数理解析研究所講究録
第 1996 巻 2016 年 113-120

113

A method of diminishing intervals andsemiclosed subspaces
Go Hirasawa
Ibaraki University

1. INTRODUCrION
Let $H$ be an infinite dimensional separable complex Hilbert space andbe the set of all bounded operators on $H$ . A bounded operator
let
on $H$ means a linear bounded operator from $H$ to $H$ . This report istreated with the Invariant Subspace Problem on $H$ . That is, for any
bounded operator , does there exist nontrivial closed subspace $M(\neq$$H)$ such that? This problem has not been completelysolved yet. But it is known that, under additional conditions with(normal, subnormal, having rich spectrum and so on), there exists a
nontrivial -invariant closed subspace.
In our study for ISP, we do not set additional conditions on . However, we set hypothesises on something which are not concerned with. Something is (a choice function’ which assigns a Hilbert norm to eachsemiclosed subspace. Among of all, we paticularly consider a choice function satisfying hypothesises (h1), (h2) and (h3) as stated in the following
section. We do not know whether the choice funtion truely exists ornot. A reason why we consider hypothesises is that, if a propositio any
bounded operator on $H$ has nontrivial -invariant closed subspace’ isindependent of ZFC $(+CH)$ , then we are disconsolate.
Our aim of this report is to state an outline of our approach to ISP,
which does not mean that ISP can be solved by this way. There aresome key words in our approach, that is, a linear dimension, a Hilbertdimension, a choice function with hypothesises, -invariant semiclosedsubspaces, $codimen\mathcal{S}ion$ and so on. The following Theoreml.1 is one ofmotivations for our study. It says that, under CH (continuous hypothesis), a separable Hilbert space is the only case the Hilbert dimension and$\mathcal{B}(H)$$T$$TM\underline{\subseteq}M$$\{O\},$

$T$

$T$

$T$

$T$

$n^{}$

$T$

$T$

$T$

$a$

114
the linear dimension are diﬀerent. We feel a kind of corresponding between Theoreml.1 and ISP from the fact that a separable Hilbert spaceis the only case ISP is not solved.
Theorem 1.1 ([2]). Let $H$ be a Hilbert space. Then,(1) $\dim H<\infty\Leftrightarrow\dim_{l}H<\infty$ . In this case, $\dim H=\dim_{l}H.$(2) If. Hence, $\dim H<\dim_{l}H.$then(3) If, then $\dim H=\dim_{l}H.$$\aleph_{0}\leq\dim H<2^{\aleph 0,}$$\dim_{l}H=2^{\aleph_{0}}$$2^{\aleph_{0}}\leq\dim H$

2. FROM SEMICLOSED SUBSPACES TO A CLOSED SUBSPACE
It is well known ([3]) that any bounded operator on $H$ have many-invariant semiclosed subspaces. If ISP is aﬃrmatively solved, then theset of all nontrivial -invariant semiclosed subspaces necessarily contains
a closed subspace. In the following, we seek (but now, can not obtain)
a nontrivial closed subspace among the set of all -invariant semiclosed
subspaces by a method of diminishing intervals of semiclosed subspaces,
which are based on a choice function as stated above.
$T$

$T$

$T$

$T$

Definition 2.1. A subspace $M$ in $H$ is said to be semiclosed if there(continuouslyexists a Hilbert norm
on $M$ such that$(M, \Vert\cdot\Vert_{M})\mapsto H$$\Vert\cdot\Vert_{M}$embedded).

It is easily shown that a semiclosed subspace is equivalent to an operator range. Clearly, a closed subspace is semiclosed.
The following theorem plays a central role for applying a method ofdiminishing intervals to ISP.
Theorem 2.1 ([4]). Let
and letandbe nontrivial
in H. Suppose that-invariant semiclosed subspaces such that$\dim_{l}M_{2}/M_{1}>1$ . Then, there exists-invariant semiclosed subspacethat$T\in \mathcal{B}(H)$$T$$M_{1}$$M_{2}$$M_{1}\subsetneq M_{2}$$T$$M_{3}$$\mathcal{S}uch$$M_{1}\wedge\subseteq M_{3}\subsetneq M_{2}.$3. A CHOICE FUNCTION$\alpha$By Definition 2.1, we can consider a concept of the following functionin Definition 3.1.

