数理解析研究所講究録1080 巻 1999 年 100-104100

ハンケル作用素の積が再びハンケル作用素になる為の条件について

東北大学理学研究科
吉野崇 (Takashi Yoshino)

constitute, then the functions
If $e_{n}(z)=z^{n}$ for $|z|=1$ and $n=0,$
constitute an
and the functions
an orthonormal basis for$n=0,1,2,$
be the set of all essentially bounded functions in. Let
orthonormal basis for
onis, the Toeplitz operator. For a function
and letonto
given by $T_{\varphi}f=P(\varphi f)$ for $f\in H^{2}$ where $P$ is the orthogonal projection fromis given by $H_{\varphi}f=J(I-P)(\varphi f)$ for $f\in H^{2}$onand the Hankel operator$\pm 1,$ $\pm 2,$$\cdots$$L^{2}$$H^{2}$$\cdots$$e_{n},$$L^{\infty}$$H^{\infty}=H^{2}\cap L^{\infty}$$L^{2}$$e_{n}$$\varphi\in L^{\infty}$$H^{2}$$T_{\varphi}$$L^{2}$$H^{2}$$H^{2}$$H_{\varphi}$

where$J$

is the unitary operator on$L^{2}$

defined by $Je_{-}=enn-1$ .

Concerning these operators, the following results are known.
Proposition 1. Ifexists an inner function

is a non-zero closed invariant subspace of , then there
uniquely,up to a unimodular constant, such that
$T_{z}$$\mathcal{M}$$g$$\mathcal{M}=T_{g}H^{2}$and$\mathcal{M}^{\perp}=H_{\overline{g}}^{*}H^{2}$

Proposition 2. If is a non-constant function inwhere
denotes the point spectrum.

.

$L^{\infty}$$\varphi$$\emptyset$, then$\sigma_{p}(T_{\varphi})\cap\overline{\sigma \mathrm{P}(T_{\varphi^{*}})}=$

and$T_{\psi^{*}}H_{\varphi}=H_{\varphi}\psi*=$$\sigma_{p}(\cdot)$

Proposition 3. For any.

$\psi\in H^{\infty},$$H_{\varphi}T\psi=H_{\varphi\psi}$$H_{\varphi}T_{\psi^{*}}$

Proposition 4.

$H\psi^{*}H_{\varphi}=T_{\overline{\psi}\varphi}-\tau\tau_{\varphi}\overline{\psi}$.

Proposition 5. The following assertions are
(1)
.
(2)
.
(3)
for some inner function andcommon non-constant inner factor.

.

$\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{i}\mathrm{v}\mathrm{a}\mathrm{l}\mathrm{e}\mathrm{n}\mathrm{t}|$$N_{H_{\varphi}}\neq\{0\}$$[H_{\varphi}H^{2}]\sim L^{2}\neq H^{2}$$\varphi=\overline{g}h$$g$$h\in H^{\infty}$

such that

$g$

and

$h$

have no101
Now we shall consider the following theorems.

Theorem 1.

$H_{\varphi}H\psi=O$

if and only if $H_{\varphi}=O$ or $H\psi=O$ .

Proof. By Proposition 4, we have
$O=H_{\varphi}H\psi=\tau-_{\psi\varphi}\varphi*-\tau-*\tau_{\psi}$or$=$.

$\varphi^{*}\in H^{\infty}$$=$$H_{\varphi}=H_{\varphi}**=O$$\psi\in H^{\infty}$

or $H\psi=O$ .

Theorem 2. The productof two non-zero Hankel operators
is also a Hankel operator if and only if$H_{\varphi}H\psi$$H\psi$$\varphi=\overline{q}h$and$k\in H^{\infty}$

such that each

$h$

and

$k$

and

$\psi=\overline{q}k$where $q(z)=(z-\overline{\lambda})(1-\lambda z)^{-}1$ for some complex number$h,$$H_{\varphi}$such as $|\lambda|<1$ andis non-zero and has no inner factor . And, in$\lambda$$q$

this case,$H_{\varphi}H_{\psi hk}=\alpha_{q}H_{\overline{q}}$where$\alpha_{q}$is the non-zero eigenvalue of$H_{\overline{q}}$.

To prove Theorem 2, we need the following lemmas.

