# Quantifying the uncertainty in the time-redshift relationship

Michael S. Turner
Kavli Institute for Cosmological Physics
University of Chicago, Chicago, IL 60637-1433

Department of Physics and Astronomy
The University of California, Los Angeles
Los Angeles, CA 90095-1547

email<EMAIL_ADDRESS>
###### Abstract

The age of the Universe at a given redshift is a fundamental relationship in
cosmology. For many years, the uncertainties in it were dauntingly large,
close to a factor of 2. In this age of precision cosmology, they are now at
the percent level and dominated by the uncertainty in the Hubble constant. The
uncertainties due to the parameters that describe the current cosmological
model, $\Lambda$CDM, are much less important. In decreasing order they are:
uncertainty due to the matter density $\Omega_{M}$ around 0.9% for $z>3$;
uncertainty due to the dark energy equation-of-state parameter $w$, less than
0.3% for $z>3$; and uncertainty due to the curvature parameter $\Omega_{k}$,
at most 0.07%.

With only the assumption of the Robertson-Walker metric it is simple to show
that the age of the Universe at a given redshift is given by (see e.g., Ref.
[1])

$\displaystyle t(z)$ $\displaystyle=$
$\displaystyle\int_{z}^{\infty}{dz\over(1+z)H(z)}$ (1) $\displaystyle=$
$\displaystyle
H_{0}^{-1}\int_{z}^{\infty}{dz\over(1+z)\left[(1-\Omega_{M}+\Omega_{k})(1+z)^{3(1+w)}-\Omega_{k}(1+z)^{2}+\Omega_{M}(1+z)^{3}\right]^{1/2}}$
$\displaystyle=$ $\displaystyle H_{0}^{-1}\int_{(1+z)}^{\infty}{dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{1/2}}$

where $x\equiv 1+z$, $\Omega_{M}$ is the fraction of critical density in
matter, $w$ is the dark energy equation-of-state parameter, and $\Omega_{k}$
is the curvature parameter. Here it is assumed that the Universe comprises
matter and dark energy and the small, about 0.01% fraction of critical density
in radiation has been neglected here (see footnote below).

Twenty-five years ago, the uncertainties in the age-redshift were very large,
both because of uncertainty in the Hubble constant and in the cosmological
model. Until 2001, when the HST Key Project pinned down $H_{0}$ with a
reliable error budget [2], one could argue for any value between 50 km/sec/Mpc
and 100 km/sec/Mpc (or even beyond [3]) and the use of “little $h$” was
universal, $h\equiv H_{0}/$100/km/sec/Mpc. Likewise, values for the density
parameter $\Omega_{0}$ ranged from $0.1$ to $1.0$ and even above. Together,
this meant that $t(z)$ was uncertain by around a factor of two, making
quantifying the uncertainty in the age-redshift relationship a fool’s errand.

In the current era of precision cosmology [4], the situation couldn’t be more
different: We have a standard model, $\Lambda$CDM, well supported by data,
with a small set of cosmological parameters that are well determined.
Likewise, the Hubble constant is known to a precision of better than 5% [8,
9]. For $\Lambda$CDM, $\Omega_{k}=0$ and $w=-1$, the age-redshift relationship
is given by:

$t(z)={1\over
3}H_{0}^{-1}\Omega_{\Lambda}^{-1/2}\ln\left[{[1+x^{3}\Omega_{M}/\Omega_{\Lambda}]^{1/2}+1\over[1+x^{3}\Omega_{M}/\Omega_{\Lambda}]^{1/2}-1}\right]\\\
$

where as before $x=1+z$.

The purpose of this note is to estimate the uncertainty in the age-redshift
relationship for the $\Lambda$CDM model. In particular, with the exception of
$H_{0}$ (see below), I will be guided by the cosmological parameters and
uncertainties from Planck [5], supplemented by measurements of $w$ by Pantheon
[6] and the Dark Energy Survey [7]:

  * •

$\Omega_{M}=0.3111\pm 0.0056$

  * •

$\Omega_{\Lambda}=0.6889\pm 0.0056$

  * •

$\Omega_{k}=0.0007\pm 0.0019$

  * •

$w=-1\pm 0.05$

where I have inflated the quoted uncertainty in $w$ by a factor of 2.

Adopting these cosmological parameters, i.e., $\Omega_{M}=0.3111$,
$\Omega_{k}=0$ and $w=-1$, the age-redshift relationship above is

$\displaystyle t(z)$ $\displaystyle=$ $\displaystyle
0.402H_{0}^{-1}\ln\left[{(1+0.452x^{3})^{1/2}+1\over(1+0.452x^{3})^{1/2}-1}\right]$
(2) $\displaystyle=$ $\displaystyle 5.616(0.7/h)\,{\rm
Gyr}\ln\left[{(1+0.452x^{3})^{1/2}+1\over(1+0.452x^{3})^{1/2}-1}\right]$

The determination of the Hubble constant is once again of pressing interest
and controversy, with apparent discrepancies between the value directly
measured and that inferred from CMB anisotropy measurements and the assumption
of $\Lambda$CDM [8, 9]. Since the Hubble constant factors out in determining
the age-redshift relationship it is easy to separate its contribution to the
uncertainty in $t(z)$ from the other cosmological parameters, remembering that
there may be correlations with the other parameters. Without making a judgment
on the current Hubble tension, and for our purposes here, I think it is fair
to say the current uncertainty is in the range of 1% to 5%.

Taking as the independent cosmological parameters, $H_{0}$, $\Omega_{M}$, $w$
and $\Omega_{k}$, it follows by differentiation that

$\displaystyle{\delta t\over t(z)}$ $\displaystyle=$ $\displaystyle-{\delta
H_{0}\over H_{0}}$ (3) $\displaystyle{\delta t\over\delta\Omega_{M}}$
$\displaystyle=$ $\displaystyle-{1\over
2}H_{0}^{-1}\int_{1+z}^{\infty}{(x^{3}-1)dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{3/2}}$
(4) $\displaystyle{\delta t\over\delta w}$ $\displaystyle=$
$\displaystyle-{3\over
2}H_{0}^{-1}\int_{1+z}^{\infty}{\ln(x)(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{3/2}}$
(5) $\displaystyle{\delta t\over\delta\Omega_{k}}$ $\displaystyle=$
$\displaystyle{1\over 2}H_{0}^{-1}\int_{1+z}^{\infty}{(x^{2}-1)dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{3/2}}$
(6)

Bringing everything together, the uncertainty in the age-redshift relationship
can be written as:

${\delta t\over t(z)}=\left\\{-{\delta H_{0}/H_{0}},\
\alpha_{M}\delta\Omega_{M},\ \alpha_{w}\delta_{w},\
\alpha_{k}\delta\Omega_{k}\right\\},$

where

$\displaystyle\alpha_{M}$ $\displaystyle=$ $\displaystyle-{1\over
2\alpha_{0}}\int_{1+z}^{\infty}{(x^{3}-1)dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{3/2}}$
(7) $\displaystyle\alpha_{w}$ $\displaystyle=$ $\displaystyle-{3\over
2\alpha_{0}}\int_{1+z}^{\infty}{\ln(x)(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{3/2}}$
(8) $\displaystyle\alpha_{k}$ $\displaystyle=$ $\displaystyle{1\over
2\alpha_{0}}\int_{1+z}^{\infty}{(x^{2}-1)dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{3/2}}$
(9) $\displaystyle\alpha_{0}$ $\displaystyle=$
$\displaystyle\int_{1+z}^{\infty}{dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{1/2}}$
(10)

I have not added the uncertainties in quadrature, which would require a choice
for the $H_{0}$ uncertainty and taking proper account of correlated error in
the cosmological parameters. Rather, I have shown the individual contributions
separately to illustrate their relative importance and to allow the reader to
re-scale if they feel it is necessary.

Fig. 1 shows the relative uncertainties in the age-redshift relationship as a
function of redshift. From the least important to the most important they are:
$\Omega_{k}$, at most 0.07%; $w$, at most 0.2% for $z>3$; $\Omega_{M}$, around
0.9% for $z>3$; and $H_{0}$, between 1% and 5% (which follows directly from
its assumed uncertainty).111Had one included the radiation component,
$\Omega_{R}\simeq 0.00008$ whose major uncertainty owes to that in $H_{0}$,
its contribution to the uncertainty in the age-redshift relationship would be
at most 0.01%, and of course, correlated with $H_{0}$.

The redshift dependence of the uncertainty upon $\Omega_{M}$ and $w$ is easy
to understand. It is most dependent upon $w$ at $z<1$ where dark energy
dominates and the age is close to that of a dark energy dominated universe.
The uncertainty due to $w$ drops rapidly with redshift because for $z>1$ the
Universe is increasingly matter-dominated and the age approaches that of a
matter only universe. In particular, $\delta t(z)/t(z)\rightarrow-{1\over
2}\delta\Omega_{M}/\Omega_{M}=0.93\%$ for $z\gg 1$..

Figure 1: Contributions to the uncertainty in the age of the Universe at
redshift $z$ from the cosmological parameters $H_{0}$, $w$, $\Omega_{M}$ and
$\Omega_{k}$.

For completeness, now consider the look-back time:

$\displaystyle t_{LB}(z)$ $\displaystyle\equiv$
$\displaystyle\int_{0}^{z}{dz\over(1+z)H(z)}$ (11) $\displaystyle=$
$\displaystyle
H_{0}^{-1}\int_{0}^{z}{dz\over(1+z)\left[(1-\Omega_{M}+\Omega_{k})(1+z)^{3(1+w)}-\Omega_{k}(1+z)^{2}+\Omega_{M}(1+z)^{3}\right]^{1/2}}$
$\displaystyle=$ $\displaystyle H_{0}^{-1}\int_{1}^{(1+z)}{dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{1/2}}$

The analogous formulae for the uncertainty in it are:

$\displaystyle{\delta t_{LB}\over t_{LB}(z)}$ $\displaystyle=$
$\displaystyle-{\delta H_{0}\over H_{0}}$ (12) $\displaystyle{\delta
t_{LB}\over\delta\Omega_{M}}$ $\displaystyle=$ $\displaystyle-{1\over
2}H_{0}^{-1}\int_{1}^{1+z}{(x^{3}-1)dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{3/2}}$
(13) $\displaystyle{\delta t_{LB}\over\delta w}$ $\displaystyle=$
$\displaystyle-{3\over
2}H_{0}^{-1}\int_{1}^{1+z}{\ln(x)(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{3/2}}$
(14) $\displaystyle{\delta t_{LB}\over\delta\Omega_{k}}$ $\displaystyle=$
$\displaystyle{1\over 2}H_{0}^{-1}\int_{1}^{1+z}{(x^{2}-1)dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{3/2}}$
(15) ${\delta t_{LB}\over t_{LB}(z)}=\left\\{-{\delta H_{0}/H_{0}},\
\beta_{M}\delta\Omega_{M},\ \beta_{w}\delta_{w},\
\beta_{k}\delta\Omega_{k}\right\\},$

where

$\displaystyle\beta_{M}$ $\displaystyle=$ $\displaystyle-{1\over
2\beta_{0}}\int_{1}^{1+z}{(x^{3}-1)dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{3/2}}$
(16) $\displaystyle\beta_{w}$ $\displaystyle=$ $\displaystyle-{3\over
2\beta_{0}}\int_{1}^{1+z}{\ln(x)(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{3/2}}$
(17) $\displaystyle\beta_{k}$ $\displaystyle=$ $\displaystyle{1\over
2\beta_{0}}\int_{1}^{1+z}{(x^{2}-1)dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{3/2}}$
(18) $\displaystyle\beta_{0}$ $\displaystyle=$
$\displaystyle\int_{1}^{1+z}{dx\over
x\left[(1-\Omega_{M}+\Omega_{k})x^{3(1+w)}-\Omega_{k}x^{2}+\Omega_{M}x^{3}\right]^{1/2}}$
(19)

Fig. 2 shows the relative uncertainties in the look-back time as a function of
redshift. From the least important to the most important they are:
$\Omega_{k}$, at most 0.07%; $\Omega_{M}$, at most 0.5%; $w$, at most 0.9%;
and $H_{0}$, between 1% and 5%. Here too the dependence of the uncertainties
is easy to understand. For small redshifts, $z<1$, the uncertainty due to
knowledge of $\Omega_{M}$ is less important because look back only encompasses
the dark energy dominated era. The uncertainty due to $\Omega_{M}$ rises at
higher redshift because look back includes the matter-dominated era as well.

