DOI 10.1007/s11868-017-0189-9
Benedicks–Amrein–Berthier type theorem related to Opdam–Cherednik transform
Azzedine Achak1 · R. Daher1
Received: 6 October 2016 / Revised: 16 January 2017 / Accepted: 18 January 2017
© Springer International Publishing 2017
Abstract The aim of this paper is to prove a new uncertainty principle for the Opdam–Cherednik transform. This result is an analogue of a result Benedicks–
Amrein–Berthier, it states that a non zero function f and its Opdam–Cherednik transformHα,β(f)cannot both have support of finite measure. We also prove Donoho–
Strak’s local uncertainty principle to the Opdam–Cherednik transform.
Keywords Opdam–Cherednik transform·Benedicks–Amrein–Berthier’s theorem· Donoho–Stark’s uncertainty principle
Mathematics Subject Classification 42A38·44A35·34B30 1 Introduction
Uncertainty principles are mathematical results that give limitations on the simulta- neous concentration of a function and its Fourier transform. They have implications in two main areas: quantum physics and signal analysis. There are numerous mathe- matical formulations for this principle as well as extensions to other transforms (e.g.
Fourier type transforms on various types of Lie groups, other integral transforms…) and we refer to the book [8] and the surveys [4,6] for further references. The concept of concentration has taken different interpretations in different contexts.
B
Azzedine Achak [email protected] R. Daher1 Department of Mathematics, Faculty of Sciences Aïn Chock, University of Hassan II, 20100 Casablanca, Morocco
Our aim here is to consider uncertainty principles in which concentration is mea- sured in sense of smallness of the support and when the transform under consideration is the Opdam–Cherednik transform.
In order to describe our results, we first need to introduce some facts about har- monic analysis related to the Opdam–Cherednik transform. We cite here, as briefly as possible, some properties. For more details we refer to [9].
• Aα,β(x)=(2 sinh|x|)2α+1(2 cosh|x|)2β+1, whereα≥β ≥ −21andα > −21.
• dσα,β(λ)= 1− ρ
iλ
dλ
8π|Cα,β(λ)|2, where
Cα,β(λ)= 2ρ−iλ(α+1)(iλ) 1
2(ρ+iλ) 1
2(α−β+1+iλ) ρ=α+β+1,λ∈C.
• Lp(R,Aα,β)(resp.Lp(R, σα,β), p ∈ [1,∞[,the space of measurable functions gonRsuch that
gLp( ,Aα,β)=
|g(t)|pAα,β(t)dt 1p
<∞, respgLp( ,σα,β) =
|g(t)|pdσα,β(t) 1p
<∞ .
The Opdam hypergeometric functionsGα,βλ onRare eigenfunctions α,βGα,βλ (x)= iλGα,βλ (x)of the differential-difference operator
α,βu(x)= du(x)
d x +[(2α+1)coth(x)+(2β+1)tanh(x)]
×u(x)−u(−x)
2 −ρf(−x).
that are normalized such thatGα,βλ (0)=1.The eigenfunctionGα,βλ is given by Gα,βλ (x)=ϕα,βλ (x)− 1
ρ−iλ d
d xϕλα,β(x)
=ϕα,βλ (x)+ iλ
4(α+1)sinh(2x)ϕα+λ 1,β+1(x), whereϕα,βλ is the Jacobi function given by
ϕα,βλ (x)= 2F1
ρ+iλ
2 ,ρ−iλ
2 , α+1,−(sinh(x))2
.
For everyλ∈Candx∈R, the eigenfunctionGα,βλ satisfy Gα,βξ (x)≤ce(|I mg(ξ)|−(α+β+1))|x| whereca positive constant (see [2]).
Without loss of generality we can assume thatc=1,therefor when|I mg(ξ)| ≤ α+β+1,
Gα,βξ (x)≤1.
The Opdam–Cherednik transform of function f ∈L1(R,Aα,β)is defined by Hα,β(f)(λ)=
f(x)Gα,βλ (−x)Aα,β(x)d x, ∀λ∈R.
We have the following inversion formula f(x)=
Hα,β(f)(λ)Gα,βλ (x)dσα,β(λ), ∀x∈R.
The Plancherel formula is given by
|f(x)|2Aα,β(x)d x=
Hα,βf(λ)Hα,βfˇ(λ)dσα,β(λ)
where fˇ(x)= f(−x).
