Positivity of the gap in (??) is exactly Theorem 5.1.6. The Loewner inequality (??) is its order-theoretic reformulation.

Since \(A\) is normal, diagonalize \(A=UDU^\dagger \) with \(D\) diagonal and \(U\) unitary. The spectral projectors of \(A\) generate a commutative \(*\)-subalgebra. The restriction of \(T\) to this subalgebra is completely positive, so the Kadison–Schwarz inequality for completely positive subunital maps gives \(T(A^\dagger )T(A)\le T(A^\dagger A)\).

Embed \(A\) as the northwest corner of a normal operator \(N\) on \(\mathbb {C}^D\oplus \mathbb {C}^E\). Extend \(T\) to a positive subunital map \(\widetilde{T}\) on the larger space, acting as \(T\) on the northwest block and as zero elsewhere. Apply Theorem 20.1.1.2 to \(\widetilde{T}\) and \(N\), then compress the resulting inequality back to \(\mathbb {C}^D\).

First take \(B{\gt}0\). Since \(B\) commutes with \(A\) and \(A^\dagger A\le B\), also \(AA^\dagger \le B\), so \(D_R=(B-A^\dagger A)^{1/2}\) and \(D_L=(B-AA^\dagger )^{1/2}\) are positive semidefinite. The block matrix

is normal. As in the subnormal case, extend \(T\) to a positive subunital map on \(M_{2D}(\mathbb {C})\), apply the normal-operator Schwarz inequality (Theorem 20.1.1.2) to that extension and \(N\), and compress back to \(\mathbb {C}^D\). Since \(N\) is normal, both \(T(A^\dagger )T(A)\le T(B)\) and \(T(A)T(A^\dagger )\le T(B)\) follow. For general \(B\ge 0\), apply this to \(B_\varepsilon =B+\varepsilon \mathbb {1}{\gt}0\) and let \(\varepsilon \to 0\).

Apply Theorem 20.1.2.2 to the adjoint Kraus map, which is unital under the trace-preservation hypothesis.

Write \(A=U\operatorname{diag}(\mu )U^\dagger \) by the spectral theorem and set \(w=U^\dagger v\). Since \(U\) is unitary, \(p_i:=\lvert w_i\rvert ^2\) satisfies \(\sum _i p_i=\lVert v\rVert ^2=1\), so \((p_i)\) is a probability distribution on \(n\). The eigenvalues \(\mu _i\) lie in \([0,\infty )\) because \(A\) is positive semidefinite. A direct computation yields

so the scalar Jensen inequality applied to the weights \((p_i)\) and points \((\mu _i)\) gives the conclusion. See [ Bha97 , Chapter V ] .

Let \(A=tA_1+(1-t)A_2\), and let \(\{ \psi _j\} \) be an orthonormal eigenbasis of \(A\) with eigenvalues \(\mu _j\ge 0\). A direct computation gives

Scalar concavity of \(x\mapsto x^p\) on \([0,\infty )\) yields \(\mu _j^p\ge t\, a_j^p+(1-t)b_j^p\). Applying Lemma 20.2.1 to the convex function \(x\mapsto -x^p\) and multiplying by \(-1\) gives

Summing over \(j\) and using \(\sum _j\operatorname{Re}\langle \psi _j,M\psi _j\rangle =\operatorname{Re}\operatorname{tr}M\) yields the conclusion. See [ Bha97 , Chapter V ] .

Use the eigenbasis reduction from Theorem 20.3.1, with scalar convexity rather than concavity of \(x\mapsto x^p\) on \([0,\infty )\) for \(p\ge 1\) and the convex form of the diagonal Jensen inequality.

For \(p\in (1,2)\) one uses the integral representation \(a^p=\int _0^\infty f_{p,s}(a)\, d\mu (s)\) of Carlen [ Car10 , Lemma 2.8 ] , whose integrand \(f_{p,s}(x)=s^{p-1}(s^{-1}x+s(s+x)^{-1}-1)\) is operator convex: it decomposes into a linear term, a non-negative multiple of the operator convex resolvent \(x\mapsto (s\cdot 1+x)^{-1}\), and a constant. Since \(f_{p,s}(ta+(1-t)b)\le t\, f_{p,s}(a)+(1-t)f_{p,s}(b)\) for every \(s{\gt}0\), integrating this pointwise Loewner inequality against \(\mu \) gives

The endpoint \(p=1\) is the identity, and \(p=2\) is \(a\mapsto a^2\), operator convex because \(t\, a^2+(1-t)b^2-(ta+(1-t)b)^2=t(1-t)(a-b)^2\ge 0\). See [ Bha97 , Chapter V ] .

