PhyslibAlpha.AlgebraicFramework.HilbertSpace.Unbounded.SpectralIntegral.Construction

For a natural number $n$, the function $f_n: \mathbb{R} \to \mathbb{C}$ is the identity function restricted to the closed interval $[-n, n]$ and zero elsewhere. Specifically, it is defined as $f_n(x) = x$ if $x \in [-n, n]$ and $f_n(x) = 0$ otherwise.

For a given natural number $n \in \mathbb{N}$, the function $f_n: \mathbb{R} \to \mathbb{R}$ is the identity function restricted to the interval $[-n, n]$ and zero elsewhere. That is, $$f_n(x) = \begin{cases} x & \text{if } x \in [-n, n] \\ 0 & \text{if } x \notin [-n, n] \end{cases}$$ This is the product of the indicator function $\mathbb{1}_{[-n, n]}$ and the identity function $\text{id}_{\mathbb{R}}$.

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

Let $S_n = [-n, n] \subseteq \mathbb{R}$ denote the spectral cutoff set for a natural number $n \in \mathbb{N}$. The sequence of sets $S_n$ is monotone increasing with respect to set inclusion; that is, for any $n, m \in \mathbb{N}$, if $n \le m$, then $S_n \subseteq S_m$.

The union of the spectral cutoff sets $S_n = [-n, n]$ for all natural numbers $n \in \mathbb{N}$ is equal to the set of all real numbers $\mathbb{R}$: $$\bigcup_{n \in \mathbb{N}} [-n, n] = \mathbb{R}$$

For any natural number $n$, the truncated identity function $f_n: \mathbb{R} \to \mathbb{C}$, defined as $$f_n(x) = \begin{cases} x & \text{if } x \in [-n, n] \\ 0 & \text{if } x \notin [-n, n] \end{cases}$$ is measurable.

For any natural number $n \in \mathbb{N}$, the truncation function $f_n: \mathbb{R} \to \mathbb{C}$ (defined as $f_n(x) = x$ if $x \in [-n, n]$ and $f_n(x) = 0$ otherwise) is bounded. That is, there exists a constant $C \in \mathbb{R}$ such that for all $r \in \mathbb{R}$, $\|f_n(r)\| \le C$.

For any natural number $n \in \mathbb{N}$, the real truncation function $f_n: \mathbb{R} \to \mathbb{R}$, defined as $$f_n(x) = \begin{cases} x & \text{if } x \in [-n, n] \\ 0 & \text{if } x \notin [-n, n] \end{cases}$$ is measurable.

For any natural number $n \in \mathbb{N}$, the real truncation function $f_n: \mathbb{R} \to \mathbb{R}$, defined as $$f_n(x) = \begin{cases} x & \text{if } x \in [-n, n] \\ 0 & \text{if } x \notin [-n, n] \end{cases}$$ is bounded. That is, there exists a constant $C \in \mathbb{R}$ such that $|f_n(x)| \le C$ for all $x \in \mathbb{R}$.

For any natural number $n \in \mathbb{N}$, let $f_{n, \mathbb{R}}: \mathbb{R} \to \mathbb{R}$ be the real-valued truncation function defined as $$f_{n, \mathbb{R}}(x) = \begin{cases} x & \text{if } x \in [-n, n] \\ 0 & \text{if } x \notin [-n, n] \end{cases}$$ and let $f_{n, \mathbb{C}}: \mathbb{R} \to \mathbb{C}$ be the complex-valued truncation function defined as $$f_{n, \mathbb{C}}(x) = \begin{cases} x & \text{if } x \in [-n, n] \\ 0 & \text{if } x \notin [-n, n] \end{cases}$$ Then, the function mapping $x$ to the complex coercion of $f_{n, \mathbb{R}}(x)$ is equal to $f_{n, \mathbb{C}}$.

Let $f_n: \mathbb{R} \to \mathbb{C}$ be the truncation function defined by $f_n(x) = x$ for $x \in [-n, n]$ and $f_n(x) = 0$ otherwise. For any real number $r \in \mathbb{R}$, there exists a natural number $N$ such that for all $n \geq N$, $f_n(r) = r$.

For any real number $r$, the sequence of values $f_n(r)$ of the truncated identity functions $f_n$ (defined as $f_n(x) = x$ if $x \in [-n, n]$ and $f_n(x) = 0$ otherwise) converges to $r$ (viewed as a complex number) as $n \to \infty$.