We have also the following estimate (see [9])
fL2( ,Aα,β)≥ Hα,β(f)L2( ,σα,β). (1) Definition 1 Let S, be two measurable subsets of R. Then (S, ) is called a strong annihilating pair for the Opdam–Cherednik transform if there exists a constant C(S, )such that for all function f ∈L2(R,Aα,β),withsuppHα,β(f)⊂,
fL2( ,Aα,β)≤C(S, )fL2(Sc,Aα,β) (2) where Sc =R\Sandsupp f = {x : f(x) =0}. (S, )is a weak annihilating pair if,supp f ⊂SandsuppHα,β(f)⊂implies f =0.
Of course, every strong annihilating pair is also a weak one. There are several examples of the Uncertainty Principle of the form (2) for the Euclidean Fourier transform . One of them is the Amrein–Berthier theorem [1] which is a quantitative version of a result due to Benedicks [3] showing that a pair of sets of finite measure is an annihilating pair.
Lemma 1 Let1≤ p<2and Dpbe the strip in the complexξ-plane defined by Dp=
ξ ∈C: |I mg(ξ)|< (α+β+1) 2
p −1
.
For any function f ∈ Lα,βp (R,Aα,β)its Opdam–Cherednik transformHα,β is well defined and holomorphic in Dpand for allξ ∈ Dp,
|Hα,β(f)(ξ)| ≤ fLp( ,Aα,β)Gα,βξ
Lq( ,σα,β)
where 1p+1q =1.
Proof See Lemma 3.1 in [9].
AsGα,βξ (x)≤ 1, then if f ∈ L1α,β(R,Aα,β),Hα,β(f)is continuous also in the closureD1ofD1and for allξ ∈ D1
Hα,β(f)∞≤ fL1
α,β( ,Aα,β), (3)
where.∞is the usual essential supremum norm.
2 Uncertainty principles
In this section we will give some remarks about Annihilating sets. We shall use Ghob- ber’s techniques [7].
Proposition 1 Let f ∈L2(R,Aα,β)has non empty support, then σα,β(suppHα,β(f))Aα,β(supp f)≥1.
In particular, if :
Aα,β(supp f)σα,β(suppHα,β(f)) <1then f =0.
Proof If the function f ∈ L2(R,Aα,β)has non empty support, by the Cauchy–
Schwartz inequality and (3) we have:
Hα,β2L2( ,σα,β)≤σα,β(suppHα,β(f))Hα,β(f)2∞
≤σα,β(suppHα,β(f))f2L1( ,Aα,β)
≤σα,β(suppHα,β(f))Aα,β(supp f)f2L2( ,Aα,β). Using (1) we have the following quantitative uncertainty inequality connecting the support of f and the support of its Opdam–Cherednik transformHα,β:
σα,β(suppHα,β)Aα,β(supp f)≥1. (4)
It follows that if :
Aα,β(supp f)σα,β(suppHα,β(f)) <1 then f =0.
We consider a pair of orthogonal projections onL2(R,Aα,β). The first is the time- limiting operator defined by:
ESf =χSf,
and the second is the frequency-limiting operator defined by:
Ff =H−α,β1
χHα,β(f) ,
whereSandare measurable subsets ofR, andχSdenote the characteristic function ofS.
Lemma 2 Let(S, )be two measurable subsets ofR.Then the following assertions are equivalent
(i) ESF = ESFL2( ,Aα,β)<1.
(ii) (S, )is strongly annihilating pair for the Opdam–Cherednik transform. More- over we have
f2L2( ,Aα,β)≤(1− ESF)−2
EScf2L2( ,Aα,β)+ Fcf2L2( ,Aα,β)
.
Proof Firstly we show the following implication(i)⇒(ii). The identity operatorI satisfies
I =ES+ESc =ESF+ESFc+ESc, we have from the orthogonality ofESandESc
f −ESFf2L2( ,Aα,β)= ESFcf +EScf2L2( ,Aα,β)
= ESFcf2L2( ,Aα,β)+ EScf2L2( ,Aα,β).
It follows byES =1 that f −ESFfL2( ,Aα,β)≤
Fcf2L2( ,Aα,β)+ EScf2L2( ,Aα,β)
1
2 . (5)
On the other hand, we have
f −ESFfL2( ,Aα,β)≥ fL2( ,Aα,β)− ESFfL2( ,Aα,β)
≥ fL2( ,Aα,β)− ESF.fL2( ,Aα,β).
