Physlib.QuantumMechanics.DDimensions.Operators.SpectralTheory.Basic

Let $H$ be a complex Hilbert space and $T: D(T) \subseteq H \to H$ be a partial linear operator. For any complex number $z \in \mathbb{C}$, the resolvent of $T$ at $z$ is the partial linear operator defined as the inverse of $T - zI$, where $I$ denotes the identity operator on $H$. This is denoted as $(T - zI)^{-1}$.

The symbol $\mathcal{R}$ is the notation used to represent the resolvent operator of a linear operator $T$ at a complex number $z$. In spectral theory, this usually refers to the operator $(T - zI)^{-1}$.

For a partially defined linear operator $T$ on a complex Hilbert space $H$, a complex number $z \in \mathbb{C}$, and a real number $c \in \mathbb{R}$, the property `IsLowerBound T z c` states that for all $x$ in the domain of $T$, the following inequality holds: $$c \|x\| \leq \|Tx - zx\|$$ where $\| \cdot \|$ denotes the norm in $H$.

Let $T$ be a partially defined linear operator on a complex Hilbert space $H$, $z$ be a complex number, and $c$ be a real number. If the inequality $c \|x\| \leq \|Tx - zx\|$ holds for all $x$ in the domain of $T$, then the inequality $c \|x\| \leq \|(-T)x - (-z)x\|$ holds for all $x$ in the domain of $-T$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. For any complex number $z \in \mathbb{C}$ and real numbers $c_1, c_2 \in \mathbb{R}$, if $c_1 \le c_2$ and the operator satisfies the lower bound condition $c_2 \|x\| \le \|Tx - zx\|$ for all $x$ in the domain of $T$, then it also satisfies the lower bound condition $c_1 \|x\| \le \|Tx - zx\|$ for all $x$ in the domain of $T$.

Let $H$ be a complex Hilbert space. Let $T_1$ and $T_2$ be partially defined linear operators on $H$ such that $T_1 \subseteq T_2$ (that is, $T_2$ is an extension of $T_1$, meaning $\text{dom}(T_1) \subseteq \text{dom}(T_2)$ and $T_1 x = T_2 x$ for all $x \in \text{dom}(T_1)$). If $T_2$ satisfies a lower bound of the form $c \|x\| \le \|T_2 x - zx\|$ for some $c \in \mathbb{R}$ and $z \in \mathbb{C}$ for all $x \in \text{dom}(T_2)$, then $T_1$ satisfies the same lower bound $c \|x\| \le \|T_1 x - zx\|$ for all $x \in \text{dom}(T_1)$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. For any complex number $z \in \mathbb{C}$ and real number $c \in \mathbb{R}$, if $T$ satisfies the lower bound $c \|x\| \leq \|(T - z I)x\|$ for all $x$ in the domain of $T$, then the closure of the operator $\bar{T}$ also satisfies the lower bound $c \|x\| \leq \|(\bar{T} - z I)x\|$ for all $x$ in the domain of $\bar{T}$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$ with domain $D(T) \subseteq H$. The regularity domain of $T$ is the set of complex numbers $z \in \mathbb{C}$ such that there exists a constant $c > 0$ satisfying $$\|Tx - zx\| \geq c \|x\|$$ for all $x \in D(T)$, where $\|\cdot\|$ denotes the norm on $H$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$ with domain $D(T) \subseteq H$. Let $\text{reg}(T)$ denote the regularity domain of $T$, which is the set of complex numbers $z \in \mathbb{C}$ such that there exists a constant $c > 0$ satisfying $\|Tx - zx\| \geq c \|x\|$ for all $x \in D(T)$. Then the regularity domain of the negated operator $-T$ is the set of the negatives of the elements in the regularity domain of $T$: $$\text{reg}(-T) = \{ -z \mid z \in \text{reg}(T) \}.$$

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$ with domain $D(T)$. For any non-zero complex scalar $w \in \mathbb{C}$, the regularity domain of the scaled operator $wT$ is equal to the scaling of the regularity domain of $T$. That is, $$\text{regularityDomain}(wT) = w \cdot \text{regularityDomain}(T),$$ where $w \cdot \text{regularityDomain}(T) = \{ wz \mid z \in \text{regularityDomain}(T) \}$.

