα ), and for all measurable subsets E ⊂ Γ+ ; 0 < γα (E) < +∞, we have Z Z 2 |Fα (f )(µ, λ)| dγα (µ, λ) < Kα,ξ 2ξ γα (E) 2α+3 +∞ Z E 0 Z (r2 + x2 )ξ |f (r, x)|2 dνα (r, x). R , there exists a positive constant Mα,ξ such that for all • For all real number ξ; ξ > 2α+3 2 f ∈ L2 (dνα ), and for all measurable subsets E ⊂ Γ+ ; 0 < γα (E) < +∞, we have Z Z |Fα (f )(µ, λ)|2 dγα (µ, λ) < Mα,ξ γα (E) Z E +∞ Z 0 Z +∞ R Z × 0 2ξ−2α−3 2ξ |f (r, x)|2 dνα (r, x) 2α+3 2ξ (r + x ) |f (r, x)| dνα (r, x) , 2 2 ξ 2 R where Mα,ξ is the best (the smallest) constant satisfying this inequality. J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ 4 S. O MRI AND L.T. R ACHDI 2. T HE F OURIER T RANSFORM A SSOCIATED WITH O PERATOR THE R IEMANN -L IOUVILLE It is well known [2] that for all (µ, λ) ∈ C2 , the system ∆1 u(r, x) = −iλu(r, x), ∆2 u(r, x) = −µ2 u(r, x), u(0, 0) = 1, ∂u (0, x) = 0, ∀x ∈ R, ∂r admits a unique solution ϕµ,λ , given by (2.1) ∀(r, x) ∈ R2 ; where jα (x) = 2α Γ(α + 1) p ϕµ,λ (r, x) = jα r µ2 + λ2 e−iλx , +∞ x 2n X (−1)n Jα (x) = Γ(α + 1) , xα n!Γ(α + n + 1) 2 n=0 and Jα is the Bessel function of the first kind and index α [6, 11, 12, 24]. The modified Bessel function jα has the following integral representation [1, 11], for all µ ∈ C, and r ∈ R we have R √2Γ(α+1)1 1 (1 − t2 )α− 12 cos(rµt)dt, if α > − 1 ; 2 πΓ(α+ 2 ) 0 jα (rµ) = cos(rµ), if α = − 12 . In particular, for all r, s ∈ R, we have (2.2) Z 1 1 2Γ(α + 1) |jα (rs)| ≤ √ (1 − t2 )α− 2 | cos(rst)|dt 1 πΓ α + 2 0 Z 1 1 2Γ(α + 1) (1 − t2 )α− 2 dt = 1. ≤√ 1 πΓ(α + 2 ) 0 From the properties of the Bessel function, we deduce that the eigenfunction ϕµ,λ satisfies the following properties • (2.3) sup |ϕµ,λ (r, x)| = 1, (r,x)∈R2 if and only if (µ, λ) belongs to the set Γ = R2 ∪ {(it, x); (t, x) ∈ R2 ; |t| ≤ |x|}. • The eigenfunction ϕµ,λ has the following Mehler integral representation R1 R1 √ 1 απ −1 −1 cos(µrs 1 − t2 )e−iλ(x+rt) (1 − t2 )α− 2 (1 − s2 )α−1 dtds; if α > 0, ϕµ,λ (r, x) = 1 R 1 cos(rµ√1 − t2 )e−iλ(x+rt) √ dt ; if α = 0. π −1 1−t2 In [2], using this integral representation, the authors have defined the Riemann-Liouville integral transform associated with ∆1 , ∆2 by √ α R1 R1 1 π −1 −1 f rs 1 − t2 , x + rt (1 − t2 )α− 2 (1 − s2 )α−1 dtds; if α > 0, Rα (f )(r, x) = √ R 1 1 f r 1 − t2 , x + rt √ dt ; if α = 0. π −1 2 (1−t ) J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ H EISENBERG -PAULI -W EYL U NCERTAINTY P RINCIPLE 5 where f is a continuous function on R2 , even with respect to the first variable. The transform Rα generalizes the "mean operator" defined by Z 2π 1 R0 (f )(r, x) = f (r sin θ, x + r cos θ)dθ. 2π 0 In the following we denote by • dνα the measure defined on R+ × R, by dνα (r, x) = r2α+1 √ dr ⊗ dx. 2α Γ(α + 1) 2π • Lp (dνα ) the space of measurable functions f on R+ × R such that Z +∞ Z p1 p kf kp,να = |f (r, x)| dνα (r, x) < ∞, if p ∈ [1, +∞[, 0 R kf k∞,να = ess sup |f (r, x)| < ∞, if p = +∞. (r,x)∈R+ ×R • h / iνα the inner product defined on L2 (dνα ) by Z +∞ Z hf /giνα = f (r, x)g(r, x)dνα (r, x). 0 R • Γ+ = R+ × R ∪ (it, x); (t, x) ∈ R+ × R; t ≤ |x| . • BΓ+ the σ-algebra defined on Γ+ by BΓ+ = {θ−1 (B), B ∈ B(R+ × R)}, (2.4) where θ is the bijective function defined on the set Γ+ by p θ(µ, λ) = µ2 + λ2 , λ . • dγα the measure defined on BΓ+ by ∀ A ∈ BΓ+ ; γα (A) = να (θ(A)) (2.5) p • L (dγα ) the space of measurable functions f on Γ+ , such that Z Z p1 kf kp,γα = |f (µ, λ)|p dγα (µ, λ) < ∞, if p ∈ [1, +∞[, Γ+ kf k∞,γα = ess sup |f (µ, λ)| < ∞, if p = +∞. (µ,λ)∈Γ+ • h / iγα the inner product defined on L2 (dγα ) by Z Z hf /giγα = f (µ, λ)g(µ, λ)dγα (µ, λ). Γ+ Then, we have the following properties. Proposition 2.1. i) For all non negative measurable functions g on Γ+ , we have Z +∞ Z Z Z 1 √ (2.6) g(µ, λ)dγα (µ, λ) = g(µ, λ)(µ2 + λ2 )α µdµdλ α 2 Γ(α + 1) 2π Γ+ 0 R ! Z Z |λ| + g(iµ, λ)(λ2 − µ2 )α µdµdλ . R J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. 