QuantumInfo.ForMathlib.HayataGroup.TraceInequality.LownerHeinzTheorem

Given a complex inner product space $\mathcal{H}$ (which is also a normed additive commutative group), $L(\mathcal{H})$ denotes the space of bounded (continuous) linear operators from $\mathcal{H}$ to itself, commonly denoted as $\mathcal{B}(\mathcal{H})$.

The space of bounded linear operators $\mathcal{B}(\mathcal{H})$ on a complex inner product space $\mathcal{H}$ is nontrivial, meaning it contains at least two distinct elements.

Let $\mathcal{H}$ be a complex inner product space and $\mathcal{B}(\mathcal{H})$ be the algebra of bounded linear operators on $\mathcal{H}$. This theorem states that $\mathcal{B}(\mathcal{H})$ belongs to the class of algebras where the spectrum of any non-negative operator is contained in the set of non-negative real numbers. That is, for any operator $A \in \mathcal{B}(\mathcal{H})$ such that $A \geq 0$, its spectrum $\sigma(A)$ satisfies $\sigma(A) \subseteq [0, \infty)$.

Let $\mathcal{H}$ be a complex inner product space and $\mathcal{B}(\mathcal{H})$ denote the space of bounded linear operators on $\mathcal{H}$. There exists a continuous functional calculus for the set of self-adjoint operators in $\mathcal{B}(\mathcal{H})$ associated with continuous functions from $\mathbb{R}$ to $\mathbb{R}$.

Given a real-valued function $f: \mathbb{R} \to \mathbb{R}$ and a bounded linear operator $A \in \mathcal{B}(\mathcal{H})$ acting on a complex Hilbert space $\mathcal{H}$, this definition represents the operator $f(A) \in \mathcal{B}(\mathcal{H})$ obtained via the continuous functional calculus. This is a specialization of the general $C^*$-algebraic functional calculus to the algebra of bounded operators on a Hilbert space.

A function $f: \mathbb{R} \to \mathbb{R}$ is operator monotone on the space of bounded linear operators $\mathcal{B}(\mathcal{H})$ if for any pair of non-negative operators $A, B \in \mathcal{B}(\mathcal{H})$ such that $0 \le B \le A$ in the Loewner order, the inequality $f(B) \le f(A)$ holds, where $f(A)$ and $f(B)$ are the operators obtained via the continuous functional calculus.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. A function $f: \mathbb{R} \to \mathbb{R}$ is said to be operator monotone on a set $s \subseteq \mathbb{R}$ if for any two self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ such that their spectra $\sigma(A)$ and $\sigma(B)$ are contained in $s$, the inequality $B \le A$ implies $f(B) \le f(A)$. Here, $\le$ denotes the Loewner order (i.e., $B \le A$ if $A - B$ is a positive semi-definite operator), and the operators $f(A)$ and $f(B)$ are defined via the continuous functional calculus.

A function $f: \mathbb{R} \to \mathbb{R}$ is operator antitone on the space of bounded linear operators $\mathcal{B}(\mathcal{H})$ (where $\mathcal{H}$ is a complex Hilbert space) if for any two self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$, the operator inequality $B \le A$ implies $f(A) \le f(B)$ in the Loewner order, where $f(A)$ and $f(B)$ are defined via the continuous functional calculus.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the algebra of bounded linear operators on $\mathcal{H}$. For a set $s \subseteq \mathbb{R}$, a function $f: \mathbb{R} \to \mathbb{R}$ is said to be operator antitone on $s$ if for all $A, B \in \mathcal{B}(\mathcal{H})$ whose spectra satisfy $\sigma(A) \subseteq s$ and $\sigma(B) \subseteq s$, the operator inequality $B \le A$ implies $f(A) \le f(B)$ in the Loewner order, where $f(A)$ and $f(B)$ are defined via the continuous functional calculus.

