Proof. Let \(f(r)\) denote the bracket. Equal trace gives \(f(r)=O(r^2)\) at zero, while \(f(r)=O(\ln r)\) at infinity. Hence the integral converges absolutely and \(f(r)/r\) vanishes at both endpoints. Integration by parts reduces it to \(\int_0^\infty f'(r)\,\mathrm dr/r\). If \(a_i\) and \(b_j\) are the positive eigenvalues of \(U\) and \(V\), then \[\frac{f'(r)}r =\sum_i\frac{a_i}{r(1+ra_i)} -\sum_j\frac{b_j}{r(1+rb_j)}.\] A primitive of the term for \(a>0\) is \(a\ln\bigl(r/(1+ra)\bigr)\). In the signed sum, its lower endpoint limit is zero because the coefficients of \(\ln r\) cancel by equal trace. Its upper endpoint contribution is \(-a\ln a\). Taking their difference gives \(\mathcal S(U)-\mathcal S(V)\). Zero eigenvalues contribute nothing. ◻

Proof. Set \[\mathcal D=\begin{pmatrix}D&0\\0&E\end{pmatrix},\qquad \mathcal C=\begin{pmatrix}D&Z\\-Z^*&E\end{pmatrix}.\] The block determinant identity gives \[\det\mathcal C=\det E\det(D+ZE^{-1}Z^*)>0, \qquad G_Z(D,E)=\ln\det\mathcal C-\ln\det\mathcal D.\] Although \(\mathcal C\) is not generally Hermitian, its determinant is positive and its logarithm here is the real logarithm. Along an affine line in \((D,E)\), let \(\mathcal H\) be the derivative of \(\mathcal D\), which is also the derivative of \(\mathcal C\). The determinant derivative formula gives \[ G_Z''=\mathop{\mathrm{Tr}}(\mathcal D^{-1}\mathcal H\mathcal D^{-1}\mathcal H) -\mathop{\mathrm{Tr}}(\mathcal C^{-1}\mathcal H\mathcal C^{-1}\mathcal H). \tag{16}\] At the point under consideration put \[L=\mathcal D^{-1/2}\mathcal H\mathcal D^{-1/2},\qquad K=\mathcal D^{-1/2}(\mathcal C-\mathcal D)\mathcal D^{-1/2}.\] Then \(L=L^*\) and \(K^*=-K\). Consequently \(\|(I+K)v\|_2^2=\|v\|_2^2+\|Kv\|_2^2\) for every vector \(v\), and \(R=(I+K)^{-1}\) exists with operator norm at most one. Cyclicity of trace rewrites (16) as \[G_Z''=\|L\|_{\mathrm{HS}}^2-\mathop{\mathrm{Tr}}((RL)^2).\] The last trace is real because both log determinants along the line are real. For any complex matrix \(B\), entrywise Cauchy–Schwarz gives \(|\mathop{\mathrm{Tr}}(B^2)|=|\sum_{i,j}B_{ij}B_{ji}|\le\|B\|_{\mathrm{HS}}^2\). Therefore \[\mathop{\mathrm{Tr}}((RL)^2)\le|\mathop{\mathrm{Tr}}((RL)^2)| \le\|RL\|_{\mathrm{HS}}^2\le\|L\|_{\mathrm{HS}}^2.\] Every second directional derivative is thus nonnegative. The positive definite domain is convex, so \(G_Z\) is convex along each line segment, as required. ◻

Proof. The nonnegative entries in the two lists \((x,y,z)\) and \((\lambda_+,\lambda_-)\) have the same sum. Applying 4 to the corresponding diagonal matrices gives \[\begin{align*} F(x,y,z) &=\int_0^\infty \ln\frac{(1+rx)(1+ry)(1+rz)} {(1+r\lambda_+)(1+r\lambda_-)}\frac{\,\mathrm dr}{r^2}\\ &=\int_0^\infty\left[\ln(1+rx) -\ln\!\left(1+\frac{rx}{(1+ry)(1+rz)}\right)\right] \frac{\,\mathrm dr}{r^2}. \tag{20}\end{align*}\] The second equality uses \((1+r\lambda_+)(1+r\lambda_-)=1+r(x+y+z)+r^2yz\); the integral is absolutely convergent by the lemma. For \(x=1\) its bracket is \[\ln(1+r)-G_{\sqrt r}(1+ry,1+rz).\] By 5, this is concave in \((y,z)\) for every \(r>0\). Integrating the concavity inequality shows that \(F(1,\cdot,\cdot)\) is concave on the nonnegative quadrant.

