Bergman inner functions and $m$-hypercontractions

TL;DR

This paper characterizes Bergman inner functions in the context of $m$-hypercontractions, extending classical results from the unit disc to the unit ball in complex space.

Contribution

It provides a new operator-theoretic representation of $K_m$-inner functions associated with $m$-hypercontractions on the unit ball.

Findings

$K_m$-inner functions are characterized explicitly.

Extension of Olofsson's results from the unit disc to the unit ball.

Operator matrix conditions for $m$-hypercontractions are established.

Abstract

Let $H_m(\mathbb B,\mathcal D)$ be the $\mathcal D$-valued functional Hilbert space with reproducing kernel $K_m(z,w) = (1-\langle z,w\rangle)^{-m}1_{\mathcal D}$. A $K_m$-inner function is by definition an operator-valued analytic function $W: \mathbb B \rightarrow L(\mathcal E, \mathcal D)$ such that $\|Wx\|_{H_m(\mathbb B,\mathcal D)} = \|x\|$ for all $x \in \mathcal E$ and $(W\mathcal E) \perp M_z^{\alpha}(W\mathcal E)$ for all $\alpha \in \mathbb N^n \setminus \{0\}$. We show that the $K_m$-inner functions are precisely the functions of the form $W(z) = D + C \sum^m_{k=1}(1 - ZT^*)^{-k}ZB$, where $T \in L(H)^n$ is a pure $m$-hypercontraction and the operators $T^*, B, C,D$ form a $2 \times 2$-operator matrix satisfying suitable conditions. Thus we extend results proved by Olofsson on the unit disc to the case of the unit ball $\mathbb B \subset \mathbb C^n$.

Equations

$$H^n\to H,(x_i{)}_{1\le i\le n}\mapsto \sum _{i=1}^n{T}_i{x}_i$$

$$K:\mathbb{B}\times \mathbb{B}\to L(\mathcal{D}),(z,w)\mapsto \frac{1_{\mathcal{D}}}{1-⟨z,w⟩},$$

$$\mathrm{SOT}-\underset{k\to \infty }{\lim }{\sigma }_T^k(1_H)=0,$$

$${\theta }_T(z)=-T+D_{T^{\ast }}(1_H-Z{T}^{\ast })Z{D}_T$$

$${\mathbb{H}}_T=H(\mathbb{B},{\mathcal{D}}_{T^{\ast }})\ominus {\theta }_{T}H(\mathbb{B},{\mathcal{D}}_T).$$

$$K_m:\mathbb{B}\times \mathbb{B}\to L(\mathcal{D}),(z,w)\mapsto \frac{1_{\mathcal{D}}}{(1-⟨z,w⟩{)}^m}.$$

$$(1-{\sigma }_T{)}^k(1_H)\ge 0(k=1,\dots ,m).$$

$${\theta }_T{\mathcal{D}}_T\perp M_z^k({\theta }_T{\mathcal{D}}_T)(k\ge 1).$$

$${\int }_{\mathbb{D}}(|f{|}^2-1)z^k\mathrm{d}z=0(k\ge 0).$$

$$f\mathbb{C}\perp M_z^k(f\mathbb{C})(k\ge 1).$$

$$W{\mathcal{E}}_{\ast }\perp M_z^k(W{\mathcal{E}}_{\ast })(k\ge 1).$$

$${\sigma }_T:L(H)\to L(H),X\mapsto \sum _{i=1}^n{T}_{i}X{T}_i^{\ast }$$

$${\Delta }_T^{(k)}=(1-{\sigma }_T{)}^k(1_H)=\sum _{j=0}^k(-1{)}^j\left(\genfrac{}{}{0ex}{}{k}{j}\right)\sum _{|\alpha |=j}{\gamma }_{\alpha }{T}^{\alpha }{T}^{\ast \alpha }(k\in \mathbb{N}),$$

$${\Delta }_T^{(1)}\ge 0\text{mbox}and{\Delta }_T^{(m)}\ge 0.$$

$$\mathrm{SOT}-\underset{k\to \infty }{\lim }{\sigma }_T^k(1_H)=0.$$

$${\rho }_{\ell }(\alpha )=\frac{(\ell +|\alpha |-1)!}{\alpha !(\ell -1)!}.$$

$$H\to {\ell }^2({\mathbb{N}}^n,\mathcal{D}),x\mapsto ({\rho }_m(\alpha {)}^{1/2}C{T}^{\ast \alpha }x{)}_{\alpha \in {\mathbb{N}}^n}$$

$$(S_j^{(m)}x)(\alpha )=(\frac{{\rho }_m(\alpha )}{{\rho }_m(\alpha +e_j)}{)}^{1/2}x(\alpha +e_j)(\alpha \in {\mathbb{N}}^n,j=1,\dots ,n)$$

