Let the trace norm of $X$ be

$$\Vert X\Vert_1 := \operatorname{tr} \left(\,(X^\dagger X)^{1/2}\right)$$

If the quantum states (finite-dimensional Hermitian positive semidefinite matrices that have unit trace) $\rho$ and $\sigma$ are close to each other in the trace norm, $\Vert \rho - \sigma \Vert_1 \leq \epsilon$, for some small $\epsilon > 0$, does there exist a projector $\Pi$, such that

$$\Pi \rho \Pi \leq (1+g_1(\epsilon)) \sigma$$

where the operator inequality $A \leq B$ represents that the operator $B-A$ is positive semidefinite, and

$$\text{tr}(\Pi \sigma) \geq 1- g_2(\epsilon)$$

for some small functions $g_1(\epsilon)$ and $g_2(\epsilon)$, i.e., both $g_1(\epsilon)$ and $g_2(\epsilon)$ tend to $0$ as $\epsilon \to 0$? Note $g_1$ and $g_2$ should be independent of the dimensions of the matrices.

This is true for classical states (matrices which commute) and also true when we also sandwich $\sigma$ in the operator inequality (proofs are here).

This also seems to be true for randomly selected matrices in small dimensions (program and description are here; the Jupyter notebook also contains some more observations).

So far, I have not been able to come up with a way which avoids having a projector on $\sigma$ as well. It seems that using the assumption $\Vert \sqrt{\rho}- \sqrt{\sigma} \Vert_2 \leq \epsilon$, which is equivalent (upto the exponent of $\epsilon$) to the original assumption, is far more easier, since there does not seem to be a lot of ways one can manipulate the trace norm. I have been trying to use a strengthened version of Gresgorin Theorem (Corollary 6.1.6 of Horn & Johnson's Matrix Analysis) but no luck so far.

Any help or ideas are appreciated.

Let the trace norm of $X$ be

$$\Vert X\Vert_1 := \operatorname{tr} \left(\,(X^\dagger X)^{1/2}\right)$$

and let the operator inequality $A \leq B$ denote that the operator $B-A$ is positive semidefinite.

If the quantum states (finite-dimensional Hermitian positive semidefinite matrices that have unit trace) $\rho$ and $\sigma$ are close to each other in the trace norm, $\Vert \rho - \sigma \Vert_1 \leq \epsilon$, for some small $\epsilon > 0$, does there exist a projector $\Pi$, such that

$$ \begin{aligned} \Pi \rho \Pi &\leq (1+g_1(\epsilon)) \sigma \\ \operatorname{tr} (\Pi \sigma) &\geq 1- g_2(\epsilon) \end{aligned} $$

for some small functions $g_1(\epsilon)$ and $g_2(\epsilon)$, i.e., both $g_1(\epsilon)$ and $g_2(\epsilon)$ tend to $0$ as $\epsilon \to 0$? Note $g_1$ and $g_2$ should be independent of the dimensions of the matrices. Some observations:

1. This is true for classical states (matrices which commute) and also true when we also sandwich $\sigma$ in the operator inequality (proofs are here).

2. It seems that this is also true for randomly selected matrices in small dimensions (program and description are here; the Jupyter notebook also contains some more observations).

So far, I have not been able to come up with a way which avoids having a projector on $\sigma$ as well. It seems that using the assumption $\Vert \sqrt{\rho}- \sqrt{\sigma} \Vert_2 \leq \epsilon$, which is equivalent (upto the exponent of $\epsilon$) to the original assumption, is far more easier, since there does not seem to be a lot of ways one can manipulate the trace norm. I have been trying to use a strengthened version of Gresgorin Theorem (Corollary 6.1.6 of Horn & Johnson's 2nd edition of Matrix Analysis) but no luck so far.

Any help or ideas are appreciated.