GEOMETRIC ASPECTS OF FRAME REPRESENTATIONS OF ABELIAN GROUPS 1. Introduction ...

20 Pages·2004·0.28 MB·English

TRANSACTIONSOFTHE
AMERICANMATHEMATICALSOCIETY
Volume356,Number12,Pages4767{4786
S0002-9947(04)03679-7
ArticleelectronicallypublishedonJune22,2004
GEOMETRIC ASPECTS OF FRAME REPRESENTATIONS
OF ABELIAN GROUPS
AKRAMALDROUBI,DAVIDLARSON,WAI-SHINGTANG,ANDERICWEBER
Abstract. Weconsiderframesarisingfromtheactionofaunitaryrepresen-
tation of a discrete countable abelian group. We show that the range of the
analysisoperatorcanbedeterminedbycomputingwhichcharactersappearin
therepresentation. Thisallowsonetocomparetherangesoftwosuchframes,
which is useful for determining similarity and also for multiplexing schemes.
Our results then partially extend to Bessel sequences arising from the action
ofthegroup. Weapplytheresultstosamplingonbandlimitedfunctions and
to wavelet andWeyl-Heisenberg frames. Thisyields asu(cid:14)cient condition for
two sampling transforms to have orthogonal ranges, and two analysis opera-
tors forwavelet and Weyl-Heisenberg framesto have orthogonal ranges. The
su(cid:14)cient condition is easy to compute in terms of the periodization of the
Fouriertransformoftheframegenerators.
1. Introduction
FramesinaseparableHilbertspaceprovideredundantbutstableexpansionsfor
theHilbertspace. Framesarisenaturallyinmanysettings,suchassamplingtheory,
time-scale(wavelet)analysis,andtime-frequency(WeylHeisenbergorGabor)anal-
ysis. The \power of redundancy" o(cid:11)ered by frames has been successfully demon-
strated in such settings, such as Ron and Shen constructing compactly supported
wavelet frames with high approximation order ([23] and references), Benedetto et.
al. denoising signals, and other data ([5, 6]). The focus of the present paper is
motivatedbyanotheraspectofthepowerofredundancywhichallowsmultiplexing
of signals.
A frame for a separable Hilbert space H is a sequence X = fxjgj2J such that
there exist constants 0<C and C <1 such that for all x2H,
1 2
X
C kxk2 (cid:20) jhx;x ij2 (cid:20)C kxk2:
1 j 2
j2J
There exists a (possibly non-unique) dual frame sequence x~ such that the recon-
j
struction formula holds:
X X
x= hx;x ix~ = hx;x~ ix :
j j j j
j2J j2J
ReceivedbytheeditorsMarch4,2002.
2000 Mathematics Subject Classi(cid:12)cation. Primary 43A70, 94A20, 42C40; Secondary 43A45,
46N99.
Key words and phrases. Regular sampling, periodic sampling, multiplexing, locally compact
abeliangroup,framerepresentation, spectralmultiplicity,wavelet, Weyl-Heisenbergframe.
(cid:13)c2004AmericanMathematicalSociety
4767
4768 A.ALDROUBI,D.LARSON,W.-S.TANG,ANDE.WEBER
The analysis operator for X is
(cid:2) :H !l2(J):x7!(hx;x i):
X j
Often one wishes to know the range of the analysis operator; for instance, two
frames fxjgj2J (cid:26)H and Y =fyjgj2J (cid:26)K are similar if and only if their analysis
operators have the same range in l2(J): If the range of (cid:2) is orthogonal to the
X
range of (cid:2) , then for all x2H and y 2K,
Y
X
x= (hx;x i+hy;y i)x~ ;
j j j
Xj2J
y = (hx;x i+hy;y i)y~ ;
j j j
j2J
provided x~ and y~ are the standard duals of x and y , respectively. Such frames
j j j j
are called strongly disjoint (see [18]) and this procedure is what engineers call
multiplexing.
The purpose of this paper is to determine the range of the analysis operator for
aframeoraBesselsequencethatarisesfromthe actionofaunitaryrepresentation
of a discrete countable abelian group. Such frames occur in the setting of regular
sampling on bandlimited functions. In section 2, we show that the range of the
analysis operator can be determined by computing which characters appear in the
representation. ABesselsequencefxjgj2J (cid:26)H issuchthataconstantC2 asabove
exists,butthereisnotnecessarilyaconstantC . ABesselsequencealsode(cid:12)nes an
1
analysis operator just as a frame sequence does. Bessel sequences occur naturally
when considering a subsequence of a frame sequence, which we shall do in our
investigation of wavelet and Weyl-Heisenberg frames.
