Heisenberg uncertainty principles for an oscillatory integral operator

Heisenberg uncertainty principles for an oscillatory integral operator

Citation: 1798, 020037 (2017); doi: 10.1063/1.4972629 View online: http://dx.doi.org/10.1063/1.4972629

View Table of Contents: http://aip.scitation.org/toc/apc/1798/1

Heisenberg Uncertainty Principles for an

Oscillatory Integral Operator

L. P. Castro

, R. C. Guerra

and N. M. Tuan

1_{CIDMA - Center for Research and Development in Mathematics and Applications, Department of Mathematics,}

University of Aveiro, 3810-193 Aveiro, Portugal.

2_{Department of Mathematics, College of Education, Viet Nam National University, G7 Build., 144 Xuan Thuy Rd.,}

Cau Giay Dist., Hanoi, Vietnam.

a)_{Corresponding author: [email protected]} b)_{URL: http://sweet.ua.pt/castro/}

c)_{[email protected]} d)_{[email protected]}

Abstract.The main aim of this work is to obtain Heisenberg uncertainty principles for a specific oscillatory integral operator which representatively exhibits different parameters on their sine and cosine phase components. Additionally, invertibility theorems, Parseval type identities and Plancherel type theorems are also obtained.

INTRODUCTION

Uncertainty principles of Heisenberg type are well-known to have strong consequences in different subareas of Physics and Mathematics. More specifically, within quantum mechanics, the uncertainty principle is basically the translation of a characteristic feature of quantum mechanical systems; from the mathematical point of view, this may be real-ized by a specific inequality which, e.g., for the Fourier transform, basically states that a nonzero function and its Fourier transform cannot both be sharply localized. This fact is of significative importance e.g. in the theory of partial differential equations (cf. [2]).

Heisenberg’s original proposal was given in very generic terms. It was basically concentrated on qualitative examples which would allow an understanding of simple experiments in quantum mechanics. The first mathematically exact formulation of the Heisenberg uncertainty relations is due to Kennard [7]. The inequality of Kennard was generalized in 1929 by Robertson [8] (by englobing self-adjoint operators), and this was also generalized, later on, in 1930, by Schr¨odinger [10]. Anyway, many variations on Heisenberg’s inequality have been developed during the last century. We refer to [2, 3, 4, 5, 6, 7, 11, 12, 13] for the seminal paper of Heisenberg, as well as for some of the subsequent related fundamental works and surveys. In particular, Heisenberg uncertainty principles for integral operators are known to be important e.g. in the interpretation of the behaviour of this type of operators and have consequent influences in their applications.

The main object of this work is the oscillatory integral operator T , defined in the following way in the framework of the L2_(Rd_{) space,} (T f ) (x)= 1 (2π)d2 ∫ Rd [

2 cos(xy)+ i sin(xy)]f (y) dy, (1) where the different coefficients on the cosine and sine functions are chosen as a representative case of an unbalanced weight on the corresponding oscillation. In this work we will derive such type of uncertainty principles for the operator (1). Anyway, prior to that, we will analyse more basic properties of (1). Namely, we will prove its inversion formula, and obtain a Plancherel theorem and Parseval type identities. Moreover, we will add an example of a classical partial differential equation which can be treated in an efficient way by using our operator. In the next section, we can see

ICNPAA 2016 World Congress

AIP Conf. Proc. 1798, 020037-1–020037-10; doi: 10.1063/1.4972629 Published by AIP Publishing. 978-0-7354-1464-8/$30.00

that T is an invertible, non-unitary and asymmetric, continuous linear operator in L2(Rd), and it is also well-defined on L1_(Rd_{) with an explicit inversion formula.}

FUNDAMENTAL PROPERTIES

(2π)d/2

∫

Rde−ixyf (y) dy and (F−1f )(x)=

1 (2π)d/2

∫

Rde

ixy_{f (y) dy denote the Fourier and inverse Fourier transform, respectively. Throughout the paper we write} N0 := {0, 1, 2, . . .}. If α = (α1, . . . , αd) is a d-tuple of non-negative integersαk, k = 1, . . . , d, we call α a multi-index, and|α| := α1+ · · · + αd. For β = (β1, . . . , βd), we consider Dβx= ∂

β

∂xβ11···∂x βd

d

. The Schwartz space will be denoted by S. The multi-dimensional Hermite functions are defined byΦ_α(x) := (−1)|α|e12|x|

2

Dαxe−|x|

2

(see [15]). The Hermite functions are essential in several applications, as it is the case of the study of the celebrated harmonic oscillator in quantum mechanics. To begin with, we formulate a theorem relating those functions to our operator T .

Theorem 1 The following formula holds

TΦ_α= (−1)

|α|

2 2Φ_α, i f |α| ≡ 0, 2 (mod 4)

(−1)|α|−12 iΦ_α, i f |α| ≡ 1, 3 (mod 4).