$$K_{\ell }=K_{\ell }^{\mathcal{E}}:\mathbb{B}\times \mathbb{B}\to L(\mathcal{E}),K_{\ell }(z,w)=\frac{1_{\mathcal{E}}}{(1-⟨z,w⟩{)}^{\ell }}$$

$$H_{\ell }(\mathbb{B},\mathcal{E})=\{f=\sum _{\alpha \in {\mathbb{N}}^n}{f}_{\alpha }{z}^{\alpha }\in \mathcal{O}(\mathbb{B},\mathcal{E});\Vert f{\Vert }^2=\sum _{\alpha \in {\mathbb{N}}^n}\frac{\Vert f_{\alpha }{\Vert }^2}{{\rho }_{\ell }(\alpha )}<\infty \}$$

$$U:{\ell }^2({\mathbb{N}}^n,\mathcal{D})\to H_m(\mathbb{B},\mathcal{D}),(x_{\alpha })\mapsto \sum _{\alpha \in {\mathbb{N}}^n}{\rho }_m(\alpha {)}^{1/2}{x}_{\alpha }{z}^{\alpha }$$

$$U{S}_j^{(m)\ast }(x_{\alpha })=\sum _{\begin{array}{c}\alpha \in {\mathbb{N}}^n\\ {\alpha }_j\ge 1\end{array}}{\rho }_m(\alpha -e_j{)}^{1/2}{x}_{\alpha -e_j}{z}^{\alpha }=M_{z_j}U(x_{\alpha }).$$

$$j(x)=\sum _{\alpha \in {\mathbb{N}}^n}{\rho }_m(\alpha )(C{T}^{\ast \alpha }x)z^{\alpha }=C(1_H-Z{T}^{\ast }{)}^{-m}x$$

$$M=H_m(\mathbb{B},\mathcal{D})\ominus (\mathrm{Im}j)\subset H_m(\mathbb{B},\mathcal{D})$$

$$W(M)=M\ominus (\sum _{i=1}^n{M}_{z_i}M)$$

$$M_z{H}_m(\mathbb{B},\mathcal{D}{)}^n=\{f\in H_m(\mathbb{B},\mathcal{D});f(0)=0\},$$

$$\delta (\sum _{k=0}^{\infty }\sum _{|\alpha |=k}{f}_{\alpha }{z}^{\alpha })=f_0+\sum _{k=1}^{\infty }(\frac{m+k-1}{k}\sum _{|\alpha |=k}{f}_{\alpha }{z}^{\alpha }).$$

$$H_m(\mathbb{B},\mathcal{D})={\oplus }_{k=0}^{\infty }{\mathbb{H}}_k(\mathcal{D})$$

$$(M_z^{\ast }{M}_z{)}^{-1}(M_z^{\ast }f)=M_z^{\ast }\delta f.$$

$$H_m(\mathbb{B},\mathcal{D}{)}^n\stackrel{\delta M_z}{⟶}{H}_m(\mathbb{B},\mathcal{D})$$

Full text

[TABLE]

Im particular, the row operator

[TABLE]

defines a continuous linear extension of the operator

[TABLE]

[TABLE]

for $g=\sum_{\alpha\in\mathbb{N}^{n}}g_{\alpha}z^{\alpha}\in H_{m}(\mathbb{B},\mathcal{D})$ and $i=1,\ldots,n$. Let $f\in H_{m}(\mathbb{B},\mathcal{D})$ be arbitrary. To prove the first assertion we may suppose that $f(0)=0$. It is easy to see and well known (see e.g. Section 3 in [1]) that in this case $f=\sum^{n}_{i=1}z_{i}f_{i}$ with

[TABLE]

for $i=1,\ldots,n$. But then

[TABLE]

Since $M_{z}M^{*}_{z}\in L(H_{m}(\mathbb{B},\mathcal{D}))$ is a diagonal operator, we have $M_{z}(M^{\ast}_{z}M_{z})^{-1}M^{\ast}_{z}=M_{z}M^{\ast}_{z}\delta=\delta M_{z}M^{\ast}_{z}$. Hence also the second assertion follows. $\Box$

Let $P_{\mathcal{D}}\in L(H_{m}(\mathbb{B},\mathcal{D}))$ be the orthogonal projection onto the closed subspace consisting of all constant functions. Then ${\rm Ker}\ P_{\mathcal{D}}=\{f\in H_{m}(\mathbb{B},\mathcal{D});f(0)=0\}$.

