Let $\mathcal H(n)$ be the set of $n\times n$ Hermitian matrices, and $\mathcal S(n) \subset \mathcal H(n)$ be the subset of density matrices, i.e., $A \in \mathcal S(n)$ iff $A$ is Hermitian, positive semidefinite and of trace one. Does there exist $c>0$ such that the following inequality


$$\big| \operatorname{tr} (ALBL) - \operatorname{tr}\left(AL^2B\right) \big| \leq c \|L\|^2 \operatorname{tr} \big((I-A)B\big)$$


holds for all $A, B \in \mathcal S(n)$ and $L \in \mathcal H(n)$?


PS: By choosing a suitable basis, we may assume without loss of generality that $A$ is diagnosable, i.e.


$$A=\sum\_{k=1}^n a\_{kk}P^{kk},$$


where $a\_{ij}:=(A)\_{ij}$ denote the elements of $A$ and $P^{ij}$ is the matrix whose only non-zero element is $(P^{ij})\_{ij}=1$. Note further the above inequality is linear with respect to $A$, it suffices to deal with the case that $a\_{11}=1$ and $a\_{kk}=0$ for $k\neq 1$. Hence, the l.r.s. and r.h.s. become respectively


$$\big| (LBL)\_{11} - (L^2B)\_{11} \big| \leq c \|L\|^2 \big(1-(B)\_{11}\big)?$$


Is the above inequality true for some suitable $c$? I'm unable to carry out the computation.



On a matrix trace inequality
The answer is **false**.


Counterexample: $L$ is the 2x2 all-$1$ matrix, $\epsilon ↘ 0$,


$B=\left[\begin{array}{cc}
1-\epsilon & \sqrt{\epsilon (1-\epsilon)} \\ \sqrt{\epsilon (1-\epsilon)} & \epsilon 
\end{array}\right]$.


Then the LHS converges to $1$ and $1-B\_{11}$ converges to $0$, so there can't be such $c$.