(2)

Proof. If we consider L1(Rd) as the domain of the Fourier and inverse Fourier transform F and F−1, respectively, then the domain of T is also L1_(Rd_{). In fact, we can rewrite T as T =} 1

2F+ 3

2F−1. Since FΦα = (−i)|α|Φα and

F−1Φ_α = i|α|Φ_α (see [16]), we have TΦ_α =[1 2(−i)|α|+

3 2i|α|

]

Φα. Calculating the coefficient on the right-hand side of

this equality we obtain (2). The theorem is proved.

The following auxiliary lemma is very useful for proving some of our further theorems. Lemma 2 (cf. [16]) The formula 1

2{ f (x + 0) + f (x − 0)} = limλ→∞ 1 π ∫_+∞ −∞ f (t) sin(_λ(x−t)) x−t dt holds if f (x) 1+|x| ∈ L 1_(R).

The operator T also has its Riemann-Lebesgue lemma which can be proved in a straightforward way. Theorem 3 (Riemann-Lebesgue Lemma) T is a bounded linear operator from L1_(Rd_{) into C}

0(Rd). Namely, if f ∈ L1_(Rd_{), then T f ∈ C} 0(Rd) and∥T f ∥_∞≤ √ 5 (2_π)d2∥ f ∥1, where ∥ · ∥∞

and∥ · ∥1are the usual supremum and L1-norms,

respectively.

Proof. By using the Riemann-Lebesgue lemma for the integral transforms with each one of cosine and sine kernels, we see that T f ∈ C0(Rd), provided f ∈ L1(Rd). Moreover, by the Cauchy-Schwartz inequality we have

∥T f ∥∞= sup x∈Rd   1 (2π)d2 ∫ Rd [

2 cos(xy)+ i sin(xy)]f (y) dy = sup x∈Rd

1 (2π)d2

∫

Rd|2 cos(xy) + i sin(xy)| | f (y)| dy

≤ sup x∈Rd 1 (2π)d2 ∫ Rd √

[4+ 1][cos2_{(xy)+ sin}2_(xy)]_{| f (y)| dy =}

√ 5 (2π)d2

∥ f ∥1.

Theorem 4 T is a continuous linear operator ofS into itself, and fulfills the polynomial identity: T4_−3T2_{−4I = 0,}

where I is the identity operator. Moreover, the operator T is invertible inS.

Proof. Clearly, T is a continuous operator inS. We shall prove the polynomial identity. Let us first prove the identity (T2_{f )(x)=} 3

2f (x)+ 5

2f (−x), for every f ∈ S. For λ > 0, consider the d-dimensional box in R

d_{: B(0, λ) :=} {

y= (y1, . . . , yd)∈ Rd:|yk| ≤ λ, k = 1, . . . , d }

. Obviously,∫_B(0,λ)cos(xy) sin(xv) dx= 0. Acting inductively on d, we

obtain _∫ B(0,λ) cos (y(x− t)) dy = 2 d_{sin (λ(x} 1− t1))· · · sin (λ(xd− td)) (x1− t1)· · · (xd− td) . (3)

Since f ∈ S, we may use the Fubini’s theorem, Lemma 2, and (3) to calculate T2f , as follows (T2f )(x) = lim λ→∞ 1 (2π)d ∫ Rd [ 2 cos(xy)+ i sin(xy)]dy ∫ B(0,λ) [ 2 cos(yt)+ i sin(yt)]f (t) dt = lim λ→∞ 1 (2π)d ∫ Rd f (t) ∫ B(0,λ) [ 5 2cos (y(x+ t)) + 3

2cos (y(x− t)) + 2i sin (y(x + t)) ] dy dt = lim λ→∞ 1 (2π)d ∫ Rd f (t) ∫ B(0,λ) [ 5 2cos (y(x+ t)) + 3 2cos (y(x− t)) ] dy dt = lim λ→∞ 1 (2π)d2 ∫ Rd [ 3 2 2d_{sin (λ(y} 1− t1))· · · sin λ(yd− td) (y1− t1)· · · (yd− td) +5 2 2d_{sin (λ(y} 1+ t1))· · · sin (λ(yd+ td)) (y1+ t1)· · · (yd+ td) ] f (t) dt = 3 2f (x)+ 5 2f (−x), x ∈ R d_{. (4)} Thus, we have (T4f )(x) = T2[(T2f )(x)]= T2 [ 3 2f (x)+ 5 2f (−x) ] = 3 2 [ 3 2f (x)+ 5 2f (−x) ] +5 2 [ 3 2f (−x) + 5 2f (x) ] = 34 4 f (x)+ 30 4 f (−x), x ∈ R d_{. (5)}

Combining (4) and (5), we get T4f − 3T2f− 4 f = 0, for every f ∈ S. The polynomial identity for T is proved. By this polynomial identity, we obtain T[1₄T3₋3

4T ] =[1 4T 3₋3 4T ]

T = I , which implies that T is invertible by T−1= 1

4T 3₋3

4T . The proof of Theorem 4 is complete.