[TABLE]

[TABLE]

for the tuples consisting of the left and right multiplication operators

[TABLE]

By Lemma 1.2 in [2] the projection $P_{\mathcal{D}}$ is given by

[TABLE]

where $C(z,w)=K_{m}(z,\overline{w})$ and $(1/C)(z,w)=(1-\langle z,\overline{w}\rangle)^{m}$ is regarded as the analytic polynomial

[TABLE]

Thus the assertion follows. $\Box$

A more elementary, alternative representation of the orthogonal projection $P_{\mathcal{D}}$ follows from the observation that the operator $M_{z}(M^{\ast}_{z}M_{z})^{-1}M^{\ast}_{z}\in L(H_{m}(\mathbb{B},\mathcal{D}))$ is the orthogonal projection onto the space ${\rm Im}\ M_{z}=\{f\in H_{m}(\mathbb{B},\mathcal{D});\ f(0)=0\}={\rm Ker}\ P_{\mathcal{D}}$. The above preparations allow us to deduce a first description of the wandering subspace $W(M)=M\ominus(\sum^{n}_{i=1}M_{z_{i}}M)$.

[TABLE]

for some vectors $f_{0}\in\mathcal{D},x_{1},\ldots,x_{m}\in H$ with $(jx_{i})^{n}_{i=1}\in M^{\ast}_{z}H_{m}(\mathbb{B},\mathcal{D})$ and

[TABLE]

In this case, the identity $(jx_{i})^{n}_{i=1}=M^{*}_{z}f$ holds.

[TABLE]

Using the definition of the isometry $j:H\rightarrow H_{m}(\mathbb{B},\mathcal{D})$ we find that

[TABLE]

for all $y\in H$ and $x\in\mathcal{D}$ and hence that $j^{\ast}x=Cx$ for $x\in\mathcal{D}$ regarded as a constant function in $H_{m}(\mathbb{B},\mathcal{D})$. Using Lemma 1 and the intertwining properties of $j$ we obtain that

[TABLE]

A straightforward calulation, or the results in Section 3 of [1], show that

[TABLE]

for $z,w\in\mathbb{B}$, $x\in\mathcal{D}$ and $i=1,\ldots,n$. Using Lemma 2 we find that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

for $z,w\in\mathbb{B}$, $x\in\mathcal{D}$ and $i=1\ldots,n$. By comparing the previous two results and using the fact that the closed linear span of the functions $K_{m}(\cdot,w)x$ $(w\in\mathbb{B},x\in\mathcal{D})$ is of of $H_{m}(\mathbb{B},\mathcal{D})$, we obtain

[TABLE]

Conversely, suppose that $f\in H_{m}(\mathbb{B},\mathcal{D})$ is a function that has a representation as described in Theorem 3. We show that $f\in W(M)$. Note first that

[TABLE]

and therefore $jj^{\ast}M^{\ast}_{z_{i}}f=jx_{i}=M^{\ast}_{z_{i}}f$ for $i=1,\ldots,n$. Then exactly as in the first part of the proof it follows that

[TABLE]

Thus also the reverse implication follows. $\Box$

For future use, note that in the above proof we deduced the formula

[TABLE]

The defect operators $\Delta^{(k)}_{T}=(1-\sigma_{T})^{k}(1_{H})$ can be used to rewrite the conditions used in Theorem 3 to characterize the functions in $W(M)$.

[TABLE]

[TABLE]

and $\Sigma_{0}=0$. Then

[TABLE]

for $k=0,\ldots,m-1$. Hence $\sum^{m-1}_{k=0}\Delta^{(k)}_{T}=\Sigma_{m}$. $\Box$

An inspection of the proof of Theorem 3 allows us to calculate the norms of the functions $f\in W(M)$ in terms of the data $f_{0},x_{1},\ldots,x_{m}$ occurring in their standard representation.

[TABLE]

be a representation of a function $f\in W(M)$ as in Theorem 3. Then

[TABLE]

[TABLE]

As observed in Theorem 3 $(jx_{i})^{n}_{i=1}=M^{\ast}_{z}f$. Exactly as in the proof of Theorem 3, it follows that the second term in the above sum is given by

[TABLE]

[TABLE]

Thus the assertion follows. $\Box$

Let $T\in L(H)^{n}$ be an $m$-hypercontraction of class $C_{\cdot 0}$. Define $D_{T,k}=(\Delta^{(k)}_{T})^{1/2}\in L(H)$ for $k=0,\ldots,m-1$. Then

[TABLE]

defines a scalar product on $H$ such that the induced norm $\|x\|_{m}=(x,x)^{1/2}$ is equivalent to the original norm of $H$. We write $\tilde{H}$ for $H$ equipped with this norm. Then $I_{m}:H\rightarrow\tilde{H},x\mapsto x$, defines an invertible bounded operator such that

