R E S E A R C H
Open Access
On equivalent conditions of two sequences
to be R-dual
Zhitao Chuang
1*and Junjian Zhao
2*Correspondence:
1College of Mathematics and
Information Science, North China University of Water Resources and Electric Power, Zhengzhou, 450011, P.R. China
Full list of author information is available at the end of the article
Abstract
The concept of R-duals was introduced by Casazza, Kutyniok, and Lammers in 2004. In this paper, we give a condition when a Parseval frame can be dilated to an
orthonormal basis of a given separable Hilbert spaceH. This is advantageous for deriving a condition for a sequence{
ω
j}j∈Jto be an R-dual of a given frame{fj}j∈J.1 Introduction
The concept of R-duals was first introduced by Casazzaet al.in []. Although it is defined in general frame theory, there is a natural connection with Gabor frame theory. And it is still an open problem whether the duality principle in Gabor analysis actually can be derived from the theory of the R-dual. Lots of scholars have done much research in this area. Reference [] introduces various alternative R-duals and shows their relations with Gabor frames. References [] and [] consider R-dual in Banach space. In [], the authors give an equivalent condition for a sequence{ωj}j∈Jto be an R-dual of a given frame{fj}j∈J.
However, we think there is a mistake in their proof. The correction of it will be discussed in Section .
The dilation viewpoint on frames is introduced by Larson and Han in [], which has a natural relation with the R-dual. They point out that any Parseval frame can be dilated to an orthonormal basis. But given a Hilbert spaceHand a Parseval frame of a subspace ofH, can the Parseval frame be dilated to an orthonormal basis forH? This will be discussed in Section .
In the entire paper, we letH denote a separable Hilbert space, with the inner product
·,·, andJbe a countable index set.
Definition A sequence{fj}j∈J of elements inHis a frame forHif there exist constants
A,B> such that
Af≤
j∈J
f,fj
≤Bf, f ∈H.
The constantsA,Bare called a lower and upper frame bounds for the frame. A frame is A-tight, ifA=B. IfA=B= , it is called a Parseval frame (a normalized tight frame in []).
Definition A sequence{ωj}j∈JinHis a Riesz sequence if there exist constantsC,D>
such that
C
j∈J |cj|≤
j∈J
cjωj
≤D
j∈J |cj|
for all finite sequence{cj}j∈J. The numbersC,Dare called Riesz bounds. A Riesz sequence
is a Riesz basis forHif it is complete inH.
For more information as regards frames and Riesz bases we refer to the monograph []. We now state the definition of the R-dual sequence.
Definition [] Let{ei}i∈J and{hi}i∈Jdenote two orthonormal bases forH, and let{fi}i∈J
be any sequence inHfor which
i∈J
fi,ej
<∞, ∀j∈J.
The R-dual of{fi}i∈Jwith respect to the orthonormal bases{ei}i∈jand{hi}i∈Jis the sequence {ωj}j∈J given by
ωj=
i∈J
fi,ejhi, j∈J. (.)
It is well known from [] that {fi}i∈J is a frame forH with bounds A,Bif and only if {ωj}j∈Jis a Riesz sequence inHwith boundsA,B. But given two sequence{fi}i∈Jand{ωj}j∈J,
under what conditions can we find orthonormal bases{ei}i∈J and{hi}i∈J forHsuch that
(.) holds? This is the main question we want to answer in this paper. It will be discussed in Section explicitly. Assume that{fi}i∈J is a frame forH. Define a sequence{ni}i∈Jby
ni=
k∈J
ek,fi ˜ωk, i∈J, (.)
where{ ˜ωj}j∈J is the canonical dual Riesz sequence of{ωj}j∈J. The construction of{ni}i∈J
comes from []. It plays an important role in this paper.
Proposition [] Let{ωj}j∈J be a Riesz basis for the subspace W of H,with dual Riesz
basis{ ˜ωk}k∈J.Let{ei}i∈Jbe an orthonormal basis for H.Given any sequence{fi}i∈Jin H,the
following hold:
(i) There exists a sequence{hi}i∈JinHsuch that
fi=
j∈J
ωj,hiej, ∀i∈J. (.)
