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I found a proof of this problem for the case of $\varphi(x)=\sum_i{x_i\log x_i-x_i}$. If you find there is anything mistake in the proof, please let me know. Thank you!

The case of $\varphi(x)=\sum_i{x_i\log x_i-x_i}$ can be proved with the method similar to Lindblad, Completely positive maps and entropy inequalities, 1975 and Lindblad, Expectations and entropy inequalities for finite quantum systems, 1974. The inequality can be strengthened as $$ \sum_i{\mathrm{KL}\left(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_i^\top B\mathbf{x}\_i\right)}\leq D_{vN}\left(A,B\right) $$

Actually, a very similar result has been proposed in some papers about quantum information theory, such as the two papers referred above. The referred result is that for any trace preserving map $\Phi$, given by $\Phi(A)=\sum_{i=1}^n{V\_iAV\_i^\top}$ and $\sum\_{i=1}^n{V\_i^\top V\_i}=1$, we have that $\mathrm{tr}\left(\Phi(A),\Phi(B)\right)\leq D_\phi(A,B)$, where $A,B$ are both density operators which are Hermitian positive definite matrices satisfying $\mathrm{tr}A=\mathrm{tr}B=1$ and $\varphi(x)=x\log x$.

We found that if the trace constraints $\mathrm{tr}A=\mathrm{tr}B=1$ are dropped and $\varphi(x)=x\log x$ is replaced with $\varphi(x)=x\log x-x$, the inequality still holds.

The proof is outlined as following:

1. The von Neumann divergence has the following additivity property with Kronecker product: $$D_{vN}(A\otimes P,B\otimes P)=D_{vN}(A,B)\cdot\mathrm{tr}P$$

2. Using the joint convexity and the additivity, we can prove that the von Neumann divergence has the monotonicity with partial trace as \begin{equation*} \begin{split} D\_{vN}(\mathrm{tr}\_2(A),\mathrm{tr}\_2(B)) =&D\_{vN}\left(\mathrm{tr}\_2(A)\otimes\frac{\mathbf{I}\_2}{m}, \mathrm{tr}\_2(B)\otimes\frac{\mathbf{I}\_2}{m}\right) /\mathrm{tr}\left(\frac{\mathbf{I}\_2}{m}\right)\\\ =&D\_{vN}\left(\sum\_{j=1}^N{p_jW_jAW_j^+},\sum\_{j=1}^N{p\_jW\_jBW\_j^+}\right)\\\ \leq&\sum\_{j=1}^{N}{p\_jD\_{vN}\left(W\_jAW\_j^+,W\_jBW\_j^+\right)}\\\ =&D\_{vN}(A,B)\end{split} \end{equation*}

3. For any trace preserving map $\Phi$, given by $\Phi(A)=\sum_{i=1}^n{V\_iAV\_i^\top}$ and $\sum\_{i=1}^n{V\_i^\top V\_i}=1$, it can be represented as a unitary operation+partial tracing. Therefore, we have that \begin{equation*} \begin{split} D\_{vN}\left(\Phi(A),\Phi(B)\right) =&D\_{vN}\left(\mathrm{tr}\_2(W(A\otimes\mathbf{s}\mathbf{s}^\top)W^\top), \mathrm{tr}\_2(W(B\otimes\mathbf{s}\mathbf{s}^\top)W^\top)\right)\\\ \leq&D\_{vN}\left(W(A\otimes\mathbf{s}\mathbf{s}^\top)W^\top, W(B\otimes\mathbf{s}\mathbf{s}^\top)W^\top\right)\\\ =&D\_{vN}\left(A,B\right) \end{split} \end{equation*}

4. Then the sum of relative entropy of the quadratic forms could be represented as matrix divergence and bounded. \begin{equation*} \begin{split} \sum_i{\mathrm{KL}\left(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_i^\top B\mathbf{x}\_i\right)} =&\sum\_{i,j}{(\mathbf{x}\_i^\top\mathbf{x}\_j)^2 \mathrm{KL}(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_j^\top B\mathbf{x}\_j)}\\\ =&D_{vN}(\sum\_i{X\_iAX\_i^\top},\sum_i{X\_iBX\_i^\top})\\\ \leq&D_{vN}\left(A,B\right) \end{split} \end{equation*} where $X_i=\mathbf{x}_i\mathbf{x}_i^\top$.

If there is any mistake in the proof, please let me know. Any other suggestions are also welcomed. Thank you very much!