Let $\mu$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ acting on a Hilbert space $H$. For any two vectors $x, y \in H$, let $\mu_{x,y}$ be the complex scalar measure defined by $\mu_{x,y}(S) = \langle y, \mu(S)x \rangle$. Let $f_n: \mathbb{R} \to \mathbb{C}$ be the truncated identity function defined by $f_n(r) = r$ for $r \in [-n, n]$ and $f_n(r) = 0$ otherwise. If the identity function $r \mapsto r$ is integrable with respect to $\mu_{x,y}$, then the sequence of integrals of the truncated functions converges to the complex weak integral of the identity: $$\lim_{n \to \infty} \int_{\mathbb{R}} f_n(r) \, d\mu_{x,y}(r) = \int_{\mathbb{R}} r \, d\mu_{x,y}(r)$$

Let $\alpha$ be a measurable space and $\mu$ be a complex-valued vector measure on $\alpha$. For any real number $c$ and any set $s \subseteq \alpha$, the vector integral of the complex-valued indicator function $x \mapsto \mathbb{1}_s(x) \cdot c$ (where $c$ is treated as a complex number) is equal to the vector integral of the real-valued indicator function $x \mapsto \mathbb{1}_s(x) \cdot c$ coerced to the complex numbers. Both integrals are taken with respect to the scalar multiplication of $\mathbb{C}$ by $\mathbb{R}$.

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 vector integral of $g$ with respect to $\mu$ is equal to the vector integral of $g$ when $g$ is treated as a complex-valued function (i.e., by embedding the real values into $\mathbb{C}$). That is: $$\int g \, d\mu = \int (g + 0i) \, d\mu$$ where the integration is performed via the scalar multiplication of real numbers on the complex space.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak-operator spectral measure on $\mathbb{R}$. For any natural number $n \in \mathbb{N}$, the truncation integral is the bounded linear operator on $H$ defined by the spectral integral of the $n$-th truncated identity function $f_n$. The function $f_n: \mathbb{R} \to \mathbb{C}$ is defined as $f_n(x) = x$ if $x \in [-n, n]$ and $f_n(x) = 0$ otherwise. This operator is denoted as: $$\int f_n \, d\mu_S = \int_{-n}^n x \, d\mu_S(x)$$

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak-operator spectral measure on $\mathbb{R}$. For any natural numbers $n, m \in \mathbb{N}$ and any vector $x \in H$, the difference of the truncation integrals applied to $x$ satisfies: $$\left( \int f_n \, d\mu_S \right) x - \left( \int f_m \, d\mu_S \right) x = \left( \int (f_n - f_m) \, d\mu_S \right) x$$ where $f_k: \mathbb{R} \to \mathbb{C}$ is the $k$-th truncation function defined as $f_k(r) = r$ if $r \in [-k, k]$ and $f_k(r) = 0$ otherwise, and the right-hand side is the bounded spectral integral of the difference function $f_n - f_m$.

Let $H$ be a Hilbert space and $\mu_S$ be a weak spectral measure on $\mathbb{R}$ taking values in the operators on $H$. For any vector $x \in H$, let $\mu_x$ denote the diagonal measure associated with $x$, which is a positive measure defined for any measurable set $E \subseteq \mathbb{R}$ by $\mu_x(E) = \langle x, \mu_S(E) x \rangle = \|\mu_S(E) x\|^2$. For any two vectors $x, y \in H$, the diagonal measure of their sum satisfies the inequality $$\mu_{x+y} \le 2\mu_x + 2\mu_y$$ in the sense of measures (i.e., $\mu_{x+y}(E) \le 2\mu_x(E) + 2\mu_y(E)$ for every measurable set $E$).

Let $H$ be a Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ taking values in the operators on $H$. The zero vector $0 \in H$ belongs to the square-moment domain of $\mu_S$, which is the set of vectors $x$ such that $\int_{\mathbb{R}} \lambda^2 d\mu_x(\lambda) < \infty$ for the diagonal measure $\mu_x(E) = \langle x, \mu_S(E) x \rangle$.

Let $H$ be a Hilbert space and let $\mu_S$ be a weak spectral measure on $\mathbb{R}$ taking values in the operators on $H$. The **square-moment domain** of $\mu_S$, denoted here as $\mathcal{D}(\mu_S)$, is the set of vectors $x \in H$ such that the diagonal measure $\mu_x$ satisfies $\int_{\mathbb{R}} \lambda^2 d\mu_x(\lambda) < \infty$, where $\mu_x(E) = \langle x, \mu_S(E) x \rangle$. For any vectors $x, y \in H$, if $x \in \mathcal{D}(\mu_S)$ and $y \in \mathcal{D}(\mu_S)$, then their sum $x + y$ is also in $\mathcal{D}(\mu_S)$.