115
Definition 3.1. We choose a Hilbert norm from each semiclosed subspace. Denote such a choice function by$\alpha.$

Given a choice function . For a semiclosed subspace $M$ , there isthe Hilbert normby such that $(M, \Vert \Vert_{M})\mapsto H$ . It follows
from [1] thatis isometrically isomorphic to de Branges space
for a unique positive operator. Then, we denote$M=\alpha AH$. A corresponding between
and is one to one. Hence, weconsider a choice function as a choice of positive operators. Summingup, the notation a has two meanings such as a choice of Hilbert norms
or a choice of positive operators.
We explain an example of in. A subspace$\alpha$$\Vert$$\Vert_{M}$$\alpha$$(M, \Vert\cdot\Vert_{M})$$(\mathcal{M}(A), \Vert\cdot\Vert_{A})$$\mathcal{A}\in \mathcal{B}(H)$$A$$\Vert\cdot\Vert_{M}$$\alpha$$\alpha$$L^{2}(\mathbb{R}^{d})(d\geq I)$$M^{\sigma}:=\{f\in L^{2}(\mathbb{R}^{d}):(1+|\xi|^{2})^{\frac{\sigma}{2}}\hat{f}(\xi)\in L^{2}(\mathbb{R}^{d})\} (\sigma>0)$ofis semiclosed. Because, the inclusion mapping
is contraction, whereis the Fouriertransform of ). As you know,
is known as
Sobolev spaces of the order . Then, we choose Sobolev normfor
each semiclosed subspacefor $\sigma>0$ , and we choose the appropriate
Hilbert norm for other semiclosed subspaces.
On the other hand, it is proved ([5]) that Sobolev spaceisisometrically isomorphic to de Branges space, where $A_{\sigma}=(I-$(the Bessel potential of the order a). Hence we can also say that
we choose the Bessel potential, and wefrom Sobolev spacechoose the appropriate positive operator from each semiclosed subspace
except for Sobolev spaces.
$(M^{\sigma}, \Vert \Vert_{\sigma})carrow$$L^{2}(\mathbb{R}^{d})$$L^{2}(\mathbb{R}^{d})$$\Vert f\Vert_{\sigma}$$f$$:=\Vert(1+|\xi|^{2})^{\frac{\sigma}{2}}\hat{f}\Vert_{L^{2}}(\hat{f}$$H^{\sigma}(\mathbb{R}^{d})$$:=(M^{\sigma},$$\Vert$$\Vert_{\sigma}\rangle$$\Vert$$\sigma$$\Vert_{\sigma}$$M^{\sigma}$$H^{\sigma}(\mathbb{R}^{d})$$\mathcal{M}(A_{\sigma})$$\Delta)^{-\frac{\sigma}{2}}$$H^{\sigma}(\mathbb{R}^{d})$$A_{\sigma}$

4. HYPOTHESISES OF$\alpha$Ekom this section, we handle a paticular choice function for constructing a method of diminishing intervals. A $paticular^{)}$ choice functionmeans that it satisfies some hypothesises (h1), (h2) and (h3).
In the following statements (h2) and (h3), we use the notation as asubset of.
$\alpha$$\mathcal{B}(H)$116(h1) For semiclosed subspaces $M,$ $N$ such that

$M\subseteq N,$(contractively)$(M, \Vert \Vert_{M})\mapsto(N, \Vert \Vert_{N})$, where $\alpha^{2}:=\{A^{2}:A\in\alpha\}.$(h2), that is,(h3) A set is closed in$\alpha=\alpha^{2}$$\mathcal{B}(H)$$\alpha$$\alpha=\overline{\alpha}.$Remark 4.1.

