Proof Given any $z \in \mathbb {D}$ , since $(\mathscr {C} _\mu h) (z) := \left \langle {k_z} , {h} \right \rangle _{L^2 (\mu )}$ , consider the Leibniz difference quotient

$$ \begin{align*}\lim _{\epsilon \rightarrow 0} \frac{(\mathscr{C} _\mu h) (z +\epsilon ) -(\mathscr{C} _\mu h) (z)}{\epsilon } = \lim _{\epsilon \rightarrow 0} \epsilon ^{-1} \left\langle {k_{z+\epsilon } - k_z} , {h} \right\rangle_{L^2 (\mu )}.\end{align*} $$

(Here, recall that our inner products are conjugate linear in their first argument.) This limit will exist and $\mathscr {C} _\mu h$ will be holomorphic, if and only if the limit of $\overline {\epsilon } ^{-1} (k_{z+\epsilon } - k_z)$ exists in $H^2 (\mu )$ . This limit exists in supremum norm on the circle (and so belongs to $A (\mathbb {D} )$ ), and so it certainly exists in the $L^2 (\mu )$ -norm by Cauchy–Schwarz. Indeed,

$$ \begin{align*} \lim _{\epsilon \rightarrow 0} \overline{\frac{k_{z+\epsilon } (\zeta ) - k_z (\zeta )}{\overline{\epsilon }}} & = \lim \frac{1 - \overline{\zeta} z - (1 - \overline{\zeta} z - \overline{\zeta} \epsilon )}{\epsilon (1 - \overline{\zeta} z) (1-\overline{\zeta} (z +\epsilon ))} \\ & = \frac{\overline{\zeta}}{(1-\overline{\zeta} z) ^2}, \end{align*} $$

and this limit is continuous on $\partial \mathbb {D}$ for any fixed $z \in \mathbb {D}$ . Hence $\mathscr {C} _\mu h \in \mathscr {O} (\mathbb {D})$ for any $h \in H^2 (\mu )$ .

Proof To show that this inner product is well-defined, we need to check that $\mathscr {C} _\mu h \equiv 0$ in the disk implies that $h=0$ in $H^2 (\mu )$ . Indeed,

$$ \begin{align*}(\mathscr{C} _\mu h) (z) = \left\langle {k_z} , {h} \right\rangle_{H^2 (\mu )},\end{align*} $$

and since $\bigvee k_z$ is dense in $A (\mathbb {D})$ , the linear span of the Szegö kernels is also dense in $H^2 (\mu )$ as described above. Hence, this vanishes for all $z \in \mathbb {D}$ if and only if $h=0$ .

By definition, for any $z \in \mathbb {D}$ ,

$$ \begin{align*}(\mathscr{C} _\mu h) (z) = \left\langle {k_z} , {h} \right\rangle_{H^2 (\mu )} = \left\langle {\mathscr{C} _\mu k_z} , {\mathscr{C} _\mu h} \right\rangle_\mu,\end{align*} $$

so that $\mathscr {H} ^+ (\mu )$ is a RKHS in $\mathbb {D}$ with kernel vectors $k ^\mu _z := \mathscr {C} _\mu k_z$ and reproducing kernel

$$ \begin{align*}k^\mu (z,w) := \left\langle {\mathscr{C} _\mu k_z} , {\mathscr{C} _\mu k_w} \right\rangle_{L^2 (\mu )} = \int _{\partial \mathbb{D}} \frac{1}{1 -z \overline{\zeta}} \frac{1}{1-\overline{w} \zeta} \mu (d\zeta ).\end{align*} $$

Finally,

$$ \begin{align*} H_\mu (z) + \overline{H_\mu (w)} & = \int _{\partial \mathbb{D}} \frac{1 +z\overline{\zeta}}{1-z\overline{\zeta}} \mu (d\zeta ) + \int _{\partial \mathbb{D}} \frac{1 +\overline{w} \zeta}{1-\overline{w}\zeta} \mu (d\zeta ) \\ & = 2 (1 - z \overline{w} ) \int _{\partial \mathbb{D}} \frac{1}{1-z \overline{\zeta}} \frac{1}{1-\overline{w} \zeta} \mu (d\zeta ), \end{align*} $$

establishing the second formula.

Proof (Necessity) If $\mu \leq t^2 \lambda $ , then $\gamma := t^2 \lambda - \mu $ is a positive measure and

$$ \begin{align*} t^2 k^\lambda (z,w) - k^\mu (z,w) & = t^2 \int _{\partial \mathbb{D}} \frac{1}{1-z\overline{\zeta}}\frac{1}{1-\overline{w}\zeta} \lambda (d\zeta ) - \int _{\partial \mathbb{D}} \frac{1}{1-z\overline{\zeta}}\frac{1}{1-\overline{w}\zeta} \mu (d\zeta ) \\ & = \int _{\partial \mathbb{D}} \frac{1}{1-z\overline{\zeta}}\frac{1}{1-\overline{w}\zeta} \gamma (d\zeta ) = k^\gamma (z,w). \end{align*} $$

It follows that $t^2 k^\lambda - k^\mu = k^\gamma $ is a positive kernel so that $k^\mu \leq t^2 k^\lambda $ .

First proof of sufficiency. Conversely, suppose that $K:= t^2 k ^\lambda - k^\mu \geq 0$ is a positive kernel. View the analytic polynomials, $\mathbb {C} [ \zeta ]$ , as a dense subspace of the disk algebra $A (\mathbb {D} )$ , embedded isometrically in the Banach space $\mathscr {C} (\partial \mathbb {D} )$ . For any finite, positive and regular Borel measure on the complex unit circle, $\mu $ , we define the positive linear functional, $\hat {\mu }$ on $\mathscr {C} (\partial \mathbb {D})$ by

$$ \begin{align*}\hat{\mu} (f) := \int _{\partial \mathbb{D}} f d\mu.\end{align*} $$

(The map $\mu \mapsto \hat {\mu }$ is a bijection, by the Riesz–Markov theorem.) We then define a bounded linear functional, $\ell _K$ on $\mathscr {C} (\partial \mathbb {D})$ by $\ell _K := t^2 \hat {\lambda } - \hat {\mu }$ .

