PhyslibAlpha.AlgebraicFramework.HilbertSpace.Unbounded.CayleySpectralData.SpecTheorem

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear operator on $H$ with domain $D(T)$. Let $\mu_T$ denote the real spectral measure of $T$ (the spectral measure on $\mathbb{R}$ obtained via the inverse Cayley transform). For any vector $x \in D(T)$, $x$ belongs to the square-moment domain of $\mu_T$. This means that the integral of the squared identity function with respect to the diagonal measure $\mu_x$ is finite: $$\int_{\mathbb{R}} \lambda^2 d\mu_x(\lambda) < \infty$$ where $\mu_x(S) = \langle x, \mu_T(S) x \rangle$ for measurable sets $S \subseteq \mathbb{R}$.

Let $H$ be a Hilbert space and $\mu$ be a weak operator topology (WOT) spectral measure on a measurable space $\alpha$. For any measurable set $S \subseteq \alpha$ and vectors $x, y \in H$, let $\mu_{x, y}$ denote the complex scalar measure defined by $\mu_{x, y}(A) = \langle y, \mu(A) x \rangle$ for measurable sets $A$. Then the scalar measure associated with the projected vector $\mu(S)x$ and the vector $y$ is equal to the restriction of the scalar measure $\mu_{x, y}$ to the set $S$: $$\mu_{\mu(S)x, y} = \mu_{x, y} |_S$$ where $\mu_{x, y} |_S$ denotes the restriction of the measure to $S$ (i.e., $(\mu_{x, y} |_S)(A) = \mu_{x, y}(A \cap S)$).

Let $H$ be a Hilbert space and $\mu$ be a weak operator topology (WOT) spectral measure on a measurable space $\alpha$. For any two vectors $x, y \in H$, let $\mu_{x, y}$ denote the complex scalar measure defined by $\mu_{x, y}(A) = \langle y, \mu(A) x \rangle$ for every measurable set $A \subseteq \alpha$. For any measurable set $S \subseteq \alpha$, the scalar measure formed by projecting the second vector $y$ via $\mu(S)$ is equal to the restriction of the original scalar measure to $S$: $$\mu_{x, \mu(S)y} = \mu_{x, y}|_S$$

Let $H$ be a Hilbert space and $\mu$ be a weak operator topology (WOT) spectral measure on a measurable space $\alpha$. For any measurable set $S \subseteq \alpha$, any vectors $x, y \in H$, and any complex-valued function $g: \alpha \to \mathbb{C}$ that is integrable with respect to the scalar measure $\mu_{x, y}$ (defined by $\mu_{x, y}(A) = \langle y, \mu(A) x \rangle$), the complex weak integral with the projected vector $\mu(S)x$ satisfies: $$\int_{\alpha} g(z) \, d\langle y, \mu(z) (\mu(S)x) \rangle = \int_{\alpha} \mathbb{1}_S(z) g(z) \, d\langle y, \mu(z) x \rangle$$ where $\mathbb{1}_S$ is the indicator function of the set $S$.

Let $H$ be a Hilbert space and $\mu$ be a weak operator topology (WOT) spectral measure on a measurable space $\alpha$. For any two vectors $x, y \in H$, let $\mu_{x, y}$ denote the complex scalar measure defined by $\mu_{x, y}(A) = \langle y, \mu(A) x \rangle$ for every measurable set $A \subseteq \alpha$. For any measurable set $S \subseteq \alpha$ and any complex-valued function $g: \alpha \to \mathbb{C}$ that is integrable with respect to $\mu_{x,y}$, the complex weak integral satisfies: $$\int_\alpha g(z) \, d\langle \mu(S)y, \mu(z) x \rangle = \int_\alpha \mathbb{1}_S(z) g(z) \, d\mu_{x,y}(z)$$ where $\mathbb{1}_S$ is the indicator function of the set $S$. This identity shows that projecting the test vector $y$ onto the subspace associated with $S$ is equivalent to restricting the integration of $g$ to the set $S$.

Let $H$ be a Hilbert space and $\alpha$ a measurable space. Let $\mu$ be a weak operator topology (WOT) spectral measure on $\alpha$ acting on $H$. Let $f: \alpha \to \mathbb{C}$ be a complex-valued function that is integrable with respect to every scalar measure $\mu_{x,y}$ defined by $\mu_{x,y}(S) = \langle y, \mu(S) x \rangle$ for $x, y \in H$. Suppose there exists a bounded linear operator $U$ on $H$ such that for all $x, y \in H$, the relation $$\langle y, U x \rangle = \int_\alpha f(z) \, d\langle y, \mu(z) x \rangle$$ holds. Then, for any measurable set $S \subseteq \alpha$, the projection $\mu(S)$ commutes with $U$, i.e., $$\mu(S) U = U \mu(S).$$