Let $H_{\varphi}H\psi=H_{u}$ for non-zero Hankel operators$H_{\varphi}$and$H\psi$. Then we have thefollowing.
Lemma 1.

$0\in\sigma_{p}(H_{\varphi})\cap\sigma_{p}(H\psi)$.

Proof. Since$H_{\varphi}T_{z}H\psi=T_{z}^{*}H_{\varphi\psi}H=T_{z}^{*}H_{u}$$=H_{u}T_{z}=H_{\varphi}H_{\psi}\tau_{z}=H_{\varphi}T_{z\psi}^{*}H$$H_{\varphi}(T_{z}-\tau_{z}^{*})H\psi=O$. If$0\not\in\sigma_{p}(H_{\varphi})$,, then $(T_{z}-\tau_{z}^{*})H\psi=O$ and $H\psi=O$ because

by Proposition 2. This contradicts the assumption that
non-zero. Therefore.
$0\not\in\sigma_{p}(T_{z}-T_{z}*)$$0\in\sigma_{p}(H_{\varphi})$$H_{\psi}$is

102
by Proposition 5 and $H_{\varphi}(T_{z}-T_{z}*)=$
is dense in, then
Ifby Proposition 2. Andand hence $H_{\varphi}=O$ because $(T_{z}-T^{*}z)H2$ is dense in.
this also contradicts the assumption and hence$H^{2}$$H\psi H^{2}$$0\not\in\sigma_{p}(H\psi)$$H^{2}$$O$$0\in\sigma_{p}(H\psi)$

Iffor some inner function, then, by Proposition 5,
and $H_{\varphi g}=H_{h}=O$ . And we have the following.
$\varphi=\overline{g}h$$0\in\sigma_{p}(H_{\varphi})$$h\in H^{\infty}$$g$

andLemma 2. For an inner function , the following assertions are equivalent.
$g$(1) $H_{\varphi g}=O$ ,(2) $H_{\psi g}=O$and(3) $H_{ug}=O$ .

Proof. Since, by Proposition 3$\neg$$H_{\varphi g}H_{\psi}=T_{\mathit{9}^{*}}*HH\varphi\psi=\tau_{g^{*H}uu}*=HT\mathit{9}$$=H_{ug}=H_{\varphi}H_{\psi}\tau_{g}=H_{\varphi}H\psi_{\mathit{9}}$and since$H_{\varphi}$

and$H\psi$

are non-zero by the assumption, the assertion follows fromTheorem 1.

Lemma 3.

$\dim[H_{u}*H^{2}]\sim L2=1$

Proof. Since$N_{H_{u}}\neq\{\mathit{0}\}$.

by Lemma l,we have, by Proposition 1,$[H_{u}*H^{2}]\sim L^{2}=H_{\overline{q}}^{*}H^{2}$

and$N_{H_{u}}=T_{q}H^{2}$for some inner function . If $\dim[H_{u}*H^{2}]\sim L2\geq 2$ , then$q$$\dim[T_{q}H^{2}]^{\perp}=\dim[H_{u}*H^{2}]\sim L2\geq 2$. Sincesuch asand there exists a closed invariant subspacefor some non-constant inner function by Proposition 1, $q=q_{1}q_{2}$ forsome non-constant inner function . Since, by Proposition 3,$\mathcal{M}\mathrm{o}\mathrm{f}T_{z}$$\mathcal{M}=T_{q_{1}}H^{2}$$T_{q}H^{2}\subset \mathcal{M}\subset H^{2}$$q_{1}$$q_{2}$$O=H_{u}T_{q}=H_{u}\tau_{q_{1}}\tau_{q_{2}}=T_{q_{1}}*H*T_{q}u2$$=T_{q_{1}}*H_{\varphi\psi}*H\tau_{q2}=H_{\varphi}T_{q_{1}}H\psi T_{q2}=H_{\varphi q_{1}}H\psi q_{2}$’or $H_{\psi q_{2}}=O$ by Theorem 1 and, by Lemma 2, $H_{uq_{1}}=O$ or $H_{uq_{2}}=O$ .
If $H_{uq_{1}}=O$ , then, by Proposition 3, $T_{q_{1}}H^{2}\subseteq N_{H_{u}}=T_{q}H^{2}=T_{q1}T_{q2}H^{2}$ and$H_{\varphi q_{1}}=O$103
is an isometry and this contradicts that is a non-constantbecauseinner function. Hence $H_{uq_{1}}\neq O$ . By the same reason, $H_{uq_{2}}\neq O$ . These contradict
the abovethat $H_{uq_{1}}=O$ or $H_{uq_{2}}=O$ . Therefore $\dim[H_{u}*H^{2}]\sim L2\leq 1$ . By
are-zero by the assumption andTheorem 1, $H_{u}\neq O$ because
andand hence $\dim[H_{u}*H^{2}]\sim L2=\dim[N_{H_{u}}]^{\perp}\geq 1$ . Therefore $\dim[H_{u}*H^{2}]\sim L2=1$ .
$H^{2}\subseteq\overline{T}_{q_{2}}H^{2}$$T_{q_{1}}$$q_{2}$$\mathrm{r}\mathrm{e}^{1}\mathrm{s}\mathrm{u}\mathrm{l}\mathrm{t}$$\sim$‘$H\psi$$H_{\varphi}$$N_{H_{u}}\neq$$\mathrm{n}\mathrm{o}\mathrm{n}$$H^{2}$Proof of Theorem 2.