By Weierstraß approximation, $\mathbb {C} [ \zeta ] + \overline {\mathbb {C} [ \zeta ] }$ is supremum-norm dense in the continuous functions, $\mathscr {C} (\partial \mathbb {D} )$ . Since the Fejér kernel is positive semi-definite, the partial Cesàro sums of any positive semi-definite $f \in \mathscr {C} (\partial \mathbb {D} )$ will be a positive trigonometric polynomial, i.e., a positive semi-definite element of $\mathbb {C} [\zeta ] + \overline {\mathbb {C} [\zeta ]}$ and, by Fourier theory, it follows that the positive cone of $\mathbb {C} [ \zeta ] + \overline {\mathbb {C} [ \zeta ] }$ is supremum norm-dense in the positive cone of $\mathscr {C} (\partial \mathbb {D})$ . Moreover, by the Fejér–Riesz theorem, any positive trigonometric polynomial, $p + \overline {q} \geq 0$ , on $\partial \mathbb {D}$ factors as $|g| ^2$ for an analytic $g \in \mathbb {C} [\zeta ]$ (and necessarily, $p =q$ , $\mathrm {deg} (p) = \mathrm {deg} (g)$ ). Hence, to check that $\ell _K$ is a positive linear functional on $\mathscr {C} (\partial \mathbb {D} )$ , it suffices to check that $\ell _K ( p + \overline {p} ) \geq 0$ for any $p + \overline {p} = |g| ^2 \geq 0$ , $p, g \in \mathbb {C} [ \zeta ]$ . If $p = \sum _{j=0} ^n \hat {p} _j \zeta ^j$ and $g = \sum _{j=0} ^n \hat {g} _j \zeta ^j$ , then by construction

$$ \begin{align*}\left\langle {\mathscr{C} _\mu g} , {\mathscr{C} _\mu g} \right\rangle_\mu = \left\langle {g} , {g} \right\rangle_{H^2 (\mu) } = \left\langle {p} , {1} \right\rangle_{H^2 (\mu)} + \left\langle {1} , {p} \right\rangle_{H^2 (\mu )} = \left\langle {\mathscr{C} _\mu p} , {\mathscr{C} _\mu 1} \right\rangle_\mu + \left\langle {\mathscr{C} _\mu 1} , {\mathscr{C} _\mu p} \right\rangle_{\mu}.\end{align*} $$

Since

$$ \begin{align*}\mathscr{C} _\mu p = \sum _{j=0} ^n \hat{p} _j k ^\mu _j,\end{align*} $$

where the $k^\mu _j$ , $j \in \mathbb {N} \cup \{ 0 \}$ are the Taylor coefficient evaluation vectors, and since similar formulas hold for $\lambda $ , we obtain that

$$ \begin{align*}\ell _K (|g| ^2) = \int _{\partial \mathbb{D}} |g (\zeta ) | ^2 (t^2 \lambda (d\zeta ) - \mu (d\zeta ) ) = \sum _{i,j=0} ^n \overline{\hat{g} _i} \hat{g} _j \left\langle {K_i} , {K_j} \right\rangle_{\mathcal{H} (K)},\end{align*} $$

where the $K_i$ , $i \in \mathbb {N} \cup \{ 0 \}$ are the Taylor coefficient evaluation vectors in $\mathcal {H} (K)$ . Namely, $\mathcal {H} (K)$ is also a Taylor coefficient RKHS in $\mathbb {D}$ so that $\hat {K} (i , j) := \left \langle {K_i} , {K_j} \right \rangle _{\mathcal {H} (K)}$ defines a positive kernel function on the set $\mathbb {N} \cup \{ 0 \}$ . It follows that

$$ \begin{align*}\int _{\partial \mathbb{D}} |g (\zeta ) | ^2 (t^2 \lambda (d\zeta ) - \mu (d\zeta ) ) \geq 0,\end{align*} $$

for any $g \in \mathbb {C} [ \zeta ]$ , or, equivalently,

$$ \begin{align*}\ell _K (p + \overline{p}) = \int _{\partial \mathbb{D}} (p + \overline{p} ) (t^2 d\lambda - d\mu ) \geq 0,\end{align*} $$

for any positive semi-definite $p + \overline {p} \in \mathbb {C} [\zeta ] + \overline {\mathbb {C} [ \zeta ]}$ . By density of the positive cone of $\mathbb {C} [\zeta ] + \overline {\mathbb {C} [\zeta ] }$ in the positive cone of the continuous functions, it follows that $\ell _K$ is a bounded, positive linear functional on $\mathscr {C} (\partial \mathbb {D} )$ , with norm $\| \ell _K \| = \ell _K (1) = t^2 \lambda (\partial \mathbb {D} ) - \mu (\partial \mathbb {D} ) = K( 0, 0) \geq 0$ . By the Riesz–Markov theorem, there is then a unique, positive, finite, and regular Borel measure, $\gamma $ , on $\partial \mathbb {D}$ , so that

$$ \begin{align*}\ell _K (f) = \int _{\partial \mathbb{D}} f(\zeta ) \gamma (d\zeta ),\end{align*} $$

for any $f \in \mathscr {C} (\partial \mathbb {D} )$ , i.e., $\ell _K = \hat {\gamma }$ , and we conclude that $\gamma = t^2 \lambda - \mu \geq 0$ so that $t^2 \lambda \geq \mu $ .

Second proof of sufficiency. If $t^2 k^\lambda \geq k^\mu $ , then by Aronszajn’s inclusion theorem, $\mathscr {H} ^+ (\mu ) \subseteq \mathscr {H} ^+ (\lambda )$ and the norm of the embedding $\mathrm {e} _{\mu , \lambda } : \mathscr {H} ^+ (\mu ) \hookrightarrow \mathscr {H} ^+ (\lambda )$ is at most $t>0$ . Observe that $\mathrm {e} := \mathrm {e} _{\mu , \lambda }$ acts trivially as a multiplier by the constant function $1$ , so that

$$ \begin{align*}\mathrm{e} ^* k ^\lambda _z = k^\mu _z, \quad \mbox{and} \quad \mathrm{e} ^* k^\lambda _j = k^\mu _j,\end{align*} $$

for any $z \in \mathbb {D}$ and $j \in \mathbb {N} \cup \{ 0 \}$ . Hence,

$$ \begin{align*} \mathrm{e} ^* V^\lambda k^\lambda _z \overline{z} & = \mathrm{e} ^* ( k^\lambda _z - k ^\lambda _0 ) \\ & = k^\mu _z - k ^\mu _0 = V^\mu k^\mu _z \overline{z} \\ & = V^\mu \mathrm{e} ^* k^\lambda _z \overline{z}, \end{align*} $$

so that $\mathrm {e} ^*$ intertwines $V^\lambda $ with $V^\mu $ , $\mathrm {e} ^* V^\lambda = V^\mu \mathrm {e} ^*$ . Setting $E := \mathscr {C} _\mu ^* \mathrm {e} ^* \mathscr {C} _\lambda $ , we see that for any monomial,

