# Some New Characterizations andg-Minimality and Stability of g-Bases in the Hilbert Spaces

Authors: Xunxiang Guo
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

The concept ofg-basis in the Hilbert spaces is introduced by Guo (2012) who generalizes the Schauder basis in the Hilbert spaces. g-basis plays the similar role in g-frame theory to that the Schauder basis plays in frame theory. In this paper, we establish some important properties of g-bases in the Hilbert spaces. In particular, we obtain a simple condition under which some important properties established in Guo (2012) are still true. With these conditions, we also establish some new interesting properties of g-bases which are related to g-minimality. Finally, we obtain a perturbation result about g-bases.


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## Body

## 1. Introduction

In 1946, Gabor [1] introduced a fundamental approach to signal decomposition in terms of elementary signals. In 1952, Duffin and Schaeffer [2] abstracted Gabor’s method to define frames in the Hilbert spaces. Frame was reintroduced by Daubechies et al. [3] in 1986. Today, frame theory is a central tool in many areas such as characterizing function spaces and signal analysis. We refer to [4–10] for an introduction to frame theory and its applications. The following are the standard definitions on frames in the Hilbert spaces. A sequence {fi}i∈𝒩 of elements of a Hilbert space H is called a frame for H if there are constants A,B>0 so that
(1)A∥f∥2≤∑i∈𝒩‍|〈f,fi〉|2≤B∥f∥2.
The numbers A,B are called the lower (resp., upper) frame bounds. The frame is a tight frame if A=B and a normalized tight frame if A=B=1.In [11], Sun raised the concept of g-frame as follows, which generalizes the concept of frame extensively. A sequence {Λi∈B(H,Hi):i∈𝒩} is called a g-frame for H with respect to {Hi:i∈𝒩}, which is a sequence of closed subspaces of a Hilbert space V, if there exist two positive constants A and B such that for any f∈H(2)A∥f∥2≤∑i∈𝒩‍∥Λif∥2≤B∥f∥2.
We simply call {Λi:i∈𝒩} a g-frame for H whenever the space sequence {Hi:i∈𝒩} is clear. The tight g-frame, normalized tight g-frame, g-Riesz basis are defined similarly. Recently, g-frames in the Hilbert spaces have been studied intensively; for more details see [12–16] and the references therein.In the history of the Hilbert space theory, the concept of basis was introduced more earlier than the concept of frame. And frames are generalizations of bases. When we look back the development ofg-frame theory, it is well known that g-frames are generalizations of frames. Then it is naturally to ask what are the generalizations of bases in the similar manner? It is the motivation of our paper [17]. In [17], we put up with the concept of g-basis in the Hilbert space. We also find that g-basis really fills in the gap in the g-frame theory. g-bases play the similar role in g-frame theory to that the bases play in the frame theory. But we also find that the properties of basis cannot be adjusted trivially so that they are hold by g-bases. For more details, please see [17]. In this paper, we will do some study on g-bases in the Hilbert spaces further. And some important properties of g-bases are established. In Section 2, we will give the definitions and lemmas. In Section 3, we give some characterizations of g-bases. In Section 4, we study the property of g-minimal of g-bases. In Section 5, we discuss the stability of g-bases under perturbations.Throughout this paper, we use𝒩 to denote the set of all natural numbers, Z to denote the set of all integer numbers, and C to denote the field of complex numbers. The sequence of {Hj:j∈𝒩} always means a sequence of closed subspace of some Hilbert space V. And all Hilbert spaces are complex separable Hilbert spaces.


## 2. Definitions and Lemmas

In this section, we introduce the definitions and lemmas which will be needed in this paper.Definition 1.
{
Λ
j
∈
B
(
H
,
H
j
)
}
j
=
1
∞ is called g-complete with respect to {Hj} if {f:Λjf=0,∀j}={0}.Definition 2.
{
Λ
j
∈
B
(
H
,
H
j
)
}
j
=
1
∞ is called g-linearly independent with respect to {Hj} if ∑j=1∞‍Λj*gj=0; then gj=0, where gj∈Hj(j=1,2,…).Definition 3.
{
Λ
j
∈
B
(
H
,
H
j
)
}
j
=
1
∞ is called g-minimal with respect to {Hj} if, for any m∈𝒩 and any gm∈Hm such that Λm*gm≠0, we have Λm*gm∉span¯i≠m{Λi*(Hi)}.Definition 4.
LetM(H,K) denote the set of all maps from H to K. {Ti∈M(Hi,H):i∈𝒩} and {Li∈M(Hi,H):i∈𝒩} are called biorthonormal if
(3)〈Tjgj,Ligi〉=δj,i〈gj,gi〉,∀j,i∈𝒩,gj∈Hj,gi∈Hi.{Λj∈B(H,Hj)}j=1∞ and {Γj∈B(H,Hj)}j=1∞ are called g-biorthonormal with respect to {Hj}, if {Λj*:j∈𝒩} and {Γj*:j∈𝒩} are biorthonormal; that is,
(4)〈Λj*gj,Γi*gi〉=δi,j〈gj,gi〉,∀i,j∈𝒩,gi∈Hi,gj∈Hj.Definition 5.