Theproblemofsamplingiswellknownandsamplingtheoryinvarioussettingsis
wellestablished(see[4,19]foroverviews). Inthispaper,wewillrestrictourselvesto
samplingontotallytranslationinvariantsubspacesV ,i.e. allfunctionsf 2L2(Rd)
E
such that the support of f^is contained in the band (i.e. measurable set) E (cid:26) Rd.
We shall assume that the band has (cid:12)nite measure. A d(cid:2)d invertible matrix A is
a sampling matrix for the set E if the lattice fAZdg is a set of sampling for the
spaceV . By asetofsamplingwe meanthatthe normofthe function ispreserved
E
by the sampling transform(cid:2) , i.e. there are constants C andC suchthat for all
A 1 2
f 2V ,
E
C kfk2 (cid:20)k(cid:2) (f)k2 (cid:20)C kfk2;
1 A 2
where (cid:2) is de(cid:12)ned as
A
(cid:2) :V !l2(Zd):f 7!(f(Az)) :
A E z
The samplingtransformde(cid:12)nes aunitary representationofZd onV . We apply
E
the results of section 2 to obtain the following result regarding sampling, which is
proven in section 3:
Theorem 1. Let A be a sampling matrix for the set E, and let B be a sampling
matrix for the set F. De(cid:12)ne the sums
X X
m ((cid:24)):= (cid:31) (A(cid:3)(cid:0)1((cid:24)+k)); m ((cid:24)):= (cid:31) (B(cid:3)(cid:0)1((cid:24)+k)); (cid:24) 2Rd;
A E B F
k2Zd k2Zd
and let X and Y denote the support sets of m and m , respectively. Then we
A B
FRAME REPRESENTATIONS OF ABELIAN GROUPS 4769
have the following:
1. (cid:2) (V )=(cid:2) (V ), if and only if (cid:21)(X(cid:1)Y)=0;
A E B F
2. (cid:2) (V )?(cid:2) (V ) if and only if (cid:21)(X \Y)=0;
A E B F
3. (cid:2) (V )\(cid:2) (V )6=f0g if and only if (cid:21)(X \Y)6=0;
A E B F
4. (cid:2) (V )(cid:26)(cid:2) (V ) if and only if X (cid:26)Y modulo null sets;
A E B F
5. the two conditions that a) (cid:2) (V )\(cid:2) (V )=f0g and b) the angle between
A E B F
(cid:2) (V ) and (cid:2) (V ) is strictly positive imply that (cid:2) (V ) and (cid:2) (V ) are
A E B F A E B F
orthogonal;
6. the samples for A and B commute, i.e. the projections P and P onto
A B
(cid:2) (V ) and (cid:2) (V ), respectively, commute.
A E B F
In section 4, we turn to frames which arise from the ordered product of two
unitary groups, i.e. fG G (cid:9)g, where G and G are unitary groups, and (cid:9) (cid:26) H
1 2 1 2
is a (cid:12)nite set. If fG G (cid:9)g is a frame, then the sequences fG g and fG g are
1 2 1 j 2 j
Bessel, to which we apply results from section 2.
We de(cid:12)ne for any (cid:11) 2 Rd the translation operator T by T f = f((cid:1)(cid:0)(cid:11)) and
(cid:11) (cid:11)
the modulation operator E f = e2(cid:25)ih(cid:11);(cid:1)if((cid:1)). For an expansive matrix A (all
(cid:11)
its eigenvaplues have modulus greater than one), we de(cid:12)ne the dilation operator
D f((cid:1))= jdetAjf(A(cid:1)).
A
A wavelet frame (cid:9) = f ;:::; g for L2(Rd) has the form fD T (cid:9) : m 2
1 n Am Xz
Z;z 2 Zdg, where A is expansive and X is nonsingular. The analysis operator for
such a frame is
Mn
(cid:2) :L2(Rd)! l2(Z(cid:2)Zd):x7!(hx;D T i):
(cid:9) Am Xz j
j=1
Likewise,aWeyl-HeisenbergframehastheformfE T f :l;z 2Zd;j=1;:::;ng,
Al Xz j
where F =ff ;:::;f g, with analysis operator
1 n
Mn
(cid:2) :L2(Rd)! l2(Zd(cid:2)Zd):x7!(hx;E T f i):
F Al Xz j
j=1
Here A and X are both nonsingular (but not necessarily expansive).