$$ \begin{align*} E \zeta ^j & = \mathscr{C} _\mu ^* \mathrm{e} ^* k^\lambda _j \\ & = \mathscr{C} _\mu ^* k^\mu _j = \zeta ^j \in H^2 (\mu ), \end{align*} $$

so that $E : H^2 (\lambda ) \hookrightarrow H^2 (\mu )$ obeys $Ep =p$ for any $p \in \mathbb {C} [\zeta ] \subseteq H^2 (\lambda )$ . In particular, $E Z^\lambda = Z^\mu E$ . At this point one could argue using the Riesz–Markov theorem as above, however, here is an alternative argument. Since $Z^\lambda $ and $Z^\mu $ are contractions (they are isometries), we can apply the intertwining version of the commutant lifting theorem [Reference Paulsen19, Corollary 5.9] to conclude that E can be “lifted” to a bounded operator $\hat {E} : L^2 (\lambda ) \rightarrow L^2 (\mu )$ , with norm $\| \hat {E} \| = \| E \|$ , so that $\hat {E} M^\lambda _\zeta = M^\mu _\zeta \hat {E}$ , and $P _{H^2 (\mu )} \hat {E} | _{H ^2 (\lambda )} = E$ . Setting $\hat {T} := \hat {E}^* \hat {E} \in \mathscr {L} (L ^2 (\lambda ) )$ , $\hat {T} \geq 0$ is a positive semi-definite $M^\lambda _\zeta $ -Toeplitz operator in the sense that

$$ \begin{align*}M^{\lambda *} _\zeta \hat{T} M^\lambda _\zeta = \hat{E} ^* M^{\mu *} _\zeta M^\mu _\zeta \hat{E} = \hat{T},\end{align*} $$

since $M ^\lambda _\zeta , M^\mu _\zeta $ are isometries. Since, $M^\lambda _\zeta $ (and $M^\mu _\zeta $ ) is in fact unitary, it follows that $\hat {T}$ commutes with $M^\lambda _\zeta $ . Since $\mathbb {C} [\zeta ] + \overline {\mathbb {C} [\zeta ]} \subseteq \mathrm {Ran} \, M^\lambda _\zeta $ is a dense set in $L^2 (\lambda )$ , a simple argument shows that $\hat {T}$ acts as multiplication by $f:=T1 \in L^2 (\lambda )$ . However, since ${T = M_f}$ is a bounded and positive semi-definite operator, it is easy to check that $t ^2 \geq \| \hat {T} \| = \| f \| _{\infty }$ , $\| \hat {T} \| = \| \hat {E} \| ^2 = \| E \| ^2 = \| T \|$ , and $f \geq 0\ \lambda $ -a.e. Finally, one can also check that $T = P_{H^2 (\lambda )} \hat {T} | _{H^2 (\lambda )}$ is $Z^\lambda $ -Toeplitz, i.e., $Z^{\lambda *} T Z^\lambda =T$ . In conclusion, for any $p, q \in \mathbb {C} [\zeta ]$ ,

$$ \begin{align*} \int _{\partial \mathbb{D}} \overline{p (\zeta)} q (\zeta ) \mu (d\zeta ) & = \left\langle {p} , {q} \right\rangle_{H^2 (\mu )} = \left\langle {Ep} , {Eq} \right\rangle_{H^2 (\mu )} \\ & = \left\langle {\sqrt{T} p} , {\sqrt{T}q} \right\rangle_{H^2 (\lambda )} = \left\langle {p} , {\hat{T} q} \right\rangle_{L^2 (\lambda )} \\ & = \left\langle {p} , {M_f q} \right\rangle_{L ^2 (\lambda )} = \int _{\partial \mathbb{D}} \overline{p (\zeta )} q (\zeta) f(\zeta ) \lambda (d\zeta ). \end{align*} $$

This formula extends to elements of the form $g= \overline {p} + q \in \mathbb {C} [\zeta ] + \overline {\mathbb {C} [\zeta ]}$ since

$$ \begin{align*}\left\langle {\overline{p} _1 + q_1} , {\overline{p_2} + q_2} \right\rangle_{L^2 (\mu )} = \left\langle {q_1} , {q_2} \right\rangle_{H^2 (\mu )} + \left\langle {q_1 p_2} , {1} \right\rangle_{H^2 (\mu)} + \left\langle {p_2} , {p_1} \right\rangle_{H^2 (\mu )} + \left\langle {1} , {p_1 q_2} \right\rangle_{H^2 (\mu )}\!.\end{align*} $$

Again, by Weierstraß approximation, since $\mathbb {C} [\zeta ] + \overline {\mathbb {C} [\zeta ]}$ is supremum-norm dense in $\mathscr {C} (\partial \mathbb {D} )$ , which is in turn dense in $L^2 (\lambda )$ and $L^2 (\mu )$ , it follows that for any $g, h \in L^2 (\lambda )$ ,

$$ \begin{align*}\int _{\partial \mathbb{D}} \overline{g (\zeta )} h (\zeta ) \mu (d\zeta ) = \int _{\partial \mathbb{D}} \overline{g (\zeta )} h (\zeta ) f(\zeta ) \lambda (d\zeta ),\end{align*} $$

where $f \geq 0$ , $\lambda $ -a.e. and $\| f \| _\infty \leq t^2$ . We conclude that $\mu \leq t^2 \lambda $ and that

$$ \begin{align*}f = \frac{\mu (d\zeta )}{\lambda (d\zeta )},\end{align*} $$

is the (bounded) Radon–Nikodym derivative of $\mu $ with respect to $\lambda $ .

