Methods of Functional Analysis and Topology
Vol. 12 (2006), no. 2, pp. 170–182
CONTINUOUS FRAMES IN HILBERT SPACES
A. RAHIMI, A. NAJATI, AND Y. N. DEHGHAN
Abstract. In this paper we introduce a mean of a continuous frame which is a
generalization of discrete frames. Since a discrete frame is a special case of these
frames, we expect that some of the results that occur in the frame theory will be
generalized to these frames. For such a generalization, after giving some basic results
and theorems about these frames, we discuss the following: dual to these frames,
perturbation of continuous frames and robustness of these frames to an erasure of
some elements.
1. Introduction
The concept of frames (discrete frames) in Hilbert spaces has been introduced by
Duffin and Schaeffer [11] in 1952 to study some deep problems in nonharmonic Fourier
series, after the fundamental paper [10] by Daubechies, Grossman and Meyer, frame
theory began to be widely used, particularly in the more specialized context of wavelet
frames and Gabor frames.
Traditionally, frames have been used in signal processing, image processing, data compression and sampling in sampling theory. A discreet frame is a countable family of
elements in a separable Hilbert space which allows for a stable, not necessarily unique,
decomposition of an arbitrary element into an expansion of the frame elements. The
concept of a generalization of frames to a family indexed by some locally compact space
endowed with a Radon measure was proposed by G. Kaiser [15] and independently by Ali,
Antoine and Gazeau [2]. These frames are known as continuous frames. Gabardo and
Han in [14] called these frames frames associated with measurable spaces, Askari-Hemmat,
Dehghan and Radjabalipour in [3] called them generalized frames and in mathematical
physics they are referred to as coherent states [2]. For more details, the reader can refer
to [1, 2, 3, 9, 12, 14, 17]. If in the definition of a continuous frame, the measure space
Ω := N and µ is the counting measure, the continuous frame will be a discrete frame and
so we expect that some of the results obtained in the frame theory hold in the continuous
frame theory.
In this paper, we focus on a continuous frame with a positive measure and H a complex
Hilbert space.
The paper is organized as follows. In Section 2, we introduce some definitions and
basic facts about continuous frames such as a continuous frame operator, a pre-frame
operator, etc. In Section 3, we discuss the dual of continuous frames and further we
present a condition that shows when a continuous frame is robust to erasure of some
elements. A type of Paley-Wiener theorem about perturbation of continuous frames that
was discussed by Gabardo and Han in [14] will be generalized in Section 4.
2000 Mathematics Subject Classification. Primary 41A58, 42C15.
Key words and phrases. Frame, continuous frame, measure space, Riesz type, Riesz basis, Riesz
frame, wavelet frame, short-time Fourier transform, Gabor frame.
170
CONTINUOUS FRAMES IN HILBERT SPACES
171
2. Preliminaries
Throughout this paper, H and K will be complex Hilbert spaces.
Definition 2.1. Let H be a complex Hilbert space and (Ω, µ) be a measure space with
positive measure µ. A mapping F : Ω → H is called a continuous frame with respect to
(Ω, µ), if
(i) F is weakly-measurable, i.e., for all f ∈ H, ω → hf, F (ω)i is a measurable
function on Ω;
(ii) there exist constants A, B > o such that
Z
2
(2.1)
Akf k ≤
|hf, F (ω)i|2 dµ(ω) ≤ Bkf k2 , ∀f ∈ H.
Ω
The constants A and B are called continuous frame bounds. F is called a tight continuous
frame if A = B. The mapping F is called Bessel if the second inequality in (2.1) holds.
In this case, B is called the Bessel constant. If µ is a counting measure and Ω = N, F is
called a discrete frame.
The first inequality in (2.1) shows that F is complete, i.e.,
span{F (ω)}ω∈Ω = H.
The following are well known examples in wavelet frames and Gabor frames that are
continuous frames.
Example 2.2. If ψ ∈ L2 (R) is admissible, i.e.,
Z +∞
|ψ̂(γ)|2
dγ < +∞
Cψ :=
|γ|
−∞
and, for a, b ∈ R, a 6= 0,
ψ a,b (x) := (Tb Da ψ)(x) =
1
|a|
1
2
ψ
x − b
a
,
∀x ∈ R,
then {ψ a,b }a6=0,b∈R is a continuous frame for L2 (R) with respect to R \ {0} × R equipped
2
with the measure dadb
a2 and, for all f ∈ L (R),
Z +∞ Z +∞
dadb
f=
Wψ (f )(a, b)ψ a,b 2 ,
a
−∞
−∞
where Wψ is the continuous wavelet transform defined by
Z +∞
1 x − b
a,b
Wψ (f )(a, b) := hf, ψ i =
f (x) 1 ψ
dx.
a
|a| 2
−∞
For details, see the Proposition 11.1.1 and Corollary 11.1.2 in [8].