$\underline{(arrow)}$; By Lemma 3 and its proof, we have$\dim N_{T_{q^{*}}}=\dim[H_{u}*H2]\sim L2--1$

and hence, for somesuch as $|\lambda|<1$ ,$q(z)–(z-\overline{\lambda})(1-\lambda z)-1$ . Since $N_{H_{u}}=T_{q}H^{2},$ $H_{uq}=H_{u}T_{q}=O$ by Proposition 3
and, by Lemma 2, $H_{\varphi q}=O$ and $H_{\psi q}=O$ and hence $\varphi q=h$ and $\psi q=k$ for some. Therefore
andandbecause is inner. Since$H\psi\neq O$ by the assumption, eachand is non-zero and has no inner factor .
; Conversely, if
andwhere $h,$ $k\in H^{\infty}$ and $q(z)=(z-$for somesuch as $|\lambda|<1$ , thenis a partial isometry by

and$h,$$N_{T_{q^{*}}}$

is an eigenspace of$k\in H^{\infty}$$T_{z}^{*}$$\psi=\overline{q}k$$\varphi=\overline{q}h$$h$$\varphi=\overline{q}h$$\underline{(arrow)}$$\overline{\lambda})(1-\lambda z)^{-1}$$\lambda\in \mathbb{C}$$H_{\varphi}\neq O$

$q$

$k$

$q$

$\psi=\overline{q}k$$\lambda\in \mathbb{C}$$H_{\overline{q}}$Proposition 4 and$H_{\overline{q}}H^{2}=H_{\overline{q}}H_{\overline{q}}^{*}H^{2}=(I-\tau_{q^{*}q^{*}}\tau*)H^{2}=N_{\tau_{q^{*}}*}$Since$N_{T_{q^{*}}}*=\{\mathbb{C}(1-\overline{\lambda}z)-1\}$$\alpha_{q}(1-\overline{\lambda}z)-1$

for some$\alpha_{q}\in \mathbb{C}$.

because $q^{*}(z)=(z-\lambda)(1-\overline{\lambda}z)-1,$. Hence, for any $f\in H^{2}$ , we have, by Proposition 3,$H_{\overline{q}}(1-\overline{\lambda}z)-1=$$H_{\varphi}H\psi f=H_{\overline{q}h}H_{\overline{q}k}f=T_{h^{*}}*H_{\overline{q}}H\overline{q}\tau_{k}f$$=T_{h}*H*\{\overline{q}\mu(1-\overline{\lambda}z)-1\}$(becausefor some$\mu\in \mathbb{C}$$H_{\overline{q}}T_{k}f\in H_{\overline{q}}H^{2}=\{\mathbb{C}(1-\overline{\lambda}z)^{-1}\}$)$=T_{h}*\{*(\mu\alpha_{q}1-\overline{\lambda}\mathcal{Z})^{-}1\}=\alpha\tau_{h^{*}}*H\tau fq\overline{q}k$$=\alpha_{q}H_{\overline{q}hk}f$

andby Theorem 1.