Proof We have that for all n, $0 \leq \mu _n \leq \mu $ and $\mu _n \uparrow \mu $ . Moreover, $\mathscr {H} ^+ (\mu _n ) \subseteq \mathrm {int} (\mu , \lambda )$ for all n. If $1 \geq g_n \geq 0, \mu -a.e.$ are the Radon–Nikodym derivatives of the $\mu _n$ with respect to $\mu $ , and $p \in \mathbb {C} [\zeta ]$ , let $\mathscr {C} _n := \mathscr {C} _{\mu _n}$ and let $\mathrm {e} _n : \mathscr {H} ^+ (\mu _n ) \hookrightarrow \mathscr {H} ^+ (\mu )$ . Then,

$$ \begin{align*} \| \mathscr{C} _\mu p - \mathrm{e} _n \mathscr{C} _n p \| ^2 _\mu & = \| \mathscr{C} _\mu p \| ^2 _\mu -2 \mathrm{Re} \, \left\langle {\mathscr{C} _\mu p} , {\mathrm{e} _n \mathscr{C} _n p} \right\rangle_\mu + \| \mathrm{e} _n \mathscr{C} _n p \| ^2 _\mu \\ & = \| p \| ^2 _{H^2 (\mu)} - 2 \mathrm{Re} \, \left\langle {E_n p} , {p} \right\rangle_{H^2 (\mu _n )} + \| E_n ^* p \| ^2 _{H^2 (\mu)} \\ & \leq 2 \| p \| ^2 _{H^2 (\mu )} - 2 \| p \| ^2 _{H^2 (\mu _n )} \\ & = 2 \int _{\partial \mathbb{D}} | p (\zeta ) | ^2 (1 - g_n (\zeta ) ) \mu (d\zeta ) \rightarrow 0, \end{align*} $$

by the Lebesgue monotone convergence theorem. In conclusion,

$$ \begin{align*}\mathscr{H} ^+ (\mu ) = \bigvee _{n=1} ^\infty \mathscr{H} ^+ (\mu _n ),\end{align*} $$

where, here, $\bigvee $ denotes closed linear span.

Proof Let $\mathrm {e} := \mathrm {e} _{\mu ,\lambda }$ and observe that $\mathrm {e}$ is, trivially, a multiplier by the constant function $1$ from $\mathscr {H} ^+ (\mu )$ into $\mathscr {H} ^+ (\lambda )$ . By Proposition 2.1 and Remark 2.2, $\mathrm {e}$ is closed on its maximal domain, $\mathrm {int} (\mu , \lambda )$ , its adjoint acts as

$$ \begin{align*}\mathrm{e} ^* k^\lambda _z = k^\mu _z, \quad \mbox{and} \quad \mathrm{e} ^* k^\lambda _j = k^\mu _j,\end{align*} $$

on kernels and coefficient kernels and the linear spans of the point evaluation and Taylor coefficient kernels in $\mathscr {H} ^+ (\lambda )$ are both cores for $\mathrm {e} ^*$ . Since $E = E_{\mu , \lambda } := \mathscr {C} _\mu ^* \mathrm {e} ^* \mathscr {C} _\lambda $ , E is closed, the formulas $E p = p$ for $p \in \mathbb {C} [\zeta ]$ and $E k_z = k_z$ are easily verified, and it further follows that $\mathscr {K} _{\mathbb {D}}$ and $\mathbb {C} [\zeta ]$ are cores for E.

To check that $T:= T_\mu = E_{\mu ,\lambda } ^* E_{\mu ,\lambda } \geq 0$ is $\lambda $ -Toeplitz, consider, for any $p, q \in \mathbb {C} [\zeta ]$ ,

$$ \begin{align*} \left\langle {\sqrt{T} Z^\lambda p} , {\sqrt{T} Z^\lambda q} \right\rangle_{H^2 (\lambda )} & = \left\langle { E \zeta \cdot p} , {E \zeta \cdot q} \right\rangle_{H^2 (\mu )} \\ & = \left\langle { \zeta p} , {\zeta q} \right\rangle_{H^2 (\mu )} = \left\langle {Z^\mu p} , {Z^\mu q} \right\rangle_{H^2 (\mu )} \\ & = \left\langle {p} , {q} \right\rangle_{H^2 (\mu )} = \left\langle {Ep} , {Eq} \right\rangle_{H^2 (\mu )} \\ & = \left\langle {\sqrt{T} p} , {\sqrt{T} q} \right\rangle_{H^2 (\lambda )}. \end{align*} $$

As the polynomials are a core for $E = E_{\mu ,\lambda }$ , this calculation holds on $\mathrm {Dom} \, E = \mathrm {Dom} \, \sqrt {T}$ . Moreover, since $\mathrm {Dom} \, E = \mathrm {Dom} \, \sqrt {T}$ , by polar decomposition of closed operators, and since $\mathbb {C} [\zeta ]$ and $\mathscr {K} _{\mathbb {D}}$ are $Z^\lambda $ -invariant cores for E, they are also cores for $\sqrt {T}$ , and it follows that $\mathrm {Dom} \, \sqrt {T} _\mu = \mathrm {Dom} \, E$ is also $Z^\lambda $ -invariant.

Proof Recall that $\lambda $ is extreme if and only if $H^2 (\lambda ) = L^2 (\lambda )$ so that $Z^\lambda = M^\lambda _{\zeta }$ . In this case, the self-adjoint $\lambda $ -Toeplitz operator $T_\mu \geq 0$ is Toeplitz with respect to the unitary $M^\lambda _\zeta $ . That is,

$$ \begin{align*}\left\langle {\sqrt{T_\mu} M^\lambda _\zeta h} , {\sqrt{T_\mu} M^\lambda _\zeta h} \right\rangle_{H^2 (\lambda )} = \left\langle {\sqrt{T_\mu} h} , {\sqrt{T_\mu} h} \right\rangle_{H^2 (\lambda )},\end{align*} $$

for all $h \in \mathrm {Dom} \, \sqrt {T_\mu }$ . Hence the quadratic forms for $(M^\lambda _\zeta ) ^* T _\mu M^\lambda _\zeta $ and $T_\mu $ are the same. By uniqueness of the unbounded Riesz representation, $(M^\lambda _\zeta ) ^* T_\mu M^\lambda _\zeta = T_\mu $ , so that, by Lemma 2.6,

$$ \begin{align*}( I + (M^\lambda _\zeta) ^* T_\mu M^\lambda _\zeta ) ^{-1} = ( I + T _\mu ) ^{-1},\end{align*} $$

or, equivalently,

$$ \begin{align*}M^\lambda _\zeta (I + T_\mu ) ^{-1} = (I + T_\mu ) ^{-1} M^\lambda _\zeta.\end{align*} $$

This shows that $T_\mu $ , and hence $\sqrt {T} _\mu $ are affiliated to the commutant of the unitary operator $M^\lambda _\zeta $ . Since $\mathbb {C} [\zeta ] \subseteq \mathrm {Dom} \, \sqrt {T_\mu }$ , we conclude that $\sqrt {T _\mu } = M _{\sqrt {T_\mu } 1}$ acts as multiplication by $\sqrt {T _\mu } 1 =: f \in L^2 (\lambda )$ . Since $\sqrt {T _\mu } \geq 0$ , we necessarily have that $f\geq 0$ , $\lambda -a.e.$ , and we conclude that for any polynomials $p,q$ ,

$$ \begin{align*}\left\langle {p} , {q} \right\rangle_{H^2 (\mu )} = \left\langle {\sqrt{T_\mu}p} , {\sqrt{T_\mu }q} \right\rangle_{H^2 (\lambda )} = \int _{\partial \mathbb{D}} \overline{p (\zeta )} q(\zeta ) f (\zeta ) ^2 \lambda (d\zeta ).\end{align*} $$