The invertibility of T inS is a key ingredient to prove the inversion theorem below, simply because S is dense in L1_(Rd_{). Moreover, we will provide the explicit inversion formula in the next identity (6).}

Theorem 5 (Inversion Theorem) If f ∈ L1_(Rd_{), and if T f ∈ L}1_(Rd_{), then} f0(x) := 1 (2π)d2 ∫ Rd (T f )(y) [ 1 2cos(xy)− i sin(xy) ]

dy= f (x), for almost every x∈ Rd. (6) Proof. Let us first prove the inversion formula inS, i.e.

g(x)= 1 (2π)d2 ∫ Rd (T g)(y) [ 1 2cos(xy)− i sin(xy) ] dy, for every x∈ Rd, g ∈ S. (7) Indeed, since T g∈ S, the inner function on the right-hand side of (7) belongs to S. This means that the integral (7) is uniformly convergent onRdaccording to each variable x1, . . . , xd. Let us calculate the right-hand side of (7), using Fubini’s theorem, Lemma 2 and (3). So, we have

1 (2π)d2 ∫ Rd (T g)(y) [ 1 2cos(xy)− i sin(xy) ] dy = lim λ→∞ 1 (2π)d2 ∫ B(0,λ) (T g)(y) [ 1 2cos(xy)− i sin(xy) ] dy = lim λ→∞ 1 (2π)d ∫ Rd [ 1 2cos(xy)− i sin(xy) ] ∫ B(0_,λ) [

2 cos(yt)+ i sin(yt)]g(t) dtdy = lim λ→∞ 1 (2π)d ∫ Rd g(t) ∫ B(0,λ) [ 1 2cos(xy)− i sin(xy) ] [ 2 cos(yt)+ i sin(yt)]dy dt = lim λ→∞ 1 (2π)d ∫ Rd g(t) (∫ B(0,λ) (cos y(x− t)) dy dt ) = 1 (2π)d _λ→∞lim ∫ Rd g(t)2 d_{sin (λ(x} 1− t1))· · · sin (λ(xd− td)) (x1− t1)· · · (xd− td) dt= g(x),

for every x ∈ Rd (having in mind that g ∈ S). Identity (7) is proved. Now let f ∈ L1(Rd), and let g ∈ S. Clearly, ∫

Rd f (x)(T g)(x)dx=

∫

Rdg(y)(T f )(y)dy. Hence,

∫ Rd f (x)(T g)(x) dx = 1 (2π)d2 ∫ Rd (∫ Rd (T g)(x) [ 1 2cos(xy)− i sin(xy) ] dx ) (T f )(y) dy = ∫ Rd (T g)(x) ( 1 (2π)d2 ∫ Rd (T f )(y) [ 1 2cos(xy)− i sin(xy) ] dy ) dx = ∫ Rd f0(x)(T g)(x) dx.

By (7) the functions T g cover allS. Therefore,∫_Rd( f0(x)− f (x)) Ψ(x) dx = 0, for every Ψ ∈ S. Since S is dense in

L1_(Rd_{), it follows f}

0(x)− f (x) = 0 for almost every x ∈ Rd. Theorem 5 is proved.

Corollary 6 (Uniqueness Theorem) If f ∈ L1_(Rd_{), and if T f = 0, then f = 0.} Theorem 4 leads us to a Plancherel theorem for T (asS is dense in L2(Rd)).

Theorem 7 (Plancherel Theorem) There is a linear isomorphic operator T from L2(Rd) onto itself which is uniquely determined by T f = T f , for every f ∈ S.

The extension operator satisfies the polynomial identity T4− 3T2− 4I = 0.

Proof. Recall thatS is dense in L2(Rd). As the map f 7→ T f is continuous (relatively to the L2−metric) of the dense subspaceS of L2(Rd) ontoS, it has a unique continuous extension T : L2(Rd)→ L2(Rd) (see [9]). Therefore, the polynomial identity T4f − 3T2f − 4 f = 0 (in L2_(Rd_{)) follows directly from Theorem 4, the continuity of T , and} the circumstance thatS is dense in L2_(Rd_{). Theorem 7 is proved.}

Since the uniqueness of the extension and identity (7) holds on the spaceS, which is dense in L2(Rd), we can state the Plancherel theorem in a clearer way. We shall denote by⟨·, ·⟩2the usual inner product of L2(Rd), and by∥ · ∥2

the corresponding norm.

Theorem 8 (Plancherel Theorem) T defines a bounded linear operator in L2_(Rd_{), admitting an inverse given by} f (y)= lim k→∞ 1 (2π)d2 ∫ |y|≤k [ 1 2cos(xy)− i sin(xy) ] (T f ) (x)dx, in the sense of strong convergence, this is,

lim k→∞ 1 (2π)d2 ∫ |y|≤k [ 1 2cos(xy)− i sin(xy) ] (T f ) (x)dx− f (y) 2 = 0.

In the sequel, we will also denote by T the integral operator defined in the Hilbert space L2_(Rd_{) – as there is no} danger of confusion.