Definition 2.3. Fix a function g ∈ L2 (R) \ {0}. The short-time Fourier transform of a
function f ∈ L2 (R) with respect to the window function g is given by
Z +∞
Ψg (f )(y, γ) =
f (x)g(x − y)e−2πixγ dx, y, γ ∈ R.
−∞
Note that in terms of modulation operators and translation operators, Ψg (f )(y, γ) =
hf, Eγ Ty gi.
Example 2.4. Let g ∈ L2 (R) \ {0}. Then {Eb Ta g}a,b∈R is a continuous frame for ∈ L2 (R)
with respect to X = R2 equipped with the Lebesgue measure. Let f1 , f2 , g1 , g2 ∈ L2 (R).
Then
Z +∞ Z +∞
Ψg1 (f1 )(a, b)Ψg2 (f2 )(a, b) db da = hf1 , f2 ihg2 , g1 i.
−∞
−∞
172
A. RAHIMI, A. NAJATI, AND Y. N. DEHGHAN
For details see the Proposition 8.1.2 in [8].
The following proposition shows that it is enough to check the continuous frame condition on a dense set. One can find a discrete version of this proposition in ([8], Lemma 5.1.7).
Proposition 2.5. Suppose that (Ω, µ) is a measure space and µ is σ-finite. Let F : Ω →
H be a weakly-measurable vector-valued function and assume that there exist constants
A, B > 0 such that (2.1) holds for all f in a dense subset V of H. Then F is a continuous
frame with respect to (Ω, µ) for H with the bounds A, B.
Proof.
Let {Ωn }∞
n=1 be a family of disjoint measurable subsets of Ω such that Ω =
S∞
Ω
with
µ(Ω
n
n ) < ∞ for each n ≥ 1. Let
n=1
∆m = { ω ∈ Ω | m ≤ kF (ω)k < m + 1 }
S∞
∞
for all integers m ≥ 0. It is clear that Ω = m=0,n=1 Ωn ∩∆m where {Ωn ∩∆m }∞
n=1 m=0
is a family of disjoint and measurable subsets of Ω. We show that F is Bessel. Suppose
that there exists f ∈ H such that
Z
|hf, F (ω)i|2 dµ(ω) > Bkf k2 .
Ω
Therefore,
XZ
m,n
∆m ∩Ωn
hF (ω), f i
2
and thus there exist finite sets I, J such that
X XZ
(2.2)
hF (ω), f i
m∈I n∈J
∆m ∩Ωn
> Bkf k2
2
> Bkf k2 .
Let {fk } be a sequence in V such that fk → f as n → ∞. The assumption implies that
X XZ
2
hF (ω), fk i ≤ Bkfk k2
m∈I n∈J
∆m ∩Ωn
which is a contradiction to (2.2) (by Lebesgue’s Dominated Convergence Theorem). For
the rest of the proof, we show that
Z
Z
2
2
hF (ω), fk i dµ(ω) →
hF (ω), f i dµ(ω)
Ω
Ω
as n → ∞. For this we have
Z
2
2
hF (ω), fk i − hF (ω), f i
dµ(ω)
Ω
Z
Z
2
hF (ω), fk − f i dµ(ω) + 2
≤
hF (ω), f ihF (ω), fk − f i dµ(ω)
Ω
Ω
2
≤ Bkfk − f k + 2Bkfk − f kkf k,
the last inequality follows from Cauchy-Schwarz’ inequality. Since fk → f , the result is
proved.
Let F be a continuous frame with respect to (Ω, µ). Then the mapping
Ψ:H×H→C
defined by
Ψ(f, g) =
Z
Ω
hf, F (ω)ihF (ω), gi dµ(ω)
CONTINUOUS FRAMES IN HILBERT SPACES
173
is well defined, a sesquilinear form (i.e., linear in the first and conjugate-linear in the
second variable) and is bounded. By Cauchy-Schwarz’ inequality we get
Z
|Ψ(f, g)| ≤
|hf, F (ω)ihF (ω), gi| dµ(ω)
Ω
≤
Z
Ω
21
21 Z
≤ Bkf kkgk.
|hF (ω), gi|2 dµ(ω)
|hf, F (ω)i|2 dµ(ω)
Ω
Hence, kΨk ≤ B. By Theorem 2.3.6 in [16] there exists a unique operator SF : H → H
such that
Ψ(f, g) = hSF f, gi, ∀f, g ∈ H
and, moreover, kΨk R= kSk.