ByconsideringtheBesselsequencesfT g,fT f g,andfE f g,wegetthe
Xz j Xz j Al j
following results.
Theorem 2. Suppose fD T :j =1;:::;ng and fD T (cid:30) :j =1;:::;ng
Am Xz j Bm Yz j
are a(cid:14)ne frames. De(cid:12)ne the sums
X X
m ((cid:24)):= j ^ (X(cid:3)(cid:0)1((cid:24)+k))j2 and n ((cid:24)):= j(cid:30)^ (Y(cid:3)(cid:0)1((cid:24)+k))j2:
j j j j
k2Zd k2Zd
Let E and F denote the support sets of m and n , respectively. The following
j j j j
statements hold:
1. if (cid:2) (L2(Rd))(cid:26)(cid:2) (L2(Rd)), then (cid:21)(E nF )=0 for j =1;:::;n;
(cid:9) (cid:8) j j
2. if (cid:2) (L2(Rd))=(cid:2) (L2(Rd)), then (cid:21)(E (cid:1)F )=0 for j =1;:::;n;
(cid:9) (cid:8) j j
3. if (cid:21)(E \F ) = 0 for j = 1;:::;n, then (cid:2) (L2(Rd)) ? (cid:2) (L2(Rd)), i.e.
j j (cid:9) (cid:8)
fD T : j = 1;:::;ng and fD T (cid:30) : j = 1;:::;ng are strongly
Am Xz j Bm Yz j
disjoint.
4770 A.ALDROUBI,D.LARSON,W.-S.TANG,ANDE.WEBER
Theorem 3. Suppose that fE T f : j = 1;:::;ng and fE T g : j =
Al Xz j Bl Yz j
1;:::;ng are Weyl-Heisenberg frames. De(cid:12)ne the sums
X X
m ((cid:24)):= jf^(X(cid:3)(cid:0)1((cid:24)+k))j2 and n ((cid:24)):= jg^ (Y(cid:3)(cid:0)1((cid:24)+k))j2;
j j j j
kX2Zd kX2Zd
m~ ((cid:24)):= jf (A(cid:3)(cid:0)1((cid:24)+k))j2 and n~ ((cid:24)):= jg (B(cid:3)(cid:0)1((cid:24)+k))j2:
j j j j
k2Zd k2Zd
Let E , F , E~ , and F~ denote the support sets of m , n , m~ , and n~ , respectively.
j j j j j j j j
Then the following hold:
1. if (cid:2) (L2(Rd)) (cid:26) (cid:2) (L2(Rd)), then (cid:21)(E nF ) = 0 and (cid:21)(E~ nF~ ) = 0 for
F G j j j j
j =1;:::;n;
2. if (cid:2) (L2(Rd)) = (cid:2) (L2(Rd)), then (cid:21)(E (cid:1)F ) = 0 and (cid:21)(E~ (cid:1)F~ ) = 0 for
F G j j j j
j =1;:::;n;
3. if (cid:21)(E \F )=0 for j =1;:::;n, then (cid:2) (L2(Rd))?(cid:2) (L2(Rd));
j j F G
4. if X(cid:3)A = Y(cid:3)B and (cid:21)(E~ \F~ ) = 0 for j = 1;:::;n, then (cid:2) (L2(Rd)) ?
j j F
(cid:2) (L2(Rd)).
G
We make several remarks regarding the remainder of the paper. All Hilbert
spaces will be separable. All groups will be countable and Abelian. We will use (cid:21)
to denote Haar measure on G, on G^ (the dual group), and consequently also for
Legesgue measure on Rd (this should cause no confusion). We utilize the following
de(cid:12)nition of the Fourier transform: for f 2L1(Rd),
Z
f^((cid:24))= e(cid:0)2(cid:25)ih(cid:24);xif(x)dx:
Rd
Therefore, the inversion formula is given by
Z
f(x)= e2(cid:25)ihx;(cid:24)if^((cid:24))d(cid:24)
Rd
when the integral is valid.