I found a proof of this problem for the case of $\varphi(x)=\sum_i{x_i\log x_i-x_i}$. If you find there is anything mistake in the proof, please let me know. Thank you!

The case of $\varphi(x)=\sum_i{x_i\log x_i-x_i}$ can be proved with the method similar to Lindblad, Completely positive maps and entropy inequalities, 1975 and Lindblad, Expectations and entropy inequalities for finite quantum systems, 1974. The inequality can be strengthened as $$ \sum_i{\mathrm{KL}\left(\mathbf{x}_i^\top A\mathbf{x}_i,\mathbf{x}_i^\top B\mathbf{x}_i\right)}\leq D_{vN}\left(A,B\right) $$

Actually, a very similar result has been proposed in some papers about quantum information theory, such as the two papers referred above. The referred result is that for any trace preserving map $\Phi$, given by $\Phi(A)=\sum_{i=1}^n{V_iAV_i^\top}$ and $\sum_{i=1}^n{V_i^\top V_i}=1$, we have that $\mathrm{tr}\left(\Phi(A),\Phi(B)\right)\leq D_\phi(A,B)$, where $A,B$ are both density operators which are Hermitian positive definite matrices satisfying $\mathrm{tr}A=\mathrm{tr}B=1$ and $\varphi(x)=x\log x$.

We found that if the trace constraints $\mathrm{tr}A=\mathrm{tr}B=1$ are dropped and $\varphi(x)=x\log x$ is replaced with $\varphi(x)=x\log x-x$, the inequality still holds.

The proof is outlined as following:

1. The von Neumann divergence has the following additivity property with Kronecker product: $$D_{vN}(A\otimes P,B\otimes P)=D_{vN}(A,B)\cdot\mathrm{tr}P$$

2. Using the joint convexity and the additivity, we can prove that the von Neumann divergence has the monotonicity with partial trace as \begin{equation*} \begin{split} D_{vN}(\mathrm{tr}_2(A),\mathrm{tr}_2(B)) =&D_{vN}\left(\mathrm{tr}_2(A)\otimes\frac{\mathbf{I}_2}{m}, \mathrm{tr}_2(B)\otimes\frac{\mathbf{I}_2}{m}\right) /\mathrm{tr}\left(\frac{\mathbf{I}_2}{m}\right)\\\ =&D_{vN}\left(\sum_{j=1}^N{p_jW_jAW_j^+},\sum_{j=1}^N{p_jW_jBW_j^+}\right)\\\ \leq&\sum_{j=1}^{N}{p_jD_{vN}\left(W_jAW_j^+,W_jBW_j^+\right)}\\\ =&D_{vN}(A,B)\end{split} \end{equation*}

3. For any trace preserving map $\Phi$, given by $\Phi(A)=\sum_{i=1}^n{V_iAV_i^\top}$ and $\sum_{i=1}^n{V_i^\top V_i}=1$, it can be represented as a unitary operation+partial tracing. Therefore, we have that \begin{equation*} \begin{split} D_{vN}\left(\Phi(A),\Phi(B)\right) =&D_{vN}\left(\mathrm{tr}_2(W(A\otimes\mathbf{s}\mathbf{s}^\top)W^\top), \mathrm{tr}_2(W(B\otimes\mathbf{s}\mathbf{s}^\top)W^\top)\right)\\\ \leq&D_{vN}\left(W(A\otimes\mathbf{s}\mathbf{s}^\top)W^\top, W(B\otimes\mathbf{s}\mathbf{s}^\top)W^\top\right)\\\ =&D_{vN}\left(A,B\right) \end{split} \end{equation*}

4. Then the sum of relative entropy of the quadratic forms could be represented as matrix divergence and bounded. \begin{equation*} \begin{split} \sum_i{\mathrm{KL}\left(\mathbf{x}_i^\top A\mathbf{x}_i,\mathbf{x}_i^\top B\mathbf{x}_i\right)} =&\sum_{i,j}{(\mathbf{x}_i^\top\mathbf{x}_j)^2 \mathrm{KL}(\mathbf{x}_i^\top A\mathbf{x}_i,\mathbf{x}_j^\top B\mathbf{x}_j)}\\\ =&D_{vN}(\sum_i{X_iAX_i^\top},\sum_i{X_iBX_i^\top})\\\ \leq&D_{vN}\left(A,B\right) \end{split} \end{equation*} where $X_i=\mathbf{x}_i\mathbf{x}_i^\top$.