Figure 2: Contributions to the uncertainty in the look back time at redshift
$z$ from the cosmological parameters $H_{0}$, $w$, $\Omega_{M}$ and
$\Omega_{k}$.

Two final comments. First, other than a variation in the dark energy EOS
parameter, I did not consider modifications to the $\Lambda$CDM paradigm. The
effect of early dark energy on the age-redshift relationship was discussed in
detail in Ref. [10]. As I mentioned above, I did not combine all the parameter
uncertainties, because of the correlations between the values determined for
the parameters. But more importantly, the Hubble tension raises the issue of a
possible missing element in the current cosmological model or a systematic
error in $H_{0}$, which could impact $t(z)$ significantly.

This note was inspired by a question asked of my UCLA colleague Professor
Alice Shapley at the International Space Science Institute’s Symposium on
Cosmic Beginnings: Unveiling the first billion years with JWST at the
University of Bern on 14 March 2024. The question was, what is the uncertainty
in the age-redshift relationship. She responded that it was small but she
didn’t know precisely how small. I thought it would make a nice problem for my
UCLA Cosmology class’ final problem set. Unfortunately, I had just given out
that problem set and so I did the problem myself to share with the class.

## References

  * [1] E.W. Kolb and M.S. Turner, The Early Universe.
  * [2] W.L. Freedman et al, Astrophys. J. 553, 47 (2001).
  * [3] J.G. Bartlett et al, Science 267, 980 (1995).
  * [4] M.S. Turner, Ann. Rev. Nucl. Part. Sci. 72: 1 (2022).
  * [5] Planck Collaboration, Astron. Astrophys. 641, A1 (2020).
  * [6] D. Brout et al, Astrophys. J. 938: 100 (2022).
  * [7] T.M.C. Abbott et al (DES Collaboration), arXiv:2401.02929v2 (2024).
  * [8] A.G. Riess et al, Astrophys. J. 937: L37 (2022).
  * [9] W.L. Freedman, Astrophys. J. 919: 16 (2021).
  * [10] M. Boylan-Kolchin and D.R. Weisz, arXiv:2103.15825v2 (2021).


# Schauder Bases and Operator Theory

Yang Cao  Yang Cao, Department of Mathematics , Jilin university, 130012,
Changchun, P.R.China<EMAIL_ADDRESS>,  Geng Tian  Geng Tian, Department
of Mathematics , Jilin university, 130012, Changchun, P.R.China
<EMAIL_ADDRESS>and  Bingzhe Hou  Bingzhe Hou, Department of
Mathematics , Jilin university, 130012, Changchun, P.R.China<EMAIL_ADDRESS>
(Date: Oct. 14, 2010)

###### Abstract.

In this paper, we firstly give a matrix approach to the bases of a separable
Hilbert space and then correct a mistake appearing in both review and the
English translation of the Olevskii’s paper. After this, we show that even a
diagonal compact operator may map an orthonormal basis into a conditional
basis.

###### Key words and phrases:

.

###### 2000 Mathematics Subject Classification:

Primary 47B37, 47B99; Secondary 54H20, 37B99

##  1\. Introduction and preliminaries

In operator theory, an invertible operator on an infinite dimensional complex
Hilbert space $\mathcal{H}$ means the bounded operator which has a bounded
inverse operator, and it is well-known that, for an $n\times n$ matrix $M_{n}$
(seen as an operator on finite dimensional Hilbert space $\mathbb{C}^{n}$),
$M_{n}$ is invertible if and only if its column vectors are linearly
independent in $\mathbb{C}^{n}$. In other words, the column vectors of $M_{n}$
comprise a basis of $\mathbb{C}^{n}$. From this point of view, we could
generalize the "invertibility" of $\omega\times\omega$ matrix $M$ (the
representation of a bounded operator on an orthonormal basis of $\mathcal{H}$)
in the following manner: all column vectors of $M$ form some kind of basis of
$\mathcal{H}$. Actually, the invertible operator do have a natural
understanding in the ‘basis’ language. That is, the column (or row) vectors of
the matrix of an invertible operator always comprise a ‘Riesz basis’ (it is a
direct corollary of theorem 2, paper [1], although the authors do not state it
in this way). From the above observation, it suggests us to consider the
$\omega\times\omega$ matrix whose column vectors form more general kind of
bases.

Naturally we consider the $\omega\times\omega$ matrix whose column vectors
comprise a Schauder basis. We shall call them the Schauder matrix therefrom.
An operator which has a Schauder matrix representation under some orthonormal
basis (ONB) will be called a Schauder operator. An easy fact is that an
operator is a Schauder operator if and only if it maps some ONB into a
Schauder basis. Many scholars have studied some kind of these operators. A. M.
Olevskii gave a surprising result on the bounded operators which map some ONB
into a conditional quasinormal basis ([5], theorem 1, p479); Stephane Jaffard
and Robert M. Young proved that a Schauder basis always can be given by an
one-to-one positive transformation ([1], theorem 1, p554). I. Singer gave lots
of examples of bases of $\mathcal{H}$ which can be rewritten into a matrix
form (see, [6], p429, p497). Besides these results, as for a joint research
both on operator theory and the basis theory but not in this direction, the
paper [25], [26] by Gowers, the paper [13] by Kwapien, S. and Pelczynski, A.
and the elegant book [2] by M. Young are remarkable examples.

Nevertheless, there is still a gap between the researches in the field of
basis theory and operator theory. There are few joint works on both basis of
Hilbert space and the operators on the Hilbert space. The reason reflects on
two aspects. One is the different terminology systems and the other one is
that there are scanty common objects to study with. The main purpose of this
paper is to show that the Schauder matrix is a candidate to fill this gap. As
basic and traditional tools, the matrix representation of operators plays an
important role in the study of the operators on the Hilbert space
$\mathcal{H}$. So the matrix approach to the basis theory is a good beginning
to the joint research on the bases of the Hilbert space $\mathcal{H}$ and the
operators on it.

In this paper, the matrix representation of operators and bases will be the
bridge between basis theory and operator theory. We firstly give a matrix
approach to the bases of a separable Hilbert space and then correct a mistake
appearing in both review and the English translation of the Olevskii’s paper.
After this, we follow the Olevskii’s result to consider the operators which
can map some ONB into a conditional Schauder basis. We shall call them
conditional operators therefrom. In matrix language, it is equivalent to study
the operator $T$ which has a matrix representation $M$ under some ONB such
that the column vector sequence of $M$ comprise a conditional Schauder basis.

##  2\. An Operator Theory Description of Schauder basis

### 2.1.

Suppose that $\\{e_{k}\\}_{k=1}^{\infty}$ is an ONB of $\mathcal{H}$. An
$\omega\times\omega$ matrix $M=(m_{ij})$ automatically represents an operator
under this ONB. In more details, for a vector $x\in\mathcal{H}$ there is an
unique $l^{2}-$sequence $\\{x_{n}\\}_{n=1}^{\infty}$ such that
$x=\sum_{n=1}^{\infty}x_{n}e_{n}$ in which the series converges in the norm of
$\mathcal{H}$. Let
$y_{n}=\sum_{k=1}^{\infty}m_{ik}x_{k},y=\sum_{n=1}^{\infty}y_{n}e_{n}$, then
the operator $T_{M}$ defined by $T_{M}x=y$ is just the corresponding operator
represented by $M$. In general, $T$ is not a bounded operator. We shall
identify the $\omega\times\omega$ matrix $M$ and the operator $T_{M}$, and
denote them by the same notation $M$ if we have fixed an ONB and there is no
confusion.

Recall that a sequence $\psi=\\{f_{n}\\}_{n=1}^{\infty}$ is called a Schauder
basis of the Hilbert space $\mathcal{H}$ if and only if for every vector
$x\in\mathcal{H}$ there exists an unique sequence
$\\{\alpha_{n}\\}_{n=1}^{\infty}$ of complex numbers such that the partial sum
sequence $x_{k}=\sum_{n=1}^{k}\alpha_{n}f_{n}$ converges to $x$ in norm.

Denote by $P_{k}$ the the diagonal operator with the first $k-$th entries on
diagonal line equal to 1 and $0$ for others. Then as an operator $P_{k}$
represents the orthogonal projection from $\mathcal{H}$ to the subspace
$\mathcal{H}^{(k)}=span\\{e_{1},e_{2},\cdots,e_{k}\\}$.

######  Lemma 2.1.

Assume that $\\{e_{k}\\}_{k=1}^{\infty}$ is a fixed ONB of $\mathcal{H}$.
Suppose that an $\omega\times\omega$ matrix $F=(f_{ij})$ satisfies the
following properties:

1\. Each column of the matrix $F$ is a $l^{2}-$sequence;

2\. $F$ has an unique left inverse matrix $G^{*}=(g_{kl})$ such that each row
of $G^{*}$ is also a $l^{2}-$sequence;

3\. Operators $Q_{k}$ defined by the matrix $Q_{k}=FP_{k}G^{*}$ are well-
defined projections on $\mathcal{H}$ and converges to the unit operator $I$ in
the strong operator topology.

Then the sequence
$\\{f_{k}\\}_{k=1}^{\infty},f_{k}=\sum_{i=1}^{\infty}f_{ik}e_{i}$ must be a
Schauder basis.

Here we use the term “left reverse” in the classical means, that is, the
series $\sum_{j=1}^{\infty}g_{kj}f_{jn}$ converges absolutely to $\delta_{kn}$
for $k,n=1,2,\cdots$. $G^{*}$ does not mean the adjoint of $G$, it is just a
notation.