Let $H$ be a complex Hilbert space and $\mu$ be a weak operator topology (WOT) spectral measure on the complex plane $\mathbb{C}$ acting on $H$. Let $U: H \to H$ be a bounded linear operator such that $U$ is the weak integral of the identity function, meaning $\langle y, U x \rangle = \int_{\mathbb{C}} z \, d\langle y, \mu(z) x \rangle$ for all vectors $x, y \in H$. Assuming that the identity function $z \mapsto z$ is integrable with respect to the complex scalar measures $\mu_{x,y}$ (defined by $\mu_{x,y}(S) = \langle y, \mu(S) x \rangle$) and that these scalar measures have finite variation, then for any $x, y \in H$, the complex weak integral of the function $f(z) = 1 - z$ satisfies: $$\int_{\mathbb{C}} (1 - z) \, d\langle y, \mu(z) x \rangle = \langle y, x - U x \rangle$$

Let $H$ be a complex Hilbert space and $\mu$ be a weak operator topology (WOT) spectral measure on $\mathbb{C}$ acting on $H$. Let $U: H \to H$ be a continuous linear operator such that for all $x, y \in H$, $\langle y, U x \rangle = \int_{\mathbb{C}} z \, d\langle y, \mu(z) x \rangle$. Assume that for all $x, y$, the identity function $f(z) = z$ is integrable with respect to the complex scalar measure $\mu_{x,y}(S) = \langle y, \mu(S) x \rangle$ and that these measures have finite variation. Further, assume that $U$ commutes with the projections $\mu(S)$ for all measurable sets $S \subseteq \mathbb{C}$. Given vectors $x, v, y \in H$ such that $x = v - U v$, then for any measurable set $S \subseteq \mathbb{C}$, it holds that: $$\langle y, \mu(S) x \rangle = \int_{\mathbb{C}} (1 - z) \, d\langle y, \mu(z) \mu(S) v \rangle$$

Let $\alpha$ be a measurable space and $\mu$ be a complex-valued vector measure on $\alpha$. Let $q: \alpha \to \mathbb{C}$ be a function that is integrable with respect to $\mu$, and let $\nu = q\mu$ be the complex vector measure defined by the density $q$ with respect to $\mu$. Let $g: \alpha \to \mathbb{R}$ be a real-valued function that is integrable with respect to $\nu$. Provided that the variation of the measure $\nu$ satisfies $|\nu| = |q||\mu|$ (where $|\cdot|$ denotes the variation of a vector measure), then the integral of $g$ with respect to $\nu$ is equal to the integral of the product $gq$ with respect to $\mu$: $$\int g \, d\nu = \int g(x)q(x) \, d\mu(x)$$ In this identity, the left-hand integral views $g$ as a real scalar acting on the complex values of $\nu$, while the right-hand integral treats the product $gq$ as a complex-valued function.

Let $\mu$ be a weak operator topology (WOT) spectral measure on $\mathbb{C}$ acting on a Hilbert space $H$. Let $U$ be a continuous linear operator on $H$ such that $U$ is the integral of the identity function with respect to $\mu$, i.e., $\langle y, U x \rangle = \int_{\mathbb{C}} z \, d\langle y, \mu(z) x \rangle$ for all $x, y \in H$. Assume that: 1. The identity function $z \mapsto z$ is integrable with respect to the complex scalar measures $\mu_{x,y}$ defined by $S \mapsto \langle y, \mu(S) x \rangle$. 2. The variation of each scalar measure $\mu_{x,y}$ is finite. 3. $U$ commutes with the spectral projections $\mu(S)$ for every measurable set $S \subseteq \mathbb{C}$. For any vectors $x, v, y \in H$ such that $x = v - Uv$, the scalar measure $\mu_{x,y}$ is equal to the scalar measure $\mu_{v,y}$ with density $(1 - z)$. That is: $$\mu_{x,y} = (1 - z) \mu_{v,y}$$ or equivalently, for any measurable set $S \subseteq \mathbb{C}$: $$\langle y, \mu(S) x \rangle = \int_S (1 - z) \, d\langle y, \mu(z) v \rangle$$

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear operator on $H$. Let $U = (T - iI)(T + iI)^{-1}$ be the bounded Cayley transform of $T$, and let $E$ be the spectral measure (projection-valued measure) associated with $U$. For any measurable set $S \subseteq \mathbb{C}$, the spectral projection $E(S)$ commutes with the Cayley transform $U$: $$E(S) U = U E(S)$$

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint operator on $H$ with domain $\mathcal{D}(T)$. Let $E_U$ be the spectral measure associated with the bounded Cayley transform $U = (T - iI)(T + iI)^{-1}$ of $T$. Let $f: \mathbb{C} \to \mathbb{R}$ be the inverse scalar Cayley transform defined by: $$f(z) = \begin{cases} -\frac{\text{Im } z}{1 - \text{Re } z} & \text{if } z \neq 1 \\ 0 & \text{if } z = 1 \end{cases}$$ For any $x \in \mathcal{D}(T)$ and $y \in H$, the function $f$ is integrable with respect to the complex scalar measure $\mu_{x,y}$ defined by $\mu_{x,y}(S) = \langle y, E_U(S) x \rangle$, and the operator $T$ satisfies the following weak integral representation: $$\langle y, Tx \rangle = \int_{\mathbb{C}} f(z) \, d\langle y, E_U(z) x \rangle$$