As in the proof of sufficiency in Theorem 4.1, we conclude that the above formula holds for any $g,h \in \mathbb {C} [\zeta ] + \overline {\mathbb {C} [\zeta ]}$ , which is dense in $\mathscr {C} (\partial \mathbb {D} )$ and $L^\infty (\mu )$ . In particular, the formula holds for all simple functions and characteristic functions of Borel sets. Since $f \in L^2 (\lambda )$ , $f^2 \in L^1 (\lambda )$ and it follows that

$$ \begin{align*}f^2 = \frac{\mu (d\zeta )}{\lambda (d\zeta )}\end{align*} $$

is the Radon–Nikodym derivative of $\mu $ with respect to $\lambda $ .

Proof of Theorem 4.10

Let $(x _j )_{j=1} ^\infty \subseteq \mathrm {Dom} \, \mathfrak {q}$ be a sequence with dense linear span. Apply Gram–Schimdt orthogonalization to $(x_j)$ with respect to the $\mathfrak {q} +\mathfrak {id}$ -inner product of $\hat {\mathcal {H}} (\mathfrak {q} )$ . This yields a countable basis $(y_j )_{j=1} ^\infty \subseteq \mathrm {Dom} \, \mathfrak {q}$ , so that the sequence $(\mathrm {j} _{\mathfrak {q}} (y_j) )$ is an orthonormal basis of $\hat {\mathcal {H}} (\mathfrak {q} )$ . Hence,

$$ \begin{align*}E _{\mathfrak{q}} E _{\mathfrak{q}} ^* = E _{\mathfrak{q}} I_{\hat{\mathcal{H}} _{\mathfrak{q}}} E_{\mathfrak{q}} ^* = \sum _{j=1} ^\infty \left\langle {E_{\mathfrak{q}} \mathrm{j} _{\mathfrak{q}} (y_j)} , {\cdot} \right\rangle_{\mathcal{H}} E_{\mathfrak{q}} \mathrm{j}_{\mathfrak{q}} (y_j ) = \sum \left\langle {y_j} , {\cdot} \right\rangle_{\mathcal{H}} y_j.\end{align*} $$

By [Reference Simon22, Theorem 2.1 and Corollary 2.3], see Theorem 4.8 and Equation (4.2) above,

$$ \begin{align*}\mathfrak{q}_{ac} (x,y) + \left\langle {x} , {y} \right\rangle_{\mathcal{H}} = \mathfrak{q}_{I+T} (x,y) = \left\langle {\sqrt{I+T}x} , {\sqrt{I+T}y} \right\rangle_{\mathcal{H}} = \left\langle {\mathrm{j} _{\mathfrak{q}} (x)} , {P_{ac} \mathrm{j} _{\mathfrak{q}} (y)} \right\rangle_{\hat{\mathcal{H}} (\mathfrak{q} )},\end{align*} $$

for any $x,y \in \mathrm {Dom} \, \mathfrak {q} \subseteq \mathrm {Dom} \, \mathfrak {q} _{ac} \subseteq \mathrm {Dom} \, \sqrt {I+T}$ . Hence, for any $x,y \in \mathrm {Dom} \, T$ ,

$$ \begin{align*} q_{E_{\mathfrak{q}} E_{\mathfrak{q}} ^*} \left( (I+T ) x , (I+T) y \right) & = \left\langle { \sqrt{E_{\mathfrak{q}} E_{\mathfrak{q}} ^*} (I+T) x} , {\sqrt{E_{\mathfrak{q}} E_{\mathfrak{q}} ^*} (I+T) y} \right\rangle_{\mathcal{H}} \\ & = \sum _{j=1} ^\infty \left\langle {(I+T)x} , {y_j} \right\rangle_{\mathcal{H}} \left\langle {y_j} , {(I+T) y} \right\rangle_{\mathcal{H} } \\ & = \sum \left\langle {\sqrt{I+T} x} , {\sqrt{I+T} y_j} \right\rangle_{\mathcal{H}} \left\langle {\sqrt{I+T} y_j} , {\sqrt{I+T} y} \right\rangle_{\mathcal{H} } \\ & = \sum \left\langle {P_{ac} \, \mathrm{j} _{\mathfrak{q}} (x)} , {\mathrm{j}_{\mathfrak{q}} (y_j)} \right\rangle_{\mathfrak{q}+\mathfrak{id}} \left\langle {\mathrm{j} _{\mathfrak{q}} (y_j)} , {P_{ac} \, \mathrm{j}_{\mathfrak{q}} (y)} \right\rangle_{\mathfrak{q}+\mathfrak{id}} \\ & = \left\langle {P _{ac} \, \mathrm{j} _{\mathfrak{q}} (x)} , {P_{ac} \, \mathrm{j}_{\mathfrak{q}} (y)} \right\rangle_{q+\mathfrak{id}} = q_{I+T} (x,y ) \\ & = \left\langle {x} , {(I+T) y} \right\rangle_{\mathcal{H}}. \end{align*} $$

That is, the (closeable) quadratic forms of $I+T$ and $(I+T) E_{\mathfrak {q}} E_{\mathfrak {q}} ^* (I+T)$ agree on $\mathrm {Dom} \, T$ , which is a core for $\sqrt {I+T}$ , and a form-core for $\mathfrak {q}_{I+T}$ . Moreover, ${\mathscr {D} := \mathrm {Ran} \, \sqrt {I+T} \cap \mathrm {Dom} \, \sqrt {I+T}}$ is a core for $\sqrt {I+T}$ , so that for all $x = \sqrt {I+T} y \in \mathscr {D}$ ,

$$ \begin{align*} \left\langle {x} , {x} \right\rangle _{\mathcal{H}} & = \left\langle {\sqrt{I+T} y} , {\sqrt{I+T} y} \right\rangle_{\mathcal{H}} = \left\langle {(I+T) y} , {E_{\mathfrak{q}} E_{\mathfrak{q}} ^* (I+T) y} \right\rangle_{\mathcal{H}} \\ & = \left\langle { \sqrt{I+T} x} , {E_{\mathfrak{q}} E_{\mathfrak{q}} ^* \sqrt{I+T}x} \right\rangle_{\mathcal{H}}. \end{align*} $$