(i) A set consists of positive contractions $A(0.\leq A\leq I)$ .
(ii) For $M=\alpha AH$ and $N=\alpha BH(M\underline{\subseteq}N)$ , (hl) if and only if$\alpha$$A^{2}\leq B^{2}$(iii) For a closed subspace $M=\alpha AH$ , it deducesby (h2), we seeimplies
the orthogonal projection.
$A\in\alpha$$A=A^{\frac{1}{2}}$$A^{\frac{1}{2}}\in\alpha$H. Sinceis. That is,$AH=A^{\frac{1}{2}}$$A$

5. A METRIC ON SEMICLOSED SUBSPACES
Let
be the set of all semiclosed subspaces in $H$ . For semiclosed
on bysubspaces $M=\alpha AH$ and $N=\alpha BH$ , we define a metric
$S$

$S$

$\rho_{\alpha}$$\rho_{\alpha}(M, N)(=\rho(M, N))$ $:=\Vert A-B\Vert$(the operator norm)

Example 5.1 ([5]). The metric between Sobolev spaces are given by thefollowing. For$0<\sigma_{1}<\sigma_{2},$$\rho((H^{\sigma x}(\mathbb{R}^{d}), H^{\sigma}2(\mathbb{R}^{d}))=(\frac{\sigma_{1}}{\sigma_{2}})^{\frac{\sigma 1}{\sigma 2^{-\sigma}1}}-(\frac{\sigma_{1}}{\sigma_{2}})21\sigma.$

In particular,$\rho(H^{1}(\mathbb{R}^{d}), H^{2}(\mathbb{R}^{d}))=0.25.$

Under the hypothesises (h1), (h2) and (h3), we have the following
propositions.

Proposition 5.1. The metric space$(S, \rho_{\alpha})$is complete.

We define an interval $[M, N]$ for semiclosed subspaces$M\subseteq N.$$[M, N] :=\{L\in S : M\underline{\subseteq}L\underline{\subseteq}N\}.$

Proposition 5.2. An interval $[M, N]$ is$clo\mathcal{S}ed$

in$(S, \rho_{\alpha})$We define the diameter of $I=[M, N]$ by a usual way.

diam

$L$

$I$
$:= \sup_{L,L\in I}\rho_{\alpha}(L,$.

117
Proposition 5.3. Let$:=[M_{n}, N_{n}](n=1,2, \cdots)$$I_{n}$. Then, as$narrow\infty,$$diamI_{n}arrow 0\Leftrightarrow\rho_{\alpha}(M_{n}, N_{n})arrow 0$

From previous propositions, we have a method of diminishing intervals’ of semiclosed subspaces.
$\langle$. If
Theorem 5.4. Let $I_{n}:=[M_{n}, N_{n}](n=1,2, \cdots)$ such that, then there exists a unique semiclosed subspace$I_{n}\supset I_{n+1}$$\rho_{\alpha}(M_{n}, N_{n})arrow 0(narrow\infty)$$M_{\infty} \in\bigcap_{n=1}^{\infty}I_{n\prime}$

6. A THINKING FOR ISP. By [3], there exists many -invariant semiclosed
Given anysubspaces. We want to find a closed subspace among the set of all non. Without a loss oftrivial -invariant semiclosed subspaces
generality, we may assume that the linear dimension $\dim_{l}H/M_{\lambda}=\infty$
is closed. (Itfor all . For, $\dim_{l}H/M_{\lambda}<\infty$ for some implies thatis known that a semiclosed subspace which has finite codimension in $H$
an idea related with a codimension
is closed in $H.$ ) Roughly$T$$T\in \mathcal{B}(H)$$T$$\{M_{\lambda}\}_{\lambda\epsilon\Lambda}$$\lambda$$\lambda$$M_{\lambda}$$explaining_{\rangle}$is the following.
. The first step.
Pick non-trivial -invariant semiclosed subspaces$T$......

$(*)$$M_{1}$

and$M_{2}$

such that$\{\begin{array}{l}M_{1}\subsetneq M_{2}\dim_{l}M_{2}/M_{1}=\infty.\end{array}$