Let $H$ be a Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ acting on $H$. The square-moment domain of $\mu_S$, denoted as $\mathcal{D}$, is the set of vectors $x \in H$ such that the function $\lambda \mapsto \lambda^2$ is integrable with respect to the diagonal measure $\mu_x$, i.e., $\int_{\mathbb{R}} \lambda^2 d\mu_x(\lambda) < \infty$, where $\mu_x(S) = \langle x, \mu_S(S) x \rangle$. For any scalar $c \in \mathbb{C}$ and any vector $x \in \mathcal{D}$, the scalar multiple $c \cdot x$ is also in the square-moment domain $\mathcal{D}$.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ taking values in the operators on $H$. The **spectral square-moment submodule** is the $\mathbb{C}$-submodule of $H$ consisting of all vectors $x \in H$ such that the function $\lambda \mapsto \lambda^2$ is integrable with respect to the associated diagonal measure $\mu_x(E) = \langle x, \mu_S(E) x \rangle$. Specifically, a vector $x$ belongs to this submodule if it satisfies: $$\int_{\mathbb{R}} \lambda^2 d\langle x, \mu_S(\lambda) x \rangle < \infty.$$ This submodule serves as the domain for the maximal spectral integral of real-valued measurable functions.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak-operator spectral measure on $\mathbb{R}$. For any natural numbers $n, m \in \mathbb{N}$ and any vector $x \in H$, the squared norm of the difference between the $n$-th and $m$-th truncation integrals applied to $x$ is equal to the $L^2$ integral of the squared difference of the truncation functions $f_n$ and $f_m$ with respect to the diagonal measure $\mu_x$. That is, $$\left\| \int f_n \, d\mu_S(x) - \int f_m \, d\mu_S(x) \right\|^2 = \int_{\mathbb{R}} |f_n(r) - f_m(r)|^2 \, d\mu_x(r)$$ where $f_k: \mathbb{R} \to \mathbb{C}$ is the identity function restricted to $[-k, k]$ (and zero elsewhere), and $\mu_x$ is the positive measure defined by $\mu_x(E) = \|\mu_S(E)x\|^2$.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak-operator spectral measure on $\mathbb{R}$. For any vector $x \in H$, let $\mu_x$ denote the diagonal measure associated with $x$, where $\mu_x(E) = \langle x, \mu_S(E) x \rangle$. Let $f_n: \mathbb{R} \to \mathbb{C}$ be the $n$-th truncated identity function, defined as $f_n(r) = r$ if $r \in [-n, n]$ and $f_n(r) = 0$ otherwise. Let $\int f_n \, d\mu_S$ denote the $n$-th truncated spectral integral operator. For any $n, m \in \mathbb{N}$ and $x \in H$, the squared norm of the difference between the $n$-th and $m$-th truncated integrals applied to $x$ satisfies the inequality: $$\left\| \left( \int f_n \, d\mu_S \right) x - \left( \int f_m \, d\mu_S \right) x \right\|^2 \le 2 \int_{\mathbb{R}} |f_n(r) - r|^2 \, d\mu_x(r) + 2 \int_{\mathbb{R}} |f_m(r) - r|^2 \, d\mu_x(r)$$

Let $\mu_S$ be a weak spectral measure (WOT-spectral measure) on $\mathbb{R}$ over a Hilbert space $H$. For any vector $x \in H$, let $\mu_x$ denote the associated diagonal measure. If the function $r \mapsto r^2$ is integrable with respect to $\mu_x$, then the $L^2$-norm of the difference between the identity function and its truncation $f_n$ tends to zero as $n \to \infty$. That is, $$\lim_{n \to \infty} \int_{\mathbb{R}} |f_n(r) - r|^2 \, d\mu_x(r) = 0$$ where $f_n(r)$ is defined as $r$ if $r \in [-n, n]$ and $0$ otherwise.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak-operator spectral measure on $\mathbb{R}$. For any vector $x \in H$ belonging to the square-moment domain of $\mu_S$ (i.e., such that $\int_{\mathbb{R}} \lambda^2 \, d\mu_x(\lambda) < \infty$), the sequence of vectors $(v_n)_{n \in \mathbb{N}}$ defined by the truncated spectral integrals $$v_n = \left( \int_{-n}^n \lambda \, d\mu_S(\lambda) \right) x$$ is a Cauchy sequence in $H$.