We call{Λj∈B(H,Hj)}j=1∞ a g-basis for H with respect to {Hj:j∈𝒩} if for any x∈H there is a unique sequence {gj:j∈𝒩} with gj∈Hj such that x=∑j=1∞‍Λj*gj. The constant
(5)C=sup{∥∑k=1m‍Λk*gk∥∣∥∑k=1n‍Λk*gk∥=1,m≤n,gk∈Hk}
is called the basis constant associated with {Λj:j∈𝒩}.Remark 6.
If{xi}i∈𝒩 is a Schauder basis of Hilbert space H, then it induces a g-basis {Λxi:i∈𝒩} of H with respect to the complex number field C, where Λxi is defined by Λxif=〈f,xi〉. In fact, it is easy to see that Λxi*c=c·xi for any c∈C, so, for any x∈H, there exists a unique sequence of constants {an:n∈𝒩} such that x=∑i∈𝒩‍aixi=∑i∈𝒩‍Λi*ai.Definition 7.
If{Λj:j∈𝒩} is a g-basis of H with respect to {Hj:j∈𝒩}, then, for any x∈H, there exists a unique sequence {gj:j∈𝒩} such that gj∈Hj and x=∑j=1∞‍Λj*gj. We define a map Γj:H→Hj, by Λjx=gj for each j. Then {Γj:j∈𝒩} is well defined. We call it the g-dual sequence of {Λj:j∈𝒩}; in case that {Γj:j∈𝒩} is also a g-basis, we call it the dualg-basis of {Λj:j∈𝒩}.Lemma 8 (see [13]).
{
Λ
j
:
j
∈
𝒩
} is g-complete if and only if span¯{Λj*(Hj):j∈𝒩}=H.The following result is about pseudoinverse, which plays an important role in some proofs.Lemma 9 (see [5]).
Suppose thatT:K→H is a bounded surjective operator. Then there exists a bounded operator (called the pseudoinverse of T) T†:H→K for which
(6)TT†f=f,∀f∈H.The following is a simple property aboutg-basis, which gives a necessary condition for g-basis in terms of g-complete and g-linearly independent.Lemma 10 (see [17]).
If{Λj:j∈𝒩} is a g-basis for H with respect to {Hj:j∈𝒩}, then {Λj:j∈𝒩} is g-complete and g-linearly independent with respect to {Hj:j∈𝒩}.


## 3. Characterizations ofg-Bases

In this section, we characterizeg-bases under the condition that the sequence is bounded surjective operators.Theorem 11.
Suppose thatΛj∈B(H,Hj) and Λj is surjective for each j∈𝒩. Let Y={{gj}j=1∞∣gj∈Hj,∑j=1∞‍Λj*gj is convergent}. If, for any {gj}j=1∞∈Y, set ∥{gj}∥Y=SupN∥∑j=1N‍Λj*gj∥, then(1)
Y is a Banach space,(2)
when{Λj:j∈𝒩} is a g-basis with respect to {Hj:j∈𝒩} as well, S:Y→H, S({gj}j=1∞)=∑j=1∞‍Λj*gj is a linear bounded and invertible operator; that is, S is a homeomorphism between H and Y.Proof.
(
1
) Let {gj}j=1∞∈Y; then ∑j=1N‍Λj*gj is convergent as N→∞. Hence {∑j=1N‍Λj*gj}N=1∞ is a convergent sequence, so it is bounded. So ∥{gj}∥Y<∞. It is obvious that, for a∈C,{gj},{hj}∈Y, we have ∥{gj}+{hj}∥Y≤∥{gj}∥Y+∥{hj}∥Y and ∥a·{gj}∥Y=|a|·∥{gj}∥Y. If ∥{gj}∥Y=0, then, for any N, ∥∑j=1N‍Λj*gj∥=0. In particular, ∥Λ1*g1∥=0. Since Λ1 is surjective, there exists a bounded operator Λ1†∈B(H1,H) such that Λ1Λ1†=IH1 by Lemma 9. So (Λ1†)*Λ1*=IH1, which implies that g1=(Λ1†)*Λ1*g1. It implies that ∥g1∥≤∥(Λ1†)*∥∥Λ1*g1∥=0. So g1=0. Since ∥∑j=12‍Λj*gj∥=∥Λ2*g2∥=0, similarly, we have g2=0. Keep doing in this way; then we get gj=0 for all j=1,…,N. But N is arbitrary, so {gj}={0}. Thus ∥·∥Y is a norm on Y. Suppose {Gk}k=1∞∈Y is a Cauchy sequence, where Gk={gjk}j=1∞. Then
(7)limk,l→∞∥Gk-Gl∥Y=limk,l→∞∥{gjk-gjl}∥Y(8)=limk,l→∞SupN∥∑j=1N‍Λj*(gjk-gjl)∥=0.
For any fixed j, we have that
(9)∥Λj*(gjk-gjl)∥=∥∑t=1j‍Λt*(gtk-gtl)-∑t=1j-1‍Λt*(gtk-gtl)∥≤∥∑t=1j‍Λt*(gtk-gtl)∥+∥∑t=1j-1‍Λt*(gtk-gtl)∥≤2∥Gk-Gl∥Y.