Corollary 9 (Spectrum) (P1) For anyλ < {−i, i, −2, 2} we have (T− λI)−1= − 1

λ4− 3λ2− 4

[

T3+ λT2+ (λ2− 3)T + (λ3− 3λ)I]. (8) (P2) The spectrum of the operator T is given byσ(T) = {−i, i, −2, 2} .

Proof. Proposition (P1) is an immediate consequence of the polynomial identity of Theorem 7, and (P2) follows at once from Theorem 1 and from the fact that±i, ±2 are eigenvalues of T.

The reflection operator will be denoted by W, this is, (Wφ)(x) := φ(−x), for φ ∈ L2(Rd). Theorem 10 (Parseval type Identities) The following identities hold for any f, g ∈ L2_(Rd_):

⟨T f, Tg⟩2= 3 2⟨ f, g⟩2+ 5 2⟨ f, Wg⟩2; ⟨T −1_{f, T}−1_g⟩ 2= − 3 8⟨ f, g⟩2+ 5 8⟨ f, Wg⟩2; ⟨T f, T −1_g⟩ 2= ⟨ f, g⟩2. (9)

Proof. Let us write T = 1₂F+3₂F−1. The well-known identities⟨F f, g⟩2 = ⟨ f, Fg⟩2and⟨F−1f, g⟩2 = ⟨ f, F−1g⟩2,

and a straightforward computation yield the just presented identities (9). Theorem 11 T is not unitary in L2(Rd), being its norm∥T∥2= 2.

Proof. If f ∈ L2_(Rd_{), then F f ∈ L}2_(Rd_{) and F}−1_{f ∈ L}2_(Rd_{). In this way, using (9), we obtain} ∥T f ∥2 2= 3 2⟨ f, f ⟩2+ 5 2⟨ f, W f ⟩2= 3 2∥ f ∥ 2 2+ 5 2⟨ f, W f ⟩2≤ 3 2∥ f ∥ 2 2+ 5 2(∥ f ∥2∥W f ∥2)= 3 2∥ f ∥ 2 2+ 5 2∥ f ∥ 2 2 = 4∥ f ∥ 2 2.

On the other hand, by Theorem 1, we have TΦ_α = 2Φ_αfor the Hermite functionsΦ_αwith|α|-even. So, we deduce that∥T∥2= 2.

We end up this section by exemplifying that similarly to the use of the Fourier transform, it is possible to use T (or T−1) for solving many linear partial differential equations. We will consider just one typical case of classical partial differential equation. If x ∈ Rd and ifα ∈ Nd₀, the monomial xα is defined by xα = xα1

1 · · · x αd

d . For f ∈ S, we define g_β(x)= xβf (x), where x ∈ Rd,β ∈ Nd₀. The function Dβx(T f ) belongs toS, for all β ∈ Nd0. The following

properties of T are very convenient for using in the partial differential calculus. Theorem 12 Let f ∈ S. For β ∈ Nd

0, the following identity holds

Dβx(T f )=    (−1) |β| 2 T g_β, if |β| is even (−1)|β|−12 i S g_β, if |β| is odd. (10) where S is the transform given by

(S f )(x)= 1 (2π)d2

∫

Rd

[

cos(xy)+ 2i sin(xy)]f (y)dy, which admits the inverse

(S−1f )(x)= 1 (2π)d2 ∫ Rd [ cos(xy)− i 2sin(xy) ] f (y)dy. (11)

The proof of (10) can be done through a straightforward computation, and that of the inversion formula (11) can be completed in the same way as in the proof of Theorem 5. Therefore, we choose not to add the proof of Theorem 12 in here.

Example 13 (see [1, Example 2.12.3], [14, Chapter 5]) Consider the following equation ∂u

∂t = ∂2_u

∂x2 such that u(x, 0) = f (x), −∞ < x < ∞, t > 0.

Let U(ξ, t) = √1 2π

∫

Ru(x, t)

[

2 cos(xξ) + i sin(xξ)]dx. Integrating twice by parts and assuming that the terms at +∞ and −∞ vanish, we get

∂U ∂t = 1 √ 2π ∫ R ∂u ∂t [ 2 cos(xξ) + i sin(xξ)]dx = √1 2π ∫ R ∂2_u ∂x2 [ 2 cos(xξ) + i sin(xξ)]dx = −√ξ2 2π ∫ Ru [ 2 cos(xξ) + i sin(xξ)]dx= −ξ2U. It implies that U(ξ, t) = A(ξ)e−ξ2t_{. Putting t = 0, we obtain A(ξ) = √}1

2_π

∫

R f (x)

[

2 cos(xξ) + i sin(xξ)]dx = (T f )(ξ). Hence, U(ξ, t) = (T f )(ξ)e−ξ2t_{. Thus, the solution is}

u(x, t) = √1 2π ∫ R(T f )(ξ)e −ξ2_t [ 1 2cos(xξ) − i sin(xξ) ] dξ.

As the functions 2e−ξ2tsin(ξx) cos(ξu) and1₂e−ξ2tsin(ξu) cos(ξx) are odd corresponding to the variable ξ, the solution can be represented as u(x, t) = ₂1_π∫_R∫_Rcos (ξ(x − u)) e−ξ2tf (u) du dξ.