Let $H$ be a Hilbert space and $\mu_S$ 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 associated diagonal measure. If the function $r \mapsto r^2$ is integrable with respect to $\mu_x$, then the integral of the squared norm of the truncated identity function $f_n$ converges to the second moment of $\mu_x$ as $n \to \infty$: $$\lim_{n \to \infty} \int_{\mathbb{R}} |f_n(r)|^2 \, d\mu_x(r) = \int_{\mathbb{R}} r^2 \, d\mu_x(r),$$ where $f_n(r) = r$ for $r \in [-n, n]$ and $f_n(r) = 0$ otherwise.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ taking values in the operators on $H$. For a vector $x$ belonging to the spectral square-moment submodule (the set of vectors $x \in H$ such that the second moment $\int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu_S(\lambda) x \rangle$ is finite), the **truncation limit** is the vector in $H$ defined as the limit of the sequence of truncated spectral integrals applied to $x$: $$\lim_{n \to \infty} \left( \int_{-n}^n \lambda \, d\mu_S(\lambda) \right) x$$ where $\int_{-n}^n \lambda \, d\mu_S(\lambda)$ is the spectral integral of the identity function truncated to the interval $[-n, n]$.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$. For any vector $x$ belonging to the spectral square-moment submodule of $\mu_S$ (the set of vectors $x \in H$ such that $\int_{\mathbb{R}} \lambda^2 d\langle x, \mu_S(\lambda) x \rangle < \infty$), the sequence of truncated spectral integrals applied to $x$ converges to the truncation limit as $n \to \infty$: $$\lim_{n \to \infty} \left( \int_{-n}^n \lambda \, d\mu_S(\lambda) \right) x = \text{truncationLimit}(\mu_S, x).$$

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ taking values in the operators on $H$. Let $x$ and $y$ be vectors in the spectral square-moment submodule of $\mu_S$ (the set of vectors $v \in H$ such that $\int_{\mathbb{R}} \lambda^2 \, d\langle v, \mu_S(\lambda) v \rangle < \infty$). Then the truncation limit is additive: $$\text{truncationLimit}(\mu_S, x + y) = \text{truncationLimit}(\mu_S, x) + \text{truncationLimit}(\mu_S, y)$$ where the truncation limit for a vector $v$ is defined as the limit in $H$ of the truncated spectral integrals: $$\text{truncationLimit}(\mu_S, v) = \lim_{n \to \infty} \left( \int_{-n}^n \lambda \, d\mu_S(\lambda) \right) v.$$

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ taking values in the operators on $H$. For any vector $x$ in the spectral square-moment submodule (the set of vectors $x \in H$ such that $\int_{\mathbb{R}} \lambda^2 d\langle x, \mu_S(\lambda) x \rangle < \infty$) and any scalar $c \in \mathbb{C}$, the truncation limit is homogeneous with respect to scalar multiplication: $$\lim_{n \to \infty} \left( \int_{-n}^n \lambda \, d\mu_S(\lambda) \right) (cx) = c \lim_{n \to \infty} \left( \int_{-n}^n \lambda \, d\mu_S(\lambda) \right) x$$

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak-operator spectral measure on $\mathbb{R}$. For any natural number $n$ and vector $x \in H$, the squared norm of the image of $x$ under the $n$-th truncated spectral integral $\int f_n \, d\mu_S$ is equal to the integral of the squared magnitude of the truncation function $f_n$ with respect to the diagonal measure $\mu_x$. Mathematically, this is expressed as: $$\left\| \left( \int f_n \, d\mu_S \right) x \right\|^2 = \int_{\mathbb{R}} |f_n(r)|^2 \, d\mu_x(r)$$ where $f_n(r) = r$ for $r \in [-n, n]$ and $0$ otherwise, and $\mu_x$ is the diagonal measure such that $\mu_x(S) = \langle x, \mu_S(S) x \rangle$.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak-operator spectral measure on $\mathbb{R}$. For any natural number $n \in \mathbb{N}$, the $n$-th truncated spectral integral $T_n = \int_{-n}^n x \, d\mu_S(x)$ is a self-adjoint operator. That is, the adjoint of the truncation integral is equal to the integral itself: $$T_n^* = T_n$$ where $T_n^*$ (denoted by `star` in the formal statement) represents the adjoint of the bounded linear operator $T_n$.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak-operator spectral measure on $\mathbb{R}$. For any natural number $n \in \mathbb{N}$ and any vectors $x, y \in H$, the $n$-th truncated spectral integral (denoted as $\int f_n \, d\mu_S$) satisfies the symmetry property: $$\left\langle \left( \int f_n \, d\mu_S \right) x, y \right\rangle_{\mathbb{C}} = \left\langle x, \left( \int f_n \, d\mu_S \right) y \right\rangle_{\mathbb{C}}$$ where $\langle \cdot, \cdot \rangle_{\mathbb{C}}$ denotes the inner product on $H$.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ taking values in the operators on $H$. The **maximal spectral integral** is the partially defined linear operator $A: D(A) \subseteq H \to H$ whose domain $D(A)$ is the spectral square-moment submodule: $$D(A) = \left\{ x \in H \;\middle|\; \int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu_S(\lambda) x \rangle < \infty \right\}$$ For any vector $x \in D(A)$, the value of the operator is defined as the limit of the truncated spectral integrals: $$A x = \lim_{n \to \infty} \left( \int_{-n}^n \lambda \, d\mu_S(\lambda) \right) x$$ where $\langle \cdot, \cdot \rangle$ denotes the inner product on $H$.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ taking values in the bounded operators on $H$. The maximal spectral integral $A = \int \lambda \, d\mu_S$ is a symmetric operator. Specifically, for any vectors $x, y$ in the domain $$D(A) = \left\{ x \in H \;\middle|\; \int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu_S(\lambda) x \rangle < \infty \right\},$$ the inner product satisfies: $$\langle Ax, y \rangle = \langle x, Ay \rangle.$$

