# 数学代写|复变函数作业代写Complex function代考|MATH 716

#### Doug I. Jones

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## 数学代写|复变函数作业代写Complex function代考|Unidirectional Case

Realizing the unity coupling $\varepsilon:=\left(L^{2}(\mathfrak{K}), \mathfrak{K}, \mathfrak{K} ; U_{\mathfrak{K}}^{\times}, V_{\mathfrak{K}}, V_{\mathfrak{K}}\right)$ in the Hilbert space $L^{2}(\mathfrak{K})$, where $V_{\mathfrak{K}}$ is the inclusion operator of $\mathfrak{K}$ into $L^{2}(\mathfrak{K})$, we can speak for simplicity about channels of the coupling $\varepsilon$ as channels in $L^{2}(\mathfrak{K})$. A description of the set of all such channels was given in [12] (Theorem 7.10), and unitary couplings generated by pairs of such channels were also studied there (Theorem 7.26). Inclusions into each other of subspaces of $L^{2}(\mathfrak{K})$ reducing the operator $U_{\mathfrak{K}}^{\times}$and subspaces $\mathfrak{L}$ of $L^{2}(\mathfrak{K})$, where a bilateral channel of the form $\left(\mathfrak{L}, \mathfrak{N} ; V_{\mathfrak{N}}\right)$ can be realized, were explored in [12] as well (Theorem $7.22$ and Corollary 7.23). Some of these results can be formulated in the following form.
Theorem 4.1
(a) There exists a bijective correspondence between bilateral channels $\left(\mathfrak{L}, \mathfrak{N} ; V_{\mathfrak{N}}\right)$ in $L^{2}(\mathfrak{K})$ and isometric operator functions $\theta\left(e^{i t}\right) \in C M[\mathfrak{N}, \mathfrak{K}]$. This correspondence is established by the formulas
$$\mathfrak{L}=\operatorname{Ran} \theta, \quad V_{\mathfrak{N}}=\left.\theta\right|{\mathfrak{N}} ; \quad \theta=\left(\Phi{U_{\mathfrak{R}}^{\times}}^{\mathfrak{N}}\right)^{} .$$ (b) Let $\left(\mathfrak{L}{2}, \mathfrak{F} ; V{\mathfrak{F}}\right)$ and $\left(\mathfrak{L}{1}, \mathfrak{G} ; V{\mathfrak{K})}\right)$ be two channels in $L^{2}(\mathfrak{K}), O_{2}\left(e^{i t}\right) \in$ $C M[\mathfrak{F}, \mathfrak{K}]$ and $\theta_{1}\left(e^{i t}\right) \in C M[\mathfrak{G}, \mathfrak{K}]$ be the corresponding pair of isometric operator functions, and let $\sigma:=\left(L^{2}(\mathfrak{K}), \mathfrak{F}, \mathfrak{G} ; U_{\mathfrak{K}}^{\times}, V_{\mathfrak{F}}, V_{\mathfrak{G}}\right)$. Then
$$\hat{\theta}{\sigma}\left(e^{i t}\right)=\theta{2}^{}\left(e^{i t}\right) \hat{\theta}{1}\left(e^{i t}\right)$$ The function $\theta{\sigma}\left(e^{i t}\right)$ is an isometric (coisometric, unitary) operator function iff $\mathfrak{L}{1} \subset \mathfrak{L}{2}\left(\mathfrak{L}{1} \supset \mathfrak{L}{2}, \mathfrak{L}{1}=\mathfrak{L}{2}\right)$.
(c) Let $\mathfrak{M}$ be a subspace of $L^{2}(\mathfrak{K})$ reducing the operator $U_{\mathfrak{K}}^{\times}$. There exists a channel $\left(\mathfrak{L}, \mathfrak{N} ; V_{\mathfrak{N}}\right)$ in $L^{2}(\mathfrak{K})$ such that
$$\mathfrak{L} \subset \mathfrak{M} \text { and } \operatorname{dim} \mathfrak{N}=\gamma$$
iff $0 \leq \gamma \leq \alpha_{\mathfrak{M}}$. Moreover, the equality $\mathfrak{L}=\mathfrak{M}$ is possible iff $\alpha_{\mathfrak{M}}=\beta_{\mathfrak{M}}=\gamma$, that is, when $\rho_{\mathfrak{m}}\left(e^{i I}\right)=\gamma$ almost everywhere.

