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On page $408$ of Nielsen & Chuang in the step going from equation $(9.48)$ to $(9.49)$, I don't see how:

$$\sum\limits_i (p_i - q_i)tr(P \sigma_i) \leq D(p_i, q_i)$$

I proceed as follows:

$$\sum\limits_i (p_i - q_i)tr(P \sigma_i)\leq \sum\limits_i |p_i - q_i| tr(P \sigma_i)$$

How is this $\leq \frac{1}{2}\sum\limits_i |p_i - q_i|$?

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Let $A$ be the set of indices $i$ such that $p_i\ge q_i$. We have $$ \begin{align} \sum_i(p_i-q_i)\mathrm{tr}(P\sigma_i)\le&\,\sum_{i\in A}(p_i-q_i)\mathrm{tr}(P\sigma_i)\tag1\\ \le&\,\sum_{i\in A}(p_i-q_i)\tag2\\ \le&\,\max_S\sum_{i\in S}(p_i-q_i)\tag3\\ =&\,D(p,q)\tag4 \end{align} $$ where in the first step we drop non-positive terms, in the second we use $\mathrm{tr}(P\sigma_i)\le 1$, in the third we take maximum over subsets $S$ of the index set and in the final step we use $(9.4)$ on page $401$.

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  • $\begingroup$ Thank you Adam. How do I prove tr$(P\sigma_i) \leq 1$? It seems intuitive as P is a projection operator and tr$(\sigma_i) = 1$. $\endgroup$
    – Sam
    Nov 28, 2022 at 13:37
  • $\begingroup$ Choose a basis $\psi_i$ with $i=1,\dots,k$ of the image of $P$ and extend it to a basis $\psi_i$ with $i=1,\dots,n$ of the full Hilbert space . Then $\mathrm{tr}(P\sigma)=\sum_{i=1}^k\langle\psi_i|\sigma|\psi_i\rangle\le\sum_{i=1}^n\langle\psi_i|\sigma|\psi_i\rangle=\mathrm{tr}(\sigma)=1$. $\endgroup$
    – Adam Zalcman
    Nov 28, 2022 at 16:21

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