PhyslibAlpha.ProbabilisticTheory.HilbertSpace.Unbounded.WeakIntegral

Let $\alpha$ be a non-empty set and $H$ be a Hilbert space. Let $\mu_S$ be a spectral measure on $\alpha$. Let $f: \alpha \to \mathbb{C}$ be a complex-valued function and $\{s_n\}$ be a sequence of simple functions that converges uniformly to $f$ and is uniformly bounded (i.e., there exists $C \in \mathbb{R}$ such that $\|s_n(z)\| \leq C$ for all $n$ and $z \in \alpha$). For any vectors $x, y \in H$, let $\mu_{x,y}$ be the complex-valued scalar measure defined by the matrix coefficient $\mu_{x,y}(A) = \langle y, \mu_S(A)x \rangle$. If the total variation of $\mu_{x,y}$ is finite, then the inner product of $y$ with the vector-valued spectral integral of $f$ acting on $x$ is equal to the integral of $f$ with respect to the scalar measure $\mu_{x,y}$: $$\langle y, \left( \int_{\alpha} f \, d\mu_S \right) x \rangle = \int_{\alpha} f(z) \, d\mu_{x,y}(z)$$

For a real-valued function $f: \alpha \to \mathbb{R}$ and vectors $x, y \in H$, the weak integral is defined as the integral $\int f \, d\mu_{x,y}$ of $f$ with respect to the complex scalar measure $\mu_{x,y}$ associated with the spectral measure $\mu_S$, where $\mu_{x,y}(A) = \langle y, \mu_S(A)x \rangle$. The integration is performed using the scalar multiplication of the real values of $f$ on the complex values of the measure.

For a spectral measure $\mu_S$, a complex-valued multiplier function $f: \alpha \to \mathbb{C}$, and two vectors $x, y \in H$ in a Hilbert space, the complex weak integral is the integral of $f$ with respect to the complex-valued scalar measure $\mu_{x, y}$ associated with $\mu_S$ and the vectors $x, y$. This scalar measure is defined such that for any measurable set $E$, $\mu_{x, y}(E) = \langle y, \mu_S(E) x \rangle$. The integral is denoted as $\int f \, d\langle y, \mu_S x \rangle$.

Let $\alpha$ and $\beta$ be measurable spaces, and let $f: \alpha \to \beta$ be a measurable function. Let $\mu_S$ be a spectral measure on $\alpha$, and let $x, y$ be vectors in a Hilbert space $H$. Let $\mu_{x,y}$ denote the complex scalar measure defined by $\mu_{x,y}(E) = \langle y, \mu_S(E) x \rangle$. For a real-valued function $g: \beta \to \mathbb{R}$, if $g$ is almost everywhere strongly measurable with respect to the pushforward of the total variation of $\mu_{x,y}$ under $f$, and the composition $g \circ f$ is integrable with respect to $\mu_{x,y}$, then $$\int g \, d\langle y, (f_* \mu_S)(\cdot) x \rangle = \int (g \circ f) \, d\langle y, \mu_S(\cdot) x \rangle$$ where $f_* \mu_S$ denotes the pushforward of the spectral measure $\mu_S$ along the map $f$.

Let $\mu_S$ be a spectral measure on a measurable space $\alpha$ acting on a Hilbert space $H$, and let $f: \alpha \to \beta$ be a measurable function. For any vectors $x, y \in H$ and a complex-valued function $g: \beta \to \mathbb{C}$, if $g$ is almost everywhere strongly measurable with respect to the pushforward of the variation of the scalar measure $\mu_{x, y}$ (where $\mu_{x, y}(E) = \langle y, \mu_S(E) x \rangle$), and $g \circ f$ is integrable with respect to $\mu_{x, y}$, then the complex weak integral of $g$ with respect to the pushforward spectral measure $f_*\mu_S$ satisfies: $$\int_\beta g \, d\langle y, (f_*\mu_S) x \rangle = \int_\alpha (g \circ f) \, d\langle y, \mu_S x \rangle.$$

Let $H$ and $H'$ be complex Hilbert spaces and $u: H \cong H'$ be a unitary operator (linear isometric equivalence). For a spectral measure $\mu_S$ on $H$ and a real-valued function $f: \alpha \to \mathbb{R}$, let $u \mu_S u^*$ denote the spectral measure on $H'$ obtained by unitary conjugation. Then for any vectors $x, y \in H'$, the weak integral of $f$ with respect to the conjugated measure is given by $$\int f \, d\langle y, (u \mu_S u^*)(\cdot) x \rangle = \int f \, d\langle u^{-1} y, \mu_S(\cdot) u^{-1} x \rangle,$$ where $u^{-1}$ is the inverse of $u$.