Since Λj is surjective, ΛjΛj†=IHj by Lemma 9. So (Λj†)*Λj*=IHj. It implies that
(10)∥gjk-gjl∥=∥(Λj†)*Λj*(gjk-gjl)∥≤∥(Λj†)*∥·∥Λj*(gjk-gjl)∥(11)≤2·∥(Λj†)*∥·∥Gk-Gl∥Y.
So, for any fixed j, {gjk}k=1∞ is a Cauchy sequence. Suppose limk→∞gjk=gj. From (8), we know that, for any ε>0, there exists L0>0, such that whenever k,l≥L0, we have
(12)SupQ∥∑j=1Q‍Λj*(gjk-gjl)∥<ε.
Fix l≥L0, since limk→∞gjk=gj, so whenever l≥L0,
(13)SupQ∥∑j=1Q‍Λj*(gj-gjl)∥≤ε.
Since GL0={gjL0}j=1∞∈Y, ∑j=1∞‍Λj*gjL0 is convergent. So there exists K0>0, such that whenever M>P≥K0, we have ∥∑j=P+1M‍Λj*gjL0∥<ε. So when M>P>max{L0,K0}, we have
(14)∥∑j=P+1M‍Λj*gj∥=∥∑j=1M‍Λj*(gj-gjL0)-∑j=1P‍Λj*(gj-gjL0)+∑j=P+1M‍Λj*gjL0∥≤∥∑j=1M‍Λj*(gj-gjL0)∥+∥∑j=1P‍Λj*(gj-gjL0)∥+∥∑j=P+1M‍Λj*gjL0∥≤3ε.
So ∑j=1∞‍Λj*gj is convergent; thus G={gj}∈Y. Let l→∞ in inequality (13), we get that ∥Gl-G∥Y→0. Hence Y is a complete normed space; that is, Y is a Banach space.
(2) If{Λj:j∈𝒩} is a g-basis, then it is g-complete and g-linearly independent with respect to {Hj:j∈𝒩} by Lemma 10; then the operator S:Y→H,S({gj}j=1∞)=∑j=1∞‍Λj*gj is not only well defined but also is one to one and onto. And, for any {gj}j=1∞∈Y, we have
(15)∥S({gj}j=1∞)∥=∥∑j=1∞‍Λj*gj∥=limN→∞∥∑j=1N‍Λj*gj∥≤SupQ∥∑j=1Q‍Λj*gj∥=∥{gj}∥Y.
So S is bounded operator. Since Y is a Banach space, by the open mapping theorem we get that S is a homeomorphism.Theorem 12.
Suppose, for eachj∈𝒩, Λj∈B(H,Hj) and Λj is surjective. {Λj:j∈𝒩} is a g-basis of H with respect to {Hj:j∈𝒩} and {Γj:j∈𝒩} is its g-dual sequence. Then(1)
for allx∈H, let SNx=∑j=1N‍Λj*Γjx; then SupN∥SNx∥<∞,(2)
C=SupN∥SN∥<∞,(3)
∥|x|∥=SupN∥SNx∥ is a norm on H and ∥·∥≤∥|·|∥≤C∥·∥.Proof.
(
1
) Let Y and S be as defined in Theorem 11. Then for any x∈H, S-1x={Γjx:j∈𝒩}. So
(16)SupN∥SNx∥=SupN∥∑j=1N‍Λj*Γjx∥=∥{Γjx}∥Y=∥S-1x∥≤∥S-1∥∥x∥<∞.
(
2
) Since ∥SNx∥≤SupQ∥SQx∥≤∥S-1∥∥x∥, ∥SN∥≤∥S-1∥. Thus C=SupN∥SN∥<∞.
(3) It is obvious that∥|·|∥ is a seminorm. It is sufficient to show that ∥·∥≤∥|·|∥≤C∥·∥. For any x∈H, we have
(17)∥|x|∥=SupN∥SNx∥≤SupN∥SN∥∥x∥=C∥x∥.
On the other hand,
(18)∥x∥=∥∑j=1∞‍Λj*Γjx∥=limN→∞∥∑j=1N‍Λj*Γjx∥≤SupQ∥∑j=1Q‍Λj*Γjx∥=SupQ∥SQx∥=∥|x|∥.Theorem 13.
Suppose thatΛj∈B(H,Hj) and Λj is surjective. {Λj:j∈𝒩} is a g-basis for H with respect to {Hj:j∈𝒩} and {Γj:j∈𝒩} is the g-dual sequence of {Λj:j∈𝒩}. Then Γj is linear and bounded; that is, Γj∈B(H,Hj) for each j∈𝒩.Proof.
Suppose thatx=∑j∈𝒩‍Λj*gj and y=∑j∈𝒩‍Λj*hj; then Γjx=gj and Γjy=hj by Definition 7. Since x+y=∑j∈𝒩‍Λj*(gj+hj), Γj(x+y)=gj+hj=Γjx+Γjy and Γj(ax)=agj=aΓjx. It implies that Γj is linear, since, for any x∈H, we have
(19)∥Λj*Γjx∥=∥∑i=1j‍Λi*Γix-∑i=1j-1‍Λi*Γix∥≤∥∑i=1j‍Λi*Γix∥+∥∑i=1j-1‍Λi*Γix∥=∥Sjx∥+∥Sj-1x∥≤2C·∥x∥.