Since hSF f, f i = Ω |hf, F (ω)|2 dµ(ω), we see that SF is positive and AI ≤ SF ≤ BI.
Hence, S
R F is invertible. We call SF a continuous frame operator of F and use the notation
SF f = Ω hf, F (ω)iF (ω) dµ(ω). Thus, every f ∈ H has the representations
Z
f = SF−1 SF f = hf, F (ω)iSF−1 F (ω) dµ(ω),
ZΩ
−1
f = SF SF f = hf, SF−1 F (ω)iF (ω) dµ(ω).
Ω
The next theorem is analogous to Theorem 3.2.3 in [8].
Theorem 2.6. Let (Ω, µ) be a measure space and let F be a Bessel mapping from Ω to
H. Then the operator TF : L2 (Ω, µ) → H weakly defined by
Z
ϕ(ω)hF (ω), hi dµ(ω), h ∈ H
hTF ϕ, hi =
Ω
is well defined, linear, bounded, and its adjoint is given by
TF∗ : H → L2 (Ω, µ),
(TF∗ h)(ω) = hh, F (ω)i,
ω ∈ Ω.
The operator TF is called a pre-frame operator or synthesis operator and TF∗ is called an
analysis operator of F .
Proof. The proof is straightforward.
The converse of Theorem 2.6 holds when µ is a σ-finite measure.
Proposition 2.7. Let (Ω, µ) be a measure space, where µ is a σ-finite measure and let
F : Ω → H be a measurable function. If the mapping TF : L2 (Ω, µ) 7→ H defined by
Z
hTF ϕ, hi =
ϕ(ω)hF (ω), hi dµ(ω), ϕ ∈ L2 (Ω, µ), h ∈ H
Ω
is a bounded operator, then F is Bessel.
Proof. By Theorem 2.6, we have
(T ∗ h)(ω) = hh, F (ω)i,
Hence for each h ∈ H
Z
Ω
ω ∈ Ω.
|hh, F (ω)i|2 dµ(ω) = kT ∗ hk2 ≤ kT k2 khk2 .
We now give a characterization of continuous frames in terms of the pre-frame operators. For the next theorem we will need the following lemma that is proved in [8].
174
A. RAHIMI, A. NAJATI, AND Y. N. DEHGHAN
Lemma 2.8. Let T : K → H be a bounded operator with a closed range RT . Then there
exists a bounded operator T † : H → K for which
T T † f = f,
∀f ∈ RT .
The next theorem gives an equivalent characterization of a continuous frame. For a
discrete case of this theorem, see [5].
Theorem 2.9. Let (Ω, µ) be a measure space where µ is a σ-finite measure. The mapping
F : Ω → H is a continuous frame with respect to (Ω, µ) for H if and only if the operator
TF as defined in Theorem (2.6) is a bounded and onto operator.
Proof. Let F be a continuous frame. Then, by Theorem 2.6, TF is bounded and
TF∗ : H → L2 (Ω, µ),
(TF∗ h)(ω) = hh, F (ω)i,
Hence, for each f ∈ H,
kTF∗ f k2 =
Z
Ω
h ∈ H,
ω ∈ Ω.
|hf, F (ω)i|2 dµ(ω).
Therefore, TF∗ is one to one and so TF is onto.
Conversely, let TF be a bounded and onto operator. Then, by Lemma 2.8 there exists
a bounded operator TF† : H → L2 (Ω, µ) such that TF TF† f = f for all f ∈ H. Since TF is
bounded, by Proposition 2.7, F is Bessel and
Z
∗
2
|hf, F (ω)i|2 dµ(ω), ∀f ∈ H.
kTF f k =
Ω
Let f ∈ H, then kf k2 ≤
kTF† k2 kTF∗ f k2 .
kTF† k−2 .kf k2 ≤
Z
Ω
Hence,
|hf, F (ω)i|2 dµ(ω),
∀f ∈ H.
As in the discrete case, we have the following lemma.