We put the interval $:=[M_{1}, M_{2}].$The second step.
Since $\dim_{l}M_{2}/M_{1}=\infty$ , there exists ([4]) a -invariant semiclosed subspacesuch that$I_{1}$$T$$M_{3}$$M_{1}\subsetneq M_{3}\subsetneq M_{2}.$Then, we see that $\dim_{l}M_{3}/M_{1}=\infty$ or $\dim_{l}M_{2}/M_{3}=\infty$ . Now wesuppose that $dim,$ $M_{3}/M_{1}=\infty$ . Then we put the interval $:=[M_{1\}}M_{3}].$
The third step.
Since $\dim_{l}M_{3}/M_{1}=\infty$ , there exists a -invariant semiclosed subspacesuch that$I_{2}$$T$$M_{4}$$M_{1_{\wedge\wedge}}\subseteq M_{4}\subseteq M_{3}.$118
Then, we see that $\dim_{l}M_{4}/M_{1}=\infty$ or $\dim_{t}M_{3}/M_{4}=\infty$ . Now wesuppose that $\dim_{l}M_{3}/M_{4}=\infty$ . Then we put the interval $:=[M_{4}, M_{3}].$
By an inductive way, we get a sequenceof intervals. Note that
semiclosed subspaces $M_{i}(i=1,2, \cdots)$ are -invariant.
$I_{3}$$\{I_{n}\}_{n}$$T$(If’ diam $I_{n}arrow 0$ , then, by Theorem 5.4, we get the semiclosed subspace$\exists!M_{\infty}\in\bigcap_{n=1}^{\infty}I_{n}.$We want to expect thatis the -invariant closed subspace, so that
it is necesarry to check the following questions.
$T$$M_{\infty}$

Ql. Does there exist
satisfying $(*)$ ?
andQ2. Does there exist a sequencesuch that (diamQ3. Is-invariant?
Q4. Isclosed?
$M_{1}$$M_{2}$$\{I_{n}\}_{n}$$M_{\infty}$$I_{n}arrow 0^{)}$?

$T$$M_{\infty}$

7. ABOUT QIThere exist semiclosed subspaces
andsuch that $(*)$ . Pick any. Put. Since-invariant
Hence,
bywe define
to be the semiclosed subspace, and let, thenIf
is ( -invariant) closed subspace. This attains
our goal. Thus, without a loss of generality, we assume thatSince
is -invariant,is also -invariant. Since
there exists -invariant
satisfying. Moreover we canprove that. Then we get the interval
whichsatisfies the condition $(*)$ .
$M_{1}$$T$$M_{1}$$M_{1}=\alpha A_{1}H$$A_{1}\in\alpha,$$A^{\frac{1}{12}}\in\alpha$$A^{\frac{1}{12}}H$$M_{1}^{\frac{1}{2}}$$M_{1}=M^{\frac{1}{12}}$$M_{2}$$M_{1}$$a$$(h2)1^{\cdot}$$M_{1}^{\tilde{2}}=\alpha A^{\frac{1}{12}}H.$$T$$M_{1}\underline{\subseteq}M_{1}^{\frac{1}{2}}$$M_{1}$$T$$T$$M_{1}^{\frac{1}{2}}$$M_{2}$$\dim_{l}H/M_{1}^{\frac{1}{2}}=\infty,$$T$$M_{1}^{\frac{1}{2}}\subset\wedge M_{2}\subseteq\wedge H$$\dim_{l}M_{1}^{\frac{1}{2}}/M_{1}=\infty$$[M_{1}, M_{2}]$8. ABOUT Q2, Q3 AND Q4
We do not know that how do we try to solve these questions Q2, Q3
and Q4. But we think that these questions are linked each other. In thefollowing, we state an idea in our brain.

119
Let $M=\alpha AH$ and $N=\alpha BH$ . By (h1) and Douglas’s majorization
theorem, the interval$[M, N]=\{L\in S:M\subseteq L\subseteq N\}$is order isomorphic to the operator interval$[A^{2}, B^{2}] :=\{X^{2} :A^{2}\leq X^{2}\leq B^{2}\},$