Since Λj is surjective, ∥Γjx∥≤∥(Λj†)*∥·∥Λj*Γjx∥. It follows that
(20)∥Γjx∥≤∥(Λj†)*∥·2C·∥x∥.
So Γj is bounded for each j∈𝒩.Theorem 14.
Suppose thatΛj∈B(H,Hj) and Λj is surjective. {Λj:j∈𝒩} is a g-basis for H with respect to {Hj:j∈𝒩} and {Γj:j∈𝒩} is the dual sequence of {Λj:j∈𝒩}. Then {Γj:j∈𝒩} is also a g-basis for H and {Λj:j∈𝒩} and {Γj:j∈𝒩} are g-biorthonormal.Proof.
By Theorem13, we know that Γj∈B(H,Hj). Since {Λj:j∈𝒩} is a g-basis for H with respect to {Hj:j∈𝒩} and {Γj:j∈𝒩} is the dual sequence of {Λj:j∈𝒩}, x=∑j∈𝒩‍Λj*Γjx, for all x∈H. In particular, for each i∈𝒩, gi∈Hi, we have
(21)Λi*gi=∑j∈𝒩‍Λj*ΓjΛi*gi=Λi*ΓiΛi*gi+∑j≠i‍Λj*ΓjΛi*gi.
Since the representation is unique, ΓiΛi*gi=gi and ΓjΛi*gi=0 for j≠i. It implies that
(22)〈Λi*gi,Γj*gj〉=〈ΓjΛi*gi,gj〉=δi,j〈gi,gj〉.
So, {Λj:j∈𝒩} and {Γj:j∈𝒩} are g-biorthonormal. Since ∑j∈𝒩‍Λj*Γj=I, ∑j∈𝒩‍Γj*Λj=I. So for every x∈H, we have that x=∑j∈𝒩‍Γj*Λjx. To prove that {Γj:j∈𝒩} is a g-basis, it is sufficient to show that the representation is unique. It is sufficient to prove that {Γj:j∈𝒩} is g-linearly independent. Suppose that ∑j∈𝒩‍Γj*gj=0; then, for each i∈𝒩, we have
(23)gi=Λi∑j∈𝒩‍Γj*gj=0.
It implies that {Γj:j∈𝒩} is g-linearly independent.Theorem 15.
Suppose thatΛj∈B(H,Hj) and Λj is surjective. {Λj:j∈𝒩} is a g-basis for H with respect to {Hj:j∈𝒩} and {Γj:j∈𝒩} is the dual g-basis of {Λj:j∈𝒩}. SNx=∑j=1N‍Λj*Γjx; that is, SN is the partial sum operator associated with ({Λj},{Γj}). Then under the norm ∥|·|∥ defined in Theorem 12(3), ∥SN∥=1, for all N>0.Proof.
Since{Λj:j∈𝒩} and {Γj:j∈𝒩} are g-biorthonormal by Theorem 14, if K≤L, then
(24)SKSLx=∑j=1K‍Λ-j*Γj(SLx)=∑j=1K‍Λj*Γj∑i=1L‍Λi*Γix=∑j=1K‍∑i=1L‍Λj*ΓjΛi*Γix=∑j=1K‍Λj*Γjx=SKx;
if K≥L, then
(25)SKSLx=∑j=1K‍Λ-j*Γj(SLx)=∑j=1K‍Λj*Γj∑i=1L‍Λi*Γix=∑j=1K‍∑i=1L‍Λj*ΓjΛi*Γix=∑j=1L‍Λj*Γjx=SLx.
Hence
(26)∥|SLx|∥=supK∥SK(SLx)∥=sup{∥S1x∥,…,∥SLx∥}.
So, for any L∈𝒩, ∥|SLx|∥≥∥SLx∥, which follows that
(27)supL∥|SLx|∥≥supL∥SLx∥=∥|x|∥.
On the other hand, for any L∈𝒩, from (26), we also have that
(28)∥|SLx|∥=supK≤L∥SKx∥≤supK∈𝒩∥SKx∥=∥|x|∥,
which follows that supL∥|SLx|∥≤∥|x|∥. So supL∥|SLx|∥=∥|x|∥. So
(29)∥SL∥=sup∥|x|∥=1∥|SLx|∥=sup∥|x|∥=1∥|x|∥=1.Theorem 16.