###### Proof.

Property 1 just ensure that series
$\\{f_{k}=\sum_{j=1}^{\infty}f_{ij}e_{i}\\}_{k=0}^{\infty}$ converges to a
well-defined vector $f_{k}$ in $\mathcal{H}$ by norm. Property 2 implies that
$span\\{f_{n};n=1,2,\cdots\\}=\mathcal{H}$ by the uniqueness of the left
inverse. Moreover, the $k-$th row of the matrix $G^{*}$ is just the vector
$g_{k}^{*}$ such that $(g_{k}^{*},f_{n})=\delta_{kn}$. Therefore the vector
sequence $\\{f_{n}\\}_{n=1}^{\infty}$ must be minimal by the Hahn-Banach
theorem(cf, [7] corollary6.8, p82) and the Riesz representation theorem(see,
[7], theorem3.4, p12).

Now for each vector $x=(x_{1},x_{2},\cdots)$ denote by
$\alpha_{k}^{x}=(g_{k}^{*},x),$ it is easy to check that $Q_{k}^{2}=Q_{k}$ and

$Q_{k}x=FP_{k}G^{*}x=\sum_{n=1}^{k}\alpha^{x}_{k}f_{k}.$

By property 3, we have $Q_{k}x\rightarrow x$ since $Q_{k}$ converges to $I$ in
strong operator topology(SOT). That is, series
$\sum_{n=1}^{\infty}\alpha^{x}_{n}f_{n}$ converges to the vector $x$ in norm.
So we have proved that each vector $x$ in $\mathcal{H}$ can be represented by
the sequence $\\{f_{n}\\}_{n=1}^{\infty}$ with coefficients
$\\{\alpha_{n}^{x}\\}_{n=1}^{\infty}$.

To show that $\\{f_{n}\\}_{n=1}^{\infty}$ is a Schauder basis, we just need to
show that this representation is unique. Suppose that
$\\{\alpha_{n}\\}_{n=1}^{\infty}$ is a sequence such that the series
$\sum_{n=1}^{\infty}\alpha_{n}f_{n}$ converges to $0$ in the norm of the
Hilbert space $\mathcal{H}$. Assume that the integer $n_{0}$ is the first
number satisfying $\alpha_{n_{0}}\neq 0$. Then we have

$f_{n_{0}}=-\frac{1}{\alpha_{n_{0}}}\cdot\sum_{n=n_{0}+1}^{\infty}\alpha_{n}f_{n}$

in which the series also converges in the norm topology. It counter to the
fact that the sequence $\\{f_{n}\\}_{n=1}^{\infty}$ is a minimal sequence. ∎

Conversely, suppose that $\psi=\\{f_{n}\\}_{n=1}^{\infty}$ is a basis of
$\mathcal{H}$. For a fixed ONB $\\{e_{n}\\}_{n=1}^{\infty}$, each vector
$f_{n}$ has a representation $f_{n}=\sum_{k=1}^{\infty}f_{kn}e_{k}$. Denote
$F_{\psi}=(f_{kn})$. We shall call $F_{\psi}$ the Schauder matrix
corresponding to the basis $\psi$. The following lemma is the inverse of the
above lemma.

######  Lemma 2.2.

Assume that $\psi=\\{f_{n}\\}_{n=1}^{\infty}$ is a Schauder basis. Then the
corresponding Schauder matrix $F_{\psi}$ satisfies the following properties:

1\. Each column of the matrix $F_{\psi}$ is a $l^{2}-$sequence;

2\. $F_{\psi}$ has an unique left inverse matrix $G_{\psi}^{*}=(g_{kl})$ such
that each row of $G_{\psi}^{*}$ is also a $l^{2}-$sequence;

3\. Operators $Q_{k}$ defined by the matrix $Q_{k}=F_{\psi}P_{k}G_{\psi}^{*}$
are well-defined projections on $\mathcal{H}$ and converges to the unit
operator $I$ in the strong operator topology.

###### Proof.

Property 1 comes from the fact that $f_{n}$ is a vector in $\mathcal{H}$.

If $\\{f_{k}\\}_{k=1}^{\infty}$ is a Schauder basis, then the subspace
$\widehat{\mathcal{H}}_{k}=span\\{f_{n};n\neq k\\}$ for each $k$ satisfying
$f_{k}\notin\widehat{\mathcal{H}}_{k}$(cf, [6], p50-51). So we must have a
unique linear functional $\varphi_{k}$ such that
$\varphi_{k}(f_{n})=\delta_{kn}$. Then by the Riesz representation theorem,
there is a unique vector $g^{*}_{k}=(g^{*}_{kl})\in\mathcal{H}$ such that
$\sum_{j=1}^{n}g^{*}_{kj},f_{jn}=\delta_{kn}$ in which
$\\{g^{*}_{kl}\\}_{l=1}^{\infty}$ is a $l^{2}-$sequence. The uniqueness holds
because the sequence $\\{f_{k}\\}_{k=1}^{\infty}$ spans the Hilbert space.
Hence a Schauder matrix must have a unique left inverse matrix whose rows are
$l^{2}-$sequence. Then we have proved the property 2.

Property 3 is just a direct corollary of the definition of Schauder basis.
Denote by $G=(G^{*})^{*}=g_{nk}$ the adjoint matrix of $G^{*}$, then we have
$g_{nk}=\overline{g_{kn}}$. Moreover, denote by $g_{n}$ the $n-$th column
vector and for a vector $x=\sum_{n=1}^{\infty}x_{n}e_{n}$ denote by
$y_{n}=\sum_{k=1}^{\infty}g_{nk}^{*}x_{k}$. Then trivially we have
$y_{n}=(x,g_{n})$ and $(f_{k},g_{n})=\delta_{kn}$ Suppose that
$x=\sum_{k=1}^{\infty}\alpha_{k}f_{k}$ is the representation of the vector $x$
under the basis $\psi$. Then we must have $\alpha_{n}=y_{n}$ since

$y_{n}=(x,g_{n})=(\sum_{k=1}^{\infty}\alpha_{k}f_{k},g_{n})=\alpha_{n}.$

Therefore we have $Q_{k}x=\sum_{n=1}^{\infty}\alpha_{n}f_{n}$. Clearly we have
$Q_{k}x\rightarrow x$ in the norm topology. In other words,
$||Q_{k}x-x||\rightarrow 0$ when $k\rightarrow\infty$ which implies
$Q_{k}\rightarrow I$ in SOT(cf, [7], proposition 1.3, p262). ∎

The matrix $G_{\psi}^{*}$ is unique and decided completely by $F_{\psi}$. In
fact the matrix $G^{*}$ is also the “right inverse” of the matrix $F$ in the
classical sense. For more details, let $F=(f_{kn})_{\omega\times\omega}$,
$G^{*}=(g_{mk})_{\omega\times\omega}$, $f_{n}=\\{f_{kn}\\}_{k=1}^{\infty}$ and
$g_{m}^{*}=\\{g_{mk}^{*}\\}_{k=1}^{\infty}$. Moreover, denote their adjoint
matrices by
$F^{*}=(f^{*}_{kn})_{\omega\times\omega}=(\overline{f_{nk}})_{\omega\times\omega}$,
$G=(g^{*}_{mk})_{\omega\times\omega}=(\overline{g_{km})}_{\omega\times\omega}$.
Then both $\psi=\\{f_{n}\\}_{n=1}^{\infty}$ and
$\psi^{*}=\\{g_{m}\\}_{m=1}^{\infty}$ are biorthogonal basis to each other.
That is, $\psi$ and $\psi^{*}$ are bases and we have
$(f_{n},g_{m})=\delta_{nm}$ for all $n,m\in\mathbb{N}$. Now we show that the
series $\sum_{k=1}^{\infty}f_{nk}g^{*}_{km}$ converges to $\delta_{nm}$ as
$k\rightarrow\infty$ for all $n,m\in\mathbb{N}$. Let
$\\{e_{l}\\}_{l=1}^{\infty}$ be the corresponding ONB. We write $e_{n},e_{m}$
into the linearly combinations of basis vector in $\psi$ and
$\psi^{{}^{\prime}}$ as follows:

$e_{n}=\sum_{k=1}^{\infty}\alpha_{nk}f_{k},e_{m}=\sum_{k=1}^{\infty}\beta_{mk}g^{*}_{k}.$

Then we have $\alpha_{nk}=g^{*}_{kn}$ and
$\beta_{mk}=f^{*}_{km}=\overline{f_{mk}}$. Hence for any integer $N$

$\begin{array}[]{rl}\sum_{k=1}^{N}f_{nk}g^{*}_{km}&=(\sum_{k=1}^{N}\alpha_{nk}f_{k},\sum_{k=1}^{N}\beta_{mk}g^{*}_{k})\\\
&=(e_{n}-\sum_{k=N}^{\infty}\alpha_{nk}f_{k},e_{m}-\sum_{k=N}^{\infty}\beta_{mk}g^{*}_{k}).\end{array}$

Now given $\epsilon>0$, we choose an integer $N$ such that inequalities
$||e_{n}-\sum_{k=N}^{\infty}\alpha_{nk}f_{k}||<\frac{\epsilon}{2},||e_{m}-\sum_{k=N}^{\infty}\beta_{mk}g^{*}_{k}||<\frac{\epsilon}{2}$
hold. Then we have

$\begin{array}[]{rl}&|\sum_{k=1}^{N}f_{nk}g^{*}_{km}-(e_{n},e_{m})|\\\
=&|-(\sum_{k=N}^{\infty}\alpha_{nk}f_{k},e_{m}-\sum_{k=N}^{\infty}\beta_{mk}g^{*}_{k})\\\
&~{}~{}~{}~{}-(e_{n}-\sum_{k=N}^{\infty}\alpha_{nk}f_{k},\sum_{k=N}^{\infty}\beta_{mk}g^{*}_{k})+(\sum_{k=N}^{\infty}\alpha_{nk}f_{k},\sum_{k=N}^{\infty}\beta_{mk}g^{*}_{k})|\\\
\leq&\epsilon(|1+\frac{\epsilon}{2}|+\frac{\epsilon}{4}).\end{array}$

For this reason, we have the following definition.

######  Definition 2.3.

For a Schauder matrix $F_{\psi}$, the corresponding matrix $G_{\psi}^{*}$ is
called the inverse matrix of $F_{\psi}$.

If we do not ask that each row of $G^{*}$ is a $l^{2}-$sequence, an
$\omega\times\omega$ matrix may have a “left inverse” in the classical sense.

######  Example 2.4.

Let $F$ be the matrix

$\begin{bmatrix}1&1&0&0&\cdots\\\ 0&-1&1&0&\cdots\\\ 0&0&-1&1&\cdots\\\
0&0&0&-1&\cdots\\\ \vdots&\vdots&\vdots&\vdots&\ddots\end{bmatrix},$

and $G^{*}$ be the matrix

$\begin{bmatrix}1&1&1&1&\cdots\\\ 0&-1&-1&-1&\cdots\\\ 0&0&-1&-1&\cdots\\\
0&0&0&-1&\cdots\\\ \vdots&\vdots&\vdots&\vdots&\ddots\end{bmatrix}.$

It is trivial to check $G^{*}F=FG^{*}=I$. Then by above lemma 2.1 we know $F$
is not a Schauder matrix since the rows of its inverse matrix are not
$l^{2}-$sequence. Moreover, if we denote by $g_{n}$ the $n-th$ column vector,
then the sequence $\xi=\\{g_{n}\\}_{n=1}^{\infty}$ is a complete minimal
sequence(see [6], p24 and p50 for definitions). It is easy to check that $\xi$
is complete since the $l^{2}$-sequence
$h_{n}=\\{h_{n}(j)\\}_{j=1}^{\infty},h_{n}(j)=\delta_{nj}$ is in its range; On
the other hand, the row vector sequence $\\{f_{k}\\}_{k=1}^{\infty}$ satisfies
$(g_{n},f_{k})=\delta_{kn}$ which implies $g_{n}\notin\vee_{m\neq n}g_{m}$(or
in notations of singer, we have
$g_{n}\notin[g_{1},\cdots,g_{n-1},g_{n+1},\cdots]$) by the fact $\vee_{m\neq
n}g_{m}=\ker\varphi_{k}$ in which $\varphi_{k}(x)=(x,f_{k})$ is a bounded
functional by Riesz’s theorem. Therefore $\xi$ is an example which is complete
and minimal sequence but not a basis sequence.

