PhyslibAlpha.ProbabilisticTheory.HilbertSpace.Unbounded.SpectralIntegral.Construction

For a given natural number $n$, the truncation function $f_n: \mathbb{R} \to \mathbb{C}$ is defined as the indicator function of the interval $[-n, n]$ applied to the identity function. Specifically, for any $r \in \mathbb{R}$: $$f_n(r) = \begin{cases} r & \text{if } -n \le r \le n \\ 0 & \text{otherwise} \end{cases}$$ where the result is treated as a complex number.

For a natural number $n$, this function is the real-valued function $f_n: \mathbb{R} \to \mathbb{R}$ defined as the indicator function of the closed interval $[-n, n]$ applied to the identity function. Specifically, $f_n(x) = x$ if $x \in [-n, n]$ and $f_n(x) = 0$ otherwise.

For each natural number $n$, the spectral cutoff set is the closed interval $[-n, n]$ in the real numbers $\mathbb{R}$.

For any natural numbers $n, m \in \mathbb{N}$, if $n \leq m$, then the spectral cutoff set $S_n = [-n, n]$ is a subset of the spectral cutoff set $S_m = [-m, m]$.

The union of the spectral cutoff sets $[-n, n]$ over all natural numbers $n \in \mathbb{N}$ is the set of all real numbers $\mathbb{R}$, expressed as $\bigcup_{n \in \mathbb{N}} [-n, n] = \mathbb{R}$.

For any natural number $n$, the truncation function $f_n: \mathbb{R} \to \mathbb{C}$ defined by $$f_n(r) = \begin{cases} r & \text{if } -n \le r \le n \\ 0 & \text{otherwise} \end{cases}$$ is measurable.

For any natural number $n$, the complex-valued truncation function $f_n: \mathbb{R} \to \mathbb{C}$ (which maps $r$ to $r$ if $r \in [-n, n]$ and to $0$ otherwise) is bounded. That is, there exists a constant $C \in \mathbb{R}$ such that for all $r \in \mathbb{R}$, $\|f_n(r)\| \leq C$.

For any $n \in \mathbb{N}$, the real-valued truncation function $f_n: \mathbb{R} \to \mathbb{R}$, defined as $f_n(x) = x$ if $x \in [-n, n]$ and $f_n(x) = 0$ otherwise, is a measurable function.

For any natural number $n$, there exists a constant $C \in \mathbb{R}$ such that for all $r \in \mathbb{R}$, the real-valued truncation function $f_n(r)$ satisfies $|f_n(r)| \le C$, where $f_n(r)$ is defined as $r$ if $r \in [-n, n]$ and $0$ otherwise.

For any natural number $n \in \mathbb{N}$, the real-valued truncation function $f_n : \mathbb{R} \to \mathbb{R}$ (which acts as the identity on $[-n, n]$ and vanishes elsewhere) and the complex-valued truncation function $g_n : \mathbb{R} \to \mathbb{C}$ are equal when the values of $f_n$ are considered as complex numbers. That is, for all $r \in \mathbb{R}$, $g_n(r) = (f_n(r) : \mathbb{C})$.

For any real number $r \in \mathbb{R}$, the truncation function $f_n(r)$ (defined as $r$ if $r \in [-n, n]$ and $0$ otherwise) is equal to $r$ for all sufficiently large natural numbers $n$. That is, there exists an $N \in \mathbb{N}$ such that for all $n \ge N$, $f_n(r) = r$.

For any real number $r \in \mathbb{R}$, let $f_n(r)$ be the truncation function defined as $r$ if $-n \le r \le n$ and $0$ otherwise. The sequence $f_n(r)$ converges to $r$ (as a complex number) as $n \to \infty$.

Let $H$ be a Hilbert space and $\mu_S$ be a spectral measure on $\mathbb{R}$. For any vectors $x, y \in H$, let $\mu_{x, y}$ be the complex scalar measure defined by $\mu_{x, y}(E) = \langle y, \mu_S(E) x \rangle$ for every measurable set $E$. Suppose that the identity function $r \mapsto r$ is integrable with respect to $\mu_{x, y}$. For each natural number $n$, let $f_n: \mathbb{R} \to \mathbb{C}$ be the truncation function defined by $$f_n(r) = \begin{cases} r & \text{if } |r| \le n \\ 0 & \text{otherwise} \end{cases}$$ Then the sequence of integrals of $f_n$ with respect to $\mu_{x, y}$ converges to the complex weak integral of the identity function as $n \to \infty$: $$\lim_{n \to \infty} \int_{\mathbb{R}} f_n(r) \, d\mu_{x, y} = \int_{\mathbb{R}} r \, d\mu_{x, y}.$$

Let $\alpha$ be a measurable space and $\mu$ be a complex-valued vector measure on $\alpha$. For any real number $c \in \mathbb{R}$ and any set $s \subseteq \alpha$, the vector integral of the complex-valued indicator function $\mathbf{1}_s \cdot c$ (with respect to the real-linear scalar multiplication on $\mathbb{C}$) is equal to the vector integral of the real-valued indicator function $\mathbf{1}_s \cdot c$ coerced to the complex numbers.