Lemma 2.10. Let F : Ω → H be a Bessel function with respect to (Ω, µ). By the above
notations, SF = TF TF∗ .
Proof. For all f, g ∈ H,
hTF TF∗ f, gi = hTF∗ f, TF∗ gi =
So SF = TF TF∗ .
Z
Ω
hf, F (ω)ihF (ω), gi dµ(ω) = hSf, gi.
Proposition 2.11. Let F be a continuous frame with respect to (Ω, µ) for H with a
frame operator SF and let V : H → K be a bounded and invertible operator. Then V F
is a continuous frame for K with the frame operator V SF V ∗ .
Proof. For each f ∈ H, ω → hV ∗ f, F (ω)i = hf, V F (ω)i is measurable. Let A and B be
frame bounds for F. Therefore, for every f ∈ H,
Z
Z
∗
2
∗
2
AkV f k ≤
|hV f, F (ω)i| dµ(ω) =
|hf, V F (ω)i|2 dµ(ω) ≤ BkV ∗ f k2 .
Ω
Ω
Hence,
kf k ≤ kV −1 kkV ∗ f k
and kV ∗ f k ≤ kV ∗ kkf k,
∀f ∈ H.
CONTINUOUS FRAMES IN HILBERT SPACES
175
Therefore, V F is a continuous frame with frame bounds AkV −1 k−2 and BkV k2 . Let SV F
be the frame operator for V F . Then for each f, g ∈ H,
Z
hSV F f, gi = hf, V F (ω)ihV F (ω), gi dµ(ω)
ZΩ
= hV ∗ f, F (ω)ihF (ω), V ∗ gi dµ(ω) = hSF V ∗ f, V ∗ gi = hV SF V ∗ f, gi.
Ω
∗
So SV F = V SF V .
Corollary 2.12. Let F be a continuous frame with respect to (Ω, µ) for H with a frame
operator SF . Then for all α ∈ R, SFα F is a continuous frame for H with the frame
′
operator S = SF2α+1 . In particular, SF−1 F is a continuous frame and we call it a standard
−1/2
dual frame for F . Also, SF F is a normalized tight frame with the frame operator I.
For later reference we state a general version of Proposition 2, where we only assume
that U is a bounded operator with a closed range RU . The next proposition is analogous
to Proposition 5.3.1 in [8].
Proposition 2.13. Let F be a continuous frame with respect to (Ω, µ) for H with bounds
A, B and let U : H → K be a bounded operator with a closed range RU . Then U F is a
continuous frame for RU with the bounds AkU † k−2 , BkU k2 .
Proof. It is clear that ω → hf, V F (ω)i is measurable for all f ∈ H. We may assume that
U is onto. If f ∈ K, then
Z
Z
|hU ∗ f, F (ω)i|2 dµ(ω) ≤ BkU k2 kf k2 ,
|hf, U F (ω)i|2 dµ(ω) =
Ω
Ω
which proves that U F is Bessel. For the lower frame condition, let f ∈ K. Then
kf k ≤ kU † kkU ∗ f k and
Z
Z
|hf, U F (ω)i|2 dµ(ω) =
|hU ∗ f, F (ω)i|2 dµ(ω)
Ω
Ω
≥ AkU ∗ f k2 ≥ AkU † k−2 kf k2 ,
which gives the result.
Corollary 2.14. If F is a continuous frame with respect to (Ω, µ) for H with bounds
A, B and U : H → K is a bounded surjective operator, then U F is a continuous frame
with respect to (Ω, µ) for K with the bounds AkU † k−2 , BkU k2 .
The following proposition is a criterion for a continuous frame for a closed subspace
of H to be a continuous frame for H. For the discrete, case see ([8], Lemma 5.2.1).
Proposition 2.15. Suppose that F is a continuous frame with respect to (Ω, µ) for a
closed subspace K of H, where µ is a σ-finite measure. Let T : L2 (Ω, µ) → H be the
mapping defined by
Z
hT ϕ, hi =
ϕ(ω)hF (ω), hi dµ(ω), ϕ ∈ L2 (Ω, µ), h ∈ H.
Ω
Then F is a continuous frame for H if and only if T ∗ is injective.
Proof. It is clear that T is well defined and bounded. Let F be a continuous frame for H.
Then, by Theorem 2.6, T ∗ is injective. Conversely, suppose that T ∗ is injective. Then T
is onto and the result follows from Theorem 2.9.
The next proposition is analogous to Proposition 5.3.5 in [8] and gives a relationship
between continuous frames and orthogonal projections.
