PhyslibAlpha.AlgebraicFramework.HilbertSpace.Unbounded.Cayley.Basic

The scalar Cayley transform is a function that maps a real number $x \in \mathbb{R}$ to a complex number defined by the formula $$\frac{x - i}{x + i}$$ where $i$ is the imaginary unit. This transform maps the real line onto the unit circle in the complex plane, excluding the point $1$.

The inverse scalar Cayley transform maps a complex number $z$ to a real number. It is defined as: $$z \mapsto \begin{cases} 0 & \text{if } z = 1 \\ -\frac{\text{Im } z}{1 - \text{Re } z} & \text{if } z \neq 1 \end{cases}$$ where $\text{Re } z$ and $\text{Im } z$ denote the real and imaginary parts of $z$, respectively. The value at $z = 1$ is assigned arbitrarily, as this point corresponds to infinity under the forward Cayley transform.

For any real number $x$, the scalar Cayley transform defined by $\text{cayley}(x) = \frac{x - i}{x + i}$ is not equal to $1$.

For any real number $x$, the real part of the scalar Cayley transform $\text{cayley}(x) = \frac{x - i}{x + i}$ is given by the formula: $$\text{Re}(\text{cayley}(x)) = \frac{x^2 - 1}{x^2 + 1}$$

For any real number $x$, the imaginary part of its Cayley transform $\text{cayley}(x) = \frac{x - i}{x + i}$ is given by the formula $$\operatorname{Im}(\operatorname{cayley} x) = \frac{-2x}{x^2 + 1}$$ where $i$ is the imaginary unit.

For any real number $x$, the inverse scalar Cayley transform of the scalar Cayley transform of $x$ is equal to $x$: $$\text{cayleyInverse}(\text{cayley}(x)) = x$$

For any real number $x$, the norm (modulus) of its Cayley transform $\text{cayley}(x) = \frac{x - i}{x + i}$ is equal to $1$, where $i$ is the imaginary unit.

For any complex number $z \in \mathbb{C}$ such that $\|z\| = 1$ and $z \neq 1$, the composition of the scalar Cayley transform and its inverse satisfies $$\text{cayley}(\text{cayleyInverse}(z)) = z$$ where $\text{cayley}(x) = \frac{x - i}{x + i}$ for $x \in \mathbb{R}$, and $\text{cayleyInverse}$ is its inverse map from the unit circle (excluding $1$) back to the real line.

The scalar Cayley transform, defined as the map $x \mapsto \frac{x - i}{x + i}$ from $\mathbb{R}$ to $\mathbb{C}$, is measurable.

The inverse scalar Cayley transform $\text{cayleyInverse} : \mathbb{C} \to \mathbb{R}$, defined as $$z \mapsto \begin{cases} 0 & \text{if } z = 1 \\ -\frac{\text{Im } z}{1 - \text{Re } z} & \text{if } z \neq 1 \end{cases}$$ where $\text{Re } z$ and $\text{Im } z$ denote the real and imaginary parts of $z$, is a measurable function.

For a partial linear map $T$ on a complex Hilbert space $H$, its Cayley transform is the partial linear map defined by the expression $$(T - iI)(T + iI)^{-1}$$ where $i$ is the imaginary unit and $I$ (represented by $1$ in the formal text) denotes the identity operator on $H$. This map is the operator-theoretic analogue of the complex-valued Möbius transformation $z \mapsto \frac{z - i}{z + i}$.

Let $H$ be a complex Hilbert space. If $T$ is a self-adjoint partial linear operator on $H$, then the domain of its Cayley transform, $\text{cayleyPMap } T = (T - iI)(T + iI)^{-1}$, is the entire Hilbert space $H$ (denoted by $\top$).

Let $H$ be a complex Hilbert space. Given a partial linear map $A$ on $H$, if the domain of $A$ is the entire space (i.e., $\text{dom}(A) = \top$) and $A$ is continuous, this definition constructs the corresponding continuous (bounded) linear operator $A \in \mathcal{L}(H)$.

Let $H$ be a complex Hilbert space. Let $A$ be a partial linear map on $H$ such that its domain is the entire space $H$ and $A$ is continuous. Let $\tilde{A} \in \mathcal{L}(H)$ be the continuous linear operator constructed from $A$ (represented by `topDomainToContinuousLinearMap`). For any vector $x \in H$, the evaluation of this operator $\tilde{A}$ at $x$ is equal to the application of the partial map $A$ to $x$: $$\tilde{A}(x) = A(x)$$

Let $H$ be a complex Hilbert space. For any self-adjoint partial linear operator $T$ on $H$, the Cayley transform of $T$, denoted as $\text{cayleyPMap}(T)$, satisfies the identity: $$\text{cayleyPMap}(T) = I - 2i(T + iI)^{-1}$$ where $i$ is the imaginary unit and $I$ denotes the identity operator on $H$.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear operator on $H$. The Cayley transform of $T$, denoted $C(T)$, satisfies the identity: $$C(T) = I - 2i(T - (-i)I)^{-1}$$ where $I$ is the identity operator on $H$ and $i$ is the imaginary unit.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear operator on $H$. The Cayley transform of $T$ is the bounded (continuous) linear operator $C(T): H \to H$ defined by the formula $C(T) = I - 2i(T + iI)^{-1}$, where $I$ is the identity operator and $i$ is the imaginary unit. This operator is well-defined on the entire space $H$ because the resolvent set of a self-adjoint operator contains all non-real complex numbers, specifically $-i$.

For a continuous linear operator $L: H \to H$ on a complex Hilbert space $H$, this definition constructs a linear partial map whose domain is the entire space $H$ and whose action is given by $L$.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear operator on $H$. Then the Cayley transform of $T$ defined as a partial linear map is equal to the partial linear map associated with the bounded (continuous) linear operator $C(T) = I - 2i(T + iI)^{-1}$. That is, $\text{cayleyPMap}(T) = \text{continuousLinearMapToPMap}(C(T))$.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial operator on $H$. For any vector $y$ in the domain of $T$, the following equality holds: $$\|Ty - iy\| = \|Ty + iy\|$$ where $i$ is the imaginary unit and $\|\cdot\|$ denotes the norm on $H$.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial operator on $H$. Let $i$ denote the imaginary unit and $I$ the identity operator. For any vector $y$ in the domain of $T$ and any $x \in H$ such that $(T + iI)y = x$, the bounded Cayley transform $C(T)$ satisfies the following equality: $$C(T)x = (T - iI)y$$

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear operator on $H$. Let $C(T)$ denote the Cayley transform of $T$, which is a bounded linear operator on $H$. For any vector $x \in H$, the norm of the transformed vector $C(T)x$ is equal to the norm of $x$: $$\|C(T)x\| = \|x\|$$

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear operator on $H$. The Cayley transform of $T$, which is the bounded linear operator $C(T) = (T - iI)(T + iI)^{-1}$, is an isometry. That is, for every $x \in H$, the norm is preserved: $\|C(T)x\| = \|x\|$.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear operator on $H$. The Cayley transform of $T$, denoted as $C(T): H \to H$, is surjective.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear operator on $H$. The Cayley transform of $T$, denoted $C(T)$, is the unitary operator (an isometric linear isomorphism) from $H$ to itself defined by $C(T) = (T - iI)(T + iI)^{-1}$, where $i$ is the imaginary unit and $I$ is the identity operator.

Let $H$ be a complex Hilbert space and $T$ be a self-adjoint partial linear operator on $H$. For any vector $x \in H$, the unitary Cayley transform of $T$ applied to $x$ is equal to the bounded continuous linear Cayley transform of $T$ applied to $x$.