UNCERTAINTY PRINCIPLES

In view to derive Heisenberg uncertainty type principles for T , we will use the Hermite functions, as it is expected from the knowledge of similar cases; cf. [17]. We will start with the one dimensional case, for the real line, and later on we will consider theRd_{case (with d ≥ 1). For the real line case (see e.g. [15] and [16]), the Hermite polynomial} of degree n can be defined by

Hn(t)= (−1)net 2 ( d dt )n e−t2, n ∈ N0, (12) and we writeϕn(t) := e− 1 2t 2 Hn(t)= (−1)ne 1 2t 2(_d dt )n

e−t2, that is a solution of the following differential equation d2_y

dt2 − t

2_{y= −(2n + 1)y. (13)}

For the corresponding orthogonal system on L2_{(R), we take ψ}

n(t) := ϕn(t)/(2nn! √

π)1

2. We will be able to analyse

and use the images ofψnby our operator T . From the fact that y= ϕnis a solution of (13), we obtain the following recurrence relation for the Hermite functions:

√ 2tψn(t)= √ n+ 1ψn+1(t)+ √ nψn−1(t), n ∈ N. (14)

We will admit thatψ₋₁= 0. Moreover, by the properties of the Hermite functions, we also know that ⟨ψm, ψn⟩2= δmn, whereδmnis the Kronecker symbol.

Our first uncertainty principle of Heisenberg type is now stated in the following theorem. Theorem 14 If f ∈ L2_{(R) and g = T f , then}

∥t f (t)∥2· ∥ξg(ξ)∥2≥ ∥ f ∥22. (15)

In view of the next main goal of proving this theorem, let us take a function f ∈ L2(R) and consider g = T f . As T =1₂F+3₂F−1, and using (9), we obtain:

γn:= ⟨ f, ψn⟩2= ⟨T f, T−1ψn⟩2= ⟨g, 3 4Fψn− 1 4F −1_ψ n⟩2= 3 i−n− in 4 ⟨g, ψn⟩2. (16)

Equivalently, we have⟨g, ψn⟩2= θnγn, whereθn:=_{3 i}−n4_−in, for n= 0, 1, . . .. In a straightforward way, we obtain

θn= { (−1)n22, n ≡ 0, 2 (mod 4) (−1)n−12 i, n ≡ 1, 3 (mod 4) , θ 2 n= −θnθn+2= { 4, n ≡ 0, 2 (mod 4) −1, n ≡ 1, 3 (mod 4) (17) Using these facts, we will first prove the following lemma.

Lemma 15 If f ∈ L2(R) and g = T f , then ∥t f (t)∥2 2+ ∥tg(t)∥ 2 2 = 1 2 ∞ ∑ n=0 [ (2n+ 1)(1 + θ2_n)|γn|2 + √ n(n+ 1)(1 + θn−1θn+1) (γn−1γn+1+ γn−1γn+1) ] . (18)

Proof. We will use the recurrence relation (14), admitting thatψ₋₁= 0. So, if we consider that γ₋₁= 0, we obtain √ 2⟨t f (t), ψn(t)⟩2= √ n+ 1γn+1+ √ nγn−1, n ∈ N. (19)

Admitting thatθ₋₁= 1 and using (16)–(17), we have √ 2⟨tg(t), ψn(t)⟩2 = ⟨g(t), √ 2tψn(t)⟩2= ⟨g(t), √ n+ 1ψn+1(t)+ √ nψn−1(t)⟩2 = √n+ 1⟨g(t), ψn+1(t)⟩2+ √ n⟨g(t), ψn−1(t)⟩2= √ n+ 1θn+1γn+1+ √ nθn−1γn−1.

Hence, ∥t f (t)∥2 2 = ⟨t f (t), t f (t)⟩2 = 1 2 ∞ ∑ n₌₀  √n+ 1γn₊₁+ √ nγn₋₁ 2 = 1 2 ∞ ∑ n=0 [ (n+ 1) |γn+1|2+ n |γn−1|2+ √ n(n+ 1) (γn−1γn+1+ γn−1γn+1) ] = 1 2 ∞ ∑ n=0 (n+ 1) |γn+1|2+ 1 2 ∞ ∑ n=0 n|γn−1|2+ 1 2 ∞ ∑ n=0 √ n(n+ 1) (γn−1γn+1+ γn−1γn+1). Let k := n + 1 and i := n − 1. So, we obtain

∥t f (t)∥2 2 = 1 2 ∞ ∑ k=1 k|γk|2+ 1 2 ∞ ∑ i=0 (i+ 1) |γi|2+ 1 2 ∞ ∑ n=0 √ n(n+ 1) (γn−1γn+1+ γn−1γn+1) = 1 2 ∞ ∑ n=0 n|γn|2+ 1 2 ∞ ∑ n=0 (n+ 1) |γn|2+ 1 2 ∞ ∑ n=0 √ n(n+ 1) (γn−1γn+1+ γn−1γn+1) = 1 2 ∞ ∑ n=0 [ (2n+ 1) |γn|2+ √ n(n+ 1) (γn₋₁γn₊₁+ γn₋₁γn₊₁) ] . (20)