Apply the Schur-complement criterion to the block matrix \(\begin{psmallmatrix} \end{psmallmatrix}Y^{-1}& W\\ W^\dagger & W^\dagger YW\end{psmallmatrix}\). The lower-right Schur complement is zero, hence positive semidefinite, and compressing the resulting upper-left Schur complement by \(W\) gives (??).

Expand the block matrix multiplication entrywise. The defect identity (??) gives the isometry relation after summing the \(\iota \)-blocks and the defect block, and the weighted compression identity follows from the same expansion with the scalar diagonal inserted. Positive definiteness uses strict positivity of every diagonal entry, while the inverse formula uses the nonvanishing of every diagonal entry.

Apply the compression inequality (??) to the positive definite scalar block-diagonal matrix with entries \(w_i+t\) and defect entry \(t\). The compression identities rewrite the compressed matrix as \(\sum _iw_iC_iC_i^\dagger +t\mathbb {1}\) and its inverse compression as the upper bound in (??).

Diagonalize the Hermitian matrix \(A\) by a unitary matrix. The matrices \(P_i\) are the conjugates of the rank-one diagonal matrix units, their sum is the identity, and positive semidefiniteness of \(A\) gives \(\lambda _i\ge 0\).

Use the spectral resolution \(A=\sum _i\lambda _iP_i\). The matrices \(T(P_i)\) are positive semidefinite and sum to \(T(\mathbb {1})\le \mathbb {1}\). Write each \(T(P_i)\) as \(C_iC_i^\dagger \) using the positive square root, and write the defect \(\mathbb {1}-\sum _iT(P_i)\) as \(SS^\dagger \). The finite-POVM resolvent inequality gives the desired estimate after substituting the spectral formula for \((A+t\mathbb {1})^{-1}\).

Rewrite the continuous functional calculus in resolvent form as \(f_{p,t}(X)=t^{p-1}\mathbb {1}-t^p(t\mathbb {1}+X)^{-1}\). After applying the linear map \(T\), the difference \(f_{p,t}(T(A))-T(f_{p,t}(A))\) is exactly \(t^p\) times the positive matrix supplied by the positive-map resolvent inequality.

Rewrite the continuous functional calculus in resolvent form as \(g_{p,t}(X)=t^{p-2}X+t^p(t\mathbb {1}+X)^{-1}-t^{p-1}\mathbb {1}\). Under the linear map \(T\), the leading \(t^{p-2}\) terms cancel and the constant \(t^{p-1}\) terms combine, so the difference \(T(g_{p,t}(A))-g_{p,t}(T(A))\) is exactly \(t^p\) times the positive matrix supplied by the positive-map resolvent inequality, using \(t^pt^{-1}=t^{p-1}\).

The Loewner order has closed positive cone in finite dimension. Hence the ordered Bochner integral preserves almost-everywhere nonnegativity. The corresponding almost-everywhere monotonicity statement is the general ordered Bochner integral theorem.

For \(p=0\) the assertion is the subunitality condition \(T(\mathbb {1})\le \mathbb {1}\). For \(p=1\) it is an equality. For \(0{\lt}p{\lt}1\), use Carlen’s integral representation of \(A^p\) by the functions \(f_{p,t}(x)=t^p(t^{-1}-(t+x)^{-1})\). The pointwise inequality \(T(f_{p,t}(A))\le f_{p,t}(T(A))\) is Lemma 20.5.7. Integrating this inequality with respect to the representing measure gives the result by Lemma 20.5.9 and linearity of the Bochner integral under \(T\).

Direct application of Theorem 20.5.10.