176
A. RAHIMI, A. NAJATI, AND Y. N. DEHGHAN
Proposition 2.16. Let K be a closed subspace of H and let P : H → K be an orthogonal
projection. The the following holds:
(i) If F is a continuous frame with respect to (Ω, µ) for H with bounds A and B,
then P F is a continuous frame with respect to (Ω, µ) for K with the bounds A, B.
(ii) If F is a continuous frame with respect to (Ω, µ) for K with a frame operator
SF , then for each f, g ∈ H,
Z
hP f, gi = hf, SF−1 F (ω)ihF (ω), gi dµ(ω).
Ω
Proof. The proof is easy.
The proof of the following proposition is similar to the discrete case ([8], Proposition
5.3.6) and we omit it.
Proposition 2.17. Let F be a continuous frame with respect to (Ω, µ) for H with a
synthesis operator TF . The orthogonal projection Q from L2 (Ω, µ) onto RTF∗ is given by
Z
ϕ(ω)hSF−1 F (ω), F (ν)i dµ(ω), ν ∈ Ω,
Q(ϕ)(ν) =
Ω
where SF is the frame operator of F.
Let F : Ω −→ H be a vector-valued function and ϕ ∈ L2 (Ω, µ). It is natural to ask
whether we can find f ∈ H such that
hf, F (ω)i = ϕ(ω),
∀ω ∈ Ω.
A problem of this type is called a moment problem. It is clear that there are cases where
no solution exists: if, for example, F (ω) = F (ν) for some ω 6= ν, a solution can only
exist if ϕ(ω) = ϕ(ν). Since the moment problem has no solution in general, there is a
natural question of whether we can find an element in H which minimizes the function
Z
f→
|ϕ(ω) − hf, F (ω)i|2 dµ(ω).
Ω
The answer is given by the following theorem and is called a best approximation solution.
For the discrete case, see ([8], Theorem 6.5.2).
Theorem 2.18. Let F be a continuous frame with respect to (Ω, µ) for HRand ϕ ∈
L2 (Ω, µ). Then there exists a unique vector in H which minimizes the map f → Ω |ϕ(ω)−
hf, F (ω)i|2 dµ(ω); this vector is f = SF−1 TF ϕ. SF and TF are the frame operator and synthesis operators for F , respectively.
Proof. Let Q be the orthogonal projection of L2 (Ω, µ) onto RT ∗ . Then
min
ψ∈RT ∗
ϕ − ψ = kϕ − Q(ϕ)k.
Therefore,
min
g∈H
Z
Ω
2
ϕ(ω) − hg, F (ω)i dµ(ω) = kϕ − Q(ϕ)k2 .
Let Q(ϕ) = TF∗ (f ) for some f ∈ H. By Proposition 2.17,
Z
ϕ(ω)hSF−1 F (ω), F (ν)i dµ(ω) = hf, F (ν)i,
Ω
∀ν ∈ Ω.
Hence f = SF−1 TF ϕ. Since TF∗ is injective, the minimization is unique.
CONTINUOUS FRAMES IN HILBERT SPACES
177
3. Dual of continuous frames
In this section, we mention some important properties of continuous frames and their
dual. Gabardo and Han in [14] defined a dual frame for a continuous frame as follows.
Definition 3.1. Let F, G be a continuous frames with respect to (Ω, µ) for H. We call
G a dual frame if the following holds true:
Z
hf, gi = hf, F (ω)ihG(ω), gi dµ(ω), ∀f, g ∈ H.
(3.1)
Ω
In this case (F, G) is called a dual pair. If TF and TG denote the synthesis operators of
F and G, respectively, then (3.1) is equivalent to TG TF∗ = I.
It is certainly possible for a continuous frame F to have only one dual. In this case we
call F a Riesz-type frame. The proof of the following proposition can be found in [14].
Proposition 3.2. Let F be a continuous frame with respect to (Ω, µ) for H. Then F is
a Riesz-type frame if and only if R(TF∗ ) = L2 (Ω, µ).
Corollary 3.3. Let F : Ω → H be a Riesz-type frame with respect to (Ω, µ) for H, where
µ is σ-finite. Then F (ω) 6= 0 for all ω ∈ Ω.
The next proposition shows that every continuous frame has a dual frame, as in the
discrete case. For the discrete case, see ([11], Lemma VIII).