[TABLE]

Hence $I^{\ast}_{m}:\tilde{H}\rightarrow H$ acts as $I^{\ast}_{m}x=\sum^{m-1}_{k=0}\Delta^{(k)}_{T}x$. We write $\tilde{T}=(\tilde{T_{1}},\ldots,\tilde{T_{n}}):\tilde{H}^{n}\rightarrow H$ for the row operator with components $\tilde{T_{i}}=T_{i}\circ I^{\ast}_{m}\in L(\tilde{H},H)$. Then

[TABLE]

and hence $\tilde{T}$ is a contraction. As in [14] we shall use its defect operators

[TABLE]

and defect spaces $\mathcal{D}_{\tilde{T}^{\ast}}=\overline{D_{\tilde{T}^{\ast}}H}=\mathcal{D},\mathcal{D}_{\tilde{T}}=\overline{D_{\tilde{T}}(\tilde{H}^{n})}$ to deduce an alternative description of the wandering subspace $W(M)$. Exactly as in the single-variable theory of contractions it follows that $\tilde{T}D_{\tilde{T}}=D_{\tilde{T}^{\ast}}\tilde{T}$ and that

[TABLE]

is a well-defined unitary operator. In the following we construct an analytically parametrized family of operators $W_{T}(z)\in L(\tilde{\mathcal{D}},\mathcal{D})$ on the closed subspace

[TABLE]

such that

[TABLE]

where $W_{T}x:\mathbb{B}\rightarrow\mathcal{D}$ acts as $(W_{T}x)(z)=W_{T}(z)x.$

[TABLE]

In this case, we have $\|f\|^{2}=\|y\|^{2}_{\tilde{H}^{n}}$.

[TABLE]

where $f_{0}\in\mathcal{D},x_{1},\ldots,x_{n}\in H$ are vectors with $(jx_{i})^{n}_{i=1}\in M^{\ast}_{z}H_{m}(\mathbb{B},\mathcal{D})$ and

[TABLE]

Then $y=D_{\tilde{T}}(I_{m}x_{i})^{n}_{i=1}-\tilde{T}^{\ast}f_{0}\in\mathcal{D}_{\tilde{T}}$ is a vector with

[TABLE]

or equivalently, with

[TABLE]

But then $y\in\tilde{\mathcal{D}}$ and

[TABLE]

Conversely, if $f$ is a function of this form for some vector $y\in\tilde{\mathcal{D}}$, then by reversing the above arguments, one finds that

[TABLE]

are vectors with $(jx_{i})^{n}_{i=1}\in M^{\ast}_{z}H_{m}(\mathbb{B},\mathcal{D})$ and

[TABLE]

Hence $f\in W(M)$ by Theorem 3. By Lemma 4 and Lemma 5,

[TABLE]

This observation completes the proof. $\Box$

We define an analytic operator-valued function $W_{T}:\mathbb{B}\rightarrow L(\tilde{\mathcal{D}},\mathcal{D})$ by

[TABLE]

Note that the first term in the defining sum on the right equals

[TABLE]

[TABLE]

[TABLE]

to obtain the representations

[TABLE]

for $z\in\mathbb{B}$. A simple calculation shows that

[TABLE]

for $j\in\mathbb{N}$ and $\alpha\in\mathbb{N}^{n}$ with $|\alpha|=j$. Let $D\in L(H_{m}(\mathbb{B},\mathcal{D}))$ be the diagonal operator used in Lemma 1. Then

[TABLE]

[TABLE]

for $z\in\mathbb{B}$ and $(x_{i})^{n}_{i=1}\in H^{n}$. If $(x_{i})^{n}_{i=1}\in H^{n}$ is a tuple with $(jx_{i})^{n}_{i=1}\in{\rm Im}M^{\ast}_{z}$, then by Lemma 1

[TABLE]

Comparing the previous two results we obtain the identity

[TABLE]

for every vector $x\in\tilde{\mathcal{D}}$. Thus an application of Lemma 6 completes the proof. $\Box$

We briefly indicate an alternative description of the defect space $\tilde{\mathcal{D}}$. For a commuting tuple $S\in L(H)^{n}$, let $H^{p}(S,H)={\rm Ker}\ \delta^{p}_{S}/{\rm Im}\ \delta^{p-1}_{S}$ $(p=0,\ldots,n)$ be the cohomology groups of its Koszul complex (cf. Chapter 2 in [10])

[TABLE]

Then $H^{p}(M_{z},H_{m}(\mathbb{B},\mathcal{D}))=0$ for $p<n$ and $H^{n}(M_{z},H_{m}(\mathbb{B},\mathcal{D}))\cong\mathcal{D}$. By duality we obtain that $H^{0}(M^{\ast}_{z},H_{m}(\mathbb{B},\mathcal{D}))\cong\mathcal{D}$ and $H^{p}(M^{\ast}_{z},H_{m}(\mathbb{B},\mathcal{D}))=0$ for $p>0$.