Under scaling of \((x,y,z)\) by a positive constant, both \(\lambda_\pm\) scale by the same constant. Their sum is \(m\), so the logarithms of the scaling constant cancel in (18). This proves homogeneity, including scale zero by continuity. For a nonempty collection with all \(x_i>0\), set \(x=\sum_i x_i\). Homogeneity and the preceding concavity now give \[\begin{align*} \sum_iF(x_i,y_i,z_i) &=x\sum_i\frac{x_i}{x}F(1,y_i/x_i,z_i/x_i)\\ &\le xF\!\left(1,\frac{\sum_i y_i}{x}, \frac{\sum_i z_i}{x}\right) =F\!\left(x,\sum_i y_i,\sum_i z_i\right). \end{align*}\] For a collection containing zero first coordinates, replace every \(x_i\) by \(x_i+\varepsilon\), apply this inequality, and let \(\varepsilon\downarrow0\). Continuity of \(F\) gives (19), even when every \(x_i=0\). The empty collection gives zero on both sides. Finally, applying superadditivity to the two triples \(\theta a\) and \((1-\theta)b\) and using homogeneity gives \(F(\theta a+(1-\theta)b)\ge\theta F(a)+(1-\theta)F(b)\) for nonnegative triples \(a,b\) and \(0\le\theta\le1\). In particular, \(F\) is concave on the probability simplex. ◻

Proof of 3. First suppose that all three blocks are square of size \(N\), and temporarily allow arbitrary nonnegative masses \(x,y,z\). The matrices \[U=AA^*\oplus BB^*\oplus CC^*,\qquad V=MM^*\] have equal trace. For \(r>0\), put \(D=I+rBB^*\) and \(E=I+rC^*C\). The Schur complement of \(I+rCC^*\) in \(I+rMM^*\) yields \[ \det(I+rMM^*) =\det(I+rCC^*)\det(D+rAE^{-1}A^*), \tag{21}\] where \(I-rC^*(I+rCC^*)^{-1}C=(I+rC^*C)^{-1}\). Consequently the integrand bracket in (14) is \[ \ln\det(I+rAA^*)-G_{\sqrt r A}(D,E). \tag{22}\]

Choose a singular-value decomposition \(A=U_A\Sigma V_A^*\). Replacing \((A,B,C)\) by \((\Sigma,U_A^*B,CV_A)\) amounts to multiplying \(M\) on the left and right by block-diagonal unitaries; all entropies and masses are preserved. In these coordinates write \[A=\mathop{\mathrm{diag}}(\sqrt{x_1},\ldots,\sqrt{x_N}),\qquad y_i=(BB^*)_{ii},\qquad z_i=(C^*C)_{ii}.\] These entries are nonnegative and satisfy \(\sum_i(x_i,y_i,z_i)=(x,y,z)\). For each \(\varepsilon\in\{-1,1\}^N\), let \(S_\varepsilon=\mathop{\mathrm{diag}}(\varepsilon_1,\ldots,\varepsilon_N)\). Because \(S_\varepsilon A S_\varepsilon=A\), \[G_{\sqrt r A}(S_\varepsilon DS_\varepsilon, S_\varepsilon ES_\varepsilon) =G_{\sqrt r A}(D,E).\] Averaging simultaneously over all signs removes the off-diagonal entries of both arguments. Joint convexity from 5 therefore gives \[G_{\sqrt r A}(D,E)\ge G_{\sqrt r A}\bigl(\mathop{\mathrm{diag}}(1+ry_i),\mathop{\mathrm{diag}}(1+rz_i)\bigr).\] Because \(G\) is subtracted in (22), that bracket is bounded above by \[\begin{align*} &\sum_i\left[\ln(1+rx_i) -\ln\!\left(1+\frac{rx_i}{(1+ry_i)(1+rz_i)}\right)\right]\\ &\hspace{12mm}= \sum_i\ln\frac{(1+rx_i)(1+ry_i)(1+rz_i)} {1+r(x_i+y_i+z_i)+r^2y_iz_i}. \tag{23}\end{align*}\] By (20), each summand integrates against \(\,\mathrm dr/r^2\) to \(F(x_i,y_i,z_i)\), and these finitely many integrals converge absolutely. Integrating the inequality and applying 6 proves \[ \mathcal S(AA^*)+\mathcal S(BB^*)+\mathcal S(CC^*)-\mathcal S(MM^*) \le\sum_iF(x_i,y_i,z_i)\le F(x,y,z). \tag{24}\] When \(x+y+z=1\), the last expression is \(H(x,y,z)-g(yz)\). For \(x>0\), \(\mathcal S(AA^*)=x\mathcal S(AA^*/x)-x\ln x\), and the same identity holds for the other two blocks. Substitution gives (13). If a mass vanishes, the corresponding block is zero, and its contribution is zero. Thus the proof includes every boundary case; all inverses used above are of positive definite matrices even when \(A,B,C\) are singular.