Proposition 3.4. Let F be a continuous frame with respect to (Ω, µ) for H with a frame
operator S. Then (S α F, S −1−α F ) is a dual pair for each α ∈ R. In particular, (F, S −1 F )
is a dual pair and S −1 F is called a standard dual frame of F .
Proof. For any f, g ∈ H we have
hf, gi = hS α SS −1−α f, gi = hSS −1−α f, S α gi =
Z
Ω
hS −1−α f, F (ω)ihF (ω), S α gi dµ(ω).
The following proposition states an important property of a standard dual continuous
frame of a given continuous frame. Its proof is similar to that in the discrete case ([11],
Lemma VIII).
Proposition 3.5. Let F be a continuous frame with respect to (Ω, µ) for H with a frame
operator S and let f ∈ H. If there exists ϕ ∈ L2 (Ω, µ) such that
Z
hf, gi =
ϕ(ω)hF (ω), gi dµ(ω), ∀g ∈ H,
Ω
then
Z
Ω
2
|ϕ(ω)| dµ(ω) =
Z
Ω
|hf, S
−1
2
F (ω)i| dµ(ω) +
Z
Ω
|ϕ(ω) − hf, S −1 F (ω)i|2 dµ(ω).
The following theorem is analogous to Lemma IX in [11] and shows when we can
remove some elements from a continuous frame so that the set still remains a continuous
frame.
Theorem 3.6. Let F be a continuous frame with respect to (Ω, µ) for H with a frame
operator S and let ω0 ∈ Ω such that
1
µ({ω0 }) 6=
.
hF (ω0 ), S −1 F (ω0 )i
Then F is robust to erasure of F (ω0 ), i.e., F : Ω \ {ω0 } → H is a continuous frame
for H.
178
A. RAHIMI, A. NAJATI, AND Y. N. DEHGHAN
Proof. It is clear that the upper frame condition holds. For the lower frame bound, we
have
Z
hF (ω0 ), f i = hF (ω0 ), S −1 F (ω)ihF (ω), f i dµ(ω), ∀f ∈ H,
Ω
therefore,
hF (ω0 ), f i =
Z
Ω\{ω0 }
hF (ω0 ), S −1 F (ω)ihF (ω), f i dµ(ω)
+ hF (ω0 ), S −1 F (ω0 )ihF (ω0 ), f iµ({ω0 }).
Hence, by the assumption of theorem,
hF (ω0 ), f i =
1
1 − µ({ω0 })hS −1 F (ω0 ), F (ω0 )i
Z
Ω\{ω0 }
hF (ω0 ), S −1 F (ω)ihF (ω), f i dµ(ω).
Let A be the lower frame bound for F. For any f ∈ H, the Cauchy-Schwartz inequality
gives
Z
|hF (ω), f i|2 dµ(ω)
|hf, F (ω0 )i|2 ≤ K
Ω\{ω0 }
where
K :=
Z
Ω\{ω0 }
Therefore, for any f ∈ H,
Z
Akf k2 ≤
|hF (ω0 ), S −1 F (ω)i|2
dµ(ω).
|1 − µ({ω0 })hF (ω0 ), S −1 F (ω0 )i|2
Ω\{ω0 }
|hF (ω), f i|2 dµ(ω) + |hF (ω0 ), f i|2 µ({ω0 })
and so
Akf k2 ≤ (1 + Kµ({ω0 }))
Z
Ω\{ω0 }
|hF (ω), f i|2 dµ(ω).
Therefore, F : Ω \ {ω0 } → H is a continuous frame for H with the lower frame bound
A
1+Kµ({ω0 }) .
Corollary 3.7. Let F be a continuous frame with respect to (Ω, µ) for H with a frame
operator S and let ω0 ∈ Ω such that µ({ω0 }) 6= hF (ω0 ),S1−1 F (ω0 )i . Then F is not a Riesztype frame.
Proof. By Theorem 3.6, F : Ω\{ω0 } → H is a continuous frame for H. Let G : Ω\{ω0 } →
H be a standard dual frame for F|Ω\{ω0 } and let G(ω0 ) = 0. Then S −1 F 6= G and for all
f, h ∈ H,
Z
hf, hi =
Ω
hf, G(ω)ihF (ω), hi.
Therefore, G : Ω → H is a dual frame for F which is different from S −1 F.
4. Perturbation of continuous frames
A perturbation of discrete frames has been discussed in [4]. In this section we introduce
and extent Christensen’ works on a discrete frame, however, there is a similar theorem
about perturbation of frames for measurable spaces in [14]. The following theorem is
another version of a perturbation of continuous frames and its proof is based on the
following lemma that is proved in [4].