Besides this, we have ∥tg(t)∥2 2 = ⟨tg(t), tg(t)⟩2 = 1 2 ∞ ∑ n₌₀ [ (n+ 1)θ_n2₊₁|γn+1|2+ nθ2n−1|γn−1|2+ √ n(n+ 1)θn−1θn+1(γn−1γn+1+ γn−1γn+1) ] = 1 2 ∞ ∑ n=0 [ (2n+ 1)θ2_n|γn|2+ √ n(n+ 1)θn−1θn+1(γn−1γn+1+ γn−1γn+1) ] . (21)

Adding (20) and (21), we obtain (18).

Lemma 16 The inequality 4∥t f (t)∥2₂+ ∥tg(t)∥2₂≥ 4∥ f ∥2₂holds true for f ∈ L2(R) and g = T f . Proof. From the previous lemma, we have that

∥t f (t)∥2 2+ ∥tg(t)∥ 2 2 = 1 2 ∞ ∑ n=0 [ (2n+ 1)(1 + θ2_n)|γn|2+ √ n(n+ 1)(1 + θn−1θn+1) (γn−1γn+1+ γn−1γn+1) ] = 1 2 ∞ ∑ n=0 [ (2n+ 1)(1 + θ2_n)|γn|2+ (1 + θn₋₁θn₊₁) √ n+ 1γn₊₁+ √ nγn₋₁ 2] −1 2 ∞ ∑ n₌₀ (1+ θn−1θn+1) [ (n+ 1) |γn+1|2+ n |γn−1|2 ] . (22)

Doing n := n + 1 and n := n − 1, we obtain ∥t f (t)∥2 2+ ∥tg(t)∥ 2 2 = 1 2 ∞ ∑ n=0 [ (2n+ 1)(1 + θ2n)|γn|2+ (1 + θn−1θn+1) √ n+ 1γn+1+ √ nγn−1 2] −1 2 ∞ ∑ n=0 (2n+ 1)(1 + θnθn+2)|γn|2 = ∑∞ n=0   (2n + 1)θ2n|γn|2+ 1 2 ∞ ∑ n=0 (1− θ2_n−1)√n+ 1γn+1+ √ nγn−1 2  = 4|γ0|2+ 3|γ1|2+ ∞ ∑ n=2 (2n+ 1)θ2_n|γn|2− 1 2 ∞ ∑ n=1 (1− θ2_n−1)√n+ 1γn₊₁+ √ nγn₋₁ 2 .

As 1− θ2 n≥ −3 and 3|γ1|2≥ 4|γ1|2−3₂|γ1|2, we have ∥t f (t)∥2 2+ ∥tg(t)∥ 2 2 ≥ 4|γ0|2+ 4|γ1|2+ 4 ∞ ∑ n₌₂ |γn|2− 3 2|γ1| 2₋3 2 ∞ ∑ n₌₁  √n+ 1γn₊₁+ √ nγn₋₁ 2 = 4∑∞ n₌₀ |γn|2− 3 2 ∞ ∑ n₌₀  √n+ 1γn+1+ √ nγn−1 2 = 4 ∫ R| f (t)| 2 _{dt− 3} ∫ R t2| f (t)|2dt. (23)

In this way, we can say that 4∥t f (t)∥2

2+ ∥tg(t)∥ 2 2≥ 4∥ f ∥

2 2.

After the previous preparatory results, we are now in condition to present the proof of Theorem 14. Proof. For some p> 0, we will consider the following functions f1(t) := p−

1 2f (t

p) and g1(t) := p

1

2g(t p). We can

prove that g1= T f1. Indeed,

(T f1)(t) = 1 √ 2π ∫ R [

2 cos(ty)+ i sin(ty)]f1(y)dy =

1 2 1 √ 2π ∫ R e−ityp−12f ( y p ) dy+3 2 1 √ 2π ∫ R eityp−12f ( y p ) dy = p1 2 [ 1 2 1 √ 2π ∫ R e−itpyf (y)dy+3 2 1 √ 2π ∫ R eit pyf (y)dy ] = p1 2(T f )(t p)= p 1 2g(t p)= g₁(t).

Applying Lemma 16 for f1and g1, we obtain 4∥t f1(t)∥22+ ∥tg1(t)∥22 ≥ 4∥ f1∥22, which is equivalent to

4p2∥t f (t)∥2₂+ p−2∥tg(t)∥2₂≥ 4∥ f1∥22= 4∥ f ∥ 2

2. (24)

Taking the minimum on the left-hand side, with respect to the variable p∈ (0, +∞), we have min p_∈(0,+∞){4p 2_{∥t f (t)∥}2 2+ p−2∥tg(t)∥ 2 2} = 4∥t f (t)∥2· ∥tg(t)∥2. (25)

So, we obtain that 4∥t f (t)∥2· ∥tg(t)∥2≥ 4∥ f ∥2₂, which is equivalent to (15).

