Lemma A6.2: Let $R1 , R2 , S1 , S2 , T1, T2$ be positive operators such that $0 = [R1, R2 ] = [S1, S 2 ] = [T1, T2 ]$, and $$ R1 ≥ S1 + T1\\ R2 ≥ S2 + T2 $$ Then for all $0 ≤ t ≤ 1$, $$ R_1^t R_2^{1−t} ≥ S_1^t S_2^{1−t} + T_1^tT_2^{1−t} $$ is true as a matrix inequality.
Lemma A6.8: Let $A$ be a positive matrix. Define a superoperator (linear operator on matrices) by the equation $A(X)≡AX$. Show that $A$ is positive with respect to the Hilbert–Schmidt inner product. That is, for all $X$, $tr(X^† A(X)) ≥ 0$. Similarly, show that the superoperator defined by $A(X) ≡ XA$ is positive with respect to the Hilbert–Schmidt inner product on matrices.
Lieb's theorem states that,
Let $X$ be a matrix and $0\le t\le 1$. Then the function $$ f(A,B)=tr (X^\dagger A^t XB^{1-t}) $$ is jointly concave in positive matrices $A$ and $B$, ie., $$ tr\Big[X^\dagger\big(\lambda A_1+(1-\lambda)A_2\big)^tX\big(\lambda B_1+(1-\lambda)B_2\big)^{1-t}\Big]\ge\lambda tr(X^\dagger A_1^t XB_1^{1-t})+(1-\lambda)tr(X^\dagger A_2^t XB_2^{1-t}) $$
Proof
Let $0\le\lambda\le 1$ and define superoperators (linear maps on operators) $S_1,S_2,T_1,T_2,R_1,R_2$ as follows, $$ S_1(X)\equiv \lambda A_1 X\\ S_2(X)\equiv\lambda XB_1\\ T_1(X)\equiv (1-\lambda)A_2X\\ T_2(X)\equiv (1-\lambda)XB_2\\ R_1\equiv S_1+T_1\\ R_2\equiv S_2+T_2\\ $$ where $S_1$ and $S_2$ commute, as do $T_1$ and $T_2$, and $R_1$ and $R_2$, since $$ S_1S_2(X)=S_1(\lambda XB_1)=\lambda A_1.\lambda XB_1=\lambda^2A_1XB_1\\ S_2S_1(X)=S_2(\lambda A_1X)=\lambda.\lambda A_1X.B_1=\lambda^2A_1XB_1\\ $$ From lemma $A6.8$, \begin{align} A_1,B_1,A_2,B_2 \text{ are positive matrices}&\implies \\S_1,S_2,T_1,T_2,R_1,R_2&\text{ are positive w.r.t the Hilbert-Schmidt inner product} \end{align} Now, \begin{align} \text{Lemma A6.2}&\implies R_1^tR_2^{1-t}\ge S_1^tS_2^{1-t}+T_1^tT_2^{1-t}\\ &\implies (S_1+T_1)^t(S_2+T_2)^{1-t}\ge S_1^tS_2^{1-t}+T_1^tT_2^{1-t} \end{align}
Then, Using the Hilbert–Schmidt inner product to take the X ·X matrix element of the previous inequality gives the Lieb's theorem of joint concavity. $$ tr\Big[X^\dagger\big(\lambda A_1+(1-\lambda)A_2\big)^tX\big(\lambda B_1+(1-\lambda)B_2\big)^{1-t}\Big]\ge\lambda tr(X^\dagger A_1^t XB_1^{1-t})+(1-\lambda)tr(X^\dagger A_2^t XB_2^{1-t}) $$
How does the last statement reach the Lieb's theorem from the above inequality?
My Attempt
My understanding is that, the Hilbert–Schmidt inner product can be defined as $(A,B)_{HS}=tr(A^\dagger B)$, and $$ R_1^tR_2^{1-t}-S_1^tS_2^{1-t}-T_1^tT_2^{1-t}\ge 0\\ \implies R_1^tR_2^{1-t}-S_1^tS_2^{1-t}-T_1^tT_2^{1-t}\text{ is positive} $$ From Lemma $A6.8$, given $A$ is positive and $A(X)≡AX$ then $A$ is positive with respect to the Hilbert–Schmidt inner product, ie., $tr(X^† A(X))\ge 0\quad\forall \quad X$. Therefore, $$ tr\bigg(X^\dagger\Big(R_1^tR_2^{1-t}-S_1^tS_2^{1-t}-T_1^tT_2^{1-t}\Big)(X)\bigg)\ge 0\\ tr\bigg(X^\dagger\Big(R_1^t(X)R_2^{1-t}(X)-S_1^t(X)S_2^{1-t}(X)-T_1^t(X)T_2^{1-t}(X)\Big)\bigg)\ge 0\\ tr\bigg(X^\dagger\Big((\lambda A_1 X+(1-\lambda)A_2X)^t(\lambda XB_1+(1-\lambda)XB_2)^{1-t}- (\lambda A_1 X)^t(\lambda XB_1)^{1-t}-((1-\lambda)A_2X)^t((1-\lambda)XB_2)^{1-t}\Big)\bigg)\ge 0\\ tr\bigg(X^\dagger(\lambda A_1+(1-\lambda)A_2)^tX(\lambda B_1+(1-\lambda)B_2)^{1-t}-X^\dagger(\lambda A_1)^tX(\lambda B_1)^{1-t}-X^\dagger((1-\lambda)A_2)^tX((1-\lambda)B_2)^{1-t}\bigg)\ge 0\\ tr\Big(X^\dagger(\lambda A_1+(1-\lambda)A_2)^tX(\lambda B_1+(1-\lambda)B_2)^{1-t}\Big)-tr\Big(X^\dagger(\lambda A_1)^tX(\lambda B_1)^{1-t}\Big)-tr\Big(X^\dagger((1-\lambda)A_2)^tX((1-\lambda)B_2)^{1-t}\Big)\ge 0\\ tr\Big(X^\dagger(\lambda A_1+(1-\lambda)A_2)^tX(\lambda B_1+(1-\lambda)B_2)^{1-t}\Big)-\lambda tr\Big(X^\dagger(A_1)^tX(B_1)^{1-t}\Big)-(1-\lambda)tr\Big(X^\dagger(A_2)^tX(B_2)^{1-t}\Big)\ge 0\\ $$
But, since lemma $A6.2$ appears to prove the case for usual inner product on the Hilbert space, how can it be naturally valid for the matrix space and Hilber-Schmidt inne product?