Lemma 4.1. Let U be a linear operator on a Banach space X and assume that there
exist λ1 , λ2 ∈ [0, 1) such that
kx − U xk ≤ λ1 kxk + λ2 kU xk
CONTINUOUS FRAMES IN HILBERT SPACES
179
for all x ∈ X. Then U is bounded and invertible. Moreover,
1 + λ1
1 − λ1
kxk ≤ kU xk ≤
kxk
1 + λ2
1 − λ2
and
1 − λ2
1 + λ2
kxk ≤ kU −1 xk ≤
kxk
1 + λ1
1 − λ1
for all x ∈ X.
Theorem 4.2. Let F be a continuous frame with respect to (Ω, µ) for H, where µ is
σ-finite. Let G : Ω → H be a weakly-measurable vector-valued
and assume that
√ function
there exist constants λ1 , λ2 , γ ≥ 0 such that max λ1 + γ/ A, λ2 < 1 and
Z
ϕ(ω)hF (ω) − G(ω), f i dµ(ω)
Ω
(4.1)
Z
Z
ϕ(ω)hG(ω), f i dµ(ω) + γkϕk2
ϕ(ω)hF (ω), f i dµ(ω) + λ2
≤ λ1
Ω
Ω
for all ϕ ∈ L2 (Ω, µ) and for all f in the unit sphere in H. Then G : Ω → H is a
continuous frame with respect to (Ω, µ) for H with the bounds
"
"
√ #2
√ #2
1 − λ1 + γ/ A)
1 + λ1 + γ/ B
and B
,
A
1 + λ2
1 − λ2
where A, B are frame bounds for F.
Proof. Let S = { f ∈ H : kf k = 1} be the unit sphere in H. We first prove that G is
Bessel. By the assumption, for any f ∈ S and ϕ ∈ L2 (Ω, µ), we have
Z
ϕ(ω)hG(ω), f i dµ(ω)
Ω
Z
Z
≤
(4.2)
ϕ(ω)hF (ω) − G(ω), f i dµ(ω) +
ϕ(ω)hF (ω), f i dµ(ω)
Ω
Ω
Z
Z
≤ (1 + λ1 )
ϕ(ω)hF (ω), f i dµ(ω) + λ2
ϕ(ω)hG(ω), f i dµ(ω) + γkϕk2 ,
Ω
Ω
which implies that
Z
Z
γ
1 + λ1
ϕ(ω)hF (ω), f i dµ(ω) +
kϕk2
ϕ(ω)hG(ω), f i dµ(ω) ≤
1
−
λ
1
−
λ2
2
Ω
Ω
(4.3)
γ
1 + λ1 √
B+
kϕk2 .
≤
1 − λ2
1 − λ2
Let U : L2 (Ω, µ) → H be defined by
Z
hU ϕ, f i =
ϕ(ω)hG(ω), f i dµ(ω),
∀f ∈ H, ϕ ∈ L2 (Ω, µ).
Ω
Then
kU ϕk = sup |hU ϕ, f i| = sup
kf k=1
kf k=1
Z
Ω
ϕ(ω)hG(ω), f i dµ(ω)
γ
1 + λ1 √
B+
kϕk2 .
1 − λ2
1 − λ2
Therefore U is bounded and so, by Proposition 2.7, G is Bessel with the required upper
bound. Now we prove that G has the required lower frame bound. Let TF and SF be a
≤
180
A. RAHIMI, A. NAJATI, AND Y. N. DEHGHAN
synthesis operator and a frame operator for F , respectively. Let us define V = U TF∗ SF−1 ,
then
Z
hV f, gi = hf, SF−1 F (ω)ihG(ω), gi dµ(ω)
Ω
and
hf, gi =
Z
Ω
hf, SF−1 F (ω)ihF (ω), gi dµ(ω)
for all f, g ∈ H. For each f ∈ H, let ϕf : Ω → C be a mapping defined by ϕf (ω) =
hf, S −1 F (ω)i. Since S −1 F is a standard dual frame of F , we have ϕf ∈ L2 (Ω, µ). Therefore, using the assumption, we deduce that
|hf − V f, gi| ≤ λ1 |hf, gi| + λ2 |hV f, gi| + γkϕf k2
for all f ∈ H and g ∈ S. Hence,
kf − V f k = sup |hf − V f, gi| ≤ λ1 kf k + λ2 kV f k + γk ϕf k2
kgk=1
γ
λ1 + √
kf k + λ2 kV f k
A
for all f ∈ H. By Lemma 4.1, V is invertible and
√
1 + λ2
1 + λ1 + γ/ A
√ .