Given a complex Hilbert space $\mathcal{H}$, a function $f: \mathbb{R} \to \mathbb{R}$ is operator convex if for any two bounded linear operators $A, B \in \mathcal{B}(\mathcal{H})$ and any scalar $t \in [0, 1]$, the inequality $$f((1 - t)A + tB) \leq (1 - t)f(A) + tf(B)$$ holds, where the inequality $\leq$ denotes the standard Loewner partial order on self-adjoint elements in $\mathcal{B}(\mathcal{H})$ and $f$ is applied via the continuous functional calculus.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. A function $f: \mathbb{R} \to \mathbb{R}$ is said to be **operator convex** on a set $s \subseteq \mathbb{R}$ for the space $\mathcal{B}(\mathcal{H})$ if for all self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ whose spectra $\sigma(A)$ and $\sigma(B)$ are contained in $s$, and for any $t \in [0, 1]$, the following inequality holds in the operator (Löwner) order: $$f((1 - t)A + tB) \leq (1 - t)f(A) + tf(B)$$ where $f(A), f(B),$ and $f((1 - t)A + tB)$ are defined via the continuous functional calculus for self-adjoint operators.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. A function $f: \mathbb{R} \to \mathbb{R}$ is **operator concave** if for all self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ and any scalar $t \in [0, 1]$, the inequality $$f((1 - t)A + tB) \geq (1 - t)f(A) + tf(B)$$ holds, where the inequality $\geq$ denotes the standard Löwner partial order on self-adjoint operators and the function application $f(\cdot)$ is defined via the continuous functional calculus.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. A function $f: \mathbb{R} \to \mathbb{R}$ is said to be **operator concave** on a set $s \subseteq \mathbb{R}$ for the space $\mathcal{B}(\mathcal{H})$ if for all self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ whose spectra $\sigma(A)$ and $\sigma(B)$ are contained in $s$, and for any $t \in [0, 1]$, the following inequality holds in the operator (Löwner) order: $$f((1 - t)A + tB) \geq (1 - t)f(A) + tf(B)$$ where $f(A), f(B),$ and $f((1 - t)A + tB)$ are defined via the continuous functional calculus for self-adjoint operators.

A function $f: \mathbb{R} \to \mathbb{R}$ is operator monotone over all Hilbert spaces if for every non-trivial complex Hilbert space $\mathcal{H}$ in the universe $u$, $f$ is operator monotone on the space of bounded linear operators $\mathcal{B}(\mathcal{H})$. This means that for any pair of operators $A, B \in \mathcal{B}(\mathcal{H})$ satisfying $0 \le B \le A$ in the Loewner order, the inequality $f(B) \le f(A)$ holds, where $f(A)$ and $f(B)$ are the operators obtained via the continuous functional calculus.

Let $f: \mathbb{R} \to \mathbb{R}$ be a function and $s \subseteq \mathbb{R}$ be a set. The property `OperatorMonotoneOnAll s f` holds if for every non-trivial complex Hilbert space $\mathcal{H}$, the function $f$ is operator monotone on $s$. That is, for any complex Hilbert space $\mathcal{H}$ and any two self-adjoint bounded linear operators $A, B \in \mathcal{B}(\mathcal{H})$ with spectra $\sigma(A), \sigma(B) \subseteq s$, the inequality $B \le A$ in the Loewner order implies $f(B) \le f(A)$, where the operators $f(A)$ and $f(B)$ are defined via the continuous functional calculus.

A function $f: \mathbb{R} \to \mathbb{R}$ is operator antitone on all Hilbert spaces if for every complex Hilbert space $\mathcal{H}$ and for any two self-adjoint bounded linear operators $A, B \in \mathcal{B}(\mathcal{H})$, the inequality $B \le A$ in the Loewner order implies $f(A) \le f(B)$, where the operators $f(A)$ and $f(B)$ are defined via the continuous functional calculus.

For a set $s \subseteq \mathbb{R}$ and a function $f : \mathbb{R} \to \mathbb{R}$, this property states that $f$ is operator antitone on $s$ for every nontrivial complex Hilbert space $\mathcal{H}$ in universe $u$. Specifically, for any such $\mathcal{H}$ and any bounded linear operators $A, B \in \mathcal{B}(\mathcal{H})$ whose spectra $\sigma(A)$ and $\sigma(B)$ are contained in $s$, the inequality $B \le A$ in the Loewner order implies $f(A) \le f(B)$.

A function $f: \mathbb{R} \to \mathbb{R}$ satisfies this property if, for every non-trivial complex Hilbert space $\mathcal{H}$, the function $f$ is operator convex on $\mathcal{B}(\mathcal{H})$. That is, for any two bounded self-adjoint linear operators $A, B \in \mathcal{B}(\mathcal{H})$ and any scalar $t \in [0, 1]$, the inequality $$f((1 - t)A + tB) \leq (1 - t)f(A) + tf(B)$$ holds, where $\leq$ denotes the Loewner partial order and $f$ is applied to the operators via the continuous functional calculus.