For \(p=1\) the assertion is an equality. For \(1{\lt}p{\lt}2\), use Carlen’s integral representation of \(A^p\) by the convex integrands \(g_{p,t}(x)=t^{p-2}x+t^p(t+x)^{-1}-t^{p-1}\). The pointwise inequality \(g_{p,t}(T(A))\le T(g_{p,t}(A))\) is Lemma 20.5.8. Integrating this inequality with respect to the representing measure and using Lemma 20.5.9 together with linearity of the Bochner integral under \(T\) gives the inequality for \(1{\lt}p{\lt}2\). The endpoint \(p=2\) follows by taking the limit \(q\to 2^-\) in \((T(A))^q\le T(A^q)\): since \(M^q\to M^2\) in the operator norm as \(q\to 2^-\) for any positive semidefinite \(M\), and the Loewner order is closed, this gives \((T(A))^2\le T(A^2)\).

Direct application of Theorem 20.5.12.

Apply Theorem 20.5.10 to \(0{\lt}p{\lt}1\) and rewrite it as an inequality for \((A^p-\mathbb {1})/p\). The continuous functional calculus gives \((A^p-\mathbb {1})/p\to \log A\) as \(p\to 0^+\), and the same limit holds for \(T(A)\). Since the Loewner order graph is closed in finite dimension, the limiting inequality is \(T(\log A)\le \log (T(A))\).

Direct application of Theorem 20.5.14.

This is the value \(B(s,1-s)\) of the Beta function, which equals \(\Gamma (s)\Gamma (1-s)=\pi /\sin (\pi s)\) by Euler’s reflection formula. The substitution \(x=u/(1+u)\) carries the half-line onto the unit interval and turns the integrand into the Beta integrand \(x^{s-1}(1-x)^{-s}\).

The integrand is \(O(t^{s-1})\) as \(t\to 0^+\) and \(O(t^{s-2})\) as \(t\to \infty \), so it has integrable comparison functions on each tail.

The scaling substitution \(t=(a/b)\, u\) turns the integrand into \(a^s b^{1-s}\, u^{s-1}/(1+u)\), whose half-line integral is the reflection integral \(\pi /\sin (\pi s)\) of Lemma 20.6.1.

By Lemma 20.6.3, the integral equals \(a^s b^{1-s} \pi /\sin (\pi s)\), so

Both sides are diagonal, so the identity holds entry by entry; on each diagonal entry it is the scalar Lieb integral identity (Lemma 20.6.4).

The pair \((A\otimes \mathbb {1},\mathbb {1}\otimes B^\top )\) is simultaneously diagonalized by \(V=U_A\otimes U_{B^\top }\) (eigenbases of \(A\) and \(B^\top \)). In that basis the pair becomes a commuting pair of positive diagonal matrices, and the identity follows from the diagonal case (Lemma 20.6.5).

Write \(Z=X(X+Y)^{-1}Y\). The identity \(X(X+Y)^{-1}(X+Y)=X\) gives \((X-Z)-X(X+Y)^{-1}X=0\), so the block matrix \(\begin{psmallmatrix} \end{psmallmatrix}X-Z & X\\ X & X+Y\end{psmallmatrix}\) is positive semidefinite (its Schur complement against \(X+Y\) vanishes). Setting \(Z_i=X_i(X_i+Y_i)^{-1}Y_i\) and \(Z_\theta =\theta Z_1+(1-\theta )Z_2\), the convex combination of the two block matrices is \(\begin{psmallmatrix} \end{psmallmatrix}X_\theta -Z_\theta & X_\theta \\ X_\theta & X_\theta +Y_\theta \end{psmallmatrix}\), which is positive semidefinite, and its Schur complement against \(X_\theta +Y_\theta \) yields \(X_\theta (X_\theta +Y_\theta )^{-1}Y_\theta -Z_\theta \ge 0\), which is the claimed inequality (Ando, 1979).

The integrand equals \(t^{-1} \hat A(\hat A+t\hat B)^{-1}(t\hat B)\), a rescaled parallel sum of \(\hat A\) and \(t\hat B\), so concavity follows from Lemma 20.6.7 and the linearity of \(A\mapsto \hat A\), \(B\mapsto \hat B\).