That is, the (bounded) positive quadratic form of the identity, I, agrees with the quadratic form of $\sqrt {I+T} E_{\mathfrak {q}} E_{\mathfrak {q}} ^* \sqrt {I+T}$ on the dense subspace $\mathrm {Dom} \, \sqrt {I+T}$ . Here, if $V:= E _{\mathfrak {q}} ^* \sqrt {I+T}$ , this is a closeable operator by Lemma 4.11. In fact, V extends by continuity to an isometry, since $\mathfrak {q} _{V^*V} = \mathfrak {q} _I | _{\mathrm {Dom} \, \sqrt {I+T}}$ . Moreover, $E_{\mathfrak {q}}$ , and hence $\sqrt {E_{\mathfrak {q}} E_{\mathfrak {q}} ^*}$ have dense range, so that $\sqrt {E_{\mathfrak {q}} E_{\mathfrak {q}} ^*} \sqrt {I+T}$ extends to an isometry with dense range, i.e., a unitary. For any $x,y \in \mathrm {Dom} \, \sqrt {I+T}$ , we have that

$$ \begin{align*}\left\langle {x} , {y} \right\rangle_{\mathcal{H}} = \left\langle {\sqrt{I+T} x} , {E_{\mathfrak{q}} E_{\mathfrak{q}} ^* \sqrt{I+T} y} \right\rangle_{\mathcal{H}}.\end{align*} $$

Hence, by definition of the adjoint, for any $y \in \mathrm {Dom} \, \sqrt {I+T}$ , $E _{\mathfrak {q}} E_{\mathfrak {q}} ^* \sqrt {I+T}y \in \mathrm {Dom} \, \sqrt {I+T}$ , and

$$ \begin{align*}\sqrt{I+T} E_{\mathfrak{q}} E_{\mathfrak{q}} ^* \sqrt{I+T} y = y.\end{align*} $$

Hence, for any $x = \sqrt {I+T} ^{-1} g$ and $y = \sqrt {I+T} ^{-1} h$ ,

$$ \begin{align*} \left\langle {g} , {(I+T) ^{-1} h} \right\rangle_{\mathcal{H}} & = \left\langle {x} , {y} \right\rangle_{\mathcal{H}} = \left\langle {E_{\mathfrak{q}} ^* \sqrt{I+T}x} , {E_{\mathfrak{q}} ^* \sqrt{I+T}y} \right\rangle_{\mathcal{H}} \\ & = \left\langle {g} , {E_{\mathfrak{q}} E_{\mathfrak{q}} ^* h} \right\rangle_{\mathcal{H}}, \end{align*} $$

and we conclude, by the Riesz lemma for bounded sesquilinear forms, that

$$ \begin{align*}(I+T) ^{-1} = E_{\mathfrak{q}} E_{\mathfrak{q}} ^*.\\[-37pt] \end{align*} $$

Proof Necessity was established in Proposition 4.4 and sufficiency, in the case where $\lambda $ is extreme, was proven in Proposition 4.6. To prove sufficiency in general, assume that $\mu \ll _{RK} \lambda $ so that the intersection space $\mathrm {int} (\mu , \lambda )$ is dense in the space of $\mu $ -Cauchy transforms, $\mathscr {H} ^+ (\mu )$ . Note that $\mu \ll _{RK} \lambda $ if and only if $\mu + \lambda \ll _{RK} \lambda $ (and the same is true for $\ll $ ). This follows from Aronszajn’s “sums of kernels” theorem as stated in Section 2.2. By Lemma 4.5, $\mathrm {e} _{\mu +\lambda , \lambda } : \mathrm {int} (\mu + \lambda , \lambda ) \subseteq \mathscr {H} ^+ (\mu +\lambda ) \hookrightarrow \mathscr {H} ^+ (\lambda )$ is a closed embedding, $E_{\mu +\lambda , \lambda } := \mathscr {C} _{\mu +\lambda } ^* \mathrm {e} _{\mu +\lambda , \lambda } ^* \mathscr {C} _\lambda $ , is a closed co-embedding with $\mathscr {K} _{\mathbb {D}} = \bigvee k_z$ and $\mathbb {C} [\zeta ]$ as cores, and $T := T_{\mu +\lambda } = E_{\mu +\lambda , \lambda } ^* E_{\mu +\lambda , \lambda } \geq 0$ is self-adjoint, positive semi-definite, densely-defined, and $\lambda $ -Toeplitz.

By Remark 4.9 and Theorem 4.10 above, if $\hat {E} : L^2 (\mu + \lambda ) \hookrightarrow L^2 (\lambda )$ is the contractive co-embedding, and $\hat {T} := M_f$ , where $f \geq 0\ \lambda -a.e.$ , $f \in L^1 (\lambda )$ is the self-adjoint, positive semi-definite multiplication operator by the Radon–Nikodym derivative, ${f = \frac {d\mu }{d\lambda }}$ , then $(I +\hat {T}) ^{-1} = \hat {E} \hat {E}^*$ . On the other hand, $E:= E_{\mu + \lambda , \lambda } : \mathrm {Dom} \, E_{\mu +\lambda , \lambda } \subseteq H^2 (\lambda ) \hookrightarrow H^2 (\mu +\lambda )$ is closed and bounded below by $1$ , and so has a contractive inverse, namely, for any $h \in \mathscr {K} _{\mathbb {D}}$ or in $\mathbb {C} [\zeta ]$ ,

$$ \begin{align*}\hat{E} E h = h \in H^2 (\lambda ), \quad \mbox{and,} \quad E \hat{E} h = h \in H^2 (\mu +\lambda ).\end{align*} $$

Hence,

$$ \begin{align*}\hat{E} E | _{\mathrm{Dom} \, E} = I_{H^2 (\lambda )} |_{\mathrm{Dom} \, E}, \quad \mbox{and} \quad E \hat{E} | _{H^2 (\mu + \lambda ) } = I _{\mu +\lambda }.\end{align*} $$

Observe that the contractive co-embedding, $\hat {E}$ is necessarily injective and has dense range, so that it has a closed, potentially unbounded inverse, $\hat {E} ^{-1}$ , which is densely-defined. Since $(I +\hat {T} ) ^{-1} = \hat {E} \hat {E} ^*$ , we conclude that $I +\hat {T} = \hat {E} ^{-*} \hat {E} ^{-1}$ . Hence, for any $p, q \in \mathbb {C} [\zeta ] \subseteq H^2 (\lambda )$ ,

$$ \begin{align*} \left\langle {\sqrt{I +\hat{T}} p} , {\sqrt{I+ \hat{T}} q} \right\rangle_{L^2 (\lambda )} & = \left\langle {\hat{E} ^{-1} p} , {\hat{E} ^{-1} q} \right\rangle_{L^2 (\mu + \lambda )} \\ & = \left\langle {p} , {q} \right\rangle_{H^2 (\mu +\lambda )} = \left\langle {Ep} , {Eq} \right\rangle_{H^2 ( \mu +\lambda )} \\ & = \left\langle {\sqrt{T} p} , {\sqrt{T} q} \right\rangle_{H^2 (\lambda )}. \end{align*} $$