0 http://jipam.vu.edu.au/ 6 S. O MRI AND L.T. R ACHDI In particular µ2 + λ2 dγα (µ, λ). 2(α + 1) ii) For all measurable functions f on R+ × R, the function f ◦ θ is measurable on Γ+ . Furthermore, if f is non negative or an integrable function on R+ × R with respect to the measure dνα , then we have Z Z Z +∞ Z (2.8) (f ◦ θ)(µ, λ)dγα (µ, λ) = f (r, x)dνα (r, x). dγα+1 (µ, λ) = (2.7) Γ+ 0 R In the following, we shall define the Fourier transform Fα associated with the operator Rα and we give some properties that we use in the sequel. Definition 2.1. The Fourier transform Fα associated with the Riemann-liouville operator Rα is defined on L1 (dνα ) by Z +∞ Z (2.9) ∀(µ, λ) ∈ Γ; Fα (f )(µ, λ) = f (r, x)ϕµ,λ (r, x)dνα (r, x). 0 R By the relation (2.3), we deduce that the Fourier transform Fα is a bounded linear operator from L1 (dνα ) into L∞ (dγα ), and that for all f ∈ L1 (dνα ), we have kFα (f )k∞,γα ≤ kf k1,να . (2.10) Theorem 2.2 (Inversion formula). Let f ∈ L1 (dνα ) such that Fα (f ) ∈ L1 (dγα ), then for almost every (r, x) ∈ R+ × R, we have Z Z f (r, x) = Fα (f )(µ, λ)ϕµ,λ (r, x)dγα (µ, λ). Γ+ Theorem 2.3 (Plancherel). The Fourier transform Fα can be extended to an isometric isomorphism from L2 (dνα ) onto L2 (dγα ). In particular, for all f, g ∈ L2 (dνα ), we have the following Parseval’s equality Z Z Z +∞ Z (2.11) Fα (f )(µ, λ)Fα (g)(µ, λ)dγα (µ, λ) = f (r, x)g(r, x)dνα (r, x). Γ+ 0 R 3. H ILBERT BASIS OF THE S PACES L2 (dνα ), AND L2 (dγα ) In this section, using Laguerre and Hermite polynomials, we construct a Hilbert basis of the spaces L2 (dνα ) and L2 (dγα ), and establish some intermediate results that we need in the next section. It is well known [11, 23] that for every α ≥ 0, the Laguerre polynomials Lαm are defined by the following Rodriguez formula 1 r −α dm m+α −r Lαm (r) = er r e ; m ∈ N. m! drm Also, the Hermite polynomials are defined by the Rodriguez formula n n x2 d −x2 Hn (x) = (−1) e e ; n ∈ N. dxn Moreover, the families (s ) ) (s m! 1 α √ Hn L and Γ(α + m + 1) m 2n n! π m∈N J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. n∈N http://jipam.vu.edu.au/ H EISENBERG -PAULI -W EYL U NCERTAINTY P RINCIPLE 7 2 are respectively a Hilbert basis of the Hilbert spaces L2 (R+ , e−r rα dr) and L2 (R, e−x dx). Therefore the families ) (s ) (s x2 1 2α+1 Γ(α + 1)m! − r2 α 2 √ e− 2 Hn e 2 Lm (r ) and n Γ(α + m + 1) 2 n! π n∈N m∈N 2α+1 are respectively a Hilbert basis of the Hilbert spaces L2 R+ , 2αrΓ(α+1) dr and L2 R, √dx2π , n o α hence the family em,n defined by (m,n)∈N2 eαm,n (r, x) = 2α+1 Γ(α + 1)m! n− 12 2 ! 12 n!Γ(m + α + 1) e− r 2 +x2 2 Lαm (r2 )Hn (x), is a Hilbert basis of the space L2 (dνα ). n o α Using the relation (2.8), we deduce that the family ξm,n (m,n)∈N2 α+1 α (µ, λ) = (eαm,n ◦ θ)(µ, λ) = ξm,n ! 12 2 Γ(α + 1)m! n− 12 n!Γ(m + α + 1) 2 e− , defined by µ2 +2λ2 2 Lαm (µ2 + λ2 )Hn (λ), is a Hilbert basis of the space L2 (dγα ), where θ is the function defined by the relation (2.4). In the following, we agree that the Laguerre and Hermite polynomials with negative index are zero. Proposition 3.1. For all (m, n) ∈ N2 , (r, x) ∈ R+ × R and (µ, λ) ∈ Γ+ , we have (3.1) xeαm,n (r, x) (3.2) α λξm,n (µ, λ) (3.3) r2 eα+1 m,n (r, x) = (3.4) r n+1 α e (r, x) + 2 m,n+1 r n α e (r, x). 2 m,n−1 r n+1 α ξ (µ, λ) + 2 m,n+1 r n α ξ (µ, λ). 2 m,n−1 = = p p 2(α + 1)(α + m + 1)eαm,n (r, x) − 2(α + 1)(m + 1)eαm+1,n (r, x). α+1 (µ, λ) = (µ2 + λ2 )ξm,n p α 2(α + 1)(α + m + 1)ξm,n (µ, λ) p α − 2(α + 1)(m + 1)ξm+1,n (µ, λ). Proof. We know [11] that the Hermite polynomials satisfy the following recurrence formula Hn+1 (x) − 2xHn (x) + 2nHn−1 (x) = 0; J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. n ∈ N, http://jipam.vu.edu.au/ 8 S. O MRI AND L.T. R ACHDI Therefore, for all (r, x) ∈ R+ × R, we have xeαm,n (r, x) = 2α+1 Γ(α + 1)m! ! 