Let $H$ be a complex Hilbert space. For partially defined linear operators $T$ and $T'$ on $H$, if $T'$ is an extension of $T$ (denoted $T \leq T'$), then the regularity domain of $T'$ is a subset of the regularity domain of $T$: $$\text{regularityDomain}(T') \subseteq \text{regularityDomain}(T).$$ (Note: $T \leq T'$ means the domain $D(T)$ is a subspace of $D(T')$ and the operators agree on $D(T)$).

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$ with domain $D(T) \subseteq H$. A complex number $z \in \mathbb{C}$ belongs to the regularity domain of $T$ if and only if the operator $T - zI$ (where $I$ is the identity operator) is injective (i.e., its kernel is $\{0\}$) and its inverse $(T - zI)^{-1}$ is continuous.

Let $T$ be a partially defined linear operator on a complex Hilbert space $H$ with domain $D(T)$. If there exists a constant $c \in \mathbb{R}$ such that $c \|x\| \leq \|Tx - zx\|$ for all $x \in D(T)$ and some complex number $z \in \mathbb{C}$, then the open ball $B(z, c) = \{w \in \mathbb{C} : |z - w| < c\}$ is a subset of the regularity domain of $T$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$ with domain $D(T) \subseteq H$. The regularity domain of $T$, defined as the set of complex numbers $z \in \mathbb{C}$ for which there exists a constant $c > 0$ such that $\|Tx - zx\| \geq c\|x\|$ for all $x \in D(T)$, is an open subset of $\mathbb{C}$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$ with domain $D(T)$. The regularity domain of $T$, denoted as $\text{reg}(T)$, is the set of complex numbers $z \in \mathbb{C}$ such that there exists a constant $c > 0$ satisfying $\|Tx - zx\| \geq c \|x\|$ for all $x \in D(T)$. This theorem states that the regularity domain of $T$ is equal to the regularity domain of its closure $\overline{T}$: $$\text{reg}(\overline{T}) = \text{reg}(T)$$

Let $H$ be a complex Hilbert space and $T$ be a closable partially defined linear operator on $H$ with closure $\overline{T}$. If $z \in \mathbb{C}$ belongs to the regularity domain of $T$, then the closure of the range of $T - zI$ is equal to the range of $\overline{T} - zI$: $$\overline{\text{ran}(T - zI)} = \text{ran}(\overline{T} - zI)$$ where $I$ denotes the identity operator.

Let $H$ be a complex Hilbert space and $T$ be a closed, partially defined linear operator on $H$ with domain $D(T)$. If $z \in \mathbb{C}$ belongs to the regularity domain of $T$ (meaning there exists a constant $c > 0$ such that $\|Tx - zx\| \geq c \|x\|$ for all $x \in D(T)$), then the range of the operator $T - zI$ is a closed subspace of $H$.

Let $H$ be a Hilbert space and $T$ be an unbounded linear operator on $H$. Let $\bar{T}$ denote the closure of $T$ and $T^*$ denote its adjoint. For any complex number $z \in \mathbb{C}$, the orthogonal complement of the range of $\bar{T} - zI$ is equal to the kernel of $T^* - \bar{z}I$: $$(\operatorname{ran}(\bar{T} - zI))^\perp = \ker(T^* - \bar{z}I)$$ where $I$ is the identity operator on $H$ and $\bar{z}$ is the complex conjugate of $z$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. Let $\bar{T}$ denote the closure of $T$ and $T^\dagger$ denote the adjoint of $T$. If $z \in \mathbb{C}$ belongs to the regularity domain of $T$, then the orthogonal complement of the kernel of $T^\dagger - \bar{z}I$ is equal to the range of $\bar{T} - zI$: $$(\ker(T^\dagger - \bar{z}I))^\perp = \text{ran}(\bar{T} - zI)$$ where $I$ is the identity operator and $\bar{z}$ is the complex conjugate of $z$.