Finally, for \(d\times e\) blocks let \(N=\max(d,e)\) and choose the coordinate isometries \(J:\mathbb C^d\to\mathbb C^N\) and \(K:\mathbb C^e\to\mathbb C^N\). Replace each block \(A\) by \(JAK^*\), and likewise for \(B,C\). The resulting square block matrix is \((J\oplus J)M(K\oplus K)^*\), so this operation only adds zero singular values to \(M\) and to each block. All masses and entropies in (13) are unchanged. The square case therefore proves the rectangular case as well. ◻

Proof. For \(0<u<1/4\), put \[t=\sqrt{1-4u},\qquad L=\frac{\mathop{\mathrm{atanh}}t}{t},\qquad \mathop{\mathrm{atanh}}t=\frac12\ln\frac{1+t}{1-t}.\] Differentiating the definition of \(g\) gives \[ g'(u)=2L,\qquad g''(u)=\frac{4L-1/u}{t^2},\qquad ug''(u)+\frac12g'(u)=\frac{L-1}{t^2}\ge0. \tag{25}\] The last inequality follows from \(\mathop{\mathrm{atanh}}t\ge t\), since \((\mathop{\mathrm{atanh}}t)'=(1-t^2)^{-1}\ge1\) and both functions vanish at zero. For \(0<s<1/2\), these identities give \[\frac{\mathrm d}{\mathrm ds}g(s^2)=2s g'(s^2)\ge0,\qquad \frac{\mathrm d^2}{\mathrm ds^2}g(s^2) =4\left(s^2g''(s^2)+\frac12g'(s^2)\right)\ge0.\] Continuity supplies both conclusions at the endpoints; no boundary derivatives are required. Since \(u\mapsto\sqrt u\) is nondecreasing, \(g(u)=g((\sqrt u)^2)\) is nondecreasing as well. ◻

Proof. The completed-square formula (7) gives \[\sqrt{v(p)}= \left\|\begin{pmatrix} \sqrt{\gamma\nu(1-\nu)}\\ \sqrt{\gamma(1-\gamma)}\,(p-\nu) \end{pmatrix}\right\|_2.\] This is a convex function of \(p\), being the norm of an affine vector, even when either coefficient vanishes. By 7, \(s\mapsto g(s^2)\) is nondecreasing and convex on \([0,1/2]\), which contains the range of \(\sqrt{v}\). A nondecreasing convex function of a convex function is convex: apply the inner convexity, then monotonicity, then the outer convexity. This proves the assertion, with no parameter exclusions. ◻

Proof. Choose a finite Kraus representation \[\mathcal N(P)=\sum_{\ell=1}^r V_\ell P V_\ell^*, \qquad \sum_{\ell=1}^r V_\ell^*V_\ell=I.\] Here \(V_\ell:\mathcal H_{\mathrm{in}}\to\mathcal H_{\mathrm{out}}\). Complete positivity in finite dimensions gives such a finite family; the last identity is trace preservation. If \(d_{\mathrm{out}}=\dim\mathcal H_{\mathrm{out}}\), form \(X,Y\in\mathbb C^{d_{\mathrm{out}}\times r}\) by giving their \(\ell\)th columns the values \(V_\ell\psi_0\) and \(V_\ell\psi_1\), respectively. The common column index gives the three Gram identities and normalizations \[ \begin{split} XX^*&=\mathcal N(\lvert \psi_0\rangle\langle \psi_0\rvert),\qquad YY^*=\mathcal N(\lvert \psi_1\rangle\langle \psi_1\rvert),\\ XY^*&=\mathcal N(\lvert \psi_0\rangle\langle \psi_1\rvert),\qquad \mathop{\mathrm{Tr}}(XX^*)=\mathop{\mathrm{Tr}}(YY^*)=1. \end{split} \tag{39}\] No equality between \(d_{\mathrm{out}}\), \(r\), and the input dimension is required, and the branch vectors need not be orthogonal.