12 n− 12 e− r 2 +x2 2 Lαm (r2 )xHn (x) 2 n!Γ(m + α + 1) ! 12 r 2 2 2α+1 Γ(α + 1)m! n+1 − r +x 2 = e Lαm (r2 )Hn+1 (x) 1 n+ 2 2 2 (n + 1)!Γ(m + α + 1) ! 12 r α+1 |(r,x)|2 2 Γ(α + 1)m! n − 2 e Lαm (r2 )Hn−1 (x) + 3 n− 2 2 2 (n − 1)!Γ(m + α + 1) r r n+1 α n α = em,n+1 (r, x) + e (r, x) 2 2 m,n−1 α and it is obvious that the same relation holds for the elements ξm,n . On the other hand α+2 r2 eα+1 m,n (r, x) = 2 Γ(α + 2)m! n− 12 n!Γ(m + α + 2) 2 ! 12 e− r 2 +x2 2 2 r2 Lα+1 m (r )Hn (x). However, the Laguerre polynomials satisfy the following recurrence formulas (m + 1)Lαm+1 (r) + (r − α − 2m − 1)Lαm (r) + (m + α)Lαm−1 (r) = 0; m ∈ N, and α+1 α Lα+1 m (r) − Lm−1 (r) = Lm (r); m ∈ N. Hence, we deduce that r2 eα+1 m,n (r, x) = 2α+2 Γ(α + 2)m! n− 12 2 n!Γ(m + α + 2) ! 12 e− r 2 +x2 2 Hn (x) 2 α+1 2 α+1 2 × (α + 2m + 2)Lα+1 m (r ) − (m + 1)Lm+1 (r ) − (α + m + 1)Lm−1 (r ) ! 12 2 2 2α+2 Γ(α + 2)m! − r +x 2 (α + m + 1)e Lαm (r2 )Hn (x) = n− 12 2 n!Γ(m + α + 2) ! 12 2 2 2α+2 Γ(α + 2)m! − r +x 2 − (m + 1)e Lαm+1 (r2 )Hn (x) 1 n− 2 2 n!Γ(m + α + 2) p p = 2(α + 1)(α + m + 1)eαm,n (r, x) − 2(α + 1)(m + 1)eαm+1,n (r, x). Proposition 3.2. For all (m, n) ∈ N2 , and (µ, λ) ∈ Γ+ , we have (3.5) α Fα (eαm,n )(µ, λ) = (−i)2m+n ξm,n (µ, λ). J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ H EISENBERG -PAULI -W EYL U NCERTAINTY P RINCIPLE 9 Proof. It is clear that for all (m, n) ∈ N2 , the function eαm,n belongs to the space L1 (dνα ) ∩ L2 (dνα ), hence by using Fubini’s theorem, we get Fα (eαm,n )(µ, λ) = 2α+1 Γ(α + 1)m! ! 12 1 2n− 2 n!Γ(m + α + 1) Z +∞ p r2 × e− 2 Lαm (r2 )jα r µ2 + λ2 r2α+1 dr 2α Γ(α + 1) 0 Z 2 dx − x2 −iλx × e Hn (x) √ , 2π R and then the required result follows from the following equalities [11]: Z +∞ y r α α √ ∀m ∈ N; e− 2 Lαm (r)Jα ( ry)r 2 dr = (−1)m 2e− 2 y 2 Lαm (y), 0 and √ y2 x2 eixy e− 2 Hn (x)dx = in 2πe− 2 Hn (y), Z ∀n ∈ N; R where Jα denotes the Bessel function of the first kind and index α defined for all x > 0 by +∞ x 2n+α X (−1)n Jα (x) = . n!Γ(α + n + 1) 2 n=0 Proposition 3.3. Let f ∈ L2 (dνα ) ∩ L2 (dνα+1 ) such that Fα (f ) ∈ L2 (dγα+1 ), then for all (m, n) ∈ N2 , we have s s α + m + 1 m+1 (3.6) hf /eα+1 hf /eαm,n iνα − hf /eαm+1,n iνα , m,n iνα+1 = 2(α + 1) 2(α + 1) and s α+1 (3.7) hFα (f )/ξm,n iγα+1 = α+m+1 (−i)2m+n hf /eαm,n iνα 2(α + 1) s m+1 + (−i)2m+n hf /eαm+1,n iνα . 2(α + 1) Proof. We have hf /eα+1 m,n iνα+1 Z +∞ Z f (r, x)eα+1 m,n (r, x)dνα+1 (r, x) 0 R Z +∞ Z 1 = f (r, x)r2 eα+1 m,n (r, x)dνα (r, x) 2(α + 1) 0 R 1 = hf /r2 eα+1 m,n iνα , 2(α + 1) = hence by using the relation (3.3), we deduce that s s α + m + 1 m+1 hf /eα+1 hf /eαm,n iνα − hf /eαm+1,n iνα . m,n iνα+1 = 2(α + 1) 2(α + 1) J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ 10 S. O MRI AND L.T. R ACHDI In the same manner, and by virtue of the relation (2.7), we have Z Z α+1 α+1 hFα (f )/ξm,n iγα+1 = Fα (f )(µ, λ)ξm,n (µ, λ)dγα+1 (µ, λ) Γ+ Z Z 1 α+1 = Fα (f )(µ, λ)(µ2 + λ2 )ξm,n (µ, λ)dγα (µ, λ), 2(α + 1) Γ+ using the relations (3.4) and (3.5), we deduce that Z Z √ 1 α+1 α hFα (f )/ξm,n iγα+1 = p Fα (f )(µ, λ) α + m + 1ξm,n (µ, λ) 2(α + 1) Γ+ √ α − m + 1ξm+1,n (µ, λ) dγα (µ, λ) s α+m+1 hFα (f )/(i)2m+n Fα (eαm,n )iγα = 2(α + 1) s m+1 − hFα (f )/(i)2m+2+n Fα (eαm+1,n )iγα , 2(α + 1) hence, according to the Parseval’s equality (2.11), we obtain s α+m+1 α+1 hFα (f )/ξm,n iγα+1 = (−i)2m+n hf /eαm,n iνα 2(α + 1) s m+1 + (−i)2m+n hf /eαm+1,n iνα . 