2. Representation theory
LetGbeadiscrete,countableabeliangroup,andlet(cid:25) :G!B(H)beaunitary
b
representation of G on a separable Hilbert space H. Let G denote the dual group
of G, i.e. the group of characters on G. Let (cid:21) denote normalized Haar measure on
b b
G. By Stone’s theorem there exists a projection-valued measure p on G such that
Z
(cid:25)(g)= g((cid:24))dp((cid:24)):
b
G
Then, by the theory of projection-valuedmeasures,there exists a probability mea-
sure (cid:22) on Gb, a multiplicity function m:Gb !f0;1;:::;1g and a unitary operator
Mk
U :H ! L2(F ;(cid:22)),!L2(F ;(cid:22);Ck);
j 1
j=1
where Gb (cid:27) F (cid:27) F (cid:27) (cid:1)(cid:1)(cid:1) (cid:27) F and m((cid:24)) = #fF : (cid:24) 2 F g. By L2(F ;(cid:22);Ck),
1 2 k j j 1
we mean functions on F attaining values in Ck which are measurable and square
1
integrable with respect to the measure (cid:22). Without loss of generality, the sets F
j
may all be taken to have non-zero (cid:22) measure. Note that k above could be in(cid:12)nite,
in which case Ck is replaced by l2(G).
FRAME REPRESENTATIONS OF ABELIAN GROUPS 4771
The operatorU intertwines the projection-valuedmeasureon H andthe canon-
b
icalprojection-valuedmeasure onG, andcanbe thoughtof asa Fourier-liketrans-
form on H. Indeed, if x2 H, we will denote Ux by x^. We will call the unitary U
the decomposition operator. Moreover, we will see that for our representation (cid:25) ,
A
the Fourier transform on L2(Rd) is used in computing the unitary U.
Byutilizingthesetheorems,wehavethattherepresentation(cid:25) isunitarilyequiv-
alent, via the decomposition operator, to the representation
0 1
Mk
(cid:27) :G!U@ L2(F ;(cid:22))A
j
j=1
given by
(cid:27)(g)=M
g((cid:24))
where M is the multiplication operator with symbol g((cid:24)).
g((cid:24))
Two representations (cid:25) and (cid:26) are unitarily equivalent if there is a unitary V :
H !K,calledanintertwining operator, suchthat(cid:26)(g)V =V(cid:25)(g). Suppose (cid:25) and
(cid:26) are two representations of an Abelian group G on the Hilbert spaces H and K,
respectively. Then there is a measure (cid:22) and (cid:22) associatedto eachrepresentation,
(cid:25) (cid:26)
and a multiplicity function m and m associated to each representation. Two
(cid:25) (cid:26)
representations are unitarily equivalent if and only if the measures are equivalent
andthemultiplicityfunctions agreeuptoasetofmeasure0(withrespecttoeither
measure).
Note that the cyclic subspaces generated by ((cid:31) ((cid:24));0;:::;0) and (0;(cid:31) ((cid:24));
F1 F2
:::;0) are orthogonal. It follows that if there is a vector v 2 H such that the
collection f(cid:25)(g)v : g 2 Gg is a frame for H, then the representation (cid:25) must be
cyclic,whence the multiplicity function mustbe bounded aboveby 1 a:e:(cid:22)andthe
decomposition operator U maps H !L2(F ;(cid:22)).
1
If a group representation has a frame vector as above, we say that the repre-
sentation is a frame representation. A frame representation of a group must be
unitarily equivalent to a subrepresentation of the (left) regular representation [18,
Theorem 3.11] and [25]. This, in turn, implies that the measure (cid:22) is equivalent to
the restriction of Haar measure (cid:21) to the set F .
1
Lemma 1. Let (cid:25) be a representation of the Abelian group G on the Hilbert space
K. The vector w 2 K is a frame vector for (cid:25) if and only if the following two
conditions hold:
1. the decomposition operator U maps K !L2(F;(cid:21)j ),
F
2. there exist positive constants C and C such that for almost every (cid:24) 2F,
1 2
C (cid:20)jUw((cid:24))j2 =jw^((cid:24))j2 (cid:20)C :
1 2
Moreover, the lower and upper frame bounds are given by the supremum of all such
C andin(cid:12)mumofallsuchC , respectively. Inparticular, xgeneratesatight frame
1 2
under the action of (cid:25) if and only if jw^((cid:24))j2 =C for (cid:21) almost every (cid:24) in F.