is a positive operator.
whereLet $I_{1}=[M_{1}, M_{2}]$ be the interval satisfying the condition $(*)$ in theis -invariant, there exists aandsection 6. Since
In thesuch that-invariant semiclosed subspacesuch that
we have a -invariant semiclosed subspace
sameHexlce, we have inductively $M_{2k+1}(k=1,2, \cdots)$ such that
. On the other hand, with respect to indexesof even numbers, we choose a -invariant semiclosed subspace $M_{2k+2}$ sothat $\rho_{\alpha}(M_{2k+1}, M_{2k+2})arrow 0(karrow\infty)$ , and let $I_{k+1}=[M_{2k+1}, M_{2k+2}]$$X\in \mathcal{B}(H)$$\dim_{i}M_{1}^{\frac{1}{2}}/M_{1}=\infty$$M_{1}^{\frac{1}{2}}$$T$$M_{1}\subseteq M_{3}arrow\subseteq\sim M^{\frac{1}{12}}$$T$$M_{3}$$T$$M_{5}$$M_{3}\subsetneq$$way_{\’{i}}$$M_{5}\subset<M_{3}^{\overline{2}}.$$M_{2k-1}\subseteq M_{2k+\lambda}\subseteq M_{2k-1}^{\frac{1}{2}}$$T$$(k=0,1,2, \cdots)$ .

$M_{1}$.

$I_{1}$$M_{2}$$M_{1}^{\frac{1}{2}}$$M_{3}$.

$I_{2}$$M_{4}$$M_{3}^{\frac{1}{2}}$$M_{5}$$I_{3}$.

$M_{6}$

By $\rho_{\alpha}(M_{2k+1}, M_{2k+2})arrow 0$ , that $is_{\rangle}diamI_{k}arrow 0(karrow\infty)$ , there uniquely. For intervals $[M_{n}, ]$ orexists the semiclosed subspace$[ , M_{n}],$ $\rho_{\alpha}(M_{n}, M_{\infty})=\Vert A_{n}-A_{\infty}||arrow 0(narrow\infty)$ , where $M_{n}=\alpha A_{n}H,$. Therefore, since $A_{2k+1}^{2}\leq A_{2k-1}\leq A_{2k+2}^{2}(k=1,2, \cdots)$ ,is the orthogonal. This means that
we have that
$M_{\infty} \in\bigcap_{k=1}^{\infty}I_{k}$$M_{\infty}=\alpha A_{\infty}H$$A_{\infty}^{2}\leq A_{\infty}\leq A_{\infty}^{2}$$A_{\infty}$

120
projection, i.e.,
is closed, which is a thinking for Q2 andQ4. For Q3, we note that $M_{2k+1}$ and $M_{2k+2}$ are -invariant and$M_{\infty}=\alpha A_{\infty}H$$T$$M_{\infty}\subsetneq M_{2k+2}$. Now$M_{2k+1}\subsetneq$$\rho_{\alpha}(M_{2k+1}, M_{2k+2})arrow 0(karrow\infty)$, is$M_{\infty}$-invariant?

$T$

Ibaraki University
Colleage of Engineering
Nakanarusawa 4-12-1
Hitachi 316-8511, Japan$\ovalbox{\tt\small REJECT} \mathfrak{M}X\ovalbox{\tt\small REJECT} \mathcal{I}ae_{Q}^{R}\beta \mp\backslash ;\ovalbox{\tt\small REJECT}^{\backslash } \mathbb{R}$REFERENCES[1] T.Ando, De Branges Spaces and Analytic Operator Functions, Lecture note,Sapporo, Japan, 1990.
ス*き] (Japanese)[2][3] E.Nordgren, M.Radjabalpour, H.Radjavi and P.Rosenthal, On Invariant Operator Ranges, Trans. Amer. Math. Soc. vol.251 (1979), 389-398.
[4] L.Rodman and N.Zobin, Linear Preservers of Isomorphic Types of Lattices ofInvariant Operator Ranges, Proc. Amer. Math. Soc. vol.129 (2001), 2981-2986.
[5] G.Hirasawa, A Metric for Unbounded Linear 0perators in a Hilbert Space, Integr.
Equ. Oper. Theory vol.70 (2011), 363-378.

Hokkaido$University_{\rangle}$$\ovalbox{\tt\small REJECT}\Pi\backslash \ovalbox{\tt\small REJECT} g\neq,$$\delta’\mp/J$$\ovalbox{\tt\small REJECT}\Phi \mathscr{X}\ovalbox{\tt\small REJECT}_{J^{1)-}}\backslash ^{\backslash }$$\ovalbox{\tt\small REJECT}\Re ffiffi\lambda P\ovalbox{\tt\small REJECT}$$\Re$$\ovalbox{\tt\small REJECT}$$\mathfrak{F}$