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If we have the following upper bound on the sum of trace distances:

$$ \frac{1}{N} \sum_{a, b}||p_1(a | b) \rho_{ab} - p_2(a | b) \sigma_{ab}|| \le \epsilon, $$ where $p_1$ and $p_2$ are two probability distributions and $\rho_{ab}$ and $\sigma_{ab}$ are quantum states depending on $a, b$. Then, what is the best that we could say about the following in terms of $\epsilon$:

$$ ||\rho_{ab} - \sigma_{ab}|| \le ? $$

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In general I would imagine not much. For simplicity consider the case where $p_1 = p_2$. Then the worst case would be that for all $(i,j) \neq (a,b)$ we have $\rho_{ij} = \sigma_{ij}$. Thus the entire difference in the first equation is coming only from the difference between the pair $\rho_{ab}$ and $\sigma_{ab}$. In such a case you'd have $$ \|\rho_{ab} - \sigma_{ab}\| \leq \frac{N \epsilon}{p(a|b)}. $$ which could even be bigger than the trivial bound of $1$ unless $\epsilon < 1/N$.

Without more context to the problem I don't see how you'd get around this.

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  • $\begingroup$ I see your point, Rammus. Thanks a lot for the clarification. $\endgroup$ Commented Mar 10, 2023 at 3:06

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