1
Proof. We have established the necessity of condition 1. We now show, by contra-
positive, the necessity of the bound in condition 2. Suppose C > 0 is given and
suppose that jw^((cid:24))j2 > C for (cid:24) 2 F0 (cid:26) F, F0 a set of positive measure. Then
4772 A.ALDROUBI,D.LARSON,W.-S.TANG,ANDE.WEBER
(cid:31)F0((cid:24))w^((cid:24))2L2(F;(cid:21)), and consider the following calculation:
(cid:12)Z (cid:12)
X X(cid:12) (cid:12)2
jh(cid:31)F0((cid:24));(cid:25)d(g)w^((cid:24))ij2 = (cid:12)(cid:12) (cid:31)F0((cid:24))g((cid:24))w^i((cid:24))d(cid:21)(cid:12)(cid:12)
b
g2G g2G G
=k(cid:31)F0w^k2;
b
since G forms an orthonormal basis of L2(G;(cid:21)) [16, Corollary 4.26]. We have
Z
k(cid:31)F0w^k2 = j(cid:31)F0((cid:24))w^((cid:24))j2d(cid:21)
GbZ
>C j(cid:31) ((cid:24))j2d(cid:21)=Ck(cid:31) k2;
F F
b
G
whenceupperframeboundisgreaterthanC. Ananalogouscalculationshowsthat
if there is a set F00 such that jw^((cid:24))j2 < C for (cid:24) 2 F00, then the lower frame bound
is less than C.
We now establish the su(cid:14)ciency of conditions 1 and 2. Note that by condition
1, the collection f(cid:31) ((cid:24))g((cid:24)) : g 2 Gg is a normalized tight frame for L2(F;(cid:21)j ),
F F
since it is the image of the orthonormal basis fg((cid:24)) : g 2 Gg under the projection
M . Moreover,by condition 2, for all x2K, x^((cid:24))w^((cid:24))2L2(F;(cid:21)j ). Consider
(cid:31)F F
(cid:12)Z (cid:12)
X X(cid:12) (cid:12)2
jhx;(cid:25)(g)wij2 = (cid:12)(cid:12) x^((cid:24))g((cid:24))w^((cid:24))d(cid:21)(cid:12)(cid:12)
b
g2G g2G G
=kx^w^k2
Z
= jx^((cid:24))w^((cid:24))j2d(cid:21)((cid:24))
GbZ
(cid:20)C jx^((cid:24))j2d(cid:21)((cid:24))
2
b
G
=C kxk2:
2
The lower bound follows similarly. (cid:3)
See [22] for a similar result.
Lemma 2. Let (cid:25) be a frame representation of G on H, with x and y frame vectors
for (cid:25). Then the ranges of the analysis operators (cid:2) and (cid:2) are identical.
x y
Proof. Weshallshowthatthe framesgeneratedbyxandy aresimilar. ByLemma
1, the decomposition operator U maps H to L2(F;(cid:21)j ); and the moduli of x^ and
F
y^are uniformly bounded above and below. De(cid:12)ne the operator M :L2(F;(cid:21)j )!
F
L2(F;(cid:21)j ) given by
F
y^((cid:24))
f((cid:24))! f((cid:24)):
x^((cid:24))
Clearly, M is well de(cid:12)ned and invertible and maps x^((cid:24)) to y^((cid:24)). Moreover, M
commutes with the multiplication operators g((cid:24)). Therefore, the operator U(cid:3)MU
is the necessary similarity. (cid:3)
ForaframerepresentationofG,thereisactuallyasecondunitaryrepresentation
ofGinthespaceofcoe(cid:14)cients. If(cid:25)isaframerepresentationofGonH,withframe
vectorv 2H,then the analysis operator(cid:2) :H !l2(G) intertwines the operators
v
FRAME REPRESENTATIONS OF ABELIAN GROUPS 4773
(cid:25)(g) and L , i.e. (cid:2) (cid:25)(g) = L (cid:2) where L : l2(G) ! l2(G) : x(h) 7! x(g(cid:0)1h).
g v g v g
Indeed, if e denotes the characteristic function of fhg,
h
X
(cid:2) (cid:25)(g)x= h(cid:25)(g)x;(cid:25)(h)vie
v h
hX2G
= hx;(cid:25)(g(cid:0)1h)vie
h
hX2G
= hx;(cid:25)(h)vie
gh
h2GX
=L hx;(cid:25)(h)vie
g h
h2G
=L (cid:2) x:
g v
Denote by J the image of the analysis operator, and let (cid:26) denote the repre-
sentation of G on J by L j . The analysis operator (cid:2) : H ! J is invertible,
g J v
whence by the polar decomposition of (cid:2) , there is a unitary operator Q : H ! J
v
whichintertwinestherepresentations. Therefore,thereisadecompositionoperator
U~ : J ! L2(F0;(cid:21)jF0), but since this is unitarily equivalent to (cid:25)(g), F0 = F. We
have proven the following lemma.
Lemma 3. A frame representation (cid:25) of G is unitarily equivalent to the associated
representation (cid:26) given by the restriction of the left regular representation to the
range of the analysis operator.