This calculation shows that the “compression” of $I + \hat {T}$ to the intersection of its domain with the subspace $H^2 (\lambda )$ is equal to T, in this quadratic form sense. In particular, ${T - I \geq 0}$ is the compression of $\hat {T} = M_f$ to $H^2 (\lambda )$ , where $f \in L^1 (\lambda )$ is the Radon–Nikodym derivative of $\mu $ with respect to $\lambda $ . In conclusion, for any polynomials $p,q$ ,

$$ \begin{align*} \int _{\partial \mathbb{D}} \overline{p(\zeta)} q (\zeta) \mu (d\zeta ) & = \left\langle {p} , {q} \right\rangle_{H^2 (\mu )} \\ & = \left\langle {\sqrt{T}p} , {\sqrt{T}q} \right\rangle_{H^2 (\lambda )} - \left\langle {p} , {q} \right\rangle_{H^2 (\lambda )} \\ & = \left\langle {\sqrt{\hat{T}} p} , {\sqrt{\hat{T}}q} \right\rangle_{L^2 (\lambda )} \\ & = \left\langle {M_{\sqrt{f}} p} , {M_{\sqrt{f}} q} \right\rangle_{L^2 (\lambda )} \\ & = \int _{\partial \mathbb{D}} \overline{p(\zeta)} q(\zeta) f (\zeta ) \lambda (d\zeta ). \end{align*} $$

As in the second proof of sufficiency of Theorem 4.1, this equality can be extended to arbitrary $g,h \in \mathbb {C} [\zeta ] + \overline {\mathbb {C} [\zeta ] }$ , so that by Weierstraß approximation, $\mu \ll \lambda $ with Radon–Nikodym derivative $f \geq 0$ , $f \in L^1 (\lambda )$ .

Proof First, if $\mathscr {H} ^+ (\gamma ) = \mathscr {M} \subseteq \mathscr {H} ^+ (\mu )$ is contractively contained, then, $\mathrm {e} : \mathscr {H} ^+ (\gamma ) \hookrightarrow \mathscr {H} ^+ (\mu )$ is trivially a (contractive) multiplier so that, as before, $\mathrm {e} ^* k_z ^\mu = k_z ^\gamma $ , and

$$ \begin{align*}\mathrm{e} ^* V^\mu = V^\gamma \mathrm{e} ^*.\end{align*} $$

In conclusion,

$$ \begin{align*}V ^{\mu *} \mathrm{e} \mathrm{e} ^* V^\mu = \mathrm{e} V ^{\gamma *} V^\gamma \mathrm{e} ^* = \mathrm{e} \mathrm{e} ^* = \tau.\end{align*} $$

Conversely, if $\tau = \mathrm {e} \mathrm {e} ^*$ is $V^\mu $ -Toeplitz and contractive, then, as in the proof of Theorem 4.1, $T := \mathscr {C} _\mu ^* \mathrm {e} \mathrm {e} ^* \mathscr {C} _\mu $ is a contractive $Z ^\mu $ -Toeplitz operator and we can appeal to the Riesz–Markov theorem to show that there is a $\gamma \geq 0$ , so that $T = P_{H^2 (\mu )} M_f | _{H^2 (\mu )}$ , where $f \geq 0$ , $\| f \| _\infty \leq 1$ is the Radon–Nikodym derivative of $\gamma $ with respect to $\mu $ . Namely, one can define a linear functional, $\hat {\mu } _T$ , on $\mathbb {C} [\zeta ] + \overline {\mathbb {C} [\zeta ]} \subseteq \mathscr {C} (\partial \mathbb {D} )$ , by

$$ \begin{align*}\hat{\mu} _T (p + \overline{q} ) := \left\langle {1} , {Tp} \right\rangle_{H^2 (\mu )} + \left\langle {q} , {T1} \right\rangle_{H^2 (\mu )}.\end{align*} $$

It is easy to check that $\hat {\mu } _T$ is bounded and positive using the Fejér–Riesz theorem, as in the proof of Theorem 4.1. The fact that T is a positive semi-definite $Z^\mu $ -Toeplitz contraction ensures that $\hat {\mu } _T$ extends to a bounded, positive linear functional on $\mathscr {C} (\partial \mathbb {D} )$ , and that $\hat {\mu } _T \leq \hat {\mu }$ , so that $\hat {\mu } _T = \hat {\gamma }$ for some finite, regular, and positive Borel measure, $\gamma \leq \mu $ , by the Riesz–Markov theorem.

By Theorem 2.4, the complementary space, $\mathscr {R} ^c (\mathrm {e} )$ , of $\mathscr {R} (\mathrm {e}) = \mathscr {H} ^+ (\gamma )$ is a RKHS in $\mathbb {D}$ with reproducing kernel $k ' (z,w) = k^\mu (z,w) - k^\gamma (z,w)$ and it is contractively contained in $\mathscr {H} ^+ (\mu )$ , by the inclusion theorem. Moreover, if $\mathrm {j} : \mathscr {R} ^c (\mathrm {e} ) \hookrightarrow \mathscr {H} ^+ (\mu )$ is the contractive embedding, then it follows that $\mathrm {j} \mathrm {j} ^* = I - \mathrm {e} \mathrm {e} ^* \geq 0$ is also a positive semi-definite $V^\mu $ -Toeplitz contraction. Hence, by the first part of the proof, $\mathscr {H} = \mathscr {H} ^+ (\nu )$ for a positive measure, $\nu $ . Finally, since $k^\mu = k^\gamma + k^\nu $ , we obtain that $\mu = \gamma + \nu $ .

Proof This is immediate, by Lemma 3.4, since both $V_\mu ^*$ and $V_\lambda ^*$ act as “backward shifts” on power series.