Let $\alpha$ be a measurable space and $\mu$ be a complex-valued vector measure on $\alpha$. For any real-valued function $g: \alpha \to \mathbb{R}$ that is integrable with respect to $\mu$, the integral of $g$ against $\mu$ (interpreting the product of the function's real values and the measure's complex values via real-scalar multiplication) is equal to the integral of $g$ cast to a complex-valued function against $\mu$ (interpreting the product via standard complex multiplication). That is, $$\int_{\alpha} g(x) \, d\mu(x) = \int_{\alpha} (g(x) + 0i) \, d\mu(x).$$

Let $H$ be a Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ taking values in the bounded operators on $H$. For a natural number $n$, the **truncation integral** is the bounded linear operator defined by integrating the $n$-th truncation function $f_n: \mathbb{R} \to \mathbb{C}$ with respect to $\mu$, where $f_n$ is defined as: $$f_n(r) = \begin{cases} r & \text{if } -n \le r \le n \\ 0 & \text{otherwise} \end{cases}$$ The resulting operator $\int f_n \, d\mu$ is an element of the space of bounded operators on $H$ equipped with the weak operator topology.

Let $H$ be a Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ taking values in the bounded operators on $H$. For any natural numbers $n$ and $m$, and any vector $x \in H$, let $f_k: \mathbb{R} \to \mathbb{C}$ be the $k$-th truncation function defined by $f_k(r) = r$ if $r \in [-k, k]$ and $0$ otherwise. Then the difference between the $n$-th and $m$-th truncated spectral integrals applied to $x$ is equal to the spectral integral of the difference of the truncation functions applied to $x$: $$\left(\int f_n \, d\mu\right)x - \left(\int f_m \, d\mu\right)x = \left(\int (f_n - f_m) \, d\mu\right)x$$ where the integral on the right-hand side is the spectral integral of the bounded measurable function $g(r) = f_n(r) - f_m(r)$.

Let $H$ be a Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ taking values in the operators on $H$. For any vector $x \in H$, let $\mu_x$ be the diagonal measure associated with $x$, defined such that for any measurable set $S \subseteq \mathbb{R}$, $\mu_x(S) = \|\mu(S)x\|^2$. Then for any vectors $x, y \in H$, the following inequality holds between measures: $$\mu_{x+y} \leq 2\mu_x + 2\mu_y$$

For any spectral measure $\mu$ on $\mathbb{R}$ acting on a Hilbert space $H$, the zero vector $0 \in H$ is an element of the spectral square moment domain, which is defined as the set of vectors $x \in H$ such that $$\int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda)x \rangle < \infty.$$

Let $\mu$ be a spectral measure on $\mathbb{R}$ acting on a Hilbert space $H$. Let $\mathcal{D}_\mu$ be the set of vectors $x \in H$ with a finite second moment with respect to $\mu$, defined by $$\mathcal{D}_\mu = \left\{ x \in H \mid \int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda)x \rangle < \infty \right\}.$$ For any vectors $x, y \in H$, if $x \in \mathcal{D}_\mu$ and $y \in \mathcal{D}_\mu$, then their sum $x + y \in \mathcal{D}_\mu$.

Let $\mu$ be a spectral measure on $\mathbb{R}$ acting on a Hilbert space $H$. For any complex scalar $c \in \mathbb{C}$ and any vector $x \in H$, if $x$ belongs to the domain of finite second moment of $\mu$ (i.e., $\int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda)x \rangle < \infty$), then the scaled vector $c \cdot x$ also belongs to this domain.

Let $H$ be a Hilbert space and $\mu$ be a spectral measure on $\mathbb{R}$ acting on $H$. The **spectral square moment submodule** is the complex submodule of $H$ whose underlying set consists of all vectors $x \in H$ with a finite second moment with respect to $\mu$. Specifically, a vector $x$ is an element of this submodule if $$\int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda) x \rangle < \infty.$$ This definition explicitly bundles the domain of finite second moments with the proofs that it is closed under addition, scalar multiplication by $\mathbb{C}$, and contains the zero vector.