Set \(x=1-p\) and \(y=p\), and use \(a,b,c,d,k\) from (34). Applying the two channels to the branch decomposition gives \[(\mathcal A\otimes\mathcal N)(\lvert \psi\rangle\langle \psi\rvert)= \begin{pmatrix} axXX^*+byYY^*&k\sqrt{xy}\,XY^*\\ k\sqrt{xy}\,YX^*&cxXX^*+dyYY^* \end{pmatrix}.\] This is (29) with \[ \alpha=ax,\quad\beta=by,\quad\eta=cx,\quad\delta=dy, \qquad K=k\sqrt{xy}. \tag{40}\] The coefficients are nonnegative, sum to one, and satisfy \(\alpha\delta-\beta\eta=(ad-bc)xy=K^2\). Their column sums are \(\alpha+\eta=x\) and \(\beta+\delta=y\). Moreover, their scalar parameter is \[u=(ax+by)(cx+dy)-k^2xy=v(p).\] Thus 8 applies to the arbitrary rectangular matrices \(X,Y\) and gives (38). The theorem includes \(u=0\), and the zero-weight branch terms vanish, so this argument includes \(p=0,1\) and every boundary value of \(\gamma,\nu\). ◻

Proof. For \(n=1\) there is equality by (37). Suppose the claim holds for \(n-1\) and write an arbitrary pure input as \[ \psi=\sqrt x\,\lvert 0\rangle\otimes\psi_0 +\sqrt y\,\lvert 1\rangle\otimes\psi_1, \qquad x=1-p_1,\quad y=p_1. \tag{42}\] The branch vectors are normalized but need not be orthogonal. For a zero-weight branch choose any unit vector. The decomposition is valid for every input vector, and each conditional vector may be entangled across the remaining sites.

Apply 11 with \(\mathcal N=\mathcal A^{\otimes(n-1)}\), and write \(\sigma_i=\mathcal A^{\otimes(n-1)}(\lvert \psi_i\rangle\langle \psi_i\rvert)\) for the two branch outputs. The lemma gives the explicit recursion \[ \mathcal S\bigl(\mathcal A^{\otimes n}(\lvert \psi\rangle\langle \psi\rvert)\bigr) \ge e(p_1)+x\mathcal S(\sigma_0)+y\mathcal S(\sigma_1). \tag{43}\] Apply the induction hypothesis to both pure branch inputs. If \(p_{j|i}\) is the excitation probability of site \(j>1\) in \(\psi_i\), then \[p_j=xp_{j|0}+yp_{j|1}.\] This identity follows from \(\langle0|1\rangle=0\) on the first site; it does not require \(\psi_0\perp\psi_1\). By 10, \[x\mathcal S(\sigma_0)+y\mathcal S(\sigma_1) \ge\sum_{j=2}^n\bigl(xe(p_{j|0})+ye(p_{j|1})\bigr) \ge\sum_{j=2}^ne(p_j).\] Together with (43), this proves the induction step. All terms from zero-weight branches have zero coefficient, so the argument also includes \(p_1=0,1\). ◻

Proof. Take a finite pure-state decomposition \(\rho=\sum_aq_a\lvert \psi_a\rangle\langle \psi_a\rvert\), and let \(p_{a,j}\) be the corresponding populations. Entropy concavity, 12, and convexity from 10 give \[\mathcal S\bigl(\mathcal A^{\otimes n}(\rho)\bigr) \ge\sum_aq_a\sum_j e(p_{a,j}) \ge\sum_j e\!\left(\sum_aq_ap_{a,j}\right)=\sum_j e(p_j).\] ◻

Alternative proof of 6. In the simplex interior, set \(x=1-y-z\) and \(u=yz\). The negative Hessian of \(F(1-y-z,y,z)\) is \[J=\begin{pmatrix} \dfrac1y+\dfrac1x+z^2g''(u)&\dfrac1x+g'(u)+ug''(u)\\[5pt] \dfrac1x+g'(u)+ug''(u)&\dfrac1z+\dfrac1x+y^2g''(u) \end{pmatrix}.\] Expanding its determinant and substituting (25) and \(x+y+z=1\) yields \[\begin{align*} \det J &=\frac1{xu}+\frac{y+z-4u}{x}g''(u) -\frac2xg'(u)-g'(u)^2-2ug'(u)g''(u)\\ &=\frac{1-4uL^2}{ut^2}>0. \tag{52}\end{align*}\] Indeed, for \(0<t<1\), \[\mathop{\mathrm{atanh}}t<\frac{t}{\sqrt{1-t^2}},\] as both sides vanish at zero and their derivatives are respectively \((1-t^2)^{-1}\) and \((1-t^2)^{-3/2}\). Since \(4u=1-t^2\), this proves the strict inequality in (52). Furthermore, \[\begin{pmatrix}y&z\end{pmatrix}J\begin{pmatrix}y\\z\end{pmatrix} =\frac{y+z}{x} +4u\left(ug''(u)+\frac12g'(u)\right)>0.\] A real symmetric \(2\times2\) matrix with positive determinant has eigenvalues of the same sign; this positive quadratic form forces both to be positive. Hence \(F\) is concave in the simplex interior, and continuity extends concavity to the closed simplex.