[TABLE]

[TABLE]

it follows that $x\in{\rm Ker}\ \delta^{1}_{T^{\ast}}$. Conversely, if $x\in{\rm Ker}\ \delta^{1}_{T^{\ast}}$, then

[TABLE]

Since $H^{1}(M^{\ast}_{z},H_{m}(\mathbb{B},\mathcal{D}))=0$, it follows that $jx\in M^{*}_{z}H_{m}(\mathbb{B},\mathcal{D})$ and hence that $y\in\tilde{\mathcal{D}}$. $\Box$

Since in the setting of Theorem 7 the space $W_{T}(\tilde{\mathcal{D}})=W(M)$ is a wandering subspace for $M_{z}|_{M}\in L(M)^{n}$, it follows that $W_{T}:\mathbb{B}\rightarrow L(\tilde{\mathcal{D}},\mathcal{D})$ is an operator-valued analytic function with $W_{T}(x)\in H_{m}(\mathbb{B},\mathcal{D}),\|W_{T}x\|=\|x\|$ for all $x\in\tilde{\mathcal{D}}$ and

[TABLE]

This means precisely that $W_{T}:\mathbb{B}\rightarrow L(\tilde{\mathcal{D}},\mathcal{D})$ is a $K_{m}$-inner function in the sense of [6]. As an application of Theorem 6.2 in [6] we obtain that $W_{T}$ is a contractive multiplier from $H_{1}(\mathbb{B},\tilde{\mathcal{D}})$ to $H_{m}(\mathbb{B},\mathcal{D})$.

[TABLE]

For $m=1$, the set of $m$-hypercontractions $T\in L(H)^{n}$ of class $C_{\cdot 0}$ coincides with the class of all pure row contractions. For a pure row contraction $T\in L(H)^{n}$, the operator-valued function

[TABLE]

is called the characteristic function of $T$. It is known [7, 8] that $\theta_{T}$ induces a partially isometric multiplication operator

[TABLE]

such that $M_{\theta_{T}}M^{\ast}_{\theta_{T}}+jj^{\ast}=1_{H_{1}(\mathbb{B},\mathcal{D}_{T^{\ast}})}$. In particular, $M=({\rm Im}j)^{\perp}={\rm Im}M_{\theta_{T}}$.

[TABLE]

[TABLE]

and the $K_{1}$-inner function $W_{T}$ acts as

[TABLE]

Hence $\tilde{\mathcal{D}}\rightarrow W(M),x\mapsto\theta_{T}x$, is a unitary operator by Theorem 7. $\Box$

An elementary exercise shows (cf. Theorem 6.6 in [6]) that, in the setting of the last corollary, the space $\tilde{\mathcal{D}}$ is also given by

[TABLE]

§3   $K_{m}$-Inner functions

Let $T\in L(H)^{n}$ be an $m$-hypercontraction of class $C_{\cdot 0}$ and let $W_{T}:\mathbb{B}\rightarrow L(\tilde{\mathcal{D}},\mathcal{D})$,

[TABLE]

be the associated $K_{m}$-inner function. If we define

[TABLE]

then the above representation of $W_{T}$ becomes

[TABLE]

The operators $T\in L(H^{n},H),B\in L(\tilde{\mathcal{D}},H^{n}),C\in L(H,\mathcal{D})$ and $D\in L(\tilde{\mathcal{D}},\mathcal{D})$ satisfy the conditions

[TABLE]

Here $P_{\tilde{\mathcal{D}}}\in L(\tilde{H}^{n},\tilde{\mathcal{D}})$ denotes the orthogonal projection onto $\tilde{\mathcal{D}}$. Using the definition of the set $\tilde{\mathcal{D}}\subset\tilde{H}^{n}$ we obtain the additional condition

[TABLE]

Our next aim is to show that the $K_{m}$-inner functions $W:\mathbb{B}\rightarrow L(\mathcal{E}_{\ast},\mathcal{E})$ between arbitrary Hilbert spaces $\mathcal{E}_{\ast}$ and $\mathcal{E}$ are precisely the operator-valued functions on $\mathbb{B}$ which possess a representation of the form

[TABLE]

where $T\in L(H)^{n}$ is an $m$-hypercontraction of class $C_{\cdot 0}$ on a suitable Hilbert space $H$ and the coefficients of the matrix operator