Suppose thatΛj∈B(H,Hj) and Λj is surjective. Then the following statements are equivalent:(1)
{Λj:j∈𝒩} is a g-basis for H with respect to {Hj:j∈𝒩};(2)
there exists{Γj∈B(H,Hj):j∈𝒩} such that {Λj:j∈𝒩} is g-biorthonormal with {Γj:j∈𝒩} and
(30)∀x∈H,x=∑j∈𝒩‍Λj*Γjx,(3)
{Λj:j∈𝒩} is g-complete and there exists {Γj∈B(H,Hj):j∈𝒩} such that {Λj:j∈𝒩} is g-biorthonormal with {Γj:j∈𝒩} and
(31)∀x∈H,supN∥SNx∥<∞,(4)
{Λj:j∈𝒩} is g-complete and there exists {Γj∈B(H,Hj):j∈𝒩} such that {Λj:j∈𝒩} is g-biorthonormal with {Γj:j∈𝒩} and
(32)supN∥SN∥<∞.Proof.
(
1
)
⇒
(
2
): It is obvious that the g-dual basis {Γj:j∈𝒩} is g-biorthonormal with {Λj:j∈𝒩} by Theorem 13 and, for each x∈H, x=∑j∈𝒩‍Λj*Γjx.
(
2
)
⇒
(
1
): It is only needed, to show that the representation for x is unique. It is sufficient to show that {Λj:j∈𝒩} is g-linearly independent. Suppose that ∑j∈𝒩‍Λj*gj=0, where gj∈Hj, for all j∈𝒩. Then gi=Γi∑j∈𝒩‍Λj*gj=0, for each i∈𝒩, since {Γj:j∈𝒩} and {Λj:j∈𝒩} are g-biorthonormal. So {Λj:j∈𝒩} is g-linearly independent.
(
2
)
⇒
(
3
): It is obvious that {Λj:j∈𝒩} is g-complete. It is only needed to prove that, for all x∈H, supN∥SNx∥<∞. Since x=∑j∈𝒩‍Λj*Γjx, x=limN→∞SNx. It implies that {SNx}N=1∞ is convergent for each x∈H. Thus {SNx}N=1∞ is a bounded sequence for each x∈H. Hence supN∥SNx∥<∞.
(
3
)
⇒
(
4
): It only needed, to show that supN∥SN∥<∞. Since {SN}N=1∞ is a sequence of linear bounded operators on H and, for each x∈H, supN∥SNx∥<∞. So supN∥SN∥<∞ by the Banach-Steinhaus theorem.
(
4
)
⇒
(
2
): For all x∈span{Λj*(Hj):j∈𝒩}, suppose that x=∑n=1M‍Λn*gn, gn∈Hn. Since SN is linear and {Λj:j∈𝒩} and {Γj:j∈𝒩} are g-biorthonormal, for N≥M, we have
(33)SNx=∑n=1M‍SN(Λn*gn)=∑n=1M‍∑i=1N‍Λi*Γi(Λn*gn)=∑n=1M‍Λn*gn=x.
It implies that, for any x∈span{Λj*(Hj):j∈𝒩}, x=limN→∞SNx=∑j∈𝒩‍Λj*Γjx. Next we prove that, for all x∈H, it is also correct. Since {Λj:j∈𝒩} is g-complete, span¯{Λj*(Hj):j∈𝒩}=H by Lemma 8. For any ε>0, there exists y∈span{Λj(Hj):j∈𝒩} such that ∥x-y∥<ε(1+supN∥SN∥)-1. Since y∈span{Λj(Hj):j∈𝒩}, there exist N such that SNy=y. Since
(34)∥x-SNx∥≤∥x-y∥+∥y-SNy∥+∥SNy-SNx∥≤∥x-y∥+0+∥SN∥∥x-y∥≤(1+supN∥SN∥)∥x-y∥<ε,
thus limN→∞SNx=x.Theorem 17.
Suppose, for eachj∈𝒩, Λj∈B(H,Hj) and Λj is surjective. {Λj:j∈𝒩} is g-complete with respect to {Hj:j∈𝒩}. Then {Λj:j∈𝒩} is a g-basis with respect to {Hj:j∈𝒩} if and only if there exists a constant C such that, for any gj∈Hj, any m,n∈𝒩, and m≤n, we have
(35)∥∑j=1m‍Λj*gj∥≤C·∥∑j=1n‍Λj*gj∥.Proof.
⇒: Suppose {Λj:j∈𝒩} is a g-basis with respect to {Hj:j∈𝒩}. Then, for any x∈H, there exists a unique sequence {gj:j∈𝒩} with gj∈Hj for each j∈𝒩 such that x=∑j=1∞‍Λj*gj. Let ∥|x|∥=Supn∥∑j=1n‍Λj*gj∥. Then by Theorem 12, ∥|·|∥ is a norm on H and it is equivalent to ∥·∥. So there exists a constant C such that, for any x∈H, ∥|x|∥≤C·∥x∥. Hence, for any n∈𝒩, any gj∈Hj, j=1,2,…,n, we choose x=∑j=1n‍Λj*gj; then, for any m≤n, we have
(36)∥∑j=1m‍Λj*gj∥≤C·∥∑j=1n‍Λj*gj∥.⇐: Let 𝒜={∑k∈𝒩‍Λk*gk,gk∈Hk, and ∑k∈𝒩‍Λk*gk is convergent}. First, we show that 𝒜=H. Since {Λj∈B(H,Hj):j∈𝒩} is g-complete, 𝒜 is dense in H. It is sufficient to show that 𝒜 is closed. Suppose {yk}⊂𝒜 and limk→∞yk=y. Denote yk=∑j∈𝒩‍Λj*gj(k). Then for any j∈𝒩 and any n≥m≥j, we have, for any k,l∈𝒩,
(37)∥Λj*gj(k)-Λj*gj(l)∥≤C·∥∑s=1m‍Λs*(gs(k)-gs(l))∥≤C2·∥∑s=1n‍Λs*(gs(k)-gs(l))∥≤C2·(∥∑s=1n‍Λs*gs(k)-yk∥+∥yk-y∥)+C2·(∥y-yl∥+∥yl-∑s=1n‍Λs*gs(l)∥).