Let $H$ be a Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ taking values in the bounded operators on $H$. For any natural numbers $n$ and $m$, and any vector $x \in H$, let $f_n$ and $f_m$ be the $n$-th and $m$-th truncation functions (where $f_k(r) = r$ if $r \in [-k, k]$ and $0$ otherwise). Let $\mu_x$ be the diagonal measure associated with $x$, defined by $\mu_x(S) = \|\mu(S)x\|^2$ for any measurable set $S \subseteq \mathbb{R}$. The squared norm of the difference between the $n$-th and $m$-th truncated spectral integrals applied to $x$ is equal to the integral of the squared difference of the truncation functions with respect to $\mu_x$: $$\left\| \left( \int f_n \, d\mu \right)x - \left( \int f_m \, d\mu \right)x \right\|^2 = \int_{\mathbb{R}} |f_n(r) - f_m(r)|^2 \, d\mu_x(r)$$

Let $H$ be a Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ taking values in the bounded operators on $H$. For any vector $x \in H$ and natural numbers $n, m \in \mathbb{N}$, the squared norm of the difference between the $n$-th and $m$-th truncated spectral integrals applied to $x$ is bounded by the sum of their individual truncation errors: $$\left\| \int f_n \, d\mu(x) - \int f_m \, d\mu(x) \right\|^2 \le 2 \int_{\mathbb{R}} |f_n(r) - r|^2 \, d\mu_x(r) + 2 \int_{\mathbb{R}} |f_m(r) - r|^2 \, d\mu_x(r)$$ where $f_k: \mathbb{R} \to \mathbb{C}$ is the truncation function defined as $f_k(r) = r$ for $r \in [-k, k]$ and $0$ otherwise, and $\mu_x$ is the diagonal measure associated with the vector $x$, defined by $\mu_x(S) = \|\mu(S)x\|^2$.

Let $H$ be a Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ taking values in the operators on $H$. For any vector $x \in H$, let $\mu_x$ be the diagonal measure associated with $x$, defined by $\mu_x(S) = \|\mu(S)x\|^2$ for any measurable set $S \subseteq \mathbb{R}$. Let $f_n: \mathbb{R} \to \mathbb{C}$ be the truncation function defined by $f_n(r) = r$ if $r \in [-n, n]$ and $f_n(r) = 0$ otherwise. If the function $r \mapsto r^2$ is integrable with respect to $\mu_x$, then the truncation error tends to zero as $n$ goes to infinity: $$\lim_{n \to \infty} \int_{\mathbb{R}} |f_n(r) - r|^2 \, d\mu_x(r) = 0$$

Let $H$ be a Hilbert space and $\mu$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ taking values in the bounded operators on $H$. For any vector $x \in H$ belonging to the domain of the second moment, defined as the set of vectors $x$ such that $\int_{\mathbb{R}} \lambda^2 \, d\mu_x(\lambda) < \infty$ (where $\mu_x(A) = \langle x, \mu(A)x \rangle$ is the associated diagonal measure), the sequence of vectors $(x_n)_{n \in \mathbb{N}}$ defined by the $n$-th truncated spectral integral $$x_n = \left( \int_{-n}^n \lambda \, d\mu(\lambda) \right) x$$ is a Cauchy sequence in $H$.

Let $H$ be a Hilbert space and $\mu$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ taking values in the operators on $H$. For any vector $x \in H$, let $\mu_x$ be the associated diagonal measure. Let $f_n: \mathbb{R} \to \mathbb{C}$ be the truncation function defined by $f_n(r) = r$ if $r \in [-n, n]$ and $f_n(r) = 0$ otherwise. If the function $r \mapsto r^2$ is integrable with respect to $\mu_x$, then $$\lim_{n \to \infty} \int_{\mathbb{R}} |f_n(r)|^2 \, d\mu_x(r) = \int_{\mathbb{R}} r^2 \, d\mu_x(r).$$

Let $H$ be a Hilbert space and $\mu$ be a spectral measure on $\mathbb{R}$ acting on $H$. Given a vector $x$ in the spectral square moment submodule (the set of vectors $x \in H$ such that $\int_{\mathbb{R}} \lambda^2 \, d\mu_x(\lambda) < \infty$), the **truncation limit** is the vector in $H$ defined as the limit of the truncated spectral integrals: $$\lim_{n \to \infty} \left( \int_{-n}^{n} \lambda \, d\mu(\lambda) \right) x$$ where the integral $\int_{-n}^{n} \lambda \, d\mu$ is the spectral integral of the identity function $f(\lambda) = \lambda$ restricted to the interval $[-n, n]$. This limit exists in $H$ because the sequence of truncated integrals is Cauchy for any $x$ in the finite-second-moment domain.