By above lemma 2.1 and 2.2, we have

######  Theorem 2.5.

An $\omega\times\omega$ matrix $F$ is a Schauder matrix if and only if it
satisfies property 1, 2 and 3.

For a Schauder matrix $F$, the column vector sequence
$\\{g_{n}\\}_{n=1}^{\infty}$ of $G$ defined in above lemmas is also a Schauder
basis which is called the biorthogonal basis to the basis
$\\{f_{k}\\}_{k=1}^{\infty}$(cf [2], pp23-29, [6] pp23-25).

The projection $FP_{n}G^{*}$ is just the $n-$th “natural projection” so called
in [4](p354). It is also the $n-$th partial sum operator so called in
[6](definition 4.4, p25). Now we can translate theorem 4.1.15 and corollary
4.1.17 in [4] into the following

######  Proposition 2.6.

If $F$ is a Schauder matrix, then $M=\sup_{n}\\{||FP_{n}G^{*}||\\}$ is a
finite const.

The const $M$ is called the basis const for the basis
$\\{f_{n}\\}_{n=1}^{\infty}$.

Assume that $\psi=\\{f_{n}\\}_{n=1}^{\infty}$ is a basis. For a subset
$\Delta$ of $\mathbb{N}$, denote by $P_{\Delta}$ the diagonal matrix defined
as $P_{\Delta}(nn)=1$ for $n\in\Delta$ and $P_{\Delta}(nn)=0$ for
$n\notin\Delta$. The projection $Q_{\Delta}=F_{\psi}P_{\Delta}G_{\psi}^{*}$
defined in above lemmas is called a natural projection(see, definition 4.2.24,
[4], p378). In fact for a vector $x=\sum_{n=1}^{\infty}x_{n}f_{n}$, it is
trivial to check $Q_{\Delta}x=\sum_{n\in\Delta}x_{n}f_{n}$. Then we have a
same result for the  unconditional basis const(cf, definition4.2.28, [4],
p379):

######  Proposition 2.7.

If $F_{\psi}$ is a Schauder matrix, then the unconditional basis const of the
basis $\psi$ is
$M_{ub}=\sup_{\Delta\subseteq\mathbb{N}}\\{||F_{\psi}P_{\Delta}G_{\psi}^{*}||\\}$.

In virtue of the proposition 4.2.29 and theorem 4.2.32 in the book [4], we
have

######  Proposition 2.8.

For a Schauder basis $\psi$, it is an unconditional basis if and only if
$\sup_{\Delta\subseteq\mathbb{N}}\\{||F_{\psi}P_{\Delta}G_{\psi}^{*}||\\}<\infty$.

Following the notations in lemma 2.1, as a direct corollary of lemma 2.1 and
theorem 6 in [2](p28), we have

######  Proposition 2.9.

$F$ is a Schauder matrix if and only if the adjoint matrix (conjugate
transpose) $G$ of its left inverse $G^{*}$ is a Schauder matrix.

As well known that a sequence of operators $T_{n}$ converges to an operator
$T$ in SOT dose not imply $T_{n}^{*}$ converging to $T$ in SOT, so the above
proposition is not trivial.

######  Corollary 2.10.

$M=\sup_{n}\\{||FP_{n}G^{*}||\\}<\infty$ if and only if
$M^{{}^{\prime}}=\sup_{n}\\{||GP_{n}F^{*}||\\}$ $<\infty$.

### 2.2.

From the definition of the Schauder matrix $F_{\psi}$, basic properties of
Schauder matrix have natural relations to the Schauder basis $\psi$. This
understanding lead us to the following definition.

######  Definition 2.11.

A matrix $F$ is called an unconditional, conditional, Riesz, normalized or
quasinormal respectively if and only if the sequence of its column vectors
comprise an unconditional, conditional, Riesz, normalized or quasinormal
basis. Two Schauder matrices $F_{\psi},F_{\varphi}$ are called equivalent if
and only if the corresponding bases $\psi$ and $\varphi$ are equivalent.

Here we use the term quasinormal instead of “bounded” to avoid ambiguity(cf
[5] p476, [6] p21). Arsove use the word “similar” in the same meaning as the
word “equivalent”(cf, [10] p19, [4]p387).

Denote by $\pi_{\infty}$ the set of all permutations of $\mathbb{N}$(see [6],
p361). Denote by $U_{\pi}$ both the unitary operator which maps $e_{\pi(n)}$
to $e_{n}$ and the corresponding matrix under the ONB
$\\{e_{n}\\}_{n=1}^{\infty}$.

######  Theorem 2.12.

Assume that $F$ is a Schauder matrix and $G^{*}$ is its inverse matrix. We
have
1\. For each invertible matrix $X$, $XF$ is also a Schauder matrix. Moreover,
$XF$ is unconditional(conditional) if and only if $F$ is
unconditional(conditional);
2\. For each diagonal matrix $D=diag(\alpha_{1},\alpha_{2},\cdots)$ in which
each diagonal element $\alpha_{k}$ is nonzero, $FD$ is also a Schauder matrix.
Moreover, $FD$ is unconditional(conditional) if and only if $F$ is
unconditional(conditional);
3\. For a unconditional matrix $F$, $FU$ is also a unconditional matrix for
$U\in\pi_{\infty}$;
4\. Two Schauder matrix $F$ and $F^{{}^{\prime}}$ are equivalent if and only
if there is a invertible matrix $X$ such that $XF=F^{{}^{\prime}}$.

###### Proof.

Property 1, 2, 3 and 4 are basic facts about basis just in a matrix language.
Their counterparts are proposition 4.1.8, 4.2.14, 4.1.5, 4.2.12, and corollary
4.2.34 in [4], Theorem 1 in [10]. Some of those facts are easy to check by our
lemma 2.1. As an example, we shall prove property 1. Let $F^{{}^{\prime}}=XF$,
then clearly $G^{*^{\prime}}=GX^{-1}$ is its inverse matrix. Both properties 1
and 2 in lemma 2.1 hold immediately. To verify property 3, we know that
$FP_{n}G^{*}$ converges to $I$ in SOT if and only if $XFP_{n}G^{*}X^{-1}$
converges to $I$ in SOT. Also we have

$||XFPG^{*}X^{-1}||\leq||X||\cdot||X^{-1}||\cdot||FPG^{*}||$

for any natural projection $P$, which implies the last part of property 1(cf,
[4] theorem 4.2.32). ∎

### 2.3.

Now we turn to study the basic properties of Schauder operators. Recall that a
Schauder operator $T$ is an operator mapping some ONB into a Schauder basis.
In his paper [5], Olevskii call an operator to be generating if and only if it
maps some ONB into a quasinormal conditional basis. Hence our definition of
Schauder operator is a generalization of Olevskii’s one.

######  Theorem 2.13.

Following conditions are equivalent:
1\. $T$ is a Schauder operator;
2\. $T$ maps some ONB $\\{e_{n}\\}_{n=1}^{\infty}$ into a basis;
3\. $T$ has a polar decomposition $T=UA$ in which $A$ is a Schauder operator;
4\. Assume that $T$ has a matrix representation $F$ under a fixed ONB
$\\{e_{n}\\}_{n=1}^{\infty}$. There is some unitary matrix $U$ such that $FU$
is a Schauder matrix.

###### Proof.

$2\Rightarrow 1$. The $k-$th column of the matrix of $T$ under the ONB
$\\{e_{n}\\}_{n=1}^{\infty}$ is just the $l^{2}-$coefficients of $Te_{k}$.

$1\Rightarrow 3$. Assume that $\\{f_{n}\\}_{n=1}^{\infty}$ is a basis in which
$f_{n}$ is the $n-$th column of the matrix $F$ of $T$ under some ONB. Then if
we denote the matrix of $U$ and $A$ also by the same notations, we have
$UA=F$. Property 1 of lemma 2.12 tell us $U^{*}F=A$ is also a Schauder matrix.

$3\Rightarrow 4$. Assume that $\\{g_{n}\\}_{n=1}^{\infty}$ is an ONB such that
the matrix of $A$ under it is a Schauder matrix. Then the operator $U$ defined
as $Ue_{n}=g_{n}$ is a unitary operator and the $n-$th column of its matrix
under the ONB $\\{e_{n}\\}_{n=1}^{\infty}$ is just the $l^{2}-$coefficients of
$g_{n}$. Hence we have $AU$ is a Schauder matrix.

$4\Rightarrow 1$. The column vector sequence of the unitary matrix $U$ is an
ONB. The matrix of $T$ under this ONB is just $U^{*}FU$. Property 1 of lemma
2.12 shows that $U^{*}FU$ is a Schauder matrix since $FU$ is a Schauder matrix
itself. ∎

The equivalence $1\Leftrightarrow 3$ had been used in proof of the theorem
$1^{{}^{\prime}}$ of [5], although Olevskii had not given an explanation.

######  Proposition 2.14.

A Schauder operator $T$ must be injective and has a dense range in
$\mathcal{H}$.

###### Proof.

$T$ must be injective since the representation of $0$ is unique. For a basis
$\\{f_{n}\\}_{n=1}^{\infty}$, the finite linear combination of
$\\{f_{n}\\}_{n=1}^{\infty}$ is dense in the Hilbert space $\mathcal{H}$.
Therefore the range of $T$ must be dense in $\mathcal{H}$. ∎

### 2.4.

If $T$ is a Schauder operator, does for each ONB sequence
$\\{e_{n}\\}_{n=1}^{\infty}$ the vector sequence $\\{Te_{n}\\}_{n=1}^{\infty}$
always be a basis? In this subsection, we shall show that the answer is
negative in general and it is true only in the case that $T$ is an invertible
operator.

######  Lemma 2.15.

Assume that $A$ is a positive operator satisfying
$\sigma(A)\subseteq[\lambda_{1},\lambda_{2}]$ and
$\lambda_{1},\lambda_{2}\in\sigma(A)$ for some $\lambda_{1}>0$. Then for any
const $\varepsilon>0$ small enough, there is a rank 1 projection $P$ such that
$\frac{1}{2\sqrt{2}}\frac{\lambda_{2}}{\lambda_{1}}-\varepsilon<||APA^{-1}||$.