Since limk→∞yk=y, so, for any ε>0, there exists M>0, such that whenever k≥M, we have ∥y-yk∥≤ε/2C2. In the above inequality, let n→∞; we get
(38)∥Λj*gj(k)-Λj*gj(l)∥≤εforanyj∈𝒩,k,l≥M,forany∥∑s=1m‍Λs*(gs(k)-gs(l))∥≤εCforanym∈𝒩andanyk,l≥M.
Since Λj is surjective, by inequality (10), we have that ∥gj(k)-gj(l)∥≤∥(Λj†)*∥·∥Λj*(gj(k)-gj(l))∥≤∥(Λj†)*∥ε for any j∈𝒩 and any k,l≥M. So {gj(k)}k∈𝒩 is convergent for each j∈𝒩. Suppose limk→∞gj(k)=gj. Then
(39)∥Λj*(gj(k)-gj)∥≤εforanyj∈𝒩,k≥M,forany∥∑s=1m‍Λs*(gs(k)-gs)∥≤εCforanym∈𝒩andanyk≥M.
Since
(40)∥y-∑j=1m‍Λj*gj∥≤∥y-yk∥+∥yk-∑j=1m‍Λj*gj(k)∥+∥∑j=1m‍Λj*gj(k)-∑j=1m‍Λj*gj∥,
so ∑j=1m‍Λj*gj converges to y, which implies that y∈𝒜. Thus 𝒜 is a closed set. Now we will show that {Λj:j∈𝒩} is g-linearly independent. Suppose that ∑j∈𝒩‍Λj*gj=0, where gj∈Hj for each j∈𝒩. Since, for any n∈𝒩 and any j≤n, we have ∥Λj*gj∥≤C·∥∑s=1n‍Λs*gs∥, hence ∥Λj*gj∥=0 for any j≤n. But since Λj is surjective, ∥gj∥≤∥(Λj†)*∥·∥Λj*gj∥. So, for each j≤n,gj=0. Since n is arbitrary, gj=0 for any j∈𝒩. Thus {Λj:j∈𝒩} is g-linearly independent. So {Λj:j∈𝒩} is a g-basis.


## 4. The Property ofg-Minimal of g-Bases

In this section, we studied the property ofg-minimal of g-bases and the perturbations of g-bases.Theorem 18.
SupposeΛj∈B(H,Hj) and Λj is surjective. Then(1)
if{Λj:j∈𝒩} is a g-basis, then {Λj:j∈𝒩} is g-minimal and g-complete;(2)
if{Λj:j∈𝒩} is g-minimal, then {Λj:j∈𝒩} is g-linearly independent.Proof.
(
1
) Since {Λj:j∈𝒩} is a g-basis, it is obvious that {Λj:j∈𝒩} is g-complete. We only need to prove that {Λj:j∈𝒩} is g-minimal. Suppose that {Γj:j∈𝒩} is the unique dual g-basis of {Λj:j∈𝒩}. Then {Λj:j∈𝒩} and {Γj:j∈𝒩} are g-biorthonormal by Theorem 16; that is, 〈Λj*gj,Γi*gi〉=δi,j〈gj,gi〉 for any i,j∈𝒩 and gj∈Hj,gi∈Hi. For any m∈𝒩 and any gm∈Hm such that Λm*gm≠0, let Em=span¯i≠m{Λi*gi}. Then, for any x∈Em, 〈x,Γm*gm〉=0, but 〈Λm*gm,Γm*gm〉=〈gm,gm〉≠0, so Λm*gm∉Em. Hence {Λj:j∈𝒩} is g-minimal.
(
2
) Suppose {Λj:j∈𝒩} is g-minimal and ∑j∈𝒩‍Λj*gj=0. Since Λj is surjective for any j∈𝒩, then ∥gj∥≤∥(Λj†)*∥·∥Λj*gj∥ for any j∈𝒩 and gj∈Hj. If there exists j0∈𝒩 such that gj0≠0, then ∥Λj0*gj0∥>0. It follows that Λj0*gj0≠0. Since Λj0*gj0=-∑j≠j0‍Λj*gj, Λj0*gj0∈span¯j≠j0{Λj*gj}, which contradicts with the fact that {Λj:j∈𝒩} is g-minimal.Theorem 19.
