PhyslibAlpha.ProbabilisticTheory.HilbertSpace.Unbounded.CayleySpectralData.SpecTheorem

Let $H$ be a Hilbert space and $T$ be a self-adjoint operator on $H$ with domain $\mathcal{D}(T)$. Let $\mu$ be the spectral measure on $\mathbb{R}$ associated with $T$ (via the Cayley transform). For any vector $x$ in the domain $\mathcal{D}(T)$, the second moment of the scalar measure $\mu_x$ is finite. That is, $x$ satisfies: $$\int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda)x \rangle < \infty,$$ where $\mu_x$ is defined by $\mu_x(A) = \langle x, \mu(A)x \rangle$ for measurable sets $A \subseteq \mathbb{R}$.

Let $\mu$ be a projection-valued measure on a measurable space $\alpha$ taking values in the operators on a Hilbert space $H$. For any vectors $x, y \in H$ and any measurable set $S \subseteq \alpha$, let $\mu_{x,y}$ be the complex scalar measure defined by $A \mapsto \langle y, \mu(A)x \rangle$. 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 $S$. That is, $$\mu_{\mu(S)x, y} = \mu_{x,y}|_S$$ where the restricted measure is defined by $\mu_{x,y}|_S(A) = \mu_{x,y}(A \cap S)$ for any measurable set $A \subseteq \alpha$.

Let $H$ be a Hilbert space and $\mu$ be a projection-valued measure on a measurable space $\alpha$. For any two vectors $x, y \in H$, let $\mu_{x,y}$ be the complex scalar measure defined by $\mu_{x,y}(A) = \langle y, \mu(A)x \rangle$ for measurable sets $A \subseteq \alpha$. Then for any measurable set $S \subseteq \alpha$, the scalar measure associated with the pair $(x, \mu(S)y)$ is equal to the restriction of the measure $\mu_{x,y}$ to $S$: $$\mu_{x, \mu(S)y} = \mu_{x,y}|_S$$ where $\mu_{x,y}|_S(A) = \mu_{x,y}(A \cap S)$.

Let $\mu$ be a spectral measure on a measurable space $\alpha$ taking values in the operators on a Hilbert space $H$. For any measurable set $S \subseteq \alpha$, vectors $x, y \in H$, and a complex-valued function $g: \alpha \to \mathbb{C}$ that is integrable with respect to the scalar complex measure $\mu_{x, y}$ defined by $E \mapsto \langle y, \mu(E)x \rangle$, the following equality holds: $$\int g \, d\langle y, \mu(\cdot) (\mu(S)x) \rangle = \int \mathbb{1}_S g \, d\langle y, \mu(\cdot) x \rangle$$ where $\mathbb{1}_S$ denotes the indicator function of the set $S$.

Let $H$ be a Hilbert space and $\mu$ be a spectral measure (projection-valued 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 $E \mapsto \langle y, \mu(E)x \rangle$), the complex weak integral satisfies: $$\int g \, d\langle \mu(S)y, \mu x \rangle = \int_S g \, d\langle y, \mu x \rangle$$ where the right-hand side is the integral of $g$ over the set $S$ with respect to the scalar measure $\mu_{x, y}$.

Let $H$ be a complex Hilbert space and $\mu$ be a spectral measure (projection-valued measure) on a measurable space $\alpha$. Let $U: H \to H$ be a bounded linear operator and $f: \alpha \to \mathbb{C}$ be a complex-valued function. Suppose that for every pair of vectors $x, y \in H$, $f$ is integrable with respect to the scalar complex measure $\mu_{x,y}$ (defined by $\mu_{x,y}(E) = \langle y, \mu(E)x \rangle$) and the integral satisfies the relation $$\int_{\alpha} f \, d\langle y, \mu x \rangle = \langle y, U x \rangle.$$ Then, for any measurable set $S \subseteq \alpha$, the projection operator $\mu(S)$ commutes with $U$, that is: $$\mu(S) U = U \mu(S).$$

Let $H$ be a complex Hilbert space and $\mu$ be a spectral measure on $\mathbb{C}$ with values in the operators on $H$. Suppose $U: H \to H$ is a bounded linear operator such that for all $x, y \in H$, the identity function $z \mapsto z$ is integrable with respect to the scalar complex measure $\mu_{x, y}$ (defined by $\mu_{x, y}(S) = \langle y, \mu(S)x \rangle$) and satisfies the reconstruction property $\int z \, d\langle y, \mu x \rangle = \langle y, Ux \rangle$. If the total variation of each scalar measure $\mu_{x, y}$ is finite, then for any $x, y \in H$, it holds that: $$\int (1 - z) \, d\langle y, \mu x \rangle = \langle y, x - Ux \rangle$$

