QuantumInfo.ForMathlib.HayataGroup.TraceInequality.LiebAndoTrace

Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$ be the algebra of bounded linear operators on $\mathcal{H}$. Let $\mathcal{B}(\mathcal{H})^{op}$ denote the multiplicative opposite algebra of $\mathcal{B}(\mathcal{H})$. This theorem states that $\mathcal{B}(\mathcal{H})^{op}$ is a non-negative spectrum class over $\mathbb{R}$, which implies that for any positive element in the algebra, its spectrum is contained in the interval $[0, \infty)$.

Let $\mathcal{H}$ be a complex inner product space and $B(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. The multiplicative opposite algebra $B(\mathcal{H})^{op}$ admits an isometric continuous functional calculus over the complex numbers $\mathbb{C}$ for star-normal elements.

Let $\mathcal{H}$ be a complex Hilbert space and let $\mathcal{B}(\mathcal{H})$ denote the algebra of bounded linear operators on $\mathcal{H}$. The opposite algebra $\mathcal{B}(\mathcal{H})^{op}$ admits a continuous functional calculus for self-adjoint elements with respect to real-valued continuous functions.

Given a bounded linear operator $T \in \mathcal{B}(\mathcal{H})$ on a finite-dimensional complex inner product space $\mathcal{H}$, this function returns the real part of its trace, denoted as $\text{Re}(\text{Tr}(T))$.

Given a complex inner product space $\mathcal{H}$ and the space of bounded linear operators $\mathcal{B}(\mathcal{H})$, this function takes a real number $s \in \mathbb{R}$ and three bounded linear operators $K, A, B \in \mathcal{B}(\mathcal{H})$, and returns the real part of the trace of the operator product $A^s K^* B^{1-s} K$, denoted as: $$\text{Re}(\text{Tr}(A^s K^* B^{1-s} K))$$ where $K^*$ is the adjoint of the operator $K$. This functional is the primary trace functional utilized in Lieb's concavity theorem.

Let $\mathcal{H}$ be a finite-dimensional complex inner product space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. For real numbers $q, p \in \mathbb{R}$ and operators $K, A, B \in \mathcal{B}(\mathcal{H})$, this functional is defined as the real part of the trace: $$\text{Re}(\text{Tr}(A^q K^* B^p K))$$ where $K^*$ denotes the adjoint of the operator $K$, and the operator powers $A^q$ and $B^p$ are defined via the continuous functional calculus. This functional appears in the context of Lieb's extension theorem.

Given real numbers $q$ and $r$, and bounded linear operators $A, B, K \in \mathcal{B}(\mathcal{H})$ on a complex inner product space $\mathcal{H}$, this function computes the real part of the trace of the operator product $A^q K^* B^{1-r} K$, where $K^*$ is the adjoint of $K$. The powers of the operators $A^q$ and $B^{1-r}$ are defined via the continuous functional calculus. This specific functional appears in the context of Lieb's and Ando's trace inequalities (specifically Corollary 1.3).

Given a complex inner product space $\mathcal{H}$, let $\mathcal{B}(\mathcal{H})$ denote the space of bounded linear operators on $\mathcal{H}$. For real numbers $q, r \in \mathbb{R}$ and operators $K, A, B \in \mathcal{B}(\mathcal{H})$, this function calculates the real part of the trace of the operator product $A^q K^* B^{-r} K$. The expression is given by: $$\text{Re}\left(\text{Tr}\left(A^q K^* B^{-r} K\right)\right)$$ where $K^*$ is the adjoint of $K$, and the powers $A^q$ and $B^{-r}$ are defined via the continuous functional calculus. This functional appears in the context of Ando's convexity theorem.

Let $\mathcal{H}$ be a complex inner product space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. Let $A$ and $B$ be strictly positive operators in $\mathcal{B}(\mathcal{H})$ (meaning they are self-adjoint and their spectra are contained in $(0, \infty)$). For any real number $t$ such that $0 \le t \le 1$, the convex combination $(1 - t)A + tB$ is also a strictly positive operator.