For a partial linear map $T$ on a Hilbert space $H$ and a complex number $z \in \mathbb{C}$, the deficiency subspace is the closed subspace of $H$ defined as the orthogonal complement of the range of the operator $T - zI$, where $I$ is the identity map on $H$. It is denoted as $\text{Ran}(T - zI)^\perp$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. For any complex number $z \in \mathbb{C}$, the deficiency subspace of $T$ at $z$ is equal to the orthogonal complement of the range of the operator $T - zI$, where $I$ is the identity operator on $H$: $$\text{deficiencySubspace}(T, z) = (\text{ran}(T - zI))^\perp$$

For a partial linear map $T$ on a complex Hilbert space $H$ and a complex number $z \in \mathbb{C}$, the defect number is the dimension (rank) of the deficiency subspace at $z$. This deficiency subspace is the orthogonal complement of the range of the operator $T - zI$, where $I$ is the identity map. The defect number is expressed as $\dim(\text{Ran}(T - zI)^\perp)$.

Let $H$ be a complex Hilbert space, $T$ be a partially defined linear operator on $H$, and $z \in \mathbb{C}$. The defect number of $T$ at $z$ is equal to the dimension (rank) of the deficiency subspace of $T$ at $z$ over the field $\mathbb{C}$.

Let $H$ be a complex Hilbert space and $T$ be a closed partially defined linear operator on $H$. For any complex number $z$ in the regularity domain of $T$, the defect number of $T$ at $z$ is equal to $0$ if and only if the range of the operator $T - zI$ is the entire space $H$, where $I$ is the identity operator on $H$.

Let $H$ be a Hilbert space and $T$ be a partially defined linear operator on $H$ with closure $\bar{T}$. If a complex number $z \in \mathbb{C}$ belongs to the regularity domain of $T$, then the defect number of $\bar{T}$ at $z$ is equal to the defect number of $T$ at $z$. That is, $$\text{dim}(\text{Ran}(\bar{T} - zI)^\perp) = \text{dim}(\text{Ran}(T - zI)^\perp)$$ where $I$ denotes the identity operator and $\perp$ denotes the orthogonal complement.

Let $E$ and $F$ be submodules of a complex Hilbert space $H$. Suppose $E$ admits an orthogonal projection. If the rank (dimension) of $E$ is strictly less than the rank of $F$, then the intersection of the orthogonal complement of $E$ (denoted $E^\perp$) and $F$ is non-trivial, i.e., $E^\perp \cap F \neq \{0\}$.

Let $H$ be a complex Hilbert space and $T$ be a closed, partially defined linear operator on $H$ with domain $D(T)$. Suppose $z_1 \in \mathbb{C}$ is in the regularity domain of $T$ (meaning there exists $c > 0$ such that $\|Tx - z_1 x\| \geq c \|x\|$ for all $x \in D(T)$). If the defect number of $T$ at $z_1$ is strictly less than the defect number of $T$ at $z_2$, where the defect number is defined as the dimension of the orthogonal complement of the range of $T - zI$, then there exists a non-zero vector $x \in D(T)$ such that the inner product of $(T - z_1 I)x$ and $(T - z_2 I)x$ is zero: $$\langle (T - z_1 I)x, (T - z_2 I)x \rangle = 0.$$

Let $H$ be a complex Hilbert space and $T$ be a closable partially defined linear operator on $H$. Suppose there exists a real number $c$ such that for all $x$ in the domain of $T$, the inequality $c \|x\| \leq \|(T - z_1 I) x\|$ holds. If $z_2 \in \mathbb{C}$ is a complex number such that $|z_2 - z_1| < c$, then the defect number of $T$ at $z_1$ is equal to the defect number of $T$ at $z_2$, where the defect number at $z$ is the dimension of the orthogonal complement of the range of $T - zI$, denoted by $\dim(\text{Ran}(T - zI)^\perp)$.

Let $H$ be a complex Hilbert space and $T$ be a closable partially defined linear operator on $H$. If $z_1$ and $z_2$ are complex numbers belonging to the same connected component of the regularity domain of $T$, then the defect numbers of $T$ at $z_1$ and $z_2$ are equal: $$\text{defectNumber}(T, z_1) = \text{defectNumber}(T, z_2)$$ where the defect number at $z$ is the dimension of the orthogonal complement of the range of the operator $T - zI$.