###### Proof.

Let $e_{1},e_{2}$ be two normalized vectors in $\mathcal{H}$ such that

$e_{1}\in E_{[\lambda_{1},\lambda_{1}+\delta]},e_{2}\in
E_{[\lambda_{2}-\delta,\lambda_{2}]}.$

in which $E_{[\lambda_{1},\lambda_{1}+\delta]}$ and
$E_{[\lambda_{2}-\delta,\lambda_{2}]}$ is the spectral projection of $A$ on
the interval $[\lambda_{1},\lambda_{1}+\delta]$ and
$[\lambda_{2}-\delta,\lambda_{2}]$ respectively(cf, [7], pp269-272). Then for
$\delta<\frac{\lambda_{2}-\lambda_{1}}{2}$, we have $(e_{1},e_{2})=0$ and

$\lambda_{1}\leq||Ae_{1}||\leq\lambda_{1}+\delta,\lambda_{2}-\delta\leq||Ae_{2}||\leq\lambda_{2}.$

Consider the vector $e=\frac{1}{\sqrt{2}}e_{1}+\frac{1}{\sqrt{2}}e_{2}$ and
the operator $P=e\otimes e$ defined as: $Px=(x,e)e.$ It is trivial to check
that $P$ is a rank 1 orthogonal projection. Now we have
$APA^{-1}(x)=(A^{-1}x,e)Ae,$ hence $||APA^{-1}||=\sup_{||x||=1}||APA^{-1}x||$.
Then we have

$\begin{array}[]{rl}(A^{-1}e,e)=&\frac{1}{\sqrt{2}}(A^{-1}e_{1},e)+\frac{1}{\sqrt{2}}(A^{-1}e_{2},e)\\\
=&\frac{1}{2}\\{(A^{-1}e_{1},e_{1})+(A^{-1}e_{2},e_{2})\\}\\\
\geq&\frac{1}{2}\\{\frac{1}{\lambda_{1}+\delta}+\frac{1}{\lambda_{2}}\\}\end{array}$

and

$||Ae||^{2}\geq\frac{1}{2}\lambda_{1}^{2}+\frac{1}{2}(\lambda_{2}-\delta)^{2}.$

Therefore the following inequality holds:

$\begin{array}[]{rl}||APA^{-1}e||\geq&\frac{1}{2}\\{\frac{1}{\lambda_{1}+\delta}+\frac{1}{\lambda_{2}}\\}\sqrt{\frac{1}{2}\lambda_{1}^{2}+\frac{1}{2}(\lambda_{2}-\delta)^{2}}\\\
\geq&\frac{1}{2\sqrt{2}}\frac{\lambda_{2}-\delta}{\lambda_{1}+\delta}.\end{array}$

Let $\varepsilon$ be a const satisfying $\varepsilon<\frac{1}{2\sqrt{2}}$.
Hence for the positive number
$\delta<\frac{2\sqrt{2}\lambda_{1}^{2}\varepsilon}{(1-2\sqrt{2}\varepsilon)\lambda_{1}+\lambda_{2}}$
the required inequality holds. ∎

######  Theorem 2.16.

If an operator $A$ maps every ONB sequence into a basis, then $A$ must be an
invertible operator.

###### Proof.

A direct result of 2.12 is that if an operator $A$ maps every ONB into a basis
then it maps each ONB into a unconditional basis. By virtue of theorem 2.13,
we can assume that $T$ is a positive operator. We need to show that
$0\notin\sigma(A)$. Firstly, we have $0\notin\sigma_{p}(A)$ by above
proposition 2.14 since $A$ is a Schauder operator. If $0\in\sigma(p)$ then $0$
must be an accumulation point of $\sigma(T)$. Hence we can choose a sequence
$\\{\lambda_{k}\\}_{k=1}^{\infty}$ such that:
1\. $\\{\lambda_{k}\\}_{k=1}^{\infty}\subseteq\sigma(A)$; and
2\. $\lambda_{k+1}<\lambda_{k}$ and
$\frac{\lambda_{2n}}{\lambda_{2n-1}}<\frac{1}{n+1}$.
Denote by $I_{0}=\sigma(A)-\cup_{n=1}^{\infty}[\lambda_{2n},\lambda_{2n-1}]$
and $A_{0}=AE_{I_{0}}$. Let $A_{n}=AE_{[\lambda_{2n},\lambda_{2n-1}]}$, then
we have $A=A_{0}\oplus A_{1}\oplus A_{2}\oplus A_{3}\cdots$. And each operator
$A_{n}$ is an invertible positive operator for $n\geq 1$. Now by above lemma
2.15, we can choose a vector $e_{1}^{(n)}\in
RanE_{[\lambda_{2n},\lambda_{2n-1}]}$ such that the projection
$P^{(n)}_{1}=e_{1}^{(n)}\otimes e_{1}^{(n)}$ satisfying

$AP^{(n)}_{1}A^{-1}=A_{n}P^{(n)}_{1}A_{n}^{-1}>n$

for each n. Here we use the fact

$E_{[\lambda_{2n},\lambda_{2n-1}]}P^{(n)}_{1}=P^{(n)}_{1}E_{[\lambda_{2n},\lambda_{2n-1}]}=P^{(n)}_{1}.$

Now for each subspace $RanE_{[\lambda_{2n},\lambda_{2n-1}]}$ we choose an ONB
$\\{f_{k}^{(n)}\\}_{k=1}^{\alpha_{k}}$ such that $e^{(n)}_{1}=f^{(n)}_{1}$.
Moreover, choose an ONB $\\{e^{(0)}_{k}\\}_{k=1}^{\alpha_{0}}$ of the subspace
$RanE_{I_{0}}$. Here $\alpha_{k}$ is a finite number or the countable cardinal
which is equal to the dimension of the subspace
$RanE_{[\lambda_{2n},\lambda_{2n-1}]}$ and $RanE_{I_{0}}$ respectively.
Clearly the set $\\{f^{(n)}_{k};n=0,1,2,\cdots\hbox{ and
}k=1,2,\cdots,\alpha_{k}\\}$ is an ONB for $\mathcal{H}$ itself. It is a
countable set and each its arrangement $\psi$ give an ONB sequence of
$\mathcal{H}$. In more details, denote by
$\Delta=\\{(n,k);n=0,1,2,\cdots\hbox{ and }k=1,2,\cdots,\alpha_{k}\\}$. For
any bijection $\sigma:\Delta\rightarrow\mathbb{N}$, define
$g_{n}=f_{t}^{(s)},(s,t)=\sigma^{-1}(n)$. Then
$\psi_{\sigma}=\\{g_{n}\\}_{n=1}^{\infty}$ is an ONB sequence.

######  Claim 2.17.

For each ONB sequence $\psi_{\sigma}$, $\\{Ag_{n}\\}_{n=1}^{\infty}$ is not a
basis.

We have shown that if $\\{Ag_{n}\\}_{n=1}^{\infty}$ is a basis it must be a
unconditional one. So it is enough to show that it is not a unconditional
basis, which can be verified by its unconditional const. Assume that the claim
is not true, that is, $\\{Ag_{n}\\}_{n=1}^{\infty}$ is a basis. It is trivial
to check that $A_{n}P^{(n)}_{1}A_{n}^{-1}$ is a natural projection
corresponding to the basis $\\{Ag_{n}\\}$. In fact, we have

$A_{n}P^{(n)}_{1}A_{n}^{-1}=P_{\sigma(n,1)}-P_{\sigma(n,1)-1}.$

Here we denote by $P_{n}$ the $n-th$ partial sum operator so called in the
book [6]. But now we have $||A_{n}P^{(n)}_{1}A_{n}^{-1}||\rightarrow\infty$
which counters to the fact that a unconditional basis must have a finite
unconditional const (cf, [4], corollary4.2.26). ∎

######  Corollary 2.18.

If an operator $T$ is not invertible, then there is some ONB
$\\{e_{n}\\}_{n=1}^{\infty}$ such that the sequence
$\\{Te_{n}\\}_{n=1}^{\infty}$ is not a basis.

By the theorem 1 of [5], a generating operator never be invertible. Hence we
have

######  Corollary 2.19.

For a generating operator $T$, there is some ONB $\\{e_{n}\\}_{n=1}^{\infty}$
such that the sequence $\\{Te_{n}\\}_{n=1}^{\infty}$ is not a basis.

Both the English translation and the review(MR0318848) of the paper [5] by A.
M. Oleskii make a pity clerical mistake:

Review(MR0318848):“The author obtains a spectral characterization for the
linear operators that transform $\mathbf{every}$ complete orthonormal system
into a conditional basis in a Hilbert space.”

The English translation: “Definition. A bounded noninvertible linear operator
$T:\mathcal{H}\rightarrow\mathcal{H}$ is said to be generating if it maps
$\mathbf{every}$ orthonormal basis $\varphi$ into a quasinormed basis $\psi$.”

The word “every” should be “some” in both of them. Note that in the proof of
the theorem 1 ([5]), Olevskii had shown that an operator never can maps every
ONB into a conditional basis. Even the theorem 1 of [5] itself shows it, but
need a little operator theory discussion.

Since in the Hilbert space $\mathcal{H}$ all quasinormal unconditional bases
are equivalent(cf, Theorem 18.1, [6], p529) and in addition with theorem 2.12,
we have

######  Proposition 2.20.

An $\omega\times\omega$ matrix $F$ is a Riesz matrix if and only if it
represents an invertible operator.

Above result also can be obtained directly form theorem 2 of the paper [1].

######  Corollary 2.21.

An operator $T$ is invertible if and only if there is some ONB such that the
matrix $F$ under this ONB of $T$ is a Riesz matrix.

######  Corollary 2.22.

For an invertible operator $T$, its matrix always be a Riesz matrix under any
ONB.

### 2.5.

Conditional and unconditional bases have very different behaviors. On the
other side, properties of operators given by Schauder matrices are strongly
dependent on the related bases. Both the theorem 1 of the paper [5] and the
behaviors of Riesz matrix(cf, proposition 2.20) support this observation. In
this subsection, we give a same classification of operators dependent on their
matrix representation(Or equivalently, on their actions on ONBs). And then we
give some more remarks on Olevskii’s paper.

######  Definition 2.23.

A Schauder operator $T$ will be called a conditional operator if and only if
there is some ONB $\\{e_{n}\\}_{n=1}^{\infty}$ such the column vector sequence
of its matrix representation $F$ of $T$ under the ONB comprise a conditional
basis. Otherwise, $T$ will be called a unconditional operator.

By the theorem 2.13, we have

######  Corollary 2.24.

A Schauder operator $T$ is conditional if and only if it maps some ONB
$\\{e_{n}\\}_{n=1}^{\infty}$ into a conditional basis
$\\{Te_{n}\\}_{n=1}^{\infty}$.

For convenience, we correct the error appearing in the translation and rewrite
Olevskii’s definition as follows:

######  Definition 2.25.

A bounded operator $T\in\mathcal{L}(\mathcal{H})$ is said to be generating if
and only if it maps some ONB into a quasinormal conditional basis.

Above definition modifies slightly from the original form on the Olevskii’s
paper. We write down the original one to compare them in details:

######  Definition 2.26.

([5], p476) A bounded non-invertible operator
$T:\mathcal{H}\rightarrow\mathcal{H}$ is said to be generating if and only if
it maps some ONB into a quasinormal basis.