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint operator on $H$ with domain $\mathcal{D}(T)$. Let $E_U$ be the spectral measure associated with the bounded Cayley transform of $T$, and let $f: \mathbb{C} \to \mathbb{R}$ be the inverse scalar Cayley transform. Suppose that for every $x \in \mathcal{D}(T)$ and $y \in H$, the function $f$ is integrable with respect to the complex scalar measure $S \mapsto \langle y, E_U(S) x \rangle$ and satisfies the identity: $$\langle y, Tx \rangle = \int_{\mathbb{C}} f(z) \, d\langle y, E_U(z) x \rangle$$ Then, the real spectral measure $E_{\mathbb{R}}$ of $T$ (defined as the push-forward of $E_U$ via $f$) is a weak spectral resolution of $T$. That is, for every $x \in \mathcal{D}(T)$ and $y \in H$, the identity function $\lambda \mapsto \lambda$ is integrable with respect to the scalar measure $S \mapsto \langle y, E_{\mathbb{R}}(S) x \rangle$ and satisfies: $$\langle y, Tx \rangle = \int_{\mathbb{R}} \lambda \, d\langle y, E_{\mathbb{R}}(\lambda) x \rangle$$

Let $T$ be a self-adjoint linear operator on a complex Hilbert space $H$ with domain $\mathcal{D}(T)$. Let $\mu_T$ be the real spectral measure associated with $T$ (constructed via the Cayley transform). Then $\mu_T$ is a weak spectral resolution of $T$. That is, for every $x \in \mathcal{D}(T)$ and $y \in H$, the identity function $\lambda \mapsto \lambda$ is integrable with respect to the complex scalar measure $\langle y, \mu_T(\cdot) x \rangle$, and the following reconstruction formula holds: $$\langle y, Tx \rangle = \int_{\mathbb{R}} \lambda \, d\langle y, \mu_T(\lambda) x \rangle$$

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint operator on $H$. Then $T$ satisfies the spectral theorem with respect to its real spectral measure $\mu_T$, which is the spectral measure on the real line $\mathbb{R}$ obtained by pulling back the spectral measure of the Cayley transform of $T$ to $\mathbb{R}$.

Let $H$ be a complex Hilbert space and $T: D(T) \subseteq H \to H$ be a self-adjoint partially defined linear operator. Let $\mu_T$ be the real spectral measure of $T$ obtained via the Cayley transform. Then $T$ is a restriction of the maximal spectral integral of $\mu_T$. That is, $$T \le \int_{\mathbb{R}} \lambda \, d\mu_T(\lambda)$$ where the relation $\le$ for partially defined linear operators denotes that $D(T)$ is contained in the domain of the maximal spectral integral and the operators agree on $D(T)$.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear operator on $H$. Let $\mu_T$ be the real spectral measure associated with $T$ (obtained via the Cayley transform). Then $T$ is equal to the maximal spectral integral of the identity function with respect to $\mu_T$: $$T = \int_{\mathbb{R}} \lambda \, d\mu_T(\lambda)$$ This equality implies that both operators have the same domain and agree on all elements of that domain.

Let $H$ be a complex Hilbert space and $T: \mathcal{D}(T) \subseteq H \to H$ be a self-adjoint partial linear operator. Let $\mu_T$ be the real spectral measure of $T$ (the spectral measure on $\mathbb{R}$ obtained by pulling back the spectral measure of its Cayley transform). Then $T$ satisfies the domain-aware spectral theorem with respect to $\mu_T$, which characterizes $T$ as the spectral integral of the identity function $\lambda \mapsto \lambda$ and accounts for the correspondence between the domain $\mathcal{D}(T)$ and the integrability conditions of the spectral measure.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint linear operator on $H$ (which may be unbounded). Then $T$ satisfies the spectral theorem with respect to its real spectral measure $\mu_T$ (constructed via the Cayley transform). Specifically, this theorem provides a spectral decomposition where $T$ is represented as the spectral integral of the identity function with respect to $\mu_T$, formally satisfying the `DomainAwareSelfAdjointSpectralTheorem` interface which relates the operator $T$ to its measure $\mu_T$ while accounting for the operator's domain.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. If $T$ is essentially self-adjoint, then its closure $\bar{T}$ is self-adjoint and satisfies the unbounded spectral theorem with respect to the real spectral measure $\mu$ derived from the Cayley transform of $\bar{T}$. This theorem provides a spectral resolution for $\bar{T}$ and establishes that the domain of the operator $\bar{T}$ is exactly the set of vectors in $H$ whose square-moment with respect to $\mu$ is finite.