Theorem 17 A generalization of Theorem 14 holds true for any f ∈ L2(Rd). Namely, (∫ Rd|x − a| 2_{| f (x)|}2_dx )1/2(∫ Rd|ξ − b| 2_{|T f (ξ)|}2_dξ )1/2 ≥ d ∫ Rd| f (x)| 2_{dx, (26)}

where a, b ∈ Rd_{are arbitrary vectors.}

Proof. Notice that by a simple changing of variables (or, equivalently, a shift of the axes Oxk, Oξ, k = 1, . . . , d), we can assume that a= b = 0. Put

g(x2, . . . , xd) := (∫ Rx 2 1| f (x)| 2_dx 1 )1/2 , G(x2, . . . , xd) := (∫ Rx 2 1|T f (x)| 2_dx 1 )1/2 , (27)

in which g and G are positive functions of variables x2, . . . , xd. On the other hand, for a convenient use of some integral inequalities, we will unify the symbols x andξ in the inequality (26). Having in mind that in the integrals of (27), x2, . . . , xd are considered as parameters, and using the one-dimensional version of the uncertainty inequality already obtained in Theorem 14, we have

(∫ Rx 2 1| f (x)| 2 dx1 )1/2(∫ Rx 2 1|T f (x)| 2 dx1 )1/2 ≥ ∫ R| f (x)| 2 dx1. (28)

By considering the consequent iterated integrals, the Cauchy-Schwartz inequality (C-S inequality) and (28), we have (∫ Rd x2₁| f (x)|2dx1· · · dxd ) (∫ Rd x2₁|(T f )(x)|2dx1· · · dxd ) = (∫ Rd−1|g(x2, . . . , xd )|2dx2· · · dxd ) (∫ Rd−1|G(x2, . . . , xd )|2dx2· · · dxd ) | {z }

using the C-S inequality

≥ (∫ Rd−1 g(x2, . . . , xd) G(x2, . . . xd)dx2. . . dxd )2 =      ∫ Rd−1      (∫ Rx 2 1| f (x)| 2_dx 1 )1/2(∫ Rx 2 1|T f (x)| 2_dx 1 )1/2 | {z } using (28)     dx2· · · dxd      2 ≥ (∫ Rd| f (x)| 2_dx 1dx2· · · dxd )2 = ∥ f ∥4 2.

Similarly, we can prove that (∫ Rd x2_k| f (x)|2dx )1/2(∫ Rd x2_k|(T f )(x)|2dx )1/2 ≥ ∥ f ∥2 2, for every k = 1, . . . , d. (29)

Taking both sides of (29) and summing over k, and using the Cauchy-Schwartz inequality of an inner product of vectors inRd_{, we obtain at once the d-dimensional form of the Heisenberg’s inequality. Indeed, we have}

d∥ f ∥2₂ ≤ d ∑ k=1 (∫ Rd x2_k| f (x)|2dx )1/2 | {z } putting asAk (∫ Rd x2_k|(T f )(x)|2dx )1/2 | {z } putting asBk ≤      d ∑ k=1 [∫ Rd x2_k| f (x)|2dx ] | {z } A2 k      1/2     d ∑ k=1 [∫ Rd x2_k|T f (x)|2dx ] | {z } B2 k      1/2 = (∫ Rd|x| 2_{| f (x)|}2_dx )1/2 (∫ Rd|x| 2_{|T f (x)|}2_dx )1/2 ,

as desired. The theorem is proved.

Another uncertainty principle of Heisenberg type for our operator T , also for the Rd _{case, is presented in the} following theorem.

Theorem 18 Ifψ ∈ L2(Rd), then ∥xψ(x)∥2

[

−⟨y(Tψ)(y), y(T−1_ψ)(−y)⟩ 2 ]1 2 _≥1 2∥ψ∥ 2 2. (30)

Proof. We will start with the case d= 1. Let ψ, ψ′∈ S. Integrating by parts, we have ∥ψ∥2 2= ∫ R|ψ(x)| 2_{dx= −} ∫ R x d dx|ψ(x)| 2_{dx= −⟨xψ}′_{(x), ψ(x)⟩} 2− ⟨xψ(x), ψ′(x)⟩2.

Using the fact that|⟨xψ′(x), ψ(x)⟩2| = |⟨xψ(x), ψ′(x)⟩2| and the Cauchy-Schwartz inequality, we obtain that

∥ψ∥2 2≤ 2 ∫ R|x||ψ(x)||ψ ′_{(x)| dx ≤ 2∥xψ(x)∥} 2∥ψ′(x)∥2. (31)

By Theorem 8, we can writeψ = (T−1(Tψ)). So, ψ′_{(x)= √}1 2π ∫ R [ −y 2sin(xy)− iy cos(xy) ] (Tψ)(y) dy = √1 2π ∫ Ry [ −1 2sin(xy)− i cos(xy) ] (Tψ)(y) dy.