Let $H$ be a Hilbert space and $\mu$ be a spectral measure on $\mathbb{R}$ acting on $H$. For any vector $x$ in the spectral square moment submodule (the set of vectors $x \in H$ such that $\int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda) x \rangle < \infty$), the sequence of truncated spectral integrals $(\int_{-n}^{n} \lambda \, d\mu(\lambda))x$ converges to the truncation limit vector in $H$ as $n \to \infty$.

Let $H$ be a Hilbert space and $\mu$ be a spectral measure on $\mathbb{R}$ acting on $H$. For any vectors $x$ and $y$ in the spectral square moment submodule of $\mu$ (the set of vectors satisfying $\int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda) x \rangle < \infty$), the truncation limit, defined as $T(x) = \lim_{n \to \infty} \left( \int_{-n}^{n} \lambda \, d\mu(\lambda) \right) x$, satisfies the additive property: $$T(x + y) = T(x) + T(y)$$

Let $H$ be a Hilbert space and $\mu$ be a spectral measure on $\mathbb{R}$ acting on $H$. For any complex scalar $c \in \mathbb{C}$ and any vector $x$ in the spectral square moment submodule (the set of vectors $x \in H$ such that $\int_{\mathbb{R}} \lambda^2 \, d\mu_x(\lambda) < \infty$), the truncation limit of $c \cdot x$ is equal to $c$ times the truncation limit of $x$: $$\lim_{n \to \infty} \left( \int_{-n}^{n} \lambda \, d\mu(\lambda) \right) (c \cdot x) = c \cdot \lim_{n \to \infty} \left( \int_{-n}^{n} \lambda \, d\mu(\lambda) \right) x$$

Let $H$ be a Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ taking values in the bounded operators on $H$. For any vector $x \in H$ and natural number $n$, the squared norm of the $n$-th truncated spectral integral applied to $x$ satisfies the identity: $$\left\| \left( \int f_n \, d\mu \right) x \right\|^2 = \int_{\mathbb{R}} |f_n(r)|^2 \, d\mu_x(r)$$ where $f_n: \mathbb{R} \to \mathbb{C}$ is the truncation function defined as $f_n(r) = r$ if $r \in [-n, n]$ and $f_n(r) = 0$ otherwise, and $\mu_x$ is the diagonal measure associated with $x$, defined such that $\mu_x(S) = \|\mu(S)x\|^2$ for any measurable set $S \subseteq \mathbb{R}$.

Let $H$ be a Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ taking values in the bounded operators on $H$. For any natural number $n$, let $T_n = \int f_n \, d\mu$ be the $n$-th truncation integral associated with the real-valued function $f_n(r) = r \cdot \mathbb{1}_{[-n, n]}(r)$. Then $T_n$ is a self-adjoint operator, satisfying $T_n^* = T_n$.

Let $H$ be a Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ with values in the bounded operators on $H$. For any natural number $n$, let $T_n$ be the $n$-th truncation integral $\int f_n \, d\mu$, where $f_n$ is the function $f_n(r) = r$ for $r \in [-n, n]$ and $f_n(r) = 0$ otherwise. For any vectors $x, y \in H$, the inner product satisfies: $$\langle T_n x, y \rangle = \langle x, T_n y \rangle$$ This indicates that the truncation integral $T_n$ is a symmetric operator.

Let $H$ be a Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ acting on $H$. The **maximal spectral integral** is a partially defined complex linear operator from $H$ to $H$ whose domain is the spectral square-moment submodule. Specifically, the domain consists of all vectors $x \in H$ such that $$\int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda) x \rangle < \infty.$$ For any vector $x$ in this domain, the operator maps $x$ to the limit of the truncated spectral integrals: $$x \mapsto \lim_{n \to \infty} \left( \int_{-n}^{n} \lambda \, d\mu(\lambda) \right) x.$$ This operator is the maximal operator obtained from the real spectral measure by the norm convergence of its bounded truncations.

Let $H$ be a complex Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ acting on $H$. The maximal spectral integral operator $T = \int_{\mathbb{R}} \lambda \, d\mu(\lambda)$ is symmetric. That is, for every pair of vectors $x, y$ in the domain of $T$, the inner product satisfies: $$\langle Tx, y \rangle = \langle x, Ty \rangle$$ where the domain of $T$ consists of all vectors $x \in H$ such that $\int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda) x \rangle < \infty$.