If there is any mistake in the proof, please let me know. Any other suggestions are also welcomed. Thank you very much!

I found a proof of this problem for the case of $\varphi(x)=\sum_i{x_i\log x_i-x_i}$. If you find there is anything mistake in the proof, please let me know. Thank you!

The case of $\varphi(x)=\sum_i{x_i\log x_i-x_i}$ can be proved with the method similar to Lindblad, Completely positive maps and entropy inequalities, 1975 and Lindblad, Expectations and entropy inequalities for finite quantum systems, 1974. The inequality can be strengthened as $$ \sum_i{\mathrm{KL}\left(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_i^\top B\mathbf{x}\_i\right)}\leq D_{vN}\left(A,B\right) $$

Actually, a very similar result has been proposed in some papers about quantum information theory, such as the two papers referred above. The referred result is that for any trace preserving map $\Phi$, given by $\Phi(A)=\sum_{i=1}^n{V\_iAV\_i^\top}$ and $\sum\_{i=1}^n{V\_i^\top V\_i}=1$, we have that $\mathrm{tr}\left(\Phi(A),\Phi(B)\right)\leq D_\phi(A,B)$, where $A,B$ are both density operators which are Hermitian positive definite matrices satisfying $\mathrm{tr}A=\mathrm{tr}B=1$ and $\varphi(x)=x\log x$.

We found that if the trace constraints $\mathrm{tr}A=\mathrm{B}=1$ are dropped and $\varphi(x)=x\log x$ is replaced with $\varphi(x)=x\log x-x$, the inequality still holds.

The proof is outlined as following:

1. The von Neumann divergence has the following additivity property with Kronecker product: $$D_{vN}(A\otimes P,B\otimes P)=D_{vN}(A,B)\cdot\mathrm{tr}P$$

2. Using the joint convexity and the additivity, we can prove that the von Neumann divergence has the monotonicity with partial trace as \begin{equation*} \begin{split} D\_{vN}(\mathrm{tr}\_2(A),\mathrm{tr}\_2(B)) =&D\_{vN}\left(\mathrm{tr}\_2(A)\otimes\frac{\mathbf{I}\_2}{m}, \mathrm{tr}\_2(B)\otimes\frac{\mathbf{I}\_2}{m}\right) /\mathrm{tr}\left(\frac{\mathbf{I}\_2}{m}\right)\\\ =&D\_{vN}\left(\sum\_{j=1}^N{p_jW_jAW_j^+},\sum\_{j=1}^N{p\_jW\_jBW\_j^+}\right)\\\ \leq&\sum\_{j=1}^{N}{p\_jD\_{vN}\left(W\_jAW\_j^+,W\_jBW\_j^+\right)}\\\ =&D\_{vN}(A,B)\end{split} \end{equation*}

3. For any trace preserving map $\Phi$, given by $\Phi(A)=\sum_{i=1}^n{V\_iAV\_i^\top}$ and $\sum\_{i=1}^n{V\_i^\top V\_i}=1$, it can be represented as a unitary operation+partial tracing. Therefore, we have that \begin{equation*} \begin{split} D\_{vN}\left(\Phi(A),\Phi(B)\right) =&D\_{vN}\left(\mathrm{tr}\_2(W(A\otimes\mathbf{s}\mathbf{s}^\top)W^\top), \mathrm{tr}\_2(W(B\otimes\mathbf{s}\mathbf{s}^\top)W^\top)\right)\\\ \leq&D\_{vN}\left(W(A\otimes\mathbf{s}\mathbf{s}^\top)W^\top, W(B\otimes\mathbf{s}\mathbf{s}^\top)W^\top\right)\\\ =&D\_{vN}\left(A,B\right) \end{split} \end{equation*}

4. Then the sum of relative entropy of the quadratic forms could be represented as matrix divergence and bounded. \begin{equation*} \begin{split} \sum_i{\mathrm{KL}\left(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_i^\top B\mathbf{x}\_i\right)} =&\sum\_{i,j}{(\mathbf{x}\_i^\top\mathbf{x}\_j)^2 \mathrm{KL}(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_j^\top B\mathbf{x}\_j)}\\\ =&D_{vN}(\sum\_i{X\_iAX\_i^\top},\sum_i{X\_iBX\_i^\top})\\\ \leq&D_{vN}\left(A,B\right) \end{split} \end{equation*} where $X_i=\mathbf{x}_i\mathbf{x}_i^\top$.