######  Proposition 2.27.

Above two definitions are equivalent.

###### Proof.

If a bounded operator $T\in\mathcal{L}(\mathcal{H})$ maps some ONB into a
quasinormal conditional basis, then it must be non-invertible since an
invertible operator maps each ONB into a Riesz basis(hence a unconditional
basis) by proposition 2.20; On the other side, If a bounded non-invertible
operator $T:\mathcal{H}\rightarrow\mathcal{H}$ maps some ONB into a
quasinormal basis. Then the quasinormal basis must be a conditional one
otherwise $T$ must be invertible again by proposition 2.20. ∎

######  Corollary 2.28.

A generating operator is a conditional operator; An invertible operator is a
unconditional operator.

##  3\.  A Criterion for Operators to be Conditional

### 3.1.

Question:  Is $K=diag\\{1,\frac{1}{2},\frac{1}{3},\cdots\\}$ a conditional
operator?

From Olevskii’s result, we can not obtain the confirm answer. In this section,
we will improve the Olevskii’s technology and gain a confirm answer.

First, let us recall some notations in the line of Olevskii.

Let $A_{k}=\left(\begin{array}[]{c}a_{ij}\\\ \end{array}\right)\in
M_{2^{k}}(\mathbb{C})$ (where $1\leq i,j\leq 2^{k}$) be defined as follows:
$a_{i1}=2^{-\frac{k}{2}},1\leq i\leq 2^{k}$; and if $j=2^{s}+v(1\leq v\leq
2^{s})$, then

$\displaystyle a_{ij}$
$\displaystyle=\left\\{\begin{array}[]{ll}2^{\frac{s-k}{2}},\hskip
28.45274pt(v-1)2^{k-s}<i\leq(2v-1)2^{k-s-1},\\\\[5.69054pt]
-2^{\frac{s-k}{2}},\hskip 28.45274pt(2v-1)2^{k-s-1}<i\leq v2^{k-s}.\\\
\end{array}\right.$

For $\alpha,\frac{1}{\sqrt{2}}<\alpha<1$, let $T_{(k,\alpha)}\in
M_{2^{k}}(\mathbb{C})$ be defined as follows:

$T_{(k,\alpha)}=\begin{bmatrix}\begin{bmatrix}\alpha^{k}&\\\ &\alpha^{k}\\\
\end{bmatrix}\\\ &\begin{bmatrix}\alpha^{k-1}\\\ &\alpha^{k-1}\\\
\end{bmatrix}&\\\ &&\ddots&\\\ &&&\begin{bmatrix}\alpha&\\\ &\ddots\\\
&&\alpha\end{bmatrix}_{2^{k-1}\times 2^{k-1}}\\\ \end{bmatrix}.$

In this section, we will show that if the positive operator $T$ does not admit
the eigenvalue zero and $\sigma(T)$ has a decreasing sequence
$\\{\lambda_{n},n=1,2,\ldots\\}$ which converges to zero and

$\lim\limits_{n\rightarrow\infty}\frac{\lambda_{n}}{\lambda_{n+1}}=1,$

then $T$ must be a conditional operator. Thus the compact operator
$K=diag\\{1,\frac{1}{2},\frac{1}{3}$, $\cdots\\}$ is a conditional operator.

### 3.2.

Now, we give a key lemma.

######  Lemma 3.1.

Let $T$ be a diagonal operator with entries
$\\{\lambda_{1},\lambda_{2},\lambda_{3},\ldots\\}$ under the ONB
$\\{e_{k}\\}_{k=1}^{\infty}$, where $\lambda_{n}>0$. Given $\alpha$,
$\frac{1}{\sqrt{2}}<\alpha<1$. If for each $k\geq 1$, there exist positive
numbers $c_{k}\leq d_{k}$, such that

a) $sup_{k}\frac{d_{k}}{c_{k}}<\infty$,

b) there exists subset
$\triangle_{k}=\\{n^{k}_{1},n^{k}_{2},\cdots,n^{k}_{2^{k}}\\}$ of $\mathbb{N}$
such that
$c_{k}\leq\frac{\alpha^{k}}{\lambda_{n^{k}_{2^{k}-1}}},\frac{\alpha^{k}}{\lambda_{n^{k}_{2^{k}}}}\leq
d_{k}$, and $c_{k}\leq\frac{\alpha^{j}}{\lambda_{n^{k}_{i}}}\leq d_{k}$ when
$1\leq j\leq k-1,~{}2^{k}(1-\frac{1}{2^{j-1}})+1\leq i\leq
2^{k}(1-\frac{1}{2^{j}}),$

c) $max{\triangle_{k}}<min{\triangle_{k^{\prime}}}$ when $k<k^{\prime}$,

then $T$ is a conditional operator.

###### Proof.

In this proof, we shall identify the operators and the $\omega\times\omega$
matrix representation of the operators under ONB $\\{e_{k}\\}_{k=1}^{\infty}$.

We rearrange $n^{k}_{1},n^{k}_{2},\cdots,n^{k}_{2^{k}}$ into a increasing
sequence and denote it by $m^{k}_{1},m^{k}_{2}$, $\cdots,m^{k}_{2^{k}}$
($m^{k}_{1}<m^{k}_{2}<\cdots<m^{k}_{2^{k}}$). Let $m^{0}_{1}=1$ and
$\mathcal{H}_{k}=span\\{e_{m^{k}_{1}},e_{m^{k}_{1}+1}$,
$\ldots,e_{m^{k+1}_{1}-1}\\}$ for $k\geq 0$, then since
$max{\triangle_{k}}<min{\triangle_{k^{\prime}}}$ when $k<k^{\prime}$, we know
$\mathcal{H}_{k}\cap\mathcal{H}_{k^{\prime}}=(0)$ when $k\neq k^{\prime}$ and
$\oplus_{k\geq 0}\mathcal{H}_{k}=\mathcal{H}$. Moreover,
$\\{\lambda_{n^{k}_{1}},\lambda_{n^{k}_{2}},\ldots,\lambda_{n^{k}_{2^{k}}}\\}\subseteq\\{\lambda_{m^{k}_{1}},\lambda_{m^{k}_{1}+1},\ldots,$
$\lambda_{m^{k+1}_{1}-1}\\}$ for any $k\geq 1$.

Let $T_{k}\in\mathcal{L}(\mathcal{H}_{k})$ the k-th block of $T$ on
$\mathcal{H}_{k}$, i.e.

$T_{k}=\begin{bmatrix}\lambda_{m^{k}_{1}}&&&\\\ &\lambda_{m^{k}_{1}+1}&&\\\
&&\ddots\\\ &&&\lambda_{m^{k+1}_{1}-1}\\\
\end{bmatrix}\begin{matrix}e_{m^{k}_{1}}\\\ e_{m^{k}_{1}+1}\\\ \vdots\\\
e_{m^{k+1}_{1}-1}\end{matrix},$

then $\oplus_{k\geq 0}T_{k}=T$. Denote $\widetilde{T}_{0}=T_{0}$. For $k\geq
1$, let

$\widetilde{T}_{k}=\begin{bmatrix}\lambda_{n^{k}_{2^{k}}}&&&\\\
&\lambda_{n^{k}_{2^{k}-1}}&&\\\ &&\ddots\\\ &&&\lambda_{n^{k}_{1}}&\\\
&&&&S_{k}\\\ \end{bmatrix}\begin{matrix}e_{m^{k}_{1}}\\\ e_{m^{k}_{1}+1}\\\
\vdots\\\ e_{m^{k}_{1}+2^{k}-1}\\\ \widetilde{\mathcal{H}}_{k}\end{matrix},$

where
$\widetilde{\mathcal{H}}_{k}=\bigvee\\{e_{m^{k}_{1}+2^{k}},\ldots,e_{m^{k+1}_{1}-1}\\}$
and $S_{k}$ is a diagonal operator with entries
$\\{\lambda_{m^{k}_{1}},\lambda_{m^{k}_{1}+1},\ldots,$
$\lambda_{m^{k+1}_{1}-1}\\}\backslash\\{\lambda_{n^{k}_{1}},\lambda_{n^{k}_{2}}$,
$\ldots,\lambda_{n^{k}_{2^{k}}}\\}$. It is easy to see that the entries of
$\widetilde{T}_{k}$ are just a rearrangement of entries of $T_{k}$ for $k\geq
1$.

We will prove $\widetilde{T}\triangleq\oplus_{k\geq 0}\widetilde{T}_{k}$ is a
conditional operator and then show $T$ is a conditional operator.

Let $X_{0}=I\in\mathcal{L}(\mathcal{H}_{0})$. For $k\geq 1$, let

$X_{k}=\begin{bmatrix}c_{k}\cdot\begin{bmatrix}\frac{\lambda_{n^{k}_{2^{k}}}}{\alpha^{k}}&&\\\
&\frac{\lambda_{n^{k}_{2^{k}-1}}}{\alpha^{k}}\\\ &&\ddots&\\\
&&&\frac{\lambda_{n^{k}_{i}}}{\alpha^{j}}\\\ &&&&\ddots\\\
&&&&&\frac{\lambda_{n^{k}_{1}}}{\alpha}\\\ \end{bmatrix}&\\\ &I\\\
\end{bmatrix}\in\mathcal{L}(\mathcal{H}_{k}),$

since

$Sup_{k}max\\{c_{k}\frac{\lambda_{n^{k}_{2^{k}}}}{\alpha^{k}},\ldots,c_{k}\frac{\lambda_{n^{k}_{1}}}{\alpha},c_{k}^{-1}\frac{\alpha^{k}}{\lambda_{n^{k}_{2^{k}}}},\ldots,c_{k}^{-1}\frac{\alpha}{\lambda_{n^{k}_{1}}}\\}\leq
Sup_{k}max\\{1,\frac{d_{k}}{c_{k}}\\}<\infty,$

we have $X\triangleq\oplus_{k\geq 0}X_{k}$ is an invertible operator.

Moreover for $k\geq 1$,

$\widetilde{T}_{k}=X_{k}\cdot\begin{bmatrix}T_{(k,\alpha)}c_{k}^{-1}&\\\
&S_{k}\\\ \end{bmatrix},$

so

$\widetilde{T}=\oplus_{k\geq 0}\widetilde{T}_{k}=X\cdot\oplus_{k\geq
0}\begin{bmatrix}T_{(k,\alpha)}c_{k}^{-1}&\\\ &S_{k}\\\ \end{bmatrix},$

where we denote $\begin{bmatrix}T_{(k,\alpha)}c_{k}^{-1}&\\\ &S_{k}\\\
\end{bmatrix}$ by $\widetilde{T}_{0}$ when $k=0$.

