PhyslibAlpha.ProbabilisticTheory.HilbertSpace.Unbounded.UnitaryInfra.SpectralMeasure

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. For a measurable set $S \subseteq \sigma(U)$ in the spectrum of $U$, `cfcSpectralOperatorAux` is the continuous linear operator $E(S): H \to H$ obtained via the Riesz representation theorem from the sesquilinear form $B_S(x, y) = \overline{\mu_{x, y}(S)}$. It is the unique operator such that for any $x, y \in H$, the inner product $\langle y, E(S) x \rangle$ is equal to the value of the polarized complex scalar measure $\mu_{x, y}$ evaluated at $S$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. For any Borel measurable set $S \subseteq \sigma(U)$ and any vectors $x, y \in H$, the inner product of the spectral operator $E(S)$ satisfies $$\langle y, E(S) x \rangle = \mu_{x, y}(S)$$ where $\mu_{x, y}$ is the polarized complex scalar measure associated with the operator $U$ and vectors $x, y$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. For any subset $S \subseteq \sigma(U)$ of the spectrum of $U$, the spectral operator $E(S): H \to H$ is defined as follows: if $S$ is a measurable set, $E(S)$ is the unique continuous linear operator such that for all $x, y \in H$, the inner product $\langle y, E(S) x \rangle$ equals the value of the polarized complex scalar measure $\mu_{x, y}$ evaluated at $S$. If $S$ is not measurable, $E(S)$ is defined to be the zero operator $0$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$ with spectrum $\sigma(U)$. For any subset $S \subseteq \sigma(U)$ and any vectors $x, y \in H$, the inner product of $y$ with the spectral operator $E(S)$ applied to $x$ is equal to the polarized complex scalar measure $\mu_{x, y}$ evaluated at $S$: $$\langle y, E(S) x \rangle = \mu_{x, y}(S)$$

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. Let $\sigma(U)$ denote the spectrum of $U$. For any subset $S \subseteq \sigma(U)$, if $S$ is not a measurable set, then the spectral operator $E(S)$ associated with $U$ is the zero operator $0$.

Let $U$ be a bounded normal operator on a complex Hilbert space $H$ and $\sigma(U)$ be its spectrum. For any vector $v \in H$, let $\mu_v$ be the scalar spectral measure on $\sigma(U)$ associated with the vector $v$ (the regular Borel measure representing the functional $f \mapsto \text{Re} \langle v, f(U) v \rangle$ for $f \in C(\sigma(U), \mathbb{R})$). Then for any Borel set $S \subseteq \sigma(U)$, the measure of $S$ with respect to $v$ is equal to the measure of $S$ with respect to the vector rotated by the negative imaginary unit, $-iv$: $$\mu_v(S) = \mu_{-iv}(S)$$

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$ with spectrum $\sigma(U)$. For any two vectors $x, y \in H$ and any Borel set $S \subseteq \sigma(U)$, let $\mu_{x, y}$ be the polarized complex spectral measure associated with $U$. Then the measure satisfies the conjugate symmetry property: $$\mu_{y, x}(S) = \overline{\mu_{x, y}(S)}$$ where $\overline{z}$ denotes the complex conjugate of $z \in \mathbb{C}$. This identity reflects the Hermitian symmetry of the underlying sesquilinear form, matching the convention that the spectral measure represents the inner product $\langle y, E(S) x \rangle$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$ with spectrum $\sigma(U)$. For any subset $S \subseteq \sigma(U)$, the spectral operator $E(S)$ (formally `cfcSpectralOperator U hU S`) is self-adjoint. If $S$ is measurable, $E(S)$ is the unique operator such that $\langle y, E(S) x \rangle = \mu_{x, y}(S)$ for all $x, y \in H$, where $\mu_{x, y}$ is the polarized complex spectral measure; if $S$ is not measurable, $E(S) = 0$. In both cases, $E(S)$ satisfies the property $\langle y, E(S) x \rangle = \langle E(S) y, x \rangle$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. Let $\mu_{x, y}$ denote the polarized complex spectral measure associated with vectors $x, y \in H$, which is a measure defined on the spectrum $\sigma(U)$. Then the measure of the whole spectrum is equal to the inner product of the vectors: $$\mu_{x, y}(\sigma(U)) = \langle y, x \rangle$$

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. Let $E(S)$ denote the spectral operator associated with a subset $S$ of the spectrum $\sigma(U)$. Then for any vector $x \in H$, the spectral operator associated with the entire spectrum applied to $x$ is equal to $x$, i.e., $E(\sigma(U))x = x$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. Let $E(S)$ denote the spectral operator (as defined by the continuous functional calculus) for a subset $S$ of the spectrum $\sigma(U)$. Then the spectral operator evaluated on the entire spectrum is the identity operator $I$ on $H$: $$E(\sigma(U)) = I$$