Let $s \subseteq \mathbb{R}$ be a set and $f: \mathbb{R} \to \mathbb{R}$ be a function. $f$ is **operator convex on $s$ for all Hilbert spaces** if for every non-trivial complex Hilbert space $\mathcal{H}$ in the universe $u$, $f$ is operator convex on $s$ for the space of bounded linear operators $\mathcal{B}(\mathcal{H})$. This means that for every such $\mathcal{H}$, for all self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ whose spectra $\sigma(A)$ and $\sigma(B)$ are contained in $s$, and for any $t \in [0, 1]$, the following inequality holds in the Löwner order: $$f((1 - t)A + tB) \leq (1 - t)f(A) + tf(B)$$ where the operators $f(A), f(B)$, and $f((1 - t)A + tB)$ are defined via the continuous functional calculus.

Let $f: \mathbb{R} \to \mathbb{R}$ be a function. $f$ satisfies this property if, for every non-trivial complex Hilbert space $\mathcal{H}$, the function $f$ is operator concave on the space of bounded linear operators $\mathcal{B}(\mathcal{H})$. That is, for any two bounded self-adjoint linear operators $A, B \in \mathcal{B}(\mathcal{H})$ and any scalar $t \in [0, 1]$, the inequality $$f((1 - t)A + tB) \geq (1 - t)f(A) + tf(B)$$ holds, where $\geq$ denotes the Löwner partial order and the function $f$ is applied to the operators via the continuous functional calculus.

Let $s \subseteq \mathbb{R}$ be a set and $f: \mathbb{R} \to \mathbb{R}$ be a function. $f$ is **operator concave on $s$ for all Hilbert spaces** if for every non-trivial complex Hilbert space $\mathcal{H}$ in the universe $u$, $f$ is operator concave on $s$ for the space of bounded linear operators $\mathcal{B}(\mathcal{H})$. This means that for every such $\mathcal{H}$, for all self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ whose spectra $\sigma(A)$ and $\sigma(B)$ are contained in $s$, and for any $t \in [0, 1]$, the following inequality holds in the Löwner order: $$f((1 - t)A + tB) \geq (1 - t)f(A) + tf(B)$$ where the operators $f(A), f(B)$, and $f((1 - t)A + tB)$ are defined via the continuous functional calculus for self-adjoint operators.

Let $\mathcal{H}$ be a complex Hilbert space and $f: \mathbb{R} \to \mathbb{R}$ be a function. If $f$ is operator convex on the space of bounded linear operators $\mathcal{B}(\mathcal{H})$, then $f$ is a convex function on $\mathbb{R}$.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. If a function $f: \mathbb{R} \to \mathbb{R}$ is operator convex on $\mathcal{B}(\mathcal{H})$, then $f$ is continuous on the entire real line $\mathbb{R}$.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. If a function $f: \mathbb{R} \to \mathbb{R}$ is operator convex, then for any two operators $A, B \in \mathcal{B}(\mathcal{H})$, $f$ is continuous on the union of their spectra $\sigma(A) \cup \sigma(B)$.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the algebra of bounded linear operators on $\mathcal{H}$. The function $f: (0, \infty) \to \mathbb{R}$ defined by $f(x) = \frac{1}{x}$ is operator antitone. That is, for any $A, B \in \mathcal{B}(\mathcal{H})$ whose spectra satisfy $\sigma(A) \subseteq (0, \infty)$ and $\sigma(B) \subseteq (0, \infty)$, the operator inequality $B \le A$ in the Loewner order implies $A^{-1} \le B^{-1}$.