Suppose thatΛj∈B(H,Hj) and Λj is surjective for each j∈𝒩. Then(1)
there exists a sequence{Γj∈M(Hj,H):j∈𝒩}, such that {Λj*:j∈𝒩} and {Γj:j∈𝒩} are biorthonormal if and only if {Λj:j∈𝒩} is g-minimal,(2)
there exists a unique sequence{Γj∈M(Hj,H):j∈𝒩} such that {Λj*:j∈𝒩} and {Γj:j∈𝒩} are biorthonormal, if and only if {Λj:j∈𝒩} is g-minimal and g-complete.Proof.
(
1
)
⇒: Suppose that {Γj:j∈𝒩} is biorthonormal with {Λj*:j∈𝒩}; that is, 〈Λj*gj,Γigi〉=δi,j〈gj,gi〉 for any i,j∈𝒩 and gj∈Hj,gi∈Hi. For any m∈𝒩 and any gm∈Hm such that Λm*gm≠0, let Em=span¯i≠m{Λi*(Hi)}. Then, for any x∈Em, 〈x,Γmgm〉=0, but 〈Λm*gm,Γmgm〉=〈gm,gm〉≠0, so Λm*gm∉Em. Hence {Λj:j∈𝒩} is g-minimal.
⇐: Suppose that {Λj:j∈𝒩} is g-minimal. For all m∈𝒩, for all gm∈Hm such that gm≠0, let Em=span¯j≠m{Λj*(Hj)}. Since Λj is surjective, ∥(Λj†)*∥·∥Λm*gm∥≥∥gm∥>0. Since {Λj:j∈𝒩} is g-minimal, Λm*gm∉Em. So there exists xm∈H such that 〈Λm*gm,xm〉=∥gm∥2 and 〈y,xm〉=0, for all y∈Em. Keep in mind that the notation M(Hj,H) in statement (1) means just a general map with no linearity and bounded requirements, defining a map Γm:Hm→H by Γmgm=xm if gm≠0, Γmgm=0 if gm=0. Then {Γj:j∈𝒩} is biorthonormal with {Λj*:j∈𝒩}, since 〈Λj*gj,Γigi〉=δi,j〈gj,gi〉 for every i,j∈𝒩 and gj∈Hj,gi∈Hi.
(
2
)
⇒: By (1) we know that {Λj:j∈𝒩} is g-minimal. So it is only needed to show that {Λj:j∈𝒩} is g-complete. Suppose that x∈H and 〈x,Λj*gj〉=0 for any j∈𝒩 and any gj∈Hj. Since 〈Λj*gj,Γigi〉=δij〈gj,gi〉, so 〈Λj*gj,x+Γigi〉=δij〈gj,gi〉, which implies that {δx+Γj:j∈𝒩} and {Λj*:j∈𝒩} are biorthonormal, where δx∈B(Hj,H) for each j∈𝒩 defined by δx(f)=x for any f∈Hj. But it is assumed that there exists a unique sequence {Γj∈M(Hj,H):j∈𝒩} such that {Λj*:j∈𝒩} and {Γj:j∈𝒩} are biorthonormal, so δx=0; hence x=δx(f)=0, which implies that span¯{Λj*gj}=H. So {Λj:j∈𝒩} is g-complete.
⇐: By (1), we know that if {Λj:j∈𝒩} is g-minimal, then there exists a sequence {Γj∈M(Hj,H):j∈𝒩} such that {Λj*:j∈𝒩} and {Γj:j∈𝒩} are biorthonormal. Now we only need to show that such sequence is unique. If there is another sequence {Tj∈M(Hj,H):j∈𝒩} such that {Λj*:j∈𝒩} and {Tj:j∈𝒩} are biorthonormal; then 〈Λj*gj,(Γi-Ti)gi〉=0 for any i,j∈𝒩 and gi∈Hi,gj∈Hj. Since {Λj:j∈𝒩} is g-complete, span¯{Λj*(Hj):j∈𝒩}=H by Lemma 8. Thus (Γi-Ti)gi=0 for every i∈𝒩 and gi∈Hi. It follows that Ti=Γi for every i∈𝒩. So the sequence which is biorthonormal with {Λj*:j∈𝒩} is unique.Theorem 20.
Suppose thatΛj∈B(H,Hj) and Λj is surjective for each j∈𝒩. Then the following statements are equivalent:(1)
there exists a sequence{Γj∈B(H,Hj):j∈𝒩} which is g-biorthonormal with {Λj:j∈𝒩},(2)
for allM>0, there exists a constant CM≥1 such that, for any N≥M and gj∈Hj(j=1,…,N), we have
(41)∥∑j=1M‍Λj*gj∥≤CM∥∑j=1N‍Λj*gj∥.Proof.
(
1
)
⇒
(
2
): For any N>0, any x∈H, let SNx=∑j=1N‍Λj*Γjx. Then for any N≥M and any gj∈Hj(j=1,…,N), we have
(42)SM(∑j=1N‍Λj*gj)=∑j=1N‍SM(Λj*gj)=∑j=1N‍∑i=1M‍Λi*Γi(Λj*gj)=∑j=1M‍Λj*gj.
Hence
(43)∥∑j=1M‍Λj*gj∥=∥SM(∑j=1N‍Λj*gj)∥≤∥SM∥∥∑j=1N‍Λj*gj∥.
Then ∥SM∥≥1. So CM=∥SM∥ satisfies the requirement.