Let $H$ be a Hilbert space and $\mu$ be a spectral measure on $\mathbb{C}$. Let $U: H \to H$ be a bounded linear operator such that $U$ is the spectral integral of the identity function, expressed weakly as $\langle y, U x \rangle = \int z \, d\langle y, \mu(z) x \rangle$ for all $x, y \in H$. Assume that $U$ commutes with the spectral projections $\mu(S)$ for all measurable sets $S \subseteq \mathbb{C}$, and that the identity function is integrable with respect to the scalar measures $\langle y, \mu(\cdot) x \rangle$, which are assumed to have finite variation. For any vectors $v, y \in H$, let $x = v - Uv$. Then for any measurable set $S \subseteq \mathbb{C}$, the scalar measure $\langle y, \mu(S) x \rangle$ satisfies the following factorization: $$\langle y, \mu(S) x \rangle = \int (1 - z) \, d\langle y, \mu(z) (\mu(S) v) \rangle$$

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 integrable with respect to $\mu$, and let $g : \alpha \to \mathbb{R}$ be a real-valued function. Let $q\mu$ denote the vector measure obtained by taking $\mu$ with density $q$. If $g$ is integrable with respect to $q\mu$ and the variation of $q\mu$ satisfies $|\mu_q| = |q||\mu|$, then the integral of $g$ with respect to $q\mu$ satisfies: $$\int g \, d(q\mu) = \int (g \cdot q) \, d\mu$$ where on the right-hand side, $g$ is treated as a complex-valued function.

Let $H$ be a complex Hilbert space and $\mu$ be a projection-valued spectral measure. Let $U$ be a bounded linear operator on $H$ such that for all $x, y \in H$, the inner product $\langle y, U x \rangle$ is given by the complex weak integral of the identity function: $$\langle y, U x \rangle = \int z \, d\langle y, \mu z \rangle$$ Assume that for all $x, y \in H$, the identity function is integrable with respect to the scalar complex measure $\mu_{x, y}$ (where $\mu_{x, y}(S) = \langle y, \mu(S) x \rangle$) and that the variation of $\mu_{x, y}$ is finite. Suppose further that $U$ commutes with the spectral measure $\mu$, such that $\mu(S) U = U \mu(S)$ for every measurable set $S$. Then, for any vectors $x, v, y \in H$ such that $x = (I - U)v$, the scalar measure $\mu_{x, y}$ is equal to the scalar measure $\mu_{v, y}$ weighted by the density function $f(z) = 1 - z$: $$\mu_{x, y} = (1 - z) \mu_{v, y}$$ Equivalently, for any measurable set $S$, $\langle y, \mu(S) x \rangle = \int_S (1 - z) \, d\langle y, \mu v \rangle$.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint operator on $H$. Let $U_T$ be the Cayley transform of $T$ (the `cayleyBoundedOperator`) and $E$ be its associated projection-valued spectral measure (`cayleyBoundedSpectralMeasure`). For any measurable set $S \subseteq \mathbb{C}$, the projection operator $E(S)$ commutes with $U_T$: $$E(S) U_T = U_T E(S)$$

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint operator on $H$ with domain $\mathcal{D}(T)$. Let $E$ be the projection-valued spectral measure associated with the bounded Cayley transform of $T$. For any vector $x \in \mathcal{D}(T)$ and any vector $y \in H$, the inverse Cayley transform function $\Phi: \mathbb{C} \to \mathbb{R}$ is integrable with respect to the complex scalar measure $\mu_{x,y}$ (defined by $\mu_{x,y}(S) = \langle y, E(S)x \rangle$), and the following identity holds: $$\langle y, Tx \rangle = \int_{\mathbb{C}} \Phi(z) \, d\langle y, E(z)x \rangle$$ where $\Phi(z)$ is the inverse Cayley transform which maps the spectrum of the Cayley transform back to the real line.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint operator on $H$. Let $E_U$ be the projection-valued spectral measure on $\mathbb{C}$ associated with the Cayley transform of $T$ (the `cayleyBoundedSpectralMeasure`). Suppose that for every vector $x$ in the domain of $T$ and every $y \in H$, the inverse Cayley transform $\phi$ (the function mapping $z \in \mathbb{C}$ to $\mathbb{R}$, where $\phi(z) = \frac{-\text{Im} z}{1 - \text{Re} z}$ for $z \neq 1$) 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 following integral identity holds: $$\langle y, Tx \rangle = \int_{\mathbb{C}} \phi(z) \, d\langle y, E_U(z)x \rangle$$ Then the real spectral measure $\mu$ associated with $T$ (the `cayleyRealSpectralMeasure`) is a weak spectral resolution of $T$. Specifically, for every $x \in \text{dom}(T)$ and $y \in H$, the identity function $f(\lambda) = \lambda$ is integrable with respect to the scalar measure $\nu_{x,y}(A) = \langle y, \mu(A)x \rangle$ and satisfies: $$\langle y, Tx \rangle = \int_{\mathbb{R}} \lambda \, d\langle y, \mu(\lambda)x \rangle$$