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. Let $\sigma(U)$ denote the spectrum of $U$. For any measurable set $S \subseteq \sigma(U)$ and any vector $x \in H$, the inner product of $x$ and the spectral operator $E(S)$ applied to $x$ is equal to the value of the scalar spectral measure $\mu_x$ evaluated at $S$: $$\langle x, E(S) x \rangle = \mu_x(S)$$ Here, $E(S)$ (represented by `cfcSpectralOperator`) is the spectral projection associated with the set $S$, and $\mu_x$ (represented by `cfcScalarMeasure`) is the regular Borel measure on the spectrum associated with the vector $x$ via the Riesz–Markov–Kakutani representation theorem.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. For any measurable subset $S \subseteq \sigma(U)$ of the spectrum of $U$ and any vector $x \in H$, the real part of the quadratic form of the spectral operator $E(S)$ at $x$ is equal to the value of the real-valued scalar spectral measure $\mu_x$ evaluated at $S$. That is, $$\text{Re} \langle x, E(S) x \rangle = \mu_x(S)$$ where $E(S)$ is the spectral operator associated with $S$ and $\mu_x$ is the regular Borel measure on $\sigma(U)$ associated with $x$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. For any measurable subset $S \subseteq \sigma(U)$ of the spectrum of $U$, the spectral operator $E(S)$ is a positive operator, meaning that $\langle x, E(S) x \rangle \ge 0$ for all $x \in H$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. For any measurable subset $S \subseteq \sigma(U)$ of the spectrum of $U$, the spectral operator $E(S)$ is non-negative, i.e., $0 \le E(S)$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. Let $\sigma(U)$ denote the spectrum of $U$. For any measurable subsets $S, T \subseteq \sigma(U)$, if $S \subseteq T$, then the corresponding spectral operators satisfy $E(S) \le E(T)$ in the Loewner order (i.e., $E(T) - E(S)$ is a positive operator).

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. For any measurable subset $S \subseteq \sigma(U)$ of the spectrum of $U$, the spectral operator $E(S)$ satisfies $E(S) \le I$, where $I$ denotes the identity operator on $H$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. Let $E(S)$ denote the spectral operator (provided by the continuous functional calculus) associated with a subset $S$ of the spectrum $\sigma(U)$. Then, the spectral operator of the empty set is the zero operator, i.e., $$E(\emptyset) = 0.$$

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. Let $E(S)$ denote the spectral operator associated with $U$ for a subset $S \subseteq \sigma(U)$. For any two disjoint measurable subsets $S$ and $T$ of the spectrum $\sigma(U)$, the spectral operator of their union is the sum of their individual spectral operators: $$E(S \cup T) = E(S) + E(T)$$

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. For any subset $S \subseteq \sigma(U)$ of the spectrum of $U$, this function returns the spectral operator $E(S)$ (see `cfcSpectralOperator`) as an element of the space of bounded linear operators on $H$ equipped with the weak operator topology (WOT).

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$ with spectrum $\sigma(U)$. For any subset $S \subseteq \sigma(U)$ and any vectors $x, y \in H$, the inner product of $y$ with the spectral operator $E(S)$ acting on $x$ is equal to the value of the polarized complex spectral measure $\mu_{x, y}$ evaluated on the set $S$: $$\langle y, E(S) x \rangle = \mu_{x, y}(S)$$ where $E(S)$ (represented by `cfcSpectralMeasureFun`) is the projection-valued measure associated with $U$, and $\mu_{x, y}$ is the polarized complex spectral measure corresponding to the vectors $x$ and $y$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. The spectral vector measure is a vector measure $E$ that assigns to each Borel subset $S$ of the spectrum $\sigma(U)$ a bounded linear operator $E(S)$ on $H$. This measure takes values in the space of bounded linear operators equipped with the weak operator topology (WOT). Specifically, for any vectors $x, y \in H$, the set function $S \mapsto \langle y, E(S) x \rangle$ is a complex-valued Borel measure on $\sigma(U)$, and the operator-valued measure is countably additive in the weak sense.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$ with spectrum $\sigma(U)$. For any subset $S \subseteq \sigma(U)$, the spectral vector measure evaluated at $S$ is equal to the spectral operator $E(S)$ defined by the spectral measure function.

Let $\mu$ be a finite measure on a space $X$. For any compact set $K \subseteq X$ and any non-negative continuous function $f: X \to \mathbb{R}$ with compact support such that $f(x) = 1$ for all $x \in K$, the measure of $K$ is bounded by the integral of $f$: $$\mu(K) \le \int_X f \, d\mu$$

Let $\mu$ be a finite measure on a space $X$. Let $V \subseteq X$ be an open set and $f: X \to \mathbb{R}$ be a continuous function with compact support. If the support of $f$ is contained in $V$ (i.e., $\text{supp}(f) \subseteq V$) and $0 \le f(x) \le 1$ for all $x \in X$, then the integral of $f$ with respect to $\mu$ satisfies $$\int_X f \, d\mu \leq \mu(V),$$ where $\mu(V)$ denotes the measure of the set $V$.