(ii) The sequence{hi}i∈Jsatisfying(.)is characterized as
hi=mi+ni, (.)
(iii) If{ωj}j∈Jis a Riesz basis forH,then(.)has the unique solution
hi=ni, i∈J.
In [], Christensenet al.give a solution to the main question.
Theorem [] Let{ωj}j∈J be a Riesz sequence spanning a proper subspace W of H and {ei}i∈J an orthonormal basis for H.Given any frame{fi}i∈J for H,the following are
equiva-lent:
(i) {ωj}j∈Jis an R-dual of{fi}i∈Jw.r.t.{ei}i∈Jand some orthonormal basis{hi}i∈J.
(ii) There exists an orthonormal basis{hi}i∈JforHsatisfying(.).
(iii) The sequence{ni}i∈Jin(.)is a Parseval frame.
We point out that, in fact, (iii) is not equivalent to the other items in Theorem . In order to clarify this, we need the following proposition from [].
Proposition [] Let J be a countable(or finite)index set.Suppose that{xn:n∈J}is a
Parseval frame for W.Then there exist a Hilbert space K⊇W and an orthonormal basis
{en:n∈J}for K such that Pen=xn,where P is the orthogonal projection from K onto W.
2 A dilation theorem
In this section, a dilation theorem is given, which will be used in Section . Firstly, we give an example to show that Theorem is not strictly right.
Example In this example, we choose the index setJ=N, the natural number set.
Sup-pose{zi}i∈J is an orthonormal basis forH. Definefi= ziandωi= zifor alli∈J. Then
the sequence{fi}i∈Jis a Parseval frame with frame bounds and{ωj}j∈Jis a Riesz sequence
with bounds as well. The canonical dual{ ˜ωj}j∈Jof{ωj}j∈J equals{zj}j∈J. Let
ni=
k∈J
zk,fi ˜ωk=
k∈J zk, zi
zk=zi.
Obviously,{ni}i∈J is a Parseval frame, but{ωj}j∈J cannot be an R-dual of{fi}i∈J. If not, by
(ii) of Proposition , an orthonormal basis{hi}i∈J forHcan be characterized by
hi=mi+ni,
wheremi∈W⊥for alli∈J. Sinceni∈W, we have
=hi=mi+ni=mi+ni=mi+zi.
Sincezi= , one hasmi= for alli∈J. Thereforehi=ni=zi. This contradicts{hi}i∈J
being an orthonormal basis forH. Thus (iii) of Proposition is not right.
Theorem Given two separable Hilbert spaces H⊇M,suppose that{xn}n∈J is a Parseval
frame for W.Then there exists an orthonormal basis{en}n∈Jfor H s.t.Pen=xnif and only if
dim(kerT) =dimW⊥, (.)
where P is an orthogonal projection from H onto W,T is the synthesis operator of{xi}i∈J.
Proof First we treat sufficiency. Since
i∈J
cixi=
i∈J
ciPei=P
i∈J
ciei,
for any{ci}i∈J∈(J), a sequence{ci}i∈J∈kerTif and only if
i∈Jciei∈W⊥. So (.) holds.
Now we treat necessity. Suppose (.) holds, from the proof of the Proposition , there exist a Hilbert spaceK=(J), an orthogonal projectionP, and an orthonormal basis{e
i}i∈J
forK, such that
Pei=θ(xi), (.)
where θ is the analysis operator of{xi}. Sinceθ is injective, it has inverse restricted to
θ(W). For simplicity, we just denote it byθ–. For any{ci}i∈J∈(J), since
i∈J
cix,xi=
x,
i∈J
cixi ,
we have
dim kerT=dimθ(W)⊥. (.)
Together with (.), we have
dimW⊥=dim(kerT) =dimθ(W)⊥.