[TABLE]

satisfy the conditions

[TABLE]

where $j_{C}:H\rightarrow H_{m}(\mathbb{B},\mathcal{E})$ is the operator defined by

[TABLE]

Note that by condition $({\rm KI}1)$ the operator $j_{C}$ is a well-defined isometry. It is easily seen that $j_{C}$ intertwines the tuples $T^{\ast}\in L(H)^{n}$ and $M^{\ast}_{z}\in L(H_{m}(\mathbb{B},\mathcal{E}))^{n}$ componentwise. The following results extend characterizations of $K_{m}$-inner functions proved by Olofsson in [15] on the unit disc to the case of the unit ball in $\mathbb{C}^{n}$.

[TABLE]

where $T\in L(H)^{n}$ is an $m$-hypercontraction of class $C_{\cdot 0}$ and the matrix operator

[TABLE]

satisfies the conditions $({\rm KI}1)-({\rm KI}4)$. Then $W$ is a $K_{m}$-inner function.

[TABLE]

It follows that

[TABLE]

[TABLE]

By the remark following Theorem 3 we have

[TABLE]

Hence we obtain that

[TABLE]

[TABLE]

[TABLE]

where the last equality follows from condition $({\rm KI}3)$ together with Lemma 4. Since $M_{z}H_{m}(\mathbb{B},\mathcal{E})^{n}=H_{m}(\mathbb{B},\mathcal{E})\ominus\mathcal{E}$, we conclude that the map

[TABLE]

is a well-defined isometry.

Repeating the above calculations we obtain that

[TABLE]

and hence that

[TABLE]

Similarly, for $x\in\mathcal{E}_{\ast}$ and $f\in H_{m}(\mathbb{B},\mathcal{E})$ as above, we find that

[TABLE]

[TABLE]

[TABLE]

where the last identity follows from condition $({\rm KI}2)$. Thus we have shown that $W\mathcal{E}_{\ast}\subset M\ominus\sum^{n}_{i=1}z_{i}M$. Hence

[TABLE]

for all $\alpha\in\mathbb{N}^{n}\setminus\{0\}$. Thus the proof is complete. $\Box$

[TABLE]

such that its coefficients satisfy the conditions $({\rm KI}1)-({\rm KI}4)$ and such that

[TABLE]

[TABLE]

It is elementary to check and well known that the compression $T=P_{H}M_{z}|_{H}$ of $M_{z}$ to the co-invariant subspace $H=H_{m}(\mathbb{B},\mathcal{E})\ominus\mathcal{S}$ is an $m$-hypercontraction of class $C_{\cdot 0}$. Let $\mathcal{R}\subset H_{m}(\mathbb{B},\mathcal{E})$ be the smallest reducing subspace for $M_{z}\in L(H_{m}(\mathbb{B},\mathcal{E}))^{n}$ with $H\subset\mathcal{R}$. Then

[TABLE]

and the inclusion $i:H\hookrightarrow H_{m}(\mathbb{B},\mathcal{R}\cap\mathcal{E})$ is a minimal $m$-dilation for $T$. Let $j:H\rightarrow H_{m}(\mathbb{B},\mathcal{D})$ be the canonical $m$-dilation of $T$ constructed at the beginning of Section 2. It follows from Section 4 in [12] that there is a unitary operator $U:\mathcal{D}\rightarrow\mathcal{R}\cap\mathcal{E}$ such that

[TABLE]

Set $\hat{\mathcal{E}}=\mathcal{E}\ominus(\mathcal{R}\cap\mathcal{E})$. By definition the space

[TABLE]

is the largest reducing subspace for $M_{z}\in L(H_{m}(\mathbb{B},\mathcal{E}))^{n}$ which is contained in $\mathcal{S}$. In particular, one obtains the orthogonal decomposition

[TABLE]

Define

[TABLE]

and $W(M)=M\ominus(\sum^{n}_{i=1}z_{i}M)$. Using the above commutative diagram we find that

[TABLE]

defines a unitary operator which intertwines the restrictions of $M_{z}$ to both sides componentwise. Consequently, we obtain the orthogonal decomposition

[TABLE]

Let $W_{T}:\mathbb{B}\rightarrow L(\tilde{\mathcal{D}},\mathcal{D})$ be the $K_{m}$-inner function associated with $T\in L(H)^{n}$ as in Theorem 7 (see also the beginning of Section 3). Then there are bounded operators $B\in L(\tilde{\mathcal{D}},H^{n}),\ C\in L(H,\mathcal{D})$ and $D\in L(\tilde{\mathcal{D}},\mathcal{D})$ such that