Let $X$ be a space with a finite measure $\mu$. Let $D \subseteq X$ be a measurable set and $f: X \to \mathbb{R}$ be a non-negative continuous function with compact support. If $f(x) \leq \mathbf{1}_D(x)$ for all $x \in X$ (where $\mathbf{1}_D$ is the indicator function of $D$), then the integral of $f$ with respect to $\mu$ is less than or equal to the measure of $D$: $$\int_X f \, d\mu \leq \mu(D)$$

Let $\mu$ be a finite measure on a space $X$. For any subset $S \subseteq X$ and any real number $\epsilon > 0$, if the measure $\mu(S)$ is strictly less than $\epsilon$ (expressed as an extended non-negative real number), then the real-valued measure of $S$, denoted $\mu_{\mathbb{R}}(S)$, is also strictly less than $\epsilon$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. For any measurable subset $S \subseteq \sigma(U)$ of the spectrum of $U$, the spectral operator $E(S)$ is non-negative, meaning that $0 \le E(S)$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. For any measurable subset $S$ of the spectrum $\sigma(U)$, the spectral operator $E(S)$ satisfies $E(S) \le I$, where $I$ is the identity operator on $H$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$ with spectrum $\sigma(U)$. For any two measurable sets $S, T \subseteq \sigma(U)$, if $S \subseteq T$, then the corresponding spectral operators $E(S)$ and $E(T)$ (defined via the continuous functional calculus) satisfy the operator inequality $E(S) \le E(T)$. This means that for all $x \in H$, the inner products satisfy $\langle x, E(S) x \rangle \le \langle x, E(T) x \rangle$.

Let $U$ be a bounded normal operator on a complex Hilbert space $H$, and let $\sigma(U)$ denote its spectrum. For any real-valued continuous function $f$ with compact support on $\sigma(U)$ and any vector $x \in H$, the real part of the inner product $\langle x, f(U) x \rangle$ is equal to the integral of $f$ with respect to the scalar spectral measure $\mu_x$ associated with $x$: $$\text{Re} \langle x, f(U) x \rangle = \int_{\sigma(U)} f(\lambda) \, d\mu_x(\lambda)$$

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$ with spectrum $\sigma(U)$. Let $K \subseteq \sigma(U)$ be a compact set and $V \subseteq \sigma(U)$ be an open set such that $K \subseteq V$. Let $f: \sigma(U) \to \mathbb{R}$ be a continuous function with compact support such that $0 \le f(z) \le 1$ for all $z \in \sigma(U)$. If $f(z) = 1$ for all $z \in K$ and the support of $f$ is contained in $V$, then the spectral operator $E(K)$ and the operator $f(U)$ (obtained via the continuous functional calculus) satisfy the operator inequality: $$E(K) \le f(U)$$ This inequality implies that for all $x \in H$, the inner products satisfy $\langle x, E(K) x \rangle \le \langle x, f(U) x \rangle$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$ with spectrum $\sigma(U)$. Let $V \subseteq \sigma(U)$ be an open set. For any real-valued, compactly supported continuous function $f: \sigma(U) \to \mathbb{R}$ such that $0 \le f(\lambda) \le 1$ for all $\lambda \in \sigma(U)$ and whose topological support $\text{supp}(f)$ is contained in $V$, the operator $f(U)$ defined by the continuous functional calculus and the spectral projection operator $E(V)$ satisfy the operator inequality: $$f(U) \le E(V)$$ In this context, $A \le B$ denotes the standard Loewner order on self-adjoint operators, meaning $\langle x, Ax \rangle \le \langle x, Bx \rangle$ for all $x \in H$.