Let $H$ be a Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$. For any vector $x \in H$ and any non-negative real constant $C \geq 0$, the vector obtained by applying the spectral projection of the interval $[-C, C]$ to $x$, denoted by $\mu_S([-C, C])x$, belongs to the square-moment domain of $\mu_S$. This means that the integral of the squared coordinate function with respect to the diagonal measure associated with $\mu_S([-C, C])x$ is finite: $$\int_{\mathbb{R}} \lambda^2 \, d\mu_{\mu_S([-C, C])x}(\lambda) < \infty$$

Let $H$ be a Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$. For any vector $x \in H$, let $\mu_x$ denote the diagonal measure associated with $x$, defined for any measurable set $S \subseteq \mathbb{R}$ as $\mu_x(S) = \text{Re} \langle x, \mu_S(S) x \rangle$ (which is equivalent to $\|\mu_S(S) x\|^2$). Let $I_n = [-n, n]$ be the $n$-th spectral cutoff set. Then the measure of the complement of $I_n$ tends to zero as $n$ goes to infinity: $$\lim_{n \to \infty} \mu_x(\mathbb{R} \setminus I_n) = 0$$

Let $H$ be a Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on the real line $\mathbb{R}$. For any vector $x \in H$, let $I_n = [-n, n]$ be the $n$-th spectral cutoff set. Then the sequence of vectors obtained by applying the spectral projection of $I_n$ to $x$ converges to $x$ in $H$ as $n \to \infty$: $$\lim_{n \to \infty} \mu_S([-n, n]) x = x$$

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ taking values in the operators on $H$. The maximal spectral integral $A = \int \lambda \, d\mu_S$ is a densely defined operator. That is, the domain of $A$, defined as $$D(A) = \left\{ x \in H \;\middle|\; \int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu_S(\lambda) x \rangle < \infty \right\},$$ is a dense subspace of $H$.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on the real line $\mathbb{R}$. The maximal spectral integral operator $A = \int_{\mathbb{R}} \lambda \, d\mu_S(\lambda)$ is a closable operator.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ taking values in the operators on $H$. The maximal spectral integral $A = \int_{\mathbb{R}} \lambda \, d\mu_S(\lambda)$, defined on the domain $$D(A) = \left\{ x \in H \;\middle|\; \int_{\mathbb{R}} \lambda^2 \, d\langle x, \mu_S(\lambda) x \rangle < \infty \right\},$$ is a densely defined and closable operator.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ taking values in the operators on $H$. Let $A = \int \lambda \, d\mu_S$ be the maximal spectral integral operator associated with $\mu_S$. Then the closure of $A$ is a closed operator.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ taking values in the operators on $H$. Let $A = \int_{\mathbb{R}} \lambda \, d\mu_S(\lambda)$ be the maximal spectral integral associated with $\mu_S$. Then the closure of $A$, denoted $\overline{A}$, is a symmetric operator. That is, for all $x, y$ in the domain of $\overline{A}$, the inner product satisfies $\langle \overline{A}x, y \rangle = \langle x, \overline{A}y \rangle$.

Let $H$ be a complex Hilbert space and $\mu_S$ be a weak operator topology (WOT) spectral measure on $\mathbb{R}$ taking values in the operators on $H$. Let $A = \int_{\mathbb{R}} \lambda \, d\mu_S(\lambda)$ be the maximal spectral integral associated with $\mu_S$. Then $A$ is essentially self-adjoint if and only if its closure $\bar{A}$ is a self-adjoint operator.