[TABLE]

and $W(M)=\{W_{T}(x);\ x\in\tilde{\mathcal{D}}\}$. Let us denote by

[TABLE]

the orthogonal projections with respect to the above orthogonal decomposition of $\mathcal{W}$. The $K_{m}$-inner functions $W:\mathbb{B}\rightarrow L(\mathcal{E}_{\ast},\mathcal{E})$ and $W_{T}:\mathbb{B}\rightarrow L(\tilde{\mathcal{D}},\mathcal{D})$ induce unitary operators

[TABLE]

Using these unitary operators we construct bounded linear operators

[TABLE]

and

[TABLE]

By construction the column operator $(U_{1},U_{2}):\mathcal{E}_{\ast}\rightarrow\hat{\mathcal{E}}\oplus\tilde{\mathcal{D}}$ is an isometry such that

[TABLE]

[TABLE]

for $z\in\mathbb{B}$ and $x\in\mathcal{E}_{\ast}$. To complete the proof it suffices to check that the operators $T\in L(H^{n},H),\ \tilde{B}=BU_{2}\in L(\mathcal{E}_{\ast},H^{n}),\ \tilde{C}=UC\in L(H,\mathcal{E})$ and $\tilde{D}=U_{1}+UDU_{2}\in L(\mathcal{E}_{\ast},\mathcal{E})$ satisfy the conditions $({\rm KI}1)-({\rm KI}4)$. Obviously,

[TABLE]

and

[TABLE]

[TABLE]

To check condition $({\rm KI}3)$ note that $\tilde{D}$ acts as the column operator

[TABLE]

Thus we obtain that

[TABLE]

Since $j_{\tilde{C}}=Uj_{C}$, it follows that

[TABLE]

for all $x\in\mathcal{E}_{\ast}$. Thus also condition $({\rm KI}4)$ holds. This observation completes the proof. $\Box$

§4   Characteristic functions

In the previous section we saw that, for a pure row contraction $T\in L(H)^{n}$, the associated $K_{1}$-inner function $W_{T}:\mathbb{B}\rightarrow L(\tilde{\mathcal{D}},\mathcal{D})$ is obtained by restricting its characteristic function

[TABLE]

to a suitable subspace of $\mathcal{D}_{T}$. More precisely,

[TABLE]

where $\tilde{D}\subset D_{T}$ is the closed subspace given by

[TABLE]

In the one-dimensional case $n=1$, we even have the identities $\tilde{\mathcal{D}}=\mathcal{D}_{T}$ and $W_{T}=\theta_{T}$. Thus it seens natural to ask whether also for $m>1$ there is a canonically defined characteristic function for each $m$-hypercontraction $T\in L(H)^{n}$ that extends the function $W_{T}:\mathbb{B}\rightarrow L(\tilde{\mathcal{D}},\mathcal{D})$ in a natural way. In the present section we offer a possible answer to this question.

Let $T\in L(H)^{n}$ be an $m$-hypercontraction of class $C_{\cdot 0}$. We denote by

[TABLE]

its first-order defect operators and by

[TABLE]

the associated defect spaces. As in Section 2 we write $C=(\Delta^{(m)}_{T})^{1/2}\in L(H)$ and $\mathcal{D}=\overline{CH}\subset H$ for the $m$-th order defect operator and defect space of $T$.

For $k=0,\ldots,m$ the operator

[TABLE]

is a well-defined contraction. To check this, it suffices to observe that

[TABLE]

[TABLE]

for all $x\in H$. Here the above estimate follows from the fact that (Corollary 3 in [13])

[TABLE]

The inequalities

[TABLE]

imply in particular that

[TABLE]

An elementary calculation yields that

[TABLE]

As a consequence we find that

[TABLE]

[TABLE]

or equivalently,

[TABLE]

Thus we obtain the identities

[TABLE]

for $k=0,\ldots,m$.

As before we write $j=j_{m}:H\rightarrow H_{m}(\mathbb{B},\mathcal{D})$. To simplify the notation we define $\pi=j_{m-1}:H\rightarrow H_{m-1}(\mathbb{B},\mathcal{D})$. Because of

[TABLE]

there is a unique unitary operator $U:\mathcal{D}_{T^{\ast}}\rightarrow\mathcal{K}=\overline{{\rm Im}\pi}$ with $UD_{T^{\ast}}x=\pi x$ for all $x\in H$. We denote by

[TABLE]

the point evaluations. Define $M=H_{m-1}(\mathbb{B},\mathcal{D})\ominus({\rm Im}\pi)$. Then

[TABLE]

where $i_{M}:M\hookrightarrow H_{m-1}(\mathbb{B},\mathcal{D})$ is the inclusion mapping, is a unitary operator. Since