If there is any mistake in the proof, please let me know. Any other suggestions are also welcomed. Thank you very much!

I found a proof of this problem for the case of $\varphi(x)=\sum_i{x_i\log x_i-x_i}$. If you find there is anything mistake in the proof, please let me know. Thank you!

The case of $\varphi(x)=\sum_i{x_i\log x_i-x_i}$ can be proved with the method similar to Lindblad, Completely positive maps and entropy inequalities, 1975 and Lindblad, Expectations and entropy inequalities for finite quantum systems, 1974. The inequality can be strengthened as $$ \sum_i{\mathrm{KL}\left(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_i^\top B\mathbf{x}\_i\right)}\leq D_{vN}\left(A,B\right) $$

Actually, a very similar result has been proposed in some papers about quantum information theory, such as the two papers referred above. The referred result is that for any trace preserving map $\Phi$, given by $\Phi(A)=\sum_{i=1}^n{V\_iAV\_i^\top}$ and $\sum\_{i=1}^n{V\_i^\top V\_i}=1$, we have that $\mathrm{tr}\left(\Phi(A),\Phi(B)\right)\leq D_\phi(A,B)$, where $A,B$ are both density operators which are Hermitian positive definite matrices satisfying $\mathrm{tr}A=\mathrm{tr}B=1$ and $\varphi(x)=x\log x$.

We found that if the trace constraints $\mathrm{tr}A=\mathrm{tr}B=1$ are dropped and $\varphi(x)=x\log x$ is replaced with $\varphi(x)=x\log x-x$, the inequality still holds.

The proof is outlined as following:

1. The von Neumann divergence has the following additivity property with Kronecker product: $$D_{vN}(A\otimes P,B\otimes P)=D_{vN}(A,B)\cdot\mathrm{tr}P$$

2. Using the joint convexity and the additivity, we can prove that the von Neumann divergence has the monotonicity with partial trace as \begin{equation*} \begin{split} D\_{vN}(\mathrm{tr}\_2(A),\mathrm{tr}\_2(B)) =&D\_{vN}\left(\mathrm{tr}\_2(A)\otimes\frac{\mathbf{I}\_2}{m}, \mathrm{tr}\_2(B)\otimes\frac{\mathbf{I}\_2}{m}\right) /\mathrm{tr}\left(\frac{\mathbf{I}\_2}{m}\right)\\\ =&D\_{vN}\left(\sum\_{j=1}^N{p_jW_jAW_j^+},\sum\_{j=1}^N{p\_jW\_jBW\_j^+}\right)\\\ \leq&\sum\_{j=1}^{N}{p\_jD\_{vN}\left(W\_jAW\_j^+,W\_jBW\_j^+\right)}\\\ =&D\_{vN}(A,B)\end{split} \end{equation*}

3. For any trace preserving map $\Phi$, given by $\Phi(A)=\sum_{i=1}^n{V\_iAV\_i^\top}$ and $\sum\_{i=1}^n{V\_i^\top V\_i}=1$, it can be represented as a unitary operation+partial tracing. Therefore, we have that \begin{equation*} \begin{split} D\_{vN}\left(\Phi(A),\Phi(B)\right) =&D\_{vN}\left(\mathrm{tr}\_2(W(A\otimes\mathbf{s}\mathbf{s}^\top)W^\top), \mathrm{tr}\_2(W(B\otimes\mathbf{s}\mathbf{s}^\top)W^\top)\right)\\\ \leq&D\_{vN}\left(W(A\otimes\mathbf{s}\mathbf{s}^\top)W^\top, W(B\otimes\mathbf{s}\mathbf{s}^\top)W^\top\right)\\\ =&D\_{vN}\left(A,B\right) \end{split} \end{equation*}

4. Then the sum of relative entropy of the quadratic forms could be represented as matrix divergence and bounded. \begin{equation*} \begin{split} \sum_i{\mathrm{KL}\left(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_i^\top B\mathbf{x}\_i\right)} =&\sum\_{i,j}{(\mathbf{x}\_i^\top\mathbf{x}\_j)^2 \mathrm{KL}(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_j^\top B\mathbf{x}\_j)}\\\ =&D_{vN}(\sum\_i{X\_iAX\_i^\top},\sum_i{X\_iBX\_i^\top})\\\ \leq&D_{vN}\left(A,B\right) \end{split} \end{equation*} where $X_i=\mathbf{x}_i\mathbf{x}_i^\top$.

