# Proof of the Lieb's theorem

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?