2(α + 1) 4. H EISENBERG -PAULI -W EYL I NEQUALITY FOR THE F OURIER T RANSFORM Fα In this section, we will prove the main result of this work, that is the Heisenberg-Pauli-Weyl inequality for the Fourier transform Fα connected with the Riemann-Liouville operator Rα . Next we give a generalization of this result, for this we need the following important lemma. Lemma 4.1. Let f ∈ L2 (dνα ), such that k|(r, x)|f k2,να < +∞ and 1 k(µ2 + 2λ2 ) 2 Fα (f )k2,γα < +∞, then (4.1) 1 k|(r, x)|f k22,να + k(µ2 + 2λ2 ) 2 Fα (f )k22,γα = +∞ X 2α + 4m + 2n + 3 |am,n |2 , m,n=0 where am,n = hf /eαm,n iνα ; (m, n) ∈ N2 . Proof. Let f ∈ L2 (dνα ), such that ∀(r, x) ∈ R+ × R; f (r, x) = +∞ X am,n eαm,n (r, x), m,n=0 and assume that k|(r, x)|f k2,να < +∞ 1 and k(µ2 + 2λ2 ) 2 Fα (f )k2,γα < +∞, J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ H EISENBERG -PAULI -W EYL U NCERTAINTY P RINCIPLE 11 then the functions (r, x) 7−→ rf (r, x) and (r, x) 7−→ xf (r, x) belong to the space L2 (dνα ), in particular f ∈ L2 (dνα ) ∩ L2 (dνα+1 ). In the same manner, the functions 1 (µ, λ) 7−→ (µ2 + λ2 ) 2 Fα (f )(µ, λ), (µ, λ) 7−→ λFα (f )(µ, λ) and belong to the space L2 (dγα ). In particular, by the relation (2.7), we deduce that Fα (f ) ∈ L2 (dγα ) ∩ L2 (dγα+1 ), and we have Z +∞ Z 2 krf k2,να = r2 |f (r, x)|2 dνα (r, x) 0 R = 2(α + 1)kf k22,να+1 +∞ X 2 hf /eα+1 = 2(α + 1) m,n iνα+1 , m,n=0 hence, according to the relation (3.6), we obtain +∞ X √ √ α + m + 1am,n − m + 1am+1,n 2 . = krf k22,να (4.2) m,n=0 Similarly, we have kxf k22,να Z +∞ Z x2 |f (r, x)|2 dνα (r, x) = = 0 +∞ X R +∞ X hxf /eαm,n iνα 2 = hf /xeαm,n iνα 2 , m,n=0 m,n=0 and by the relation (3.1), we get kxf k22,να (4.3) 2 r +∞ r X n n+1 = am,n+1 + am,n−1 . 2 2 m,n=0 By the same arguments, and using the relations (3.2), (3.7) and the Parseval’s equality (2.11), we obtain (4.4) 2 2 k(µ + λ ) 1 2 Fα (f )k22,γα +∞ X √ √ α + m + 1am,n + m + 1am+1,n 2 , = m,n=0 and (4.5) kλFα (f )k22,γα 2 r +∞ r X n n+1 am,n+1 − am,n−1 . = 2 2 m,n=0 Combining now the relations (4.2), (4.3), (4.4) and (4.5), we deduce that 1 k|(r, x)|f k22,να + k(µ2 + 2λ2 ) 2 Fα (f )k22,γα 1 = krf k22,να + k(µ2 + λ2 ) 2 Fα (f )k22,γα + kxf k22,να + kλFα (f )k22,γα J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ 12 S. O MRI AND L.T. R ACHDI =2 +∞ X (α + m + 1)|am,n |2 + (m + 1)|am+1,n |2 m,n=0 +∞ X n+1 n 2 2 +2 |am,n+1 | + |am,n−1 | 2 2 m,n=0 +∞ +∞ X X n+1 n 2 =2 (α + m + 1)|am,n | + 2 m|am,n | + 2 |am,n | + 2 |am,n |2 2 2 m,n=0 m,n=0 m,n=0 m,n=0 +∞ X +∞ X = 2 +∞ X 2 2α + 4m + 2n + 3 |am,n |2 . m,n=0 Remark 1. From the relation (4.1), we deduce that for all f ∈ L2 (dνα ), we have (4.6) 1 k|(r, x)|f k22,να + k(µ2 + 2λ2 ) 2 Fα (f )k22,γα ≥ (2α + 3)kf k22,να , with equality if and only if ∀(r, x) ∈ R+ × R; f (r, x) = Ce− r 2 +x2 2 ; C ∈ C. Lemma 4.2. Let f ∈ L2 (dνα ) such that, k|(r, x)|f k2,να < +∞ 1 k(µ2 + 2λ2 ) 2 Fα (f )k2,γα < +∞, and then 1) For all t > 0, 1 1 k|(r, x)|f k22,να + t2 k(µ2 + 2λ2 ) 2 Fα (f )k22,γα ≥ (2α + 3)kf k22,να . 2 t 2) The following assertions are equivalent 1 2α + 3 i) k|(r, x)|f k2,να k(µ2 + 2λ2 ) 2 Fα (f )k2,γα = kf k22,να . 2 ii) There exists t0 > 0, such that 1 k|(r, x)|ft0 k22,να + k(µ2 + 2λ2 ) 2 Fα (ft0 )k22,γα = (2α + 3)kft0 k22,να , where ft0 (r, x) = f (t0 r, t0 x). Proof. 1) Let f ∈ L2 (dνα ) satisfy the hypothesis.For all t > 0 we put ft (r, x) = f (tr, tx), and then by a simple change of variables, we get 1 (4.7) kft k22,να = 2α+3 kf k22,να , t and 1 (4.8) k|(r, x)|ft k22,να = 2α+5 k|(r, x)|f k22,να . t For all (µ, λ) ∈ Γ, 1 µ λ (4.9) Fα (ft )(µ, λ) = 2α+3 Fα (f ) , , t t t and by using the relation (2.6), we deduce that 1 1 1 (4.10) k(µ2 + 2λ2 ) 2 Fα (ft )k22,γα = 2α+1 k(µ2 + 2λ2 ) 2 Fα (f )k22,γα . t J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ H EISENBERG -PAULI -W EYL U NCERTAINTY P RINCIPLE 13 Then, the desired result follows by replacing f by ft in the relation (4.6). 2) Let f ∈ L2 (dνα ); f 6= 0. • Assume that 1 k|(r, x)|f k2,να k(µ2 + 2λ2 ) 2 Fα (f )k2,γα = 2α + 3 kf k22,να . 2 1 By Theorem 2.2, we have k(µ2 + 2λ2 ) 2 Fα (f )k2,γα 6= 0, then for s k|(r, x)|f k2,να t0 = , 1 k(µ2 + 2λ2 ) 2 Fα (f )k2,γα we have 1 1 k|(r, x)|f k22,να + t20 k(µ2 + 2λ2 ) 2 Fα (f )k22,γα = (2α + 3)kf k22,να , 2 t0 and this is equivalent to 1 k|(r, x)|ft0 k22,να + k(µ2 + 2λ2 ) 2 Fα (ft0 )k22,γα = (2α + 3)kft0 k22,να . • Conversely, suppose that there exists t1 > 0, such that 1 k|(r, x)|ft1 k22,να + k(µ2 + 2λ2 ) 2 Fα (ft1 )k22,γα = (2α + 3)kft1 k22,να . This is equivalent to 1 1 (4.11) k|(r, x)|f k22,να + t21 k(µ2 + 2λ2 ) 2 Fα (f )k22,γα = (2α + 3)kf k22,να . 