If there is any mistake in the proof, please let me know. Any other suggestions are also welcomed. Thank you very much!

The case of $\varphi(x)=\sum_i{x_i\log x_i-x_i}$ can be proved with the method similar to Lindblad, Completely positive maps and entropy inequalities, 1975 and Lindblad, Expectations and entropy inequalities for finite quantum systems, 1974. The inequality can be strengthened as $$ \sum_i{\mathrm{KL}\left(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_i^\top B\mathbf{x}\_i\right)}\leq D_{vN}\left(A,B\right) $$

Actually, a very similar result has been proposed in some papers about quantum information theory, such as the two papers referred above. The referred result is that for any trace preserving map $\Phi$, given by $\Phi(A)=\sum_{i=1}^n{V\_iAV\_i^\top}$ and $\sum\_{i=1}^n{V\_i^\top V\_i}=1$, we have that $\mathrm{tr}\left(\Phi(A),\Phi(B)\right)\leq D_\phi(A,B)$, where $A,B$ are both density operators which are Hermitian positive definite matrices satisfying $\mathrm{tr}A=\mathrm{tr}B=1$ and $\varphi(x)=x\log x$.

We found that if the trace constraints $\mathrm{tr}A=\mathrm{B}=1$ are dropped and $\varphi(x)=x\log x$ is replaced with $\varphi(x)=x\log x-x$, the inequality still holds.

The proof is outlined as following:

1. The von Neumann divergence has the following additivity property with Kronecker product: $$D_{vN}(A\otimes P,B\otimes P)=D_{vN}(A,B)\cdot\mathrm{tr}P$$

2. Using the joint convexity and the additivity, we can prove that the von Neumann divergence has the monotonicity with partial trace as \begin{equation*} \begin{split} D\_{vN}(\mathrm{tr}\_2(A),\mathrm{tr}\_2(B)) =&D\_{vN}\left(\mathrm{tr}\_2(A)\otimes\frac{\mathbf{I}\_2}{m}, \mathrm{tr}\_2(B)\otimes\frac{\mathbf{I}\_2}{m}\right) /\mathrm{tr}\left(\frac{\mathbf{I}\_2}{m}\right)\\\ =&D\_{vN}\left(\sum\_{j=1}^N{p_jW_jAW_j^+},\sum\_{j=1}^N{p\_jW\_jBW\_j^+}\right)\\\ \leq&\sum\_{j=1}^{N}{p\_jD\_{vN}\left(W\_jAW\_j^+,W\_jBW\_j^+\right)}\\\ =&D\_{vN}(A,B)\end{split} \end{equation*}

3. For any trace preserving map $\Phi$, given by $\Phi(A)=\sum_{i=1}^n{V\_iAV\_i^\top}$ and $\sum\_{i=1}^n{V\_i^\top V\_i}=1$, it can be represented as a unitary operation+partial tracing. Therefore, we have that \begin{equation*} \begin{split} D\_{vN}\left(\Phi(A),\Phi(B)\right) =&D\_{vN}\left(\mathrm{tr}\_2(W(A\otimes\mathbf{s}\mathbf{s}^\top)W^\top), \mathrm{tr}\_2(W(B\otimes\mathbf{s}\mathbf{s}^\top)W^\top)\right)\\\ \leq&D\_{vN}\left(W(A\otimes\mathbf{s}\mathbf{s}^\top)W^\top, W(B\otimes\mathbf{s}\mathbf{s}^\top)W^\top\right)\\\ =&D\_{vN}\left(A,B\right) \end{split} \end{equation*}

4. Then the sum of relative entropy of the quadratic forms could be represented as matrix divergence and bounded. \begin{equation*} \begin{split} \sum_i{\mathrm{KL}\left(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_i^\top B\mathbf{x}\_i\right)} =&\sum\_{i,j}{(\mathbf{x}\_i^\top\mathbf{x}\_j)^2 \mathrm{KL}(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_j^\top B\mathbf{x}\_j)}\\\ =&\mathrm{KL}(\sum\_i{X\_iAX\_i^\top},\sum_i{X\_iBX\_i^\top})\\\ \leq&D_{vN}\left(A,B\right) \end{split} \end{equation*} where $X_i=\mathbf{x}_i\mathbf{x}_i^\top$.

If there is any mistake in the proof, please let me know. Thank you!