Lemma 4. Let (cid:25) and (cid:25) be two frame representations of G on H and K, re-
H K
spectively. The ranges of the corresponding analysis operators are identical if and
only if the multiplicity functions agree modulo null sets.
Proof. If the multiplicity functions agree, then both representations are unitarily
equivalent to L2(F;(cid:21)j ), whence by Lemmas 1 and 2 the ranges will coincide.
F
Conversely, if the ranges coincide, then both representations are equivalent to the
same representation in the coe(cid:14)cient space, whence they are equivalent to each
other. (cid:3)
Two frame sequences fxjgj2J (cid:26) H and fyjgj2J (cid:26) K are said to be similar
if there is an invertible operator S : H ! K such that Sx = y . Two frame
j j
sequences are strongly disjoint if there are two frame sequences fx(cid:3)jgj2J (cid:26) H and
fyj(cid:3)gj2J (cid:26) K with the property that fx(cid:3)j (cid:8)yj(cid:3)g is a frame for H (cid:8)K and fxjg is
similar to fx(cid:3)g and fy g is similar to fy(cid:3)g.
j j j
It can be shown that two frames are similar if and only if the ranges of their
corresponding analysis operators coincide. On the other hand, two frames are
strongly disjoint if and only if the ranges of their respective analysis operators
are orthogonal [18, Theorem 2.9]. Two frame representations (cid:25) and (cid:25) of an
H K
Abelian group G are said to be strongly disjoint if the representation (cid:25) (cid:8)(cid:25) is
H K
alsoa frame representation. This is equivalentto the frames f(cid:25) (g)x:g 2Gg and
H
f(cid:25) (g)y :g 2Gg being strongly disjoint.
K
Lemma 5. Suppose (cid:25) and (cid:25) are frame representations of the Abelian group G.
H K
Then the corresponding frames are strongly disjoint if and only if the multiplicity
functions have disjoint support.
4774 A.ALDROUBI,D.LARSON,W.-S.TANG,ANDE.WEBER
Proof. By Lemma 1, there exist intertwining operators
U :H !L2(E;(cid:21)j ); V :K !L2(F;(cid:21)j );
E F
where the corresponding multiplicity functions are
m ((cid:24))=(cid:31) ((cid:24)); m ((cid:24))=(cid:31) ((cid:24)):
H E K F
We now construct a representation of G on H (cid:8)K in the usual fashion:
(cid:25)(g)=(cid:25) (g)(cid:8)(cid:25) (g);
H K
we need to check whether (cid:25) is a frame representation. It follows from the multi-
plicity theory that the multiplicity function m for (cid:25) is given by
m((cid:24))=m ((cid:24))+m ((cid:24))=(cid:31) ((cid:24))+(cid:31) ((cid:24))
H K E F
and also that the measure (cid:22) is equivalent to (cid:21)j +(cid:21)j . Thus, if (cid:21)(E \F) 6= 0,
E F
themultiplicity functionattainsthe value2onasetofnon-zeromeasure;therefore
there is no single frame vectorfor the direct sum space. Hence, the originalframes
cannot be strongly disjoint.
On the other hand, if (cid:21)(E \F) = 0, then the multiplicity function is at most
1, whence the representation (cid:25) is cyclic. Moreover, the measure (cid:22) is equivalent to
(cid:21)jE[F. Therefore,
H (cid:8)K ’L2(E[F;(cid:21)jE[F);
so by Lemma 1, (cid:31) ((cid:24))+(cid:31) ((cid:24)) is a frame vector for (cid:25). Since there is a frame for
E F
the direct sum space, it follows that the original frames were strongly disjoint. (cid:3)
Lemma 6. Suppose (cid:25) and (cid:25) are frame representations. Then the range of the
H K
analysis operator for a frame vector v of (cid:25) contains the corresponding range of
H
the analysis operator for a frame vector w of (cid:25) if and only if E (cid:27) F, where E
K
and F are the supports of the corresponding multiplicity functions.
Proof. If E (cid:27) F, then it follows that (cid:25) is equivalent to a subrepresentation of
K
(cid:25) , i.e. there is an isometry W :K !H which intertwines the representations. It
H
follows that (cid:2) (H)(cid:27)(cid:2) (K).
v w
Conversely, let P and P be the projections onto the ranges of (cid:2) (H) and
H K v
(cid:2) (K), respectively. Let U~ and V~ be the decomposition operators, respectively.
w
The projections U~P U~(cid:3) and V~P V~(cid:3) both commute with the multiplication op-
H K
erators M . Therefore, they are multiplication operators by characteristic func-
g((cid:24))
tions of sets, precisely (cid:31) and (cid:31) respectively. Now, if (cid:2) (H) (cid:27) (cid:2) (K), then
E F v w
P P =P P =P , whence E (cid:27)F. (cid:3)
H K K H K
Lemma 7. Suppose (cid:25) and (cid:25) are frame representations. The ranges of the
H K
analysis operators for the frames have non-trivial intersection if and only if the
support of the multiplicity functions have non-trivial intersection.