It follows from inequality (5) (1− ESF)fL2( ,Aα,β)≤
EScf2L2( ,Aα,β)+ Fcf2L2( ,Aα,β)
1
2. (6) AsESF<1, then we obtain the desired result.
Let us now show the second implication(ii)⇒(i). Recall that ESF = FES =supf∈L2( ,Aα,β)
FESfL2( ,Aα,β)
fL2( ,Aα,β)
=supf:f=ESf
FfL2( ,Aα,β)
fL2( ,Aα,β)
=supf:f=Ff
ESfL2( ,Aα,β)
fL2( ,Aα,β)
<1.
We suppose that ESF = 1. Then we can find a bandlimited sequence fn ∈ L2(R,Aα,β)onof norm 1 (in particular fn=Ffn) such that
ESfnL2( ,Aα,β)−→1 asn−→ ∞.
By the orthogonality ofS, we have
EScfn2L2( ,Aα,β)= fn2L2( ,Aα,β)− ESfnL2( ,Aα,β)−→0 asn −→ ∞,
which contradicts (2).
Theorem 1 If0<Aα,β(S)σα,β() <1then for all function f ∈ L2(R,Aα,β)such that suppHα,β(f)⊂
fL2( ,Aα,β)≤(1−
Aα,β(S)σα,β())−1fL2(Sc,Aα,β).
Proof A straightforward computation shows thatESFis an integral operator with kernel
N(t,x)=χS(t)H−α,β1(χGα,βξ (t))(x).
Indeed, we have ESFf(t)=χS(t)
χ(ξ)Hα,β(f)(ξ)Gα,βξ (t)dσα,β(ξ)
=χS(t)
χ(ξ)Gα,βξ (t)
f(x)Gα,βξ (x)Aα,β(x)d x
dσα,β(ξ)
=
f(x)N(t,x)Aα,β(x)d x,
where
N(t,x)=χS(t)
χ(ξ)Gα,βξ (t)Gα,βξ (x)dσα,β(ξ).
Since σα,β() < ∞ and Gα,βξ is bounded, then for all t ∈ R, χGα,βξ (t) ∈ L2(R, σα,β).ThenESFis an integral operator with Kernel
N(t,x)=χS(t)H−α,β1(χGα,βξ (t))(x).
AsESFH S= NL2( × ,Aα,β⊗Aα,β), it follows from (1) that ESF2H S=
|χS(t)|2
|Hα,β−1(χGα,βξ (t))(x)|2Aα,β(ξ)dξ
Aα,β(t)dt
≤
χS(t)
χ(ξ)|Gα,βξ (t)|2dσα,β(ξ)Aα,β(t)dt.
We can deduce from|Gα,βξ (t)| ≤1 that
ESF ≤ ESFH S≤
Aα,β(S)σα,β(). (7) SinceAα,β(S)σα,β() <1, then we have from inequality (7) and Lemma2
f2L2( ,Aα,β)≤ 1−
Aα,β(S).σα,β()−2
×
EScf2L2( ,Aα,β)+ Fcf2L2( ,Aα,β)
.
SincesuppHα,β(f)⊂, it follows from Plancherel formula that Fcf2L2( ,Aα,β) =
cHα,βf(λ)Hα,βfˇ(λ)dσα,β(λ)=0,
which shows the desired result.
We are now in position to prove our main result.
Theorem 2 Let S andbe a pair of measurable subsets of Rwith0 < Aα,β(S), σα,β() <∞. Then the pair(S, )is a strong annihilating pair.
Proof Let f ∈ L2(R,Aα,β)a nonzero function such that supp f ⊂SandsuppHα,β(f)⊂. Using the Cauchy–Schwartz inequality, we have
f2L1( ,Aα,β)≤Aα,β(supp f)f2L2( ,Aα,β).
As f has support of finite measure, hence f belongs inL1(R,Aα,β). From Lemma 1,Hα,β is analytic in the open strip {ξ : |I m(ξ)| < α+β +1}. This contradicts that Hα,β has support of finite measure. Then(S, ) is weak annihilating pair for the Opdam–Cherednik transform. According to [8, I.1.3.2.A, p. 90], ifESFis com- pact (in particular if ESF is Hilbert–Schmidt), then if the pair(S, )is weakly
annihilating, it is also strongly annihilating.