By the integral representation (??), each fractional product equals the positive multiple \(\frac{\sin (\pi s)}{\pi }\) of the integral of the resolvent integrand against the non-negative weight \(t^{s-1}\) on \((0,\infty )\). For each \(t{\gt}0\), the integrand is jointly concave by Lemma 20.6.8, so the difference of the averaged point and the convex combination is, almost everywhere, the non-negative scalar \(t^{s-1}\) times a positive-semidefinite matrix. Integrating a positive-semidefinite integrand yields a positive-semidefinite matrix, and scaling by the positive constant \(\frac{\sin (\pi s)}{\pi }\) preserves the Loewner order, which gives the inequality.

For \(s\in (0,1)\), the concavity of the fractional product of the commuting left and right multiplication operators (Lemma 20.6.9) gives a Loewner-order inequality between positive-semidefinite matrices on the model space \(\mathbb {C}^D\otimes \mathbb {C}^D\). Writing that product as \(A^s\otimes (B^\top )^{1-s}\) and pairing it with the vectorization of \(K^\top \),

expresses the trace functional as the quadratic form of the Kronecker product at that vector. Applying this positive quadratic form to the Loewner inequality transfers the concavity to the real part of the trace. The endpoints \(s=0\) and \(s=1\) reduce the varying side to a linear function of the convex-combination argument, hence hold with equality.

For the interior exponents \(s\in (0,1)\), replace each input \(A_i\) by the positive-definite matrix \(A_i+\varepsilon \mathbb {1}\), and likewise for the \(B_i\). Theorem 20.6.10 applies to these positive-definite matrices for every \(\varepsilon {\gt}0\), and the average of the regularized inputs is the regularization of the average. Letting \(\varepsilon \to 0^+\), each regularized power \((A_i+\varepsilon \mathbb {1})^s\) converges to \(A_i^s\), since \(x\mapsto x^s\) is continuous on the non-negative reals for \(0{\lt}s{\lt}1\), including at the origin; the trace functional is continuous, so the regularized inequality passes to the limit. The endpoints \(s=0\) and \(s=1\) are linear in the convex-combination argument and hold with equality.

Put \(r=x+y\). If \(r=0\), both exponents vanish and the functional is constant. If \(0{\lt}r\le 1\), set \(s=x/r\). Then \(x=rs\) and \(y=r(1-s)\), with \(s\in [0,1]\). The functional becomes the boundary Lieb functional applied to the powered inputs \(A^r\) and \(B^r\). Operator concavity of \(C\mapsto C^r\) gives

The boundary positive-semidefinite theorem gives concavity at the left-hand powered inputs, and monotonicity of \((P,Q)\mapsto \Re \operatorname{tr}(K^\dagger P^s K Q^{1-s})\) on positive-semidefinite arguments transports the result to the powered convex combinations.

Follows directly from Theorem 20.6.10.

Specialize Theorem 20.6.13 at \(K=\mathbb {1}\) and simplify \(\mathbb {1}^\dagger =\mathbb {1}\) together with \(\mathbb {1}\cdot X=X\cdot \mathbb {1}=X\).

Specialize Theorem 20.6.13 at \(B_1=B_2=B\), so that the convex combination collapses: \(tB+(1-t)B=B\).

Specialize Theorem 20.6.13 at \(A_1=A_2=A\), so that the convex combination collapses: \(tA+(1-t)A=A\).

Compare the two Schur complements of the block matrix \(\begin{psmallmatrix} \end{psmallmatrix}A^{-1}& \mathbb {1}\\ \mathbb {1}& B\end{psmallmatrix}\). Complementing its \((1,1)\) block gives \(B-(A^{-1})^{-1}=B-A\ge 0\), so the block matrix is positive semidefinite; complementing the \((2,2)\) block of the same matrix gives \(A^{-1}-B^{-1}\), which is therefore positive semidefinite as well.

Each operator \(A\otimes \mathbb {1}+t\, (\mathbb {1}\otimes B^\top )\) is positive definite, being the sum of the positive-definite \(A\otimes \mathbb {1}\) and the positive-semidefinite \(t\, (\mathbb {1}\otimes B^\top )\). From \(A_1\le A_2\) and \(B_1\le B_2\) one gets \(A_1\otimes \mathbb {1}\le A_2\otimes \mathbb {1}\) and \(\mathbb {1}\otimes B_1^\top \le \mathbb {1}\otimes B_2^\top \); since \(t{\gt}0\), adding these yields

Inverse-antitonicity (Lemma 20.7.1) reverses this inequality on passing to the resolvents.