2 t1 However, let h be the function defined on ]0, +∞[, by 1 1 h(t) = 2 k|(r, x)|f k22,να + t2 k(µ2 + 2λ2 ) 2 Fα (f )k22,γα , t then, the minimum of the function h is attained at the point s k|(r, x)|f k2,να t0 = 1 k(µ2 + 2λ2 ) 2 Fα (f )k2,γα and 1 h(t0 ) = 2k|(r, x)|f k2,να k(µ2 + 2λ2 ) 2 Fα (f )k2,γα . Thus by 1) of this lemma, we have 1 h(t1 ) ≥ h(t0 ) = 2k|(r, x)|f k2,να k(µ2 + 2λ2 ) 2 Fα (f )k2,γα ≥ (2α + 3)kf k22,να . According to the relation (4.11), we deduce that 1 h(t1 ) = h(t0 ) = 2k|(r, x)|f k2,να k(µ2 + 2λ2 ) 2 Fα (f )k2,γα = (2α + 3)kf k22,να . Theorem 4.3 (Heisenbeg-Pauli-Weyl inequality). For all f ∈ L2 (dνα ), we have (4.12) (2α + 3) kf k22,να 2 1 k|(r, x)|f k2,να k(µ2 + 2λ2 ) 2 Fα (f )k2,γα ≥ with equality if and only if ∀(r, x) ∈ R+ × R; f (r, x) = Ce J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. −r 2 +x2 2t2 0 ; t0 > 0, C ∈ C. http://jipam.vu.edu.au/ 14 S. O MRI AND L.T. R ACHDI 1 Proof. It is obvious that if f = 0, or if k|(r, x)|f k2,να = +∞, or k(µ2 + 2λ2 ) 2 Fα (f )k2,γα = +∞, then the inequality (4.12) holds. 1 Let us suppose that k|(r, x)|f k2,να + k(µ2 + 2λ2 ) 2 Fα (f )k2,γα < +∞, and f 6= 0. By 1) of Lemma 4.2, we have for all t > 0 1 2 2 2 2 12 k|(r, x)|f k + t k(µ + 2λ ) Fα (f )k22,γα ≥ (2α + 3)kf k22,να , 2,ν α 2 t and the result follows if we pick s k|(r, x)|f k2,να t= . 1 2 k(µ + 2λ2 ) 2 Fα (f )k2,γα By 2) of Lemma 4.2, we have 1 k|(r, x)|f k2,να k(µ2 + 2λ2 ) 2 Fα (f )k2,γα = 2α + 3 kf k22,να , 2 if and only if there exists t0 , such that 1 k|(r, x)|ft0 k22,να + k(µ2 + 2λ2 ) 2 Fα (ft0 )k22,γα = (2α + 3)kft0 k22,να , and according to Remark 1, this is equivalent to ft0 (r, x) = Ce− r 2 +x2 2 which means that f (r, x) = Ce −r 2 +x2 2t2 0 ; C ∈ C, ; C ∈ C. The following result gives a generalization of the Heisenberg-Pauli-Weyl inequality. Theorem 4.4. Let a, b ≥ 1 and η ∈ R such that ηa = (1 − η)b, then for all f ∈ L2 (dνα ) we have aη 2α + 3 η 1−η a 2 2 2b k|(r, x)| f k2,να k(µ + 2λ ) Fα (f )k2,γα ≥ kf k2,να 2 with equality if and only if a = b = 1 and ∀(r, x) ∈ R+ × R; f (r, x) = Ce −r 2 +x2 2t2 0 ; t0 > 0, C ∈ C. Proof. Let f ∈ L2 (dνα ), f 6= 0, such that b k|(r, x)|a f k2,να + k(µ2 + 2λ2 ) 2 Fα (f )k2,γα < +∞. Then for all a > 1, we have 1 1 2 1 2 1 a a0 2 a0 k 20 k|(r, x)|a f k2,ν kf k2,ν = k|(r, x)|2 |f | a ka,ν α k|f | a ,να , α α where a0 is defined as usual by a0 = a . a−1 By Hölder’s inequality we get 1 1 a a0 k|(r, x)|a f k2,ν kf k2,ν > k|(r, x)|f k2,να . α α The strict inequality here is justified by the fact that if f 6= 0, then the functions |(r, x)|2a |f |2 and |f |2 cannot be proportional.Thus for all a ≥ 1, we have 1 (4.13) a k|(r, x)|a f k2,ν ≥ α k|(r, x)|f k2,να 1 . a0 kf k2,ν α with equality if and only if a = 1. J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ H EISENBERG -PAULI -W EYL U NCERTAINTY P RINCIPLE 15 In the same manner and using Plancherel’s Theorem 2.3, we have for all b ≥ 1 1 1 b b 2 k(µ + 2λ ) Fα (f )k2,γα ≥ 2 (4.14) 2 k(µ2 + 2λ2 ) 2 Fα (f )k2,γα 1 b0 kFα (f )k2,γ α 1 ≥ k(µ2 + 2λ2 ) 2 Fα (f )k2,γα 1 . b0 kf k2,ν α with equality if and only if b = 1. b Let η = a+b , then by the relations (4.13), (4.14) and for all a, b ≥ 1, we have ηa 2 2 21 b 2λ ) Fα (f )k2,γα k|(r, x)|f k2,να k(µ + k|(r, x)|a f kη2,να k(µ2 + 2λ2 ) 2 Fα (f )k1−η , 1 2,γα ≥ + b10 a0 kf k2,ν α with equality if and only if a = b = 1. Applying Theorem 4.3, we obtain a k|(r, x)| f kη2,να k(µ2 2 + 2λ ) b 2 Fα (f )k1−η 2,γα ≥ 2α + 3 2 ηa kf k2,να , with equality if and only if a = b = 1 and ∀(r, x) ∈ R+ × R; f (r, x) = Ce −r 2 +x2 2t2 0 ; t0 > 0, C ∈ C. Remark 2. In the particular case when a = b = 2, the previous result gives us the HeisenbergPauli-Weyl inequality for the fourth moment of Heisenberg 2 2α + 3 2 2 2 k|(r, x)| f k2,να k(µ + 2λ )Fα (f )k2,γα > kf k22,να . 