Let R(x) := √1 2_π ∫ R [ −1 2sin(xy)− i cos(xy) ]

f (y) dy, f ∈ L2_{(R). Thus, ψ}′_{(x) = (R(y(Tψ)(y)))(x). We can now}

rewrite R in the form R= −3i₄F−₄iF−1so that we immediately obtain⟨R f, Rg⟩2 = 3₈⟨ f, g⟩2+5₈⟨ f, Wg⟩2. Hence, we

have ∥ψ′_∥2 2 = ∥R(y(Tψ)(y))∥ 2 2 = 3 8⟨y(Tψ)(y), y(Tψ)(y)⟩2− 5 8⟨y(Tψ)(y), y(Tψ)(−y)⟩2 = 1

8⟨y(Tψ)(y), 3y(Tψ)(y) − 5y(Tψ)(−y)⟩2= 1 8 1 2π ∫ R ∫ Ry(2 cos(yt)+ i sin(yt)) ψ(t) [ 3 ∫

R(y(2 cos(yv)− i sin(yv))) ψ(v) dv − 5

∫

R(y(2 cos(yv)+ i sin(yv))) ψ(v) dv

] dtdy = −1 2π ∫ R ∫ R ∫ Ry 2 (cos(yt)+ i sin(yt)) ( 1 2cos(yv)+ i sin(yv) ) ψ(v)ψ(v) dv dt dy = −⟨y(Tψ)(y), y(T−1_ψ)(−y)⟩

2. (32)

Then,∥ψ∥2

2 ≤ 2∥xψ(x)∥2∥ψ′∥2 = 2∥xψ(x)∥2

[

−⟨y(Tψ)(y), y(T−1_ψ)(−y)⟩ 2

]1

2_{, which is equivalent to (30), in the case}

d= 1.

For d> 1, (30) follows from the case d = 1, by using the same method as in the proof of Theorem 17, and so we avoid to present the corresponding detailed full proof in here.

ACKNOWLEDGMENTS

This work was supported in part by FCT–Portuguese Foundation for Science and Technology through the Center for Research and Development in Mathematics and Applications (CIDMA) of Universidade de Aveiro, within project UID/MAT/04106/2013. R. C. Guerra also acknowledges the direct support of the Portuguese Foundation for Science and Technology (FCT) through the scholarship PD/BD/114187/2016.

REFERENCES

[1] L. Debnath and D. Bhatta, Integral Transforms and their Applications, 2nd ed. (Chapman & Hall/CRC, Boca Raton, 2007).

[2] C. Fefferman, “The uncertainty principle”,Bull. Amer. Math. Soc. (N.S.)9, 129–206 (1983).

[3] G.B. Folland and A. Sitaram, “The uncertainty principle: a mathematical survey”,J. Fourier Anal. Appl.3 (3), 207–238 (1997).

[4] D. Gabor, “Theory of communication”, J. Inst. Elec. Engr. 93, 429–457 (1946).

[5] W. Heisenberg, “ ¨Uber den anschaulichen Inhalt der quantentheoretischen Kinematic und Mechanik”,Zeit. Physik43, 172–198 (1927).

[6] M. F. E. de Jeu, “An uncertainty principle for integral operators”,J. Functional Analysis22, 247–253 (1994). [7] E. H. Kennard, “Zur Quantenmechanik einfacher Bewegungstypen”,Zeit. Physik44, 326–352 (1927). [8] H. P. Robertson, “The uncertainty principle”,Phys. Rev.34, 573–574 (1929); reprinted in Wheeler and Zurek,

127–128 (1983).

[9] W. Rudin, Functional Analysis (McGraw-Hill, New York, 1991).

[10] E. Schr¨odinger, “Zum Heisenbergschen Unsch¨arfeprinzip”, Sitzungsber. Preuß. Akad. Wiss., Phys.-Math. Kl., 296–303 (1930); Proceedings Royal Irish Acad. (in English) 39 (A), 73–81 (1930).

[11] D. Slepian, “Some comments on Fourier analysis, uncertainty and modeling”, SIAM Rev. 25, 379–393 (1983).

[12] K. T. Smith, “The uncertainty principle on groups”,SIAM J. Appl. Math.50, 876–882 (1990).

[13] N. Sochen, P. Maass, C. Sagiv and H. G. Stark, “Do uncertainty minimizers attain minimal uncertainty?”,J. Fourier Anal. Appl.16, 448–469 (2010).

[14] E. M. Stein and R. Shakarchi, Fourier Analysis. An Introduction, Princeton Lectures in Analysis, 1 (Princeton University Press, Princeton and Oxford, 2003).

[15] S. Thangavelu, Harmonic Analysis on the Heisenberg Group (Springer, New York, 1998).

[16] E. C. Titchmarsh, Introduction to the Theory of Fourier Integrals (Chelsea Publ. Co., New York, 1986). [17] N. M. Tuan and P. D. Tuan, “Operator properties and Heisenberg uncertainty principles for a un-unitary