## 数学代写|复变函数作业代写Complex function代考|Bidirectional Case

Parameterizations of the sets $\mathcal{K}{r}(\theta)$ and $\mathcal{K}{c r}(\theta)$ are described in
Theorem 4.5 Let $\theta\left(e^{i t}\right) \in C M[\mathfrak{G}, \mathfrak{F}]$.
(a) A nontrivial regular up-leftward extension of the function $\theta\left(e^{i t}\right)$ exists iff at least one of the two conditions $\alpha_{\Pi}>0$ or $\alpha_{\Sigma}>0$ is satisfied.
Let $\Xi\left(e^{i t}\right) \in C M\left[\mathfrak{B}^{(1)} \oplus \mathfrak{G}, \mathfrak{F}^{(1)} \oplus \mathbb{F}\right]$ be a function of form (3.32).
(b) $\Xi\left(e^{i t}\right) \in \mathcal{K}{r}(\theta)$ iff there exist a coisometric operator function $\omega\left(e^{\text {it }} \in\right.$ $C M\left[\mathfrak{G}, \mathfrak{F}^{(1)}\right]$ and an isometric operator function $\lambda\left(e^{i t}\right) \in C M\left[\mathfrak{G}^{(1)}, \mathfrak{F}\right]$ such that the inclusions $$\operatorname{Ran} \omega^{} \subset \overline{\Pi L^{2}(\mathfrak{G})}, \quad \operatorname{Ran} \lambda \subset \overline{\Sigma L^{2}(\mathfrak{F})}$$ hold and the functions $$\theta{12}\left(e^{i t}\right) \in C M\left[\mathfrak{G}, \mathfrak{F}^{(1)}\right], \theta_{21}\left(e^{i t}\right) \in C M\left[\mathfrak{G}^{(1)}, \mathfrak{F}\right], \theta_{11}\left(e^{i t}\right) \in C M\left[\mathfrak{G}^{(1)}, \mathfrak{F}^{(1)}\right]$$
admit the representations of the forms
$$\theta_{12}\left(e^{i t}\right)=\omega\left(e^{i t}\right) \Pi\left(e^{i t}\right), \theta_{21}\left(e^{i t}\right)=\Sigma\left(e^{i t}\right) \lambda\left(e^{i t}\right), \theta_{11}\left(e^{i t}\right)=-\omega\left(e^{i t}\right) \theta^{}\left(e^{i t}\right) \lambda\left(e^{i t}\right) .$$
(c) $\Xi\left(e^{i t}\right) \in \mathcal{K}{c r}(\theta)$ iff the corresponding pair $\left{\omega\left(e^{i t}\right), \lambda\left(e^{i t}\right)\right}$ of coisometric and isometric operator functions satisfies the conditions $$\operatorname{Ran} \omega^{} \subset \operatorname{Ker} \theta, \quad \operatorname{Ran} \lambda \subset \operatorname{Ken} \theta^{} .$$
In this case, equalities (4.9) take the forms
$$\theta{12}\left(e^{i t}\right)=\omega\left(e^{i t}\right), \quad \theta_{21}\left(e^{i t}\right)=\lambda\left(e^{i t}\right), \quad \theta_{11}\left(e^{i t}\right) \equiv 0 \quad\left(0 \in\left[\mathfrak{G}^{(1)}, \mathfrak{F}^{(1)}\right]\right)$$

## 数学代写|复变函数作业代写Complex function代考|Unidirectional Case

，其中双边通道的形式 $\left(\mathfrak{L}, \mathfrak{N} ; V_{\mathfrak{N}}\right)$ 可以实现，在[12]中也进行了探索（定理 $7.22$ 和推论 7.23)。其中一些结果可以用以下形式表示。