[TABLE]

for all $h\in H,z\in\mathbb{B}$ and $x\in\mathcal{D}$, it follows that $j^{\ast}:H_{m}(\mathbb{B},\mathcal{D})\rightarrow H$ is the unique bounded operator with

[TABLE]

In the case of row contractions, that is $m=1$, we have $\mathcal{D}=\mathcal{D}_{T^{\ast}}$, the map $\pi$ reduces to the mapping $\pi=C=D_{T^{\ast}}:H\rightarrow\mathcal{D}_{T^{\ast}}$ and $M=\{0\},U=1_{\mathcal{D}_{T^{\ast}}}$. Hence in this case, the operators

[TABLE]

both collapse to the identity operator on $\mathcal{D}_{T^{\ast}}$. Thus the only difference between the characteristic functions of row contractions and of $m$-hypercontractions as defined in the following theorem is caused by the presence of the space $M$.

[TABLE]

induces a partially isometric multiplier

[TABLE]

such that

[TABLE]

[TABLE]

obtained above shows that the column operator

[TABLE]

is an isometry. Define $L=\mathcal{D}_{T}\oplus M$ and consider the operators

[TABLE]

An elementary, but tedious, computation shows that

[TABLE]

defines a unitary matrix operator. Using the identities

[TABLE]

one easily obtains the representations

[TABLE]

for all $z\in\mathbb{B}$. According to Proposition 1.2 in [10], the map

[TABLE]

defines a contractive multiplier from $H_{1}(\mathbb{B},L)$ to $H_{1}(\mathbb{B},H_{m-1}(\mathbb{B},\mathcal{D}))$ such that

[TABLE]

holds for all $z,w\in\mathbb{B}$. Since

[TABLE]

is a positive definite $L(\mathcal{D})$-valued function of $(z,w)\in\mathbb{B}\times\mathbb{B}$, the map

[TABLE]

induces a contractive multiplier from $H_{1}(\mathbb{B},H_{m-1}(\mathbb{B},\mathcal{D}))$ to $H_{m}(\mathbb{B},\mathcal{D})$. But then also the composition

[TABLE]

defines a contractive multiplier from $H_{1}(\mathbb{B},L)$ to $H_{m}(\mathbb{B},\mathcal{D})$.

To complete the proof, note that

[TABLE]

for $z,w\in\mathbb{B}$ and $x,y\in\mathcal{D}$. On the other hand,

[TABLE]

for all $z,w\in\mathbb{B}$ and $x,y\in\mathcal{D}$. Thus

[TABLE]

and the proof is complete. $\Box$

Since the matrix operator

[TABLE]

is unitary, its second column defines a unitary operator $\rho:L\rightarrow K$ onto the orthogonal complement $K$ of the image of its first column in $H^{n}\oplus H_{m-1}(\mathbb{B},\mathcal{D})$. Using the definitions of $B$ and $D$ we find that

[TABLE]

for $(x,f)\in L=\mathcal{D}_{T}\oplus(H_{m-1}(\mathbb{B},\mathcal{D})\ominus{\rm Im}\ j_{m-1})$. Let us denote by

[TABLE]

the projections of $K$ onto its first and second component. Then

[TABLE]

defines an analytic operator-valued function such that

[TABLE]

Hence also $M_{\varphi}:H_{1}(\mathbb{B},K)\rightarrow H_{m}(\mathbb{B},\mathcal{D})$ is a partially isometric multiplier and

[TABLE]

Using the above alternative characterization of $\theta_{T}$, one can show that $\theta_{T}$ is a purely contractive multiplier (cf. Section V.2 in [16] and Section 4 in [8]).

[TABLE]

for all non-zero vectors $x\in\mathcal{D}_{T}\oplus M$.

[TABLE]

So, if $(x,f)\in K$ is a vector with $\|\varphi(0)(x,f)\|\geq\|(x,f)\|$, then

[TABLE]

Hence $x=0$ and $\|f(0)\|=\|f\|_{H_{m-1}(\mathbb{B},\mathcal{D})}$. But then $f\equiv f(0)$ is a constant function. Since $(x,f)$ is orthogonal to the image of the column operator

[TABLE]

we conclude that $f\in H_{m-1}(\mathbb{B},\mathcal{D})\ominus{\rm Im}\ j_{m-1}={\rm Ker}\ j^{\ast}_{m-1}$. Since

[TABLE]

for all $h\in H$ and $z\in\mathcal{D}$, it follows that $j^{\ast}_{m-1}z=Cz$ for $z\in\mathcal{D}$. Thus we find that also

[TABLE]

This observation completes the proof. $\Box$