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint operator on $H$. Let $\mu$ be the spectral measure on $\mathbb{R}$ associated with $T$ via its Cayley transform. Then $\mu$ is a weak spectral resolution of $T$. That is, for every $x$ in the domain of $T$ and every $y \in H$, the identity function $f(\lambda) = \lambda$ is integrable with respect to the complex measure $\mu_{x,y}$ defined by $\mu_{x,y}(A) = \langle y, \mu(A)x \rangle$, and the following identity holds: $$\langle y, Tx \rangle = \int_{\mathbb{R}} \lambda \, d\langle y, \mu(\lambda)x \rangle$$

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint operator on $H$. Let $E$ be the projection-valued measure on $\mathbb{R}$ obtained via the Cayley transform of $T$. Then the pair $(T, E)$ satisfies the spectral theorem for unbounded self-adjoint operators. This means that $T$ is the spectral integral of the identity function with respect to $E$ (i.e., $T = \int_{\mathbb{R}} \lambda \, dE(\lambda)$), and the domain of $T$ consists precisely of those vectors $x \in H$ for which $\int_{\mathbb{R}} \lambda^2 \, d\langle x, E(\lambda)x \rangle < \infty$.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint operator on $H$. Let $E$ be the spectral measure on $\mathbb{R}$ associated with $T$ via its Cayley transform. Then $T$ is a sub-operator of (i.e., its graph is contained in) the operator defined by the spectral integral $\int_{\mathbb{R}} \lambda \, dE(\lambda)$ acting on its maximal domain.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint operator on $H$. Let $\mu_T$ be the spectral measure (projection-valued measure) on $\mathbb{R}$ obtained via the Cayley transform of $T$. Then $T$ is equal to the spectral integral of the identity function $\lambda \mapsto \lambda$ with respect to $\mu_T$ defined on its maximal domain.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint operator on $H$ (which may be unbounded). Let $E$ be the spectral measure (projection-valued measure) on $\mathbb{R}$ obtained from the Cayley transform of $T$. Then the spectral theorem holds for $T$ in a domain-aware sense: the domain of $T$ is exactly the set of vectors $x \in H$ such that $\int_{\mathbb{R}} \lambda^2 \, d\langle x, E(\lambda)x \rangle < \infty$, and for any $x$ in the domain of $T$, the operator acts as $Tx = \int_{\mathbb{R}} \lambda \, dE(\lambda)x$.

Let $H$ be a complex Hilbert space and $T: D(T) \to H$ be a self-adjoint operator on $H$. Let $\mu$ be the spectral measure (projection-valued measure) on $\mathbb{R}$ obtained from the Cayley transform of $T$. Then $T$ is the integral of the identity function $\lambda \mapsto \lambda$ with respect to $\mu$. More precisely, the domain of $T$ is the set of vectors $x \in H$ with finite second moment, i.e., $D(T) = \{x \in H \mid \int_{\mathbb{R}} \lambda^2 \, d\mu_{x,x} < \infty\}$, and for any $x \in D(T)$, the operator satisfies $\langle y, T x \rangle = \int_{\mathbb{R}} \lambda \, d\langle y, \mu(\lambda) x \rangle$ for all $y \in H$.

Let $H$ be a complex Hilbert space and $T$ be an essentially self-adjoint operator on $H$. Let $\overline{T}$ denote the closure of $T$ and $E$ be the spectral measure (projection-valued measure) on $\mathbb{R}$ obtained from $\overline{T}$ via the Cayley transform. Then the spectral theorem holds for $\overline{T}$ with respect to $E$: the domain of $\overline{T}$ is the set of vectors $x \in H$ such that $\int_{\mathbb{R}} \lambda^2 \, d\langle x, E(\lambda)x \rangle < \infty$, and for all $x \in D(\overline{T})$ and $y \in H$, the operator satisfies $\langle y, \overline{T} x \rangle = \int_{\mathbb{R}} \lambda \, d\langle y, E(\lambda) x \rangle$.