(a) 双边通道之间存在双射对应 $\left(\mathfrak{L}, \mathfrak{N} ; V_{\mathfrak{N}}\right)$ 在 $L^{2}(\mathfrak{K})$ 和等距算子函数
$\theta\left(e^{i t}\right) \in C M[\mathfrak{N}, \mathfrak{K}]$. 这种对应关系是由公式建立的
$$\mathfrak{L}=\operatorname{Ran} \theta, \quad V_{\mathfrak{N}}=\theta \mid \mathfrak{N} ; \quad \theta=\left(\Phi U_{\mathfrak{R}}^{\times \mathfrak{N}}\right) .$$
(b) 让 $(\mathfrak{L} 2, \mathfrak{F} ; V \mathfrak{F})$ 和 $(\mathfrak{L} 1, \mathfrak{G} ; V \mathfrak{K}))$ 成为两个通道 $L^{2}(\mathfrak{K}), O_{2}\left(e^{i t}\right) \in C M[\mathfrak{F}, \mathfrak{K}]$ 和
$\theta_{1}\left(e^{i t}\right) \in C M[\mathfrak{G}, \mathfrak{K}]$ 是对应的等距算子函数对，并且让
$\sigma:=\left(L^{2}(\mathfrak{K}), \mathfrak{F}, \mathfrak{G} ; U_{\mathfrak{K}}^{\times}, V_{\mathfrak{F}}, V_{\mathfrak{B}}\right)$.然后
$$\hat{\theta} \sigma\left(e^{i t}\right)=\theta 2\left(e^{i t}\right) \hat{\theta} 1\left(e^{i t}\right)$$

(c) 让 $\mathfrak{M}$ 是一个子空间 $L^{2}(\mathfrak{K})$ 减少算子 $U_{\mathfrak{R}}^{\times}$. 存在通道 $\left(\mathfrak{L}, \mathfrak{N} ; V_{\mathfrak{N}}\right)$ 在 $L^{2}(\mathfrak{K})$ 这样
$$\mathfrak{L} \subset \mathfrak{M} \text { and } \operatorname{dim} \mathfrak{N}=\gamma$$ $\rho_{\mathfrak{m}}\left(e^{i I}\right)=\gamma$ 几乎无处不在。

## 数学代写|复变函数作业代写Complex function代考|Bidirectional Case

(a) 函数的非平凡规则左上扩展 $\theta\left(e^{i t}\right)$ 存在当且仅当这两个条件之一 $\alpha_{\Pi}>0$ 或者 $\alpha_{\Sigma}>0$ 很满意。

(二) $\Xi\left(e^{i t}\right) \in \mathcal{K} r(\theta)$ 当且仅当存在等轴算子函数 $\omega\left(e^{i t} \in C M\left[\mathfrak{G}, \mathfrak{F}^{(1)}\right]\right.$ 和等距算子

$$\operatorname{Ran} \omega \subset \overline{\Pi L^{2}(\mathfrak{G})}, \quad \operatorname{Ran} \lambda \subset \overline{\Sigma L^{2}(\mathfrak{F})}$$

$$\theta 12\left(e^{i t}\right) \in C M\left[\mathfrak{G}, \mathfrak{F}^{(1)}\right], \theta_{21}\left(e^{i t}\right) \in C M\left[\mathfrak{G}^{(1)}, \mathfrak{F}\right], \theta_{11}\left(e^{i t}\right) \in C M\left[\mathfrak{G}^{(1)}, \mathfrak{F}^{(i)}\right.$$

$$\theta_{12}\left(e^{i t}\right)=\omega\left(e^{i t}\right) \Pi\left(e^{i t}\right), \theta_{21}\left(e^{i t}\right)=\Sigma\left(e^{i t}\right) \lambda\left(e^{i t}\right), \theta_{11}\left(e^{i t}\right)=-\omega\left(e^{i t}\right) \theta\left(e^{i}\right.$$
(C) $\Xi\left(e^{i t}\right) \in \mathcal{K} c r(\theta)$ 当且仅当对应对
$\operatorname{Ran} \omega \subset \operatorname{Ker} \theta, \quad \operatorname{Ran} \lambda \subset \operatorname{Ken} \theta .$

$$\theta 12\left(e^{i t}\right)=\omega\left(e^{i t}\right), \quad \theta_{21}\left(e^{i t}\right)=\lambda\left(e^{i t}\right), \quad \theta_{11}\left(e^{i t}\right) \equiv 0 \quad\left(0 \in\left[\mathfrak{G}^{(1)}, \mathfrak{F}^{(1)}\right]\right)$$

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