Let $\mathcal{H}$ be a complex Hilbert space. The function $f: (0, \infty) \to \mathbb{R}$ defined by $f(x) = \frac{1}{x}$ is operator convex on the interval $(0, \infty)$. Specifically, for any self-adjoint operators $A, B$ on $\mathcal{H}$ whose spectra are contained in $(0, \infty)$, and for any $t \in [0, 1]$, the inequality $$((1 - t)A + tB)^{-1} \leq (1 - t)A^{-1} + tB^{-1}$$ holds in the operator (Löwner) order.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the algebra of bounded linear operators on $\mathcal{H}$. For any real number $t > 0$, the function $f: [0, \infty) \to \mathbb{R}$ defined by $f(x) = \frac{1}{x + t}$ is operator antitone on $[0, \infty)$. That is, for any two self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ whose spectra $\sigma(A)$ and $\sigma(B)$ are contained in $[0, \infty)$, if $B \le A$ in the Loewner order, then $f(A) \le f(B)$.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. For any real number $t > 0$, the function $f: [0, \infty) \to \mathbb{R}$ defined by $f(x) = \frac{1}{x + t}$ is operator convex on the interval $[0, \infty)$. This means that for any two self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ with spectra $\sigma(A), \sigma(B) \subseteq [0, \infty)$ and any $\lambda \in [0, 1]$, the following inequality holds in the Löwner order: $$((1 - \lambda)A + \lambda B + tI)^{-1} \le (1 - \lambda)(A + tI)^{-1} + \lambda(B + tI)^{-1}$$ where $I$ is the identity operator.

For any real number $t > 0$, the function $f: [0, \infty) \to \mathbb{R}$ defined by $f(x) = \frac{x}{x + t}$ is operator monotone on the interval $[0, \infty)$. Specifically, for any two self-adjoint bounded linear operators $A$ and $B$ on a complex Hilbert space $\mathcal{H}$ such that their spectra $\sigma(A)$ and $\sigma(B)$ are contained in $[0, \infty)$, the inequality $B \le A$ in the Loewner order (meaning $A - B$ is a positive semi-definite operator) implies that $f(B) \le f(A)$, where the operators $f(A)$ and $f(B)$ are defined via the continuous functional calculus.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. For any real number $t > 0$, the function $f: [0, \infty) \to \mathbb{R}$ defined by $f(x) = \frac{x}{x + t}$ is operator concave on the interval $[0, \infty)$. Specifically, for any two self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ whose spectra $\sigma(A)$ and $\sigma(B)$ are contained in $[0, \infty)$, and for any $\lambda \in [0, 1]$, the following inequality holds in the Löwner order: $$((1 - \lambda)A + \lambda B)((1 - \lambda)A + \lambda B + tI)^{-1} \ge (1 - \lambda)A(A + tI)^{-1} + \lambda B(B + tI)^{-1}$$ where $I$ is the identity operator on $\mathcal{H}$.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. For any $p \in [0, 1]$, the function $f(x) = x^p$ is operator monotone on the interval $[0, \infty)$. Specifically, for any two self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ such that their spectra $\sigma(A)$ and $\sigma(B)$ are contained in $[0, \infty)$, if $B \le A$ (where $\le$ denotes the Loewner order), then $B^p \le A^p$.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. For any $p \in [0, 1]$, the power function $f(x) = x^p$ is operator concave on the interval $[0, \infty)$. Specifically, for all self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ whose spectra $\sigma(A)$ and $\sigma(B)$ are contained in $[0, \infty)$, and for any $t \in [0, 1]$, the following inequality holds in the Löwner order: $$((1 - t)A + tB)^p \geq (1 - t)A^p + tB^p$$ where $A^p, B^p,$ and $((1 - t)A + tB)^p$ are defined via the continuous functional calculus for self-adjoint operators.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. For any $p \in [1, 2]$, the power function $f(x) = x^p$ is operator convex on the interval $[0, \infty)$. That is, for all self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ with spectra $\sigma(A), \sigma(B) \subseteq [0, \infty)$, and for any $t \in [0, 1]$, the following inequality holds in the Löwner order: $$((1 - t)A + tB)^p \leq (1 - t)A^p + tB^p$$

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. For any real number $p$ in the interval $[-1, 0]$, the function $f(x) = -x^p$ is operator monotone on $(0, \infty)$. Specifically, for any two self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ whose spectra $\sigma(A)$ and $\sigma(B)$ are contained in $(0, \infty)$, if $B \le A$ in the Loewner order, then $-B^p \le -A^p$. Here, $\le$ denotes the Loewner order (where $B \le A$ if $A - B$ is a positive semi-definite operator) and the operators are defined via the continuous functional calculus.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. For any real number $p \in [-1, 0]$, the function $f(x) = -x^p$ is operator concave on the interval $(0, \infty)$. That is, for any two self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ whose spectra $\sigma(A)$ and $\sigma(B)$ are contained in $(0, \infty)$, and for any $t \in [0, 1]$, the following inequality holds in the Löwner order: $$-((1 - t)A + tB)^p \geq -(1 - t)A^p - tB^p$$ where the power of the operators is defined via the continuous functional calculus.