The numerical range of a linear partially defined operator $T$ on a complex Hilbert space $H$ is the set of all complex numbers $\langle x, Tx \rangle$ for vectors $x$ in the domain $\mathcal{D}(T)$ with unit norm $\|x\| = 1$. It is defined as: $$\Theta(T) = \{ \langle x, Tx \rangle \in \mathbb{C} \mid x \in \mathcal{D}(T), \|x\| = 1 \}$$ where $\langle \cdot, \cdot \rangle$ denotes the complex inner product.

For a linear partially defined map $T$ on a complex Hilbert space $H$, the symbol $\Theta$ denotes its numerical range, which is defined as the set of all complex numbers of the form $\langle x, Tx \rangle$ where $x$ is a unit vector in the domain of $T$: $$\Theta(T) = \{ \langle x, Tx \rangle \mid x \in \text{dom}(T), \|x\| = 1 \}$$ where $\langle \cdot, \cdot \rangle$ denotes the inner product on $H$.

Let $T$ be a linear partially defined operator on a complex Hilbert space $H$. The numerical range $\Theta(T)$ is the set of all complex numbers $\langle x, Tx \rangle$ where $x$ ranges over the vectors in the domain $\mathcal{D}(T)$ with unit norm $\|x\| = 1$: $$\Theta(T) = \{ \langle x, Tx \rangle \mid x \in \mathcal{D}(T), \|x\| = 1 \}$$ where $\langle \cdot, \cdot \rangle$ denotes the complex inner product.

Let $H$ be a complex Hilbert space and $T$ be a linear partially defined operator on $H$ with domain $\mathcal{D}(T)$. For any non-zero vector $x \in \mathcal{D}(T)$, the complex number $\frac{\langle x, Tx \rangle}{\|x\|^2}$ belongs to the numerical range $\Theta(T)$, where $\langle \cdot, \cdot \rangle$ denotes the complex inner product and $\|\cdot\|$ denotes the norm.

Let $H$ be a complex Hilbert space and $T$ be a linear partially defined operator on $H$. If the domain of $T$, denoted $D(T)$, is not the zero subspace $\{0\}$, then its numerical range $\Theta(T) = \{ \langle x, Tx \rangle \in \mathbb{C} \mid x \in D(T), \|x\| = 1 \}$ is non-empty.

Let $T$ be a linear partially defined operator on a complex Hilbert space $H$. The numerical range of $T$, denoted $\Theta(T)$, is the set of all complex numbers $\langle x, Tx \rangle$ for unit vectors $x$ in the domain of $T$. This theorem states that the numerical range of the operator $-T$ is the negative of the numerical range of $T$: $$\Theta(-T) = -\Theta(T)$$ where $-\Theta(T) = \{ -z \mid z \in \Theta(T) \}$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. For any complex number $c \in \mathbb{C}$, the numerical range of the operator $cT$ satisfies the equality $\Theta(cT) = c \Theta(T)$, where $c \Theta(T) = \{ cz \mid z \in \Theta(T) \}$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$ with domain $\mathcal{D}(T)$. For any complex number $c \in \mathbb{C}$, the numerical range of the operator $T - cI$ (where $I$ is the identity operator) is the translation of the numerical range of $T$ by $-c$: $$\Theta(T - cI) = \Theta(T) - c$$ where $\Theta(T) = \{ \langle x, Tx \rangle \in \mathbb{C} \mid x \in \mathcal{D}(T), \|x\| = 1 \}$ and the subtraction on the right-hand side is defined pointwise as $\Theta(T) - c = \{ z - c \mid z \in \Theta(T) \}$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$ with domain $\mathcal{D}(T)$. Let $\Theta(T)$ be the numerical range of $T$, defined as $$\Theta(T) = \{ \langle x, Tx \rangle \in \mathbb{C} \mid x \in \mathcal{D}(T), \|x\| = 1 \}.$$ The regularity domain of $T$ is the set of complex numbers $z \in \mathbb{C}$ for which there exists a constant $c > 0$ such that $\|Tx - zx\| \geq c \|x\|$ for all $x \in \mathcal{D}(T)$. Then, the complement of the closure of the numerical range is contained in the regularity domain of $T$: $$(\overline{\Theta(T)})^c \subseteq \text{regularityDomain}(T)$$