, kV −1 k ≤
kV k ≤
1 − λ2
1 − λ1 + γ/ A
≤
Let f ∈ H. Then
hf, f i = hV V −1 f, f i =
and we obtain
Z
Ω
hV −1 f, SF−1 F (ω)ihG(ω), f i dµ(ω),
Z
2
kf k4 = |hf, f i|2 =
hV −1 f, SF−1 F (ω)ihG(ω), f i dµ(ω)
Ω
Z
Z
2
2
−1
−1
hV f, SF F (ω)i dµ(ω)
≤
hG(ω), f i dµ(ω)
Ω
≤
Therefore,
Z
Ω
for all f ∈ H.
Ω
"
1 + λ2
1
√
A 1 − λ1 + γ/ A
#2
kf k2 .
Z
Ω
2
hG(ω), f i dµ(ω).
√ #2
1 − λ1 + γ/ A
hG(ω), f i dµ(ω) ≥ A
kf k2
1 + λ2
2
"
Corollary 4.3. Let F be a continuous frame with respect to (Ω, µ) for H, where µ is
σ-finite. Let G : Ω → H be a weakly-measurable vector-valued function and let A, B be
frame bounds for F. If there exists R < A such that
Z
2
hF (ω) − G(ω), f i dµ(ω) ≤ Rkf k2 , ∀f ∈ H,
Ω
then G is a continuous frame with respect to (Ω, µ) for H with the bounds
p
p
2
2
A 1 − R/A , B 1 + R/B .
Corollary 4.4. Let F and G be as in Theorem 4.2. Then G is similar to a dual of F.
Proof. Let U and V be defined as in the proof of Theorem 4.2. Then U = TG and V SF =
TG TF∗ . Therefore TG TF∗ is invertible. Let D = (TG TF∗ )−1 . Then TDG TF∗ = DTG TF∗ = I.
Therefore, DG is a dual of F.
CONTINUOUS FRAMES IN HILBERT SPACES
181
Corollary 4.5. Let F and G be as in Theorem 4.2, and assume that F is a Riesz-type
frame for H. Then G is also a Riesz-type frame for H.
Proof. Since G is similar to SF−1 F, it follows that G is similar to F and, thus, RTF∗ = RTG∗ .
So the result follows from Proposition 3.2.
Corollary 4.6. Let F and G be as in Theorem 4.2 with γ = 0. Then G is similar to F.
Proof. Since F and G are continuous frames for H, we see that RTF∗ and RTg∗ are closed
subspaces of L2 (Ω, µ). Therefore, F and G are similar if and only if ker TF = ker TG . If
ϕ ∈ ker TF , then (4.1) implies that
kTG ϕk ≤ λ2 kTG ϕk
and thus, TG ϕ = 0, since λ2 < 1. Therefore, ker TF ⊆ ker TG . Conversely, let ϕ ∈ ker TG .
By (4.1) we have
kTF ϕk ≤ λ1 kTF ϕk.
Since λ1 < 1, TF ϕ = 0 and thus ker TG ⊆ ker TF .
The proof of the following theorem is similar to discrete case and we refer to [8].
Theorem 4.7. Let F be a continuous frame with respect to (Ω, µ) for H, where µ is
σ-finite. Let G : Ω → H be a weakly-measurable vector-valued function and assume that
K : L2 (Ω, µ) → H defined by
Z
ϕ(ω)hF (ω) − G(ω), f i dµ(ω), f ∈ H, ϕ ∈ L2 (Ω, µ)
hKϕ, f i =
Ω
is a well defined compact operator. Then G is a continuous frame with respect to (Ω, µ)
for spanω∈Ω {G(ω)}.
Acknowledgments. The authors would like to give a special thanks to the Ole Christensen for providing some useful references and discussions. The authors also wish to
thank referee(s) for their useful suggestions.
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Department of Mathematics, Tabriz University, Tabriz, Iran
E-mail address:
[email protected]
Department of Mathematics, Tabriz University, Tabriz, Iran
E-mail address:
[email protected]
Department of Mathematics, Tabriz University, Tabriz, Iran
Received 04/07/2005; Revised 28/11/2005