3 The Donoho–Strak’s uncertainty principle
The classical uncertainty principle says that if a functionf(t)is essentially zero outside an interval of lightt and its Fourier transform f(w)is essentially zero outside an interval of lengthw,then
t.w≥1.
In this section we will prove a quantitative uncertainty inequality like (4) about the essential supports of a nonzero function f ∈L2(R,Aα,β)and its Opdam–Cherednik transform. The first such inequality for the usual Fourier transform was obtained by Donoho–Stark [5].
Let 0< ε(2,S), ε(2,)<1 and let f ∈ L2(R,Aα,β)be a nonzero function. We say that f isε(2,S)-time-limited onSif:
EScfL2( ,Aα,β)≤ε2,SfL2( ,Aα,β).
Similarly we say that f isε(2,)-band-limited onfor the Opdam–Cherednik trans- form if
FcfL2( ,Aα,β)≤ε2,fL2( ,Aα,β).
Theorem 3 Let,S ⊂Rbe a pair of measurable subsets and letε(2,S), ε(2,)>0 such thatε(22,S)+ε(22,) <1.Let f ∈ L2(R,Aα,β)be a non function. If f isε(2,S)- time-limited on S andε(2,)-band-limited onfor the Opdam–Cherednik transform, Then
Aα,β(S)σα,β()≥ 1−
ε2(2,S)+ε(22,)2
.
Proof The result follows from inequalities (6) and (7).
Often the uncertainty principle is used to show that certain things are impossible, such as determining the momentum and position of a particle simultaneously or mea- suring the “instantaneous frequency” of a signal. In the next we present an example where the generalized uncertainty principle Shows something. Unexpected is possible specifically the recovery of a signal or image despite significant amounts of missing information.
The following example is prototypical. A signal f is transmitted to a receiver who know that f is bandlimited onSfor the Opdam–Cherednik transform, meaning that f is synthesized using only frequency onS; equivalently f = Ff.Suppose that the observation of f is corrupted by a noisen ∈ L2(R,Aα,β)(which is nonetheless assumed to be small) and an unregistered values onS. Thus, the observable function rsatisfies
r(x)=
f(x)+n(x), x∈Sc;
0, x∈S.
Here, we have assumed without loss of generality thatn=0 onS. Equivalently, r =(I −ES)f +n.
We say that f can be stably reconstructed fromr, if there exists a linear operatorK and a constantCsuch that:
f −K rL2( ,Aα,β)≤CnL2( ,Aα,β). (8) The estimate (8) shows that the noisenis at most amplified by a factorC.
Corollary 1 If S andare arbitrary measurable sets ofRwith0<Aα,β(S)σα,β()
<1,then f can be stably reconstructed from r . The constant C in Eq. (8) is not larger than
1−
Aα,β(S)σα,β()−1
.
Proof IfAα,β(S)σα,β() <1,using (7),ESF<1.HenceI−ESFis invertible.
Let
K =(I −ESF)−1.
Since f is bandlimited on, then(I−ES)f =(I−ESF)f. Therefore f −K r = f −K((I−ES)f +n)
= f −K(I−ESF)f −K n
= f −(I−ESF)−1(I−ESF)f −K n
=0−K n. So that
f −K rL2( ,Aα,β) = K nL2( ,Aα,β)
≤ (I−ESF)−1nL2( ,Aα,β)
≤∞
k=0
ESFknL2( ,Aα,β)
≤ ∞ k=0
(Aα,β(S)σα,β())k2nL2( ,Aα,β)
= 1−
Aα,β(S)σα,β()−1
nL2( ,Aα,β).
The constantCin Eq. (8) is therefore not larger than 1−
Aα,β(S)σα,β()−1
. The identity
K =(I −ESF)−1=∞
k=0
(ESF)k
suggest an algorithm for computingK r. Put f(n)=
n
k=0
(ESF)kr,
then
f(0)=r, f(n+1)=r+ESFf(n)and f(n)→ K r as n→ ∞.
As f is bandlimited onwe deduce that
f(n+1)− f =ESF(f(n)− f).
Theorem 4 If S andare arbitrary measurable sets ofR. If the pair(S, )is strongly annihilating for the Opdam–Cherednik transform, then f can be stably reconstructed from r . The constant C in Eq. (8) is not larger than
1−
Aα,β(S)σα,β()−1
. Proof The result follows immediately from the proof of Theorem1and from Lemma
2.
Acknowledgements The authors would like to thank the anonymous reviewers for their valuable com- ments.
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