(
2
)
⇒
(
1
): Let E=span{Λj*(Hj):j∈𝒩} and x∈E. Suppose that x=∑j=1N‍Λj*gj and M≤N. Since
(44)∥gM∥≤∥(ΛM†)*∥·∥ΛM*gM∥≤∥(ΛM†)*∥(∥∑j=1M‍Λj*gj∥+∥∑j=1M-1‍Λj*gj∥)≤∥(ΛM†)*∥(CM∥∑j=1N‍Λj*gj∥+CM-1∥∑j=1N‍Λj*gj∥)=∥(ΛM†)*∥·(CM+CM-1)∥x∥,
so, if x=0, then gj=0 for j=1,…,N, which follows that the vectors gj,j=1,…,N are uniquely defined by x=∑j=1N‍Λj*gj. Now, for each m∈𝒩, we define operator Γm:E→Hm by
(45)Γm(∑j=1N‍Λj*gj)={gm,m≤N,0,m>N.
Then Γm is linear and
(46)∥Γm(∑j=1N‍Λj*gj)∥≤∥gm∥≤∥(Λm†)*∥·(Cm+Cm-1)∥∑j=1N‍Λj*gj∥.
So Γm is bounded on E; hence it can be linearly extended to E¯. Let Γm=0 on E¯⊥; then Γm∈B(H,Hm) for each m∈𝒩 and
(47)〈Λj*gj,Γi*gi〉=〈ΓiΛj*gj,gi〉=δi,j〈gj,gi〉;
that is, {Γj:j∈𝒩} is g-biorthonormal with {Λj:j∈𝒩}.


## 5. Perturbation ofg-Bases

In this section, we give one result about the stability ofg-bases under certain perturbation.Theorem 21.
Suppose thatΛj∈B(H,Hj) and Λj is surjective for each j∈𝒩. {Λj:j∈𝒩} is a g-basis for H with {Γj:j∈N} as its g-dual basis and its basis constant is C. If {Lj∈B(H,Hj):j∈𝒩} is a sequence such that R=∑j∈𝒩‍∥Λj*-Lj*∥·∥Γj∥<1, then {Lj:j∈𝒩} is a g-basis for H with basis constant C1≤((1+R)/(1-R))C.Proof.
Supposex=∑j=mn‍Λj*gj, gj∈Hj(j=m,…,n). Then gj=Γjx(j=m,…,n), since {Λj:j∈𝒩} and {Γj:j∈𝒩} are g-biorthonormal by Theorem 16. Since
(48)∥∑j=mn‍(Λj*-Lj*)gj∥≤∑j=mn‍∥Λj*-Lj*∥·∥gj∥=∑j=mn‍∥Λj*-Lj*∥·∥Γjx∥≤∑j=mn‍∥Λj*-Lj*∥·∥Γj∥·∥x∥≤R·∥x∥=R·∥∑j=mn‍Λj*gj∥,
which implies that if ∑j∈𝒩‍Λj*gj is convergent, then ∑j∈𝒩‍(Λj*-Lj*)gj is convergent, the following operator is well defined:
(49)T:H⟶H,T(∑j∈𝒩‍Λj*gj)=∑j∈𝒩‍(Λj*-Lj*)gj.
It is obvious that T is linear and ∥T∥≤R≤1. It implies that I-T is invertible. Since, for any j∈𝒩 and gj∈Hj, we have
(50)(I-T)Λj*gj=Λj*gj-T(Λj*gj)=Λj*gj-(Λj*-Lj*)gj=Lj*gj,
so Lj=Λj(I-T)* for any j∈𝒩. Since (I-T)* is invertible, it is easy to see that {Lj:j∈𝒩} is a g-basis for H. Now we start to show the basis constants inequality. For any N≥M, we have
(51)∥∑j=1M‍Lj*gj∥≤∥∑j=1M‍Λj*gj∥+∥∑j=1M‍(Lj*-Λj*)gj∥≤∥∑j=1M‍Λj*gj∥+R·∥∑j=1M‍Λj*gj∥=(1+R)·∥∑j=1M‍Λj*gj∥.
On the other hand, we have
(52)∥∑j=1N‍Λj*gj∥≤∥∑j=1N‍Lj*gj∥+∥∑j=1N‍(Λj*-Lj*)gj∥≤∥∑j=1N‍Lj*gj∥+R·∥∑j=1M‍Λj*gj∥.
It follows that
(53)∥∑j=1N‍Lj*gj∥≥(1-R)·∥∑j=1N‍Λj*gj∥.
Since {Λj:j∈𝒩} is a g-basis for H, by Theorem 17, we have that
(54)∥∑j=1M‍Λj*gj∥≤C∥∑j=1N‍Λj*gj∥.
So, by inequalities (51), (53) and (54), we have
(55)∥∑j=1M‍Lj*gj∥≤(1+R)·∥∑j=1M‍Λj*gj∥≤(1+R)C·∥∑j=1N‍Λj*gj∥≤1+R1-RC·∥∑j=1N‍Lj*gj∥.



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*Source: 272407-2013-06-13.xml*