QuantumInfo.ForMathlib.HayataGroup.TraceInequality.JensenOperatorInequality

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. There exists a continuous functional calculus for complex-valued functions on the product algebra $\mathcal{B}(\mathcal{H}) \times \mathcal{B}(\mathcal{H})$ for elements $(A, B)$ such that both $A$ and $B$ are normal operators.

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the algebra of bounded linear operators on $\mathcal{H}$. There exists a continuous functional calculus over the real numbers $\mathbb{R}$ for self-adjoint elements in the product algebra $\mathcal{B}(\mathcal{H}) \times \mathcal{B}(\mathcal{H})$.

Let $\mathcal{H}$ be a complex Hilbert space, and let $\mathcal{H} \oplus \mathcal{H}$ be the $\ell^2$ direct sum of $\mathcal{H}$ with itself. The algebra of bounded linear operators $\mathcal{B}(\mathcal{H} \oplus \mathcal{H})$ is a non-negative spectrum class over the real numbers $\mathbb{R}$. This implies that for any non-negative (positive) operator $T \in \mathcal{B}(\mathcal{H} \oplus \mathcal{H})$, its spectrum $\sigma(T)$ is contained in the set of non-negative real numbers $[0, \infty)$.

For a complex Hilbert space $\mathcal{H}$, the space of bounded linear operators on the two-fold Hilbert sum $\mathcal{H} \oplus \mathcal{H}$, denoted as $\mathcal{B}(\mathcal{H} \oplus \mathcal{H})$, possesses the structure of a module (vector space) over the real numbers $\mathbb{R}$.

Let $f: \mathbb{R} \to \mathbb{R}$ be a function. This property states that for every non-trivial complex Hilbert space $\mathcal{H}$, and for every self-adjoint bounded linear operator $A \in \mathcal{B}(\mathcal{H})$ with a non-negative spectrum $\sigma(A) \subseteq [0, \infty)$, the inequality $$f(X^* A X) \leq X^* f(A) X$$ holds in the Loewner order for every contraction $X \in \mathcal{B}(\mathcal{H})$ (i.e., $\|X\| \leq 1$), where $f(\cdot)$ is the operator defined via the continuous functional calculus.

Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function. Suppose that for every non-trivial complex Hilbert space $\mathcal{H}$, for every bounded self-adjoint operator $A \in \mathcal{B}(\mathcal{H})$ with spectrum $\sigma(A) \subseteq [0, \infty)$, and for every contraction $X \in \mathcal{B}(\mathcal{H})$ (i.e., $\|X\| \leq 1$), the operator inequality $$f(X^* A X) \leq X^* f(A) X$$ holds. Then, for any complex Hilbert space $\mathcal{H}$, for all bounded self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ with spectra in $[0, \infty)$, and for all bounded linear operators $X, Y \in \mathcal{B}(\mathcal{H})$ satisfying $X^* X + Y^* Y \le I$, the following inequality holds: $$f(X^* A X + Y^* B Y) \le X^* f(A) X + Y^* f(B) Y.$$

Let $f: \mathbb{R} \to \mathbb{R}$ be a function. Suppose that $f(0) \leq 0$ and $f$ is operator convex on all complex Hilbert spaces $\mathcal{H}$; that is, for any two bounded self-adjoint 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 in the Loewner partial order. Then, for any complex Hilbert space $\mathcal{H}$, any bounded self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ with spectra $\sigma(A), \sigma(B) \subseteq [0, \infty)$, and any bounded linear operators $X, Y \in \mathcal{B}(\mathcal{H})$ satisfying $X^* X + Y^* Y \leq I$ (where $I$ is the identity operator), the following operator inequality holds: $$f(X^* A X + Y^* B Y) \leq X^* f(A) X + Y^* f(B) Y$$ where the function $f$ is applied to the operators via the continuous functional calculus.

Let $f: \mathbb{R} \to \mathbb{R}$ be a real-valued function. Suppose $f$ satisfies the following conditions: 1. $f$ is continuous on the interval $[0, \infty)$. 2. $f(0) \le 0$. 3. $f$ is operator convex on $[0, \infty)$, meaning for any non-trivial complex Hilbert space $\mathcal{H}$, any self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ with spectra $\sigma(A), \sigma(B) \subseteq [0, \infty)$, and any $t \in [0, 1]$, the inequality $f((1 - t)A + tB) \le (1 - t)f(A) + tf(B)$ holds in the Löwner order. Then, for any complex Hilbert space $\mathcal{H}$, for all bounded self-adjoint operators $A, B \in \mathcal{B}(\mathcal{H})$ with spectra in $[0, \infty)$, and for all bounded linear operators $X, Y \in \mathcal{B}(\mathcal{H})$ satisfying $X^* X + Y^* Y \le I$, the following operator inequality holds: $$f(X^* A X + Y^* B Y) \le X^* f(A) X + Y^* f(B) Y$$ where $f(\cdot)$ denotes the operator obtained via the continuous functional calculus.