Proof. Ifthesupportofthemultiplicityfunctionshavenon-trivialintersection,then
for any vector x2H such that supp(Ux((cid:24)))(cid:26)E\F, there exists a vector y 2K
such that Vy((cid:24))=Ux((cid:24)) on E\F. It follows that (cid:2) x=(cid:2) y.
H K
Conversely, if the ranges of the analysis operators have non-trivial intersection,
then the representation L restricted to the intersection is non-trivial. As in the
g
proof of the previous lemma, the projection onto the intersection P is given by
D
(cid:31) for some D (cid:26)Gb when conjugated by U~. It follows that D =E\F. Therefore,
D
FRAME REPRESENTATIONS OF ABELIAN GROUPS 4775
if the representation on intersection is non-trivial, then E\F must have non-zero
measure. (cid:3)
We restate the previous lemmas together in the following theorem regarding
frame representations of an Abelian group G.
Theorem 4. Suppose (cid:25) and (cid:25) are frame representations of G on H and K,
H K
respectively. Let E;F denote the supports of the multiplicity functions for (cid:25) and
H
(cid:25) , respectively. Let (cid:2) and (cid:2) denote the analysis operators for some frame
K H K
vectors for H and K, respectively. Then the following hold:
1. (cid:2) (H)=(cid:2) (K), if and only if (cid:21)(E(cid:1)F)=0;
H K
2. (cid:2) (H)?(cid:2) (K) if and only if (cid:21)(F \E)=0;
H K
3. (cid:2) (H)\(cid:2) (K)6=f0g if and only if (cid:21)(E\F)6=0;
H K
4. (cid:2) (H)(cid:26)(cid:2) (K) if and only if E (cid:26)F modulo null sets.
H K
Thefollowingtwolemmataconcerncyclicsubrepresentationsofamultipleofthe
regular representation. The proof of the (cid:12)rst may be found in [1] (see Proposition
2.1 and Theorem 2.2).
Lemma 8. Suppose that the representation (cid:25) on H is a subrepresentation of some
multiple of the regular representation. If x 2 H, then the multiplicity function
associated to the subrepresentation (cid:25) on the cyclic subspace K = spanf(cid:25)(g)x :
K
g 2Gg is given by
m ((cid:24))=(cid:31) ((cid:24)) where F =f(cid:24) 2Gb :x^((cid:24))6=0g:
K F
Lemma 9. Let (cid:25) be a representation on H which is equivalent to a subrepresen-
tation of some multiple of the regular representation. Let x 2 H and let E =
supp(x^((cid:24))). Suppose that there are positive constants C and C such that C (cid:20)
1 2 1
jx^((cid:24))j2 (cid:20) C for (cid:21)-almost every (cid:24) 2 E. Then f(cid:25)(g)x : g 2 Gg is a frame for its
2
closed linear span.
Proof. Combine Lemmas 8 and 1. (cid:3)
We now turn to a similar analysis for Bessel vectors for a frame representation.
If (cid:25) is a frame representation of G on H, then w 2 H is a Bessel vector for (cid:25) if
there exists a constant C such that for all v 2H,
2 X
jhv;(cid:25) wij2 (cid:20)C kvk2:
g 2
g2G
ForaBesselvectorw, weagainhaveananalysisoperator(cid:2) :H !l2(G), de(cid:12)ned
w
as for a frame vector, only now (cid:2) may not be one-to-one, or even have closed
w
range. We havethe followinglemma, whichis animmediate corollaryofLemma 9.
Lemma 10. Suppose (cid:25) is a frame representation on H. Then w 2 H is a Bessel
vector for (cid:25) provided there exists a constant C such that jw^((cid:24))j(cid:20)C (cid:21) a:e:(cid:24).