$H_{\varphi}H\psi=\alpha_{q}H_{\overline{q}hk}=H_{\alpha_{q}\overline{q}hk}$$\mathrm{C}_{\mathrm{o}\mathrm{r}}\mathrm{o}\mathrm{l}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{y}\urcorner$. Therefore

is a Hankel operator and$\alpha_{q}\neq 0$. Every non-zero idempotent Hankel operator is of the form

where $q(z)=(z-\overline{\lambda})(1-\lambda z)-1$ for someeigenvalue of$H_{\varphi}H\psi$$H_{\overline{q}}$.

$\lambda\in \mathbb{C}$such as $|\lambda|<1$ and$\alpha_{q}$$\frac{1}{\alpha_{q}}H_{\overline{q}}$is the non-zero104, then, by Theorem 2,$q(z)=(z-\overline{\lambda})(1-\lambda z)-1$ for some complex numberIfProof.

$H_{\varphi}2=H_{\varphi}$where $h\in H^{\infty}$ andsuch as $|\lambda|<1$ and$\varphi=\overline{q}h$$\lambda$$H_{\overline{q}h}=H_{\varphi}=H_{\varphi}2\alpha=qH_{\overline{q}}h^{2}=H_{\alpha_{q}\overline{q}h}2$

and henceis the non-zero eigenvalue of
where
Therefore, by using Theorem 2 again, we have
$H_{\overline{q}}$$\alpha_{q}$$H_{\overline{q}h(1\alpha_{q}h)}---H_{\overline{q}h^{-}}H_{\alpha_{q}\overline{q}h}2=O$.

$H_{\overline{q}h}H_{\overline{q}(}1-\alpha h)q=\alpha_{q}H_{\overline{q}(1\alpha_{q}h)}h-=O$

and$H_{\overline{q}}-\alpha_{q\overline{q}h}H=H_{\overline{q}(1-\alpha h)q}=O$

the assumption and hence
Theorem 2.
$\mathrm{R}\mathrm{e}\mathrm{m}\mathrm{a}\mathrm{r}_{\tau}\mathrm{k}$. Conversely$H_{\overline{q}h}=H_{\varphi}\neq O$$( \frac{1}{\alpha_{q}}H_{\overline{q}})^{2}=\frac{1}{\alpha_{q}}H_{\overline{q}}$$H_{\varphi}=H_{\overline{q}h}= \frac{1}{\alpha_{q}}H_{\overline{q}}$. The concrete value oftion, as follows: Since

by Theorem 1 because$\alpha_{q}$$H_{\overline{q}}H^{2}=\{\mathbb{C}(1-\overline{\lambda}_{\mathcal{Z}})-1\}$$H_{\overline{q}}(1-\overline{\lambda}z)-1=\alpha_{q}(1-\overline{\lambda}z)-1$$(1-\overline{\lambda}\overline{z})^{-}1(1-\overline{\lambda}z)-1$$\infty$$\infty$$\infty$$\infty$$= \sum\sum\overline{\lambda}^{m+n_{\overline{Z}^{m}Z}}+1n-\sum\sum\lambda\overline{\lambda}^{m+n_{\overline{\mathcal{Z}}^{m}}}z^{n}$$m=0n=0$$m=0n=0$$= \sum\infty\sum\infty\overline{\lambda}^{2n+k}\overline{z}^{k+}1-\sum\infty\sum\infty\lambda\overline{\lambda}^{2n+k}\overline{z}^{k}$$n=0k=0$$+$$n=0k=1$(analytic part),$H_{\overline{q}}(1-\overline{\lambda}z)-1$$=J(I-P)(_{\overline{Z}}-\lambda)$byin Theorem 2 is given, by the direct calculabecause

and since$(\overline{z}-\lambda)$by$(1-\overline{\lambda}\overline{z})^{-}1(1-\overline{\lambda}z)-1$$0$$= \sum_{n=0}\sum_{k=0}\overline{\lambda}^{\angle n+\kappa}z^{k}-\sum_{n=0}\sum_{k=1}\lambda\overline{\lambda}^{zn+\kappa}\mathcal{Z}k-1$$=(1-|\lambda|^{2})(1-\overline{\lambda})^{-1}2(1-\overline{\lambda}z)-1$.