2 5. T HE L OCAL U NCERTAINTY P RINCIPLE Theorem 5.1. Let ξ be a real number such that 0 < ξ < 2α+3 , then for all f ∈ L2 (dνα ), f 6= 0, 2 and for all measurable subsets E ⊂ Γ+ ; 0 < γα (E) < +∞, we have ZZ 2ξ (5.1) |Fα (f )(µ, λ)|2 dγα (µ, λ) < Kα,ξ γα (E) 2α+3 k|(r, x)|ξ f k22,να , E where Kα,ξ = 2ξ ! 2α+3 2α + 3 − 2ξ 5 ξ 2 2α+ 2 Γ α + 3 2 2α + 3 2α + 3 − 2ξ 2 . Proof. For all s > 0, we put Bs = (r, x) ∈ R+ × R ; r2 + x2 < s2 . Let f ∈ L2 (dνα ). By Minkowski’s inequality, we have Z Z 12 2 (5.2) |Fα (f )(µ, λ)| dγα (µ, λ) E = kFα (f )1E k2,γα ≤ kFα (f 1Bs )1E k2,γα + kFα (f 1Bsc )1E k2,γα 1 ≤ γα (E) 2 kFα (f 1Bs )k∞,γα + kFα (f 1Bsc )k2,γα . J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ 16 S. O MRI AND L.T. R ACHDI Applying the relation (2.10), we deduce that for every s > 0, we have Z Z 21 1 2 (5.3) |Fα (f )(µ, λ)| dγα (µ, λ) ≤ γα (E) 2 kf 1Bs k1,να + kFα (f 1Bsc )k2,γα . E On the other hand, by Hölder’s inequality we have (5.4) kf 1Bs k1,να ≤ k|(r, x)|ξ f k2,να k|(r, x)|−ξ 1Bs k2,να s ξ = k|(r, x)| f (r, x)k2,να (5.5) α+ 12 2 Γ α+ 2α+3−2ξ 2 3 2 12 . (2α + 3 − 2ξ) By Plancherel’s theorem 2.3, we have also kFα (f 1Bsc )k2,γα = kf 1Bsc k2,να (5.6) ≤ k|(r, x)|ξ f k2,να k|(r, x)|−ξ 1Bsc k∞,να = s−ξ k|(r, x)|ξ f k2,να . Combining the relations (5.3), (5.5) and (5.6), we deduce that for all s > 0 we have Z Z 21 (5.7) |Fα (f )(µ, λ)|2 dγα (µ, λ) ≤ gα,ξ (s)k|(r, x)|ξ f (r, x)k2,να , E where gα,ξ is the function defined on ]0, +∞[ by gα,ξ (s) = s−ξ + α+ 12 2 γα (E) Γ α + 32 (2α + 3 − 2ξ) ! 12 s 2α+3−2ξ 2 . Thus, the inequality (5.7) holds for s0 = ! 1 5 ξ 2 2α+ 2 Γ( α + 32 ) 2α+3 , γα (E)(2α + 3 − 2ξ) however 1 ξ 2 gα,ξ (s0 ) = γα (E) 2α+3 Kα,ξ , where Kα,ξ = 2α + 3 − 2ξ 2ξ ! 2α+3 2α + 3 2α + 3 − 2ξ 2 . 5 ξ 2 2α+ 2 Γ(α + 32 ) Let us prove that the equality in (5.1) cannot hold. Indeed, suppose that Z Z 2ξ |Fα (f )(µ, λ)|2 dγα (µ, λ) = Kα,ξ γα (E) 2α+3 k|(r, x)|ξ f k22,να . E Then Z Z |Fα (f )(µ, λ)|2 dγα (µ, λ) = gα,ξ (s0 )2 k|(r, x)|ξ f k22,να , E and therefore by the relations (5.2), (5.3) and (5.4), we get 1 (5.8) kFα (f 1Bs0 )1E k2,γα = γα (E) 2 kFα (f 1Bs0 )k∞,γα , (5.9) kf 1Bs0 k1,να = kFα (f 1Bs0 )k∞,γα , J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ H EISENBERG -PAULI -W EYL U NCERTAINTY P RINCIPLE 17 and kf 1Bs0 k1,να = k|(r, x)ξ f k2,να k|(r, x)−ξ 1Bs0 k2,να . (5.10) However, if f satisfies the equality (5.10), then there exists C > 0, such that |(r, x)|2ξ f (r, x)2 = C|(r, x)|−2ξ 1Bs0 , hence ∀(r, x) ∈ R+ × R; (5.11) f (r, x) = CeiΦ(r,x) |(r, x)|−2ξ 1Bs0 , where Φ is a real measurable function on R+ × R. But if f satisfies the relation (5.9), then there exists (µ0 , λ0 ) ∈ Γ+ , such that kf k1,να = kFα (f )k∞,γα = Fα (f )(µ0 , λ0 ). So, there exists θ0 ∈ R satisfying Fα (f )(µ0 , λ0 ) = eiθ0 kf k1,να , and therefore q Z +∞ Z −2ξ iθ0 iΦ(r,x)−iλ0 x−iθ0 |(r, x)| 1Bs0 (r, x) e Ce jα r µ20 + λ20 − 1 dνα (r, x) = 0. 0 R This implies that for almost every (r, x) ∈ R+ × R, q iΦ(r,x)−iλ0 x−iθ0 2 2 e jα r µ0 + λ0 = 1. Hence, we deduce that for all r ∈ R+ , q jα r µ20 + λ20 = 1. Using the relation (2.2), it follows that µ20 + λ20 = 0, or µ0 = i|λ0 |, and then eiΦ(r,x) = eiθ0 +iλ0 x . Replacing in (5.11), we get f (r, x) = C 0iλ0 x |(r, x)|−2ξ 1Bs0 (r, x). Now, the relation (5.8) means that for almost every (µ, λ) ∈ E, we have Fα (f )(µ, λ) = kFα (f )k∞,γα = Fα (f )(µ0 , λ0 ), which implies that for almost every (µ, λ) ∈ E, we have Fα (f )(µ, λ) = eiψ(µ,λ) Fα (f )(µ0 , λ0 ), where ψ is a real measurable function on E, and therefore Z +∞ Z p 0iψ(µ,λ) C |(r, x)|−2ξ 1Bs0 (r, x) e−iλx+iλ0 x−iψ(µ,λ) jα r µ2 + λ2 − 1 dνα (r, x) = 0. 