Let $\mathcal{H}$ be a complex inner product space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. Let $\mathcal{P} \subseteq \mathcal{B}(\mathcal{H})$ be the set of strictly positive operators (self-adjoint operators with spectrum in $(0, \infty)$). For any real number $s$ such that $0 < s < 1$ and any operator $K \in \mathcal{B}(\mathcal{H})$, the Lieb trace map $(A, B) \mapsto \text{Re}(\text{Tr}(A^s K^* B^{1-s} K))$ is jointly concave on $\mathcal{P} \times \mathcal{P}$. That is, for all $A_1, A_2, B_1, B_2 \in \mathcal{P}$ and $\theta \in [0, 1]$, the following inequality holds: $$(1 - \theta) \text{Re}(\text{Tr}(A_1^s K^* B_1^{1-s} K)) + \theta \text{Re}(\text{Tr}(A_2^s K^* B_2^{1-s} K)) \le \text{Re}(\text{Tr}(((1 - \theta) A_1 + \theta A_2)^s K^* ((1 - \theta) B_1 + \theta B_2)^{1-s} K))$$

Let $\mathcal{H}$ be a complex inner product space and $L(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. Let $\mathcal{P}_+$ be the set of strictly positive operators in $L(\mathcal{H})$, which are self-adjoint operators with spectrum contained in $(0, \infty)$. For any real number $s$ such that $1 \le s \le 2$ and any fixed operator $K \in L(\mathcal{H})$, the Lieb trace map defined by $$(A, B) \mapsto \text{Re}(\text{Tr}(A^s K^* B^{1-s} K))$$ is jointly convex for $A, B \in \mathcal{P}_+$. That is, for all $A_1, A_2, B_1, B_2 \in \mathcal{P}_+$ and $\theta \in [0, 1]$, the inequality $$\Phi((1 - \theta)A_1 + \theta A_2, (1 - \theta)B_1 + \theta B_2) \le (1 - \theta)\Phi(A_1, B_1) + \theta\Phi(A_2, B_2)$$ holds, where $\Phi(A, B) = \text{Re}(\text{Tr}(A^s K^* B^{1-s} K))$.

Let $\mathcal{H}$ be a finite-dimensional complex inner product space and $L(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. Let $\mathcal{P} \subset L(\mathcal{H})$ be the set of strictly positive operators (self-adjoint operators with strictly positive spectrum). For any real numbers $p, q$ such that $p > 0, q > 0$, and $p + q \le 1$, and for any fixed operator $K \in L(\mathcal{H})$, the map $$(A, B) \mapsto \text{Re}(\text{Tr}(A^q K^* B^p K))$$ is jointly concave for $A, B \in \mathcal{P}$. That is, for any $A_1, A_2, B_1, B_2 \in \mathcal{P}$ and $\theta \in [0, 1]$, $$(1 - \theta) \text{Re}(\text{Tr}(A_1^q K^* B_1^p K)) + \theta \text{Re}(\text{Tr}(A_2^q K^* B_2^p K)) \le \text{Re}(\text{Tr}(( (1 - \theta) A_1 + \theta A_2 )^q K^* ( (1 - \theta) B_1 + \theta B_2 )^p K)).$$

Let $\mathcal{H}$ be a complex inner product space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. Let $\mathcal{P}_+$ denote the set of strictly positive operators in $\mathcal{B}(\mathcal{H})$, consisting of self-adjoint operators whose spectrum is contained in $(0, \infty)$. For any real numbers $q$ and $r$ satisfying the conditions $1 \le q \le 2$, $0 \le r \le 1$, and $q - r \ge 1$, and for any fixed operator $K \in \mathcal{B}(\mathcal{H})$, the Ando trace map defined by $$(A, B) \mapsto \text{Re}(\text{Tr}(A^q K^* B^{-r} K))$$ is jointly convex on $\mathcal{P}_+ \times \mathcal{P}_+$. That is, for all $A_1, A_2, B_1, B_2 \in \mathcal{P}_+$ and $\theta \in [0, 1]$, the following inequality holds: $$\text{Re}(\text{Tr}(((1 - \theta)A_1 + \theta A_2)^q K^* ((1 - \theta)B_1 + \theta B_2)^{-r} K)) \le (1 - \theta)\text{Re}(\text{Tr}(A_1^q K^* B_1^{-r} K)) + \theta \text{Re}(\text{Tr}(A_2^q K^* B_2^{-r} K)).$$

Let $\mathcal{H}$ be a complex inner product space and $\mathcal{B}(\mathcal{H})$ be the space of bounded linear operators on $\mathcal{H}$. Let $\mathcal{P}_+ \subset \mathcal{B}(\mathcal{H})$ denote the set of strictly positive operators (self-adjoint operators whose spectrum is contained in $(0, \infty)$). For any real numbers $q$ and $r$ such that $1 < r \le q \le 2$, and for any fixed operator $K \in \mathcal{B}(\mathcal{H})$, the functional $$(A, B) \mapsto \text{Re}(\text{Tr}(A^q K^* B^{1-r} K))$$ is jointly convex for $A, B \in \mathcal{P}_+$. Here, $K^*$ denotes the adjoint of $K$, and the operator powers $A^q$ and $B^{1-r}$ are defined via the continuous functional calculus.