Let $U$ be a bounded normal operator on a complex Hilbert space $H$, and let $\sigma(U)$ denote its spectrum. For any real-valued, compactly supported continuous function $f: \sigma(U) \to \mathbb{R}$, let $f(U)$ denote the operator obtained via the continuous functional calculus. Then the operator corresponding to the pointwise product $f^2$ satisfies $(f^2)(U) = f(U)^2$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$ with spectrum $\sigma(U)$. For any continuous real-valued function with compact support $f \in C_c(\sigma(U), \mathbb{R})$, the operator $(f - f^2)(U)$ obtained via the continuous functional calculus satisfies the identity: $$(f - f^2)(U) = f(U) - f(U)^2$$ where $f(U)$ denotes the continuous functional calculus of $U$ associated with the function $f$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. Let $\sigma(U)$ denote the spectrum of $U$, and for any subset $K \subseteq \sigma(U)$, let $E(K)$ denote the spectral operator associated with $K$. If $K$ is a compact set, then $E(K)$ is idempotent, satisfying the identity $$E(K)^2 = E(K)$$ where the multiplication denotes the composition of operators.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. For any measurable set $S \subseteq \sigma(U)$ in the spectrum of $U$, let $E(S)$ denote the spectral operator (formally `cfcSpectralOperatorAux`). Then $E(S)$ is idempotent, satisfying the identity $$E(S)^2 = E(S)$$ where the multiplication denotes the composition of operators.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. For any subset $S \subseteq \sigma(U)$ of the spectrum of $U$, let $E(S)$ denote the spectral operator (formally `cfcSpectralOperator`) associated with $S$. Then $E(S)$ is an idempotent operator, satisfying the identity $$E(S)^2 = E(S)$$ where the multiplication denotes the composition of operators. If $S$ is not a measurable set, $E(S)$ is defined to be the zero operator, which is trivially idempotent.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. For any subset $S \subseteq \sigma(U)$ of the spectrum of $U$, the spectral operator $E(S)$ (formally `cfcSpectralOperator U hU S`) is a star-projection. This means that $E(S)$ is both self-adjoint, satisfying $E(S)^* = E(S)$, and idempotent, satisfying $E(S)^2 = E(S)$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. The spectral measure of $U$ is a mapping $E$ from the Borel subsets of the spectrum $\sigma(U)$ to the space of bounded linear operators on $H$ equipped with the weak operator topology. For any Borel set $S \subseteq \sigma(U)$, the operator $E(S)$ is a star-projection, meaning it is both idempotent ($E(S)^2 = E(S)$) and self-adjoint ($E(S)^* = E(S)$). This measure is countably additive in the weak operator topology and satisfies $E(\sigma(U)) = I$, where $I$ is the identity operator on $H$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. Let $E$ denote the spectral measure of $U$ defined on its spectrum $\sigma(U)$. For any subset $S \subseteq \sigma(U)$, the value of the spectral measure $E$ evaluated at $S$ is equal to the spectral operator $E(S)$ (represented by `cfcSpectralMeasureFun`).

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$ with spectrum $\sigma(U)$. Let $E$ be the spectral measure of $U$ (denoted by `cfcSpectralMeasure`). For any two vectors $x, y \in H$, the scalar complex measure defined by $S \mapsto \langle y, E(S)x \rangle$ for Borel sets $S \subseteq \sigma(U)$ is equal to the polarized complex spectral measure $\mu_{x, y}$ associated with $x$ and $y$. (This measure $\mu_{x, y}$ is the unique complex measure such that $\int_{\sigma(U)} f \, d\mu_{x, y} = \langle y, f(U)x \rangle$ for every continuous function $f$ on the spectrum).

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$ with spectrum $\sigma(U)$. Let $E$ be the spectral measure associated with $U$. For any continuous real-valued function $f: \sigma(U) \to \mathbb{R}$ and any vectors $x, y \in H$, the integral of $f$ with respect to the complex-valued scalar measure $\mu_{x, y}(S) = \langle y, E(S) x \rangle$ is equal to the inner product of $y$ with the operator $f(U)$ acting on $x$. That is, $$\int_{\sigma(U)} f(\lambda) \, d\langle y, E(\lambda) x \rangle = \langle y, f(U) x \rangle,$$ where $f(U)$ is the operator defined by the continuous functional calculus for $U$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. Let $T$ be a bounded linear operator on $H$. If $T$ commutes with $U$ (i.e., $UT = TU$) and $T$ commutes with the adjoint operator $U^*$ (i.e., $U^*T = TU^*$), then for any continuous real-valued function $f$ defined on the spectrum $\sigma(U)$, $T$ commutes with the operator $f(U)$ obtained via the continuous functional calculus.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. Let $T$ be a bounded linear operator on $H$ that is an isometry (i.e., $T^* T = I$) and commutes with both $U$ and its adjoint $U^*$. Then for any vector $v \in H$, the scalar spectral measure $\mu_{Tv}$ of $U$ associated with the vector $Tv$ is equal to the scalar spectral measure $\mu_v$ associated with $v$.

Let $H$ be a complex Hilbert space and $U$ be a bounded normal operator on $H$. Let $T$ be a unitary operator on $H$ that commutes with both $U$ and its adjoint $U^*$ (i.e., $UT = TU$ and $U^*T = TU^*$). For any two vectors $x, y \in H$, the polarized complex spectral measure $\mu_{x, y}$ associated with $U$ satisfies the identity: $$\mu_{Tx, y} = \mu_{x, T^* y}$$ where $T^*$ denotes the adjoint of $T$. This indicates that the sesquilinear form of the spectral measure intertwines with the unitary operator $T$ such that testing at $(Tx, y)$ is equivalent to testing at $(x, T^* y)$.