0 R Consequently for all (r, x) ∈ R+ × R, p e−iλx+iλ0 x−iψ(µ,λ) jα r µ2 + λ2 = 1, which implies that λ = λ0 and µ = µ0 . However, since γα (E) > 0, this contradicts the fact that for almost every (µ, λ) ∈ E, Fα (f )(µ, λ) = Fα (f )(µ0 , λ0 ), and shows that the inequality in (5.1) is strictly satisfied. J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ 18 S. O MRI AND L.T. R ACHDI Lemma 5.2. Let ξ be a real number such that ξ > R+ × R we have 2− 2α+3 , 2 (2α+3) ξ kf k21,να ≤ Mα,ξ kf k2,να (5.12) then for all measurable function f on (2α+3) k|(r, x)|ξ f k2,ναξ , where Mα,ξ = π α+ 12 2 Γ α+ 3 2 (2ξ − 2α − 3) 2ξ−2α+3 2ξ (2α + 3) 2α+3 2ξ . 2α+3 sin π 2ξ with equality in (5.12) if and only if there exist a, b > 0 such that |f (r, x)| = (a + b|(r, x)|2ξ )−1 . Proof. We suppose naturally that f 6= 0. It is obvious that the inequality (5.12) holds if kf k2,να = +∞ or k|(r, x)|ξ f k2,να = +∞. Assume that kf k2,να + k|(r, x)|ξ f k2,να < +∞.From the hypothesis 2ξ > 2α + 3, we deduce that for all a, b > 0, the function (r, x) 7−→ (a + b|(r, x)|2ξ )−1 belongs to L1 (dνα ) ∩ L2 (dνα ) and by Hölder’s inequality, we have 2 1 2 2ξ − 12 kf k21,να ≤ (1 + |(r, x)|2ξ ) 2 f (5.13) (1 + |(r, x)| ) 2,να 2,να 1 2 ≤ kf k22,να + k|(r, x)|ξ f k22,να (1 + |(r, x)|2ξ )− 2 . 2,να However, by standard calculus, we have 2 2ξ − 21 = (1 + |(r, x)| ) 2,να α+ 32 ξ2 π . Γ α + 32 sin π 2α+3 2ξ Thus (5.14) kf k21,να ≤ α+ 32 ξ2 π kf k22,να + k|(r, x)|ξ f k22,να , Γ α + 32 sin π 2α+3 2ξ with equality in (5.14) if and only if we have equality in (5.13), that is there exists C > 0 satisfying 1 1 (1 + |(r, x)|2ξ ) 2 |f (r, x)| = C(1 + |(r, x)|2ξ )− 2 , or (5.15) |f (r, x)| = C(1 + |(r, x)|2ξ )−1 . For t > 0, we put as above, ft (r, x) = f (tr, tx), then we have (5.16) kft k21,να = 1 t4α+6 kf k21,να , and (5.17) k|(r, x)|ξ ft k22,να = 1 t2ξ+2α+3 k|(r, x)|ξ f k22,να . Replacing f by ft in the relation (5.14), we deduce that for all t > 0, we have π 2 2α+3 2 2α+3−2ξ ξ 2 t kf k2,να + t k|(r, x)| f k2,να . kf k1,να ≤ 3 ξ2α+ 2 Γ α + 32 sin π 2α+3 2ξ J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ H EISENBERG -PAULI -W EYL U NCERTAINTY P RINCIPLE 19 In particular, for t = t0 = (2ξ − 2α − 3)k|(r, x)|ξ f k22,να (2α + 3)kf k22,να ! 2ξ1 , we get 2− (2α+3) ξ kf k21,να ≤ Mα,ξ kf k2,να (2α+3) k|(r, x)|ξ f k2,ναξ , where Mα,ξ = π α+ 12 2 Γ α+ 3 2 (2ξ − 2α − 3) 2ξ−2α+3 2ξ (2α + 3) 2α+3 2ξ . 2α+3 sin π 2ξ Now suppose that we have equality in the last inequality. Then we have equality in (5.14) for ft0 and by means of (5.15), we obtain |ft0 (r, x)| = C(1 + |(r, x)|2ξ )−1 , and then |f (r, x)| = (a + b|(r, x)|2ξ )−1 . Theorem 5.3. Let ξ be a real number such that ξ > 2α+3 . Then for all f ∈ L2 (dνα ), f 6= 0, 2 and for all measurable subsets E ⊂ Γ+ ; 0 < γα (E) < +∞, we have Z Z 2α+3 2− 2α+3 (5.18) |Fα (f )(µ, λ)|2 dγα (µ, λ) < Mα,ξ γα (E)kf k2,να ξ k|(r, x)|ξ f k2,νξα , E where Mα,ξ = π α+ 12 2 Γ α+ 3 2 (2ξ − 2α − 3) 2ξ−2α+3 2ξ (2α + 3) 2α+3 2ξ . 2α+3 sin π 2ξ Moreover, Mα,ξ is the best (the smallest) constant satisfying (5.18). Proof. • Suppose that the right-hand side of (5.18) is finite. Then, according to Lemma 5.2, the function f belongs to L1 (dνα ) and we have Z Z |Fα (f )(µ, λ)|2 dγα (µ, λ) ≤ γα (E)kFα (f )k2∞,γα E ≤ γα (E)kf k21,να 2− (2α+3) ξ ≤ γα (E)Mα,ξ kf k2,να (2α+3) k|(r, x)|ξ f k2,ναξ , where Mα,ξ = π 1 2α+ 2 Γ α + 3 2 (2ξ − 2α − 3) 2ξ−2α+3 2ξ (2α + 3) 2α+3 2ξ . sin π 2α+3 2ξ • Let us prove that the equality in (5.18) cannot hold. Indeed, suppose that Z Z 2α+3 2− 2α+3 |Fα (f )(µ, λ)|2 dγα (µ, λ) = Mα,ξ γα (E)kf k2,να ξ k|(r, x)|ξ f k2,νξα . E Then Z Z (5.19) |Fα (f )(µ, λ)|2 dγα (µ, λ) = γα (E)kFα (f )k2∞,γα , Γ+ (5.20) kFα (f )k∞,γα = kf k1,να , J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ 20 S. O MRI AND L.T. R ACHDI and 2− 2α+3 (2α+3) kf k21,να = Mα,ξ kf k2,να ξ k|(r, x)|ξ f k2,ναξ . (5.21) Applying Lemma 5.2 and the relation (5.21), we deduce that ∀(r, x) ∈ R+ × R; (5.22) f (r, x) = ϕ(r, x)(a + b|(r, x)|2ξ )−1 , with |ϕ(r, x)| = 1; a, b > 0. On the other hand, there exists (µ0 , λ0 ) ∈ Γ+ such that kFα (f )k∞,γα = |Fα (f )(µ0 , λ0 )| = eiθ0 Fα (f )(µ0 , λ0 ); (5.23) θ0 ∈ R. Combining now the relations (5.20), (5.22) and (5.23), we get q Z +∞ Z iθ0 −iλ0 x 2 2 1 − e ϕ(r, x)jα r µ0 + λ0 e (a + b|(r, x)|2ξ )−1 dνα (r, x) = 0. 