Let

$U=\oplus_{k\geq 0}\begin{bmatrix}A_{k}^{*}&\\\ &I\\\ \end{bmatrix},$

where we denote $\begin{bmatrix}A_{k}^{*}&\\\ &I\\\
\end{bmatrix}=I\in\mathcal{L}(\mathcal{H}_{0})$ when $k=0$, then it is an
unitary operator and

$\displaystyle\widetilde{T}U$ $\displaystyle=$ $\displaystyle
X\cdot\oplus_{k\geq 0}\begin{bmatrix}T_{(k,\alpha)}c_{k}^{-1}&\\\ &S_{k}\\\
\end{bmatrix}\cdot\oplus_{k\geq 0}\begin{bmatrix}A_{k}^{*}&\\\ &I\\\
\end{bmatrix}$ $\displaystyle=$ $\displaystyle X\cdot\oplus_{k\geq
0}\begin{bmatrix}T_{(k,\alpha)}A_{k}^{*}c_{k}^{-1}&\\\ &S_{k}\\\
\end{bmatrix}$ $\displaystyle=$ $\displaystyle X\cdot\oplus_{k\geq
0}\begin{bmatrix}T_{(k,\alpha)}A_{k}^{*}&\\\ &S_{k}\\\
\end{bmatrix}\cdot\oplus_{k\geq 0}\begin{bmatrix}c_{k}^{-1}I&\\\ &I\\\
\end{bmatrix}.$

To show $\widetilde{T}$ is conditional, from theorem 2.12, it suffices to show
that

$F\triangleq\oplus_{k\geq 0}\begin{bmatrix}T_{(k,\alpha)}A_{k}^{*}&\\\
&S_{k}\\\ \end{bmatrix}$

is a conditional matrix.

We will deal with it by theorem 2.5 and proposition 2.8. First, one can easily
see that $F$ has an unique left inverse matrix

$G^{*}=\oplus_{k\geq 0}\begin{bmatrix}A_{k}T_{(k,\alpha)}^{-1}&\\\
&S_{k}^{-1}\\\ \end{bmatrix}$

where each row is a $l^{2}-$ sequence.

Second, $Q_{n}=FP_{n}G^{*}$ are obviously projections. Let

$\displaystyle\Lambda_{1}=\\{m^{k}_{1},m^{k}_{1}+1,\ldots,m^{k}_{1}+2^{k}-1;~{}k\geq
1\\}\subseteq\mathbb{N},$
$\displaystyle\Lambda_{2}=\\{m^{k}_{1}+2^{k},m^{k}_{1}+2^{k}+1,\ldots,m^{k+1}_{1}-1;~{}k\geq
1\\}\subseteq\mathbb{N}.$

For any $x\in\mathcal{H}$, we have

$x=\sum\limits_{j=1}^{\infty}x_{j}e_{j}=\sum\limits_{j\in\Lambda_{1}}x_{j}e_{j}+\sum\limits_{j\in\Lambda_{2}}x_{j}e_{j},$

and

$\displaystyle FP_{n}G^{*}(x)$
$\displaystyle=FP_{n}G^{*}(\sum\limits_{j\in\Lambda_{1}}x_{j}e_{j}+\sum\limits_{j\in\Lambda_{2}}x_{j}e_{j})$
$\displaystyle=(\oplus_{k\geq
0}T_{(k,\alpha)}A_{k}^{*})P^{(1)}_{n}(\oplus_{k\geq
0}A_{k}T_{(k,\alpha)}^{-1})(\sum\limits_{j\in\Lambda_{1}}x_{j}e_{j})+P^{(2)}_{n}(\sum\limits_{j\in\Lambda_{2}}x_{j}e_{j}),$

where $\oplus_{k\geq 0}T_{(k,\alpha)}A_{k}^{*}$ and $P^{(1)}_{n}$ are the
operators on $\mathcal{H}^{(1)}=\bigvee_{j\in\Lambda_{1}}\\{e_{j}\\}$,
$P^{(1)}_{n}$ converges to $I$ in the strong operator topology; $P^{(2)}_{n}$
is the operator on $\mathcal{H}^{(2)}=\bigvee_{j\in\Lambda_{2}}\\{e_{j}\\}$
and also converges to $I$ in the strong operator topology.

It follows from the result of Olevskii that $\oplus_{k\geq
0}T_{(k,\alpha)}A_{k}^{*}$ is quasinormal conditional matrix. Then from
theorem 2.5, we have

$\lim\limits_{n\rightarrow\infty}(\oplus_{k\geq
0}T_{(k,\alpha)}A_{k}^{*})P^{(1)}_{n}(\oplus_{k\geq
0}A_{k}T_{(k,\alpha)}^{-1})(\sum\limits_{j\in\Lambda_{1}}x_{j}e_{j})=\sum\limits_{j\in\Lambda_{1}}x_{j}e_{j}.$

Thus $FP_{n}G^{*}(x)$ converges to $x$ as $n\rightarrow\infty$ and $F$ is a
Schauder matrix.

Moreover, since the unconditional basis const of $\oplus_{k\geq
0}T_{(k,\alpha)}A_{k}$ is smaller than the unconditional basis const of $F$
and the unconditional basis const of $\oplus_{k\geq 0}T_{(k,\alpha)}A_{k}$ is
infinity, we have that the unconditional basis const of $F$ is infinity. Thus
from proposition 2.8, we know that $F$ is a conditional matrix and
$\widetilde{T}U$ is a conditional matrix.

Since the entries of $\widetilde{T}$ is just a rearrangement of $T$, one can
easily find an unitary matrix (operator) $\widetilde{U}$ such that
$\widetilde{U}\widetilde{T}\widetilde{U}^{*}=T$, it follows that
$\widetilde{U}^{*}T\widetilde{U}U$ is a conditional matrix. Again from theorem
2.12, $T\widetilde{U}U$ is a conditional matrix. Thus $T$ is a conditional
operator, since it maps orthonormal basis $\\{(\widetilde{U}U)e_{1}$,$\ldots$,
$(\widetilde{U}U)e_{n}$,$\ldots\\}$ into a conditional basis. ∎

Now, we come to the main results.

######  Theorem 3.2.

Let $T\geq 0$ belong to $\mathcal{L}(\mathcal{H})$ which does not admit the
eigenvalue zero. If there exists a constant $\delta>1$ such that

$\lim\limits_{t\rightarrow
0^{+}}Card\\{[\frac{t}{\delta},t]\cap\sigma(T)\\}=\infty,$

then $T$ is a conditional operator.

###### Proof.

First step, we choose a sequence $\\{\lambda_{n}\\}\subseteq\sigma(T)$
satisfying the conditions of lemma 3.1. We will find it by induction.

For $k=1$, $\Delta_{1}=\\{\lambda_{1},\lambda_{2}\\}\subseteq\sigma(T)$ and
$c_{1},d_{1}$ can be easily chosen such that

$\displaystyle\frac{d_{1}}{c_{1}}\leq\delta~{}{\rm
and}~{}c_{1}\leq\frac{\alpha}{\lambda_{1}},\frac{\alpha}{\lambda_{2}}\leq
d_{1}.$

Suppose we have found
$\Delta_{k-1}=\\{\lambda_{2^{k-1}-1},\lambda_{2^{k-1}},\lambda_{2^{k-1}+1},\cdots,\lambda_{2^{k}-2}\\}\subseteq\sigma(T)$
which satisfies

$\Delta_{k-1}\cap\bigcup_{1\leq j\leq k-2}\Delta_{j}=\emptyset,$

and $c_{k-1},d_{k-1}$ such that the first two conditions of lemma 3.1 are
satisfied. Since

$\lim\limits_{t\rightarrow
0^{+}}Card\\{[\frac{t}{\delta},t]\cap\sigma(T)\\}=\infty,$

we can find $t_{0}<min\\{\lambda;~{}\lambda\in\bigcup_{1\leq j\leq
k-1}\Delta_{j}\\}$ such that $t\leq t_{0}$,

$Card\\{[\frac{t}{\delta},t]\cap\sigma(T)\\}\geq 2^{k}.$

Choose arbitrary two elements
$\\{\lambda_{2^{k+1}-3},\lambda_{2^{k+1}-2}\\}\subseteq\sigma(T)\cap[\frac{t_{0}\alpha^{k}}{\delta},t_{0}\alpha^{k}]$,
then choose one after one as follows,

$\displaystyle\\{\lambda_{2^{k+1}-5},\lambda_{2^{k+1}-4}\\}\subseteq\\{\sigma(T)\cap[\frac{t_{0}\alpha^{k-1}}{\delta},t_{0}\alpha^{k-1}]\\}\backslash\\{\lambda_{2^{k+1}-3},\lambda_{2^{k+1}-2}\\}$
$\displaystyle\hskip 142.26378pt\vdots$
$\displaystyle\\{\lambda_{(2^{j}-1)2^{k-j+1}-1},\lambda_{(2^{j}-1)2^{k-j+1}},\lambda_{(2^{j}-1)2^{k-j+1}+1},\ldots,\lambda_{(2^{j+1}-1)2^{k-j}-2}\\}\subseteq\\{\sigma(T)$
$\displaystyle\cap[\frac{t_{0}\alpha^{j}}{\delta},t_{0}\alpha^{j}]\\}\backslash\\{\lambda_{(2^{j+1}-1)2^{k-j}-1},\lambda_{(2^{j+1}-1)2^{k-j}},\lambda_{(2^{j+1}-1)2^{k-j}+1},\ldots,\lambda_{2^{k+1}-2}\\}$
$\displaystyle\hskip 142.26378pt\vdots$
$\displaystyle\\{\lambda_{2^{k}-1},\lambda_{2^{k}},\ldots,\lambda_{3\cdot
2^{k-1}-2}\\}\subseteq\\{\sigma(T)\cap[\frac{t_{0}\alpha}{\delta},t_{0}\alpha]\\}\backslash\\{\lambda_{3\cdot
2^{k-1}-1},\lambda_{3\cdot 2^{k-1}},\ldots,$
$\displaystyle\lambda_{2^{k+1}-2}\\}.$

Since
$Card\\{[\frac{t_{0}\alpha^{j}}{\delta},t_{0}\alpha^{j}]\cap\sigma(T)\\}$ is
more than $2^{k}$, the above process is reasonable.

Denote $c_{k}=t_{0}^{-1},d_{k}=\delta t_{0}^{-1}$, then obviously

$\displaystyle
c_{k}\leq\frac{\alpha}{\lambda_{2^{k}-1}},\ldots,\frac{\alpha}{\lambda_{3\cdot
2^{k-1}-2}},\frac{\alpha^{2}}{\lambda_{3\cdot
2^{k-1}-1}},\ldots,\frac{\alpha^{2}}{\lambda_{7\cdot 2^{k-2}-2}},$
$\displaystyle\cdots\cdots,\frac{\alpha^{k-1}}{\lambda_{2^{k+1}-5}},\frac{\alpha^{k-1}}{\lambda_{2^{k+1}-4}},\frac{\alpha^{k}}{\lambda_{2^{k+1}-3}},\frac{\alpha^{k}}{\lambda_{2^{k+1}-2}}\leq
d_{k}.$

Thus we have found a sequence $\\{\lambda_{n}\\}\subseteq\sigma(T)$ satisfying
the conditions of lemma 3.1. Obviously, $\lambda_{n}$ converges to zero as
$n\rightarrow\infty$.

Second step, we will complete the proof.

We rearrange the sequence $\\{\lambda_{n}\\}\subseteq\sigma(T)$ into a
decreasing sequence $\\{\mu_{n}\\}$. Fix a constant $M>\frac{||T||}{\mu_{1}}$.