Let $\mu$ be a spectral measure on $\mathbb{R}$ acting on a Hilbert space $H$. For any vector $x \in H$ and any non-negative real number $C \ge 0$, the vector $\mu([-C, C])x$ belongs to the domain of finite second moment of $\mu$. That is, if $y = \mu([-C, C])x$, then $$\int_{\mathbb{R}} \lambda^2 \, d\langle y, \mu(\lambda)y \rangle < \infty.$$

Let $\mu$ be a spectral measure on $\mathbb{R}$ acting on a Hilbert space $H$, and let $x \in H$ be a vector. Let $\mu_x$ be the diagonal measure associated with $x$, defined such that for any measurable set $S \subseteq \mathbb{R}$, $\mu_x(S) = \|\mu(S)x\|^2$. As $n$ tends to infinity, the measure of the complement of the interval $[-n, n]$ under $\mu_x$ tends to zero. That is, $$\lim_{n \to \infty} \mu_x(\mathbb{R} \setminus [-n, n]) = 0.$$

Let $\mu_S$ be a spectral measure on the real numbers $\mathbb{R}$ acting on a Hilbert space $H$. For any vector $x \in H$, the sequence of vectors obtained by applying the spectral projection of the interval $[-n, n]$ to $x$ converges to $x$ in the norm topology as $n \to \infty$. That is, $$\lim_{n \to \infty} \mu_S([-n, n])x = x$$ where $[-n, n]$ is the spectral cutoff set for a natural number $n$.

Let $H$ be a Hilbert space and $\mu$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ acting on $H$. The maximal spectral integral operator $\int \lambda \, d\mu$ is densely defined. That is, the domain of the operator, defined as the set of vectors $x \in H$ such that $$\int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda) x \rangle < \infty,$$ is a dense subspace of $H$.

Let $H$ be a Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ acting on $H$. Let $A = \int_{\mathbb{R}} \lambda \, d\mu$ be the maximal spectral integral operator, which is a partially defined linear operator on $H$ with the domain $$\mathcal{D}(A) = \left\{ x \in H : \int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda) x \rangle < \infty \right\}.$$ Then the operator $A$ is closable.

Let $H$ be a complex Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ acting on $H$. The maximal spectral integral operator $T = \int_{\mathbb{R}} \lambda \, d\mu(\lambda)$ is densely defined and closable.

Let $H$ be a Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ acting on $H$. Let $T = \int \lambda \, d\mu$ be the maximal spectral integral operator (the operator that maps vectors $x \in H$ in the spectral square-moment domain to the limit of truncated spectral integrals). Then the closure of $T$ is a closed operator.

Let $H$ be a complex Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ acting on $H$. Let $T$ be the maximal spectral integral operator $\int_{\mathbb{R}} \lambda \, d\mu$, which is the partially defined operator mapping $x \in H$ to $\int_{\mathbb{R}} \lambda \, d\mu(\lambda)$ on the domain of vectors $x$ satisfying $\int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu(\lambda) x \rangle < \infty$. Then the closure of $T$ is a symmetric operator.

Let $H$ be a complex Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ acting on $H$. Let $T = \int_{\mathbb{R}} \lambda \, d\mu(\lambda)$ be the maximal spectral integral operator. Then $T$ is essentially self-adjoint if and only if its closure $\overline{T}$ is a self-adjoint operator.

Let $H$ be a Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ acting on $H$. Let $T = \int_{\mathbb{R}} \lambda \, d\mu$ be the maximal spectral integral operator. If the ranges of the operators $T + iI$ and $T - iI$ are both equal to the entire space $H$, where $I$ is the identity operator, then $T$ is self-adjoint.

Let $H$ be a complex Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ acting on $H$. Let $A$ be the maximal spectral integral operator $\int_{\mathbb{R}} \lambda \, d\mu$. If the ranges of the operators $A + iI$ and $A - iI$ are both equal to the entire space $H$, where $i$ is the imaginary unit and $I$ is the identity operator, then $A$ is a closed operator, such that its closure $\overline{A}$ satisfies $\overline{A} = A$.

Let $H$ be a complex Hilbert space and $\mu$ be a weak spectral measure on $\mathbb{R}$ acting on $H$. Let $A = \int \lambda \, d\mu$ denote the maximal spectral integral operator. If the ranges of the operators $A + iI$ and $A - iI$ (where $I$ is the identity operator and $i$ is the imaginary unit) are both equal to the entire space $H$, then $A$ is essentially self-adjoint.