To prove (19), omit zero triples and put \(m_i=x_i+y_i+z_i\), \(m=\sum_i m_i\). For \(m>0\), homogeneity and simplex concavity give \[\sum_i F(x_i,y_i,z_i) =m\sum_i\frac{m_i}{m} F\!\left(\frac{x_i}{m_i},\frac{y_i}{m_i},\frac{z_i}{m_i}\right) \le F\!\left(\sum_i x_i,\sum_i y_i,\sum_i z_i\right).\] If \(m=0\), all terms vanish. ◻

Alternative proof of 5. At any point \((D_0,E_0)\), use the fixed congruences \[D=D_0^{1/2}\widetilde D D_0^{1/2},\qquad E=E_0^{1/2}\widetilde E E_0^{1/2},\qquad W=D_0^{-1/2}ZE_0^{-1/2}.\] The determinant factors cancel to give \(G_Z(D,E)=G_W(\widetilde D,\widetilde E)\). These invertible linear changes preserve affine lines. Independent unitary changes in the two spaces then put \(W\) in singular-value form \(Y=\mathop{\mathrm{diag}}(v_1,\ldots,v_N)\) with \(v_i\ge0\). It therefore suffices to prove nonnegativity of the second derivative at \((I,I)\) for this \(Y\) and arbitrary Hermitian directions \(P,Q\).

Set \(R=(I+YY^*)^{-1}\) and \(N_0=P-YQY^*\). Using \((I+sQ)^{-1}=I-sQ+s^2Q^2+O(s^3)\) gives \[ \left.\frac{\mathrm d^2}{\mathrm ds^2} G_Y(I+sP,I+sQ)\right|_{s=0} =\mathop{\mathrm{Tr}}\bigl(P^2+2RYQ^2Y^*-RN_0RN_0\bigr). \tag{53}\] For completeness, the log-determinant expansion used here is \[\ln\det(H+sK+s^2L_0) =\ln\det H+s\mathop{\mathrm{Tr}}(H^{-1}K) +s^2\left(\mathop{\mathrm{Tr}}(H^{-1}L_0) -\frac12\mathop{\mathrm{Tr}}(H^{-1}KH^{-1}K)\right)+O(s^3),\] for \(H>0\) and Hermitian \(K,L_0\). It follows by factoring \(H^{1/2}\) and applying the scalar logarithm expansion to the eigenvalues of the resulting Hermitian increment. No commutativity is assumed.

Write \(r_i=(1+v_i^2)^{-1}\) and \(c_i=v_i/(1+v_i^2)\). Expanding the trace in (53) in entries gives \[\begin{align*} \sum_{i,j}\bigl[&(1-r_ir_j)|P_{ij}|^2 +\{2(1-r_i)-(1-r_i)(1-r_j)\}|Q_{ij}|^2\\ &\hspace{34mm}+2c_ic_j\operatorname{Re} (P_{ij}\overline{Q_{ij}})\bigr]. \end{align*}\] Since \(Q\) is Hermitian, \(|Q_{ij}|^2=|Q_{ji}|^2\). Pairing the \((i,j)\) and \((j,i)\) terms replaces their coefficient by \[\frac12\bigl[2(1-r_i)+2(1-r_j) -2(1-r_i)(1-r_j)\bigr]=1-r_ir_j.\] This identity also holds when \(i=j\). Completing the square therefore rewrites the second derivative as \[ \sum_{i,j}\left[ (1-r_ir_j-c_ic_j)(|P_{ij}|^2+|Q_{ij}|^2) +c_ic_j|P_{ij}+Q_{ij}|^2\right]. \tag{54}\] Every coefficient is nonnegative: \(c_ic_j\ge0\), and \[1-r_ir_j-c_ic_j =\frac{(v_i-v_j)^2+v_iv_j+v_i^2v_j^2} {(1+v_i^2)(1+v_j^2)}\ge0.\] Thus the second derivative is nonnegative in every Hermitian direction at every positive definite pair. The domain is convex, so the asserted joint convexity follows along each line segment. ◻