I found a proof of this problem for the case of $\varphi(x)=\sum_i{x_i\log x_i-x_i}$. If you find there is anything mistake in the proof, please let me know. Thank you!

The case of $\varphi(x)=\sum_i{x_i\log x_i-x_i}$ can be proved with the method similar to Lindblad, Completely positive maps and entropy inequalities, 1975 and Lindblad, Expectations and entropy inequalities for finite quantum systems, 1974. The inequality can be strengthened as $$ \sum_i{\mathrm{KL}\left(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_i^\top B\mathbf{x}\_i\right)}\leq D_{vN}\left(A,B\right) $$

Actually, a very similar result has been proposed in some papers about quantum information theory, such as the two papers referred above. The referred result is that for any trace preserving map $\Phi$, given by $\Phi(A)=\sum_{i=1}^n{V\_iAV\_i^\top}$ and $\sum\_{i=1}^n{V\_i^\top V\_i}=1$, we have that $\mathrm{tr}\left(\Phi(A),\Phi(B)\right)\leq D_\phi(A,B)$, where $A,B$ are both density operators which are Hermitian positive definite matrices satisfying $\mathrm{tr}A=\mathrm{tr}B=1$ and $\varphi(x)=x\log x$.

We found that if the trace constraints $\mathrm{tr}A=\mathrm{B}=1$ are dropped and $\varphi(x)=x\log x$ is replaced with $\varphi(x)=x\log x-x$, the inequality still holds.

The proof is outlined as following:

1. The von Neumann divergence has the following additivity property with Kronecker product: $$D_{vN}(A\otimes P,B\otimes P)=D_{vN}(A,B)\cdot\mathrm{tr}P$$

2. Using the joint convexity and the additivity, we can prove that the von Neumann divergence has the monotonicity with partial trace as \begin{equation*} \begin{split} D\_{vN}(\mathrm{tr}\_2(A),\mathrm{tr}\_2(B)) =&D\_{vN}\left(\mathrm{tr}\_2(A)\otimes\frac{\mathbf{I}\_2}{m}, \mathrm{tr}\_2(B)\otimes\frac{\mathbf{I}\_2}{m}\right) /\mathrm{tr}\left(\frac{\mathbf{I}\_2}{m}\right)\\\ =&D\_{vN}\left(\sum\_{j=1}^N{p_jW_jAW_j^+},\sum\_{j=1}^N{p\_jW\_jBW\_j^+}\right)\\\ \leq&\sum\_{j=1}^{N}{p\_jD\_{vN}\left(W\_jAW\_j^+,W\_jBW\_j^+\right)}\\\ =&D\_{vN}(A,B)\end{split} \end{equation*}

3. For any trace preserving map $\Phi$, given by $\Phi(A)=\sum_{i=1}^n{V\_iAV\_i^\top}$ and $\sum\_{i=1}^n{V\_i^\top V\_i}=1$, it can be represented as a unitary operation+partial tracing. Therefore, we have that \begin{equation*} \begin{split} D\_{vN}\left(\Phi(A),\Phi(B)\right) =&D\_{vN}\left(\mathrm{tr}\_2(W(A\otimes\mathbf{s}\mathbf{s}^\top)W^\top), \mathrm{tr}\_2(W(B\otimes\mathbf{s}\mathbf{s}^\top)W^\top)\right)\\\ \leq&D\_{vN}\left(W(A\otimes\mathbf{s}\mathbf{s}^\top)W^\top, W(B\otimes\mathbf{s}\mathbf{s}^\top)W^\top\right)\\\ =&D\_{vN}\left(A,B\right) \end{split} \end{equation*}

4. Then the sum of relative entropy of the quadratic forms could be represented as matrix divergence and bounded. \begin{equation*} \begin{split} \sum_i{\mathrm{KL}\left(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_i^\top B\mathbf{x}\_i\right)} =&\sum\_{i,j}{(\mathbf{x}\_i^\top\mathbf{x}\_j)^2 \mathrm{KL}(\mathbf{x}\_i^\top A\mathbf{x}\_i,\mathbf{x}\_j^\top B\mathbf{x}\_j)}\\\ =&D_{vN}(\sum\_i{X\_iAX\_i^\top},\sum_i{X\_iBX\_i^\top})\\\ \leq&D_{vN}\left(A,B\right) \end{split} \end{equation*} where $X_i=\mathbf{x}_i\mathbf{x}_i^\top$.

If there is any mistake in the proof, please let me know. Any other suggestions are also welcomed. Thank you very much!