On a matrix trace inequality

Question

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.

Answer 1

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 $c$.

Answer 2

Your second displayed inequality will usually fail to hold.

Indeed, if $B_{ij}=1(i=j=1)$ for all $i,j$, then $(LBL)_{11}=|L_{11}|^2$ and $(L^2B)_{11}=\sum_{i=1}^n|L_{1i}|^2$, so that the left-hand side of that inequality will be $\sum_{i=2}^n|L_{1i}|^2$, whereas the right-hand side of the inequality will be $0$.