0 R This implies that for almost every (r, x) ∈ R+ × R, q −iλ0 x iθ0 e e ϕ(r, x)jα r µ20 + λ20 = 1. Since |ϕ(r, x)| = 1, we deduce that for all r ∈ R+ , q jα r µ20 + λ20 = 1. Using the relation (2.2), it follows that µ20 + λ20 = 0, or µ0 = i|λ0 |, and therefore ϕ(r, x) = e−iθ0 eiλ0 x . Replacing in (5.22), we get f (r, x) = Ceiλ0 x (a + b|(r, x)|2ξ )−1 ; |C| = 1. Now, the relation (5.19) means that Z Z kFα (f )k2∞,γα − |Fα (f )(µ, λ)|2 dγα (µ, λ) = 0. E Hence, for almost every (µ, λ) ∈ E, |Fα (f )(µ, λ)| = kFα (f )k∞,γα = eiθ0 Fα (f )(µ0 , λ0 ). (5.24) Let Ψ(µ, λ) ∈ R, such that |Fα (f )(µ, λ)| = eiΨ(µ,λ) Fα (f )(µ, λ). Then from (5.24), for almost every (µ, λ) ∈ E, eiΨ(µ,λ) Fα (f )(µ, λ) = eiθ0 Fα (f )(µ0 , λ0 ), and therefore Z +∞ Z 0 h p i (a + b|(r, x)|2ξ )−1 1 − ei(λ0 −λ)x eiΨ(µ,λ) jα r µ2 + λ2 dνα (r, x) = 0. R Consequently, for all (r, x) ∈ R+ × R, i(λ0 −λ)x iΨ(µ,λ) e e p 2 2 jα r µ + λ = 1, which implies that λ = λ0 and µ = µ0 . However, since γα (E) > 0, this contradicts the fact that for almost every (µ, λ) ∈ E, eiΨ(µ,λ) Fα (f )(µ, λ) = eiθ0 Fα (f )(µ0 , λ0 ), J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ H EISENBERG -PAULI -W EYL U NCERTAINTY P RINCIPLE 21 and shows that the inequality in (5.18) is strictly satisfied. • Let us prove that the constant Mα,ξ is the best one satisfying (5.18). Let A be a positive constant, such that for all f ∈ L2 (dνα ), Z Z 2α+3 2− 2α+3 (5.25) |Fα (f )(µ, λ)|2 dγα (µ, λ) ≤ Aγα (E)kf k2,να ξ k|(r, x)|ξ f k2,νξα . E We assume that k|(r, x)|ξ f k2,να < +∞. Then by Lemma 5.2, f belongs to L1 (dνα ). Replacing f by ft (r, x) = f (tr, tx) in (5.25), and using the relations (4.7), (4.9) and (5.17), we deduce that for all t > 0 2 Z Z 2α+3 2α+3 ξ ξ ξ Fα (f ) µ , λ dγα (µ, λ) ≤ Akf k2− k|(r, x)| f k . 2,ν 2,ν α α t t E Using the dominate convergence theorem and the relations (2.1), (2.3), (2.9) and (2.10), we deduce that 2 Z Z µ λ 2 lim Fα (f ) t , t dγα (µ, λ) = |Fα (f )(0, 0)| γα (E). t→+∞ E Consequently, for all f ∈ L2 (dνα ), such that k|(r, x)|ξ f k2,να < +∞, (5.26) 2− 2α+3 2α+3 |Fα (f )(0, 0)|2 ≤ Akf k2,να ξ k|(r, x)|ξ f k2,νξα . Now, let f ∈ L2 (dνα ), such that k|(r, x)|ξ f k2,να < +∞, and let (µ, λ) ∈ Γ. Putting p g(r, x) = jα r µ2 + λ2 e−iλx f (r, x), then g ∈ L2 (dνα ), and k|(r, x)|ξ gk2,να < +∞. Moreover, Fα (g)(0, 0) = Fα (f )(µ, λ), and by (5.26), it follows that 2− 2α+3 2α+3 |Fα (f )(µ, λ)|2 ≤ Akf k2,να ξ k|(r, x)|ξ f k2,νξα . Thus, for all f ∈ L2 (dνα ) such that k|(r, x)|ξ f k2,να < +∞, we have (5.27) 2− 2α+3 2α+3 kFα (f )k2∞,γα ≤ Akf k2,να ξ k|(r, x)|ξ f k2,νξα . Taking f0 (r, x) = (1 + |(r, x)|2ξ )−1 , we have kFα (f0 )k2∞,γα = kf0 k21,να = kf0 k22,να = α+ 32 ξ2 π , Γ α + 32 sin π 2α+3 2ξ (2ξ − 2α − 3)π , 3 2α+3 Γ α + 2 sin π 2ξ α+ 52 ξ22 k|(r, x)|ξ f0 k22,να = 2 α+ 52 ξ22 (2α + 3)π . Γ α + 32 sin π 2α+3 2ξ Replacing f by f0 in the relation (5.27), we obtain A ≥ Mα,ξ . This completes the proof of Theorem 5.3. J. Inequal. Pure and Appl. Math., 9(3) (2008), Art. 88, 23 pp. http://jipam.vu.edu.au/ 22 S. O MRI AND L.T. R ACHDI R EFERENCES [1] G. ANDREWS, R. ASKEY AND R. ROY, Special Functions. Cambridge University Press, 1999. [2] C. BACCAR, N.B. HAMADI AND L.T. RACHDI, Inversion formulas for the Riemann-Liouville transform and its dual associated with singular partial differential operators, International Journal of Mathematics and Mathematical Sciences, 2006 (2006), Article ID 86238, 1-26. [3] G. BATTLE, Heisenberg inequalities for wavelet states, Appl. and Comp. Harmonic Anal., 4(2) (1997), 119–146. [4] M.G. COWLING AND J.F. 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