Therefore, there is an unitary operatorηfromW⊥onto (θ(W))⊥. Combining withθ, we can define a unitary operatorUfromHontoK:
Ut=U(t+t) =θt+ηt, t∈W,t∈W⊥. One can easily get
U–y=U–(y+y) =θ–y+η–y, y∈θ(W),y∈θ(W)⊥. Therefore,U∗=U–. In fact, fort∈Handy∈K,
Ut,y=U(t+t),y+y
=t,θ–y
+t,η–y
=t,θ–y+η–y
=t,U–y
=t,U∗y,
where the third equation is due to the Parseval frame property of{xn}n∈Jand unitarity ofη.
Because of the unitarity ofU, alsoi=U–eiis an orthonormal basis forH.
Now, takingU–on the two sides of (.), we have
U–Pei=U–PUU–ei=U–PUi=xi.
We claim that U–PU is also an orthogonal projection. In fact, by the properties ofU andP, we have
U–PU=U–PU=U–PU
and
U–PU∗=U∗PU∗=U∗PU=U–PU.
Thus we get as desired the complete proof.
3 Conditions of R-dual
In this section, we discuss under what conditions{ωi}i∈Jcan be an R-dual of{fi}i∈J. At first,
we give two lemmata which will be used later.
Lemma Let{ni}i∈Jbe defined as(.),W the close span of{ωj}j∈J,thenspan{ni}i∈J=W.
Proof Sinceni=
k∈Jek,fi ˜ωk, we have span{ni}i∈J⊆span{ ˜ωi}i∈J=W.
In the opposite direction, since{fi}i∈Jis a frame forH, there exists a sequence{c}∈J∈(J)
such thatem=
∈Jcfform∈J. Then one has
∈J
cn=
∈J
c
k∈J
ek,f ˜ωk=
k∈J
ek,
∈J
cf ω˜k=
k∈J
ek,em ˜ωk=ω˜m.
ThusW⊆span{ni}i∈J. We have the desired result.
DefineSωf =
k∈Jf,ωkωkandSω˜f =
k∈Jf,ω˜k ˜ωk, forf ∈W. ThenS
–
˜
ω ω˜kis an or-thonormal basis forW. Sinceωk,S–ωω=δk,by [], one hasω˜k=Sω–ωk. Furthermore,
we have
Sω˜f =
k∈J
f,S–ωωk
S–ωωk=S–ωSωSω–f =S–ωf, ∀f∈W.
Let k=S
–
˜
ω ω˜, thenω˜k=S
˜
ωk. Let{ek}k∈J be an orthonormal basis forH, define an antiunitary operator:H−→Wby
f =
k∈J
ckek
=
k∈J
ckk, forf=
k∈J
ckek∈H.
Obviously, the inverse ofis also an antiunitary operator and
–g=–
k∈J
ckk
=
k∈J
ckek, ∀g∈W.
Furthermore, the antiunitary operatorhas the following property.
Lemma Letbe defined as above,thenf,g=–g,ffor any f∈H and g∈W. Proof By the definition of, one has
f,g=
k∈J
ek,fk,g =
k∈J
ek,fk,g=
k∈J
k,gek,f =
–g,f.
Theorem There exists an orthonormal basis{ei}i∈Jsuch that{ni}i∈Jis a Parseval frame
if and only if there exists an antiunitary operatorsuch that Sω=S–,where S is the frame operator of{fi}i∈J.
Proof By the definition of{ni}i∈J and Lemma , we have
i∈J
f,ni
=
i∈J
f,
k∈J
ek,fi ˜ωk
=
i∈J
k∈J
fi,ekf,ω˜k
=
k∈J
∈J
i∈J
fi,eke,fi
f,ω˜k ˜ω,f
=
k∈J
∈J e,Sek
f,S
˜
ωek
S
˜
ωe,f
=
k∈J
∈J e,Sek
ek,–S
˜ ωf –S ˜
ωf,e
=
k∈J
ek,–S
˜ ωf –S ˜
ωf,Sek
=
k∈J
–S
˜
ωf,Sek
ek,–S
˜
ωf
=S–S
˜
ωf, –S
˜
ωf
=f,S
˜
ωS –S
˜
ωf
Suppose{ni}i∈J is a Parseval frame; then we have
i∈J
f,ni
=f, ∀f ∈W. (.)