Let $H$ be a complex Hilbert space and $T$ be a linear partially defined operator on $H$ with domain $\mathcal{D}(T)$. The numerical range of $T$, denoted by $\Theta(T)$ and defined as the set of values $$\Theta(T) = \{ \langle x, Tx \rangle \in \mathbb{C} \mid x \in \mathcal{D}(T), \|x\| = 1 \},$$ is a convex subset of the complex plane $\mathbb{C}$. This result is known as the Toeplitz-Hausdorff theorem.

Let $H$ be a complex Hilbert space and $T$ be a linear partially defined operator on $H$ with domain $D(T)$. The resolvent set $\rho(T)$ is the set of complex numbers $z \in \mathbb{C}$ such that the operator $T - zI$ (where $I$ is the identity operator) is a bijection from $D(T)$ to $H$, and its inverse $(T - zI)^{-1}$ is a continuous linear operator. Specifically, $z \in \rho(T)$ if and only if the kernel of $T - zI$ is trivial, the range of $T - zI$ is the entire space $H$, and the resolvent operator $(T - zI)^{-1}$ is continuous.

The notation $\rho$ denotes the resolvent set of a linear partially defined operator $T$ on a Hilbert space $H$. A complex number $z$ belongs to the resolvent set $\rho(T)$ if the operator $T - z I$ (where $I$ is the identity) has a continuous inverse $(T - z I)^{-1}$ with a domain equal to the entire space $H$.

Let $H$ be a complex Hilbert space and $T: D(T) \subseteq H \to H$ be a partial linear operator. The resolvent set $\rho(T)$ is the set of all complex numbers $z \in \mathbb{C}$ such that the operator $T - zI$ (where $I$ is the identity operator) has a trivial kernel, its range is the entire space $H$, and the resolvent operator $(T - zI)^{-1}$ is continuous. In symbols: $$\rho(T) = \{z \in \mathbb{C} \mid \ker(T - zI) = \{0\}, \text{ran}(T - zI) = H, \text{ and } (T - zI)^{-1} \text{ is continuous}\}.$$

Let $H$ be a complex Hilbert space and $T$ be a partial linear operator on $H$ with domain $D(T)$. For any complex number $z \in \mathbb{C}$, $z$ belongs to the resolvent set $\rho(T)$ if and only if the following three conditions are satisfied: 1. The kernel of the operator $T - zI$ is trivial, i.e., $\ker(T - zI) = \{0\}$. 2. The range of the operator $T - zI$ is the entire space $H$, i.e., $\text{ran}(T - zI) = H$. 3. The resolvent operator $R(T, z) = (T - zI)^{-1}$ is continuous.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$. If $T$ is not a closed operator, then its resolvent set $\rho(T)$ is the empty set $\emptyset$.

Let $H$ be a complex Hilbert space and $T$ be a partially defined linear operator on $H$ with domain $D(T)$. The resolvent set $\rho(T)$ is a subset of the regularity domain of $T$. That is, for any $z \in \rho(T)$, there exists a constant $c > 0$ such that $$\|Tx - zx\| \geq c \|x\|$$ for all $x \in D(T)$.

Let $H$ be a complex Hilbert space and $T$ be a closed, partially defined linear operator on $H$. The resolvent set $\rho(T)$ consists of all complex numbers $z \in \mathbb{C}$ such that the operator $T - zI$ is a bijection from its domain to $H$. Specifically, for a closed operator $T$, $z \in \rho(T)$ if and only if the kernel of $T - zI$ is trivial and its range is the entire space $H$: $$\rho(T) = \{z \in \mathbb{C} \mid \ker(T - zI) = \{0\} \text{ and } \text{ran}(T - zI) = H\}.$$ (Note: For closed operators, the requirement that the inverse $(T - zI)^{-1}$ be continuous is automatically satisfied by the bounded inverse theorem and is thus redundant in this characterization.)