2 2
Lemma11. Suppose(cid:25) isaframerepresentationon H. Supposew 2H is aBessel
vector for (cid:25), and v 2H is a frame vector for (cid:25). Then (cid:2) (H)(cid:26)(cid:2) (H).
w v
Proof. Let h2H, we shall construct an h0 2H such that hh;(cid:25) wi=hh0;(cid:25) vi. By
g g
Lemma 2, we may choose the frame vector v, which will be speci(cid:12)ed below. We
have Z Z
hh;(cid:25) wi= h^((cid:24))g((cid:24))w^((cid:24))d(cid:24) = ^h((cid:24))g((cid:24))w^((cid:24))d(cid:24):
g
b
G E
4776 A.ALDROUBI,D.LARSON,W.-S.TANG,ANDE.WEBER
Note that since w is a Bessel vector, w^((cid:24)) 2 L1(Gb;(cid:21)). We let v = U(cid:3)(cid:31) and let
E
h0 =U(cid:3)h^w^. Then
Z Z
hh0;(cid:25) vi= h^((cid:24))w^((cid:24))g((cid:24))(cid:31) ((cid:24))d(cid:24) = h^((cid:24))w^((cid:24))g((cid:24))d(cid:24):
g E
b
G E
(cid:3)
Lemma 12. Suppose (cid:25);(cid:26) are frame representations of G on H;K, respectively.
Suppose v 2H and w 2K are Bessel vectors for (cid:25) and (cid:26). Let E and F denote the
supports of v^((cid:24)) and w^((cid:24)), respectively. The following hold:
1. if (cid:2) (H)(cid:26)(cid:2) (K), then (cid:21)(EnF)=0;
v w
2. if (cid:2) (H)=(cid:2) (K), then (cid:21)(E(cid:1)F)=0;
v w
3. if (cid:21)(E\F)=0, then (cid:2) (H)?(cid:2) (K).
v w
Proof. We prove item 1 by contrapositive. Let E0 = E nF, (cid:21)(E0) 6= 0. Consider
the vector x0 =U(cid:3)(cid:31)E0 2H; we compZute (cid:2)v(x0):
hx0;(cid:25)gvi= (cid:31)E0((cid:24))g((cid:24))v^((cid:24))d(cid:24):
b
G
Since fg((cid:24))g is an orthonormal basis, for (cid:2) (y ) = (cid:2) (x ), we must have that
w 0 v 0
y^ ((cid:24))=v^((cid:24)) a.e., which is not possible.
0
For item 3, we compute the multiplicity functions of the frame subrepresenta-
tions of G on the subspaces generated by v and w. By Lemma 8, the multiplicity
functions are given by the supports of v^((cid:24)) and w^((cid:24)), whence the frame vectors in
the subrepresentationsare strongly disjoint if and only if the supports are disjoint.
Item 3 now follows from Lemma 5. (cid:3)
In [18], frame sequences fx g (cid:26) H and fy g (cid:26) K are de(cid:12)ned to be strongly
n n
disjoint if fx (cid:8) y g is again a frame for H (cid:8) K. It is then proven that they
n n
are strongly disjoint if and only if the analysis operators have orthogonal ranges.
However, with Bessel sequences, fx (cid:8)y g will always be a Bessel sequence for
n n
H (cid:8)K. Thus, we de(cid:12)ne Bessel sequences to be strongly disjoint if their analysis
operatorshaveorthogonalranges. Arestatement,then, ofthe abovelemma isthat
theBesselsequencesf(cid:25)(G)vg andf(cid:26)(G)wg arestronglydisjointifandonlyiftheir
corresponding multiplicity functions have disjoint support.
3. Sampling theory
In this section, we set out to proveTheorem 1. We de(cid:12)ne the function (cid:30) 2V
E E
by (cid:30)^ =(cid:31) . The Fourier inversion formula is valid for f 2V , therefore
E E E
f(Az)=hf((cid:1));(cid:30) ((cid:1)(cid:0)Az)i;
E
whence our stability criterion translates into the condition that the collection
f(cid:30) ((cid:1)(cid:0)Az):z 2Zdg forms a frame for V .
E E
Since V is totally translation invariant, for any matrix A, we have a represen-
E
tation(cid:25) of the integersZd onthe space V givenby (cid:25) (z)=T . Therefore,our
A E A Az
frame criterion above now becomes the question of when (cid:30) is a frame vector for
E
the representation(cid:25) .
A
Weremarkherethatothershaveusedabstractharmonicanalysisininvestigating
problems in sampling. Kluvanek [21] was the (cid:12)rstto do so; Dodson and Beaty [12]
have an excellent overview of the main ideas. See also Behmard and Faridani
[14, 3] and Feichtinger and Pandey [15]. Note that the techniques in those papers