Proof By Lemma 3.4, if $h \in \mathrm {int} (\mu , \lambda )$ , then $V_\mu h \in \mathscr {H} ^+ (\mu )$ and

(5.1) $$ \begin{align} (V_\mu h) (z) = z h(z) + (V_\mu h) (0) 1= (V_\lambda h) (z) - (V_\lambda h) (0) 1 + (V_\mu h) (0) 1. \end{align} $$

Hence, if $c:= (V_\lambda h) (0) - (V_\mu h) (0) \in \mathbb {C}$ and $\lambda $ is non-extreme, then both $V_\lambda h$ and $c1$ belong to $\mathscr {H} ^+ (\lambda )$ so that $V_\mu h \in \mathscr {H} ^+ (\lambda ) \cap \mathscr {H} ^+ (\mu ) = \mathrm {int} (\mu , \lambda )$ . Recall that $\lambda $ is extreme if and only if $\mathscr {H} ^+ (\lambda )$ does not contain the constant functions. Hence, if $\lambda $ is extreme then $\mathrm {int} (\mu ,\lambda )$ will be $V_\mu $ -reducing if and only if $(V_\mu h) (0) = (V_\lambda h) (0)$ for all $h \in \mathrm {int} (\mu , \lambda )$ . By Equation (5.1), this happens if and only if $V_\mu h = V_\lambda h$ .

Proof Let P be the orthogonal projection of $\mathscr {H} ^+ (\mu )$ onto $\mathscr {M}$ . Then if $\mathrm {e} : \mathscr {M} \hookrightarrow \mathscr {H} ^+ (\mu )$ is the isometric embedding, $P= \mathrm {e} \mathrm {e} ^*$ . Hence,

$$ \begin{align*}V_\mu ^* \mathrm{e} \mathrm{e} ^* V_\mu = V_\mu ^* P V_\mu = V_\mu ^* V_\mu P = P = \mathrm{e}\mathrm{e} ^*,\end{align*} $$

so that $\tau = \mathrm {e} \mathrm {e} ^*$ is $V ^\mu $ -Toeplitz and $\mathscr {M} = \mathscr {H} ^+ (\gamma )$ for some $0 \leq \gamma \leq \mu $ , and $\mathscr {M} ^\perp = \mathscr {H} ^+ (\gamma ' )$ , by Theorem 5.1.

Proof By Theorem 4.12, we have that $\mathrm {int} (\mu _{ac}, \lambda ) \subseteq \mathrm {int} (\mu , \lambda )$ is dense in $\mathscr {H} ^+ (\mu _{ac})$ . Since we assume that $\mathrm {int} (\mu , \lambda )$ is $V_\mu $ -reducing, its closure, $\mathrm {int} (\mu , \lambda ) ^{-\mu }$ , is also $V_\mu $ -reducing and then $\mathrm {int} (\mu , \lambda ) ^{-\mu } = \mathscr {H} ^+ (\gamma )$ for some $0 \leq \gamma \leq \mu $ by Corollary 5.4. By construction $\gamma \ll _{RK} \lambda $ so that $\gamma \ll \lambda $ by Theorem 4.12. By maximality, $\gamma \leq \mu _{ac}$ and by construction $\mathrm {int} (\mu , \lambda ) \subseteq \mathrm {int} (\gamma , \lambda )$ . However, we also have that $\mathrm {int} (\mu _{ac} , \lambda ) \subseteq \mathrm {int} (\mu , \lambda ) \subseteq \mathrm {int} (\gamma , \lambda )$ . Hence, if $\mathrm {e} : \mathscr {H} ^+ (\mu _{ac} ) \hookrightarrow \mathscr {H} ^+ (\mu )$ is the contractive embedding, then $\mathrm {e}$ , restricted to the dense subspace $\mathrm {int} (\mu _{ac} , \lambda ) \subseteq \mathscr {H} ^+ (\mu _{ac} )$ defines a contraction into $\mathscr {H} ^+ (\gamma )$ , which is isometrically contained in $\mathscr {H} ^+ (\mu )$ . It follows that $\mathrm {e}$ extends by continuity to a contractive embedding of $\mathscr {H} ^+ (\mu _{ac} )$ into $\mathscr {H} ^+ (\gamma )$ . Hence, by Theorem 4.1, $\mu _{ac} \leq \gamma $ and we conclude that $\mu _{ac} =\gamma $ .

Proof The intertwining formulas are easily verified. The range of $U _\wedge : \mathscr {H} ^+ (\mu + \lambda ) \rightarrow \mathscr {H} ^+ (\mu ) \oplus \mathscr {H} ^+ (\lambda )$ is $V_\mu \oplus V_\lambda $ -reducing if and only if, for any $h \oplus -h \in \mathrm {Ran} \, U _\wedge $ , $h \in \mathrm {int} (\mu , \lambda )$ ,

$$ \begin{align*}V _\mu h \oplus - V_\lambda h = g \oplus -g,\end{align*} $$

for some $g \in \mathrm {int} (\mu , \lambda )$ . Clearly, this happens if and only if $V_\mu | _{\mathrm {int} (\mu , \lambda )} = V_\lambda | _{\mathrm {int} (\mu , \lambda )}$ .

Proof of Proposition 5.7

If $\lambda + \mu $ is extreme, then $\mathscr {H} ^+ (\mu + \lambda ) = \mathscr {H} ^+ (\mu ) + \mathscr {H} ^+ (\lambda )$ does not contain the constant functions. Hence, both $\lambda $ and $\mu $ must also be extreme. In this case, $V_\mu , V_\lambda $ , and $V_{\mu + \lambda }$ are all unitary operators. We know that $\mathrm {Ran} \, U_\vee $ is always $V_\mu \oplus V_\lambda $ -invariant. On the other hand, since $\mu + \lambda $ is extreme, $V_{\mu + \lambda }$ is unitary, hence surjective, and

$$ \begin{align*}\mathscr{H} ^+ ( \mu + \lambda ) = \bigvee (k_z ^{\mu +\lambda } - k_0 ^{\mu + \lambda } ),\end{align*} $$

so that

$$ \begin{align*}\mathrm{Ran} \, U_\vee = \bigvee (k_z ^{\mu} - k_0 ^{\mu} ) \oplus (k_z ^\lambda - k_0 ^\lambda ).\end{align*} $$

Hence,

$$ \begin{align*}(V_\mu ^* \oplus V_\lambda ^*) \mathrm{Ran} \, U_\vee = \bigvee k_z ^\mu \overline{z} \oplus k_z ^\lambda \overline{z} \subseteq \mathrm{Ran} \, U_\vee.\end{align*} $$

It follows that $\mathrm {Ran} \, U_\vee $ is $V_\mu \oplus V_\lambda $ -reducing, so that $\mathrm {Ran} \, U_\wedge = \mathrm {Ran} \, U_\vee ^\perp $ is also reducing. The previous lemma now implies that $\mathrm {int} (\mu , \lambda )$ is $V_\mu $ -reducing.