For $n\geq 1$, cut each segment $[\mu_{n+1},\mu_{n}]$ into smaller subsegments
(many enough and we denote them by
$[\nu_{m^{n}_{j+1}},\nu_{m^{n}_{j}}],~{}1\leq j\leq k(n)-1$,
$\nu_{m^{n}_{1}}=\mu_{n}$, $\nu_{m^{n}_{k(n)}}=\mu_{n+1}$) in order that

$\dfrac{\nu_{m^{n}_{j}}}{\nu_{m^{n}_{j+1}}}\leq M,~{}1\leq j\leq
k(n)-1,n=1,2,\ldots.$

From the spectral decompose theorem of self-adjoint operator, we have

$T=\oplus_{n\geq 0}\oplus_{1\leq j\leq k(n)-1}T_{(n,j)},$

where $T_{(n,j)}$ is the operator on the subspace $\mathcal{H}_{(n,j)}$
corresponding to $[\nu_{m^{n}_{j+1}},\nu_{m^{n}_{j}}]\cap\sigma(T)$ for $n\geq
1$ and $T_{(0)}$ is the operator on the subspace $\mathcal{H}_{(0)}$
corresponding to $[\mu_{1},\infty)\cap\sigma(T)$.

Denote

$X=\oplus_{n\geq 0}\oplus_{1\leq j\leq k(n)-1}\xi_{(n,j)}^{-1}T_{(n,j)},$

where $\xi_{(0)}=\mu_{1}$,
$\xi_{(n,j)}\in[\nu_{m^{n}_{j+1}},\nu_{m^{n}_{j}}]\cap\sigma(T)$ and
$\xi_{(n,1)}=\mu_{n}$. Then since

$\displaystyle||\xi_{(n,j)}^{-1}T_{(n,j)}||\leq M~{}{\rm
and}~{}||(\xi_{(n,j)}^{-1}T_{(n,j)})^{-1}||\leq M,~{}1\leq j\leq
k(n)-1,~{}n\geq 0,$

we have $X$ is an invertible operator. Moreover,

$S\triangleq\oplus_{n\geq 0}\oplus_{1\leq j\leq
k(n)-1}\xi_{(n,j)}I_{(n,j)}=X^{-1}T,$

where $I_{(n,j)}$ is the identity operator on $\mathcal{H}_{(n,j)}$.
Obviously, $S$ is a diagonal operator with $\\{\lambda_{n}\\}$ its
subsequence. Thus $S$ satisfies the conditions of lemma 3.1 and hence it is a
conditional operator. From theorem 2.12, we obtain that $T$ is a conditional
operator. ∎

Following is a easier criterion for an operator to be conditional.

######  Theorem 3.3.

Let $T\geq 0$ belong to $\mathcal{L}(\mathcal{H})$ which does not admit the
eigenvalue zero. If $\sigma(T)$ has a decreasing sequence $\\{\lambda_{n}\\}$
which converges to zero such that

$\lim\limits_{n\rightarrow\infty}\frac{\lambda_{n}}{\lambda_{n+1}}=1,$

then $T$ is a conditional operator.

###### Proof.

It suffices to show that there exists a constant $\delta>1$ such that

$\lim\limits_{t\rightarrow
0^{+}}Card\\{[\frac{t}{\delta},t]\cap\\{\lambda_{n},n\geq 1\\}\\}=\infty.$

If not, then there exists $N>0$, such that for any $t_{0}>0$, there is a
$t\leq t_{0}$,

$Card\\{[\frac{t}{\delta},t]\cap\\{\lambda_{n},n\geq 1\\}\\}<N.$

Thus there exist sequences $a_{k},b_{k}$ converge to zero, such that for all
$k$

$\displaystyle\frac{b_{k}}{a_{k}}=\delta,Card\\{[a_{k},b_{k}]\cap\\{\lambda_{n},n\geq
1\\}\\}<N,$ $\displaystyle
b_{k+1}<a_{k},Card\\{[b_{k+1},a_{k}]\cap\\{\lambda_{n},n\geq 1\\}\\}\geq 1.$

Choose $\lambda_{n_{1}}$ such that
$\lambda_{n_{1}}=min\\{\lambda_{n};~{}\lambda_{n}\geq b_{1}\\}$, choose
$\lambda_{n_{2}}$ such that
$\lambda_{n_{2}}=max\\{\lambda_{n};~{}\lambda_{n}\leq a_{1}\\}$. Generally,
choose $\lambda_{n_{2k-1}}=min\\{\lambda_{n};~{}\lambda_{n}\geq b_{k}\\}$ and
$\lambda_{n_{2k}}=max\\{\lambda_{n};~{}\lambda_{n}\leq a_{k}\\}$. It is easy
to see that $n_{2k}-n_{2k-1}\leq N$.

On the other hand, since

$\lim\limits_{n\rightarrow\infty}\frac{\lambda_{n}}{\lambda_{n+1}}=1,$

we have

$\lim\limits_{n\rightarrow\infty}\frac{\lambda_{n}}{\lambda_{n+j}}=1,$

for any $1\leq j\leq N$ and hence

$\lim\limits_{k\rightarrow\infty}\frac{\lambda_{n_{2k-1}}}{\lambda_{n_{2k}}}=1.$

But

$\frac{\lambda_{n_{2k-1}}}{\lambda_{n_{2k}}}\geq\frac{b_{k}}{a_{k}}=\delta>1$

for any $k$, it is a contradiction.

Thus $T$ is a conditional operator. ∎

######  Remark 3.4.

Actually, suppose the limit of $\frac{\lambda_{n}}{\lambda_{n+1}}$ exists,
then

$\lim\limits_{n\rightarrow\infty}\frac{\lambda_{n}}{\lambda_{n+1}}=1$

if and only if there exists a constant $\delta>1$ such that

$\lim\limits_{t\rightarrow
0}Card\\{[\frac{t}{\delta},t]\cap\\{\lambda_{n},n\geq 1\\}\\}=\infty.$

One can easily prove it. Thus the condition of theorem 3.3 is a little
stronger than theorem 3.2.

######  Corollary 3.5.

Let $T\in\mathcal{L}(\mathcal{H})$ such that $T$ and $T^{*}$ do not admit the
eigenvalue zero. If $\sigma((T^{*}T)^{\frac{1}{2}})$ has a decreasing sequence
$\lambda_{n}$ which converges to zero such that

$\limsup\limits_{n\rightarrow\infty}\frac{\lambda_{n}}{\lambda_{n+1}}=1,$

then $T$ is a conditional operator.

###### Proof.

From the polar decomposition theorem,

$T=U(T^{*}T)^{\frac{1}{2}},$

where $U$ is a unitary operator. Thus from theorem 3.3 and theorem 2.12, we
obtain the result. ∎

######  Corollary 3.6.

Compact operator $K=diag\\{1,\frac{1}{2},\frac{1}{3},\cdots\\}$ is a
conditional operator.

## References

  * [1] Stephane Jaffard and Robert M. Young, A Representation Theorem for Schauder Bases in Hilbert Space, Proc. AMS., Vol. 126, No. 2 (Feb., 1998), pp. 553-560.
  * [2] Young, Robert M. An introduction to nonharmonic Fourier series. Pure and Applied Mathematics, 93. Academic Press, Inc. , New York-London, 1980.
  * [3] Joachim Weidmann, Linear Operators in Hilbert Space, GTM68, Springer-Verlag, 1980.
  * [4] Robert E. Megginson, An introuduction to Banach Space Theory, GTM183, Springe-Verlag, 1998.
  * [5] A. M. Olevskii, On operators generating conditional bases in a Hilbert space, Translated from Matematicheskie Zametki, Vol(12), No.1, pp. 73-84, July, 1972.
  * [6] I. Singer, Bases in Banach Space I, Springer-verlag, 1970.
  * [7] John B. Conway, A course in functional analysis, GTM96, Springer-Verlag, 1985.
  * [8] Allen L. Shields, “Weighted shift operators and analytic function theory”, in: Topics in Operator Theory, Math. Surveys No. 13, 49-128, Amer. Math. Soc., Providence (1974).
  * [9] Garling, D. J. H., Symmetric bases of locally convex spaces, Studia Math. 30, 1968, 163-181.
  * [10] Arsove, Maynard G. Similar bases and isomorphisms in Fr chet spaces. Math. Ann. 135, 1958, 283-293.
  * [11] Beurling, Arne On two problems concerning linear transformations in Hilbert space. Acta Math. 81, (1948).
  * [12] Halmos, Paul Richard, A Hilbert space problem book. Second edition. Graduate Texts in Mathematics, 19. Springer-Verlag, New York-Berlin, 1982.
  * [13] Kwapien, S.; Pelczynski, A. The main triangle projection in matrix spaces and its applications. Studia Math. 34 1970 43-68.
  * [14] Wright, J. D. Maitland, All operators on a Hilbert space are bounded. Bull. Amer. Math. Soc. 79 (1973), 1247-1250.
  * [15] Douglas, Ronald G., Banach algebra techniques in operator theory. Pure and Applied Mathematics, Vol. 49. Academic Press, New York-London, 1972.
  * [16] Jiang, Chunlan; Ji, Kui, Similarity classification of holomorphic curves. Adv. Math. 215 (2007), no. 2, 446-468.
  * [17] Jiang, Chunlan Similarity classification of Cowen-Douglas operators. Canad. J. Math. 56 (2004), no. 4, 742-775.
  * [18] Jiang, Chunlan; Wang, Zongyao Structure of Hilbert space operators. World Scientific Publishing Co. Pte. Ltd., Hackensack, NJ, 2006.
  * [19] Jiang, Chunlan; Wang, Zongyao Strongly irreducible operators on Hilbert space. Pitman Research Notes in Mathematics Series, 389. Longman, Harlow, 1998.
  * [20] Zhu, Kehe, Operators in Cowen-Douglas classes. Illinois J. Math. 44 (2000), no. 4, 767-783.
  * [21] Cowen, M. J.; Douglas, R. G. Equivalence of connections. Adv. in Math. 56 (1985), no. 1, 39-91.
  * [22] Cowen, M. J.; Douglas, R. G. Complex geometry and operator theory. Acta Math. 141 (1978), no. 3-4, 187-261.
  * [23] Elias M. Stein and Rami Shakarchi, Complex analysis, Princeton Lectures in Analysis, Princeton University Press, 2003.
  * [24] McCarthy, John E. Boundary values and Cowen-Douglas curvature. J. Funct. Anal. 137 (1996), no. 1, 1-18.
  * [25] Gowers, W. T.; Maurey, B. The unconditional basic sequence problem. J. Amer. Math. Soc. 6 (1993), no. 4, 851-874.
  * [26] Gowers, W. T.; Maurey, B. Banach spaces with small spaces of operators. Math. Ann. 307 (1997), no. 4, 543-568.
  * [27] Cao Yang and el, On unitary operators spread $\mathbb{N}$, preprint.
  * [28] Niven, Ivan; Zuckerman, Herbert S.; Montgomery, Hugh L. An introduction to the theory of numbers. Fifth edition. John Wiley and Sons, Inc., New York, 1991.