By (.), it becomes
i∈J
f,ni
=f,S
˜
ωS –S
˜
ωf
=f,f. (.)
For arbitrary complex numbersaandb, we have
S–(af +bg) =Sa–f+b–g=aS–f+bS–g.
ThusS–is a linear operator, so is the operatorS
˜
ωS–S
˜
ω. This meansS
˜
ωS–S
˜
ω= Iby (.),i.e.
Sω=S–.
On the other hand, assume there exists an antiunitary operatorsuch thatSω=S–. Defineej=–j=–S
–
ω ω˜k, then (.) means
ni=
k∈J
ek,fi ˜ωk
is a Parseval frame.
Theorem Suppose{fi}i∈Jis a frame for a separable Hilbert space H and{ωj}j∈Jis a Riesz
sequence in H.{fi}i∈Jis an R-dual of{ωj}j∈Jif and only if the following two conditions hold:
(i) there exists an antiunitary operators.t.Sw=S–;
(ii) dim(kerT) =dim(W⊥).
Proof By Proposition ,{fi}i∈Jis an R-dual of{ωj}j∈J if and only if{ni}i∈J can be dilated to
an orthonormal basis forH. By Theorem , this is equivalent to{ni}i∈J being a Parseval
frame and (ii) holding. Using Theorem , we see that{fi}i∈J is an R-dual of{ωj}j∈J if and
only (i) and (ii) hold.
We appreciate one reviewer having pointed out that Theorem is of exactly the same type as the characterizations of type II/III in []. In the special case, if{fi}i∈Nis an A-tight
frame for a separable Hilbert spaceHwith infinite dimension and{ωj}j∈N is an A-tight
Riesz sequence whereNdenotes the natural number set, then there must be an antiunitary operatorformHontoW. So we haveS=AIH,SW=AIW, and
SW=AIW=AIH–=S–.
Corollary []Let{fi}i∈J be a tight frame for H and let{ωj}j∈J be a tight Riesz sequence
in H with the same bound.Denote the synthesis operator for{fi}i∈Jby T.Then{ωj}j∈Jis an
R-dual of{fi}i∈Jif and only ifdim(kerT) =dim(W⊥)holds.
Remark SinceSWf =
j∈Jf,ωjωjand
S–f =
j∈J
fj,–f
fj=
j∈J
f,fjfj,
(i) of Theorem is equivalent to there existing an antiunitary operator such that
j∈J
f,ωjωj=
j∈J
f,fjfj.
Remark For parametersa,b∈R, define the operatorsTaandEbonL(R)byTaf(x) =
f(x–a) and Ebf(x) = eπibxf(x), respectively. From [], we know that if ab< and {EmbTnag}m,n∈Z is a frame, then {EmbTnag}m,n∈Z has an infinite excess. If ab> , then {EmbTna}m,n∈Zhas an infinite deficit. This demonstrates that, if we want to solve the open
problem, we only need (i) of Theorem to hold. By Remark , this is equivalent to finding an antiunitary operatorsuch that
m,n
f,EmbTnagEmbTnag=
m,n
f,√
abEm/aTn/bg
√
abEm/aTn/bg.
Competing interests
The authors declare that they have no competing interests.
Authors? contributions
All authors contributed to each part of this work equally and read and approved the final manuscript.
Author details
1College of Mathematics and Information Science, North China University of Water Resources and Electric Power,
Zhengzhou, 450011, P.R. China. 2Department of Mathematics, Tianjin Polytechnic University, Tianjin, 300387, P.R. China.
Acknowledgements
This study was partially supported by the National Science Foundation for Young Scientists of China (Grant No. 11401433). The authors would like to express their thanks to the reviewers for their helpful comments and suggestions.
Received: 28 September 2014 Accepted: 13 December 2014 References
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